A xrightarrow[{CHCl_3}]{{KOH}} B xrightarrow[ | vedclass

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(A) The final product is $CH_3CH_2NHCH_3$ ($N$-methylethanamine).Step $1$: The reaction $B \xrightarrow[{H^\oplus}]{{LiAlH_4}} CH_3CH_2NHCH_3$ involve

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$CH_3CH_2NH_2$

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(A) The final product is $CH_3CH_2NHCH_3$ ($N$-methylethanamine).Step $1$: The reaction $B \xrightarrow[{H^\oplus}]{{LiAlH_4}} CH_3CH_2NHCH_3$ involves the reduction of an amide or an imine. Since $LiAlH_4$ reduces amides to amines,$B$ must be $CH_3CH_2CONHCH_3$.Step $2$: The reaction $A \xrightarrow[{CHCl_3}]{{KOH}} B$ is a carbylamine-like reaction or an alkylation. However,looking at the structure of $B$ $(CH_3CH_2CONHCH_3)$,it is an $N$-substituted amide. This suggests $A$ is the primary amine $CH_3CH_2NH_2$ reacting with an acylating agent,but the reagents $CHCl_3/KOH$ are characteristic of the Carbylamine test for primary amines.Step $3$: Re-evaluating the sequence: $CH_3CH_2NH_2$ $(A)$ reacts with $CHCl_3/KOH$ to form an isocyanide $(CH_3CH_2NC)$,which is not $B$. Given the options and the transformation,the reaction sequence likely represents the formation of an amide from an amine and an acid derivative,or a specific pathway. Based on standard chemistry problems of this type,$A$ is $CH_3CH_2NH_2$.

$A$ $\xrightarrow[{CHCl_3}]{{KOH}} B$ $\xrightarrow[{H^\oplus}]{{LiAlH_4}} CH_3CH_2NHCH_3$
Identify compound $A$.

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$A$ $\xrightarrow[{CHCl_3}]{{KOH}} B$ $\xrightarrow[{H^\oplus}]{{LiAlH_4}} CH_3CH_2NHCH_3$Identify compound $A$.