In the given reactions,find the sum of (x + y | vedclass

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(B) For reaction $(1)$: The starting material is a gem-dibromide $Ph-C(Br)_2-CH_3$. Treatment with $NaNH_2$ (a strong base) causes dehydrohalogenation

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$4$

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(B) For reaction $(1)$: The starting material is a gem-dibromide $Ph-C(Br)_2-CH_3$. Treatment with $NaNH_2$ (a strong base) causes dehydrohalogenation to form an alkyne. Two equivalents of $NaNH_2$ are required to remove two molecules of $HBr$ to form the terminal alkyne $Ph-C \equiv CH$ (which is $(A)$). Thus,$x = 2$.For reaction $(2)$: The starting material is $CH_3-CH_2-C(Br)_2-CH_3$. Treatment with $NaNH_2$ causes dehydrohalogenation to form the internal alkyne $CH_3-C \equiv C-CH_3$. Two equivalents of $NaNH_2$ are required to remove two molecules of $HBr$. Thus,$y = 2$.The sum $(x + y) = 2 + 2 = 4$.

In the given reactions,find the sum of $(x + y)$.
$(1)$ $Ph-C(Br)_2-CH_3$ $\xrightarrow{xNaNH_2} (A)$ $\xrightarrow{CH_3I} Ph-C \equiv C-CH_3$
$(2)$ $CH_3-CH_2-C(Br)_2-CH_3 \xrightarrow{yNaNH_2} CH_3-C \equiv C-CH_3$
Find the value of $(x + y)$.

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In the given reactions,find the sum of $(x + y)$.$(1)$ $Ph-C(Br)_2-CH_3$ $\xrightarrow{xNaNH_2} (A)$ $\xrightarrow{CH_3I} Ph-C \equiv C-CH_3$$(2)$ $CH_3-CH_2-C(Br)_2-CH_3 \xrightarrow{yNaNH_2} CH_3-C \equiv C-CH_3$Find the value of $(x + y)$.