Question

The neopentyl halide in ethanol yields alkenes by E1 mechanism due to

• A. low concentration of solvent
• B. absence of base
• C. it is a primary halide
• D. steric factor which prevents E2 mechanism
• E. solvation effect

Hint:

Neopentyl systems are "primary halides that act like tertiary ones" in terms of mechanism because of the massive steric bulk of the adjacent quaternary carbon.

Answer 1

The Correct Option is D

Concept: Neopentyl halides are primary halides but are exceptionally bulky. The presence of a quaternary carbon ($\text{C}$ atom bonded to four other $\text{C}$ atoms) adjacent to the carbon bearing the halogen prevents standard bimolecular reactions.

Step 1: {The structure of neopentyl halide.} Neopentyl halide ($\text{(CH}_3)_3\text{CCH}_2\text{X}$) has significant steric hindrance. Even though the leaving group is on a primary carbon, the back-side approach for $S_N2$ or the removal of a $\beta$-hydrogen for $E2$ is blocked by the bulky tert-butyl group.

Step 2: {Examine the E1 pathway.} Because the $E2$ mechanism is sterically hindered, the reaction proceeds via the $E1$ mechanism. This involves the ionization of the halide to form a primary carbocation ($\text{(CH}_3)_3\text{C}-\text{CH}_2^+$).

Step 3: {Consider carbocation rearrangement.} The resulting primary carbocation undergoes a 1,2-methyl shift to form a much more stable tertiary carbocation, which then loses a proton to form an alkene (2-methylbut-2-ene). The driving force is the steric hindrance that makes $E2$ impossible.