Question

Optically active isomer (A) of $\left(C_5{H}_{9}Cl\right)$ on treatment with one mole of $H_2$ gives an optically inactive compound (B) compound (A) will be :

• A. $C{H}_3-\underset{C{H}_{2}Cl}{\underset{|}{C}}H-CH=C{H}_2$
• B. $Cl-\underset{C{H}_3}{\underset{|}{C}}H-CH=CH-C{H}_3$
• C. $C{H}_3-\underset{Cl}{\underset{|}{C}}H-C{H}_2-CH=C{H}_2$
• D. $C{H}_3-C{H}_2-\underset{Cl}{\underset{|}{C}}H-CH=C{H}_2$

Answer 1

Answer: (D)

$C{H}_3-C{H}_2-\underset{Cl}{\underset{|}{C}}H-CH=C{H}_2$

As refer to the above image optically active compound gives optically inactive isomers.