Organic Chemistry Question Thread

[QofQuimica]

All users may post questions about MCAT, DAT, OAT, or PCAT organic chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what organic topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

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

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

********

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

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

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

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

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[wannabedocta]

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

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

 

[iA-MD2013]

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

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

 

[Kaustikos]

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

Actually...

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

 

[iA-MD2013]

Actually...

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

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

 

[Kaustikos]

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

Yeah, I was just tossing that up for fun.

 

[iA-MD2013]

Yeah, I was just tossing that up for fun.

haha i love doing that too

 

[wes431]

CH3OOCH2CH2CH3

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

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

 

[tncekm]

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

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

 

[RSAgator]

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

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

final CH3 is a triplet not a quartet.

 

[tncekm]

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

 

[olygt]

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

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

 

[olygt]

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

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

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

 

[RSAgator]

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

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

 

[olygt]

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

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

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

 

[iA-MD2013]

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

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

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

Is the answer B...?

 

[bluemonkey]

Is the answer B...?

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

 

[talkalot24]

Ahh, please help!

Hydroboration:

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

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

 

[tncekm]

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

 

[Kaustikos]

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

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

 

[talkalot24]

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

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

 

[Kaustikos]

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

Believe me, it took me at least a day to come to terms with this annoying reaction (the one you speak of). Considering almost every other reaction is explained except boration, it pissed me off considerably.

 

[olygt]

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

 

[Kaustikos]

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

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

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

 

[olygt]

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

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

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

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

 

[bluemonkey]

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

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

 

[olygt]

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

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

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

 

[olygt]

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

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

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

 

[sleepy425]

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

 

[olygt]

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

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

 

[bluemonkey]

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

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

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

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

 

[Foghorn]

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

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

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

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

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

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

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

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

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

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

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

 

[bluemonkey]

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

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

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

 

[olygt]

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

 

[olygt]

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


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

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

 

[Kaustikos]

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


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

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

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

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

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

 

[TJames]

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


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

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

Answer is A.

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

 

[TJames]

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

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

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

The answer is B.

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

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

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

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

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

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

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

 

[BerkReviewTeach]

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

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

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

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

 

[BerkReviewTeach]

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

Berkeley Review Organic Chemistry Book I Example 2.11

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

 

[tawaqul]

Hi,

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

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

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

Thanks!! A prompt response is appreciated!!

 

[DrMattOglesby]

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

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

 

[unsung]

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

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

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

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

 

[tncekm]

Saponification

Okay, I just want to clarify something.

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

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

Thanks!

 

[Kaustikos]

Saponification

Okay, I just want to clarify something.

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

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

Thanks!

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

 

[sleepy425]

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

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

so tncekm, you're exactly right.

 

[Kaustikos]

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

so tncekm, you're exactly right.

My mistake, I had the completely turned around.

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

 

[sleepy425]

My mistake, I had the completely turned around.

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

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

 

[nateriver]

can i just ask how do you convert a dot and wedge to a fischer projection and vice versa, sorry if this isn't the right format of question but i haven't been able to adequately figure it out by watching khan and my prof is so mean so yeahh

 

[ELTURC0]

Q: Which of the following compounds contain the fewest tertiary carbons?
A-(CH3)3 C (CH2)2 CH3
B-4-isobutylheptane
C- 2-methyl hexane
D-1-isopropyl cyclohexane

Based on the answer key, the answer is A. And it says "Choice A contains no tertiary carbons".
How could this be? Those 3 methyl groups in choice A are all attached to the same carbon, no? I am confused, please help!
Thanks.

 

[Cawolf]

Tertiary implies 3 carbons and a hydrogen bound to the carbon in question.

I think you are confusing tertiary and quaternary - which is what the central carbon in the t-butyl is.