Question

Neopentyl bromide undergoes dehydrohalogenation to give alkenes even though it has no \(\beta\)-hydrogen. This is due to

• A. E2 mechanism
• B. E1 mechanism
• C. Rearrangement of carbocations by E1 mechanism
• D. E1cB mechanism
• E. Ei mechanism

Hint:

Neopentyl systems are "red flags" for rearrangements. Whenever you see a primary halide next to a quaternary carbon, expect a methyl shift if a carbocation is formed.

Answer 1

The Correct Option is C

Concept: Dehydrohalogenation usually requires a \(\beta\)-hydrogen for a base to abstract. Neopentyl bromide \((CH_3)_3CCH_2Br\) lacks these, yet it still reacts. This happens via a carbocation intermediate that can change its structure to become more stable.

Step 1: Formation of the primary carbocation.
Under E1 conditions, the leaving group (Bromide) departs first, leaving behind a primary neopentyl carbocation: \[ (CH_3)_3C-CH_2-Br \xrightarrow{-Br^-} (CH_3)_3C-CH_2^+ \] This primary carbocation is relatively unstable.

Step 2: 1,2-Methyl Shift.
To gain stability, a methyl group from the adjacent \(\gamma\)-carbon migrates with its bonding pair of electrons to the positive carbon. This is a rearrangement: \[ (CH_3)_3C-CH_2^+ \xrightarrow{1,2-CH_3 \text{ shift}} (CH_3)_2C^+-CH_2-CH_3 \] The result is a tertiary (3\(^\circ\)) carbocation, which is much more stable.

Step 3: Elimination.
The tertiary carbocation now has \(\beta\)-hydrogens on the adjacent carbons. A base abstracts one of these hydrogens to form a double bond (alkene): \[ (CH_3)_2C^+-CH_2-CH_3 \xrightarrow{-H^+} (CH_3)_2C=CH-CH_3 \] The final product is 2-methyl-2-butene.