Question

What is the major product $P$ in the above reaction?

$C{H}_{3}C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}-}\underset{OH}{\underset{|}{C}H}-C{H}_3\underset{heat}{\overset{H_{2}S{O}_4}{\to }}\underset{Major}{P}$

• A. $\stackrel{C{H}_3}{\stackrel{|}{C}{H}_2}-C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{C}H}-CH=C{H}_2$
• B. $C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{C}H}-\stackrel{C{H}_3}{\stackrel{|}{C}H}-CH=C{H}_2$
• C. $C{H}_{3}C{H}_2-\underset{C{H}_3}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}-}CH=C{H}_2$
• D. $C{H}_{3}C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{C}=}\underset{C{H}_3}{\underset{|}{C}-}C{H}_3$

Answer 1

The correct option is D $C{H}_{3}C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{C}=}\underset{C{H}_3}{\underset{|}{C}-}C{H}_3$

Due to 1,2-methyl shift, more stable tertiary cation forms and gives 2,3-dimethylpent-2-ene as major product.

Therefore option D is correct.