8.2 Organic chemistry isomerism — Chemistry STD 11 Science — Question

$(i)$ $C{H_3} - C{H_2} - CH = C{H_2}$ : Its possible mono-chloro derivatives are :

$C{H_3} - C{H_2} - CH = CH - Cl$

$2$ isomers : cis and trans forms

$C{H_3} - \mathop {\mathop {\mathop C\limits^ \bullet }\limits_| H}\limits_{Cl\;} - CH = C{H_2}$

optically active (exists in two forms)

$ClC{H_2} - C{H_2} - CH = C{H_2}$ (one form)

${H_3}C - C{H_2} - \mathop {\mathop C\limits^| = }\limits^{Cl\;} C{H_2}$ (one form)

$(ii)$ $C{H_3} - CH = CH - C{H_3}$ : Its possible monochloro derivatives are :

$C{H_3} - CH = \mathop {\mathop C\limits_| - }\limits_{Cl\;} C{H_3}$

Exists in two geometrical forms

$C{H_3} - CH = CH - C{H_2}Cl$

Exists in two geometrical forms

$(iii)$ $\begin{array}{*{20}{c}}
  {C{H_3} - C = C{H_2}} \\ 
  {\,|\,\,} \\ 
  {\,\,\,C{H_3}} 
\end{array}$ : Its possible monochloro derivatives are

$\begin{array}{*{20}{c}}
  {C{H_3} - C = CH - Cl} \\ 
  {|\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_{3\,\,\,\,\,\,\,\,}}} 
\end{array}$

Only one form

$\begin{array}{*{20}{c}}
  {ClC{H_2} - C = C{H_2}} \\ 
  {\,\,\,|} \\ 
  {\,\,\,\,\,\,C{H_3}} 
\end{array}$

Only one form

Thus, the total acylic isomers forms of ${C_4}{H_7}Cl$ are $12$.