This question is about 2-bromopropane - AQA - A-Level Chemistry - Question 5 - 2020 - Paper 2

The C–Br bond in 2-bromopropane is polar due to the difference in electronegativity between carbon and bromine. Bromine is more electronegative than carbon, which means that it attracts the shared pair of electrons more strongly. This results in a partial negative charge (δ-) on the bromine atom and a partial positive charge (δ+) on the carbon atom.

The reaction mechanism proceeds as follows:

1. A nucleophile (NH₃) attacks the carbon atom bonded to bromine, forming a new bond.
2. A curly arrow shows the movement of an electron pair from the C–Br bond to the bromine, resulting in the loss of Br as Br⁻.
3. This leads to the formation of a cationic intermediate where the nitrogen now has a positive charge, due to it forming three bonds.
4. A proton (H⁺) is lost from the nitrogen to restore neutrality, resulting in the formation of the amine product.

The skeletal formula for the organic product formed in this reaction is:

CH₃-CH(NH₂)-CH₃

This represents isopropylamine, which is the main organic species produced when ammonia reacts with 2-bromopropane.

The organic product, isopropylamine, is commonly used as a (hair) conditioner, a cationic surfactant, or as a fabric softener.