Question

The total number of acyclic isomers including the stereoisomers with the molecular formula $C_4{H}_{7}Cl$ is:

• A. $9$
• B. $10$
• C. $11$
• D. $12$

Answer 1

Answer: (D)

$C_4{H}_{7}Cl$ is ammonochloro derivative of $C_4{H}_{10}$ which it self exists in three ispmeric forms. (i) $C{H}_3-C{H}_2-CH=C{H}_2$ : Its possible monochloro derivatives are: $\underset{2\text{isomers cis and trands (2 from)}}{C{H}_3-C{H}_{2}CH=CH-Cl}$ $\underset{\text{one isomer (one form)}}{C{H}_3-C{H}_2-\underset{Cl}{\underset{|}{C}}=C{H}_2}$ $\underset{\text{optically active (exists in two forms)}}{C{H}_3-\underset{Cl}{\underset{|}{C}}H-CH=C{H}_2}$ (ii) $C{H}_3-CH=CH-C{H}_3$ : Its possible monochloro derivatives are: $\underset{(\text{Exists in two geometrical forms)}}{C{H}_3-CH=\underset{Cl}{\underset{|}{C}}-C{H}_3}$ $\underset{(\text{Exists in two geometrical forms)}}{C{H}_3-CH=CH-C{H}_{2}Cl}$ (iii) $C{H}_3-\underset{C{H}_3}{\underset{|}{C}}=C{H}_2$ : Its possible mono- chloro derivatives are: $\underset{Onlyoneform}{C{H}_3-\underset{C{H}_3}{\underset{|}{C}}=CH-Cl;}$ $\underset{Onlyoneform}{Cl-C{H}_2-\underset{C{H}_3}{\underset{|}{C}}=C{H}_2}$ Thus, the total acyclic isomeric forms of $C_4{H}_{7}Cl$ are twelve (12).