Question

Predict the major product(s) of the following reactions:

(a) phenylacetylene + 2 HBr

(b) hex-1-yne + 2 HCl

(c) cyclooctyne + 2 HBr

(d) hex-2-yne + 2 HCl

Short answer

(a)

(b)

(c)

(d)

Step-by-step solution

Step-by-Step SolutionStep 1: Alkyne Reaction

Alkyne is chemically reactive due to a pi bond between two carbon atoms, which is bonded loosely, and the pi bond of the alkyne is rich with electrons.

Alkyne shows a nucleophilic (triple bond) attack on the reagent, which is electrophilic.

Reaction of alkyne with hydrogen halide

Hydrogen halide, e.g., HCl, HBr, etc. reacts with the triple bond of alkyne; the first product formed is a vinyl halide.

If hydrogen halide is added to the terminal alkyne, the major product is Markovnikov’s predicted orientation (stability of carbocation). The second molecule of hydrogen halide adds the same manner as the first one giving germinal dihalide the major product.

Predicting the major product

When alkyne is reacted with hydrogen halide, the major product formed is the one which is having the stable carbocation intermediate.

(a) When phenylacetylene reacts with two moles of HBr, the major product formed is (1,1-bromomethyl) benzene orientation predicted by Markovnikov’s rules.

Formation of (1,1-bromomethyl) benzene

(b) When hex-1-ene reacts with two moles of HCl, the major product formed is 2,1- dichlorohexane orientation predicted by Markovnikov’s rules.

Formation of 2,2-dichlorohexane

(c) When cyclooctyne (symmetrical molecule) reacts with two moles of HBr, the major product formed is 1,1-dibromocyclooctane.

Formation of 1,1-dibromocyclooctane

(d) When hex-2-ene reacts with two moles of HCl, the acetylene carbon atom are equally substituted; thus, the mixture of major products formed are 2,2-dichlorohexane and 3,3-dichlorohexane.

Formation of 2,2-dichlorohexane and 3,3-dichlorohexane