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Question-211828




Question Number 211828 by Spillover last updated on 22/Sep/24
Answered by Spillover last updated on 22/Sep/24
   THE ONLY REACTION THAT WON'T PROCEED IS D. Sodium hydroxide (NaOH) is not strong enough to deprotonate an alkyl acetylene  because the pKa of a terminal alkyne is around 25. For an acid-base reaction to proceed, the base must be stronger than the conjugate base of the acid being deprotonated.    NaOH has a conjugate acid, water (H₂O), with a pKa of about 15.72. Since the pKa of water is significantly lower than that of a terminal alkyne (i.e., 10 to the 9th power), the equilibrium of the deprotonation reaction would not favor the formation of the acetylide anion. Instead, a much stronger base, such as sodium amide (NaNH₂), which has a conjugate acid (NH₃) with a pKa of around 38, is required to effectively deprotonate a terminal alkyne.
$$ \\ $$ THE ONLY REACTION THAT WON'T PROCEED IS D. Sodium hydroxide (NaOH) is not strong enough to deprotonate an alkyl acetylene because the pKa of a terminal alkyne is around 25. For an acid-base reaction to proceed, the base must be stronger than the conjugate base of the acid being deprotonated.

NaOH has a conjugate acid, water (H₂O), with a pKa of about 15.72. Since the pKa of water is significantly lower than that of a terminal alkyne (i.e., 10 to the 9th power), the equilibrium of the deprotonation reaction would not favor the formation of the acetylide anion. Instead, a much stronger base, such as sodium amide (NaNH₂), which has a conjugate acid (NH₃) with a pKa of around 38, is required to effectively deprotonate a terminal alkyne.

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