CUET · CHEMISTRY · PYQ PAPER 2023
An alkyl halide with molecular formula \(\text{C}_5\text{H}_{11}\text{Br}\) on dehydrohalogenation give two isomeric alkenes X and Y with formula \(\text{C}_5\text{H}_{10}\). On reductive ozonolysis X and Y gave four compounds: \(\text{CH}_3\text{CHO}\) (ethanal), \(\text{CH}_3\text{COCH}_3\) (propanone), \((\text{CH}_3)_2\text{CHCHO}\) (3-methylbutanal), and \(\text{HCHO}\) (methanal). The alkyl halide is:
- A 3-Bromopentane
- B 2-Bromo-3-methylbutane
- C 2-Bromo-2-methylbutane
- D 1-Bromo-2,2-dimethylpropane
Answer & Solution
Correct Answer
(B) 2-Bromo-3-methylbutane
Step-by-step Solution
Detailed explanation
\(\text{Alkenes X, Y from ozonolysis products:}\) \(\text{CH}_3\text{CHO} + \text{CH}_3\text{COCH}_3 \rightarrow \text{CH}_3\text{CH=C(CH}_3\text{)}_2\) (2-methylbut-2-ene) \((\text{CH}_3)_2\text{CHCHO} + \text{HCHO} \rightarrow (\text{CH}_3)_2\text{CHCH=CH}_2\)…
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