Consider the following reaction sequence:CH_3 | vedclass

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(B) Step $1$: Dehydrohalogenation of $1,2-$dibromopentane with $2NaNH_2$ gives pent$-1-$yne $(X = CH_3CH_2CH_2C \equiv CH)$.Step $2$: Treatment of pen

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$5-$methylhex$-2-$yne

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(B) Step $1$: Dehydrohalogenation of $1,2-$dibromopentane with $2NaNH_2$ gives pent$-1-$yne $(X = CH_3CH_2CH_2C \equiv CH)$.Step $2$: Treatment of pent$-1-$yne with $NaNH_2$ forms the sodium acetylide salt $(CH_3CH_2CH_2C \equiv C^-Na^+)$.Step $3$: The acetylide ion acts as a nucleophile and attacks isopropyl bromide $((CH_3)_2CHBr)$ via an $S_N2$ mechanism.$CH_3CH_2CH_2C \equiv C^- + (CH_3)_2CHBr \rightarrow CH_3CH_2CH_2C \equiv C-CH(CH_3)_2$.The resulting product is $5-$methylhex$-2-$yne.

Consider the following reaction sequence:
$CH_3CH_2CH_2CH(Br)CH_2Br + 2NaNH_2 \rightarrow X$
$X \xrightarrow[(ii) (CH_3)_2CHBr]{(i) NaNH_2} Y$
The product $Y$ formed is:

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Consider the following reaction sequence:$CH_3CH_2CH_2CH(Br)CH_2Br + 2NaNH_2 \rightarrow X$$X \xrightarrow[(ii) (CH_3)_2CHBr]{(i) NaNH_2} Y$The product $Y$ formed is: