When methyl iodide is heated with ammonia, the product obtained is

The correct option is D. All of the above.

When Ammonia is treated with alkyl halides undergo a nucleophilic substitution reaction. Since Iodine is a good leaving group, an amino group ($-N{H}_2$) replaces the halogen atom. The primary amine formed acts as a nucleophile and again reacts with an alkyl halide to give secondary and tertiary amine. On reacting with an excess alkyl halide, a quaternary ammonium salt is obtained.

$C{H}_{3}I\stackrel{N{H}_3}{⟶}C{H}_{3}N{H}_2\stackrel{C{H}_{3}I}{⟶}(C{H}_3{)}_{2}NH\stackrel{C{H}_{3}I}{⟶}(C{H}_3{)}_{3}N$