HC equiv CH xrightarrow[{excess}]{{CH_3MgBr}} | vedclass

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(B) $1$. Reaction of $HC \equiv CH$ with excess $CH_3MgBr$ (Grignard reagent) acts as a base,removing acidic protons: $HC \equiv CH + 2CH_3MgBr \right

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Compound $D$ gives Tollen's test

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(B) $1$. Reaction of $HC \equiv CH$ with excess $CH_3MgBr$ (Grignard reagent) acts as a base,removing acidic protons: $HC \equiv CH + 2CH_3MgBr \rightarrow BrMgC \equiv CMgBr (A) + 2CH_4 (B)$.$2$. Here,$(A)$ is $BrMgC \equiv CMgBr$ and $(B)$ is Methane $(CH_4)$.$3$. Reaction of $(A)$ with excess $CH_3I$ (nucleophilic substitution): $BrMgC \equiv CMgBr + 2CH_3I \rightarrow CH_3-C \equiv C-CH_3 (C) + 2MgBrI$.$4$. $(C)$ is But$-2-$yne. But$-2-$yne does not have acidic protons,so it does not react with $NaNH_2$ to form a terminal alkyne or give Tollen's test.$5$. The question options seem to imply a different pathway or typo in the reagents. Given the standard chemistry of terminal alkynes,$(D)$ would be the product of the reaction. Since $(C)$ is an internal alkyne,it does not give Tollen's test.

$HC \equiv CH$ $\xrightarrow[{excess}]{{CH_3MgBr}} \mathop {(A)}\limits_{\text{salt}} + \mathop {(B)}\limits_{\text{hydrocarbon}}$
$(A)$ $\xrightarrow[{excess}]{{CH_3I}} (C)$ $\xrightarrow[{(ii) H^{+} }]{{(i) NaNH_2}} (D)$

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$HC \equiv CH$ $\xrightarrow[{excess}]{{CH_3MgBr}} \mathop {(A)}\limits_{\text{salt}} + \mathop {(B)}\limits_{\text{hydrocarbon}}$$(A)$ $\xrightarrow[{excess}]{{CH_3I}} (C)$ $\xrightarrow[{(ii) H^{+} }]{{(i) NaNH_2}} (D)$