The product (K) of the following reaction is: | vedclass

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(C) $1$. The reaction of $CH_3CH_2C \equiv CH$ with $NaNH_2$ in liquid $NH_3$ produces the acetylide ion $CH_3CH_2C \equiv C^-Na^+$,which is intermedi

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(C) $1$. The reaction of $CH_3CH_2C \equiv CH$ with $NaNH_2$ in liquid $NH_3$ produces the acetylide ion $CH_3CH_2C \equiv C^-Na^+$,which is intermediate $I$.$2$. This nucleophilic acetylide ion attacks the carbonyl carbon of the cyclopentanone ring.$3$. The resulting alkoxide intermediate $J$ is then protonated by $H^+$ to yield the final product $(K)$,which is $1-(but-1-ynyl)cyclopentanol$.$4$. The structure of the product is a cyclopentane ring with an $-OH$ group and a $-C \equiv C-CH_2-CH_3$ group attached to the same carbon atom.

The product $(K)$ of the following reaction is:

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