Dehydration of tertiary alcohol via E1 mechanism produces alkene isomers

Steps:

• Form tertiary carbocation at C3 from 3-methylhexan-3-ol: CH3CH2C+(CH3)CH2CH2CH3.
• Deprotonate from adjacent positions: C2 (CH2), C4 (CH2), methyl on C3 (CH3).
• Yield alkenes: 3-methylhex-2-ene (E/Z isomers), 3-methylhex-3-ene (E/Z isomers), 3-methylenehexane (no E/Z).
• Total isomers: 2 (from 2-ene) + 2 (from 3-ene) + 1 (methylene) = 5.

Why C is correct:

• E1 mechanism allows deprotonation from three sites, with two alkenes exhibiting E/Z stereoisomerism due to distinct substituents on each carbon of the double bond.

Why the others are wrong:

• A: Counts only structural isomers, ignoring E/Z stereoisomers.
• B: Misses one E/Z pair, undercounting stereoisomers.
• D: Includes improbable carbocation rearrangements, exceeding actual products.

Final answer: C