The products (A) and (B) respectively are: | vedclass

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(C) The reaction sequence for $(A)$ involves the Hofmann elimination of an amine. The amine is first methylated using $CH_3I$ to form a quaternary amm

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(C) The reaction sequence for $(A)$ involves the Hofmann elimination of an amine. The amine is first methylated using $CH_3I$ to form a quaternary ammonium salt,which is then treated with $Ag_2O/H_2O$ to form the hydroxide,followed by heating $(\Delta)$ to undergo anti-elimination. In anti-elimination,the leaving group and the $\beta$-hydrogen must be anti-periplanar. Given the stereochemistry of the starting material,the $H$ atom is anti to the leaving group,so $H$ is eliminated,yielding the product with a double bond between the carbons that had the amine and the $H$ atom.The reaction sequence for $(B)$ involves the Cope elimination of an amine oxide. The amine is treated with $H_2O_2$ to form the amine oxide,followed by heating $(\Delta)$ to undergo syn-elimination. In syn-elimination,the leaving group and the $\beta$-hydrogen must be syn-periplanar. Given the stereochemistry,the $D$ atom is syn to the leaving group,so $D$ is eliminated,yielding the product with a double bond between the carbons that had the amine and the $D$ atom.Therefore,$(A)$ is the product of anti-elimination (loss of $H$) and $(B)$ is the product of syn-elimination (loss of $D$). Comparing this with the given options,the correct choice is $(C)$.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A) CH_3CH_2Br + Mg \xrightarrow{\text{dry ether}}$	$(i) \text{Cyclohexylmagnesium bromide}$
$(B) 2RX + 2Na \xrightarrow{\text{dry ether}}$	$(ii) CH_3CH_2MgBr$
$(C) CH_3CH_2Br \xrightarrow{\text{alc. KOH}}$	$(iii) R-R + 2NaX$
$(D) \text{Bromocyclohexane} + Mg \xrightarrow{\text{dry ether}}$	$(iv) CH_2=CH_2 + H_2O + KBr$

The products $(A)$ and $(B)$ respectively are:

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