E2 elimination and base-catalyzed isomerization produce three C4H8 alkene isomers

Steps:

• 1-Bromobutane (CH3CH2CH2CH2Br) reacts with NaOH in ethanol via E2 elimination, where base abstracts β-hydrogen from C2 and Br leaves from C1, forming 1-butene (CH2=CHCH2CH3).
• 1-Butene undergoes base-catalyzed isomerization: NaOH abstracts allylic hydrogen from C3, forming a resonance-stabilized allylic anion that reprotonates to give 2-butene (CH3CH=CHCH3).
• 2-Butene exhibits stereoisomerism due to restricted rotation around the C=C bond, yielding cis-2-butene and trans-2-butene.
• The mixture thus contains 1-butene (one structural isomer) and two stereoisomers of 2-butene, totaling three isomeric alkenes.

Why C is correct:

• C4H8 has multiple isomers, but this reaction yields exactly three: 1-butene and the cis/trans stereoisomers of 2-butene, per E2 mechanism and allylic isomerization under basic conditions.

Why the others are wrong:

• A. 1: Considers only 1-butene, ignoring isomerization to 2-butene.
• B. 2: Includes 1-butene and 2-butene but overlooks cis-trans stereoisomerism in 2-butene.
• D. 4: Includes extraneous isomers like 2-methylpropene, which are not formed here.

Final answer: C