JEE Chemistry Alcohols Phenols and Ethers MCQs Set 33

Answer: (b) 1-Bromopropane

Explanation:
1. Alcoholic KOH eliminates HBr and gives propene (A).
2. In the presence of peroxide, HBr adds against Markovnikov's rule, so Br goes to the terminal carbon.
3. B is 1-bromopropane.
4. Answer: (b)

Teacher's Note:
1. Check which functional group reacts first, then consider rearrangements and protecting groups.

Answer: (b) CH₃CH(OCH₃)CH₂C≡CCH₃

Explanation:
1. The acetylide attacks the less hindered CH₂ carbon of the epoxide and opens the ring.
2. The alkoxide formed is then methylated by CH₃I (Williamson synthesis).
3. The alkyne is retained.
4. Answer: (b)

Teacher's Note:
1. Check which functional group reacts first, then consider rearrangements and protecting groups.

Answer: (d) CH₃CH=CH₂, CH₃CH₂CH₂Br, CH₃CH₂CH₂OCH₃

Explanation:
1. Elimination gives propene (X).
2. Peroxide effect gives 1-bromopropane (Y).
3. CH₃ONa then substitutes Br in a Williamson synthesis, giving methyl propyl ether (Z).
4. Answer: (d)

Teacher's Note:
1. Check which functional group reacts first, then consider rearrangements and protecting groups.

Answer: (d) C₆H₅OH and CH₃COCH₃

Explanation:
1. Air oxidation gives cumene hydroperoxide.
2. Acid-catalysed rearrangement, with migration of phenyl to oxygen, gives phenol and acetone (the cumene process).
3. Answer: (d)

Teacher's Note:
1. Check which functional group reacts first, then consider rearrangements and protecting groups.

Answer: (b) Benzene

Explanation:
1. Loss of water from the cyclohexadienol gives a ring with three double bonds.
2. The product is aromatic benzene, which is the driving force for the elimination.
3. Answer: (b)

Teacher's Note:
1. Check which functional group reacts first, then consider rearrangements and protecting groups.