This question is about amines. 1.1. Give an equation for the preparation of 1,6-diaminohexane by the reaction of 1,6-dibromohexane with an excess of ammonia. 1.2. Complete the mechanism for the reaction of ammonia with 6-bromohexylamine to form 1,6-diaminohexane. Suggest the structure of a cyclic secondary amine that can be formed as a by-product in this reaction. Mechanism NH₃ Br- Cyclic secondary amine 1.3. 1,6-Diaminohexane can also be formed in a two-stage synthesis starting from 1,4-dibromobutane. Suggest the reagent and a condition for each stage in this alternative synthesis. Stage 1 reagent and condition Stage 2 reagent and condition 1.4. Explain why 3-aminopentane is a stronger base than ammonia. 1.5. Justify the statement that there are no chiral centres in 3-aminopentane.

A-Level AQA Chemistry Introduction to Organic Chemistry: This question is about amines.

This question is about amines - AQA - A-Level Chemistry - Question 1 - 2019 - Paper 2

Question 1

This question is about amines. 1.1. Give an equation for the preparation of 1,6-diaminohexane by the reaction of 1,6-dibromohexane with an excess of ammonia. 1.2. ... show full transcript

Worked Solution & Example Answer:This question is about amines - AQA - A-Level Chemistry - Question 1 - 2019 - Paper 2

Step 1

Give an equation for the preparation of 1,6-diaminohexane by the reaction of 1,6-dibromohexane with an excess of ammonia.

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Answer

The reaction can be represented as follows:

$ext{Br-CH}_2 ext{-CH}_2 ext{-CH}_2 ext{-CH}_2 ext{-CH}_2 ext{-Br} + 2 ext{NH}_3 \rightarrow ext{NH}_2 ext{-CH}_2 ext{-CH}_2 ext{-CH}_2 ext{-CH}_2 ext{-CH}_2 ext{-NH}_2 + 2 ext{HBr}$

Step 2

Complete the mechanism for the reaction of ammonia with 6-bromohexylamine to form 1,6-diaminohexane.

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Answer

The mechanism starts with the nucleophilic attack of ammonia (NH₃) on the carbon bonded to the 6-bromo group, leading to the formation of a tetrahedral intermediate. This intermediate subsequently loses a bromide ion, resulting in the formation of 1,6-diaminohexane. The cyclic secondary amine that could be formed as a by-product is:

$ext{Cyclic Secondary Amine:} \ ext{N}_2 ext{-CH}_2 ext{C}_5H_{10}$

Step 3

Stage 1 reagent and condition

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Answer

Reagent: KCN (Potassium cyanide) Condition: Aqueous solution with heating.

Step 4

Stage 2 reagent and condition

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Answer

Reagent: Lithium aluminum hydride (LiAlH₄) Condition: Dry ether as solvent under reflux.

Step 5

Explain why 3-aminopentane is a stronger base than ammonia.

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Answer

3-aminopentane has an alkyl group that donates electron density to the nitrogen atom, increasing its basicity compared to ammonia. This electron-donating effect enhances the availability of the lone pair on nitrogen for protonation, making 3-aminopentane a stronger base.

Step 6

Justify the statement that there are no chiral centres in 3-aminopentane.

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Answer

In 3-aminopentane, the nitrogen atom is bonded to two hydrogen atoms and only one carbon chain (the rest are hydrogen atoms), resulting in no carbon centers attached to four different substituents. Therefore, there are no chiral centers in the molecule.

This question is about amines - AQA - A-Level Chemistry - Question 1 - 2019 - Paper 2

Worked Solution & Example Answer:This question is about amines - AQA - A-Level Chemistry - Question 1 - 2019 - Paper 2

Give an equation for the preparation of 1,6-diaminohexane by the reaction of 1,6-dibromohexane with an excess of ammonia.

Complete the mechanism for the reaction of ammonia with 6-bromohexylamine to form 1,6-diaminohexane.

Stage 1 reagent and condition

Stage 2 reagent and condition

Explain why 3-aminopentane is a stronger base than ammonia.

Justify the statement that there are no chiral centres in 3-aminopentane.

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