I was curious on the basic functions of primary vs. secondary vs. tertiary amines. I understand that tert. amines are the most basic due to the electron-donating methyl groups.

Well, in class Kaplan was trying to make a point in basisity in aq. vs. gaseous phase amines:

1) gaseous amine basisity decreased in the following order:

3>2>1>NH3

2) aqeuos amine basity (due to hydrogen bonding):

2>1>3>NH3

Can someone explain the descrepancies? The Kaplan teacher mentioned this class but did not know the explanation.

Thank you.

I was curious on the basic functions of primary vs. secondary vs. tertiary amines. I understand that tert. amines are the most basic due to the electron-donating methyl groups.

Well, in class Kaplan was trying to make a point in basisity in aq. vs. gaseous phase amines:

1) gaseous amine basisity decreased in the following order:

3>2>1>NH3

2) aqeuos amine basity (due to hydrogen bonding):

2>1>3>NH3

Can someone explain the descrepancies? The Kaplan teacher mentioned this class but did not know the explanation.

Thank you.


There are solvation effects that have to be considered in aqueous solutions. In the gas phase, the solvent effects are not a factor.

We have competing factors at work here: inductive electron donation due to alkyl groups making an amine a stronger base and steric hindrance making them weaker bases.

In gas phase, inductive effect wins out.

In an aqueous solution there will be hydrogen bonding between water and amine. This will make steric hindrance more of a factor than inductive effect for tertiary amines.

Anyhow, these discoveries were probably made experimentally. I doubt that MCAT will require this level of detail.

There are solvation effects that have to be considered in aqueous solutions. In the gas phase, the solvent effects are not a factor.

We have competing factors at work here: inductive electron donation due to alkyl groups making an amine a stronger base and steric hindrance making them weaker bases.

In gas phase, inductive effect wins out.

In an aqueous solution there will be hydrogen bonding between water and amine. This will make steric hindrance more of a factor than inductive effect for tertiary amines.

Anyhow, these discoveries were probably made experimentally. I doubt that MCAT will require this level of detail.

well, in order to remember this I have to understand it. How does steric hindrance will affect the basisity? Will hydrogen stabilize the lone electrones on the nitrogen making it a worse base? But then still not clear why the amines are listed in that particular order.

I was surprised the pKa of ethanol (15.9) was higher than for methanol (15.5)-- so methanol deprotonates more readily than ethanol.

I'd had thought the ethyl chain could distribute oxy's negative charge following deprotonation BETTER than methanol-- in which case, ethanol would be more acidic.

Could someone explain why methanol is more acidic?

First post. Thanks for this resource, all.

I was surprised the pKa of ethanol (15.9) was higher than for methanol (15.5)-- so methanol deprotonates more readily than ethanol.

I'd had thought the ethyl chain could distribute oxy's negative charge following deprotonation BETTER than methanol-- in which case, ethanol would be more acidic.

Could someone explain why methanol is more acidic?

First post. Thanks for this resource, all.

Alkyl groups are considered to be electron-donating through the inductive effect (or hyperconjugation). The additional methyl substituent on ethanol compared to methanol makes the hydroxyl group electron richer, which reduces the partial positive on hydrogen and thereby lowers its acidity.

This is the same reason why alkyl groups stabilize radicals and carbocations.

There are solvation effects that have to be considered in aqueous solutions. In the gas phase, the solvent effects are not a factor.

We have competing factors at work here: inductive electron donation due to alkyl groups making an amine a stronger base and steric hindrance making them weaker bases.

In gas phase, inductive effect wins out.

In an aqueous solution there will be hydrogen bonding between water and amine. This will make steric hindrance more of a factor than inductive effect for tertiary amines.

Anyhow, these discoveries were probably made experimentally. I doubt that MCAT will require this level of detail.

Bglass: I must admit that every time I read your explanations, I am so impressed with your logic and rationale. Stay sane a few more weeks and you will own this exam. I can't tell you how much I'm rooting for you to kick the crap out of this little quiz.

The Kaplan teacher mentioned this class but did not know the explanation.

It's not a significant point really, so it's not necessarily something you should stress about. You'll find it in the TBR organic chemistry book (where we try to be thorough and list most everything), but we don't mention this example in lecture.

well, in order to remember this I have to understand it. How does steric hindrance will affect the basisity? Will hydrogen stabilize the lone electrones on the nitrogen making it a worse base? But then still not clear why the amines are listed in that particular order.

If you consider a base by the Lewis definition, then the strongest base is the one with the most readily donated lone pair. Both steric hindrance and hydrogen-bonding crowd the lone pair on nitrogen, and thus make it harder for nitrogen of a tertiary amine to donate it's lone pair to a proton. By not being able to donate its lone pair to an acidic proton, the amine becomes less basic.

Inductive effect predicts: 3 > 2 > 1 > methyl
Sterics/H-bonding predicts: methyl > 1 > 2 > 3

Compromise of the two opposing effects in a protic solvent: 2 > 1 > 3 > methyl

The trend, as Bglass mentioned, is empirical.

What is the relationship between basicity and nucleophilicity? Are there exceptions to their relationship?

What is the relationship between basicity and nucleophilicity? Are there exceptions to their relationship?

A base accepts a hydrogen, while a nucleophile usually binds to a carbon. So the difference between the 2 is in functionality.
A good base is usually a good nuceophile.

If I remember this correctly, in polar protic solvents, atomic radius effect is reversed as far as nucleophiles are concerned, but basicity is not affected by solvent's protic properties.

Sometimes I tend to get confused with extractions..

In general what are the rules for what goes into the aqueous part and what goes into the organic layer

Sometimes I tend to get confused with extractions..

In general what are the rules for what goes into the aqueous part and what goes into the organic layer
Organic layer - organic compounds not soluble in water (think oily bases...naphthalene...benzoic acid...etc). Aqueous layer - inorganic compounds (think hydrochloric acid....sodium hydroxide...etc).

HCl completely.. meaning into its ions

Can anyone link me to a good online demo of extraction.. using different acids and bases. A visual diagram or something would help alot!

How is the oxygen (or sulfer) in furan (or thiophene) is sp2 while HNR2 is sp3? I can understand HNR2 being sp3 because lone pair of electrons. But why only count one pair of lone electrons from oxygen or sulfur in the cyclic compounds?

How is the oxygen (or sulfer) in furan (or thiophene) is sp2 while HNR2 is sp3? I can understand HNR2 being sp3 because lone pair of electrons. But why only count one pair of lone electrons from oxygen or sulfur in the cyclic compounds?O/S in furan/thiophene are both sp2 because they are aromatic compounds. One set of lone pair electrons on O/S participates in the pi system in order to fulfill the Huckel # of electrons that is required for aromaticity.

O/S in furan/thiophene are both sp2 because they are aromatic compounds. One set of lone pair electrons on O/S participates in the pi system in order to fulfill the Huckel # of electrons that is required for aromaticity.

I realize that. But they each still have two sigma bonds (to neighboring carbons) and two lone pairs of electrons. So are we only to use the lone pair of electrons participating in the cyclic system? Thank you for your help!

Let's say you're doing an extraction.. lets say you have a polar substance settling in the H2O layer..

then you add HCl.. what happens?

Wouldn't the HCl disassociate in the water, turning it into a base? How can it still react with the base?

I've been looking online for any sort of "extraction" flow chart.. showing which product ends up where.. does anyone know where to find one? I feel that thats the only way for me to understand extraction properly. Everywhere I look, I see a "fill in the blanks" flow chart and that doesn't help =(

Let's say you're doing an extraction.. lets say you have a polar substance settling in the H2O layer..

then you add HCl.. what happens?

Wouldn't the HCl disassociate in the water, turning it into a base? How can it still react with the base?

I've been looking online for any sort of "extraction" flow chart.. showing which product ends up where.. does anyone know where to find one? I feel that thats the only way for me to understand extraction properly. Everywhere I look, I see a "fill in the blanks" flow chart and that doesn't help =(

Hey Axp, I don't know why you would add HCl to water, I don't think it would do much. You would add HCl to an ether solution so that the slightly basic compounds would get protonated and therefore resulting in a charge - then they would go towards the aqueous part of the solution.

As for flow charts, I know TPR orgo section has a decent one.

I guess..

so do you add stuff to the organic layer most of the time? esp.. w/ acids/bases

When they say... chemical X was extracted with NaOCH3... they mean that NaOCH3 was added to the organic layer right? And deprotonated chemical X, causing it to be charged and move to the aqueous layer.

But how do you get something to go to the organic layer then? I wish I could find a flowchart online.. it'd make understanding this 999x easier .. b/c I learned about it a looong time ago and remember it being easy.

I realize that. But they each still have two sigma bonds (to neighboring carbons) and two lone pairs of electrons. So are we only to use the lone pair of electrons participating in the cyclic system? Thank you for your help!

HNR2 is sp3 hybridized to minimize repulsion between atoms attached to N and the lone pair on N. This arrangement create maximum stability.

Now, suppose you have a N atom in the aromatic ring. Don't just blindly use the rule which says "add number of sigma bonds and number of lone pairs to get the sum of superscripts for the hybridized orbitals."

If a N is in the aromatic ring, it will be sp2 hybridized if this makes the molecule aromatic (i.e. if this helps create 4n+2 Pi electrons in the Pi system of the ring). The bonding electrons will reside in sp2 orbitals, while the lone pair will be in a p orbital. This too happens for maximum stability. Aromaticity buys you a lot of stability. Stability controls a lot of things in O-Chem.

Having said that, I am retiring from this board for now. If I do return someday, my new moniker shall be "Chairman of the Bored."

HNR2 is sp3 hybridized to minimize repulsion between atoms attached to N and the lone pair on N. This arrangement create maximum stability.

Now, suppose you have a N atom in the aromatic ring. Don't just blindly use the rule which says "add number of sigma bonds and number of lone pairs to get the sum of superscripts for the hybridized orbitals."

If a N is in the aromatic ring, it will be sp2 hybridized if this makes the molecule aromatic (i.e. if this helps create 4n+2 Pi electrons in the Pi system of the ring). The bonding electrons will reside in sp2 orbitals, while the lone pair will be in a p orbital. This too happens for maximum stability. Aromaticity buys you a lot of stability. Stability controls a lot of things in O-Chem.

Having said that, I am retiring from this board for now. If I do return someday, my new moniker shall be "Chairman of the Bored."

We'll miss ya. I'm considering doing the same thing too. I waste too much time on here for my own good (and the September test). Good luck on your August MCAT. You should kill it. :thumbup:

Basic question:

Why is para-nitro phenol more acidic than ortho-nitro phenol?

thanks

Basic question:

Why is para-nitro phenol more acidic than ortho-nitro phenol?

thanksThe close proximity of H, on the hydroxyl group of ortho-phenol, allows it to participate in intramolecular hydrogen bonding with O atom of the nitro group. This type of hydrogen bonding, makes the H less dissociable compared to the para structure. Experimentally, this is validated by proton NMR analysis where the hydrogens, on the hydroxyl groups in the ortho and para structures, will have different chemical shifts.

The endings -oate and -oic acid both refer to carboxylic acids right?

Is one preferred over the other.. or can they both be used interchangeably.

The endings -oate and -oic acid both refer to carboxylic acids right?

Is one preferred over the other.. or can they both be used interchangeably.

oate is used when you have an ester or when you form an anion from carboxylic acid like C6H5COO- Na+ would be sodium benzoate.

Thanks for everything guys. Read the thread while "relaxing-studying" today. Three days to go... :eek:

1) How many cis-trans diastereomers are possible for 1,2,3,4,5-pentachlorocyclopentane?

The answer is 4 (after drawing them out)... is there an easy way to predict this?

2) How to find most shielded proton? What is the chemical shift??

Thanks

1) How many cis-trans diastereomers are possible for 1,2,3,4,5-pentachlorocyclopentane?

The answer is 4 (after drawing them out)... is there an easy way to predict this?

2) How to find most shielded proton? What is the chemical shift??

Thanks

1 - OK, so this might help or totally confuse you (or you just might think I'm stupid), but this is how I think about these problems, if you don't want to/have time to draw it out.

Imagine you have a nice yellow, 3D circle (since this is a cyclo problem) with ten slots in it, five on each side. You've got five red sticks and five blue sticks. How many different ways can you stick them in the slots? So you can do:
- 5 red on one side
- 4 red, 1 blue
- 3 red, 2 blue next to each other
- 3 red, 2 blue one slot away from each other

and that's all the combinations you can get without degenerate ones.

Obviously this doesn't apply to your non-cyclic systems since free rotation about C-C bonds is in place and the substituents aren't ring-locked. If you're asked a question about how many diastereomers/enantiomers there are for a molecule with x chiral centers, the simple formula to figure that out is 2^x.

2 - do you mean deshielded? protons become increasingly deshielding when they're near electronegative elements (O and N, most commonly), double bonds, and triple bonds. you can think deshielding as what happens when a proton gets its electron density sucked away by its greedy neighboring atoms. So when you're looking at an NMR spectrum, the protons you see on the left (4-10 ppm) are deshielded, while the ones on the right (0-3 ppm) are shielded because their electron density isn't being sucked away by electronegative components of the molecule of interest.

how do you determine the strengths of a nucleophile?

is it more or less stable as an anion?

if I is stable as I-, it would mean
a) it's a good LG
b) weak nucleophile because it is happy alone right?

does than mean something like F- is considered a strong nucleophile?

a nucleophile with electron withdrawing groups are weaker than ones with electron donating groups right?

how do you determine the strengths of a nucleophile?

is it more or less stable as an anion?

if I is stable as I-, it would mean
a) it's a good LG
b) weak nucleophile because it is happy alone right?

does than mean something like F- is considered a strong nucleophile?

a nucleophile with electron withdrawing groups are weaker than ones with electron donating groups right?

A strong nucleophile is basically (no pun intended) a strong base. Strong bases are unstable as anions, hence their ease of reaction. Yes, I is stable as I- because, compared to the halogens above it in the periodic table, it has a large atomic radius, hence a negative charge would be diffused over a relatively large space compared to the distribution of a negative charge on, say, F-. Like you said, I- is a weak nucleophile compared to F- for precisely this reason.

Yes, a nucleophile with electron-withdrawing groups is weaker than one without them. This is because when considering their conjugate acids, acid strength increases as you add more electron-withdrawing groups to the compound (think acetic acid (weak) vs. trichloroacetic acid (comprable to sulfuric acid, as I found out today after spilling some on myself)). The stronger the acid (the more willing it is to give up its protons and exist as an anion), the weaker its conjugate base (the harder it fights NOT to reaccept those protons).

how do you determine reactivity?

If I'm givin a list of 5 skeletal structures, and asked which 2 react similarly, and why...

what criteria what I need to determine this?

The H-N-H and H-O-H bond angles in ammonia and water are 107 and 104.5 degrees respectively. What is the s-character in the lone-pair hybrids in these molecules?

Thanks in advance for anyone that can help :cool:

The H-N-H and H-O-H bond angles in ammonia and water are 107 and 104.5 degrees respectively. What is the s-character in the lone-pair hybrids in these molecules?

Thanks in advance for anyone that can help :cool:


I think they have the same degree of s-character. The hybridization of the O in H20 and N in NH3 are both sp3 and hence we expect the same 25% s character in both.

I think they have the same degree of s-character. The hybridization of the O in H20 and N in NH3 are both sp3 and hence we expect the same 25% s character in both.

Thanks! That's what I was thinking, but wasn't 100% sure.

I Just bought The Nuts and Bolts of Organic Chemistry by Karty. has anyone found this text to be useful in studying for the MCAT? I bought it because it teaches you to understand not to memorize. And to cover my bases im taking TPR in october, waiting for EK to come in the mail, and already done with Kaplan big book (however it didnt help much with Ochem). Will TPR give me enough practice? I want to totally pwn the MCAT.

Hello

I have a question for you all regarding incorporating an isotope into an alcohol. I just have to show the mechanism. I checked 2 textbooks and google but no help. You all have been very helpful in the past, so I turn here for help.

T-butyl alcohol is treated with H218O and sulfuric acid. The end product looks just like the beginning, but now the t-butyl alcohol has a 18 in front of the O for the isotope. How do you think this was done? How should I draw it?
__|__OH H218O
| H2SO4
--> __|__18OH
|
Thank you for your time and help.

I have question on what is the definition of Acyl Transfer and why is acid chloride+H2O->carboxylic acid+HCl an Acyl Transfer?

Thanks

I am currently studying 5hrs/day, 5 days a week for the organic chemistry portion of the MCAT. I have only been studying for the MCAT for about a week and would like to know what you would recommend to be an appropiate amout of hours/days studying on the organic chemistry portion of the MCAT. I want to spend at least a month on the oranic chemisty section do you think this will suffice.
thanks

What happened to this thread? No one wants to answer orgo questions anymore?

Edit: I won't answer the previously asked questions because it's already been three weeks, so I'm sure your questions have been answered. If you guys still want answers to those questions, let me know.


5 hrs/day x 5 days/week x 4 weeks = :eek::scared::eek:

Orgo seems to be under-represented in the MCAT (URM :p).

If I remember correctly, they only had simple hydrolysis and acid/base reactions. Maybe some ether stuff, and a few mechanisms as well, but nothing major. My MCAT only had 1 orgo passage, and a few discretes.

Obviously, everyone has strengths and weaknesses, so it's up to you to decide how much time to spend on it.

But in general, I would suggest this breakdown in dividing up your time:
30% Gen Chem
30% Physics
30% Bio
10% Orgo

i've been googling for the past hr and can't find an answer to this. when you have a cycloalkane substituent attached to an alkene, how do you go about assigning priority if you have to extend the comparison out to the ring?

The bigger ring wins (cyclohexane > cyclopentane > cyclobutane).

If the rings have the same size, but one of the rings has a substituent, the ring with the substituent wins.

If the rings have the same size, and both of the rings have substituents, the ring with the heavier substituents wins.


Suppose you have this molecule:

...............B.......C
.................\..../
...................=
................./
...............A


Basically, you have C=C in the middle.
A and B are substituents on the left side, and C is a methyl group on the right side.

If A is a cyclohexane ring and B is a butane ring, A wins. It would be E.
If A is a cyclohexane ring, and B is 1-methyl-cyclohexane, B wins. It would be Z.


I hope this answers your question. If not, post a picture of a problem if possible.

great! thanks so much!


The bigger ring wins (cyclohexane > cyclopentane > cyclobutane).

If the rings have the same size, but one of the rings has a substituent, the ring with the substituent wins.

If the rings have the same size, and both of the rings have substituents, the ring with the heavier substituents wins.


Suppose you have this molecule:

...............B.......C
.................\..../
...................=
................./
...............A


Basically, you have C=C in the middle.
A and B are substituents on the left side, and C is a methyl group on the right side.

If A is a cyclohexane ring and B is a butane ring, A wins. It would be E.
If A is a cyclohexane ring, and B is 1-methyl-cyclohexane, B wins. It would be Z.


I hope this answers your question. If not, post a picture of a problem if possible.

My questions are related to organic chemistry.... let's say you have 3 cyclic compounds that are only different by one group (COOH, CH2CH3, and OCH3) and you are doing column chromatography with a silica gel stationary phase and a 100% hexane mobile phase. the COOH compound will have the largest tR and the CH2CH2 compound will have the shortest tR. however, let's say you change the mobile phase to hexane/3% ammonium hydroxide. which compound will have the biggest change in tR in these 2 experiments with different mobile phases? and, will the compound that has the biggest change in tR be have a longer or shorter tR in the presence of ammonium hydroxide? thanks so much.

My questions are related to organic chemistry.... let's say you have 3 cyclic compounds that are only different by one group (COOH, CH2CH3, and OCH3) and you are doing column chromatography with a silica gel stationary phase and a 100% hexane mobile phase. the COOH compound will have the largest tR and the CH2CH2 compound will have the shortest tR. however, let's say you change the mobile phase to hexane/3% ammonium hydroxide. which compound will have the biggest change in tR in these 2 experiments with different mobile phases? and, will the compound that has the biggest change in tR be have a longer or shorter tR in the presence of ammonium hydroxide? thanks so much.

This question is pretty much the same ol' one you see with extraction. Ammonium hydroxide is a base, so it will remove acidic protons. Only one compound of the three in the mixture gets affected,and hopefully you see which one that is right away. The question now centers around the impact of deprotonation and the subsequent negative charge on the interactions with the polar column. The stronger the affinity for the column, the slower the migration down the column, which ultimately increases the elution time.

I think I'll leave this open ended for you to fill in the final blanks, because this sounds like a lab course question and not an MCAT question.

How does one assign R/S designation to chiral molecules without stereogenic molecules, eg. hexahelicene?

The R/S system is not applicable in this case. Instead, the Plus/Minus system is used:

(+)hexahelicene
(-)hexahelicene

For helix-type molecules:
If a molecule decends clockwise, it is going to be (+)
If it decends counter-clockwise, it will be (-)

This site explains it better than I can: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/stereochemie/weitere_chiralitaetselem.vlu/Page/vsc/en/ch/12/oc/stereochemie/helicale_chiralitaet/helicale_chiralitaet.vscml.html

Can someone help me understand concept of absolute Vs relative configuration?
I tried to read 2 prep books and still don't get it...

Thanks

Can someone help me understand concept of absolute Vs relative configuration?

The absolute and relative configurations are used to identify chiral molecules.

The relative configuration refers to a case where only the R/S configurations are assigned to a molecule.

Example: 2-butanol
Two possibilities exist for this molecule's configuration:
R-2-butanol, and S-2-butanol

http://www.angelo.edu/faculty/kboudrea/molecule_gallery/05_alcohols/butanol-2_02.gif

The R/S system gives you the configuration of the -OH group on the carbon relative to the molecule. If the priorities assigned are clockwise, the molecule will have an R configuration. (note: I'm assuming you already know how to assign R and S to molecules).

On the other hand, the absolute configuration means that the (+) and (-) have been assigned experimentally. An example of this would be if I told you that the (R)-2-butanol was actually (R)-(-)-2-butanol, and the (S)-2-butanol was (S)-(+)-2-butanol. The (+) and (-) designations must be determined experimentally, whereas the R/S can be assigned by just looking at the relative configurations.

In short,
If the (+) and (-) are known, the absolute configuration of the molecule is written.
If the (+) and (-) are not assigned, the relative configuration can be determined by inspection of the molecule using R/S.

Kapeesh?

Thanks Mundane...

Quite different explanation from some textbooks. How does your explanation apply to SN1 and SN2 reaction where it produces racemate and inversion (of relative configuration), respectively?

How do you apply this to a problem? If they give you a chiral molecule and ask you to determine the configuration (R/S), then will this be called relative because we don't know to which direction it will rotate plane-polarized light?












The absolute and relative configurations are used to identify chiral molecules.

The relative configuration refers to a case where only the R/S configurations are assigned to a molecule.

Example: 2-butanol
Two possibilities exist for this molecule's configuration:
R-2-butanol, and S-2-butanol

http://www.angelo.edu/faculty/kboudrea/molecule_gallery/05_alcohols/butanol-2_02.gif

The R/S system gives you the configuration of the -OH group on the carbon relative to the molecule. If the priorities assigned are clockwise, the molecule will have an R configuration. (note: I'm assuming you already know how to assign R and S to molecules).

On the other hand, the absolute configuration means that the (+) and (-) have been assigned experimentally. An example of this would be if I told you that the (R)-2-butanol was actually (R)-(-)-2-butanol, and the (S)-2-butanol was (S)-(+)-2-butanol. The (+) and (-) designations must be determined experimentally, whereas the R/S can be assigned by just looking at the relative configurations.

In short,
If the (+) and (-) are known, the absolute configuration of the molecule is written.
If the (+) and (-) are not assigned, the relative configuration can be determined by inspection of the molecule using R/S.

Kapeesh?

Thanks Mundane...

Quite different explanation from some textbooks. How does your explanation apply to SN1 and SN2 reaction where it produces racemate and inversion (of relative configuration), respectively?

How do you apply this to a problem? If they give you a chiral molecule and ask you to determine the configuration (R/S), then will this be called relative because we don't know to which direction it will rotate plane-polarized light?

Racemic mixtures cannot be identified with (+) and (-) because their specific rotation would be 0. The specific rotation is given by this equation:

Specific rotation = Observed rotation / (Length (dm) x Concentration)

One pure enantiomer would give a positive value. The other enantiomer would give the same value, but it would be negative. So if you have a racemic solution with 50/50 of the enantiomers, the values will cancel out. You will have a specific rotation of 0.

With regard to your SN1 and SN2 question, it is not possible to determine the (+) and (-). Here is why: When you replace the leaving group with another group, the optical activity will change.

For example: Suppose a SN2 reaction causes the -OH group to leave the (R)-2-butanol molecule, and another group, X, attaches to the carbon, inverting the molecule. This inversion will cause the R/S to be changed. But for the (+) and (-), it would have to be determined experimentally. You cannot predict the (+) and (-) of a molecule.

For a given problem, they cannot expect you to identify whether the molecule is going to be (+) or (-) without giving you information like its observed rotation.

Wikipedia lists some values for the specific rotation of molecules.
http://en.wikipedia.org/wiki/Specific_rotation

During oxymercuration, an intermediate is the mercurinium ion (Hg+(OAc) bonded to both Cs across the double bond). Then, water attacks the more highly substituted C (which can hold more + charge) from the backside, breaking the ring and leading to an organomercurial alcohol (following deprotonation of water).

This mechanism seems to greatly resemble Sn2 in that both utilize backside attack on an electrophilic C.

My question is:

Why can water attack a tertiary C (bonded to Hg as part of the mercurinium ion) in oxymercuration, whereas in SN2, the nucleophile CANNOT attack a tertiary C due to steric hinderance ? Isn't there steric hinderance also in the nucleophilic attack of mercurinium ion?

I hope that question made sense. Thanks in advance!

During oxymercuration, an intermediate is the mercurinium ion (Hg+(OAc) bonded to both Cs across the double bond). Then, water attacks the more highly substituted C (which can hold more + charge) from the backside, breaking the ring and leading to an organomercurial alcohol (following deprotonation of water).

This mechanism seems to greatly resemble Sn2 in that both utilize backside attack on an electrophilic C.

My question is:

Why can water attack a tertiary C (bonded to Hg as part of the mercurinium ion) in oxymercuration, whereas in SN2, the nucleophile CANNOT attack a tertiary C due to steric hinderance ? Isn't there steric hinderance also in the nucleophilic attack of mercurinium ion?

I hope that question made sense. Thanks in advance!

Consider the fact that when the Hg atom is bridged, it has three resonance forms. They are the bridged form, along with two minor contributing forms.

Looking at the minor form with the partial positive charge on the substituted carbon, it is essentially a carbocation. Obviously, it's not a "traditional" carbocation, because the double bridged form is the dominant form.

Still, at the instance where the ring opens up to leave the partial positive charge on the highly substituted carbon, consider the geometry of the molecule. A carbon with only three bonds, and no lone pairs --> somwhere between tetrahedral, and trigonal planar (which would have decreased steric hindrance). So the water molecule attacks at this instance, but anti to the mercury (which would still provide steric hindrance).

Let me pre-empt any question you might have about why the same process does not occur on the other bridged carbon. The contribution of that resonance form is negligible, so it does not have a positive charge.

In an SN2 reaction, the nucleophile has to attack first, but it can't due to the hindrance. In this reaction, the nucleophile has to wait for the bridge to open up, so that the carbon will have a partial positive charge. This is actually closer to the behavior of an SN1 reaction.

Consider the fact that when the Hg atom is bridged, it has three resonance forms. They are the bridged form, along with two minor contributing forms.

Looking at the minor form with the partial positive charge on the substituted carbon, it is essentially a carbocation. Obviously, it's not a "traditional" carbocation, because the double bridged form is the dominant form.

Still, at the instance where the ring opens up to leave the partial positive charge on the highly substituted carbon, consider the geometry of the molecule.

Molecules dont flip between resonance forms though, they are in an averaged form of all of them. My professor explained it like this:
Dr. Jekyll is sometimes Dr. Jekyll and sometimes Mr. Hyde. This is an equilbrium between two forms.

Frankenstein is half man and half monster, but he is this all the time, he does not alternate between man and monster. He is a resonance form.

Molecules dont flip between resonance forms though, they are in an averaged form of all of them. My professor explained it like this:
Dr. Jekyll is sometimes Dr. Jekyll and sometimes Mr. Hyde. This is an equilbrium between two forms.

Frankenstein is half man and half monster, but he is this all the time, he does not alternate between man and monster. He is a resonance form.

Haha.. my professor explained it by showing us a picture of Counselor Troi in the episode where she was altered to be a Romulan... and asking us-- "So, anyone know who this is?"

(The idea being that Counselor Troi is half Betazoid, half human... but she doesn't change between being a Betazed and a human- she's half and half all the time.)

I love it.

First of all, you guys have fun orgo teachers.

Second, you're right. The resonance structures don't flip from one to another. But the partial charges still exist on the carbons, as if the molecule did flip from one to another.

And one thing I found in my orgo book...they say that computer modeling shows that the bond between the more substituted carbon and the mercury is longer than the other carbon-mercury bond, which also makes the substituted carbon easier to attack.

Thanks for the replies... and, I have yet another question. This one has been bugging me!

I get that absolute configuration is R/S assignment for naming chiral centers. But the concept of "relative configuration" (D/L, not to be confused with "d/l = +/-")
is endlessly confusing to me... and a question on an EK exam clearly showed me that I did not grasp the concept . :laugh:

Not to repeat the question here, but basically the problem told us a reaction proceeds with "retention of configuration", and we're told the product is (S). Then we're asked to pick the compound that could be a reactant.

I thought, oh, I just have to look for the (S) reactant... but no, that was TOTALLY wrong. The answer was actually the reactant with (R) configuration. In the answer key, the explanation was something like "retention of configuration does not mean absolute configuration is retained; it means there's no inversion."

So basically, we were supposed to draw the (S) product, then get the structure of the reactant from it (which actually ends up being (R) ), and match it to the answer choice.

Can anyone shed some light on the topic? What does "retention of configuration" mean if it doesn't refer to retention of absolute configuration ???

I've tried researched, googled, etc. to no avail. :confused:

When you need to determine organic products for organic reactions (Sn1, Sn2, E1, E2) when you aren't given which reaction is taking place, how can you tell if the compound written above the yield arrow is going to be a nucleophile that will attack the substrate resulting in a substitution reaction or if it is a base that will pull off a Hydrogen resulting in an elimination reaction.

This is my second time around in Organic, doing better this time, but I screwed up the second test, because I couldn't tell if it was a nucleophile or a base. Our professor gave us a chart to determine what type of reaction it was, but it seemed to always come down to know if it was a nucleophile or a base.

Check out the chart:

http://img142.imageshack.us/img142/1193/reactionsxi6.th.jpg (http://img142.imageshack.us/my.php?image=reactionsxi6.jpg)

how can u tell how many a compound can have stereoisomers??

When you need to determine organic products for organic reactions (Sn1, Sn2, E1, E2) when you aren't given which reaction is taking place, how can you tell if the compound written above the yield arrow is going to be a nucleophile that will attack the substrate resulting in a substitution reaction or if it is a base that will pull off a Hydrogen resulting in an elimination reaction.

This is my second time around in Organic, doing better this time, but I screwed up the second test, because I couldn't tell if it was a nucleophile or a base. Our professor gave us a chart to determine what type of reaction it was, but it seemed to always come down to know if it was a nucleophile or a base.

Check out the chart:

http://img142.imageshack.us/img142/1193/reactionsxi6.th.jpg (http://img142.imageshack.us/my.php?image=reactionsxi6.jpg)


Substitution and elimination almost always go hand in hand. Say you have methoxide CH3O- , it's both a strong nucleophile and a strong base. This makes it likely to participate in Sn2 (as a nucleophile) and in E2 (as a base).

Otoh, methanol CH3OH is a comparative weaker nucleophile/base. This makes it likely to participate in Sn1 (as a nucleophile) and in E1 (as a base).

So, okay, the part of determining the order of reaction is easy... but how do you determine whether substitution or elimination predominates?

1) look at the substrate

If the substrate is really hindered (for ex: halide is bonded to a tertiary carbon, or worse yet neopentyl group), Sn2 basically is NOT going to happen.

If there's too much steric hinderance, the nucleophile can't get close enough to the C to participate in substitution. So instead, it's going to abstract a H from the neighboring C instead, allowing elimination to predominate over substitution.

So to state that rule of thumb more clearly: if the reaction involves a *strong* nucleophile/base, you've narrowed down your possible reactions to Sn2/E2. Next, if the substrate is hindered, Sn2 becomes unlikely, so the reaction is going to be E2.

(Note: hinderance of the substrate doesn't affect Sn1/E1 in the same way, as a tertiary carbocation is readily attacked by the nucleophile in Sn1, because it is a charged species, despite the fact that it is tertiary.

This actually has to do with the last question I asked, which I later spoke to my professor about, and that was his explanation for why steric hinderance matters for Sn2, but not for Sn1 or oxymercuration. Basically, steric hinderance is a huge consideration for Sn2, because the tertiary C is a neutral species. Whereas for Sn1, a planar tertiary carbocation carries a positive charge, which is enough to get the nucleophile to overcome steric hinderance and substitute.)


2. look at the nucleophile/base

Is it big and bulky? (like t-butoxide, or triethylamine) Or is it something small like methoxide? Something big and bulky is going to favor elimination- again, because of steric considerations. That is, something big and bulky is more likely to act as a base than as a nucleophile.

So that's also something helpful to keep in mind for synthesis problems, if you want to make the elimination product, while minimizing substitution products: use a bulky base.

Hope that helps.

That does help, but my problem is I always look at what's written above the arrow and treat everything as a nucleophile only (which you're saying is okay anyway) and refer to that chart.

But to my understanding, not all strong nucleophiles can also be strong bases and vice versa. Same for good nuc/bases, or poor nuc/bases.

This is how I determine if the nucleophile is strong:

weak = no negative charge present
moderate = negative charge on a halogen or resonance stabilized
strong = negative charge present, but not on a halogen

Is it okay to do this, or is there going to be a point where I'm going to actually have to sit down and determine if that's a base or a nucleophile right off the bat?

That's a good start, but, as unsung said, you have to pay attention to the "bulkiness" of the nucleophile/base.

One more thing to watch out for with these reactions is temperature. High temperature favors elimination while room temperature (and below) favor substitution.

That's a good start, but, as unsung said, you have to pay attention to the "bulkiness" of the nucleophile/base.

I get what you're saying, but that shouldn't matter right? Look at the chart, Strong nucleophiles pair w/ strong bases, good pairs w/ good, and poor pairs w/ poor.

One more thing to watch out for with these reactions is temperature. High temperature favors elimination while room temperature (and below) favor substitution.

I read that in an o-chem help book, but our professor didn't teach us that or throw any temps on the test. Also he completely dropped the use of polar aprotic solvents (strong indicators of Sn2) this year, so determining reaction was difficult.

It's not always that way though. Ethoxide is a good nucleophile but a weak base, while t-butyoxide is a bad nucleophile but a strong base. A compound won't always be good a nucleophile AND a good base.

I get what you're saying, but that shouldn't matter right? Look at the chart, Strong nucleophiles pair w/ strong bases, good pairs w/ good, and poor pairs w/ poor.


Okay, I don't know about that chart... but do you want to post some problems that you're working on? It's much easier to talk in terms of specifics than in the abstract.

Anyway, I thought of a really crappy metaphor ;). Say you're really really hungry. And there's a delicious cake in the next room through a door. There's also a ham sandwich in the same room as you. If you're really hungry, you might grab both the ham sandwich and the cake. (Or if you're not really very hungry at all, you'll probably just sit there and not grab either.)

Fine. But let's say there's three henchmen guarding the door to the next room... you might be really really hungry- but it's unlikely you're getting through that guarded door to get to the cake. So you'll probably grab the ham sandwich.

Alternatively, say the door isn't guarded, BUT, you're really really obese. So you can't fit through that damn door. So- same result, even though you're really hungry, you're going to have to grab that ham sandwich, instead of the cake.

So, whether or not you eat the cake (act as a nucleophile- Sn2) or eat the ham sandwich (act as a base- E2) has to do with two things: 1) how hungry are you? (really hungry = 2nd order reaction, not so hungry = 1st order reaction) and 2) what do you have access to? (elimination vs. substitution)

You've got part 1) figured out-- how to determine whether a species is potentially "strong" or "weak" (i.e. hungry or not hungry).

BUT just because you're really hungry (potential strong nucleophile/base) doesn't mean you're going to end up grabbing either the cake or the sandwich (*acting* as a nucleophile or base)... perhaps it's hard to get near the cake because of the "henchman" (steric hinderance)... or perhaps you're too fat to get through the door (bulky nucleophile = bad nucleophile).

To determine whether substitution actually occurs or whether elimination actually occurs, it's necessary to consider the circumstances limiting (or promoting) "access". It's not enough to just look at the species itself...

To determine whether substitution actually occurs or whether elimination actually occurs, it's necessary to consider the circumstances limiting (or promoting) "access". It's not enough to just look at the species itself...

If this is the case, then this might just solve my problems, because I'm putting too much thinking into this.

How about let me ask this:

When you're given an organic reaction and you must predict the product, if you just look at what's written above the yield arrow (which will be the nucleophile or the base)... if you cross EVERYTHING else out and just look above the yield arrow... Is there a way to tell if that molecule will be a nucleophile or a base right off the bat?

Also this might be helpful to bring up as well...

So I always thought this was the way to determine strength of the nucleophile:

poor = no negative charge present

good = negative charge on a halogen or a resonance stabilized structure (lone pair, single bond, double bond)

strong = negative charge is present, but not on a halogen

So let me ask this:

If you're looking at the nucleophile (or base) written above the yield arrow and don't take any other part of the problem into account, what are the rules you use to determine the strength of it? Are my rules accurate or do I need to amend them in some way? I know for sure there have been times when my definition of a strong nucleophile have failed me. I should get this straight.

Oh and thanks for your analogy, haha that was great! :laugh:

And I will try to post some example problems that are giving me trouble, but maybe after I get an answer to my above couple of questions, because I might be all right after that.

Thanks kindly.

if you are doing gas chromatography and get overlapping peaks, how would you adjust the following variables to resolve the overlap:
1. sample volume
2. column length
3. column temperature
4. injection rate
5. mobile phase flow rate
i don't understand what would cause the peaks to overlap beside the fact that the gc cannot distinguish them properly. thanks for your help.

Can someone tell me about orbitals and unpaired electrons?

For example, does a lone pair of electrons occupy an orbital? I.e., Is NH3 (with a lone pair attached to N) sp2 or sp3 hybridized?

What about a radical? Is CH3 (radical) sp3 or sp2?

Can someone tell me about orbitals and unpaired electrons?

sp3 with 0 lone pairs = tetrahedral
sp3 with 1 lone pair = trigonal pyramidal
sp3 with 2 lone pairs = bent
sp2 with 0 lone pairs = trigonal planar
sp2 with 1 lone pair = bent
sp with 0 lone pairs = linear

For example, does a lone pair of electrons occupy an orbital?

Yes. Lone pairs and bonds occupy orbitals.

Is NH3 (with a lone pair attached to N) sp2 or sp3 hybridized?

In Ammonia (NH3), the nitrogen atom is sp3 hybridized... meaning all four orbitals are arranged in a tetrahedral structure as you would initially suspect... BUT only three of the orbitals in this arrangement are responsible for bonds, so if you look at how the atoms are connected, you don't see a tetrahedron... INSTEAD you see a trigonal pyramidal arrangement.

What about a radical? Is CH3 (radical) sp3 or sp2?

A radical is just a single electron, not bonded to anything. Basically it's just a lone pair (2 electrons not bonded to anything) minus one of the electrons.



Hope all that helps you some. By the way to anyone viewing this thread, I'm still desperately seeking an answer to my question about three posts up from this one. Thanks kindly.

When you're given an organic reaction and you must predict the product, if you just look at what's written above the yield arrow (which will be the nucleophile or the base)... if you cross EVERYTHING else out and just look above the yield arrow... Is there a way to tell if that molecule will be a nucleophile or a base right off the bat?


I don't think there's a foolproof way right off the bat. Here's what I usually do: I do what you do, which is first, look above at the yield arrow. And from that, what I can determine is "strong" or "weak".

If it's strong, I write preliminarily Sn2/E2. If it's weak, I write preliminarily Sn1/E1. Then I look at the other factors, and I eliminate possibilities that are impossible. (Ex: tertiary C means Sn2 not going to happen, so only E2 is possible).

Although, SOMETIMES, you can get some clues about whether it's going to act as a nucleophile or base JUST from the species itself. And that's when it's something really bulky, like t-butoxide. So yes, if you see t-butoxide, you know first of all, that it's strong. And secondly, you can sort of know that it's more likely to eliminate than substitute, because of it's "bulk". But, this is still not a guarantee, because if you have a really easily approachable substrate (say methyl chloride), I don't think it's impossible for a bulky nucleophile to approach and substitute (please correct me if I'm wrong). Although elimination is still going to be more easy than substitution. Again though, oftentimes, both occur.



Also this might be helpful to bring up as well...

So I always thought this was the way to determine strength of the nucleophile:

poor = no negative charge present

good = negative charge on a halogen or a resonance stabilized structure (lone pair, single bond, double bond)

strong = negative charge is present, but not on a halogen

So let me ask this:

If you're looking at the nucleophile (or base) written above the yield arrow and don't take any other part of the problem into account, what are the rules you use to determine the strength of it? Are my rules accurate or do I need to amend them in some way? I know for sure there have been times when my definition of a strong nucleophile have failed me. I should get this straight.



Okay... so, your rules look good to me. I don't know if this helps any, but basically I just do this:

If what's above the yield arrow is basically a solvent (i.e. methanol CH3OH, or ethanol or whatever), I call that "weak", and write preliminarily: Sn1/E1.

If it's the conjugate base of the solvent (i.e. methoxide CH3O-, or CH3CH2O-, etc.), I call that "strong", and write preliminarily Sn2/E2.

That's pretty much it. It gets trickier if the problem asks to compare between two reagents and determine which one is *better* at Sn1, or Sn2. For ex, all else being equal, is the reagent with the N as the nucleophile the stronger nucleophile, or the reagent that has O? O is more electronegative than N, so that means it holds on to its electrons more tightly, so it's a weaker nucleophile, and N is the stronger nucleophile.

So... it really depends on the type of problem, what my strategy is. It usually involves thinking about how stable the species is. More electronegative = more stable = weaker nucleophile. etc. etc.

Okay... so, your rules look good to me.

Something isn't right with my rules and I think that's what screwing me up. -CN and -N3 are both considered good nucleophiles, but according to my definition they would be strong (negative charge on something other than a halogen).

How about:

strong = any -O or -N... exception being -N3.

Suggestions??

I'll post some of the nucelophiles we are given when I get a chance to give you a better idea what I'm working with, because usually what he gives us are not solvents or conjugate bases of solvents. What type of solvents are you speaking of anyway? The only solvents that come to mind at the moment are polar aprotic (DMF, DME, etc..) and polar protic and these are written under the arrow.

Hi would someone mind helping me with 1H NMR. I tried reading the post in the MCAT forum, but thats very basic... Im lost when it comes to actual problems when you have to figure out the structure of the compound when your given the NMR data, plays alot of mind games.

Can someone please explain the upfield, downfield, N+1, different H's etc

THANKS~!!!

1H NMR

1H NMR is used to identify the positions of hydrogen atoms within a given molecule.

Rule for getting peaks:
2(n)(I) + 1

This means that if there are (n) atoms of hydrogen surrounding a specific hydrogen atom, and they have a spin of (I) (all hydrogens have a spin of 1/2), the signal of that specific hydrogen will contain 2(n)(I) + 1 peaks.

Example:
Ethanol (CH3CH2OH)

http://www.indycar.com/tech/images/ethanol.jpg

In the molecule of ethanol, there are three hydrogens attached to the carbon on the left (A), and two hydrogens attached on the other carbon (B). There is also a hydrogen on the oxygen which is attached to (B).

(A) has three equivalent hydrogens. (Let me know if you don't understand this statement). In a 1H NMR, each of the hydrogens on (A) will have see that there are two neighboring hydrogens on (B). The (A) hydrogens will not see the other (A) hydrogens, because they are all equivalent. Hydrogens can see other hydrogens that are up to three bond lengths away. (There is a bond between the three hydrogens to (A), a bond between (A) and (B), and a bond between (B) and the other hydrogens. Three bonds.

Using the 2(n)(I) + 1 rule, the (A) hydrogens see two neighboring hydrogens, each with a spin of 1/2.
2(2 hydrogens on (B))x(1/2) + 1 = 3. Triplet.

Repeat this rule for the (B) hydrogens:
2(3 hydrogens on (A)) x (1/2) + 1 = 4. Quartet.

Before we do the OH group, let me tell you (very simplisticly) that the oxygen acts as a barrier for the OH hydrogen from seeing the other hydrogens. So if you see a hydrogen connected to an oxygen, it will not be able to "communicate" with the other hydrogens. It will see 0 neighboring hydrogens as a result.

One more time for the hydrogen on the OH group:
2(0 hydrogens) x (1/2) + 1 = 1. Singlet.

Thus, there will be a quartet due to three hydrogens attached to (A), a triplet from the two hydrogens attached to (B), and a singlet for the hydrogen attached to the oxygen.

Now. Upfield/Downfield.
Oxygen has a lot of electon density (look for lone pairs, and double/pi bonds). Electron-rich atoms will shift connected hydrogens downfield (to the left of the 0ppm starting point). Thus, you can expect that the hydrogen singlet on the OH group will be more downfield than the other hydrogen signals.

On the other hand, things with less electron density will not be as downfield as the oxygen. An example would be the transition metals, which are very upfield.

Take a look at the chart in your orgo text book (every book has one) for the upfield/downfield shifts of different groups. You'll notice trends such as the COOH group being very downfield (think about it, a double bonded oxygen, AND a OH group. ahhhh! :scared:).

The key to NMR problems is being able to see different hydrogen environments. In this problem, there are three hydrogen environments, which produce three different signals. If a molecule is symmetric, (propane), there will be one signal from the middle carbon, and one (not two) signals from the terminal carbons. Why? The two terminal carbons are perfectly symmetrical, so the NMR only see's one type of hydrogen environment.

Hope that helps. BTW, you're right, NMR seems like a game. Once you get good at it, I would go as far as to say it's pretty fun. NMR was always my favorite part of orgo.

I finally developed a pretty solid list of different types nucleophiles and found a common trend so that you can determine the strength of any given nucleophile.

I'll upload it here when I get a chance, it should definitely help anyone struggling when predicting organic products when substitution and elimination are happening at the same time and you are trying to figure out which one predominates.

Okay here it is... for students struggling with substitution/elimination reactions, this may solve your problems and frustrations.

If you're asking yourself questions like:

-How can I determine the order of this reaction?
-How can I determine if this reaction is substitution or elimination?
-How can I predict the organic product(s) of this sub./elim. reaction?
-What is the strength of this nucleophile?
-How do I know if this molecule is acting as a nucleophile for substitution or a base for elimination?

The key to predicting if sub. or elim. will predominate so you can predict the outcome, is knowing two things. The substrate (very easy to determine). And the strength of the nucleophile.

For me, my professor did not hand out a list of nucleophiles to memorize and at times the book was difficult to follow, so I was very lost. I have been where you are at right now and after seeing my professor numerous times, discussing with friends, and doing a lot of my own studying/practicing... I organized some very helpful tools that will help you succeed with these reactions.

Here is a chart. You match up your substrate and strength of your nucleophile to determine what type of reaction will predominate.

http://img101.imageshack.us/img101/9664/reactionsgridse2.th.jpg (http://img101.imageshack.us/my.php?image=reactionsgridse2.jpg)

Here are the rules for determining nucleophile strength (I came up with myself after looking at trends and had them confirmed by my professor). Also on this page are tips for differentitating between sub. and elim. and some example nucleophiles drawn out in kekule (lewis dot) form. Just a quick note, the lewis dot structures are not drawn to match molecular geometry.

Forgot to throw temp. rules on this page. 100 degrees C or higher will favor elim. products. Room Temp. (18-23 degrees C) or lower will favor sub. products.

http://img81.imageshack.us/img81/8518/nucstrengthst2.th.jpg (http://img81.imageshack.us/my.php?image=nucstrengthst2.jpg)

Finally here is just a page on some things you might want to know about sub. reactions. Like what a polar aprotic solvent looks like, leaving group (LG) information, and what a pseudohalogen is and what they look like.

http://img209.imageshack.us/img209/4470/extratipsuo0.th.jpg (http://img209.imageshack.us/my.php?image=extratipsuo0.jpg)

I hope this all helps you as it did me. I was able to go back to my old test and correctly do all the problems I missed. I understand this stuff like never before and thanks to my better understanding, I am actually able to help out a couple friends in my class (and you guys here at sdn) that need some clarity.

If you're trying to print these off and are having problems, get into print preview, try landscape view and choosing the shrink image to fit option.

If you have any questions at all about my scans (i.e. how do i use this chart) or any of the info written on the pages, I will be glad to help. Also if you just have any random question about sub./elim. I will be happy to help you the best I can. :)

Okay here it is...

you freakin' rock...I can't begin to tell you how helpful that was...thank you so much...:love:

1H NMR

1H NMR is used to identify the positions of hydrogen atoms within a given molecule.

Rule for getting peaks: 2(n)(I) + 1What if spin-spin splitting isn't 1st order, how do you figure that out?

What if spin-spin splitting isn't 1st order, how do you figure that out?

You figure that out by buying a NMR instrument with a higher resolution. :p

Seriously though, I have never encountered any questions about higher orders of splitting in my Orgo class, my MCAT studies, or even my upper-level chemistry classes. This is because only the older NMR instruments had this problem. Higher resolution NMR instruments have eliminated this problem.

Have you come across a problem about different orders of NMR spectra? Or does your question stem from intellectual curiosity?

foghorn is the mack daddy of chemistry. he was probably asking out of amusement.

Is there a list somewehere of concepts/rxns we don't need to know for the MCAT? I was scanning this thread and saw that alkenes, alkynes & hydroborations are things that won't be tested...but can anyone make/does anyone have a comprehensive list? Thanks!

And I wanted to ask if it's actually true that we don't need to know anything about alkynes (eg, H2/Lindlar's, H2/surface metal) or alkenes (eg, conversions to alcohols or halohydrins)? It seems like it would be focusing mostly on Organic 2 and skipping a lot of organic 1...

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.

http://img230.imageshack.us/img230/2002/organomercdu1.jpg

You figure that out by buying a NMR instrument with a higher resolution. :p

Seriously though, I have never encountered any questions about higher orders of splitting in my Orgo class, my MCAT studies, or even my upper-level chemistry classes. This is because only the older NMR instruments had this problem. Higher resolution NMR instruments have eliminated this problem.

Have you come across a problem about different orders of NMR spectra? Or does your question stem from intellectual curiosity?I'm messing with you dude. cheezer is right.

You don't need a higher resolution NMR apparatus though at the cost of signal peaks not being well-defined. Using different (magnetic) pulse sequences works and also at different magnetization angles. For example, Nuclear Overhauser Effects (NOEs) can help figure out stereochemistry at a specific chiral center. These techniques are taught in upper div/grad O-Chem classes and you'll never see them on the MCAT.

Pascal's (n +1) rule, aka 1st order splitting, taught in basic O-Chem are good for H-atoms separated by not more than 3-bonds i.e., H-atoms are attached on C-atoms adjacent to each other. Therefore the maximum number of peaks observed from this formula is 7. A general rule of thumb is when more than 7 NMR-peaks are observed for a particular proton, it's an indicator of higher ordered splitting.

foghorn is the mack daddy of chemistry. he was probably asking out of amusement.It's the Lounge effect :laugh:
Nah that would be QofQuimica. She has a Ph.D. in Chem. I've only taken one graduate course in NMR theory and that's because I couldn't find another class to fill my schedule last year. And stupid me didn't realize how much work was involved. I was the only undergrad in there so no curve for me :(

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.

http://img230.imageshack.us/img230/2002/organomercdu1.jpg
Hmmmm, I'm pretty sure those 2 are the same compound. I'm too lazy to build a model, but imagine flipping it. All that matters is that the OH is trans to the hydrogen. Also, either way the OH carbon is numbered 1, and the next one would be #2, so no matter how you flip it, it should be the same.

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.You've got to keep in mind Markovnikov's rule for this type of addition reaction plus the fact that the mechanism of the reaction produces two different types of 3-membered ring mercurinium ion intermediate. There's no preference which intermediate forms and both occur with equal probability; the reaction produces a racemic mixture.

Hmmmm, I'm pretty sure those 2 are the same compound. I'm too lazy to build a model, but imagine flipping it. All that matters is that the OH is trans to the hydrogen. Also, either way the OH carbon is numbered 1, and the next one would be #2, so no matter how you flip it, it should be the same.The alcohol molecules are non-superposable mirror images of each other with a plane of symmetry "cutting" though all the C-atoms. A more accurate figure, for either enantiomer, is drawing 2 chair conformations that are fused together to form the bicyclic molecule.

I'm messing with you dude. cheezer is right.

Bah. :thumbdown

I knew something was fishy. You're usually the one answering questions, not asking them.

the reaction produces a racemic mixture.

You are correct sir!

Which is correct? If you have an ester with an ethyl group attached to the oxygen attom and a group XYZ on the other side, should it be named:


A. ethyl XYZ carboxylate

OR

B. XYZ ethanoate?


Kaplan says that A is correct, but according to my textbook, a carboxylate is used to refer to the conjugate base of a carboxylic acid and not an ester.

Which is correct? If you have an ester with an ethyl group attached to the oxygen attom and a group XYZ on the other side, should it be named:


A. ethyl XYZ carboxylate

OR

B. XYZ ethanoate?


Kaplan says that A is correct, but according to my textbook, a carboxylate is used to refer to the conjugate base of a carboxylic acid and not an ester.

Both wrong. Group attached to oxygen-yl, carbon chain on side of carbonyl-oate.

should be ethyl XYZoate

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.

http://img230.imageshack.us/img230/2002/organomercdu1.jpg

Wait a minute, unless I'm not thinking clearly, I believe oxymercuration-demercuration gives the Markovnikov product (i.e. the more highly substituted alcohol). So, the OH should be attached to the tertiary C, not the secondary carbon.

In terms of the stereochemistry of the outcome, in THIS case, you'll only get one product. But in the picture of the products that you drew (which would be the products for hydroboration-oxidation), the two alcohols are enantiomers. Although in this case, oxymercuration-demercuration would give only 1 product, in general, sometimes it gives 1 product, and sometimes it gives 2 products (pair of enantiomers). Although, I'm finding that with my professor at least, he doesn't require us to give both enantiomers for the product (just one of the pair)... but, ymmv.

Wait a minute, unless I'm not thinking clearly, I believe oxymercuration-demercuration gives the Markovnikov product (i.e. the more highly substituted alcohol). So, the OH should be attached to the tertiary C, not the secondary carbon.

In terms of the stereochemistry of the outcome, in THIS case, you'll only get one product. But in the picture of the products that you drew (which would be the products for hydroboration-oxidation), the two alcohols are enantiomers. Although in this case, oxymercuration-demercuration would give only 1 product, in general, sometimes it gives 1 product, and sometimes it gives 2 products (pair of enantiomers). Although, I'm finding that with my professor at least, he doesn't require us to give both enantiomers for the product (just one of the pair)... but, ymmv.
Yeah you just caught my error! The OH would attach at the linkage point rather than the secondary carbon.

edit: Answered my own question.

edit: Answered my own question.
hydride is preferred, if a hydride shift isn't possible, then methyl.

and based on your nomenclature the carbon in the 2nd position has one methyl group and a hydrogen attached to it prior to any rearrangement occurring


*edit I want my two minutes back.

hydride is preferred, if a hydride shift isn't possible, then methyl.

and based on your nomenclature the carbon in the 2nd position has one methyl group and a hydrogen attached to it prior to any rearrangement occurring


*edit I want my two minutes back.

In 2-methylpentan-3-ol, the 2 carbon has two methyl groups attached to it and the hydrogen. For whatever reason, I was thinking of a scenario in which moving either a methyl or a hydride would form a tertiary carbocation.

But in the molecule I provided, moving the methyl off the 2 carbon would create a secondary carbocation at the 2 carbon. So, I deleted it since it was erroneous to begin with.

But thanks for answering my question about which shift would be preferred.

In 2-methylpentan-3-ol, the 2 carbon has two methyl groups attached to it and the hydrogen. For whatever reason, I was thinking of a scenario in which moving either a methyl or a hydride would form a tertiary carbocation.
http://elchem.kaist.ac.kr/jhkwak/OkanaganPdb97/nomenclature/pdb/al2040.gif
I hope Foghorn catches this, but I'll spout off what I think: 2-methylpentan-3-ol has one methyl and one hydrogen attached to the second carbon, otherwise the nomenclature would be 2,2 -dimethylpentan-3-ol and the 2nd carbon would not have a hydrogen. What you're proposing is five bonds to the second carbon and that is impossible. In 2,2 -dimethylpentan-3-ol the only way to form a tertiary carbocation is through the methyl shift, however we're dealing with the above structure, correct?


But in the molecule I provided, moving the methyl off the 2 carbon would create a secondary carbocation at the 2 carbon. So, I deleted it since it was erroneous to begin with.

But thanks for answering my question about which shift would be preferred.
Yeah. Hydride first, methyl second.

http://elchem.kaist.ac.kr/jhkwak/OkanaganPdb97/nomenclature/pdb/al2040.gif
I hope Foghorn catches this, but I'll spout off what I think: 2-methylpentan-3-ol has one methyl and one hydrogen attached to the second carbon, otherwise the nomenclature would be 2,2 -dimethylpentan-3-ol and the 2nd carbon would not have a hydrogen. What you're proposing is five bonds to the second carbon and that is impossible. In 2,2 -dimethylpentan-3-ol the only way to form a tertiary carbocation is through the methyl shift, however we're dealing with the above structure, correct?

Eh.

2-methylopentan-3-ol is the same as saying 1,1-dimethylbutan-2-ol (but calling it this does not follow IUPAC rules).

So it does, in fact, have two methyl groups attached, although one of them is part of the pentane chain. Carbon #2 in the pentane chain does have two methyl groups, technically.

Eh.

2-methylopentan-3-ol is the same as saying 1,1-dimethylbutan-2-ol (but calling it this does not follow IUPAC rules).

So it does, in fact, have two methyl groups attached, although one of them is part of the pentane chain. Carbon #2 in the pentane chain does have two methyl groups, technically.
oh, i see what he was getting at...duh!

Thanks for the replies... and, I have yet another question. This one has been bugging me!

I get that absolute configuration is R/S assignment for naming chiral centers. But the concept of "relative configuration" (D/L, not to be confused with "d/l = +/-")
is endlessly confusing to me... and a question on an EK exam clearly showed me that I did not grasp the concept . :laugh:

Not to repeat the question here, but basically the problem told us a reaction proceeds with "retention of configuration", and we're told the product is (S). Then we're asked to pick the compound that could be a reactant.

I thought, oh, I just have to look for the (S) reactant... but no, that was TOTALLY wrong. The answer was actually the reactant with (R) configuration. In the answer key, the explanation was something like "retention of configuration does not mean absolute configuration is retained; it means there's no inversion."

So basically, we were supposed to draw the (S) product, then get the structure of the reactant from it (which actually ends up being (R) ), and match it to the answer choice.

Can anyone shed some light on the topic? What does "retention of configuration" mean if it doesn't refer to retention of absolute configuration ???

I've tried researched, googled, etc. to no avail. :confused:

Okay geniuses... surely somebody can shed some light on my question? Please?

Okay geniuses... surely somebody can shed some light on my question? Please?

Can you post the exam question that you are refering to (if it's allowed on SDN)?

In the meantime, this might help:

(Note: I stole the following information from http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch07configurations.html)


http://www.mhhe.com/physsci/chemistry/carey/student/olc/graphics/carey04oc/ch07/figures/ch76a.gif
In this diagram, they say that the absolute configuration for the parent molecule is known.

After the reaction with the TsCl, only the relative configuration is known.

In this case, since the parent molecule was R, and the product is R, it is known as retention of the configuration (with respect to the relative configuration).

The image goes on to show that since the middle molecule reacts with KCN, the configuration is inverted, so the configuration is not retained.

Absolute is +/-, and R/S
Relative is only R/S

I have a question re: identifying the most stable isomer and exactly how do 'draw' it so that comparisons can be made.

For example.

Molecule A = cyclohexane with an isopropyl group on C1 projecting towards the viewer, and a methyl group projecting towards the viewer on C2.

Molecule B = exact same except that the C2-Methyl projects away from the viewer....


What are the rules in determining what is equitorial and what is axial when transforming them into chair conformations?

I have a question re: identifying the most stable isomer and exactly how do 'draw' it so that comparisons can be made.

For example.

Molecule A = cyclohexane with an isopropyl group on C1 projecting towards the viewer, and a methyl group projecting towards the viewer on C2.

Molecule B = exact same except that the C2-Methyl projects away from the viewer....


What are the rules in determining what is equitorial and what is axial when transforming them into chair conformations?Learn the differences, how to manipulate and what type of information you can get from chair conformations, Newman projections and Fisher projections. If you can do that, you can answer your question(s).

How can I systematically break down an NMR graph to identify a compound? Thanks again.

How can I systematically break down an NMR graph to identify a compound? Thanks again.

Sorry, but that requires a lot of explanation, and visual aids. I'd suggest venturing into the learning process on your own and then ask clarifying questions. You have access to a text book, right?

I found a great tutorial online long ago. I'll dig around for it.

How can I systematically break down an NMR graph to identify a compound? Thanks again.If this is a serious question and you're not just messing with me cheesehead, I can come with some basics and post it on the thread in a couple of days.

If this is a serious question and you're not just messing with me cheesehead, I can come with some basics and post it on the thread in a couple of days.
I think the question might be way to broad...BUT if you have time to post some basics for me that'd be cool. A few things to narrow it down:

1. I'm in Ochem 1
2. I'm covering Proton NMR right now.
3. Assume that I understand deshielding and the relationship spin-spin splitting has with the number of peaks on a graph (neighboring protons, n+1 etc.)

After that I don't know. I get to the point where I count the number of peaks, I identify the neigboring protons...and then I look at the ppm of the peak, compare it to some chart, and try to look for a group with the same ppm. That's when things start getting hazy and I waste a bunch of time. Anyways, is it just a matter of practice? Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)

I think the question might be way to broad...BUT if you have time to post some basics for me that'd be cool. A few things to narrow it down:

1. I'm in Ochem 1
2. I'm covering Proton NMR right now.
3. Assume that I understand deshielding and the relationship spin-spin splitting has with the number of peaks on a graph (neighboring protons, n+1 etc.)

After that I don't know. I get to the point where I count the number of peaks, I identify the neigboring protons...and then I look at the ppm of the peak, compare it to some chart, and try to look for a group with the same ppm. That's when things start getting hazy and I waste a bunch of time. Anyways, is it just a matter of practice? Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)
Can you integrate the number of protons under a graph, etc? If you can, it seems you have sufficient knowledge and you just need to bust out some abstract thinking. It takes practice.

Maybe post an example and we can walk you through it (although its been over a year for me).

Can you integrate the number of protons under a graph, etc? If you can, it seems you have sufficient knowledge and you just need to bust out some abstract thinking. It takes practice.

Maybe post an example and we can walk you through it (although its been over a year for me).
I see. Perhaps I haven't done enough practice problems.

Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)

Err..I don't think so.


Isopropyl group = (CH3)2-CH-R

R may have it's own hydrogens, which would not give a septet.


There is also the possibility that you get a septet + doublet, but it is not an isopropyl group. Try this NMR:

CH(CH2OH)3

This would also give a septet and a doublet (along with a singlet from the OH groups). But it's not an isopropyl group.


I guess it's good to remember that a septet + doublet might be an isopropyl group, but there are always exceptions. Always double check to make sure that your assumption is right. I used tricks like that for starting points, but I always check to make sure that they made sense after I was done with the whole problem.

I still remember a few tricks:
triplet + quartet @ 1-4 ppm = ethyl group
singlet @ 5-6 ppm = OH group
multiplet @ 7 ppm = phenyl group

So yeah, always double check. :thumbup:

can someone explain to me the reactivity rules for Aldehydes/ketones/carboxylic acids --> alcohols...furthermore the steric effects and the electric effects

thanks

I think the question might be way to broad...BUT if you have time to post some basics for me that'd be cool. A few things to narrow it down:

1. I'm in Ochem 1
2. I'm covering Proton NMR right now.
3. Assume that I understand deshielding and the relationship spin-spin splitting has with the number of peaks on a graph (neighboring protons, n+1 etc.)

After that I don't know. I get to the point where I count the number of peaks, I identify the neigboring protons...and then I look at the ppm of the peak, compare it to some chart, and try to look for a group with the same ppm. That's when things start getting hazy and I waste a bunch of time. Anyways, is it just a matter of practice? Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)Here's my handy dandy H-NMR problem solving guide. (http://mihd.net/849dps)

Yeah, recognizing splitting patterns helps deduce what types of adjacent protons are present. The absence of proton signals in a spectrum also helps exclude out (some) functional groups.

Use the info available whether it's from H-NMR, C-NMR, IR or mass spec. They all complement each other & help confirm whether or not specific functional groups are absent/present in the molecule.

Here are the other spectral analysis:
Mass Spec (http://img248.imageshack.us/img248/6113/msgt8.jpg)
IR (http://img248.imageshack.us/img248/3407/irpp4.jpg)
C-NMR (http://img248.imageshack.us/img248/3811/carbonoh6.jpg)

Here's my handy dandy H-NMR problem solving guide. (http://mihd.net/zvrugd)

Yeah, recognizing splitting patterns helps deduce what types of adjacent protons are present. The absence of proton signals in a spectrum also helps exclude out (some) functional groups.

Use the info available whether it's from H-NMR, C-NMR, IR or mass spec. They all complement each other & help confirm whether or not specific functional groups are absent/present in the molecule.
you rule. did i ever tell you that you rule? well you do.

can someone explain to me the reactivity rules for Aldehydes/ketones/carboxylic acids --> alcohols...furthermore the steric effects and the electric effects

thanks
I'm not sure what the question is asking exactly. Do you want to know how reactive they are relative to alcohols or in terms of conversion to alcohols?

I have a question re: identifying the most stable isomer and exactly how do 'draw' it so that comparisons can be made.

For example.

Molecule A = cyclohexane with an isopropyl group on C1 projecting towards the viewer, and a methyl group projecting towards the viewer on C2.

Molecule B = exact same except that the C2-Methyl projects away from the viewer....


What are the rules in determining what is equitorial and what is axial when transforming them into chair conformations?

The wording of this question is a little confusing, but I'm going to try and take a crack at it. With regard to equitorial and axial, if I'm remembering the rules with stability and cyclohexanes correctly, cyclohexanes are always more stable when their constituent groups are in the equitorial position. That is, the methyl goup and isopropyl group are positioned parallel to the 'equator' of the chair conformation. I'm not really sure how this translates into projecting to the line of vision of the viewer. In this case, it really depends on how you're looking at the molecule. I suggest making a molecular model, if you have those little building blocks thingy's.
It may be easier if you think about this question in terms of steric hindrance rather than thinking about axial, equitorial conformations, and chair conformations. If both bulky groups are facing you, then the methyl and isoproply group are on the same side and they each experience mutual repelling forces from one another. If, however, one group faces you and the other projects away from you, then they are on opposite sides, and the repelant force that each group experiences is reduced. Hope this helps !

you rule. did i ever tell you that you rule? well you do.I added the other spectra. See above. Also, I misnamed the actual molecule on the NMR guide example. It's supposed to be 3-pentanone NOT 3-propanone. I made the change on the pdf and posted a new link.

deleted

How do you know what the product looks like after an internal rearrangement that results in ring contraction or expansion? How can you systematically know where the carbocation ends up, and where the substituents go? Take this example below:

http://img266.imageshack.us/img266/8388/organicquestionha3.jpg

How did they know the carbocation went in the top right corner? I would have drawn the alkyl bromide on the top right corner and put the carbocation in the bottom right. And I really don't see where the 3-membered ring from the left side disappeared to. Can anyone demystify this for me?

Hi DCODY,
Here is my explanation, which might or might not help you (I hope it does!). I had a difficult time describing what happened in words and without drawings.

1. "And I really don't see where the 3-membered ring from the left side disappeared to."-->You see the triangle on the left (in the first step of the mechanism)? Where you see an arrow drawn, the electrons from that leg of the triangle attacked the bottown left corner of the triangle on the right(forming a bond there instead). Count the number of carbons now in the ring=4. Therefore, the triangle (3-membered ring) became a box (4-membered ring).

2. "How did they know the carbocation went in the top right corner? I would have drawn the alkyl bromide on the top right corner and put the carbocation in the bottom right"--> This is due to the attack of the electrons. Number the carbons in the 4 membered ring. Do you see that the electrons attacked carbon number 4 (if you numbered them clockwise)? Carbon number 4 is also where the alkyl bromide is located. That is why the alkyl bromide is on the bottom right (because that is where carbon number 4 is in the 4 membered ring after the attack).

Afterwards, there had to be a rearrangement because carbocations are more stable if they are tertiary than if they are secondary. I'm not sure why the proton is the one that did the rearrangement, but that is what I see more commonly than alkyl groups moving. Since the alkyl bromide did not move, the bottom right corner of the 4 membered ring would be a tertiery carbocation due to the rearrangement, and therefore that is where the carbocation is supposed to be located.

How do you know what the product looks like after an internal rearrangement that results in ring contraction or expansion?

In reality, it must be done empirically via experiments. (Organic) reactions are rarely 100% efficient, that's why purification steps are done in order to separate reaction components and products from each other. Among other things, the type of experiment(s) done can reveal what's occurring thermodynamically, kinetically and chemical structure(s) formed.

One of the driving forces for ring opening/rearrangement(s) is to relieve torsional and angle strain. The combination of these two is called ring strain and 3/4-membered rings are more "rigid" chemical structures relative to other larger cycloalkane molecules.

In the example, the cyclopropane analog relative to the cyclobutane analog has greater angle strain even though both are sp3 hybridized at all ring C atoms. If you recall, sp3 hybridization produces molecular orbitals with tetrahedral geometric angles of 109.5 degrees. It's preferable to maintain these angles in sigma bond formation rather than distorting, ~60 degrees (3-membered ring) or ~90 degrees (4-membered ring), because maximum overlap between orbitals is attained. Deviation from 109.5 degrees results in sub-optimal orbital overlap and a weakened sigma bond.

With regards to torsional strain, a 3-membered ring has greater torsional potential energy relative to a 4-membered ring. This energy results from the structure having one conformation. Viewing any two adjacent ring C atoms via Newman projection(s), cyclopropane's ring substituents have a completely eclipsed conformation similar to this. (http://img105.imageshack.us/img105/5391/prop1lg8.gif) Keep in mind that another C atom is bonded to both C atoms in order to form a cyclopropane ring with internal angle ~60 degrees and all C atoms in the ring are coplanar. A "rigid" cyclopropane structure essentially "locks" substituents in a completely eclipsed conformation therefore there's little or no twisting/torsion along a C-C sigma bond to minimized repulsive forces between substituents unlike 4/5/6-membered rings where "puckering/folding" or (a) chair conformation(s) allows for partial or complete relief from angle and/or torsional strain.

In cyclobutane, ring C-atoms aren't all coplanar and structurally has more "wiggle" room since bond angles aren't as distorted relative to cyclopropane. As consequence, the internal bond angle between two adjacent sigma bonds in the ring "folds", allowing ring substituents to attain a slightly non-eclipsed conformation thus reducing torsional strain/potential energy. The "folding" comes at the energetic cost of having a bond angle slightly less than 90 degrees, but this is more than offset by a gain in overall stabilization energy from torsional relief.

For more in depth material on how to propose organic reaction mechanisms, the book authored by R.B. Grossman (http://img233.imageshack.us/img233/1462/reasonableorganicreactijb7.jpg) is an excellent reference.

I'm not sure what the question is asking exactly. Do you want to know how reactive they are relative to alcohols or in terms of conversion to alcohols?

Yes relative to alcohols, and general reactions of them please
thanks

Why does benzene not undergo X2 halogenation? I understand the bonds are delocalized, but there are still pi orbital electrons ready for bonding, no?

What does being "soluble" in dilute acid/base mean?

Does a substance have to be a base to be soluble in dilute acid?

What does being "soluble" in dilute acid/base mean?


Soluble means that a decent amount of the compound can dissolve in the solvent. Diethyl ether is soluble in toluene, but not in water for example. Solubility is in general, related to how polar the molecule is.


Does a substance have to be a base to be soluble in dilute acid?

No.

The reason why I was asking was b/c I encountered a question asking us to choose whether a compound had a carboxylic acid or an amine... and the "condition" was that it was soluble in dilute acid

The answer was amine.. the explanation was that the amine was basic.

Soluble means that a decent amount of the compound can dissolve in the solvent. Diethyl ether is soluble in toluene, but not in water for example. Solubility is in general, related to how polar the molecule is.



No.



i dont think soo..=):sleep:

i dont think soo..=):sleep:

A neutral compound might need to be protonated to dissolve into some polar phase. Its done all the time in extractions.

The reason why I was asking was b/c I encountered a question asking us to choose whether a compound had a carboxylic acid or an amine... and the "condition" was that it was soluble in dilute acid

The answer was amine.. the explanation was that the amine was basic.
the assumption i will make is that dilute acid means dilute acid in a polar solvent (like water). The carboxylic acid will remain neutral (as in, protonated or not deprotonated) which reduces it's solubility while the amine, being basic, will easily become protonated (and thus, charged) which makes it more soluble.

The question is poorly written if that is indeed the wording of it. For MCAT purposes...you can think of "like dissolves like" for solubility purposes (unless it's one of those big Ksp type questions, but those are usually in the Physical section). When they say "...because it's basic"...interpret that as "because it gets charged as a result of protonation". If the same question was asked but it was in a dilute base, the answer would be the carboxylic acid because it is an acid and when put in an alkaline solution, the acid becomes deprotonated and therefore, it becomes charged, and therefore, more soluble in a polar solvent.

What role does heat play with acidification?

For example, in alcohol dehydration to alkene by addition of concentrated acid and heat.

What role does heat play with acidification?

For example, in alcohol dehydration to alkene by addition of concentrated acid and heat.

heat...so beautiful. Oftentimes used for thermodynamic control where you would otherwise get a kinetic product. Heat makes atoms vibrate faster, thus increasing the rate at which reactants come in contact with each other. In thermodynamic control, the thermodynamic product is more stable, but the kinetic product is formed faster due to having a lower activation energy, thus heat is used to make sure that thermodynamic product formation is more favored.

1>. Can someone help me with disproportionation reaction? What is it and what are some examples?

2>. Regarding interconversion of: gauche/anti/eclipsed (conformational isomers) and chair/boat/twist (cyclohexanes), do they ALL exist in room temperature?


I am reading some posts and some of them relate to alkenes. I don't know when this starts, but alkene is no longer on the MCAT.

1>. Can someone help me with disproportionation reaction? What is it and what are some examples?

2>. Regarding interconversion of: gauche/anti/eclipsed (conformational isomers) and chair/boat/twist (cyclohexanes), do they ALL exist in room temperature?


I am reading some posts and some of them relate to alkenes. I don't know when this starts, but alkene is no longer on the MCAT.

A disproportion reaction is when not all of one reactant ion can be found 100% in one product. An example would be:

Cl2(g) + 6 OH-(aq) --> 5 Cl- + ClO3-(aq) + 3 H2O(l)

The chloride ion exists both in aqueous phase and as part of chlorate.

As for your second question, I'm not an expert on this but if I recall from organic chemistry all conformations exist in room temperature, just some more than others. The most common is obviously the chair because it is the least sterically hindered, with twist boat being more hindered than chair and regular boat conformation more hindered than both. For the same reason of steric hindrance, you find the conformation of atoms in a bond as anti, then gauche, then eclipse. It's all about which conformation has the lowest energy. Think lower energy to form = more stable.

http://content.answers.com/main/content/wp/en/b/bf/Conformers2.jpgI think you're on the spot SB100. Here's an image that will help you visualize in case u ever forget though:
http://http://content.answers.com/main/content/wp/en/b/bf/Conformers2.jpg

A disproportion reaction is when not all of one reactant ion can be found 100% in one product.

Another definition is that a molecule is disproportionated when part of it is reduced and the other part is oxidized.

Another definition is that a molecule is disproportionated when part of it is reduced and the other part is oxidized.

That too, although I never learned this in general chemistry and haven't come across it in my MCAT practice. I don't think this would be on the exam unless it was a passage.

A have a question regarding acid catalysts:

During, for example, ester synthesis, when reacting the carboxylic acid and the alcohol, it is wise to react this under acidic conditions. This protonates the carbonyl oxygen, making the carbonyl carbon even more partial positive, and therefore more available to nucleophilic attack by the hydroxyl oxygen. My question:

How is a hydroxyl carbon more partial positive than a carbonyl carbon? I've always thought that a carbonyl group is more polar than a hydroxyl group. Thanks-

A have a question regarding acid catalysts:

During, for example, ester synthesis, when reacting the carboxylic acid and the alcohol, it is wise to react this under acidic conditions. This protonates the carbonyl oxygen, making the carbonyl carbon even more partial positive, and therefore more available to nucleophilic attack by the hydroxyl oxygen. My question:

How is a hydroxyl carbon more partial positive than a carbonyl carbon? I've always thought that a carbonyl group is more polar than a hydroxyl group. Thanks-
one thing you must remember about carbonxyl groups is that the negative charge is "shared" between the oxygens which means that the electron density is also slightly spread onto the carbonyl carbon. By protonating the oxygen, there is no longer a negative charge which can distribute over the carbon which causes it to become more electropositive and therefore more nucleophilic.

I remember a few different reasons for making the conditions acidic, this being one of them. Other reasons i've heard was that the acid is used to stabilize the intermediate and to help form a good leaving group.

...the negative charge is "shared" between the oxygens which means that the electron density is also slightly spread onto the carbonyl carbon...

Ahh, I do recall the dotted line spanning from oxygen to oxygen, going across the carbonyl carbon. Thanks.

Alcohol + HCl = Alkyl Halide

This can happen via a SN1 or SN2 reaction

According to an AAMC question, when we were given no additional information, the question basically gave a buncha compounds (primary, secondary, tertiary) and asked us which one was most readily prepared from an alcohol + hcl.

The answer was a tertiary compound (tert butyl chloride).... but couldn't it be a primary compound too (methyl chloride)? we don't know if a SN1 or SN2 reaction occured...

Alcohol + HCl = Alkyl Halide

This can happen via a SN1 or SN2 reaction

According to an AAMC question, when we were given no additional information, the question basically gave a buncha compounds (primary, secondary, tertiary) and asked us which one was most readily prepared from an alcohol + hcl.

The answer was a tertiary compound (tert butyl chloride).... but couldn't it be a primary compound too (methyl chloride)? we don't know if a SN1 or SN2 reaction occured...

The universe gravitates towards stability. The question asks which is more readily prepared, not which ones can be prepared.

Alcohol relative reactivity order : 3o > 2o > 1o > methyl.

Preparation of a 3o alkyl halide is always faster than an a 1o alkyl halide because formation of a tertiary carbocation is far more favored. Also, recall the rates of SN1 vs. SN2:

SN1 rate = k [R-LG]
SN2 rate = k [Nu][R-LG]

Only the rate of SN2 depends on the nature of the nucleophiles, and better nucleophiles favor SN2. The reactivity trend of hydrogen halides is HI > HBr > HCl > HF.

Both the fact that a 3o alcohol is more reactive than a 1o alcohol and the fact that HCl is not as strong of a nucleophile to react faster with a 1o alcohol by SN2 than it reacts with a 3o alcohol by SN1 mean that a 3o alkyl halide is the one more readily formed by a reaction of a 3o alcohol with hydrogen chloride.

Alcohol + HCl = Alkyl Halide

This can happen via a SN1 or SN2 reaction

According to an AAMC question, when we were given no additional information, the question basically gave a buncha compounds (primary, secondary, tertiary) and asked us which one was most readily prepared from an alcohol + hcl.

The answer was a tertiary compound (tert butyl chloride).... but couldn't it be a primary compound too (methyl chloride)? we don't know if a SN1 or SN2 reaction occured...
This reaction is almost exclusively Sn1, because -OH is a bad leaving group for Sn2 reactions. Reaction with acid converts OH to a better leaving group. It leaves, and a carbocation is formed. Because 3o carbocations are most stable (compared to primary and secondary), the tertiary substitution product is formed most readily.

Hey guys… I’ve got 3 questions for the organic chem. geniuses out there =)

1) A carboxylic acid is polar … and favors water…. Why is it that when you do an extraction, the carboxylic acid is in the ETHER solvent… in the first place? I understand that when it is DEPROTONATED, it will belong in water… but why is it in the ether? Or am I wrong?

2) I’m in OChem lab right now… and my professor said that a carboxylate anion is “IONIC” …. I thought the term ionic only referred to compounds with both metals and nonmetals… can it really be used for covalent compounds

3) What exactly does a “salt” mean… I thought it just meant something that disassociates in water into a nonmetal and a metal (NaCl) for example

Why is a carboxylate anion considered a salt… can anyone define a salt in an easy to understand manner

Thanks a lot!

1) A carboxylic acid is polar … and favors water…. Why is it that when you do an extraction, the carboxylic acid is in the ETHER solvent… in the first place? I understand that when it is DEPROTONATED, it will belong in water… but why is it in the ether? Or am I wrong?

The neutral carboxylic acid is amphopathic, meaning it can dissolve into both water and organic solvent. Because we typically use saltwater (brine solution) in this experiment, the solubility of the neutral organic acid in water is reduced. As a result, it shows a preference for the organic solvent. But the reality of extraction is that it depends on the partition coefficient (the ratio of how the solute distributes itself between the two solvents), so the organic acid is actually dissolving into both solvents, just not equally.


2) I’m in OChem lab right now… and my professor said that a carboxylate anion is “IONIC” …. I thought the term ionic only referred to compounds with both metals and nonmetals… can it really be used for covalent compounds

Ionic means it's an ion, be it a cation or anion. The carboxylate anion carries a negative charge, which makes it an ion. For every anion (negative charge), there must be a cation (positive charge). In organic chemistry, we just don't focus on it. But for every carboxylate anion, there is likely a sodium cation (or whatever counterion you used with the based added to deprotonate the carboxylate).

3) What exactly does a “salt” mean… I thought it just meant something that disassociates in water into a nonmetal and a metal (NaCl) for example

Why is a carboxylate anion considered a salt… can anyone define a salt in an easy to understand manner

That is exactly what a salt means: a compound that generates an anion and a cation. The carboxylate is half of the salt, with the unnamed cation being the second part of the salt.

The salt is NaO2CR

The neutral carboxylic acid is amphoteric, meaning it can dissolve into both water and organic solvent. Because we typically use saltwater (brine solution) in this experiment, the solubility of the neutral organic acid in water is reduced. As a result, it shows a preference for the organic solvent. But the reality of extraction is that it depends on the partition coefficient (the ratio of how the solute distributes itself between the two solvents), so the organic acid is actually dissolving into both solvents, just not equally.



Ionic means it's an ion, be it a cation or anion. The carboxylate anion carries a negative charge, which makes it an ion. For every anion (negative charge), there must be a cation (positive charge). In organic chemistry, we just don't focus on it. But for every carboxylate anion, there is likely a sodium cation (or whatever counterion you used with the based added to deprotonate the carboxylate).



That is exactly what a salt means: a compound that generates an anion and a cation. The carboxylate is half of the salt, with the unnamed cation being the second part of the salt.

The salt is NaO2CR

thanks a lot.. I really appreciate it

.

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?


Read this. Page 170

http://books.google.com/books?id=ZCXPte7PL0EC&pg=RA1-PA170&lpg=RA1-PA170&dq=acidity+alkenes&source=web&ots=pjQxc5RbqO&sig=FfIfhzc5qWXNC8KFe7LE9NM8p0c#PRA1-PA170,M1

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?

That's for "just" an orbital. An S orbital is more stable than a P orbital, so an SP orbital is more stable than a P orbital because it has more S character. So as far as stability (via low energy) of "orbitals" go, S > SP Hybridized > P.

A HCCH molecule has its carbons in an SP state. BUT, there are a total of total of 4 orbitals, and SP only accounts for 2! The other two orbitals are P orbitals, which are high energy orbitals, and thus more reactive.

So, since an SP3 has 0 P orbitals, and an SP2 has 1 P orbital, and an SP has 2 P orbitals, in order of stability hybridized atoms we have: SP3> SP2 > SP.

Those extra P orbital, which want electrons really bad, are what cause alkenes and alkynes to try really hard to pull electrons, pulling them away from the -H, making those H more acidic.

Its been a year since O-Chem, so I hope I got this right, but I believe that's the case.

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?

A much easier way to think about this problem can be read on page 34 of the BR Organic Chemistry book. Basically, it states that we need to consider that the concept of acidity deals with bonds breaking in a heterolytic fashion (into ions). With an sp-orbital, the electrons are closer to the carbon nucleus than they are with the other hybird orbitals. Being that they are so close to carbon, the carbon holds on to both electrons more tightly than other hybrids (in essence, it's more electronegative). An sp-hybridized carbon can take the electron pair from the C–H bond and form a C:- and H+ more readily than longer hybrid orbitals.

Why are substances with double bonds UV active? Or has that already been discussed?

Why are substances with double bonds UV active? Or has that already been discussed?
I'm not sure "why" is as important as knowing over what range of wavelengths that conjugated systems are photoactive--that would be the UV range, 200nm-400nm.

But, I'm not sure what level of detail you're looking for in an explanation as to "why", but it has to do with the fact that it just so happens that photons in the UV region have just the right of energy to promote an electron from a conjugated system into a higher energy state. Electrons and photons experience a 1:1 collision, and unless the photon has "just the right" amount of energy (it can be too little or too much), the electron won't be promoted to a higher energy level. If it does have the right amount of energy it will promote an electron to a higher energy level (after which time it will drop to a different energy level emitting a photon of light)

Hi - I have a couple of quick orgo questions and I was hoping someone could help me out.

1. Nucleophiles can often attack unhybridized p-orbitals from either top or bottom side. Is there a way to know which side the nucleophile will attack from?

2. Acidic protons (protons attached to alpha carbons in carbonyl groups) are acidic because the lone pair that's left once alpha carbon is deprotonated can create resonance with the carbonyl. This provides stability..I understand that part but why does that mean that the proton is acidic (what is the relationship between being stable and acidic?)

3. Why is it that when an aldehyde or a ketone is protonated (carbonyl oxygen), the carbonyl carbon becomes more electrophilic than it was before?

Thanks so much!

pyran with a carbocation para to the oxygen?

1. No. Purely kinetics, if it is not being sterically hindered.

2. You first need to think conceptually about what an acid is. It is something capable of donating a H+. Now the strength of an acid is based on how likely it is to donate a H+. The more likely it donates, the stronger it is. So, if the molecule that donates a proton is able to assume a stable conformation, it is more likely to 'let go' of the proton. It's an energy thing. Everything wants to assume the lowest energy state possible.

3. edited, because it was wrong!

To answer question number 3:

The reason why the cabonyl carbon becomes more electrophilic (more prone to nucleophilic attack) is because you now have an oxygen with a positive charge (+) that is going to be even more electron withdrawing thus decreasing the electron density around the carbonyl carbon. You can also look at it in terms of resonance. When carbonyl oxygen gets portonated, you can see that you can delocalize the pi electrons onto the oxygen and you get a positive charge on the carbon in the other contributor.

good answer for #3. That sounds much better! :)

Hi - I have a couple of quick orgo questions and I was hoping someone could help me out.

1. Nucleophiles can often attack unhybridized p-orbitals from either top or bottom side. Is there a way to know which side the nucleophile will attack from?

2. Acidic protons (protons attached to alpha carbons in carbonyl groups) are acidic because the lone pair that's left once alpha carbon is deprotonated can create resonance with the carbonyl. This provides stability..I understand that part but why does that mean that the proton is acidic (what is the relationship between being stable and acidic?)

3. Why is it that when an aldehyde or a ketone is protonated (carbonyl oxygen), the carbonyl carbon becomes more electrophilic than it was before?

Thanks so much!


1. In many cases, the nucleophile will tend to attack from the site that is less sterically hindered (cluttered by large groups such as a tert-butyl substituent). The reason for this is based on kinetics...it is faster to attack a group that is minimally hindered and has only protons attached than it is to attack a carbocation, for instance, that has two bulky groups adjacent to it. In the case of a carbonyl group such as an aldehyde or ketone, both of which contain sp2 hybridized carbons, the nucleophile can theoretically attack the electrophilic carbonyl from both sides and therefore has a 50% of attacking one side and 50% the other. This would result in the formation of a pair of enantiomers. In this particular case, it comes down to equal probability of attacking the sp2 carbonyl center.

2. Resonance and acidity are related because the former stabilizes the negative charge throughout the molecule, thereby giving the molecule the "tendency" to donate its proton with more ease and incur a negative charge. In other words, by losing a proton and incurring a negative charge, the molecule is still very stable because this charge can be evenly dispersed throughout several atoms. This is why the molecule is acidic.

3. Remember that oxygen has an electronegativity of 3.5 and as a result, it is quite electronegative (up there with fluorine). When protonated, it is highly positively charged and now a second resonance structure can be drawn for the carbonly, in which the double bond of the carbonyl donates its electrons to the charged oxygen...this leaves an even more positive charge on the carbon (it is now beyond being delta positive) and thus more electrophilic, susceptible to nucleophilic attack.

Hope this helps.

pyran with a carbocation para to the oxygen?

Aromatic, I believe. One of the lone pairs on the oxygen would participate in the conjugated pi system, while the other would sit it out. The cation on the C para to the oxygen ensures that every atom in the ring participates in the pi electron delocalization. The total of 6 pi electrons means it's 4N + 2, by Huckel's Rule, making it an aromatic compound.

Why are substances with double bonds UV active? Or has that already been discussed?

I think it has to do with the fact that the pi to pi* transition occurs at UV wavelengths, whereas it takes more energy to make other transitions such as n to pi*, which would have to occur at even shorter wavelengths, that aren't registered by the UV equipment. Someone correct me if I'm wrong.

Aromatic, I believe. One of the lone pairs on the oxygen would participate in the conjugated pi system, while the other would sit it out. The cation on the C para to the oxygen ensures that every atom in the ring participates in the pi electron delocalization. The total of 6 pi electrons means it's 4N + 2, by Huckel's Rule, making it an aromatic compound.
Ditto.

Tertiary carbocation = sp2, therefore, has p orbital available, so the system is conjugated and huckels is satisfied..

Ditto.

Tertiary carbocation = sp2, therefore, has p orbital available, so the system is conjugated and huckels is satisfied..
ugh, i missed that on my quiz. what the hell was i thinking

ugh, i missed that on my quiz. what the hell was i thinking
Its easy to miss... but you won't next time :)

Hi - I have a couple of quick orgo questions and I was hoping someone could help me out.

1. Nucleophiles can often attack unhybridized p-orbitals from either top or bottom side. Is there a way to know which side the nucleophile will attack from? Yes, flip a coin :) jk

why does that mean that the proton is acidic (what is the relationship between being stable and acidic?)
!


You might want to think of this one in terms of equilibrium and acid/base chemistry. When something is acidic, it is in part because it has a stable conjugate base (via resonance in our case with the alpha carbon) and thus the ratio of conj. base to protonated acid (the Keq) gives it a pKa value that is relatively acidic compared to neighboring carbons that are farther away from the carbonyl group.

You might want to think of this one in terms of equilibrium and acid/base chemistry. When something is acidic, it is in part because it has a stable conjugate base (via resonance in our case with the alpha carbon) and thus the ratio of conj. base to protonated acid (the Keq) gives it a pKa value that is relatively acidic compared to neighboring carbons that are farther away from the carbonyl group.
Yes, like you said its conjugate base is stable because of the resonance, which delocalizes electrons. When a carbanion can have its lone pair delocalized that immediately increases its stability considerably because carbon doesn't like to have a lone pair at all.

Why do saturated fats have higher heat of combustion than unsaturated fats?

This is from EK Organic question 76 lecture 4. They just say that saturated fats have the highest energy, twice of carbs and proteins.

Why do saturated fats have higher heat of combustion than unsaturated fats?

This is from EK Organic question 76 lecture 4. They just say that saturated fats have the highest energy, twice of carbs and proteins.
This is a guess... couldn't find any info.

CH4 + 2O2 -> CO2 + 2H20 + heat

I'm guessing that when a hydrocarbon is unsaturated it limits how much combustion can take place because of the lesser number of Hydrogens available to reduce O2 in the redox reaction.

e.g. 2-propene = CH2=CH-CH3, C3H6 vs CH3-CH2-CH3, C3H8.


If we simplify it to just H2 and O2, it makes more sense:
6H (or 3H2) + excess O2 -> 3H2O + 3 Heat units
8H (or 4H2) + excess O2 -> 4H20 + 4 Heat units

Thanks so much! that is really helpful!

I have another question about nucleophilic addition reactions. When an aldehyde or a ketone reacts with a primary amine in an acidic solution, it produces an imine. What if the primary amine structure is such that there are two primary amines on both sides and a carbonyl group in the middle. Which N would act as the nucleophile? The one that's closer to the carbonyl or farther away? and what's the reason behind it? Thanks!!

Thanks so much! that is really helpful!

I have another question about nucleophilic addition reactions. When an aldehyde or a ketone reacts with a primary amine in an acidic solution, it produces an imine. What if the primary amine structure is such that there are two primary amines on both sides and a carbonyl group in the middle. Which N would act as the nucleophile? The one that's closer to the carbonyl or farther away? and what's the reason behind it? Thanks!!
Sorry, I'm not quite understanding the question. From what I gather you want to know what would happen if:

HN-C(O)-NH reacted with R-C(O)R (R=C, H), and which HN-R would act as the Nu, but its symmetrical?

If you mean HN-N-C(O)H then there is still only one primary amine.

If you had HN-C(O)-CH2-N*H then the NH labeled (*) would be more likely to react because it would have more e- density and would be a stronger nucleophile. I don't know if it would actually react though.

I need to see what this mechanism looks like because i have never seen one like this before its all new i just started organic chem at usyd

1. The chemical action of 48%HBr on trans-1,2-dimethyloxirane results in a stereoselective reaction via protonation of the oxirane oxygen followed by ring opeing via SN2 mechanism

i need some one to show me what this mechanism would look like because i cant write it out thank you,

ps. also what would the diff be in the stereochemical outcome if it was to be an SN1 mechanism

can somebody please tell me how 4-chloro-1-cylcohexene has a chiral carbon? Examkrackers says carbon 4 is chiral but doesn't it have two CH2 groups attached to it? I must be having a crazy moment or something.

can somebody please tell me how 4-chloro-1-cylcohexene has a chiral carbon? Examkrackers says carbon 4 is chiral but doesn't it have two CH2 groups attached to it? I must be having a crazy moment or something.
it is chiral. you're right...next to c4, there are 2 ch2 groups. but, when determining chirality, you look at the next substituent. on one side we have CH2-CH=CH-R while the other side has CH2-CH2-CH=CH-R. so, there are 4 different substituents attached to the carbon, making it chiral.
hope that helps
:luck:

I need to see what this mechanism looks like because i have never seen one like this before its all new i just started organic chem at usyd

1. The chemical action of 48%HBr on trans-1,2-dimethyloxirane results in a stereoselective reaction via protonation of the oxirane oxygen followed by ring opeing via SN2 mechanism

i need some one to show me what this mechanism would look like because i cant write it out thank you,

ps. also what would the diff be in the stereochemical outcome if it was to be an SN1 mechanism

Here's the mechanism for SN2:

http://img155.imageshack.us/img155/5619/mechse3.jpg

For SN1, you would have a carbocation intermediate, so the reaction would not be stereoselective and you would have both R and S configurations.

Here's the mechanism for SN2:

http://img155.imageshack.us/img155/5619/mechse3.jpg

For SN1, you would have a carbocation intermediate, so the reaction would not be stereoselective and you would have both R and S configurations.


Also note that the sterochemistry is different when you open the ring in base. Also, you can do the Sn2 even with tertiary substrates here, because of the dipole.








Question: How the hell is it possible to remember all of the carbonyl reactions?

Also note that the sterochemistry is different when you open the ring in base. Also, you can do the Sn2 even with tertiary substrates here, because of the dipole.








Question: How the hell is it possible to remember all of the carbonyl reactions?
theyre all of the same format...makes it really easy to remember. write them all down on one piece of paper. youll find that you will memorize them much faster than you expected.

Between these two which is a stronger nucleophile?

H3CO- and Cl-

I'm inclined to pick Cl-. I thought CN- and the halide anions made better nucleophiles. The answer is H3CO- because it is a stronger base. I did not think a stronger base necessarily made it a better nucleophile, what am I missing here?
Though the passage is about SN1 vs SN2, there is no mention of mechanism in this question. Regardless, I thought Cl- would be better for SN2 and neither would be good for SN1 since SN1 would require a weak Nu and a weak base

Thank you in advance.

Stronger bases generally make better nucleophiles unless they are sterically hindered (preventing a nucleophillic attack). If something is a stronger base, it essentially means that it more strongly seeks to form a bond in order to become more stable. This is why you have nucleophillic attack. The nucleophile, an electron donor, seeks to become more stable by forming a bond with the electrophile, an electron acceptor. Another way to think of it is in terms of the leaving group. Cl- is a great leaving group because it is very stable in solution.

Also consider HCl. HCl is a strong acid because it has a strong tendency to disassociate into H+ and Cl-. This is because the Cl- ion is very stable in solution. The conjugate of a strong acid is a weak base, and we know a weak base has less tendency to form a new bond (which is why H+ and Cl- don't spontaneously reform HCl). This is also why a strong base such as OH- has a high tendency to act as a nucleophile and is a poor leaving group. A lot of organic chemistry can be explained by stabilities, and when you consider how strong a nucleophile something is you can also draw conclusions about how weak a leaving group it is.

What is the difference between carbonyl group and acetyl group? Both have C=O double bond. Whats the difference between the two?

An acetyl group is different from a carbonyl group, although it does contain a carbonyl group.

C=O is a carbonyl, as you know.

So, something with R-CH₂-(C=O)-R is a ketone with a carbonyl group.

Now, something with R-CH[(C=O)-CH₃)]-R would have an acetyl group. The acetyl group is like having an acetone hanging off the edge of a larger parent carbon chain.

http://www.crscientific.com/diagram%20-%20Ac%20groups1a.jpg

So Acetyl group is:

http://www.chemheritage.org/EducationalServices/pharm/glossary/acetyl.gif


And Carbonyl group is simply C=O double bond.

so, in other words, an acetyl group HAS a carbonyl group in it??

So Acetyl group is:

http://www.chemheritage.org/EducationalServices/pharm/glossary/acetyl.gif


And Carbonyl group is simply C=O double bond.

so, in other words, an acetyl group HAS a carbonyl group in it??
yep, exactly. :)

Couple questions for you guys.

1) For absolute configuration of an organic compound, can you determine it from a line/non-3d arrangement? If so, how do you go about doing it? Do you use the fischer method and assume horizontal is sticking out and vertical is sticking in the background? I get easily tripped up on this and would appreciate an explanation.

2) This is more of a confirmation than question, but the only way for a carbon to be chiral other than having 4 different substituents is to have 3 or less substituents with a double bond/triple bond, correct?

Thanks.

From a fischer or hawthorn projection you can find the absolute configuration, but IMHO those would be classified as "3D" just as the dashes and wedges would. You cold also look at boat/chair/etc conformations. However, other than I can't think of any others.

But really, you need to understand whether or not the depiction you're looking at give you 3D information, even if it isn't in the wedge-dash form.

This is about Michael Additions and Aldo Condensation.

In Aldol Condensation, we get the usual product that, at high temps, can further condense into water and a form of an aldehyde with a double bond alpha to the carbonyl. They state that aldol condensation primarily occurs with similar aldehydes/ketones (acetaldehyde with acetaldehyde).
Remember that product.


In Enol formation, one can make an enolate anion with a strong base (like the one we form in aldol condensation), and that the enolate can attack alkene groups alpha/beta to carbonyl carbons and form an entirely different product without the double bond.

So my question; would an aldehyde/ketone form an enolate and then attack the final product of a aldol condensation and make an entirely different product? Or is there something restricting the aldehyde (acetaldehyde in this case, since reactants are the same) from doing that and just going into aldol condensation instead without forming the michael addition?:confused:

Here's the mechanism for SN2:

http://img155.imageshack.us/img155/5619/mechse3.jpg

For SN1, you would have a carbocation intermediate, so the reaction would not be stereoselective and you would have both R and S configurations.


That's actually SN1, nto SN2. That's an acid-based cleavage of an oxirane which causes a carbocation intermediate to form on the most stable carbon. The bromide anion would then attack that cation and form the product.

For an SN2, you would start with the bromide anion attacking the least sterically hindered carbon and then have the OH form from the resulting H in solution.

This is about Michael Additions and Aldo Condensation.

In Aldol Condensation, we get the usual product that, at high temps, can further condense into water and a form of an aldehyde with a double bond alpha to the carbonyl. They state that aldol condensation primarily occurs with similar aldehydes/ketones (acetaldehyde with acetaldehyde).
Remember that product.


In Enol formation, one can make an enolate anion with a strong base (like the one we form in aldol condensation), and that the enolate can attack alkene groups alpha/beta to carbonyl carbons and form an entirely different product without the double bond.

So my question; would an aldehyde/ketone form an enolate and then attack the final product of a aldol condensation and make an entirely different product? Or is there something restricting the aldehyde (acetaldehyde in this case, since reactants are the same) from doing that and just going into aldol condensation instead without forming the michael addition?:confused:
After the aldol has formed and condensed it is conjugated, so I don't see why you wouldn't get michael addition products if you were to react react it with something like HBr. And, there doesn't seem to be anything that would further stop you from reacting a ketone or aldehyde (after the enolate is formed) with your product, either, in another aldol addition, condensation.

This is about Michael Additions and Aldo Condensation.

In Aldol Condensation, we get the usual product that, at high temps, can further condense into water and a form of an aldehyde with a double bond alpha to the carbonyl. They state that aldol condensation primarily occurs with similar aldehydes/ketones (acetaldehyde with acetaldehyde).
Remember that product.


In Enol formation, one can make an enolate anion with a strong base (like the one we form in aldol condensation), and that the enolate can attack alkene groups alpha/beta to carbonyl carbons and form an entirely different product without the double bond.

So my question; would an aldehyde/ketone form an enolate and then attack the final product of a aldol condensation and make an entirely different product? Or is there something restricting the aldehyde (acetaldehyde in this case, since reactants are the same) from doing that and just going into aldol condensation instead without forming the michael addition?:confused:
You're right...it can happen. This is the reason why these reactions are so messy. It is a mixture of thermodynamics and kinetics that allows one to determine what are the best reaction conditions to forming the desired product (in this case, the michael addition product). It may require much higher temperatures for the side reaction to occur...therefore, running the reaction at 0 deg may allow for a minimal amount of side product. It must be determined experimentally.

Thanks guys:thumbup:

why is pyridine aromatic? are all rings with n's in them aromatic?

why is pyridine aromatic? are all rings with n's in them aromatic?

That wouldn't always be the case, I think. Pyridine and Pyrrole are the only ones I know of off the top of my head. Pyridine because it doesn't donate its electrons to the group, thus making the 4n+2 pi rule fit for the aromatic and pyrrole DOES donate, making the rule fit for it as well. I wouldn't be surprised if there was a ring structure with an N that didn't fit this and wasn't aromatic.

Exactly.

Nitrogen's lone pair of electrons can contribute to the resonance forms, so its electrons are included in 4n + 2 rule.

That means there are 6 conjugated, contributing electrons. So, 4n + 2 = 6 --> n = 1, so its aromatic.

http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Pyrrole_structure.svg/454px-Pyrrole_structure.svg.png

Exactly.

Nitrogen's lone pair of electrons can contribute to the resonance forms, so its electrons are included in 4n + 2 rule.

That means there are 6 conjugated, contributing electrons. So, 4n + 2 = 6 --> n = 1, so its aromatic.

http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Pyrrole_structure.svg/454px-Pyrrole_structure.svg.png

Just for clarification; that above is pyrrole, not pyridine.

Yeah, my bad :D I had to google pyrrole to get the image, and I still wrote the wrong name :laugh:

Hey just wondering if there is a good rule of thumb to have when you have withdrawing groups and donating groups on the same compound. Take for example and alkoxide ion. It is a stronger base than water but you have the carbonyl group on the end of the molecule which is electron withdrawing which is 2 carbons from the basic oxygen. You also have the electron donating alky group which is directly attached to the carbon with the basic oxygen on it.

How do you know that the carbonyl group wouldn't withdraw enough of the charge as compared to the alkyl group donating it? Is this solely by the distance? If so is that how we can say its a stronger base than water?

Hey just wondering if there is a good rule of thumb to have when you have withdrawing groups and donating groups on the same compound. Take for example and alkoxide ion. It is a stronger base than water but you have the carbonyl group on the end of the molecule which is electron withdrawing which is 2 carbons from the basic oxygen. You also have the electron donating alky group which is directly attached to the carbon with the basic oxygen on it.

How do you know that the carbonyl group wouldn't withdraw enough of the charge as compared to the alkyl group donating it? Is this solely by the distance? If so is that how we can say its a stronger base than water?
That's a good question. Generally, the carbonyl group is more withdrawing than the alkyl's donating group. The distance is what factors into the basicity of the oxygen in that how far away that withdrawing group is. The more alkyl groups, the more basic the oxygen becomes.
I also tend to look at it from the perspective of activating/deactivating for aromatics which goes by the rule that if the oxygen/nitrogen is directly connected to the aromatic, it can donate the electrons to the pi orbitals and if the oxygen is one carbon away, it tends to take the electron. So, if you have the O=CH2-CH2-O (example), then the withdrawing capabilities of the carbonyl group isn't so great because of how far the oxygen is from the oxygen.

To answer your second question, I don't know. It seems to me that the alkyl group's electron-donating capabilities are what make the alkoxides more basic than water.

Thanks for the tip.

One more question, as far as equilibrium goes with organic reactions, the reaction tends to form the product that has the most reactive base as a leaving group so it won't leave and react in an aqueous solution? This would make the overall compound with the highly reactive basic leaving group less reactive in general?

So comparing acyl chloride R-C-O-Cl with an amide R-C-O-NR2, the amide is less likely to be reactive in a nucleophilic substitution because if the O-NR2 bound is broken then NR2 anion is a much stronger base than Cl anion.

And I guess to further up on that thought, NR2 is a strong electron donating group so the dipole that is created around the electrophillic carbonyl carbon would be less positive because the nitrogen's electrons would donate into that dipole and cancel the positive out?

Thanks for the help, April 18th MCAT'er here, gotta get these last trends ironed out.

Thanks for the tip.

One more question, as far as equilibrium goes with organic reactions, the reaction tends to form the product that has the most reactive base as a leaving group so it won't leave and react in an aqueous solution? This would make the overall compound with the highly reactive basic leaving group less reactive in general?

So comparing acyl chloride R-C-O-Cl with an amide R-C-O-NR2, the amide is less likely to be reactive in a nucleophilic substitution because if the O-NR2 bound is broken then NR2 anion is a much stronger base than Cl anion.

And I guess to further up on that thought, NR2 is a strong electron donating group so the dipole that is created around the electrophillic carbonyl carbon would be less positive because the nitrogen's electrons would donate into that dipole and cancel the positive out?

Thanks for the help, April 18th MCAT'er here, gotta get these last trends ironed out.
Yes the chloride withdraws electron density from the carbonyl carbon, making it more electrophilic and vulnerable to nucleophilic attack. When comparing Cl- vs NR2- as leaving groups, Cl is better because it the charge is less polarized. It can be spread over the large electronegative element, whereas the nitrogen is more basic in its anionic form. These are two reasons why R-CO-Cl is more susceptible to nucleophilic attack than R-CO-NR2.

The third, most important, reason is that the amine lone pair can resonate with the carbonyl pi bond. Disruption of the resonance stabilization is not favored, and this is why nucleophilic attack on the carbonyl or protonation of the amine may not be the preferred reaction. This is the same reason for why a carboxylic acid is unlikely to undergo attack by a nucleophile.

Say we have aldehyde and a ketone. How would we determine which would become the nucleophile and electrophile? I'm assuming the molecule with the more acidic alpha hydrogen would be the nucleophile. What is more acidic, ketone or aldehyde??

Say we have aldehyde and a ketone. How would we determine which would become the nucleophile and electrophile? I'm assuming the molecule with the more acidic alpha hydrogen would be the nucleophile. What is more acidic, ketone or aldehyde??

Well, first off, aldehydes and ketones don't tend to react with each other, but instead with their own kinds. So there wouldn't be any adol condensations for these two, I don't think.

In terms of nucleo/electrophilicity; I would have to guess that the H- bonded to the carbonyl carbon is what determines how nucleophilie/electrophilic the compounds are. The fact that the carbonyl compound for the ketone is more substituted probably has something to factor into the acidity of the alpha hydrogen. My guess is that since an aldehyde has no alkyl group (electron donating and which would thus decreas the acidity) makes it a better nucleophile than a ketone. A ketone, on the other hand, has an alkyl group to which donates electrons to decrease the nucleophilic capabilities of the ketone. But that's just guessing.

I'm almost positive that like reacts with like for aldehydes and ketones. So if you put the two together, you would have aldol condensation of likes instead of nucleophilic attack by one over the other.

Well, first off, aldehydes and ketones don't tend to react with each other, but instead with their own kinds. So there wouldn't be any adol condensations for these two, I don't think.
Well...that's not exactly true. You can have mixed aldol condensations, but it's not emphasized as much in ochem because we're more concerned with an aldol condensation occurring between the same molecule in one container. They're both acidic and the differences in their acidity is negligible, which is why you will get 4 different products (even more because the products can also undergo aldol condensation with the SM)...but, in general, an aldehyde is more acidic than an ketone. This is because the ketone has methyl groups to stabilize the positive charge on the carbonyl. The aldehyde has more of a positive charge on its carbonyl...which makes a negative charge on the alpha carbon more stable.

Well...that's not exactly true. You can have mixed aldol condensations, but it's not emphasized as much in ochem because we're more concerned with an aldol condensation occurring between the same molecule in one container. They're both acidic and the differences in their acidity is negligible, which is why you will get 4 different products (even more because the products can also undergo aldol condensation with the SM)...but, in general, an aldehyde is more acidic than an ketone. This is because the ketone has methyl groups to stabilize the positive charge on the carbonyl. The aldehyde has more of a positive charge on its carbonyl...which makes a negative charge on the alpha carbon more stable.

Sorry, I worded tha incorrectly. I didn't mean to say that they never. :D And again, like he said also, the alkyl group on the ketone is what affects the acidity.

Ok, great thanks. I realized where I went wrong. For an aldol condensation with an aldehyde and a ketone, I deprotonated the lone H on the aldehyde, rather than the alpha hydrogen bound to the R group of the aldehyde so my resultant molecule came out really funky.

Why is that an alkene reacts with HBr to form the anti-markovnikov product in the presence of peroxide yet if it reacts with HI or HCl it forms the markovnikov product instead?

Why is that an alkene reacts with HBr to form the anti-markovnikov product in the presence of peroxide yet if it reacts with HI or HCl it forms the markovnikov product instead?
who cares why...just know it :p
It's actually because the bond dissociation energies of the homolytic cleavages makes it more favorable to form the antimarkovnikov product. I think it's important to know peroxide indicates that a radical reaction is going on...I don't think anything else is important.

who cares why...just know it :p
It's actually because the bond dissociation energies of the homolytic cleavages makes it more favorable to form the antimarkovnikov product. I think it's important to know peroxide indicates that a radical reaction is going on...I don't think anything else is important.

Yea I'll just memorize it then. I just don't see why Bromine is more special than Iodine or Chlorine when it comes to addition w/ peroxide.

Yea I'll just memorize it then. I just don't see why Bromine is more special than Iodine or Chlorine when it comes to addition w/ peroxide.
Well, when you start using Cl or I, the reaction is no longer spontaneous. It's all energy. You could calculate the overall enthalpy using the bond energies to prove it to yourself, but it's not worth it. Easier to memorize :p

Well, when you start using Cl or I, the reaction is no longer spontaneous. It's all energy. You could calculate the overall enthalpy using the bond energies to prove it to yourself, but it's not worth it. Easier to memorize :p
Actually...

And this is hypothetically speaking;
Chlorine tends to react faster than Bromine in halogenation of an alkane, tending to attack the least substituted carbon. So, would it make sense to think that the reason we have markovnikov addition (cation on most substituted) is because the halogen attacks the least substituted because it wants to proceed to the end of the reaction so fast, just like in halogenation of alkanes?

Actually...

And this is hypothetically speaking;
Chlorine tends to react faster than Bromine in halogenation of an alkane, tending to attack the least substituted carbon. So, would it make sense to think that the reason we have markovnikov addition (cation on most substituted) is because the halogen attacks the least substituted because it wants to proceed to the end of the reaction so fast, just like in halogenation of alkanes?
The reason why bromine attacks the more substituted carbon is because the intermediate carbocation is more stable. Similarly, markovnikiv addition occurs to form the most stable carbocation...which is why those 2 are similar. But when you add the peroxide, we're looking at stability of radicals. If you draw out the mechanism, it's the exact same idea...we want the most stable intermediate (radical on the most substituted carbon). All similarities exist for that reason. But when you look at homolytic versus heterolytic cleavages, it screws up the reactivity of the different halogens. So, it's hard to compare the reactivity of the different halogens in those 3 reactions. But it could be true that Cl would react faster and with lower yield in anti-markovnikov addition...it's hard to tell without actually looking at the energitics.
Either way, I doubt it would matter on the mcat...unless there was a passage

The reason why bromine attacks the more substituted carbon is because the intermediate carbocation is more stable. Similarly, markovnikiv addition occurs to form the most stable carbocation...which is why those 2 are similar. But when you add the peroxide, we're looking at stability of radicals. If you draw out the mechanism, it's the exact same idea...we want the most stable intermediate (radical on the most substituted carbon). All similarities exist for that reason. But when you look at homolytic versus heterolytic cleavages, it screws up the reactivity of the different halogens. So, it's hard to compare the reactivity of the different halogens in those 3 reactions. But it could be true that Cl would react faster and with lower yield in anti-markovnikov addition...it's hard to tell without actually looking at the energitics.
Either way, I doubt it would matter on the mcat...unless there was a passage

Yeah, I was just tossing that up for fun.:D

Yeah, I was just tossing that up for fun.:D
haha i love doing that too :)

CH3OOCH2CH2CH3

What are the peaks(singlet, doublet, etc.) in H nmr for this compound?

I know there's four but what is each one?

From left to right, singlet, triplet, multiplet (6, hextet), triplet

That is, the number of hydrogens on the immediate neighbor carbons, plus one.

From left to right, singlet, triplet, multiplet (6, hextet), quartet

That is, the number of hydrogens on the immediate neighbor carbons, plus one.

final CH3 is a triplet not a quartet.

Yeah, my bad :) Just rambled those off too quick, I guess.

What is the hybridization of each carbon in propene?
A) sp, sp and sp^3
B) sp^2, sp and sp^3
C) sp^2, sp^2, and sp^3
D) sp^3, sp^3, and sp^3

I know there must be an easy way to figure this out. How do you tell the difference b/w the hybridizations?

What is the hybridization of each carbon in propene?
A) sp, sp and sp^3
B) sp^2, sp and sp^3
C) sp^2, sp^2, and sp^3
D) sp^3, sp^3, and sp^3

I know there must be an easy way to figure this out. How do you tell the difference b/w the hybridizations?

I just figured it out. Count # sigma bonds and add number of lone pairs on CENTRAL atom and the total should = total of superscripts on orbitals, right?

I just figured it out. Count # sigma bonds and add number of lone pairs on CENTRAL atom and the total should = total of superscripts on orbitals, right?

hybridization is the number of sigma bonds yes. An alkane would have 4 sigma bonds and thus is sp3 (4 hybrid sp orbitals). An alkene has 3 sigma bonds and 1 pi bond. The sigma bonds are formed by 3 sp2 bonds, and the pi bond is formed by the overlapping p orbitals. Alkynes, similarly, would be sp due to the 2 sigma bonds and the 2 pi bonds (2 p orbitals overlapping).

Cell membranes are composed of many molecules, including phospholipids. A phospholipid is a molecule with a glycerol backbone plus two fatty acids and a phosphate attached to the oxygen atoms of glycerol. A cell membrane would be most rigid if both its fatty acids were:

A) completely saturated and short molecules
B) completely saturated and long molecules
C) unsaturated and short molecules
D) unsaturated and long molecules

I would have though to be rigid it would be unsaturated and long b/c it would have many pi bonds which makes the structure less flexible and more rigid

Cell membranes are composed of many molecules, including phospholipids. A phospholipid is a molecule with a glycerol backbone plus two fatty acids and a phosphate attached to the oxygen atoms of glycerol. A cell membrane would be most rigid if both its fatty acids were:

A) completely saturated and short molecules
B) completely saturated and long molecules
C) unsaturated and short molecules
D) unsaturated and long molecules

I would have though to be rigid it would be unsaturated and long b/c it would have many pi bonds which makes the structure less flexible and more rigid
Is the answer B...?

Is the answer B...?

Definitely B...long chain saturated means no kinks in the tail to interfere with dispersion forces.

Ahh, please help!

Hydroboration:

In the last step, NaOH converts the trialkylborate to an alcohol.

My question: HOW?? WHY?? How does a base convert it to an alcohol?

You don't need to know this kind of info for the MCAT. But, wikipedia has a decent explanation IMHO.

You don't need to know this kind of info for the MCAT. But, wikipedia has a decent explanation IMHO.


Yeah. Exactly the same as how a phenol is formed from a benzyl sulfonate. The reaction isn't important, just know what the reaction conditions are and what the reactants/products are.

Yeah, I know I don't have to know it. I just remember things better if I understand how and why they happen.

Yeah. Exactly the same as how a phenol is formed from a benzyl sulfonate. The reaction isn't important, just know what the reaction conditions are and what the reactants/products are.

Yeah, I know I don't have to know it. I just remember things better if I understand how and why they happen.
Believe me, it took me at least a day to come to terms with this annoying reaction (the one you speak of). Considering almost every other reaction is explained except boration, it pissed me off considerably.

C7H9N3O2Cl2 would has four units of saturation which means that there are eight less electrons. My book is saying that there will be 3 pi bonds and I'm thinking there would be four pi bonds b/c 3 pi bonds only yeild 6 electrons and four pi bonds yeild 8 electrons.

C7H9N3O2Cl2 would has four units of saturation which means that there are eight less electrons. My book is saying that there will be 3 pi bonds and I'm thinking there would be four pi bonds b/c 3 pi bonds only yeild 6 electrons and four pi bonds yeild 8 electrons.
I stopped trying to draw it out and just looked at it and it does make sense. 4 saturated carbons leaves 3 carbons for unsaturation. That would most likely make a single pi bond for each, so 3 pi bonds.

Someone can feel free to correct me on this, but I think that's what they want you to do.

I stopped trying to draw it out and just looked at it and it does make sense. 4 saturated carbons leaves 3 carbons for unsaturation. That would most likely make a single pi bond for each, so 3 pi bonds.

Someone can feel free to correct me on this, but I think that's what they want you to do.

I learn better visualizing and I still can't see how all the hydrogens add up with 3 pi bonds instead of 4. Here's another one that I stumped on: Which of the following compunds with the formula C5H10O and does not have an IR absorbance peak b/w 1700 and 1750.
A) aldehyde
B) ketone
C) cyclic ester

The answer is a cyclic ester, which I got correct, but the degrees of saturation is 1 which means that two hydrogens are knocked off leaving you with one pi bond. When I draw this out the hydrogens aren't right. I'm getting only 8 hydrogens when I draw this out, but the formula is saying there needs to be 10. Here's how I'm drawing it out: a cyclic structure with 5 carbons and an oxygen (acting as the ether b/c two carbons) and then two hydrogens coming off each hydrogen except where the pi bond is I only put one hydrogen. This all together yields 8 hydrogens and not 10. Am I drawing this wrong?

On an additional note, the only reason you don't have a 1700-1750 stretch is because this is a 4-membered lactone. If this were an non-cyclic ester or a 6-membered lactone, you would definitely have a 1700-1750 stretch.

Is this actually a practice MCAT question? This seems WAY beyond the scope of the MCAT!

On an additional note, the only reason you don't have a 1700-1750 stretch is because this is a 4-membered lactone. If this were an non-cyclic ester or a 6-membered lactone, you would definitely have a 1700-1750 stretch.

Is this actually a practice MCAT question? This seems WAY beyond the scope of the MCAT!

Good to hear that...it's actually from the berkeley review mcat book.

Can somebody explain how to approach this problem. The explanation seems a little out there so I don't fully understand it.

The addition of alkyl magnesium bromide (RMgBr) to a carbonyl in ether adds a new alkyl substituent to the carbonyl carbon, resulting in conversion of the carbonyl into an alcohol. The addition of H3CMgBr to R-2-methylcyclohexanone in diethyl ether yields which products?

A) one meso compound
B) two diasteriomers
C) two enantiomers
D) two epimers

ok, so the answer should be that you get a pair of diastereomers (B). Ok, so the first thing to do is to draw out the molecule. It's just a cyclohexanone with a methyl group in the 2 position. You really don't need to know what the R configuration is to do this problem. Ok, so the methyl Grignard can attack from either above or below the plane of the carbonyl. If it attacks one way, the methyl group will end up cis to the existing methyl group. If it attacks from the other side, the methyl group will end up trans to the existing methyl group. Since one of them has methyl groups cis and one has methyl groups trans, they are different compounds so you can eliminate A because it's not a meso compound. Cis/trans isomerism can never produce enantiomers. The reason for this is that cis/trans isomerism only occurs with 2 or more stereocenters in a molecule. In order to produce an enantiomer of a molecule with multiple stereocenters, you have to invert configuration about every stereocenter. So if the configuration of one molecule is R,R its enantiomer would have the configuration S, S. If it's R, S, its enantiomer would have the configuration S, R. Since cis/trans isomerism involves an inversion of configuration around only one stereocenter (so like R, R, to S, R, or R, S to S, S or something), you can never have enantiomers. So C can be eliminated. You can't have an epimer without three or more stereocenters, because an epimer is a specific type of diastereomer where the configuration around only ONE stereocenter is changed, but again, only when you have three or more stereocenters. So that eliminates D.
So you're left with B, which makes sense, because cis/trans isomerism gives you diastereomers, which are stereoisomers that aren't mirror images of each other.

ok, so the answer should be that you get a pair of diastereomers (B). Ok, so the first thing to do is to draw out the molecule. It's just a cyclohexanone with a methyl group in the 2 position. You really don't need to know what the R configuration is to do this problem. Ok, so the methyl Grignard can attack from either above or below the plane of the carbonyl. If it attacks one way, the methyl group will end up cis to the existing methyl group. If it attacks from the other side, the methyl group will end up trans to the existing methyl group. Since one of them has methyl groups cis and one has methyl groups trans, they are different compounds so you can eliminate A because it's not a meso compound. Cis/trans isomerism can never produce enantiomers. The reason for this is that cis/trans isomerism only occurs with 2 or more stereocenters in a molecule. In order to produce an enantiomer of a molecule with multiple stereocenters, you have to invert configuration about every stereocenter. So if the configuration of one molecule is R,R its enantiomer would have the configuration S, S. If it's R, S, its enantiomer would have the configuration S, R. Since cis/trans isomerism involves an inversion of configuration around only one stereocenter (so like R, R, to S, R, or R, S to S, S or something), you can never have enantiomers. So C can be eliminated. You can't have an epimer without three or more stereocenters, because an epimer is a specific type of diastereomer where the configuration around only ONE stereocenter is changed, but again, only when you have three or more stereocenters. So that eliminates D.
So you're left with B, which makes sense, because cis/trans isomerism gives you diastereomers, which are stereoisomers that aren't mirror images of each other.


Thanks so much. That made it a little clearer, I still am not too certain I understand the enantiomer explanation. I'll have to go over it a few times before it actually becomes second nature to me.

Can somebody explain how to approach this problem. The explanation seems a little out there so I don't fully understand it.

The addition of alkyl magnesium bromide (RMgBr) to a carbonyl in ether adds a new alkyl substituent to the carbonyl carbon, resulting in conversion of the carbonyl into an alcohol. The addition of H3CMgBr to R-2-methylcyclohexanone in diethyl ether yields which products?

A) one meso compound
B) two diasteriomers
C) two enantiomers
D) two epimers

In addition to what the above poster said, keep in mind that planar species (ie a carbonyl) can be attacked from the top or the bottom, often yielding different products. This is why optically active compounds that undergo SN1 yield a racemic (or nearly racemic) mixture if optically active products are formed. In actuality, inversion is slightly favored for SN1 because the leaving group can block the incoming nucleophile. The sp2 hybridized carbocation is a planar species...

I learn better visualizing and I still can't see how all the hydrogens add up with 3 pi bonds instead of 4. Here's another one that I stumped on: Which of the following compunds with the formula C5H10O and does not have an IR absorbance peak b/w 1700 and 1750.
A) aldehyde
B) ketone
C) cyclic ester

The answer is a cyclic ester, which I got correct, but the degrees of saturation is 1 which means that two hydrogens are knocked off leaving you with one pi bond. When I draw this out the hydrogens aren't right. I'm getting only 8 hydrogens when I draw this out, but the formula is saying there needs to be 10. Here's how I'm drawing it out: a cyclic structure with 5 carbons and an oxygen (acting as the ether b/c two carbons) and then two hydrogens coming off each hydrogen except where the pi bond is I only put one hydrogen. This all together yields 8 hydrogens and not 10. Am I drawing this wrong?If you chose (C) it's wrong if it's a cyclic ester. With the given molecular formula, there are 2 degrees of unsaturation. Are you sure it's not cyclic ETHER, because all the choices including the cyclic ESTER will have a C=O IR stretch between 1700-1750.

Can somebody explain how to approach this problem. The explanation seems a little out there so I don't fully understand it.

The addition of alkyl magnesium bromide (RMgBr) to a carbonyl in ether adds a new alkyl substituent to the carbonyl carbon, resulting in conversion of the carbonyl into an alcohol. The addition of H3CMgBr to R-2-methylcyclohexanone in diethyl ether yields which products?

A) one meso compound
B) two diasteriomers
C) two enantiomers
D) two epimers(A) is not a good choice because the Grignard rx will not result in a (major) product that has an internal mirror plane rendering it optically inactive i.e., a meso compound.

(C) is not a good choise, as sleepy425 explained, since both stereocenters in the Grignard rx product would have to interchange (R/S) configurations which is not possible since configuration at C-2 is R before and after the rx. Also, as previously explained, carbonyl carbons when undergoing nucleophilic additions as in a Grignard, are prochiral i.e., can produce both R & enantiomers since the nucleophile can approach either from "above" or "below" with equal probability. Because of this, the resulting molecule at the C-1 stereocenter will be "racemic".

For the MCAT, epimer type questions will be limited to sugars/carbohydrates which the product is not, making (D) a poor choice.

One way to double check your answer in to use the formula 2^n, n = # of stereocenters, for the total # of stereoisomers possible. The caveat in the problem is that one stereocenter, C-2, has to be R. Given that the C-1 carbonyl can produce both R & S configurations for the Grignard rx, this information i.e. C-1 alternating between R & S configuration, along with C-2 being only R configuration leaves (B) as the best choice.

In essence for this problem, there are 2 stereocenters but for only one of the stereocenters are both configurations possible.

If you chose (C) it's wrong if it's a cyclic ester. With the given molecular formula, there are 2 degrees of unsaturation. Are you sure it's not cyclic ETHER, because all the choices including the cyclic ESTER will have a C=O IR stretch between 1700-1750.



This molecular formula definitely does not work for an ester as esters must have 2 oxygens. That said, if the formula were C5H10O2, then it could be a cyclic ester.In that case, it has to be a 4-membered ring with methyl substituents on the alpha and beta carbons to give the molecular formula C5H10O2. Additionally, 4-membered lactones, while indeed containing a carbonyl group, show this C=O stretch around 1840 cm-1. As I mentioned in an above post, this is WAY beyond the scope of the MCAT (the IR stretch that is). I picked this up in an upper division organic synthesis class...

I agree with Foghorn, this has to be a cyclic ether, not a cyclic ester.

How do you determine where to put the substituents on a dash-line-wedge formla? When I took organic 3 years ago, I was taught the dash formula so I'm not use to drawing the structures this way...that and it's hard to visualize.

Can somebody give me the step by step breakdown (what attacks what and how do you know what to look out for):


What are the two products formed when propanoyl chloride, H3CCH2COCl, is treated with methyl amine, H3CNH2?

A) N-methyl propanamide and hydrochoric acid
B) 1-amino-2-butanone and hydrochloric acid
C) Propanal and chloromethylamine
D) Carbon dioxide and N,N-ethylmethylamine

Can somebody give me the step by step breakdown (what attacks what and how do you know what to look out for):


What are the two products formed when propanoyl chloride, H3CCH2COCl, is treated with methyl amine, H3CNH2?

A) N-methyl propanamide and hydrochoric acid
B) 1-amino-2-butanone and hydrochloric acid
C) Propanal and chloromethylamine
D) Carbon dioxide and N,N-ethylmethylamine

It has to be A. I'm thinking this because you form amides from the reaction of acyl chlorides and amines *most preferred way of forming amides I think* The methyl amine attacks the carbonyl carbon and this forces the chloride ion to leave (The essence of this whole reaction - Remember here that acids/acyl chlorides all undergo substitution reactions, not addition reactions when attacked - nucleophilic substitution)
When the amine attacks with its electron pair, it forces the chloride ion to dissociate (better leaving group) and form the amide. But I am at a lost as to why they don't say that the HCl doesn't react with the methylamines to form the chloromehtylamine. I thought that was the case?

Honestly, I know for sure that it's an amide formed, so that eliminates B and C. Carbon Dioxide would definitely not form from this reaction, because that would be a completely different reaction mechanism (I don't even know what it is. lol)

- edit- decarboxylation of a carboxylic acid would occur only in rare cases with beta-keto acides as a reactant.

Can somebody give me the step by step breakdown (what attacks what and how do you know what to look out for):


What are the two products formed when propanoyl chloride, H3CCH2COCl, is treated with methyl amine, H3CNH2?

A) N-methyl propanamide and hydrochoric acid
B) 1-amino-2-butanone and hydrochloric acid
C) Propanal and chloromethylamine
D) Carbon dioxide and N,N-ethylmethylamine

Answer is A.

Amine attacks C=O. Cl- leaves as leaving group. The N has a + formal charge now, and loses an H+ to become neutral again. This gives N-methyl propanamide and HCl.

Can somebody explain how to approach this problem. The explanation seems a little out there so I don't fully understand it.

The addition of alkyl magnesium bromide (RMgBr) to a carbonyl in ether adds a new alkyl substituent to the carbonyl carbon, resulting in conversion of the carbonyl into an alcohol. The addition of H3CMgBr to R-2-methylcyclohexanone in diethyl ether yields which products?

A) one meso compound
B) two diasteriomers
C) two enantiomers
D) two epimers

The answer is B.

The molecule you start with has one chiral center and it is a single enantiomer.

The reaction adds an R group to the C=O carbon and an H to the C=O oxygen. This creates a second chiral center. The reaction will give products that have both R and S configurations at this center.

You generate R-R and R-S products - since it cannot be meso (no mirror plane can be drawn through the compound) you can eliminate A.

It is not an epimer because this is a new molecule with an extra group attached, not merely a stereoisomer. D is out.

Even if the alkyl group attacks from only one face, it cannot give two enantiomers - this would require that the starting material consists of both R and S and it does not.

The only answer left is B. The alkyl group adds on both faces of the molecule.

Even if you don't know that addition is to both faces you can still get the answer through process of elimination (as I've done here).

I learn better visualizing and I still can't see how all the hydrogens add up with 3 pi bonds instead of 4. Here's another one that I stumped on: Which of the following compunds with the formula C5H10O and does not have an IR absorbance peak b/w 1700 and 1750.
A) aldehyde
B) ketone
C) cyclic ester

The answer is a cyclic ester, which I got correct, but the degrees of saturation is 1 which means that two hydrogens are knocked off leaving you with one pi bond. When I draw this out the hydrogens aren't right. I'm getting only 8 hydrogens when I draw this out, but the formula is saying there needs to be 10. Here's how I'm drawing it out: a cyclic structure with 5 carbons and an oxygen (acting as the ether b/c two carbons) and then two hydrogens coming off each hydrogen except where the pi bond is I only put one hydrogen. This all together yields 8 hydrogens and not 10. Am I drawing this wrong?

The question, Example 2.12 from Berkeley Review Oragnic Chemistry Book I is actually:
Which of the following compunds with the formula C5H10O cannot have an IR absorbance peak b/w 1700 and 1750?
A) aldehyde
B) ketone
C) cyclic ether
D) all of the above compounds have an absorbance b/w 1700 and 1750

Be very careful not to mix up ethers and esters. An ester would not be possible for that formula, because it requires two oxygens. If you look more closely at that questions, you'll see that the best answer is the cyclic ether, because it lacks a carbonyl.

C7H9N3O2Cl2 would has four units of saturation which means that there are eight less electrons. My book is saying that there will be 3 pi bonds and I'm thinking there would be four pi bonds b/c 3 pi bonds only yeild 6 electrons and four pi bonds yeild 8 electrons.

Berkeley Review Organic Chemistry Book I Example 2.11

The answer explanation actually says that "there could be 3 pi-bonds and one ring." The ring also counts as a unit of unsaturation. And the point of the question is to demonstrate the easy formula for solving these types of questions.

Hi,

I'm taking my DAT tomorrow morning and just found this site. I am having the most trouble understanding how to rotate about a C-C bond. For example, if you are given a 2 C backbone with 3 substituens on each C complete with dashes and wedges, how would you convert this to fischer if you want, say the halogens that are on each C to be in a particular spot (i.e. the top most point on the fisher and the bottom most point on the fischer?)?

Or, what if I was given the fischer projection and was told to rotate it to the correct dash/wedge conformation, but it wasn't immediately obvious (i.e. it needed to be rotated about the C-C bond) ?

I'm in need of a DETAILED, step by step explanation of this if possible. The explanations I've been getting from Kaplan are not much help unfortuantely. They just keep telling me to rotate about the C-C bond, but I can't mentally do this without steps!!

Thanks!! A prompt response is appreciated!!

when you see (+) in front of the name of a molecule...that means it is "R" correct?
*where R means order of decreasing priorities runs CCW

and likewise when you see (-) , then it is "S" ??
*where S means order of decreasing priorities runs CW

when you see (+) in front of the name of a molecule...that means it is "R" correct?
*where R means order of decreasing priorities runs CCW

and likewise when you see (-) , then it is "S" ??
*where S means order of decreasing priorities runs CW

Nope, (+) = d (small, not capitol) = dextrorotatory = molecule rotates plane polarized light CW, and (-) = l = levorotatory = molecule rotates plane polarized light CCW.

R/S (or D/L for that matter) do NOT predict which way a molecule will rotate light. For example, a D series sugar can rotate light CW (+) or CCW (-). It's impossible to predict unless you do polarimetry experiments.

Saponification

Okay, I just want to clarify something.

1 Triglyceride + 3NaOH --> 1 Glycerol + 3 Na+-carboxylate FA's

Now, -OH nucleophile should expel the glycerol with a negative charge, right? However, being that the a carboxylic acid is formed it will easily lose its proton to the strong conjugate base of glycerol, and that's why we actually get 3 carboxylate anions and a glycerol, correct?

Thanks!

Saponification

Okay, I just want to clarify something.

1 Triglyceride + 3NaOH --> 1 Glycerol + 3 Na+-carboxylate FA's

Now, -OH nucleophile should expel the glycerol with a negative charge, right? However, being that the a carboxylic acid is formed it will easily lose its proton to the strong conjugate base of glycerol, and that's why we actually get 3 carboxylate anions and a glycerol, correct?

Thanks!

Actually, the carboxylate anion is an anion from the get-go of the product of substitution and never has/will give a proton to the glycerol because the glycerol is already protonated. The Sodium basically stablizes the charge on the carboxylate and the -OH goes through the nucleophilic substitution that carboxylic acids are so fond of.

Actually, the carboxylate anion is an anion from the get-go of the product of substitution and never has/will give a proton to the glycerol because the glycerol is already protonated. The Sodium basically stablizes the charge on the carboxylate and the -OH goes through the nucleophilic substitution that carboxylic acids are so fond of.

What? No, sorry, that's not correct. -OH attacks, electrons come down off the oxyanion and kick off the glycerol anionic leaving group giving a carboxylic acid and an oxyanion on the glycerol. because the pKa of a hydroxyl group is about 15 while the pKa of a carboxylic acid is like 4 or 5, the glycerol oxyanion pulls the proton off the fatty acid leaving the carboxylate anion and neutral glycerol.

so tncekm, you're exactly right.

What? No, sorry, that's not correct. -OH attacks, electrons come down off the oxyanion and kick off the glycerol anionic leaving group giving a carboxylic acid and an oxyanion on the glycerol. because the pKa of a hydroxyl group is about 15 while the pKa of a carboxylic acid is like 4 or 5, the glycerol oxyanion pulls the proton off the fatty acid leaving the carboxylate anion and neutral glycerol.

so tncekm, you're exactly right.
My mistake, I had the completely turned around.

Interesting that I said the nucleophilic substitution and thought the product I said happened....wtf

My mistake, I had the completely turned around.

Interesting that I said the nucleophilic substitution and thought the product I said happened....wtf

lol, it's cool, I do that all the time...