Question

reaction of propene with acetic amalgam in presence of NABH4​

Answer 1

Answer:

When propene reacts with acetic amalgam in presence of NaBH4, propan-1-ol is formed.

Mechanism:

1. CH(triple bond)CH-CH3 + Hg(OAc)2, H2O → CH(OH)CH(HgOAc)CH3

(OH gets attached to first Carbon and HgOAc gets attached to second Carbon)

2. CH(OH)CH(HgOAc)CH3 + NaBH4 → CH(OH)-CH2-CH3

[NaBH4, a reducing agent, reduces the intermediate to Propan-1-ol]

Hg(OAc)2 - Acetic amalgam

CH(OH)CH(HgOAc)CH3 - Organomercury intermediate

CH(OH)-CH2-CH3 - Propan-1-ol

This reaction is called Oxymercuration reaction where alkenes get reduced to alcohol.