An alkyl halide with molecular formula \(C_5H_{11}Br\) on dehydrohalogenation gives two isomeric alkenes \(X\) and \(Y\) with formula \(C_5H_{10}\). On reductive ozonolysis \(X\) and \(Y\) gave four compounds \(CH_3CHO\), \(CH_3COCH_3\), \((CH_3)_2CHCHO\) and \(HCHO\). The alkyl halide is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-Bromo-3-methylbutane
Reductive ozonolysis cuts a C=C bond and turns each carbon of it into a C=O group. So the four products can be paired back into the two alkenes:
\( \ce{CH3CHO} \) + \( \ce{CH3COCH3} \) come from X = (CH3)2C=CHCH3 (2-methylbut-2-ene)
\( \ce{(CH3)2C=CHCH3} \xrightarrow[\ce{(ii) Zn/H2O}]{\ce{(i) O3}} \ce{(CH3)2C=O + CH3CHO} \)
\( \ce{(CH3)2CHCHO} \) + HCHO come from Y = (CH3)2CHCH=CH2 (3-methylbut-1-ene)
\( \ce{(CH3)2CHCH=CH2} \xrightarrow[\ce{(ii) Zn/H2O}]{\ce{(i) O3}} \ce{(CH3)2CHCHO + HCHO} \)
Both alkenes must come from one bromide by loss of HBr. The bromide has Br on the carbon between the \( \ce{CH(CH3)2} \) group and the terminal \( \ce{CH3} \), that is CH3CHBrCH(CH3)2:
\( \ce{CH3CHBrCH(CH3)2} \xrightarrow[-\ce{HBr}]{\text{alc. }\ce{KOH}} \underset{\text{X}}{\ce{(CH3)2C=CHCH3}} + \underset{\text{Y}}{\ce{CH2=CHCH(CH3)2}} \)
The alkyl halide is 2-Bromo-3-methylbutane.
NCERT (2026-27 edition):
Elimination Reactions – “… If there is possibility of formation of more than one alkene due to the availability of more than one β-hydrogen atoms, usually one alkene is formed as the major product.”
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 05-06 S1 22-05 S3 | – |