Question

Sum of $X , Y$ and $Z$ is :-
(a) $C _{4} H _{8} O $ No. of aldehydes $= X$
(b) $C _{5} H _{12} $ No. of chain isomers $= Y$
(c) $C _{4} H _{8} $ Maximum no. of alkenes $= Z$

• A. 10
• B. 7
• C. 9
• D. 8

Answer 1

Answer: (C)

(a) $C _{4} H _{8} O$ :
$H _{7} C _{3}- CHO$
$\Rightarrow 2$ aldehyde (monovalent radical of $C _{3} H _{8}=\underset{(X)}{2}$ )
(b) $C _{5} H _{1}$ : No. of chain isomers $=\underset{(Y)}{3}$
$CH _{3} CH _{2} CH _{2} CH _{2} CH _{3} ; CH _{3} CH _{2} \underset{\overset{|}{C}H_3}{CH}CH _{3}$
$CH _{3}- \underset{\overset{|}{C}H_3}{\overset{\underset{|}{C}H_3}{C}} - CH _{3}$
(c) $C _{4} H _{8}$ :- Maximum no. of alkenes $\Rightarrow 4_{( Z )}$
(i) $CH _{3}- CH _{2}- CH = CH _{2}$

Hence $X + Y + Z $
$\Rightarrow 2+3+4=9$