Question

Briefly explain the following experimental results

d. The optical rotation of (+)-2-iodooctane slowly goes to zero when heated in water. e. Treating 1-bromopropane with potassium t-butoxide gives mainly substitution while 2-bromopropane gives only elimination.

Answer 1

Answer to d:

(+)-2-iodo octane when heated in water slowly turns into a racemic mixture with equal concentration of dextro and levo rotatory products. The racemization takes place because of SN¹ reaction. During hydrolysis of (+)-2-iodo octane, a carbocation is formed which is a sp2 hybridized planar molecule. the attack of nucleophile then takes place from either side resulting in a mixture of products. Due to this the optical activity of the product is reduced to zero.

Нc— нс- -он нс

Answer to e:

1-bromopropane is a primary alkyl halide with no steric hindrances and undergoes substitution reaction easily.

In case of 2 bromopropane, two beta hydrogens are available for elimination reaction. Potassium t- butoxide is a strong base and also a bulkier nucleophile, hence it abstracts a proton instead of attacking the carbon atom for substitution of Br group.

KOt-Bu - H₃C7 Br 1-bromopropane Butanol H307 -OH propan-1-ol HBr KOt-Bu • H₃CT CH2 - H 2-bromopropane prop-1-ene