The product is:

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1$. Friedel-Crafts acylation of ethoxybenzene with butanoyl chloride in the presence of $AlCl_3$ gives a mixture of ortho and para isomers. The p

Vedclass · Accepted Answer

methyl $4-$ethoxybenzoate

Vedclass · Answer

(C) $1$. Friedel-Crafts acylation of ethoxybenzene with butanoyl chloride in the presence of $AlCl_3$ gives a mixture of ortho and para isomers. The para isomer is the major product due to steric hindrance at the ortho position.$2$. Oxidation of the alkyl side chain of the para-substituted product with $KMnO_4$ converts the propyl group $(-CH_2CH_2CH_3)$ into a carboxylic acid group $(-COOH)$,yielding $4-$ethoxybenzoic acid.$3$. Esterification of $4-$ethoxybenzoic acid with methanol $(MeOH)$ in the presence of an acid catalyst $(H^ )$ and heat $(\Delta)$ converts the carboxylic acid group into a methyl ester group $(-CO_2Me)$,resulting in methyl $4-$ethoxybenzoate.

Similar Questions

What combination of acid chloride or anhydride and arene would you choose to prepare the given compound?

The major product of the following reaction sequence is

What is the ratio of $(x/y)$ in the above reactions?

Identify the correct option based on the following reaction sequence:
Acetic anhydride $\xrightarrow[(ii) EtOH/H^+/\Delta]{(i) H_3O^+} (A)$ $\xrightarrow[(ii) H_3O^+/\Delta]{(i) NaOEt} (B)$

Assertion: Acetamide has a more polar $C=O$ group than ethyl acetoacetate.
Reason: $-\ddot{N}H_{2}$ is more electron-donating than $-\ddot{O}C_{2}H_{5}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module