Question

List-I contains physical or stereomeric relation of the products formed and List-II contains number of relation or compounds.
$\begin{array}{cc}\text{List -I}& \text{List -II}\\ \text{(I) Enantiomeric Pairs}& \left(P\right)20\\ \text{(II) Diastereomeric Pairs}& \left(Q\right)34\\ \text{(III) Optically acitve compounds}& \left(R\right)18\\ \text{(IV) Number of fractions of compounds that can be separated}& \left(S\right)11\\ \text{by fractional distillation}& \\ & \left(T\right)9\\ & \left(U\right)32\end{array}$
Which of the following options has the correct combination considering List-I and List-II

• A. (I),(T) and (III),(R)
• B. (I),(T) and (II),(S)
• C. (IV),(T) and (II),(U)
• D. (I),(S) and (II),(Q)

Answer 1

The correct option is A (I),(T) and (III),(R)

Monochlorinated products are as follow :

The stereochemical informations of products formed are tabulated below.
$$\begin{array}{cccccccc}\text{Carbon Chlorinated}& C1& C2& C3& C6& C7& C8& \text{Total}\\ \text{Carbon becomes chiral}& C3,C7& C2,C3,C7& C7& C6,C7& \text{none}& \text{none}& \\ \text{No. Of isomeric compound formed}& 4& 8& 2& 4& 1& 1& 20\\ \text{No. of optically active compound}& 4& 8& 2& 4& 0& 0& 18\\ \text{No. of Enantiomeric pairs}& 2& 4& 1& 2& 0& 0& 9\\ \text{No of Diastereomeric pairs}& {}^4{C}_2-2=4& {}^8{C}_2-4=24& 0& {}^4{C}_2-2=4& 0& 0& 32\\ \text{No. of fractions formed}& 2& 4& 1& 2& 1& 1& 11\end{array}$$
No. of diastereomeric pair= ${}^E{C}_2-n$
(where E is no of isomers formed and $n$ is no of enantiomeric pair).
Enantiomer has identical physical properties so can be seperated in one fraction using fractional distillation.