The product (C) in the below mentioned reaction is : CH3-CH2-CH2-Br [ ]KOH(alc) A HBr B [KOH(aq)] C
- Propan-1-ol
- Propyne
- Propan-2-ol
- Propene
Answer
(3) Propan-2-ol
Solution
CH3-CH2-CH2-Br [ ]KOH(alc) (A)CH3-CH=CH2
∵ alc. KOH ⇒ dehydrohalogenation (β-elimination)
CH3-CH=CH2 HBr (B)CH3- Br |CH-CH3
∵ Markovnikov addition: H+ adds to =CH2 ⇒ more stable 2° carbocation
CH3- Br |CH-CH3 [KOH(aq)] (C)CH3- OH |CH-CH3
∵ aq. KOH ⇒ substitution of Br by OH ⇒ C = propan-2-ol
Answer: (3)