When (R)-2-chlorobutane is chlorinated, we obtain some 2,3-dichlorobutane. It consists of 71% meso isomer and 29% racemic isomers. Explain why the mixture need not be 50:50 meso and (2R,3R)-2,3-dichlorobutane
I have been doing quite a lot of brainstorm after I saw your question. The mixture is supposed to be a 71% meso-product and RR and SS products, 14,5% each.
Please view the attachment for easy understanding.
We may explain it by relatively large volume of chlorine atom. Chlorine molecule tends to approach to the alfa carbon atom from the other side of methyl group because of the steric hindrance; this is the cause of formation of 2,3-disubstituted product. Two 14,5% mixtures of racemic products are formed also. The yield of meso product is quite higher than the racemic ones, and this is because symmetrical diastereomers are less likely to occur than the asymmetric ones; thus "symmetric" racemic mixture is expected to be formed less than the "asymmetric" meso-product.
My approach is not the only explanation; I will read other comments also.
If the conditions for the chlorination are Cl2 and light, then it's possible that the chlorine atom in the 2-chlorobutane is exchanging. The light can cleave the relatively weak Cl-C bond forming a secondary radical, which has no stereochemical information. That radical can then combine with a chlorine radical (or chlorine molecule) to for racemic 2-chlorobutane. Then the second reaction can take place at the 3 position of the butane to give the observed racemic product.
It is possible, however, that this chlorine doesn't exchange. If I were really going to run this reaction, I would expect it to however.
The reason the meso is more prevalent than the racemic is because the addition of the chlorine radical to the 3 carbon is diastereoselective (it depends on the stereochemistry of the 2 carbon, to an extent). I think that's essentially what chem_tr is saying too.