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
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.