2-Bromo-3-methylpentane, $CH X_{3} - CH (Br) - CH (CH X_{3}) - CH X_{2} - CH X_{3}$ , undergoes E2 elimination with a base, with no rearrangement. How many distinct alkenes (all constitutional isomers of $C X_{6} H X_{12}$ that can form) are possible? Count cis and trans (E and Z) isomers separately, but do not count enantiomers separately.

Numerical answer type, tolerance 0.5.

Answer: 3

Beta hydrogens are on C-1 (a methyl group) and C-3. Loss from C-1 gives 3-methylpent-1-ene, one alkene (its enantiomers are not counted separately). Loss from C-3 gives 3-methylpent-2-ene, which exists as E and Z isomers, so two alkenes. The total is

1 + 2 = 3

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Answer: 3

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