Alkenes — practice questions

Step 1: The H-Br molecule is polar ($H^{\delta+}-Br^{\delta-}$). The electron pair from the π bond of ethene attacks the partially positive hydrogen atom. The H-Br bond breaks heterolytically, with both electrons going to the bromine atom.

Diagram showing curly arrow from C=C bond to H of H-Br, and another curly arrow from the H-Br bond to the Br atom.

This forms an ethyl carbocation ($CH_3CH_2^+$) and a bromide ion ($Br^−$).

Step 2: The bromide ion acts as a nucleophile. A lone pair of electrons on the $Br^−$ ion attacks the positively charged carbon atom of the carbocation.

Diagram showing curly arrow from a lone pair on the $Br^−$ ion to the $C^+$ of the carbocation.

This forms the product, bromoethane ($CH_3CH_2Br$).

But-1-ene is an unsymmetrical alkene: $CH_3CH_2CH=CH_2$. HCl is an unsymmetrical reagent.

There are two possible carbocation intermediates that can be formed when $H^+$ adds across the double bond:

Pathway 1: $H^+$ adds to C1. This forms a secondary carbocation: $CH_3CH_2\overset{+}{C}HCH_3$. This is relatively stable as the positive charge is stabilised by two electron-donating alkyl groups (an ethyl group and a methyl group).

Pathway 2: $H^+$ adds to C2. This forms a primary carbocation: $CH_3CH_2CH_2\overset{+}{C}H_2$. This is less stable as the positive charge is only stabilised by one electron-donating alkyl group (a propyl group).

Since the secondary carbocation is more stable, it is the preferred intermediate. The chloride ion ($Cl^−$) will then attack this carbocation.

Therefore, the major product is 2-chlorobutane ($CH_3CH_2CHClCH_3$). The minor product would be 1-chlorobutane.