Will the following reaction give a racemic mixture or not?

enter image description here

According to me, it should give. But my teacher disagrees, but I cannot understand the reasons he gave. Shouldn't there be equal approaches from above and below the molecules?

I am confused when he says that the adjacent 3° Carbon can change its configuration too.


Original question:-

enter image description here

Solution provided:-

enter image description here

  • $\begingroup$ The answer hinges on whether the starting material is enantiomerically pure or not. As drawn, I would assume it is a racemate. Can you confirm? $\endgroup$ – jerepierre Jan 4 '17 at 17:29
  • $\begingroup$ @jerepierre I assumed a racemic mixture. The answer given in the book is 'No racemic mixture will be formed'. I don't know what to assume as I am confused with my assumptions. $\endgroup$ – Reeshabh Ranjan Jan 4 '17 at 17:32
  • $\begingroup$ Is the methyl group bolded (or dashed)? If not, then this is a poorly designed question, given the book's answer. $\endgroup$ – jerepierre Jan 4 '17 at 17:43
  • $\begingroup$ @jerepierre I have added the original question as well as the solution provided. $\endgroup$ – Reeshabh Ranjan Jan 4 '17 at 17:50

IUPAC advises against using the term racemic mixture because of its ambiguity.

racemic mixture (obsolete) The term is confusing since it has been used as a synonym for both racemate and racemic conglomerate. Usage strongly discouraged.

The term that applies in this case is racemate.

racemate An equimolar mixture of a pair of enantiomers. It does not exhibit optical activity.

There are three qualitatively different initial conditions to consider.

1. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is not $\ce{OH-}$

Both enantiomers behave similarly, each yielding a stereocenter at the carboxylic carbon.

1st initial condition. (Chemsketch)

Verdict: not optically active but contains two pairs of equimolar enantiomers: $\mathrm{a), d)}$ and $\mathrm{b), c)}$. Even three if you count the reagents. Is a racemate if one relaxes the single pair criterium.

2. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is $\ce{OH-}$

Verdict: not optically active, equimolarity is ensured, a pair of enantiomers. Is a racemate.

3. Either (3S)-methylpentan-2-one or (3R)-methylpentan-2-one, nucleophile is / is not $\ce{OH-}$

Verdict: optically active, does not contain enantiomers. Not a racemate.

| improve this answer | |
  • $\begingroup$ In 1, shouldn't that be two pairs of diastereomers instead? And what has that to do with optical activity? In 3, this is the step where my teacher says that since a pair of diastereomers are formed, it is not optically active and we should not talk of racemisation, and I am confused. $\endgroup$ – Reeshabh Ranjan Jan 2 '17 at 19:16
  • $\begingroup$ @ReeshabhRanjan 1. Enantiomers are a type of diastereomer. In this case, if you compare $\mathrm{a), d)}$, you see that both stereocenters are in fact reverses of each other. Same applies for $\mathrm{b), c)}.$ Thus, these in effect "cancel" polarisation of light. 3. If the nucleophile is $\ce{OH-}$, then indeed you get a mixture of diasteromers that are not enantiomers. It should still be more likely to be optically active than not. But since they are already not enantiomers, optical activity is extraneous. $\endgroup$ – Linear Christmas Jan 2 '17 at 19:36
  • $\begingroup$ how are enantiomers a type of diastereomers? We say diastereomers are compounds which are neither the mirror image, nor superimposable on each other, and enantiomers are the compounds which ARE mirror image but NOT superimposable. $\endgroup$ – Reeshabh Ranjan Jan 3 '17 at 9:07
  • $\begingroup$ @ReeshabhRanjan Depends on your definition of diastereomer. If you define it as stereoisomers which are not enantiomers, then you are correct. You can also define a diastereomer as a stereoisomer that differs from its isomer in at least one stereocenter. And so if all sterocenters are different, we have enantiomer as a special case. But still: $\mathrm{b), c);\ \ \ a), d)}$ are enantiomers; similarly, $\mathrm{a), b);\ \ \ c), d)}$ are diasteromers in either definition. $\endgroup$ – Linear Christmas Jan 3 '17 at 9:53

When you add to the planar carbonyl group, attack from both sides of the plane is equally probable. If there are no other carbons bearing four different groups in the molecule (chiral centres), a pair of enantiomers will be formed in equal amounts (racemate). Keep in mind that diastereomers and enantiomers apply to relationships between pairs of molecules. If you create two pairs of enantiomers, then the molecules in each pair will be diastereomers of the other pair. So you also create two pairs of diastereomers.

Now, there is already a chiral centre in your molecule - but its configuration is not given. Without knowing that, the question cannot be answered properly. The molecules could be either all R or all S; assuming it is R addition of the nucleophile will give (R,S) and (R,R) in equal amounts. These are diastereomers because the mirror image of (R,S) is (S,R) and the mirror image of (R,R) is (S,S). The same applies if all your molecules would have an S chiral centre before addition. The molecules that are formed are not mirror images.

If your molecule would be present as 50% R and 50% S beforehand (racemate), addition will create equal amounts of (R,R), (R,S), (S,R) and (S,S). Now, you have two pairs of enantiomers: (R,R)/(S,S) and (R,S)/(S,R). Every stereoisomer in the mixture has one mirror image (together they are enantiomers) and 2 others that are diastereomers. I'd also like to add that enantiomers are definitely not diastereomers. In fact, that's how IUPAC defines diastereomers: they are stereoisomers that are not enantiomers. This mixture is not a racemate because there are four stereoisomers. According to IUPAC, a racemate is a pair of enantiomers present in equal amounts.

If the nucleophile is hydroxide, then addition will not create an additional chiral centre. So then it would depend on whether you had a single stereoisomer to start with or a mixture, whatever stereochemistry you started with will be retained since no bonds to the chiral centre are formed or broken.

| improve this answer | |
  • $\begingroup$ Oh, the racemic mixture is only defined for a pair of enantiomers only? Even if n (>1) pairs of enantiomers present making the net optical activity zero? $\endgroup$ – Reeshabh Ranjan Jan 3 '17 at 5:46

In this reactant one stereogenic carbon is already present and its configuration will not change during reaction. When Nu- attack on carbonyl carbon two products are formed due to top side and bottom attack but these two products are Diastereomers.

| improve this answer | |
  • $\begingroup$ And what if a racemic mixture of the substrate had been given? $\endgroup$ – Reeshabh Ranjan Jan 2 '17 at 16:54
  • $\begingroup$ Two pairs of diastereomers will be formed $\endgroup$ – narendra kumar Jan 2 '17 at 17:12
  • $\begingroup$ Why in the the other case too will a disteriomeric pair will be formed? $\endgroup$ – Reeshabh Ranjan Jan 2 '17 at 18:04

Out the two sites, one site is adjacent to the Me group So attack to that site is hindered because of Steric Reason. Only one site is available for attack. Therefore, enantiomers are not formed in equal ratio.

| improve this answer | |
  • $\begingroup$ Your answer seems much practical, but what if we had been given an equimolar mixture of all the 4 possibilities? Will it still be NOT a racemic mixture? $\endgroup$ – Reeshabh Ranjan Jan 4 '17 at 12:44
  • $\begingroup$ You are saying that due to steric hindrance, there will not be an equimolar mixture of any pair of enantiomers. I agree. But what if we were already given equimolar mixture of the enantiomeric pairs? $\endgroup$ – Reeshabh Ranjan Jan 4 '17 at 12:47
  • $\begingroup$ Pair of enantiomers. $\endgroup$ – Reeshabh Ranjan Jan 4 '17 at 12:49
  • $\begingroup$ I am saying that I myself add same amount of each enantiomer into a common container. $\endgroup$ – Reeshabh Ranjan Jan 4 '17 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.