The end product of the above reaction is:

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

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(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.

Column-$I$ (formula)	Column-$II$ (name)
$A$. $RCH=NH$	$i$. Diisobutyl aluminium hydride
$B$. $AlH(i-Bu)_2$	$ii$. Benzalchloride
$C$. $C_6H_5-CH(OCOCH_3)_2$	$iii$. Imine
$D$. $C_6H_5CHCl_2$	$iv$. Benzylidene diacetate

Similar Questions

The general formula for monocarboxylic acids is

$CH_3COOH \xrightarrow{LiAlH_4} A$
$A + CH_3COOH \xrightarrow{H_3O^+} B + H_2O$
In the above reactions,'$A$' and '$B$' respectively are:

Which $\beta$-keto acid shown will not undergo decarboxylation?

Match the formulas in Column-$I$ with their names in Column-$II$.

Column-$I$ (formula) Column-$II$ (name)

$A$. $RCH=NH$ $i$. Diisobutyl aluminium hydride

$B$. $AlH(i-Bu)_2$ $ii$. Benzalchloride

$C$. $C_6H_5-CH(OCOCH_3)_2$ $iii$. Imine

$D$. $C_6H_5CHCl_2$ $iv$. Benzylidene diacetate

What will be the product obtained from the following reaction?
$CH_3CH_2C \equiv N + C_2H_5OH + H_2O \xrightarrow{\text{conc. } H_2SO_4}$

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