Provide the products of the following reactions: 13) CH3Br + NH3 -> 14) (CH3CH2)2NH + CH3Br + NH3 ->

Okay, let's solve these organic chemistry reactions step by step.

Reaction 13:

CH₃Br + NH₃ → CH₃NH₂ + HBr

The reaction of methyl bromide (CH₃Br) with ammonia (NH₃) results in the formation of methylamine (CH₃NH₂) and hydrogen bromide (HBr). The ammonia acts as a nucleophile, attacking the carbon atom in methyl bromide and displacing the bromide ion.

Here's a more detailed view using the principle of nucleophilic substitution:

CH₃Br + NH₃ → CH₃NH₂ + HBr

Reaction 14:

(CH₃CH₂)₂NH + CH₃Br + NH₃ → (CH₃CH₂)₂N(CH₃) + NH₄Br

The reaction of diethylamine ((CH₃CH₂)₂NH) with methyl bromide (CH₃Br) in the presence of ammonia (NH₃) results in the formation of diethylmethylamine ((CH₃CH₂)₂N(CH₃)) and ammonium bromide (NH₄Br). Diethylamine reacts with methyl bromide to form a tertiary amine, and the ammonia acts as a base to neutralize the HBr formed, leading to ammonium bromide.

Here's a breakdown:

(CH₃CH₂)₂NH + CH₃Br + NH₃ → (CH₃CH₂)₂N(CH₃) + NH₄Br

Final Answer:

13) CH₃Br + NH₃ → CH₃NH₂ + HBr

14) (CH₃CH₂)₂NH + CH₃Br + NH₃ → (CH₃CH₂)₂N(CH₃) + NH₄Br