$C H_{3} - C H_{2} - C H = C H - C H B r - C H = C H - C H_{2} - C H_{3}$
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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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?

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

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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?