When 3,3-dichloropentane is treated with excess sodium amide in liquid

When 3,3-dichloropentane is treated with excess sodium amide in liquid ammonia, the initial product is 2-pentyne:

нн CI нн Н—с—с—с—С—с—Н нн xs NANH2 Н-с—с-с-С-С—Н нн CI нн нн 2-Pentyne IICIH


However, under these conditions, this internal alkyne quickly isomerizes to form a terminal alkyne that is subsequently deprotonated to form an alkynide ion:


XS NaNH2 XS нн Н ннн ннн Н—с—с—с—С NaNH, Н—с—с—сЕС-с—Н Н-с—с-с-сЕс-н н?


The isomerization process is believed to occur via a mechanism with the following four steps: 

(1) Deprotonate

(2) Protonate

(3) Deprotonate

(4) Protonate.

Using these four steps as a guide, try to draw the mechanism for isomerization using resonance structures whenever possible. Explain why the equilibrium favors formation of the terminal alkyne.

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