The end product of the above reaction is:

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

Question

(A) $1$. Esterification: Naphthalene$-1-$acetic acid reacts with $EtOH/H^+$ to form ethyl naphthalene$-1-$acetate.$2$. Alkylation: Treatment with $NaH

Vedclass · Accepted Answer

Vedclass · Answer

(A) $1$. Esterification: Naphthalene$-1-$acetic acid reacts with $EtOH/H^+$ to form ethyl naphthalene$-1-$acetate.$2$. Alkylation: Treatment with $NaH$ followed by $MeI$ results in the formation of ethyl $2-$(naphthalen$-1-$yl)propanoate.$3$. Hydrolysis: Basic hydrolysis followed by acidification gives $2-$(naphthalen$-1-$yl)propanoic acid.$4$. Cyclization: Treatment with $SOCl_2$ converts the carboxylic acid to an acid chloride,which then undergoes intramolecular Friedel-Crafts acylation in the presence of $AlCl_3$ to form the cyclic ketone,$2$-methyl$-2,3-$dihydro-1H-cyclopenta[a]naphthalen$-1-$one.

Similar Questions

The major product of the following reaction sequence is

$CH_3-CH_2-CH=O$ $\xrightarrow[(2) H^{+}]{(1) \text{Tollen's reagent}} X$ $\xrightarrow[(2) \text{dry distillation}]{(1) Ca(OH)_2} Y$; $Y$ is

Select the $CORRECT$ increasing order of boiling points of alcohols,amines,and carboxylic acids of comparable molar mass from the following.

Ethyl ester $\xrightarrow{CH_3MgBr \text{ (excess)}} P$. The product $P$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module