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List -IList -II(I) Enantiomeric Pairs(P)20(II) Diastereomeric Pairs(Q)34(III) Optically acitve compounds(R)18(IV) Number of fractions of compounds that can be separated(S)11by fractional distillation(T)9(U)32
Which of the following options has the correct combination considering List-I and List-II

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

The correct options are
B (II),(U) and (IV),(S)
Monochlorinated products are as follow :

The stereochemichal informations of products formed are tabulated below.
Carbon ChlorinatedC1C2C3C6C7C8TotalCarbon becomes chiralC3,C7C2,C3,C7C7C6,C7nonenoneNo. Of isomeric compound formed48241120No. of optically active compound48240018No. of Enantiomeric pairs2412009No of Diastereomeric pairs4C22=48C24=2404C22=40032No. of fractions formed24121111
No. of diastereoisomeric pair= EC2n
(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.

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