The product (B) of the above reaction is: | vedclass

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(D) $1$. The reaction starts with the nucleophilic addition of the Grignard reagent $(CH_3MgI)$ to the carbonyl group of the ketone,followed by proton

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(D) $1$. The reaction starts with the nucleophilic addition of the Grignard reagent $(CH_3MgI)$ to the carbonyl group of the ketone,followed by protonation with $NH_4Cl$ to form an alcohol intermediate $(A)$.$2$. The intermediate $(A)$ is a tertiary alcohol.$3$. Upon treatment with acid $(H^{+})$ and heating $(\Delta)$,the alcohol undergoes dehydration to form a carbocation.$4$. This carbocation is stabilized by resonance,leading to the formation of a fully conjugated aromatic system.$5$. The final product $(B)$ is $9-methylanthracene$,formed after the loss of a proton,which restores the fully conjugated anthracene ring system.

Column-$A$	Column-$B$
$(1)$ $CH_3-CH_2-Cl + KOH_{(aq)} \rightarrow$	$(A)$ $1,2-\text{dichloroethane}$
$(2)$ $CH_3-CH_2-CH_2-CH_3 \xrightarrow{AlCl_3/HCl}$	$(B)$ $\text{Chloromethane}$
$(3)$ $CH_3-CH_2-Br + \text{alc. } KOH \rightarrow$	$(C)$ $\text{But-2-ene}$
$(4)$ $CH_2=CH_2 + Cl_2 \rightarrow$	$(D)$ $\text{Ethanol}$
	$(E)$ $\text{Chloroethane}$
	$(F)$ $\text{Ethene}$
	$(G)$ $\text{Isobutane}$

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

The product $(B)$ of the above reaction is:

Similar Questions

Assertion : Alkylbenzene is not prepared by Friedel-Crafts alkylation of benzene.
Reason : Alkyl halides are less reactive than acyl halides.

Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

The product $A$ is
$4-methylcyclohexyl bromide + OH^- \xrightarrow{S_N2} A$

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