Question

Neopentyl alcohol on reaction with conc. $\large H_{2} SO_{4}$ forms two alkenes A & B in the ratio 85:15. then alkenes A & B respectively are:?​

Answer 1

Explanation:

Neopentyl alcohol (2,2-dimethylpropan-1-ol) on reaction with conc. H2SO4 forms two Alkene A & B.

The two Alkene formed in the reaction are -

Alkene A = 2-methyl But-2-ene (major) [saytzeff's product]

Alkene B = 2-methyl But-1-ene (minor)

[Hoffman's product]

Hoffman's product -

In this product Hydrogen goes from that β-C, where H is more.

Saytzeff's product -

In this product Hydrogen goes from that β-C, where H is less.

Answer 2

• Both give same major products on treatment with HBr

• Both give different major products on treatment with HBr in presence of peroxide