Question

Draw the alkene formed when 1-heptyne is treated with one equivalent of HBr

Answer 1

Concepts and reason

Hydrogen halide in addition to the alkyne leads to form an alkenyl halide product. This addition reaction follows the Markovnikov’s addition path way.

Fundamentals

Markovnikov’s addition: Markovnikov’s addition of hydrogen halide to alkyne produces the most stable alkene.

Markovnikov’s rule says that negatively charged addendum part in hydrogen halide goes to the carbon position, which is connected to less hydrogen atoms.

Given alkyne structure is drawn below.

addition reaction to the hept-1-yne is shown below.

In this addition reaction, nucleophile is (bromide ion).

A nucleophilic attack on the intermediate is shown below.

The overall reaction is as follows.

Ans:

The produced alkene during the addition reaction is shown below:

Answer 2

Answer 3

Concepts and reason

The concepts used in this question are based on Markovnikov’s rule and the electrophilic addition reaction.

Initially, the mechanism of the addition of one equivalent of hydrogen bromide $\left( {{\rm{HBr}}} \right)$ to $1 - {\rm{heptyne}}$ is shown by using the concepts of Markovnikov’s rule and the electrophilic addition reaction. Then, the product formed on the addition of one equivalent of hydrogen bromide to $1 - {\rm{heptyne}}$ is drawn.

Fundamentals

Markovnikov’s Rule: The halide of ${\rm{H}} - {\rm{X}}$ attaches with the more substituted carbon and hydrogen attaches with carbon having more number of hydrogen of a double or triple bond of alkenes or alkynes.

Electrophilic addition reaction: A chemical reaction in which an electrophile is added on an electron rich carbon atom to form a product is termed as an electrophilic addition reaction. In this reaction, a double bond is broken and two new sigma bonds are formed.

The electrophilic addition reaction of alkynes with a hydrogen halide is illustrated as follows.

The mechanism of the addition of one equivalent of hydrogen bromide (HBr) to $1 - {\rm{heptyne}}$ is shown below.

The addition of one equivalent of hydrogen bromide to $1 - {\rm{heptyne}}$ gives $2 - {\rm{bromohept}} - 1 - {\rm{ene}}$ which is drawn below.

Ans:

The product formed on the addition of one equivalent of hydrogen bromide (HBr) to ${\bf{1 - heptyne}}$ is shown as follows.