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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.

User Somjit
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1 Answer

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Answer:

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

Step-by-step explanation:


pK_(a) of methanol is around 15 and
pK_(a) of terminal alkyne is around 26.


pK_(a)=-logK_(a), where
K_(a) is acid dissociation constant

So, higher the acidity of an acid, higher will its
K_(a) value and thereby lower will be its
pK_(a) 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.

User Thiagoss
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