# Esterification of racemic acid with optically active alcohol

An excess of racemic acid ($\ce{CH3CHClCOOH}$) is allowed to react with (S)-2-methyl-1-butanol to form the ester, $\ce{CH3CHClCOOCH2CH(CH3)CH2CH3}$, and the reaction mixture is carefully distilled. Three fractions are obtained, each of which is optically active. Draw stereochemical formulas of the compound or compounds making up each fraction.

I know that two fractions obtained will be of the diastereomers (R,S) and (S,S), what is the other fraction?

Source: Morison and Boyd

• What is lost in the ester formation? Perhaps there was no workup step! – user55119 Mar 4 '18 at 4:00
• Did you consider that possibly the leftover excess of the racemic acid $\ce{CH3CHClCOOH}$ can react wit each other forming a cyclic product in which the oxygen from carboxylic group of the first molecule substitutes the chlorine from the second molecule, bringing the close enough so that the oxygen from the carboxylic group from the second molecule substitutes the chlorine from the first one? – TheHeartless Mar 4 '18 at 6:01
• This question is in a chapter before the chapter on substitution – papiya pal Mar 4 '18 at 6:39

It's a pretty basic concept. Two diastereoisomeric esters will be formed. Now diastereoisomers have different physical and chemical properties just like normal isomers. Hence, obviously, their rate of formation will be different. So, different amount of R and S acid will react

Thus, the third fraction will contain unequal amount of the acids which will show optical activity, and the other two fractions are the two diastereoisomers which are also optically active.

I think the other fraction should be the ether ($\ce{CH_3CH(OCH_2C^*HMeCH_2CH_3)COOH}$). This compound can be formed by substitution of the $\ce{Cl}$ of the racemic acid by the $\ce{OH}$ group of the optically active alcohol, which was actually allowed to form an ester.

The acid is present in the solution both in its both S and R forms. So, substitution will generally favour SN2, and thus the ether will have its both R and S configuration only at that substitution centre. But the ether formed will also contain the part coming form (S)-2-methyl-1-butanol, which has a fixed chiral centre (marked with asterisk), which will show optical activity.

So, the ether formed will actually consists of two sterochemical compound, which will be diastereomeric with respect to each other. But when distilled, due to almost similar physical properties, the third fraction obtained will consist of that mixture of ether, which will have a chiral centre and thus be optically active.

• Where did you get the C* from? (in your first sentence) What does it depict? – Gaurang Tandon Mar 15 '18 at 15:01
• That is the chiral Carbon. – Soumik Das Mar 15 '18 at 15:40
• There is another chiral carbon but you didn't label it anything. – Gaurang Tandon Mar 15 '18 at 15:41