JEE MainChemistryHydrocarbons
When 2-bromo-3-methylbutane is heated with alcoholic KOH, it gives a major product A. Product A is then subjected to ozonolysis with O ₃ followed by boiling H ₂ O . The final organic products formed are:
Options
- APropan-2-one and ethanal
- B3-Methylbutanoic acid and carbon dioxide
- C3-Methylbutanal and methanal
- DPropan-2-one and ethanoic acid
Correct answer
D. Propan-2-one and ethanoic acid
Step-by-step solution
The starting material is 2-bromo-3-methylbutane: CH ₃- CH ( Br )- CH ( CH ₃)- CH ₃ . Heating with alcoholic KOH causes dehydrohalogenation (an elimination reaction). According to Zaitsev's rule, the more substituted alkene is formed as the major product. Elimination of HBr yields 2-methylbut-2-ene as the major product A: CH ₃- C ( CH ₃)= CH - CH ₃ Product A is then reacted with O ₃ followed by boiling H ₂ O . The absence of a reducing agent (like Zn or dimethyl sulfide) means this is an oxidative ozonolysis. During