Carboxylic Acids and their Derivatives - Result Question 5-1

5. The decreasing order of ease of alkaline hydrolysis for the

following esters is

(2019 Main, 10 Jan I)

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/snip_images_JAuHgCQYsBCyaBxTds1PCiFNvVTjJY2ZLoal9DPIdtw_original_fullsize_png.jpg"width="490"/>

(a) III $>$ II $>$ IV $>$ I

(b) III $>$ II $>$ I $>$ IV

(c) II $>$ III $>$ II $>$ IV

(d) IV $>$ II $>$ III $>$ I

Show Answer

Answer:

Correct Answer: 5. (b)

Solution:

Alkaline hydrolysis of an ester (carboxylic acid derivative) follows acyl $\mathrm{S}_{\mathrm{N}} 2$ mechanism.

Rate of $\mathrm{S}{\mathrm{N}} 2$ mechanism depends on the polarity of >$\mathrm{C}=\mathrm{O}$ group of $-COOR $ group. Electron withdrawing group $(-R>-I)$ increases the rate of $\mathrm{S}{\mathrm{N}} 2$ reaction whereas electron donating group $(+R>+I)$ decreases the rate of $\mathrm{S}_{\mathrm{N}} 2$ reaction.

Here, the nature of functional groups attached para to the benzene ring are:

$ -\underset{(-R)}{\mathrm{NO}_2}>\underset{(-I)}{-\mathrm{Cl}}>-\underset{(+R)}{\mathrm{OCH}_3} $

So, the order of hydrolysis will be,

$ \underset{(-R)}{\text { III }}>\underset{(-I)}{\mathrm{II}}>\mathrm{I}>\underset{(+R)}{\mathrm{IV}} $