Question

Show how a mixture of n-butylamines prepared from n-butyl bromide and an excess of n-butylamine might be resolved into its components by reaction with succinic anhydride, separation of the products through advantage of their solubility properties in acid or base, and regeneration of the corresponding amines. Write equations for the reactions involved.

Short answer

A mixture of n-butylamines prepared from n-butyl bromide and an excess of n-butylamine can be resolved into components by reacting with succinic anhydride to give N-(n-butyl)-n-butylsuccinamic acid. The products can then be separated by exploiting their solubility properties in acidic and basic solutions. The N-(n-butyl)-n-butylsuccinamic acid can be regenerated back to secondary amine, and the butylammonium bromide can be regenerated back to n-butylamine via their reactions with a base. The overall process involves the following reactions: 1. \(C_4H_9Br + 2 C_4H_9NH_2 \rightarrow C_4H_9NH_2C_4H_9Br + C_4H_9NH_3Br\) 2. \(C_4H_9NH_2C_4H_9Br + (CH_2CO)_2O \rightarrow C_4H_9NH(C_4H_9Br)(CH_2COOH) + CH_3COOH\) 3. \(C_4H_9NH(C_4H_9Br)(CH_2COOH)_{(s)} + H^+_{(aq)} \rightarrow C_4H_9NH(C_4H_9Br)(CH_2COOH)H^+_{(s)}\) 4. \(C_4H_9NH(C_4H_9Br)(CH_2COOH)H^+_{(s)} + OH^-_{(aq)} \rightarrow C_4H_9NH(C_4H_9Br)(CH_2COO^-)_{(aq)} + H_2O_{(l)}\) 5. \(C_4H_9NH(C_4H_9Br)(CH_2COOH) + KOH \rightarrow C_4H_9NH_2C_4H_9Br + KCH_2COO\) 6. \(C_4H_9NH_3Br + KOH \rightarrow C_4H_9NH_2 + KBr + H_2O\)

Step-by-step solution

Reaction of n-butyl bromide and n-butylamine

First, let's write the reaction between n-butyl bromide and an excess of n-butylamine: \[ C_4H_9Br + 2 C_4H_9NH_2 \rightarrow C_4H_9NH_2C_4H_9Br + C_4H_9NH_3Br \] In this reaction, n-butyl bromide reacts with an excess of n-butylamine to give a mixture of secondary amine (N-butylethanamine) and a quaternary ammonium salt (butylammonium bromide).

Reaction with succinic anhydride

Now, let's add succinic anhydride to the reaction mixture: \[ C_4H_9NH_2C_4H_9Br + (CH_2CO)_2O \rightarrow C_4H_9NH(C_4H_9Br)(CH_2COOH) + CH_3COOH \] The secondary amine reacts with succinic anhydride to form an N-(n-butyl)-n-butylsuccinamic acid and acetic acid.

Separation of products based on solubility properties

After obtaining the N-(n-butyl)-n-butylsuccinamic acid, we can separate the products based on their solubility properties in acids or bases. When an acidic solution is added, the butylammonium bromide remains soluble whereas the N-(n-butyl)-n-butylsuccinamic acid becomes insoluble. Thus, by adding an acidic solution, we can separate N-(n-butyl)-n-butylsuccinamic acid as a precipitate: \[ C_4H_9NH(C_4H_9Br)(CH_2COOH)_{(s)} + H^+_{(aq)} \rightarrow C_4H_9NH(C_4H_9Br)(CH_2COOH)H^+_{(s)} \] Upon adding a basic solution, the N-(n-butyl)-n-butylsuccinamic acid becomes soluble again, allowing for its extraction: \[ C_4H_9NH(C_4H_9Br)(CH_2COOH)H^+_{(s)} + OH^-_{(aq)} \rightarrow C_4H_9NH(C_4H_9Br)(CH_2COO^-)_{(aq)} + H_2O_{(l)} \]

Regeneration of the corresponding amines

Finally, we will regenerate the corresponding amines from the separated compounds. First, let's regenerate the secondary amine from N-(n-butyl)-n-butylsuccinamic acid: \[ C_4H_9NH(C_4H_9Br)(CH_2COOH) + KOH \rightarrow C_4H_9NH_2C_4H_9Br + KCH_2COO \] Then, let's regenerate the n-butylamine from butylammonium bromide by addition of a base: \[ C_4H_9NH_3Br + KOH \rightarrow C_4H_9NH_2 + KBr + H_2O \] With these four steps, we have successfully separated the components of the n-butylamine mixture, reacted the mixture with succinic anhydride, separated the products based on their solubility properties, and regenerated the corresponding amines.

Key concepts

Reaction with Succinic Anhydride

The reaction between n-butylamine and succinic anhydride is a key step in the separation of n-butylamine mixtures. This reaction involves the formation of an amide linkage. When n-butylamine comes into contact with succinic anhydride, the anhydride's carbonyl carbon is attacked by the amine. This results in the opening of the anhydride ring and the release of acetic acid. Consequently, an N-(n-butyl)-n-butylsuccinamic acid is formed.
This transformation is particularly useful because it modifies the amine, making it amenable to further separation based on solubility. Not only does it play a crucial role in separation, but it is also an exemplary reaction of amine chemistry in organic synthesis. In essence, this reaction exploits the reactivity of amines with anhydrides to introduce a new functionality that aids in the separation process.

Solubility Properties in Acid and Base

The solubility properties of the compounds formed in the reaction are pivotal. When we talk about solubility in acids or bases, it refers to how a substance dissolves depending on the pH of the solution. This property is leveraged to separate the N-(n-butyl)-n-butylsuccinamic acid from the reaction mixture.
In acidic conditions, the N-(n-butyl)-n-butylsuccinamic acid becomes less soluble. This is because it forms a precipitate, which makes it easier to filter out and separate from soluble butylammonium bromide. On the other hand, in basic conditions, the N-(n-butyl)-n-butylsuccinamic acid becomes more soluble again. The basic solution deprotonates the acid, rendering it soluble in the aqueous phase. These shifts in solubility are crucial for isolating the desired compound.

Regeneration of Amines

Once the separation is complete, the next step involves regenerating the original amines from the separated products. Regeneration is a chemical process that reverts a compound back to its original form. For N-(n-butyl)-n-butylsuccinamic acid, this involves treating it with a base such as potassium hydroxide (KOH). This treatment releases the secondary amine and yields a corresponding salt.
Similarly, the quaternary ammonium salt (butylammonium bromide) can be treated with a base to regenerate n-butylamine. The reaction with base enables the release of the amine, while the anionic part of the salt is left behind. These steps are essential for reclaiming the individual amines that were initially part of the mixture.

Organic Chemistry Reactions

Understanding organic chemistry reactions is central to the separation and regeneration process described. Reactions such as nucleophilic substitution, amide formation, and acid-base interactions are all at play in this exercise. Each step of the process illustrates fundamental concepts of reactivity and acid-base chemistry.
The initial formation of the amide through the reaction with succinic anhydride demonstrates how electrophilic carbon centers can be attacked by nucleophiles like amines. The solubility-driven separation relies on manipulation of pH conditions to alter ionic states and solubility profiles. Finally, the regeneration of amines through basic hydrolysis underscores the utility of reversible chemical reactions. Each reaction type showcases a different aspect of organic reactivity and illustrates the interconnectedness of chemical principles to achieve desired outcomes.