Question

Neopentyl alcohol, (CH3)3CCH2OH, reacts with concentrated HBr to give 2-bromo-2-methylbutane, a rearranged product. Propose a mechanism for the formation of this product.

Short answer

Neopentyl alcohol(CH₃)₃CCH₂OHis a primary alcohol, but the bulky group has hinderedit. The backside attack is not possible, so instead of the S_N2 mechanism S_N1followedto get the desired rearranged product.

The reaction forms a rearranged stable carbocation from primary carbocation to tertiary carbocation.

Step-by-step solution

Step 1: Rearranged Product

Neopentyl alcohol(CH₃)₃CCH₂OHis a primary alcohol, but the bulky group has hinderedit. The backside attack is not possible, so instead of the S_N2 mechanism S_N1followedto get the desired rearranged product.

The reaction forms a rearranged stable carbocation from primary carbocation to tertiary carbocation.

 Mechanism for the formation of 2-bromo-2-methylbutane

Protonation of Neopentyl alcoholconverts the hydroxy group to a good leaving group.

Water leaves form a primary carbocation rearranged through methyl shift, thus forming a stable tertiary carbocation.

Bromide ion attacks the carbocation to form the product 2-bromo-2-methylbutane

Mechanism showing the formation of 2-bromo-2-methylbutane