$C H_{3} - C H_{2} - C H = C H - C H B r - C H = C H - C H_{2} - C H_{3}$
When the mixture of all stereoisomers of the given compound is subjected to fractional distillation, in how many fractions they will get separated?

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Solution

There are 4 stereoisomers in which (I) and (II) are achiral. (III) and (IV) are enantiomers and can be seperated into single fraction. Hence total of three fraction will be obtained.

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Similar questions

Q. Find total number of stereoisomers for each compound given below.

P h - O | | S - C H = C H - C H_{2} - C H = C = C = C H - C H = C H - C H_{3}

Which are free radical reactions

$I . C H_{3} - C H = C H_{2} + H B r P e r o x i d e - ---- \to C H_{3} - C H_{2} - C H_{2} B r I I . C H_{3} - C H = C H_{2} + H B r \to C H_{3} - C H B r - C H_{3} I I I . C H_{3} - C H = C H_{2} + C l_{2} \to C l C H_{2} - C H = C H_{2} I V . C H_{3} - C H_{3} + C l_{2} \to C l C H_{2} - C H_{3}$

Q. The correct order of C - H bond length in

C H_{3} - C H_{3}, C H_{2} = C H_{2}

and

C H \equiv C H

is:

C H_{3} - C H = C H_{2} + H B r \to

-------- the product is

Q. Which will undergo reaction with ammonical

A g N O_{3}

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CH3−CH2−CH=CH−CHBr−CH=CH−CH2−CH3 When the mixture of all stereoisomers of the given compound is subjected to fractional distillation, in how many fractions they will get separated?

There are 4 stereoisomers in which (I) and (II) are achiral. (III) and (IV) are enantiomers and can be seperated into single fraction. Hence total of three fraction will be obtained.

$C H_{3} - C H_{2} - C H = C H - C H B r - C H = C H - C H_{2} - C H_{3}$
When the mixture of all stereoisomers of the given compound is subjected to fractional distillation, in how many fractions they will get separated?