Amine Reactions - Online Tutor, Practice Problems & Exam Prep

Amines, acting as weak bases, undergo acid-base reactions by accepting a proton (H⁺) from an acid, resulting in the formation of an ammonium ion. This process is well-illustrated by the reaction between methylamine and hydrochloric acid. According to the Brønsted-Lowry definition, the acid (hydrochloric acid) donates a proton to the base (methylamine), transforming it into a positively charged ammonium ion, specifically methylammonium ion. This reaction highlights the change in the amine's structure as the nitrogen atom now forms four bonds, carrying a positive charge, indicative of the ammonium ion's formation.

Acid-Base Reaction

In this video we're going to take a look at a mean reactions through the prism of acid based reactions. Now under this type of reaction, an acid reacts with an amine, which represents a weak base. Now recall when a base accepts an H+, it transforms into a conjugate acid. This conjugate acid of an amine is called the ammonium ion.

Now here we're going to say for the ammonium ion we modify the ending of a mean, which is the neutral form of our mean to ammonium ion. So if we take a look here at this generic amino reaction, we have methylamine as our starting amine. It's reacting with hydrochloric acid, which is an acid. We know, following the Bronsted-Lowry definition, that the acid donate to an H+ to the base.

So when it accepts that H+, it's now accepting H+. So now nitrogen is making four bonds. So it's going to be positively charged. And we're going to say when a nitrogen is possibly charged by making four bonds, that's when the moniker of ammonium ion comes into play. So we're going from methylamine to methyl ammonia ion. So this would be the name of our newly created ammonium ion product.

Amine Reactions Example

Predictive products formed from the following reaction between nitric acid and ammonia. So nitric acid is our acid, Ammonia is our neutral mean. So it's a weak base. Here we're using the Bronsted-lowry definition. We know that the acid donates in H. So when that acid donates an H plus, it forms a conjugate base.

So it's going to form the nitrate ion and then ammonia gains in H plus to become NH₄⁺, so it becomes the ammonium ion. In addition to this, notice that there is a solid arrow going forward because we have a strong species mixing with a weak species. Nitric acid is a strong acid mixing with ammonia which is a weak base.

When a strong species acid or base is involved, we have complete formation of our products. So we have a solid arrow going forward, OK. But in this particular example here, we're going to say our nitrate ion and ammonium ions are our products created between the reaction of nitric acid and ammonia.

Determine the ammonium ion formed in the following reaction between diethylamine and hydrobromic acid.

• A.
• B.
• C.
• D.

Here’s what students ask on this topic:

Why is it so difficult to control the alkylation of primary amines in an SN2-type reaction manifold? What is the best answer?

Alkylation of primary amines in an SN2-type reaction is difficult to control due to the high reactivity of amines. Primary amines are nucleophilic, meaning they readily donate a pair of electrons to an electrophile. When a primary amine is alkylated, it transforms into a secondary amine, which is still nucleophilic and can further react to form a tertiary amine, and so on, often leading to a mixture of products with varying degrees of alkylation.

Moreover, the alkylated amines are often stronger bases than the starting primary amine. As the basicity increases, so does the nucleophilicity, which enhances the rate of further alkylation. This phenomenon is known as the "alkylation cascade" or "over-alkylation."

To control the alkylation of primary amines, chemists often use protecting groups to temporarily mask the amine functionality, steric hindrance to limit the access of the electrophile, or employ more selective reagents and reaction conditions that favor the formation of only the desired monoalkylated product.