Haloalkanes and Haloarenes: Common Mistakes and Fixes (5)

$2$-bromobutane: $\text{CH}_3\text{-CHBr-CH}_2\text{-CH}_3$ — a secondary alkyl halide.

• Alcoholic KOH → strong base, weak nucleophile (alcohol solvates K⁺ better than OH⁻) → favours elimination (E2).
• Aqueous KOH → strong nucleophile (water solvates K⁺ less strongly, OH⁻ acts free) → favours substitution (S$_\text{N}$2).

E2 removes a $\beta$-hydrogen and the leaving group ($\text{Br}^-$) simultaneously, forming a C=C double bond. There are two possible $\beta$-positions:

• Remove H from $\text{CH}_3$ (left side): $\text{CH}_2=\text{CH-CH}_2\text{-CH}_3$ ($1$-butene)
• Remove H from $\text{CH}_2$ (right side): $\text{CH}_3\text{-CH=CH-CH}_3$ ($2$-butene)

Zaitsev’s rule: the more substituted alkene is the major product. $2$-butene has the double bond between two carbons each bearing $1$ alkyl group → more substituted.

Major product: $2$-butene (mainly trans by stability).

S$_\text{N}$2: OH⁻ attacks the carbon bearing Br from the back side, displacing Br⁻ in one concerted step.

Major product: $2$-butanol, $\text{CH}_3\text{-CH(OH)-CH}_2\text{-CH}_3$, with inversion of configuration at the chiral carbon.