1-Bromopropane is heated with alcoholic KOH to give compound X. X is treated with HBr to give compound Y. Y is then heated with aqueous KOH to give compound Z. What is Z?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Propan-2-ol
The sequence uses three reactions: β-elimination with alcoholic KOH, addition of HBr by Markovnikov’s rule, and nucleophilic substitution by OH− from aqueous KOH. The net result is that the functional group moves from C-1 to C-2.
Step 1 – 1-Bromopropane → X (β-elimination): alcoholic KOH removes H from the β-carbon and Br from the α-carbon, so X is propene:
\( \displaystyle \ce{CH3CH2CH2Br + KOH} \xrightarrow{\text{alc.},\ \Delta} \ce{CH3CH=CH2 + KBr + H2O} \)
Step 2 – X → Y (addition of HBr): by Markovnikov’s rule, H adds to the CH2 end and Br to the middle carbon, so Y is 2-bromopropane:
\( \displaystyle \ce{CH3CH=CH2 + HBr -> CH3CHBrCH3} \)
Step 3 – Y → Z (substitution): in aqueous KOH, OH− replaces Br, so Z is propan-2-ol:
\( \displaystyle \ce{CH3CHBrCH3 + KOH(aq) -> CH3CH(OH)CH3 + KBr} \)
Why the other options are incorrect
Option (2) Propan-1-ol — Incorrect.
Propan-1-ol would need Y to be 1-bromopropane, that is, Br adding to the CH2 end of propene, which is against Markovnikov’s rule.
Propan-1-ol is what 1-bromopropane itself gives with aqueous KOH.
Option (3) Propene — Incorrect. Propene is X. An alkene is the product with alcoholic KOH; with aqueous KOH the main reaction is substitution, which gives an alcohol.
Option (4) 2-Bromopropane — Incorrect. 2-Bromopropane is Y. Heating it with aqueous KOH replaces the bromine, so it does not remain as the final product.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. As a result, an alkene is formed as a product.”
“…The products formed by the reaction of haloalkanes with some common nucleophiles are given in Table 6.4.”
Table 6.4: Nucleophilic Substitution of Alkyl Halides (R–X)
| Reagent | Nucleophile (Nu–) | Substitution product R–Nu | Class of main product |
|---|---|---|---|
| NaOH (KOH) | HO– | ROH | Alcohol |
| H2O | H2O | ROH | Alcohol |
| NaOR′ | R′O– | ROR′ | Ether |
| NaI | I– | R—I | Alkyl iodide |
| NH3 | NH3 | RNH2 | Primary amine |
| R′NH2 | R′NH2 | RNHR′ | Sec. amine |
| R′R″NH | R′R″NH | RNR′R″ | Tert. amine |
| KCN | :C≡N:– | RCN | Nitrile (cyanide) |
| AgCN | Ag-CN: | RNC (isocyanide) | Isonitrile |
| KNO2 | O=N—O– | R—O—N=O | Alkyl nitrite |
| AgNO2 | Ag—Ö—N=O | R—NO2 | Nitroalkane |
| R′COOAg | R′COO– | R′COOR | Ester |
| LiAlH4 | H | RH | Hydrocarbon |
| R′– M+ | R′– | RR′ | Alkane |
6.4.2 From Hydrocarbons – “(i) Addition of hydrogen halides: An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide.” … “Propene yields two products, however only one predominates as per Markovnikov’s rule.”
\( \displaystyle \ce{CH3CH=CH2 + H-I ->} \underset{\text{minor}}{\ce{CH3CH2CH2I}} + \underset{\text{major}}{\ce{CH3CHICH3}} \)