Answer
The products would be $H−C≡C:^-$ $Li^+$ + $CH_3$
Work Step by Step
$CH_3Li$ exists as $Li^+$ and $CH_3^-$. So the $CH_3^-$ attacks on the H on the alkyne. Thus, the products are $H−C≡C:^-$ $Li^+$ + $CH_3$.
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