Organic Chemistry Question Thread

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

QofQuimica

Seriously, dude, I think you're overreacting....
Moderator Emeritus
Lifetime Donor
15+ Year Member
Joined
Oct 12, 2004
Messages
18,899
Reaction score
4,290
All users may post questions about MCAT, DAT, OAT, or PCAT organic chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what organic topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
-general, MCAT-level organic
-particular MCAT-level organic problems, whether your own or from study material
-what you need to know about organic for the MCAT
-how best to approach to MCAT organic passages
-how best to study MCAT organic
-how best to tackle the MCAT biological sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

********

If you really know your organic, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the Organic Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university organic chemistry TA teaching experience. In addition, I teach organic chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 15 on BS, 43 overall.

P.S. If you shorten "organic chemistry" to "orgo," not only will I not answer your questions, but during the BS section, your test form will backside attack you with a zillion strong nucleophiles (via the SN2 mechanism, of course).

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the BS section of the MCAT, and 36 overall.

Members don't see this ad.
 
Sulfur has empty d orbitals which allow it to exceed standard valency. Al, Si, P, and Cl can also access the empty d orbitals ( Chlorine can form large chlorate polyatomic counterions- like in bleach... phosphorous can form large trigonal bipyrimidal POCl3 -a chlorinating reagent ...

Sulfur can exists as thioethers (two bonds like oxygen) and can be higher oxidations states such as sulfonates and sulfonic acids (which contains a lot more bonds than just oxygen)


This answers an old questions... I screwed up and put it in the wrong place... but maybe it'll help someone...
 
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.
 
Members don't see this ad :)
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.
 
Glass that is no longer intact said:
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?
 
Members don't see this ad :)
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
The 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.
 
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?
 
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

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

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!
 
Top