Benzene and Alkyl Benzene - Result Question 10

14. The compounds $P, Q$ and $S$

alt text

were separately subjected to nitration using $\mathrm{HNO}_3 / \mathrm{H}_2 \mathrm{SO}_4$ mixture. The major product formed in each case respectively, is

(2010)

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/snip_images_Pz9QwdeHHzhSUUfENkc6FTn81rO0k4Byvi1XacQmE3s_original_fullsize_png.jpg"width="400"/>

alt text

Show Answer

Answer:

Correct Answer: 14. (c)

Solution:

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/snip_images_p4rjbG_Jr6GQObMkdaXb7s2SrbsJoahIMlSDd6fCQyQ_original_fullsize_png.jpg"width="400"/>

$-OH$ is activating while $-COOH$ is deactivating group in $\mathrm{S}_{\mathrm{E}} \mathrm{Ar}$ reaction. Therefore, electrophile attack to ortho of the activating $-OH$ group.

alt text

Both $-\mathrm{OCH}_3$ and $-\mathrm{CH}_3$ are activating ortho/para directing groups but $-\mathrm{OCH}_3$ is stronger activator, electrophile attack to ortho of $-\mathrm{OCH}_3$.

alt text

Ring II is activated while ring I is deactivated in $\mathrm{S}_{\mathrm{E}}$ Ar reaction. Therefore, electrophile attack at para to ring-II, the less hindered position.