Question

Propose a mechanism for the entire reaction of pent-1-yne with 2 moles of HBr. Show why Markovnikov’s rule should be observed in both the first and second additions of HBr.

Short answer

When a hydrogen halide is added to an asymmetric alkene or alkyne, the acidic hydrogen attaches itself to the carbon having a greater number of hydrogen substituents whereas the halide group is attached to the carbon atom which has a greater number of alkyl substituents or a smaller number of hydrogen atoms.

Step-by-step solution

Markovnikov’s Rule

When a hydrogen halide is added to an asymmetric alkene or alkyne, the acidic hydrogen attaches itself to the carbon having a greater number of hydrogen substituents whereas the halide group is attached to the carbon atom which has a greater number of alkyl substituents or a smaller number of hydrogen atoms.

Mechanism for the reaction of pent-1-yne with 2 moles of HBr.

When a hydrogen halide (HX) adds to a terminal alkyne, the product has the orientation predicted by Markovnikov’s rule.

As seen below, the 2 moles of a hydrogen halide (HBr) when added to an alkyne, the hydrogen atom goes to the carbon with the greater number of hydrogens. The product formed is an alkene.

The second mole is also added with the same orientation as the first. Now the product formed is an alkane.

The addition of HBr to the two pi-bonds of an alkyne goes two times as shown below.