Question

The correct decreasing order of their reactivity towards hydrolysis is:

• A. $\left(ii\right)>\left(iv\right)>\left(i\right)>\left(iii\right)$
• B. $\left(ii\right)>\left(iv\right)>\left(iii\right)>\left(i\right)$
• C. $\left(i\right)>\left(ii\right)>\left(iii\right)>\left(iv\right)$
• D. $\left(iv\right)>\left(ii\right)>\left(i\right)>\left(iii\right)$

Answer 1

The correct option is A $\left(ii\right)>\left(iv\right)>\left(i\right)>\left(iii\right)$

Hydrolysis of carbonyl carbon is nucleophilic addition reaction. For that $-C^{\oplus }$ carbocation has to formed.

More easily carbocation will form, more reactive towards hydrolysis.

$N{O}_2$ and $-CHO$ are two electron-withdrawing groups which favor the formation of a carbocation. $N{O}_2$ is deactivate more than $-CHO$ group, hence $\left(ii\right)$ is more reactive than $\left(iv\right)$.

$C{H}_3$ is an electron releasing group and hence less favor towards formation of carbocation.

In $\left(i\right)$ , there is no EWG or ERG present, so $\left(i\right)$ is more reactive than $\left(iii\right)$.

Hence the decreasing order is :- $\left(ii\right)>\left(iv\right)>\left(i\right)>\left(iii\right)$