Question

In the hydrocarbon $\underset{6}{{CH}_3}-\stackrel{{CH}_3}{\underset{{CH}_2}{C}-CH=\underset{3}{CH}}-\underset{2}{C}=\underset{1}{CH}$ the state of hybridisation of carbon $2,4$ and $5$ respectively are:

• A. $sp,s{p}^2,s{p}^3$
• B. $sp,s{p}^2,s{p}^2$
• C. $s{p}^2,s{p}^3,sp$
• D. $s{p}^2,sp,s{p}^3$

Answer 1

The correct option is A $sp,s{p}^2,s{p}^3$

$\to$ If two double bonds for a carbon then $sp$ its hybridisation.

$\to$ If a single double bond for a carbon then ${sp}^2$ its hybridisation.

$\to$ If no double or triple bonds present then it is ${sp}^3$ hybridisation.

$\therefore$ $C2,C4,C5$ respectively has $sp,{sp}^2,{sp}^3$

Ans :- Option A.