Question

$\underset{C_7{H}_{14}}{\left(A\right)}\stackrel{O_3}{\underset{Zn/AcOH}{⟶}}\left(B\right)+\left(C\right)$
Compound (A) exist in geometrical isomers and (B) gives Cannizaro reaction.
(A) will be:

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

Answer 1

The correct option is C $(C{H}_3{)}_{3}C-CH=CH-C{H}_3$

Refer to image

Compound B is $\left(C{H}_3\right)C_{3}CHO$. As it doesn't contain $\alpha$ hydrogen and it is aldehyde, so, it give cannizaro reaction.

And Compound A can show cis and trans geometrical reaction.