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(A) When an alkyl halide $(R-X)$ is reacted with an excess of alcoholic ammonia $(NH_3)$,the reaction is an ammonolysis reaction.Since ammonia is in l

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primary amine

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(A) When an alkyl halide $(R-X)$ is reacted with an excess of alcoholic ammonia $(NH_3)$,the reaction is an ammonolysis reaction.Since ammonia is in large excess,the probability of the alkyl halide molecule colliding with an ammonia molecule is much higher than colliding with the formed amine.Therefore,the reaction stops at the formation of the primary amine $(R-NH_2)$:$R-X + NH_3 (	ext{alc.}) \xrightarrow{\Delta} R-NH_2 + HX$Thus,the correct option is $A$.

When alkyl halide is boiled with a large excess of alcoholic ammonia,it forms:

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