Haloalkanes and Haloarenes Part 2

Concept:The substitution of bromine by hydroxyl group using aqueous KOH gives the corresponding alcohol, and optical isomerism occurs if the alcohol product has at least one chiral carbon (carbon with four different groups).Chemical Equation / Formula:$\mathrm{C}_5\mathrm{H}_{11}\mathrm{Br} + \mathrm{KOH(aq)} \rightarrow \mathrm{C}_5\mathrm{H}_{11}\mathrm{OH} + \mathrm{KBr}$ (no rearrangement).Explanation:There are 8 structural isomers of $\mathrm{C}_5\mathrm{H}_{11}\mathrm{Br}$. For each, the alcohol formed is identified and checked for chirality:1. $1$-bromopentane → $1$-pentanol: $\mathrm{CH_3CH_2CH_2CH_2CH_2OH}$ – no chiral carbon.2. $2$-bromopentane → $2$-pentanol: $\mathrm{CH_3CH_2CH_2CHOHCH_3}$ – carbon $2$ has four different groups (H, OH, $\mathrm{CH_2CH_2CH_3}$, $\mathrm{CH_3}$) → chiral.3. $3$-bromopentane → $3$-pentanol: $\mathrm{CH_3CH_2CHOHCH_2CH_3}$ – carbon $3$ has two identical ethyl groups → achiral.4. $1$-bromo-$3$-methylbutane → $3$-methyl-$1$-butanol: $(\mathrm{CH_3})_2\mathrm{CHCH_2CH_2OH}$ – primary alcohol, no chiral carbon.5. $1$-bromo-$2$-methylbutane → $2$-methyl-$1$-butanol: $\mathrm{CH_3CH_2CH(CH_3)CH_2OH}$ – carbon $2$ (with $\mathrm{CH_3}$) is chiral (H, $\mathrm{CH_3}$, $\mathrm{CH_2CH_3}$, $\mathrm{CH_2OH}$) → chiral.6. $2$-bromo-$2$-methylbutane → $2$-methyl-$2$-butanol: $(\mathrm{CH_3})_2\mathrm{C(OH)CH_2CH_3}$ – quaternary carbon with two methyl groups → achiral.7. $2$-bromo-$3$-methylbutane → $3$-methyl-$2$-butanol: $(\mathrm{CH_3})_2\mathrm{CHCHOHCH_3}$ – carbon $2$ (bearing OH) has four different groups (H, OH, $\mathrm{CH(CH_3)_2}$, $\mathrm{CH_3}$) → chiral.8. $1$-bromo-$2,2$-dimethylpropane → $2,2$-dimethyl-$1$-propanol: $(\mathrm{CH_3})_3\mathrm{CCH_2OH}$ – no chiral carbon.Thus only three alcohols are chiral: $2$-pentanol, $2$-methyl-$1$-butanol, and $3$-methyl-$2$-butanol.Answer:$3$ (Option A).