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Select the best explanation for why methanol, CH3OH, cannot be used as a solvent for the deprotonation of a terminal alkyne by sodium amide, NaNH2.

A. Methanol is more acidic than the alkyne and will be deprotonated instead.

B. Methanol is toxic, and should be avoided when possible.

C. Sodium amide in methanol reduces alkynes to alkenes.

D. Methanol is a poor solvent for dissolving alkynes.

E. Sodium amide is not a strong enough base to deprotonate the alkyne.

Respuesta :

Answer:

Methanol is more acidic than the alkyne and will be deprotonated instead.

Explanation:

[tex]pK_{a}[/tex] of methanol is around 15 and [tex]pK_{a}[/tex] of terminal alkyne is around 26.

[tex]pK_{a}=-logK_{a}[/tex], where [tex]K_{a}[/tex] is acid dissociation constant

So, higher the acidity of an acid, higher will its [tex]K_{a}[/tex] value and thereby lower will be its [tex]pK_{a}[/tex] value.

So, methanol is certainly stronger acid than terminal alkyne.

Hence sodium amide preferably deprotonates methanol instead of terminal alkyne.

Hence, option (A) is correct.

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