Question

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

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

Answer 1

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

$C_n{H}_{2n}$ is the grneral formula of alkene.
Alkene undergo ozonolysis to form aldehydes and ketones.

Since compound (A) shows geometrical isomerism, they must be different groups attached to the doubly bonded carbons.
Thus, option (b) and (d) does not show geometrical isomerism.

Product (B) formed from ozonolysis gives cannizzaro reaction. Compound which lacks $\alpha$-hydrogen will give cannizzaro reaction.
Hence, carbonyl formed from ozonolysis should not have a $\alpha$-hydrogen.

${\left(C{H}_3\right)}_{3}C-\underset{\left(A\right)}{CH}=CH-C{H}_3\underset{Zn/AcOH}{\overset{O_3}{\to }}{\left(C{H}_3\right)}_3\underset{\left(B\right)}{C}-CHO+\underset{\left(C\right)}{C{H}_{3}CHO}$

Hence, option (c) is correct.