Explanation

(b)
$C_{4}H_{7}Cl$ is a monochloro derivative of $C_{4}H_{8}$ which itself exists in three isomeric forms.
(i)$\ CH_{3} - CH_{2} - CH = CH_{2}$: Its possible mono-chloro derivatives are
$\ \ \ \ \ \ \ \ \ \ \ CH_{3} - CH_{2} - CH = CH - Cl$
2 isomers $cis$ and $trans$ forms
[image:12_image397.emf]
Optically active (exists in two forms)
$ClCH_{2} - CH_{2} - CH = CH_{2}$(one form)
Cl
|
$H_{3}C - CH_{2} - C = CH_{2}$ (one form)
(ii) $CH_{3} - CH = CH - CH_{3}$: Its possible monochloro derivatives are
$\ \ \ \ \ \ \ CH_{3} - CH = C - CH_{3}$
|
Cl
Exists in two geometrical forms
$CH_{3} - CH = CH - CH_{2}Cl$
Exists in two geometrical forms
(iii)$CH_{3} - C = CH_{2}$: Its possible monochloro derivatives are
|
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ CH_{3}$
$CH_{3} - C = CH - Cl,$ (Only one form)
|
$\ \ \ \ \ \ \ \ \ \ \ \ CH_{3}$
$ClCH_{2} - C = CH_{2}$(only one form)
|
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ CH_{3}$
Thus, the total acylic isomeric forms of $C_{4}H_{7}Cl$ are 12.