Methyl bromide reacts with silver nitrate to form

Nitromethane

The correct answer is option 1. Nitromethane.

When methyl bromide ($CH_3Br$) reacts with silver nitrate ($AgNO_3$), the primary product formed is nitromethane.

The Reaction Mechanism

This reaction is a nucleophilic substitution. In silver nitrate, the nitrate group has two potential sites for bonding (it is an ambident nucleophile): the oxygen atom and the nitrogen atom.

However, silver salts have a strong covalent character. The silver ion ($Ag^+$) has a high affinity for the bromide ion ($Br^-$), which helps facilitate the breaking of the $C-Br$ bond. When the nitrate group attacks the methyl group, it primarily does so through the nitrogen atom, leading to the formation of a nitroalkane.

The balanced chemical equation for the reaction is:

$CH_3Br + AgNO_3 \rightarrow CH_3NO_2 + AgBr \downarrow$

Why the other options are incorrect:

• Methylamine ($CH_3NH_2$): This would require a reaction with ammonia ($NH_3$), not silver nitrate.
• Ethanamine ($C_2H_5NH_2$): This would require a two-carbon starting material and a nitrogen-containing nucleophile.
• Grignard reagent ($RMgX$): This is formed by reacting an alkyl halide with metallic magnesium in dry ether, not with a silver salt.