Organic Chemistry 1 · Carbocation rearrangements and ring expansions
Silver-assisted hydrolysis of neopentyl iodide
Conditions: AgNO3, H2O
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The mechanism, step by step
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Iodine binds silver
Silver ion from AgNO3 is a strong iodophile; nitrate is a spectator and is not drawn. An iodine lone pair bonds to Ag+, making iodide a much better leaving group.
The curved arrows
- An iodine lone pair forms a bond to the silver cation.
Loss of AgI with a methyl shift
A primary neopentyl cation is too unstable to form, so the C–I bond breaks only as a methyl group moves in from the neighbouring quaternary carbon. The concerted step skips the primary cation and leaves the tertiary 2-methylbutan-2-yl cation, while silver iodide precipitates.
The curved arrows
- The C–CH3 bonding pair carries the methyl to the CH2 carbon as the C–I bond weakens.
- The C–I bonding pair moves onto iodine, which leaves bonded to silver as AgI.
Water adds to the cation
A water lone pair bonds to the tertiary cationic carbon, giving an alkyloxonium ion.
The curved arrows
- A water lone pair forms the new C–O bond to the tertiary carbon.
Deprotonate the oxonium ion
A second water molecule removes the proton from the oxonium oxygen, giving the neutral alcohol and hydronium.
The curved arrows
- A water lone pair takes the proton.
- The O–H bonding pair stays on the alcohol oxygen.
2-Methylbutan-2-ol
The OH sits on the former quaternary carbon, and the migrated methyl has turned the CH2 into an ethyl group. Silver iodide precipitates, which pulls the reaction forward.
What you'll practice
Draw the silver-assisted hydrolysis of neopentyl iodide: iodine coordinating Ag+, loss of AgI concerted with a methyl shift, water capture and deprotonation to 2-methylbutan-2-ol.
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