Wait, I thought for C-NMR they show you the carbons (odd # isotope, C 13 for example). Because in that graph, the benzene has 6 peaks, and benzene has 6 carbons. Carbonyl has 1 carbon. I hope I'm looking at this right :(

Also, how can we tell the number of carbons in a molecule if say every carbon has another carbon just like it (same chemical shift). So the graph would only show each TYPE of carbon, not how many of each type if a few are the same type (substituents and orientation)? Will there be more info given?

I just want to make sure I have this straight. For C-NMR we just see how many hydrogens are attached to that particular Carbon and add 1. For H-NMR we would look at the Carbon and see how many neighbours that carbon has and add one. Yes? So, for example, 1,1,2-tricholoropropane (first carbon has two Cl's, one H; second carbon has 1 Cl and 1 H; last Carbon has 3 H's) we would say that the first Carbon has 2 peaks because it's neighbour has 1 H add one, second has the 1 H from the first Carbon plus the 3 H's from last Carbon and add 1 for a total of 5 peaks and the third one has 2 peaks as well from the H's of the middle Carbon. Yes?

Thanks!
Assuming that the splitting due to carbons 1 and 3 were equivalent, yes.

Wait, I thought for C-NMR they show you the carbons (odd # isotope, C 13 for example). Because in that graph, the benzene has 6 peaks, and benzene has 6 carbons. Carbonyl has 1 carbon. I hope I'm looking at this right :(
Right. The one farthest upfield is the methyl from the ester.

Also, how can we tell the number of carbons in a molecule if say every carbon has another carbon just like it (same chemical shift). So the graph would only show each TYPE of carbon, not how many of each type if a few are the same type (substituents and orientation)? Will there be more info given?
Yes. This happens if your molecule is symmetrical. So if, for example, you had methyl para-hydroxybenzoate instead of ortho like the one I posted, you would wind up only seeing four carbon peaks, because carbon 2 = carbon 6 and carbon 3 = carbon 5. In that case, you'd need something like the mass spec or the molecular formula to solve the structure. Alternatively, you would probably be able to do it with the integration ratio on the proton NMR. But you almost certainly couldn't do it with just the decoupled C-13 NMR.

Hey! Does anyone have any tips on memorizing the ortho-para-meta directing groups for aromatic reactions? I've seen questions on these on practice tests and I understand they're pretty imp, but I'm having a hard time getting them all memorized without confusing them. Any mnemonics? Thanks!

Hey! Does anyone have any tips on memorizing the ortho-para-meta directing groups for aromatic reactions? I've seen questions on these on practice tests and I understand they're pretty imp, but I'm having a hard time getting them all memorized without confusing them. Any mnemonics? Thanks!


This is what worked for me:

(I just memorized the ortha/para deactivators and meta deactivators) - so far, its worked. :thumbup:

1) ortho/para activating - everything else

2) ortho/para deactivating - Halogens

3) meta deactivating - NO2, SO3H, nitrile, N+R3, carbonyl compounds

This is what worked for me:

(I just memorized the ortha/para deactivators and meta deactivators) - so far, its worked. :thumbup:

1) ortho/para activating - everything else

2) ortho/para deactivating - Halogens

3) meta deactivating - NO2, SO3H, nitrile, N+R3, carbonyl compounds

thats very helpful, thanks a lot!

how come tert-butoxide can be used as a nucleophile in Williamson-Ether synthesis? I thought it was SN2, and SN2 is not favorable when there are steric hindrances...

does sterics just apply to the substrate, not the nucleophile?

how come tert-butoxide can be used as a nucleophile in Williamson-Ether synthesis? I thought it was SN2, and SN2 is not favorable when there are steric hindrances...

does sterics just apply to the substrate, not the nucleophile?
If you want to form a t-butoxyether, you MUST use t-butoxide as the nucleophile. If you try to do it backward, you'll wind up getting an elimination instead. See the organic explanations post about the four mechanisms if you don't understand that point. It is true that the hindrance of the Nu can be a problem in Sn2 reactions, but the major issue with Sn2 is about the hindrance of the substrate more than the hindrance of the Nu.

Is there a difference between the cis/trans and z/e designation for isomerism...or do they refer to the same thing and can be used interchangeably?

Is there a difference between the cis/trans and z/e designation for isomerism...or do they refer to the same thing and can be used interchangeably?
It depends on how complicated the molecule is. If you're talking about something with two substituents (one on each carbon), the cis and the Z are the same thing, as are the trans and the E. If you're talking about something with more than two substituents, you should use the E/Z designation.

too much stereochemistry! can anyone see anything wrong with all this:


enantiomers are mirror images and have the same phys/chem properties, but maybe different physiological properties.

diastereomers have different phys/chem/physiological properties.

meso compounds have chiral centers but their symmetry cancels out any optical activity.

anomers are cyclic carbohydrates that are diastereomers (differ at one chiral carbon) and are subject to flipping via mutarotation

epimers are aldoses (and other carbohydrates?) that differ only at chiral carbon (and are hence diastereomers)

D and L configurations are just based on the D and L glyceraldehydes (stupid conventions... argh) and so on diagrams where the carbonyl is at the top of the molecule they have an OH on the lowest left position.

D and L are independent of actual optical activity, where the real indication is +/- or d/l?

R and S are independent of actual optical activity.

too much stereochemistry! can anyone see anything wrong with all this:


enantiomers are mirror images and have the same phys/chem properties, but maybe different physiological properties.

diastereomers have different phys/chem/physiological properties.

meso compounds have chiral centers but their symmetry cancels out any optical activity.

anomers are cyclic carbohydrates that are diastereomers (differ at one chiral carbon) and are subject to flipping via mutarotation

epimers are aldoses (and other carbohydrates?) that differ only at chiral carbon (and are hence diastereomers)

D and L configurations are just based on the D and L glyceraldehydes (stupid conventions... argh) and so on diagrams where the carbonyl is at the top of the molecule they have an OH on the lowest left position.

D and L are independent of actual optical activity, where the real indication is +/- or d/l?

R and S are independent of actual optical activity.
Looks mostly good to me, except that I'd add the following clarifications:

Diastereomers tend to have similar chemical properties; it's the difference in their physical properties (ex. solubility, affinity, etc.) that is significant.

Epimers are a subset of diastereomers that differ in stereochemistry at only a single chiral center when multiple chiral centers are present. Anomers are a subtype of epimers that are formed when a sugar cyclizes to form a new stereocenter with alpha and beta designations. (That's why anomers can mutarotate, as you said.) So, for example, glucose and galactose are epimers (at C-4), but they are not anomers. Alpha-D-glucopyranose and beta-D-glucopyranose are both anomers and epimers.

Only the L-isomer of a sugar has the OH group on the left at the highest-numbered (bottom) chiral carbon. If the OH group of that C is on the right, then the configuration is analogous to D-glyceraldehyde, and the designation is therefore D.

Looks mostly good to me, except that I'd add the following clarifications:Thank you!

Hi, I have a question about acidity.

One of the trends is that acidity increases as you go down a group because the greater the size of the atom, the longer the bond between the atom and H, and therefore, the more breakable it is.

However, can you explain to me why sp hybridized C-H bonds are more acidic than sp2 hybridized C-H bonds, is more acidic than sp3 hybridized C-H bonds? I initially explained to myself that sp has more s character than sp2, than sp3, and therefore the electrons are held closer to the carbon nucleus, which is why H+ can fall off so easily. But it seems to contradict the first trend (where acidity correlates to size) because a sp hybridized C-H bond would be the shortest of the three bonds, and acording to the first trend, the longer the bond between the atom and H, the more breakable it is.

Hi, I have a question about acidity.

One of the trends is that acidity increases as you go down a group because the greater the size of the atom, the longer the bond between the atom and H, and therefore, the more breakable it is.

However, can you explain to me why sp hybridized C-H bonds are more acidic than sp2 hybridized C-H bonds, is more acidic than sp3 hybridized C-H bonds? I initially explained to myself that sp has more s character than sp2, than sp3, and therefore the electrons are held closer to the carbon nucleus, which is why H+ can fall off so easily. But it seems to contradict the first trend (where acidity correlates to size) because a sp hybridized C-H bond would be the shortest of the three bonds, and acording to the first trend, the longer the bond between the atom and H, the more breakable it is.
You can't use the bond length rule here because you are comparing like atoms. In other words, you are comparing C-H to C-H to C-H, not, say, H-Cl to H-Br to H-I. In this case, it is more helpful to think about it in terms of electronegativity instead of in terms of bond length, because every C atom is the same size as every other C atom, and the bond length isn't changing due to atom size. It's changing due to differences in electronegativity of the various hybrids. Basically, if the conjugate base being left behind after proton removal is more electronegative, that means it can better stabilize the negative charge it will hold once the proton is removed. So, since an sp-hybridized C is more electronegative than an sp2-hybridized C, it will be better able to stabilize the negative charge, and it will be a stronger acid. Same reasoning for why sp2-hybridized carbons are stronger acids than sp3-hybridized carbons. The differences are large: assuming no heteroatoms are present, an sp-hybridized carbon has a pKa of about 25, an sp2-hybridized carbon as a pKa of about 40, and an sp3-hybridized carbon has a pKa greater than 50. That is one H that ain't coming off. ;)

You can't use the bond length rule here because you are comparing like atoms. In other words, you are comparing C-H to C-H to C-H, not, say, H-Cl to H-Br to H-I. In this case, it is more helpful to think about it in terms of electronegativity instead of in terms of bond length, because every C atom is the same size as every other C atom, and the bond length isn't changing due to atom size. It's changing due to differences in electronegativity of the various hybrids. Basically, if the conjugate base being left behind after proton removal is more electronegative, that means it can better stabilize the negative charge it will hold once the proton is removed. So, since an sp-hybridized C is more electronegative than an sp2-hybridized C, it will be better able to stabilize the negative charge, and it will be a stronger acid. Same reasoning for why sp2-hybridized carbons are stronger acids than sp3-hybridized carbons. The differences are large: assuming no heteroatoms are present, an sp-hybridized carbon has a pKa of about 25, an sp2-hybridized carbon as a pKa of about 40, and an sp3-hybridized carbon has a pKa greater than 50. That is one H that ain't coming off. ;)


Wow, thanks for the reply :)

what does "stereogenic" mean?
Thanks :)

I think stereogenic just means a stereocentre, or a centre that is chiral (usually C with 4 different groups bonded to it).

I have a question: Why is a C bonded to OH more polar than a C double bonded to O? i.e. you are comparing polarities of cyclohexane with ketone vs. cyclohexanol

Thanks!

Hey! Does anyone have any tips on memorizing the ortho-para-meta directing groups for aromatic reactions? I've seen questions on these on practice tests and I understand they're pretty imp, but I'm having a hard time getting them all memorized without confusing them. Any mnemonics? Thanks!

A prof taught me this: The meta directing groups carry a partial or full positive charge, while the ortho/para groups carry unshared pairs of electrons. The exception you have to watch out for is halogens because they don't follow the rule. It's worked for me so far.

A prof taught me this: The meta directing groups carry a partial or full positive charge, while the ortho/para groups carry unshared pairs of electrons. The exception you have to watch out for is halogens because they don't follow the rule. It's worked for me so far.



A decent rule of thumb, but doesn't take into account the fact that alkyl groups are ortho-para activators. Make sure you remember that.

what does "stereogenic" mean?
Thanks :)
Stereogenic atoms are ones where the interchange of two of the groups attached to them would invert the stereochemistry of the molecule. They are NOT necessarily chiral centers, although all chiral atoms are also stereogenic. For example, if you had cis-2-butene, the two carbons in the double bond are both stereogenic, but they are not chiral. They are stereogenic because if you exchange the methyl group (C1) attached to C2 with the H attached to C2, you will get the trans isomer. (The same thing happens for C3, which therefore is also stereogenic.) They are achiral because double bonds are flat, and therefore have an internal plane of symmetry.

I think stereogenic just means a stereocentre, or a centre that is chiral (usually C with 4 different groups bonded to it).

I have a question: Why is a C bonded to OH more polar than a C double bonded to O? i.e. you are comparing polarities of cyclohexane with ketone vs. cyclohexanol

Thanks!
The size of a dipole depends on a lot of things, including the shape of the molecule and the bond length. But I would wager to say that in the example you've given, it would have to do with the fact that cyclohexanone is polar aprotic, while cyclohexanol is polar protic. (Read my explanations post about these terms if you're not familiar with them.)

Regarding radical additions to alkenes, I know that Anti-markovnikov addition with HBr is a radical addition of the Br radical to the C-radical. But are all anti-markovnikov additions radical additions (like hydroboration, for example), or does this only apply to HBr addition?

Regarding radical additions to alkenes, I know that Anti-markovnikov addition with HBr is a radical addition of the Br radical to the C-radical. But are all anti-markovnikov additions radical additions (like hydroboration, for example), or does this only apply to HBr addition?
Please do not start new threads in the subforum. Just post your question in the appropriate already existing thread. We check these threads daily.

In answer to your question, yes, you can have anti-Markovnikov additions without radicals. Hydroboration in particular does not go by a radical mechanism. Basically, any time you have a scenario where you are adding the heteroatom first before adding the H, you will get anti-Markovnikov addition. Instead of trying to memorize what reagents go Markovnikov versus anti-Markovnikov, try to understand the rationale for what's happening. The reason that these reactions have this regioselectivity is because they always want to form the more stable carbocation (or radical; remember that radical stability trends parallel carbocation stability trends). So whatever atom you add to the double bond first will always go on the less substituted carbon atom, leaving the carbocation (or radical) on the more substituted carbon because that forms the more stable intermediate. With radical addition of HBr, you're adding the Br first, so it goes to the less substituted atom. With BH3, you're adding the B first, and the same thing happens.

Please do not start new threads in the subforum. Just post your question in the appropriate already existing thread. We check these threads daily.

In answer to your question, yes, you can have anti-Markovnikov additions without radicals. Hydroboration in particular does not go by a radical mechanism. Basically, any time you have a scenario where you are adding the heteroatom first before adding the H, you will get anti-Markovnikov addition. Instead of trying to memorize what reagents go Markovnikov versus anti-Markovnikov, try to understand the rationale for what's happening. The reason that these reactions have this regioselectivity is because they always want to form the more stable carbocation (or radical; remember that radical stability trends parallel carbocation stability trends). So whatever atom you add to the double bond first will always go on the less substituted carbon atom, leaving the carbocation (or radical) on the more substituted carbon because that forms the more stable intermediate. With radical addition of HBr, you're adding the Br first, so it goes to the less substituted atom. With BH3, you're adding the B first, and the same thing happens.

Sorry about that, I think I jumped ahead of myself when I posting..did it to fast.

I still dont understand what defines a radical reaction over forming a carbocation intermediate? Why doesn't the HBr/peroxide reaction proceed with a carbocation?

Sorry about that, I think I jumped ahead of myself when I posting..did it to fast.

I still dont understand what defines a radical reaction over forming a carbocation intermediate? Why doesn't the HBr/peroxide reaction proceed with a carbocation?
You should assume that there is not a radical mechanism, unless you see one of two things: peroxides, which you've already seen form radicals; or light (often written as hv over the reaction arrow). Peroxides have an unstable O-O single bond, and that is why they tend to homolytically cleave (meaning one electron from the bond goes with one oxygen, and the other goes with the other oxygen, instead of both going with the same oxygen) into radicals. Light is very high energy, and so it is able to cause homolytic bond splitting that would not normally happen in the absence of light.

i have a last minute question about op/meta directors. i know most things are op, and i need to just remember which are meta (for some reason, this is the hardest thing EVER for me!)

is it safe to say carbonyl's are meta directors, or is there an exception to that? any help is appreciated.

sasha

You should assume that there is not a radical mechanism, unless you see one of two things: peroxides, which you've already seen form radicals; or light (often written as hv over the reaction arrow). Peroxides have an unstable O-O single bond, and that is why they tend to homolytically cleave (meaning one electron from the bond goes with one oxygen, and the other goes with the other oxygen, instead of both going with the same oxygen) into radicals. Light is very high energy, and so it is able to cause homolytic bond splitting that would not normally happen in the absence of light.

So this is generally true for reactions with peroxide or light, but in hydroboration, this is an exception, since hydroboration, although it uses peroxide, doesnt use a radical mechanism, correct?

I thought aromatics aren't on the MCAT. Why do we need to know anything about ortho meta directors?

So this is generally true for reactions with peroxide or light, but in hydroboration, this is an exception, since hydroboration, although it uses peroxide, doesnt use a radical mechanism, correct?
Ah, ok, NOW I see why you're so confused! :laugh: You're right; peroxide does get used with hydroboration, but not until AFTER the borane addition to the double bond is over. If you take a look at the reaction mechanism, you will see that the two compounds (borane and peroxide) are added in two separate steps. That's why this mechanism is different compared to the HBr reaction, where peroxides are added to the reaction WITH the HBr, and only in a catalytic amount.

I thought aromatics aren't on the MCAT. Why do we need to know anything about ortho meta directors?
They're not directly tested, but they can be tested in other indirect ways, such as nomenclature, bonding/stability questions, and reactions. Alkenes can also appear in the guise of stereoisomerism (E vs. Z) or addition reactions. I assume that's why people are asking.

It's also possible that we have some DATers and PCATers in here, or even just some people who are taking organic now and want to ask questions.

These are probably the last questions I have. :)

For chirality, how do you determine if a carbon is chiral if the adjacent carbon is exactly like another.

i.e. say a carbon has a -H, -OH, -CH2CH3, and -CH2CH2CH3. So even though the last two groups are different, an ethyl and propyl, the carbons attached to the single carbon they share are both -CH2-. Is that "chiral" carbon really chiral because the four groups are different or is it achiral because two of the groups attach to the "chiral" carbon with the same exact -CH2- part?

For a chiral carbon in a ring, would it automatically be chiral if the ring was even numbered. i.e. say cyclohexane has a carbon with an -OH and an -H group (cyclohexanol). To that carbon, there are 5 others in the ring. So if you were to follow the ring from that carbon, you could pass 3 carbons on one direction and 2 in another until you get to the same point. Would the carbon be chrial because it's like an ethyl group and propyl group susbstituents, but just had a ring closure?

Now say it's a 7 membered carbon ring, leaving 6 other carbons in the ring besides the one with the hydroxyl group. If you do the same thing again by following along both side of the ring, you can split it into 2 3-Carbon chains, so two propyl groups. Since there would be two propyl groups (but just closed in a ring), would they be considered the same as on the carbon with the -OH making it achiral?

Any help would be appreciated. Thanks!

These are probably the last questions I have. :)

For chirality, how do you determine if a carbon is chiral if the adjacent carbon is exactly like another.

i.e. say a carbon has a -H, -OH, -CH2CH3, and -CH2CH2CH3. So even though the last two groups are different, an ethyl and propyl, the carbons attached to the single carbon they share are both -CH2-. Is that "chiral" carbon really chiral because the four groups are different or is it achiral because two of the groups attach to the "chiral" carbon with the same exact -CH2- part?

For a chiral carbon in a ring, would it automatically be chiral if the ring was even numbered. i.e. say cyclohexane has a carbon with an -OH and an -H group (cyclohexanol). To that carbon, there are 5 others in the ring. So if you were to follow the ring from that carbon, you could pass 3 carbons on one direction and 2 in another until you get to the same point. Would the carbon be chrial because it's like an ethyl group and propyl group susbstituents, but just had a ring closure?

Now say it's a 7 membered carbon ring, leaving 6 other carbons in the ring besides the one with the hydroxyl group. If you do the same thing again by following along both side of the ring, you can split it into 2 3-Carbon chains, so two propyl groups. Since there would be two propyl groups (but just closed in a ring), would they be considered the same as on the carbon with the -OH making it achiral?

Any help would be appreciated. Thanks!
1) Yes, that carbon is chiral, because a propyl is not equivalent to an ethyl. Here's what you should do when you have two carbon substituents like that:

Since the first atom out (C in this case) is the same for both substituents, you must go out to the second atoms of each. For the propyl group, the first carbon is attached to a second carbon as well as two hydrogens: C, H, H. For the ethyl group, the same is true: carbon one is again attached to C, H, H. Therefore, you must go out to the third atoms. For the propyl, the second carbon is attached to one carbon plus two hydrogens, again giving: C, H, H. However, the second carbon of the ethyl is attached to three hydrogens, giving: H, H, H. You can cross out two H's from each (since they are identical), leaving you to compare C from the propyl with H from the ethyl. Since C has a higher atomic number than H, you would prioritize propyl before ethyl. Be careful here, because if you had a fluorine on your ethyl (CH2CH2F), then you would be comparing a C from propyl to a F from ethyl, and your higher priority group would then be ethyl.

In general: you should continue working your way outward from the chiral center until you find a difference in substituents. If you never do find a difference, the carbon is not chiral.

2) and 3) I'm sure you can now figure out from what I told you in part one that the answer to these questions is that chirality depends on whether the ring is symmetrical or not. :)

Thank you!!! Thanks for pointing out the CH2F beats out a CH2CH3!

2) and 3) I'm sure you can now figure out from what I told you in part one that the answer to these questions is that chirality depends on whether the ring is symmetrical or not. :)

So for cyclohexanol, the C with -OH will have 5 other ring carbons. If I do what you're saying, it would be like two ethyl groups, but both attached by a -CH2-. Would the MCAT ask me to determine if the chiral carbon is R or S? Wouldn't that depend on which chain (when ring broken) I put the CH2 on? Or maybe I got this all wrong...

Thank you!!! Thanks for pointing out the CH2F beats out a CH2CH3!



So for cyclohexanol, the C with -OH will have 5 other ring carbons. If I do what you're saying, it would be like two ethyl groups, but both attached by a -CH2-. Would the MCAT ask me to determine if the chiral carbon is R or S? Wouldn't that depend on which chain (when ring broken) I put the CH2 on? Or maybe I got this all wrong...
Ok, take a look at this picture of 4-methylcyclohexanol. You might be tempted to say that carbons 1 and 4 are chiral, right? But actually they aren't, because the ring is identical from either carbon. Contrast that with 3-methylcyclohexanol, which is asymmetrical and chiral.

http://www.cheric.org/research/kdb/hcprop/molimg/906.gif

http://www.dfmg.com.tw/member/chemical/cas/591-23-1.files/CA7DTYQJ.gif

Thank you!!! Thanks for pointing out the CH2F beats out a CH2CH3!



So for cyclohexanol, the C with -OH will have 5 other ring carbons. If I do what you're saying, it would be like two ethyl groups, but both attached by a -CH2-. Would the MCAT ask me to determine if the chiral carbon is R or S? Wouldn't that depend on which chain (when ring broken) I put the CH2 on? Or maybe I got this all wrong...

Yes you do

cyclohexanol http://upload.wikimedia.org/wikipedia/en/thumb/7/71/Cyclohexanol.png/100px-Cyclohexanol.png

this is not chiral at all because there is not 4 steriogenic attachments to the carbon or any carbon... no chirality a or other wise...

listen, think of it like this... like the other gentlemen said... Take the initial C and start your problem like this C(mark here whatever is conected to the C in order) for the case of this molicule it would look like this

C(OH)
C(H)
C(CHHCHH(CH2)
C(CHHCHH(CH2) this is equal and there for is not a steriogenic carbon...

Alright, I got it now!!! Thanks! So if there's a plane of symmetry like in cyclohexanol and 4-methylcyclohexanol, it's the same as cutting it right down the middle. Which makes two equal sides so it's achiral. Why is it always the basics that screw me up?! haha, thanks again. :)

Ok, take a look at this picture of 4-methylcyclohexanol. You might be tempted to say that carbons 1 and 4 are chiral, right? But actually they aren't, because the ring is identical from either carbon. Contrast that with 3-methylcyclohexanol, which is asymmetrical and chiral.

http://www.cheric.org/research/kdb/hcprop/molimg/906.gif

http://www.dfmg.com.tw/member/chemical/cas/591-23-1.files/CA7DTYQJ.gif

Hi Q,

What might confuse some people on 4 methylcyclohexanol is that it does have 2 isomers, a cis isomer and a trans isomer.(Ok, so that caught me at first when you were explaining it.)

Hi Q,

What might confuse some people on 4 methylcyclohexanol is that it does have 2 isomers, a cis isomer and a trans isomer.(Ok, so that caught me at first when you were explaining it.)
True. But cis/trans isomerism is not the same as stereoisomerism.

True. But cis/trans isomerism is not the same as stereoisomerism.

Quimica I freaking love your ICON... I'm Diene... I do it all the time now... IT is so freaking NERDY I love it... LMAO..

Quimica is correct...

but to be honest I think my way of working it out is absolute best... The quimica explains it is vital to understanding but per the method of working it out my way sure cuts down on silly mistakes...

What is a cross product? Also, when judging which of two proposed mechanisms would produce a cross product how would we do that?

What is a cross product? Also, when judging which of two proposed mechanisms would produce a cross product how would we do that?


Cross product of what? be specific... for example there is something called the crossed Adol reaction as well as the Crossed Claisen reaction... Which basically means that there isn't two identical molecules but rather one that fits some rule and another that doesn't... given some crossed result...

Cross product of what? be specific... for example there is something called the crossed Adol reaction as well as the Crossed Claisen reaction... Which basically means that there isn't two identical molecules but rather one that fits some rule and another that doesn't... given some crossed result...
I thought maybe s/he meant a side product, but I'm not sure either. wannabe, the only cross product I know of is for vectors, not for organic reactions. ;) Can you explain what you mean exactly? I don't think we really understand your question.

Cross product was mentioned in a practice exam passage. The passage is about the Claisen rearragement. One of the questions was something about which mechanism shows an absence of a cross-product. I guessed that the one where the substituent never broke off was whenn no "cross-products" were formed and it was correct. The other mechanism showed the substituent breaking off the ring, which then showed two separate compounds existing at once, and then reattaching. So I guessed that one was when a cross-product was formed since it could probably reattach anywhere it wanted. So was a cross product the part that breaks free off the original compound?

Yea, thanks WilliamsF1 - indeed that was the question I was referring to (I just wasn't sure if I was allowed to post it on here).
Also, what's the difference between bond energy and bond dissociation energy? Are they the same thing? More stable bonds would mean lower energy bonds with higher dissociation energies right?

Oh sorry...more thing...I'm not sure how to explain this clearly, but say you have an trans-1,2 dibromoethene molecule. In the proton NMR, would the hydrogens on either carbon be equivalent to each other?

Yea, thanks WilliamsF1 - indeed that was the question I was referring to (I just wasn't sure if I was allowed to post it on here).
Also, what's the difference between bond energy and bond dissociation energy? Are they the same thing? More stable bonds would mean lower energy bonds with higher dissociation energies right?

Oh sorry...more thing...I'm not sure how to explain this clearly, but say you have an trans-1,2 dibromoethene molecule. In the proton NMR, would the hydrogens on either carbon be equivalent to each other?
Technically we're not supposed to post any specific AAMC questions on SDN, but it's ok to ask general questions. Just don't get too specific. Clear as mud, right? :)

1) Ok, so is your passage about a mixed Claisen condensation (where you preferentially react an enolized ketone with a non-enolizable aldehyde and get an unsaturated ketone)? I checked in one of my books, and that mixed Claisen condensation product can also be called a cross product. That is different, however, than the Claisen rearrangement, where you have a 3,3-sigmatropic rearrangement of an allyl vinyl ether to a carbonyl compound. Take a look at your passage, and tell me if you see an unsaturated ether undergoing a cyclic rearrangement to a carbonyl, or two carbonyl compounds being joined together to form an unsaturated ketone. I don't understand what you guys mean by things breaking off and reattaching.

Here is a picture of a Claisen rearrangement; is that what you're seeing:

http://content.answers.com/main/content/wp/en-commons/thumb/c/c2/400px-Claisen_Rearrangement_Scheme.png

2) Bond energy is the energy required to break a covalent bond homolytically (into neutral fragments). Bond energies are called bond dissociation energies when given for specific bonds, or average bond energies when summarized for a given type of bond over many kinds of compounds. A more stable bond will have a larger standard bond energy and therefore be more difficult to break.

3) If it's trans, then the molecule is symmetrical, so yes.

1) Ok, so is your passage about a mixed Claisen condensation (where you preferentially react an enolized ketone with a non-enolizable aldehyde and get an unsaturated ketone)? I checked in one of my books, and that mixed Claisen condensation product can also be called a cross product. That is different, however, than the Claisen rearrangement, where you have a 3,3-sigmatropic rearrangement of an allyl vinyl ether to a carbonyl compound. Take a look at your passage, and tell me if you see an unsaturated ether undergoing a cyclic rearrangement to a carbonyl, or two carbonyl compounds being joined together to form an unsaturated ketone. I don't understand what you guys mean by things breaking off and reattaching.

Here is a picture of a Claisen rearrangement; is that what you're seeing:

http://content.answers.com/main/content/wp/en-commons/thumb/c/c2/400px-Claisen_Rearrangement_Scheme.png



Yes, the passage was referring to a Claisen rearrangement; there were two proposed mechanisms for this rearrangement - one was the concerted rearrangement you illustrated above, and the other had the oxygen-C1 bond undergoing a heterolytic cleavage to form a carbocation and a phenolate anion. Then, the carbocation attacks the ortho-C of the ring to give a ketone that enolized to the product.
The question asked which mechanism would NOT produce a cross product. So yea, I think the second mechanism is a Claisen condensation where the condensation cross product is possible so that's why the 1st mechanism wouldn't produce the cross product.
Thanks for your help!

Yes, the passage was referring to a Claisen rearrangement; there were two proposed mechanisms for this rearrangement - one was the concerted rearrangement you illustrated above, and the other had the oxygen-C1 bond undergoing a heterolytic cleavage to form a carbocation and a phenolate anion. Then, the carbocation attacks the ortho-C of the ring to give a ketone that enolized to the product.
The question asked which mechanism would NOT produce a cross product. So yea, I think the second mechanism is a Claisen condensation where the condensation cross product is possible so that's why the 1st mechanism wouldn't produce the cross product.
Thanks for your help!

what scares me is that you seem not to be understanding the difference between crossed and regular claisen... In order to be crossed one of the molecules isn't going to have a Beta acid hydrogen and the other is... The cleavage of the leaving group happens to the molecule that doesn't have the Beta acidic hydrogens...

Does that help at all?

Yes, the passage was referring to a Claisen rearrangement; there were two proposed mechanisms for this rearrangement - one was the concerted rearrangement you illustrated above, and the other had the oxygen-C1 bond undergoing a heterolytic cleavage to form a carbocation and a phenolate anion. Then, the carbocation attacks the ortho-C of the ring to give a ketone that enolized to the product.
The question asked which mechanism would NOT produce a cross product. So yea, I think the second mechanism is a Claisen condensation where the condensation cross product is possible so that's why the 1st mechanism wouldn't produce the cross product.
Thanks for your help!
Ok, I think that I follow you. Basically, you are being presented with two alternative mechanisms for a Claisen rearrangement. The 3,3 sigmatropic rearrangement is the generally accepted mechanism, so that's why I was so confused. :p

In that case then, yes, I would agree that the first mechanism (the 3,3 sigmatropic) would NOT form a cross product, because basically you are just rearranging the bonds in that mechanism, and it is concerted. In other words, you do not form any intermediates. Therefore, the only other thing that the reaction could do would be to reverse (i.e., reform the starting material). However, since the product carbonyl is more stable than the starting material allyl ether, this is not not likely to occur.

The alternative mechanism that they gave you is not concerted. Here, you are forming a reactive, high-potential intermediate (the carbocation), and since highly reactive compounds are not very selective, it is definitely possible that you could wind up with that carbocation reacting with some other electron pair besides the one you want it to react with. In that case, it would be reasonable to expect to see multiple products.

P.S. FYI, the second mechanism is NOT a Claisen condensation. See below for what the Claisen condensation looks like. This example is not mixed, and so no mixed product (or cross product) will form. To get a cross product from the Claisen condensation, you would need to have two different esters here, one of which is enolizable and the other of which is not.

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch21/claisen.gif

Okay, I just went over my EK Orgo again today and I think they confused me with a chiral problem, geez!

Here's the struture they had (at least as close as I could find).

http://img291.imageshack.us/img291/1554/nicotinepl0.jpg which they said was (-)-nicotine

I tried to find all the carbons where it would differ from (+)-nicotine. I figured that this would be where all the chiral ones are since the enantiomer rule in the book said all the absolute configurations of each chiral carbon must be the opposite (according to our text).

To me, it looks like Carbon 4 (the red arrow) is chiral and so is the one across with, Carbon 5 (black question mark).

Edit: I think I see where I messed up. C-5 doesn't have 4 substituents since it will always be double bonded to one of the Carbons next to it, making 3 substituents (achiral). Nevermind!!! :D I was using the technique from yesterday by assuming each Carbon flanking C-5 was identical meaning I had to go to the one connected to that one. Glad I realized this before Saturday! :)

Okay, I just went over my EK Orgo again today and I think they confused me with a chiral problem, geez!

Here's the struture they had (at least as close as I could find).

http://img291.imageshack.us/img291/1554/nicotinepl0.jpg which they said was (-)-nicotine

I tried to find all the carbons where it would differ from (+)-nicotine. I figured that this would be where all the chiral ones are since the enantiomer rule in the book said all the absolute configurations of each chiral carbon must be the opposite (according to our text).

To me, it looks like Carbon 4 (the red arrow) is chiral and so is the one across with, Carbon 5 (black question mark).

Edit: I think I see where I messed up. C-5 doesn't have 4 substituents since it will always be double bonded to one of the Carbons next to it, making 3 substituents (achiral). Nevermind!!! :D I was using the technique from yesterday by assuming each Carbon flanking C-5 was identical meaning I had to go to the one connected to that one. Glad I realized this before Saturday! :)



I used the same thought process on that very same question the first time around. Then it suddenly dawned on me that carbons in an aromatic ring can't be chiral.

I used the same thought process on that very same question the first time around. Then it suddenly dawned on me that carbons in an aromatic ring can't be chiral.
Right. Only the red marked atom is chiral. :)

Get off SDN and go get a good night's sleep. Good luck tomorrow, everyone. :)

I'd just like to say thanks to QofQuimica and everyone else that helped for answering all my questions! It really helped out a lot today!

Are alcohols too volatile a solvent to use instead of ketones in reactions involving an allylic molecule and a phenyllithium?

Are alcohols too volatile a solvent to use instead of ketones in reactions involving an allylic molecule and a phenyllithium?
No. But they are too acidic. ;)

does anyone know what the MCAT Student Manual wants from us here?:

Kinetic onctrol versus thermodynamic control of a reaction
(this on the 3rd pg under kinetics line)

does this mean that kinetics deals with how an rxn got to equil. and that thermodynamics deals with equilibrium?

I really dont know, plz help!

does anyone know what the MCAT Student Manual wants from us here?:

Kinetic onctrol versus thermodynamic control of a reaction
(this on the 3rd pg under kinetics line)

does this mean that kinetics deals with how an rxn got to equil. and that thermodynamics deals with equilibrium?

I really dont know, plz help!
Yes, in a nutshell. A reaction under kinetic control is one that forms the fastest product, even if that product isn't the most stable. A reaction under thermodynamic control forms the most stable product, even if it takes longer. One example you are probably familiar with is double bond formation after elimination reactions. Remember how you can have the Zaitzev product versus the Hoffman product? The Zaitzev product is the thermodynamic product: it is more substituted and more stable. The Hoffman product is the kinetic product: it forms faster, but it is less stable than the Zaitzev product.

I just started organic this semester, so this is a really simple question.
In C6H10, are all 6 carbons in the same plane? When I used a model set to build the moleclue, it appears like one is out of the plane of the other 5? Is this true, or are they assumed to be all in the same plane?

I just started organic this semester, so this is a really simple question.
In C6H10, are all 6 carbons in the same plane? When I used a model set to build the moleclue, it appears like one is out of the plane of the other 5? Is this true, or are they assumed to be all in the same plane?

That depends on what isomer you're talking about.

I just started organic this semester, so this is a really simple question.
In C6H10, are all 6 carbons in the same plane? When I used a model set to build the moleclue, it appears like one is out of the plane of the other 5? Is this true, or are they assumed to be all in the same plane?
It depends. You can have several molecules all having that same molecular formula. These are called isomers. Some of the isomers are coplanar, and others aren't. You probably haven't covered isomers yet, but after you do, this concept will make more sense to you.

It depends. You can have several molecules all having that same molecular formula. These are called isomers. Some of the isomers are coplanar, and others aren't. You probably haven't covered isomers yet, but after you do, this concept will make more sense to you.

Q, you think it's too early to tell him about conformational isomers? :D

Q, you think it's too early to tell him about conformational isomers? :D
And people think it's EASY to teach organic chemistry. :rolleyes: Luckily, it won't be a problem for a fully unsaturated molecule. :)

It depends. You can have several molecules all having that same molecular formula. These are called isomers. Some of the isomers are coplanar, and others aren't. You probably haven't covered isomers yet, but after you do, this concept will make more sense to you.

Well.. How about if I try to describe it? I know what an isomer is, but we havent covered it in class..sorry for being a little ambiguous about the structure. Well, C6H10, cyclic molecule. One C=C doublebond, the rest of the carbons attached by a single bond. So basically benzene, but instead of three C=C double bonds and resonance, only one C=C double bond. Do you know what I mean now?
Heres my problem. When I build the molecule with my nice molecular model set, it looks as if the one carbon is not in the same plane as the other 5 carbons. Its hard to explain. If anyone knows what I mean, or can affirm whether or not all the C's are planar, the help is appreciated! Thanks!

Well.. How about if I try to describe it? I know what an isomer is, but we havent covered it in class..sorry for being a little ambiguous about the structure. Well, C6H10, cyclic molecule. One C=C doublebond, the rest of the carbons attached by a single bond. So basically benzene, but instead of three C=C double bonds and resonance, only one C=C double bond. Do you know what I mean now?
Heres my problem. When I build the molecule with my nice molecular model set, it looks as if the one carbon is not in the same plane as the other 5 carbons. Its hard to explain. If anyone knows what I mean, or can affirm whether or not all the C's are planar, the help is appreciated! Thanks!

Ok, so you know the specific molecule which turns out to be cyclohexene. I'm just pointing that out since a molecule with a formula of C6H10 is one of the following

Has a triple bond
Has one double bond and one ring
Has 2 rings(I'm thinking dicyclopropane but I'm not sure if that's real)
Has 2 double bonds.

Anyway back to cyclohexene. I found it here
some 3d molecules (http://www.ch.cam.ac.uk/magnus/molecules/misc/s3.html)

Looking at that I've got to go with no, it doesn't have 5 carbons in one plane. Q, is this site accurate in this case?

Ok, so you know the specific molecule which turns out to be cyclohexene. I'm just pointing that out since a molecule with a formula of C6H10 is one of the following

Has a triple bond
Has one double bond and one ring
Has 2 rings(I'm thinking dicyclopropane but I'm not sure if that's real)
Has 2 double bonds.

Anyway back to cyclohexene. I found it here
some 3d molecules (http://www.ch.cam.ac.uk/magnus/molecules/misc/s3.html)

Looking at that I've got to go with no, it doesn't have 5 carbons in one plane. Q, is this site accurate in this case?
Right-o. Cyclohexene will not be planar, t2005. In fact, cyclohexene has very different physical and chemical properties versus benzene.

Right-o. Cyclohexene will not be planar, t2005. In fact, cyclohexene has very different physical and chemical properties versus benzene.

Thank you!

1. How many isomers are possible for isomeric pentanols (C5H12OH)?

2. Are there any known relationship between the number of possible isomers of a certain compound and the number of C or O or any element for that matter?

Danke! Thanks!:)

1. How many isomers are possible for isomeric pentanols (C5H12OH)?

2. Are there any known relationship between the number of possible isomers of a certain compound and the number of C or O or any element for that matter?

Danke! Thanks!:)
1). Assuming they must be straight chain alcohols, you have three pentanol isomers possible: 1-pentanol, 2-pentanol, and 3-pentanol. If you're allowed to have some branched ones, you can also have some methyl-substituted butanols (2-methyl-2-butanol, 2-methyl-1-butanol, 3-methyl-1-butanol, and 3-methyl-2-butanol).

2) I don't know. Probably. But my mind doesn't work that way; I just sit and draw them all out. :)

1). Assuming they must be straight chain alcohols, you have three pentanol isomers possible: 1-pentanol, 2-pentanol, and 3-pentanol. If you're allowed to have some branched ones, you can also have some methyl-substituted butanols (2-methyl-2-butanol, 2-methyl-1-butanol, 3-methyl-1-butanol, and 3-methyl-2-butanol).

2) I don't know. Probably. But my mind doesn't work that way; I just sit and draw them all out. :)

Q, isn't 1 a trick question? I mean pentane has 12 hydrogens and the formula he has is C5H12OH. Isn't the answer Zero?

What would determine the various organic compounds' solubility in toluene and methanol? Toluene is nonpolar so any nonpolar organic compounds such as two cyclohexanes connected by a single bond or any symmetric org compounds are soluble?
How about in methanol? Will the alcohol group in methanol attract organic compounds with polarity or with other OH's? In which solution will Succinic acid be soluble?

Thanks for your help!

BTW is Ph in any molecular structure a shorthand for phenol?

Q, isn't 1 a trick question? I mean pentane has 12 hydrogens and the formula he has is C5H12OH. Isn't the answer Zero?
No. Oxygens don't affect the unsaturation count like nitrogens do. Pentanol isomers also have twelve H's. :)

What would determine the various organic compounds' solubility in toluene and methanol? Toluene is nonpolar so any nonpolar organic compounds such as two cyclohexanes connected by a single bond or any symmetric org compounds are soluble?
How about in methanol? Will the alcohol group in methanol attract organic compounds with polarity or with other OH's? In which solution will Succinic acid be soluble?

Thanks for your help!
You've got the general idea. Like dissolves like, so polar molecules will dissolve in a polar solvent, and nonpolar ones will dissolve in a nonpolar solvent. I wrote a post about polar versus nonpolar in the organic explanations thread.

In answer to your second question, no, Ph means a phenyl group attached to something as a substituent. That is different than phenol, which is a benzene having an OH on it.

I thought you took the Aug. MCAT! Are you re-studying even before scores come out?

You've got the general idea. Like dissolves like, so polar molecules will dissolve in a polar solvent, and nonpolar ones will dissolve in a nonpolar solvent. I wrote a post about polar versus nonpolar in the organic explanations thread.

In answer to your second question, no, Ph means a phenyl group attached to something as a substituent. That is different than phenol, which is a benzene having an OH on it.

I thought you took the Aug. MCAT! Are you re-studying even before scores come out?

Thanks QofQuimica. And yes I took the MCAT but I cancelled it after the test because I wasn't entirely feeling well. I am taking again on April. Meanwhile another question....

In the liquid-liquid extraction experiment, you separate two different solutes based on their solubility in two different solvents. There are extraction of an organic mixture with HCl, NaHCO3 and NAOH. I know that during the extraction, there will be a buildup of carbon dioxide because of the mixing sodium bicarbonate and strong acid. What other compounds/gas would build so that we will need to vent the solution while mixing? How will any heat be produced? I am not sure how to predict release of heat without any free-energy number or enthalpy/entropy... Any help would be appreciated!
Thanks again!!!

Thanks QofQuimica. And yes I took the MCAT but I cancelled it after the test because I wasn't entirely feeling well. I am taking again on April. Meanwhile another question....

In the liquid-liquid extraction experiment, you separate two different solutes based on their solubility in two different solvents. There are extraction of an organic mixture with HCl, NaHCO3 and NAOH. I know that during the extraction, there will be a buildup of carbon dioxide because of the mixing sodium bicarbonate and strong acid. What other compounds/gas would build so that we will need to vent the solution while mixing? How will any heat be produced? I am not sure how to predict release of heat without any free-energy number or enthalpy/entropy... Any help would be appreciated!
Thanks again!!!
It's always a good practice to vent frequently while performing an extraction. You should assume that any extraction will need to be vented. Even if you're just washing your organic layer with water or brine, your organic solvent is probably volatile, and like you said, heat can be an issue. (You will commonly get heat produced when you mix an acid with a base.) So for example, if you're using diethyl ether as your organic layer, you really need to vent, and often, due to the volatility of the ether. The heat from your hands is enough to cause it to vaporize.

I always would start an extraction by capping the sep funnel, turning it upside down, and venting it first before any shaking to "test" the pressure. Make sure to point it at an angle toward the back of the hood and away from you or other people. If it vents audibly, give it one or two gentle shakes, then vent again. Repeat as many times as necessary until you no longer hear the hiss of escaping gas. As the pressure starts decreasing, you can shake harder and vent less often between shakes.

Carbon dioxide is probably the most common gas that you'd produce while extracting with bicarb. But you could conceivably produce hydrogen gas too, if you had certain metals in acid, for example. Again, if you follow my technique above, you won't have to worry about a sep funnel popping on you because you didn't consider some source of heat or gas production.

Hope this helps, and good luck in April. :)

<<<<<<<<--------------------------------------

Please refer to my avitar picture on the left for the skeletal structure I am talking about. I think it's a tricky question. The question asks which bond that connects C and O is shorter. I think there can be no resonance because you can't just move electrons due to octet. The carbon in the hexane (the right middle one) won't be able to have octet. Do you agree with me that there can be no resonance and hence the double bond to oxygen will be shorter than the single bond to oxygen? I am really anxious about this so please let me know!!! Thanks!

No. Oxygens don't affect the unsaturation count like nitrogens do. Pentanol isomers also have twelve H's. :)

Wow, I'm slow at getting back to this. Anyway I still think it's a typo. The formula is C5 H12 OH so this compound has 13 hydrogens when it should have 12. Wouldn't C5 H11 OH actually be the correct formula?(Unless we're talking about a penta-coordinate carbon in Sn2 reaction.)

Damn I'm not even currently in college and I still think about this stuff:D

<<<<<<<<--------------------------------------

Please refer to my avitar picture on the left for the skeletal structure I am talking about. I think it's a tricky question. The question asks which bond that connects C and O is shorter. I think there can be no resonance because you can't just move electrons due to octet. The carbon in the hexane (the right middle one) won't be able to have octet. Do you agree with me that there can be no resonance and hence the double bond to oxygen will be shorter than the single bond to oxygen? I am really anxious about this so please let me know!!! Thanks!
Sorry, but that structure DOES have a resonance form. It's basically an enolate ion. You can make a double bond and put a lone pair on the far C of the double bond. That being said, I agree with you that the permanent C=O should be shorter, b/c its bond order is higher (i.e., a full 2 instead of one point something).

Wow, I'm slow at getting back to this. Anyway I still think it's a typo. The formula is C5 H12 OH so this compound has 13 hydrogens when it should have 12. Wouldn't C5 H11 OH actually be the correct formula?(Unless we're talking about a penta-coordinate carbon in Sn2 reaction.)

Damn I'm not even currently in college and I still think about this stuff:D
Heh, yeah, you're right. I was thinking H12 total; didn't notice that it was H12OH. :)

QofQuimica, thanks for your help. Do you know any website where they have a good summary and list of reactions and synthesis? (Like reactions/equations showing how nerol would turn into alpha-terpineol when H+ was put in) Thanks!

QofQuimica, thanks for your help. Do you know any website where they have a good summary and list of reactions and synthesis? (Like reactions/equations showing how nerol would turn into alpha-terpineol when H+ was put in) Thanks!
I assume you're asking about this for class, because synthesis is basically not covered on the MCAT at all. What I recommend you do if you're really interested in learning synthesis is to get a hold of a book and workbook called "Organic synthesis: the disconnection approach" by Stuart Warren. If you work through the exercises in the book and workbook, it will teach you how to do retrosynthesis and come up with synthetic plans in a logical way. The first chapters cover easy molecules, and it gets harder and harder until you're doing some pretty complex molecules. You'd definitely ace all of your synthesis tests if you work through these books. Even doing the first few chapters would help you. See if your school library has them and check them out.

If you're actually asking about learning to write mechanisms (and your question makes it sound like maybe you are), you should get yourself a mechanism book and do the same thing. I used "Electron Flow in Organic Chemistry" by Paul Scudder, but there are some others too. I've seen several people on SDN say they liked "Pushing electrons" by Daniel Weeks, but for the record I've never read that book, so I can't really recommend it myself.

hi, QofQuimica. i have question about the Br2 addtion to alkene
i know that the first answer is right, but not sure the rest..pls help

hi, QofQuimica. i have question about the Br2 addtion to alkene
i know that the first answer is right, but not sure the rest..pls help
Your answer looks good to me. Whenever you do an electrophilic addition to a double bond like this, you always should consider the possibility of rearrangements to more stable carbocations occuring after the electrophile adds. However, since Br2 preferentially forms a cyclic bromonium ion once the electrophilic Br adds, we wouldn't predict that any rearrangements would occur in this case. None of the other double bonds is in the correct location to give you that bromination pattern except for the first one.

I have a lab this week and have to understand the procedure and exp.("dehydration of a tertiary alcohol") before teh actual experiment. we will be using IR spectroscopy. The procedure asks to think about the quesitons below but I can't understand them.. my lab book only has a brief summary of oh group having a broad band above 1500 cm^-1 but nothing else. any help would be appreciated, QofQuimica! Thanks!!

(a) The O-H portion of the spectrum changes with concentration. As the
solution becomes more dilute, the broad band centered at roughly 3300
cm-1 sharpens. Explain.
(b) The O-H stretching band for tert-butyl alcohol is much sharper than that for
methyl alcohol. Explain.

I have a lab this week and have to understand the procedure and exp.("dehydration of a tertiary alcohol") before teh actual experiment. we will be using IR spectroscopy. The procedure asks to think about the quesitons below but I can't understand them.. my lab book only has a brief summary of oh group having a broad band above 1500 cm^-1 but nothing else. any help would be appreciated, QofQuimica! Thanks!!

(a) The O-H portion of the spectrum changes with concentration. As the
solution becomes more dilute, the broad band centered at roughly 3300
cm-1 sharpens. Explain.
(b) The O-H stretching band for tert-butyl alcohol is much sharper than that for
methyl alcohol. Explain.

Intermolecular forces impact stretching frequencies. Alcohol molecules tend to hydrogen bond with each other, which impacts the O-H stretch. Hydrogen bonding is kind of like adding an extra "mass" onto the hydroxy group "spring", so a hydrogen-bonded O-H stretches at a lower frequency than a non-hydrogen bonded (free) molecule. The difference is significant. The attraction of the neighboring oxygen "pulls" on the hydrogen of the adjacent hydroxy group electrostatically, which lengthens the O-H bond and lowers the frequency of vibration. Capeesh?

Here's the kicker: not all molecules are hydrogen bonded, and/or hydrogen-bonded at the same distance and orientation (the strength of hydrogen-bonds are sensitive to these factors), at any given point in time, so the O-H stretch absorption band for the entire solution broadens as a kind of "blurred" average of all the different O-H stretches of the alcohol molecules in their various states.

For alcohols, the strength of the hydrogen bond is roughly about 5 kcal/mol; therefore, you would probably get fairly rapid association, dissociation and re-orientation of the hydrogen bond at room temperature in a particular bonding network. The IR machine usually isn't able to resolve all these different temporal vibrational states (the changes in hydrogen bonding between molecules occur in picoseconds), so what you get when you do IR on this hydrogen-bonding phenomenon is something akin to exaggerated, overlapped photographic blurring at low shutter speed. While this occurs to some degree normally, hydrogen bonding adds a larger range of vibrational states and frequencies, resulting in significant broadening of the IR stretching band. Cool?

What things might interfere with hydrogen bonding, you ask? Well, how about steric factors? If an alcohol molecule, or group, is sterically hindered, it wouldn't be able to hydrogen bond as effectively or as favorably. So what would that do to the O-H stretch absorption band for the solution? It would tend to narrow it, right (i.e., make it sharper)?

What other factors could influence hydrogen bonding? How about concentration? If you dilute an alcohol solution with a solvent (non-polar), you would increase the mean distance between the alcohol molecules, consequently decreasing the prevalence of hydrogen bonding in the solution at any given point in time, I would think. Alcohol molecules need to be in relatively close proximity to each other in order to hydrogen bond, no? With increasing dilution, the solution becomes less and less dense with alcohol. Thus, the absorption band of the O-H stretch would tend to narrow as the solution becomes more dilute.

I'm no expert, but after I read your questions, I started to think about the answers because I was curious about them myself. Take them with a grain of salt... But I think this should help get you started on answering the questions above.

Q, did I miss anything?

Q, did I miss anything?
I think I'm going to retire and turn this thread over to you. :thumbup: Would you mind if I edited your post a bit for uniformity and added it to the organic explanations thread? Also, would you be interested in writing some more posts like it?

Would you mind if I edited your post a bit for uniformity and added it to the organic explanations thread?

I don't mind at all. Feel free to edit it and add it to the explanations thread.


Also, would you be interested in writing some more posts like it?


Yes, I'd be glad to write more posts like it. I enjoy helping out when I can, and Organic chemistry was among my most favorite premedical subjects. :)

Wow, thanks I really appreciate it. It was so insightful... and thanks to QofQuimica as well.

Intermolecular forces impact stretching frequencies. Alcohol molecules tend to hydrogen bond with each other, which impacts the O-H stretch. Hydrogen bonding is kind of like adding an extra "mass" onto the hydroxy group "spring", so a hydrogen-bonded O-H stretches at a lower frequency than a non-hydrogen bonded (free) molecule. The difference is significant. The attraction of the neighboring oxygen "pulls" on the hydrogen of the adjacent hydroxy group electrostatically, which lengthens the O-H bond and lowers the frequency of vibration. Capeesh?

Here's the kicker: not all molecules are hydrogen bonded, and/or hydrogen-bonded at the same distance and orientation (the strength of hydrogen-bonds are sensitive to these factors), at any given point in time, so the O-H stretch absorption band for the entire solution broadens as a kind of "blurred" average of all the different O-H stretches of the alcohol molecules in their various states.

For alcohols, the strength of the hydrogen bond is roughly about 5 kcal/mol; therefore, you would probably get fairly rapid association, dissociation and re-orientation of the hydrogen bond at room temperature in a particular bonding network. The IR machine usually isn't able to resolve all these different temporal vibrational states (the changes in hydrogen bonding between molecules occur in picoseconds), so what you get when you do IR on this hydrogen-bonding phenomenon is something akin to exaggerated, overlapped photographic blurring at low shutter speed. While this occurs to some degree normally, hydrogen bonding adds a larger range of vibrational states and frequencies, resulting in significant broadening of the IR stretching band. Cool?

What things might interfere with hydrogen bonding, you ask? Well, how about steric factors? If an alcohol molecule, or group, is sterically hindered, it wouldn't be able to hydrogen bond as effectively or as favorably. So what would that do to the O-H stretch absorption band for the solution? It would tend to narrow it, right (i.e., make it sharper)?

What other factors could influence hydrogen bonding? How about concentration? If you dilute an alcohol solution with a solvent (non-polar), you would increase the mean distance between the alcohol molecules, consequently decreasing the prevalence of hydrogen bonding in the solution at any given point in time, I would think. Alcohol molecules need to be in relatively close proximity to each other in order to hydrogen bond, no? With increasing dilution, the solution becomes less and less dense with alcohol. Thus, the absorption band of the O-H stretch would tend to narrow as the solution becomes more dilute.

I'm no expert, but after I read your questions, I started to think about the answers because I was curious about them myself. Take them with a grain of salt... But I think this should help get you started on answering the questions above.

Q, did I miss anything?

So I am curious if esters always have to have a form "O=C-O-C"

For example,
O
||
a) CH3CH2 C CH2CH3

O
|
b) CH3CH2O S OCH2CH3
|
O

c)
O
|
CH3CH2 S CH2CH3
|
O


d)
OCH2CH3
|
CH3CH2O B OCH2CH3

e)
CH2CH3
|
CH3CH2 P CH2CH3
|
O
Will any of the choices listed above esters despite not having typical "O=C-O-C"?

So will O=C-C or any forms other than O=C-O-C be used for esters?

So I am curious if esters always have to have a form "O=C-O-C" ?


Technically, no. The form you list above is a carboxylate ester, the form you see most often in organic chemistry. In general, in introductory organic chemistry, esters, as a class, are oxy-acids where the proton of the -OH group of the oxy-acid is replaced by an alkyl (or organic group), usually via reaction of the oxy-acid with an alcohol. The oxy-acid in the ester doesn't have to be organic compound (for example, phosphoric or sulfuric acids).

There are also thio- (sulfur) analogs... Sulfur is similar to oxygen, so you can easily imagine a thio-acid like a thiocarboxylic acid (O=C-SH) forming a thioester (O=C-S-R).

i learned in class last week that primary OH from R-OH cannot be released in a reaction with H+ (strong acid) to have R- because primary carbocation is unlikely (unstable). So my questions are:
1. Is it correct?
2. How about R-Br in a reaction with H+? Will Br be released? If not, will using Ag2+ or K+ make a difference?

Thank you everyone! Have a nice day! :)

So are you saying that all the choices I listed above (a - e) are classified as esters?

Technically, no. The form you list above is a carboxylate ester, the form you see most often in organic chemistry. In general, in introductory organic chemistry, esters, as a class, are oxy-acids where the proton of the -OH group of the oxy-acid is replaced by an alkyl (or organic group), usually via reaction of the oxy-acid with an alcohol. The oxy-acid in the ester doesn't have to be organic compound (for example, phosphoric or sulfuric acids).

There are also thio- (sulfur) analogs... Sulfur is similar to oxygen, so you can easily imagine a thio-acid like a thiocarboxylic acid (O=C-SH) forming a thioester (O=C-S-R).

So are you saying that all the choices I listed above (a - e) are classified as esters?

No. If I am interpreting your drawings correctly, I don't think choices (a), (b), (c), or (e) are esters as they appear to be drawn in your post. Choice (a) looks like a ketone, choices (b) and (c) don't look like valid structures to me, and choice (e) isn't an oxy-acid derivative. Choice (d) looks like it could be a tri-ester of boric acid, B(OEt)3, triethyl borate; therefore I'd say choice (d) could be classified as an ester.

I'm thinking that your drawings got butchered by the justification of the text editor and also have errors. Perhaps you drew structure (b) incorrectly, and it was actually supposed to be (O=)2S(OEt)2, in which case it would be an diester of sulfuric acid, diethyl sulfate; if that is what you meant, then choice (b) could be classified as an ester, too. I think perhaps you meant to draw structure (c) as (O=)2S(Et)2, in which case it would be called 1,1'-sulfonyldiethane, and it is not an ester. Lastly, perhaps you meant to draw structure (e) as O=P(Et)3, in which case it would be called triethylphosphine oxide, and it is not an ester.

In general, I think you misunderstand what an ester is. As I mentioned before, it is a derivative of an oxy-acid, an acid that contains (-OH). When the oxy-acid is esterified, the proton on the (-OH) group is replaced by an alkyl group. Let's stick with simpler examples, shall we?

An example of an oxy-acid would be sulfuric acid (H2SO4). An example of an ester of sulfuric acid would be:

(EtO)SO3H, ethyl sulfate

Another example of an oxy-acid would be phosphoric acid (H3PO4). An example of an ester of phosphoric acid would be:

(EtO)PO3H2, ethyl phosphate

Yet another example would be boric acid (B(OH)3). An example of an ester of boric acid would be:

(EtO)B(OH)2, ethyl borate

A very common example of an oxy-acid would be a carboxylic acid (RC(=O)OH). An example of an ester of carboxylic acid would be:

(EtO)C(=O)CH3, ethyl acetate

A thioester would look like this:

(EtS)C(=O)CH3, s-ethyl ethanethioate

i learned in class last week that primary OH from R-OH cannot be released in a reaction with H+ (strong acid) to have R- because primary carbocation is unlikely (unstable). So my questions are:
1. Is it correct?
2. How about R-Br in a reaction with H+? Will Br be released? If not, will using Ag2+ or K+ make a difference?

Thank you everyone! Have a nice day! :)

You are on the right track. Primary, and even secondary, alcohols generally do not form carbocation intermediates in reaction with acid, primarily due to thermodynamic considerations (as you stated, a primary carbocation is considerably unstable). They can undergo slow nucleophilic substitution with a mineral acid, however. Remember that -OH is a poor leaving group, but becomes a much better one when it is protonated by an acid, creating an oxonium ion intermediate (R-OH2+). Halide ion (X-), the conjugate base of a mineral acid, is a very decent nucleophile and can then displace water via an SN2 mechanism, resulting in an alkyl halide (R-X) substitution product and water. This is not the best way to synthesize an alkyl halide from a primary alcohol, however. It would be better, and more common, to react a primary (or secondary) alcohol with phosphorus tribromide (PBr3) or thionyl chloride (SOCl2).

Tertiary alcohols do react readily with mineral acids and dilute aqueous acid solutions. Tertiary carbocations are the intermediates in both cases. For the reaction of tertiary alcohols with mineral acids, the mechanism can proceed through an SN1 pathway. Like before, the hydroxyl group is a poor leaving group, but after protonation, it becomes a much better one. Following the SN1 pathway, water leaves the oxonium ion (loss of water), resulting in a relatively stable tertiary carbocation. The halide nucleophile then attacks the carbocation, resulting in a tertiary alkyl halide. In dilute acid solution like dilute aqueous sulfuric acid, however, acid-catalyzed elimination is the favored pathway, rather than substitution because the conjugate base of sulfuric acid is a pretty weak nucleophile. The elimination reaction occurs through the E1 mechanism.

In response to your second question, primary alkyl halides (R-X) are generally unreactive to acids (as you have written it above). The type of reactions that alkyl halides, as a class, tend to undergo, as studied in general organic chemistry, are nucleophilic substitution and elimination reactions with an appropriate nucleophile; exactly which type of reaction, substitution or elimination, occurs, and by what kind of mechanism (SN1, SN2, E1, or E2) it undergoes, depends on steric factors, reaction conditions, leaving group, and nucleophile present. Primary alkyl halides do not generally form carbocations by the same reasoning that primary alcohols do not. They can react with strong nucleophiles like cyanide CN-, alkoxide (RO-), or hydroxide ions, etc., through an SN2 mechanism to form the resultant substitution product; this type of reaction is sensitive to steric factors (for example, a beta-substituted primary alkyl halides are sterically hindered from back-side attack).

I do not think adding silver ion or potassium ion will make a significant impact on the reactivity of primary alkyl halides with acid. Maybe you are thinking of some other kind of reaction? Adding a small amount of silver ion to a tertiary alkyl halide, for example, undergoing an SN1 reaction, or solvolysis-type reaction, with an appropriate nucleophile, can catalyze the reaction. Perhaps this is what you are thinking about? The silver (or mercury) ion can help facilitate the dissociation of the halide leaving group in an SN1-type reaction. Obviously, when, say, silver nitrate is used to catalyze the reaction, the SN1 product is favored; it seems like primary or secondary alkyl halide may undergo solvolysis in the presence of silver nitrate given the potential for rearrangement to form a more stable carbocation intermediate. I couldn't tell you if it is appreciable or practical, however. I don't know much about this at all. Q, any comments about this? In general, primary alkyl halides do not typically undergo SN1 reactions because primary carbocations are particularly unstable intermediates, as discussed above.

These are sweeping generalization, however, just to give you an idea of what is going on here. You will need to study the exact conditions, mechanism, and reactions for more specific and accurate results.

SM, you're awesome. :thumbup: Thanks for your help. When I get a chance, I will update the organic explanations thread.

Ok, I am confused with this really simple concept. In lab, we are going to be asked to make an equimolar solution of two reagents. How do I figure out how much of each to use. I get the whole idea of coming up with a molar ratio, however, once I get that molar ration, what exactly does it tell me? I need to make a 50mL solution, and my molar ratio is close to .95mol x/1mol y. Thanks!

Ok, I am confused with this really simple concept. In lab, we are going to be asked to make an equimolar solution of two reagents. How do I figure out how much of each to use. I get the whole idea of coming up with a molar ratio, however, once I get that molar ration, what exactly does it tell me? I need to make a 50mL solution, and my molar ratio is close to .95mol x/1mol y. Thanks!

I'm not sure I understand your question, but I'll give it a shot; it's been awhile since I completed any labs. I forgot a lot about laboratory technique. If I don't get it right, perhaps someone will step in with the correct answer. I'm confused by whether you want two separate solutions that are equimolar in concentration to use for a reaction sequence (or titration), or one solution with equimolar amounts of A and B. Maybe it won't matter for your lab.

In essence, when you are saying one solution is equimolar to another in concentration, you are saying that they have the same molarity (number of moles of given solute per total liters of solution). Does that make sense? But the question is, how are these two reagents related in your reaction scheme, or sequence? More information is required to answer your question accurately, procedure-wise. However, I'll approach it generally.

The first step in your process should be calculating the respective molecular weights of your two reagents, A and B (let's call them mA and mB), you can do that by using a periodic table and the molecular fomulae of the two compounds. You will use the molecular weights and the number of moles (nA and nB) required to find out how many grams (gA and gB) of each you will need to mix into solution. Then, you will look in your lab manual for the number of moles of the limiting reagent required to generate the amount of desired product, or the molarity that is indicated (they should give you one of the two, or some way to determine them). You need to have some reference point for the number of moles required, however it is that it is indicated.

Let's assume that the manual indicates you need x moles of A; the mass of A required to make the solution is then determined by mass of A = (mA * nA) = X(mA). So by definition of equimolar (the ratio of nA to nB = 1), you will need also need x moles of B; the mass of B required to make the solution is then determined by mass of B = (mB * nB) = X(mB). Now you have the mass of each, A and B, required to make two separate solutions of equimolar concentration, or make one solution with equimolar amounts of each.

To make two separate solutions of equimolar concentration, just measure the masses indicated above and place in two separate volumetric flasks (or other appropriate container). Fill both with dH2O (or whatever solvent is indicated) to the same final volume. Now you have two solutions that are equimolar in concentration. Depending on your reaction, you will use/mix together whatever volume of each you require into the reaction mixture to generate the amount of product or intermediate you want.

If your lab experiment simply wants you to use equimolar amounts of A and B for the reaction sequence, just measure in the masses indicated in the fourth paragraph into a single flask and mix with the desired, or appropriate volume of solvent.

Hope this helps.

Wow, SM you are really smart...
anyways I'm learning about Grignard reagent and am aware that it reacts with all protons more acidic than alkenes or alkanes and any carbonyl becomes alcohol. But I am not sure what would happen to the following problems:

1)
R-Mg-X -----------------------------------> ???????
````````(i)``O
````````````||
````````H3C-C-C-OEt
(ii) H+

My guess is that it will be something like this:

````OH
````|
H3C-C-OCH2CH4 + OH-MgX
````|
````R


2) R-Mg-X ----------------------------------> ???
i) C2H2O (cycloethane with an oxygen in the ring and one hydrogen coming out from each carbon)
ii) H+

My guess:
Something will change with the O in the cyclomethane due to its two pairs of lone bonds but not sure....

3) R-Mg-X ------------------> ??
i) CO2
ii) H+

My guess:
``````R
``````|
```OH-C-OH
``````|
``````R
Is it reasonable to think that there will be two same reactions for this case? (That is, =O(carbonyl group) changing to -OH )
Please let me know if each one is right.
As always, thanks SM!:)

Wow, SM you are really smart...
anyways I'm learning about Grignard reagent and am aware that it reacts with all protons more acidic than alkenes or alkanes and any carbonyl becomes alcohol. But I am not sure what would happen to the following problems:

1)
R-Mg-X -----------------------------------> ???????
````````(i)``O
````````````||
````````H3C-C-C-OEt
(ii) H+

My guess is that it will be something like this:

````OH
````|
H3C-C-OCH2CH4 + OH-MgX
````|
````R


2) R-Mg-X ----------------------------------> ???
i) C2H2O (cycloethane with an oxygen in the ring and one hydrogen coming out from each carbon)
ii) H+

My guess:
Something will change with the O in the cyclomethane due to its two pairs of lone bonds but not sure....

3) R-Mg-X ------------------> ??
i) CO2
ii) H+

My guess:
``````R
``````|
```OH-C-OH
``````|
``````R
Is it reasonable to think that there will be two same reactions for this case? (That is, =O(carbonyl group) changing to -OH )
Please let me know if each one is right.
As always, thanks SM!:)
1) You're adding a Grignard to an ester, so you would get a double addition there. What you've shown is the first step. In other words, once the first R group has added, the O- formed from the carbonyl will kick the ethoxy group out, reforming the carbonyl in the process. That carbonyl can then be attacked by a second R- group.

2) That's not a cycloethane; that's an epoxide. It's a type of cyclic ether, and because of its small size, its bonds are constrained, meaning it is very reactive to attacks by nucleophiles like Grignards. When you react it with a Grignard, it will open up the ring. But it doesn't attack on oxygen; remember, since the Nu has its own lone pair, it doesn't want another lone pair from the oxygen. It wants to attack somewhere that's electron deficient. That means it will attack one of the carbons.

3) Nope, not quite. Once the first R group adds to CO2, you've got a deprotonated carboxylic acid on your hands. You won't be able to add another equivalent of R- to it because the carboxylic acid that you've formed is already carrying a negative charge.

It looks to me like you need to work on your understanding of nucleophiles and electrophiles. Keep in mind whenever you do a reaction that you always want to have a nucleophile (something that is electron rich) react with an electrophile (something that is electron poor). Hope this helps, and best of luck to you. :)

I have some questions about synthesis.

1) 1-butyne -> (+/-) 3,4-dipentanol

2) Trans-3-hexene -> cis-3-hexene

3) CH3CH2OH -> 2-butanol

For each synthesis, what should be the reagents (multiple steps probably)? Any help would be greatly appreciated!

I have some questions about synthesis.

1) 1-butyne -> (+/-) 3,4-dipentanol

2) Trans-3-hexene -> cis-3-hexene

3) CH3CH2OH -> 2-butanol

For each synthesis, what should be the reagents (multiple steps probably)? Any help would be greatly appreciated!
Come on, pezzang, you've been a member long enough to know that we aren't going to do your HW for you. You need to at least make an honest effort. Whenever you're doing synthesis, start with the product, and try to work your way backward to the starting material. (This is called retrosynthesis.) So for example, you have 3,4-dipentanol. What can you make a diol from? Think of a few ideas, and keep going until you get back to your butyne. You're almost certainly not going to have to do more than a few steps at a sophomore organic level.

If t-butyl bromide is added to Mg in anh. ether, what would be the impurity formed?

If t-butyl bromide is added to Mg in anh. ether, what would be the impurity formed?

The principle side-reaction in the prepartion of alkyl magnesium halides is called "Wurtz Coupling", in which you get R-R as the side-product, R- being the original alkyl group of the alkyl halide used in the preparation. You can think of this reaction sort of like, "R-Mg-X + R-X ---> R-R + Mg(X)2".


Here is a good review of Grignard reactions; it's where I got my answer (pg. 7):

http://www.joe-harrity.staff.shef.ac.uk/meetings/GrignardReagentsReviewMeeting.ppt

Why do higher temperatures favor E1 over SN1? Can somebody please explain that in terms of reaction energy diagrams?

Thanks

Why do higher temperatures favor E1 over SN1? Can somebody please explain that in terms of reaction energy diagrams?

Thanks

Here's my crude guess:

Let's look at a simple solvolysis reaction.

First, note that per given reaction where both mechanisms are possible, both E1 and SN1 reactions share a common carbocation intermediate, which is generated through of the dissociation of the leaving group. This rate determining first step (highest activation energy) is generally referred to as the "ionization" step.

R-LG <---> R+ + LG- (1)

Second, note that the remaining steps for the substitution reaction (addition) are reversible and have lower activation energies than the first step.

R+ + OH2 <---> R(OH2)+ (2)

R(OH2)+ + OH2 <----> ROH + H3O+ (3)

In elimination, the solvent or base extracts the beta-hydrogen from the carbocation in (1), resulting in the formation of a stable double bond. I think that this step probably has a higher activation energy than the addition step of the substitution reaction. The resulting alkene product is generally more stable (i.e., lower energy), however, than the substitution product, and the remaining steps are generally not as readily reversible because of the higher activation energy required to break the double bond (614 kj/mol). The energy required to break the C-O bond is 360 kj/mol, in comparison.

If you increase the temperature, you place the reaction into thermodynamic control; the more stable product of a reaction is favored. Increasing temperature allows the products of a reaction to better overcome the activation energies for the reverse steps. The substitution product will reverse more readily to the common carbocation intermediate than the more stable elimination product (if it occurs). Futhermore, the increased temperature allows the second step of the elimination reaction (breaking of the C-H bond) to occur more readily than at room temperature to generate the more stable alkene. Thus, increasing temperature favors the elimination reaction in two ways.

Does E2 happen faster at a bromine attached to a primary carbon or at a bromine attached to a tertiary carbon?

Also, when doing R,S notation and the smallest priority group attached to the chiral center is not sticking in or out, am i supposed to treat it like it is sticking in?

Does E2 happen faster at a bromine attached to a primary carbon or at a bromine attached to a tertiary carbon?

Also, when doing R,S notation and the smallest priority group attached to the chiral center is not sticking in or out, am i supposed to treat it like it is sticking in?
First question: E2 doesn't care too much about sterics of the substrate, as long as the molecules can get into an antiperiplanar geometry. That being said, tertiary compounds often react faster just because of the relief of steric strain when you go from the hindered sp3 carbon to the planar sp2 carbon in the double bond.

Second question: No. You will either have to rotate the molecule in your mind, or else swap two of the substituents. If you do the latter, don't forget to switch the designation, because every time you make a swap, you change the configuration.

My question is: Without learning the structures by heart, how could i
distinguish the structures of the following common names?
t-butyl,
neopentyl,
isopropyl,
isobutyl,
sec-butyl

Thanks.

Thanks Q.

I also have one other question:

What is the product when you treat 1,1-dibromopentane with t-butoxide?

I know t-butoxide is the base for E2 reactions, so i said the product is 1-pentene, but I don't know if that's right because that C1 is still missing a hyrdogen.

My question is: Without learning the structures by heart, how could i
distinguish the structures of the following common names?
t-butyl,
neopentyl,
isopropyl,
isobutyl,
sec-butyl

Thanks.

memorize what the prefixes mean. neo, t (aka tert), iso, sec all indicate specific structures. every organic chemistry student should have memorized all the names for n-hydrocarbons for 1-10 carbons long. dont memorize all them, but just break them down by prefix and suffix.

Thanks Q.

I also have one other question:

What is the product when you treat 1,1-dibromopentane with t-butoxide?

I know t-butoxide is the base for E2 reactions, so i said the product is 1-pentene, but I don't know if that's right because that C1 is still missing a hyrdogen.

hint: 1-pentene is wrong, you need to change an e somewhere in the name to a y.

My question is: Without learning the structures by heart, how could i
distinguish the structures of the following common names?
t-butyl,
neopentyl,
isopropyl,
isobutyl,
sec-butyl

Thanks.
Yeah, I'd agree with nova that the best thing to do is to learn what the prefixes and suffixes all mean. If you know what tertiary, iso, and sec are, and you also know the hydrocarbon chains, you can figure out the structures based on that.

Thanks Q.

I also have one other question:

What is the product when you treat 1,1-dibromopentane with t-butoxide?

I know t-butoxide is the base for E2 reactions, so i said the product is 1-pentene, but I don't know if that's right because that C1 is still missing a hyrdogen.
You're halfway there. Go through the reaction step by step. The first thing that happens is that you lose one HBr due to E2, right? That gives you a pentene as you said. But you didn't start with just one Br; you started with two. So what pentene intermediate do you have left after losing that HBr, and what happens to it? That's the hint that nova is trying to give you. ;)

P.S. I'm not sure that t-butoxide is a strong enough base to do what you're trying to do here; the reaction might actually stop after one step.

I was a little confused about electrophoresis. In electrophoresis is the anode positive or negative? (I thought it was negative because in a galvanic cell it is negative, but I'm guessing electrophoresis is like a electrolytic cell so the anode is positive?)

I was a little confused about electrophoresis. In electrophoresis is the anode positive or negative? (I thought it was negative because in a galvanic cell it is negative, but I'm guessing electrophoresis is like a electrolytic cell so the anode is positive?)
You're guessing right. :) Don't forget, you have to put energy into an electrophoresis apparatus to get it to work. So, that means it's an electrolytic cell, not a galvanic cell.

You're guessing right. :) Don't forget, you have to put energy into an electrophoresis apparatus to get it to work. So, that means it's an electrolytic cell, not a galvanic cell.

thanks Q. wishing i was a chem master like you...:)

Is there any free software online that will let me draw 3D molecules and manipulate them (rotate it, etc.). A plus would be if I could draw them in Fischer diagrams. I'm really bad at stereoisomerism and this would really help me to visualize it.

Thanks.

Is there any free software online that will let me draw 3D molecules and manipulate them (rotate it, etc.). A plus would be if I could draw them in Fischer diagrams. I'm really bad at stereoisomerism and this would really help me to visualize it.

Thanks.

Yes. Try, ACD/ChemSketch. It's freeware.

www.acdlabs.com

Good luck!

Thanks a bunch. I actually figured out how to manipulate the Fischer diagrams on paper in regards to chirality, but this will help me for other formats.

I've got another general question. Where can I find a list of functional groups and their relative inductive effects. My O-Chem book doesn't have a concise list and I've lost my intuition on inductive effects.

My organic manual says that an alcohol behaves as an electrophile if a C-O bond is broken and behaves as a nucleophile if the O-H bond is broken. (the molecule shown is in the form R-O-H2 with the oxygen having a +1 charge)

I do not understand this at all. I don't understand the idea of an alcohol being electrophilic or nucleophilic. The oxygen gains the electrons regardless of if C-O or O-H is broken. I don't understand how it is an electrophile if the C-O bond is broken and a nucleophile when the O-H bond is broken.

Any help on this would be much appreciated. Thanks.

My organic manual says that an alcohol behaves as an electrophile if a C-O bond is broken and behaves as a nucleophile if the O-H bond is broken. (the molecule shown is in the form R-O-H2 with the oxygen having a +1 charge)

I do not understand this at all. I don't understand the idea of an alcohol being electrophilic or nucleophilic. The oxygen gains the electrons regardless of if C-O or O-H is broken. I don't understand how it is an electrophile if the C-O bond is broken and a nucleophile when the O-H bond is broken.

Any help on this would be much appreciated. Thanks.

It's just a general guideline...

ROH2+ -> R+ & H2O, R+ is electrophile (carbon is electrophilic)
ROH2+ -> ROH & H+, ROH is nucleophile (oxygen is nucleophilic)

This helps you see what is happening in a reaction. For example, if an alcohol is protonated, and then the O attacks another molecule, and solution is acidified (H+ ions), then you know nucleophilic attack has occurred. If, on the other hand, an alcohol is protonated and then water is released, you know carbon has acted as an electrophile... e.g.

ROH2+ & Br- -> RBr + H2O

The reaction of Compound IV with HBr primarily differs from that of Compound II in that it proceeds through which of the following intermediates?
A secondary carbocation
A vinylic anion
A vinylic cation
A tertiary carbocation
This is a kaplan question that I am getting a little confused on
I know that B and D are definitely wrong and C is right, but I dont know why A is wrong, because I thought that when we add HBr to an alkene it adds the H to the less substituted and then the carbocation that is formed is secondary and so choice A would be true too

please help! thanks :)

I was wondering if you could go over what makes a certain chair confirmation stable...? Thanks!

I was wondering if you could go over what makes a certain chair confirmation stable...? Thanks!
The mantra of organic chemistry and stereochemistry:
Steric hindrance.

If you have a cyclohexane ring substituted at the C1 & C4 in the cis conformation then there will be one equatorial and one axial in either chair conformation so it will interconvert between the two chairs...
but if it is in the trans conformation then they will either be both axial or both equatorial... in THAT situation the conformation with both subsitutents in the equatorial postition is energetically favored because if they were both in the axial position there would be greater steric interactions.

Maybe it would help if you could get your hands on some models and build the ring. Then you would be able to see the idea behind what I'm describing.

Hope that answered your question.

The reaction of Compound IV with HBr primarily differs from that of Compound II in that it proceeds through which of the following intermediates?
A secondary carbocation
A vinylic anion
A vinylic cation
A tertiary carbocation
This is a kaplan question that I am getting a little confused on
I know that B and D are definitely wrong and C is right, but I dont know why A is wrong, because I thought that when we add HBr to an alkene it adds the H to the less substituted and then the carbocation that is formed is secondary and so choice A would be true too

please help! thanks :)

The transition structure is cyclic with Br binding two carbons. The next nucleophile attacks "below" (anti) with respect to the first Br and adds with regular Markovnikov regioselectivity.

If you drew the mechanism through a regular SN1 with the positive charge residing on one carbon to form a carbocation, you'll still get the same answer. I have no idea why C is right since I do not have the compounds in front of me.

Why do the O-H bonds of alcohols have higher absorption frequency (measured in wavenumbers) than the O-H bonds of caboxylic acids in IR Spectroscopy?

Why do the O-H bonds of alcohols have higher absorption frequency (measured in wavenumbers) than the O-H bonds of caboxylic acids in IR Spectroscopy?

This post should lead you in the right general direction to answer the question:

http://forums.studentdoctor.net/showpost.php?p=4253801

Also, in addition to the factors introduced in the aforementioned post, consider that the hydrogen in the hydroxyl group of a carboxylic acid is considerabley more acidic than the hydrogen of the hydroxyl group of an alcohol. Part of the reason for this increase in acidity is due to resonance stabilization and delocalization of the negative charge on the conjugate base of the carboxylic acid; also, the presence of two electron withdrawing groups (the two oxygens) pull electron density away from the hydrogen atom. In comparison, the conjugate base of the alcohol has a fairly localized negative charge. In general, an acid is stronger, has a higher Ka value(tendency to lose a hydrogen ion), if the conjugate base of that acid can be stabilized by resonance. Thus, it can be said that the hydrogen atom on the carboxylic acid is more "weakly" bonded to the oxygen than the hydrogen atom on the hydroxyl group of the alcohol. I bet this effects the oscillation tendency of the -OH stretch on the carboxylic acid.

Yet another factor to consider is that carboxylic acids can dimerize. You can have -OH--O=C-, times two, for a pair of molecules, right? This phenomenon, I would guess, further effects the oscillation.

These are just my best guesses.

Thank you spicedmanna. That really helps!

O-H alcohol is about 3400 cm-1 (given)
C-H alkene is about 3000 cm-1 (given)
N-H amine (need to determine)
Relative bond polarities determine relative absorption intensities.
The greater the change in dipole moment, the more intense the absorption.
Changes in dipole moments depend on respective electronegativity values.
But since we are interested in absorption frequencies, not absorption
intensities, relative EN values of C,N,O are not the answer.
Stronger bonds and lighter atoms give rise to higher absorption frequencies.
N is lighter than O but heavier than C. so based on atomic mass alone, N-H
should absorb at a higher frequency than O-H and at a lower frequency than C-H.
N is smaller than C but larger than O. So N-H is shorter than C-H, but longer
than O-H. shorter bonds are stronger than longer bonds. So based on bond
strength alone, N-H should absorb at a higher frequency than C-H and at a lower
frequency than O-H.
So looks like we have two competing factors here. How
do we know that bond strength effect outcompetes the atomic mass effect here?
W