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Use the Data Booklet to help you answer this question - AQA - A-Level Chemistry - Question 6 - 2017 - Paper 2
Question 6
Use the Data Booklet to help you answer this question. This question is about amino acids and peptide (amide) links. 1. Draw the structure of the zwitterion formed... show full transcript
Worked Solution & Example Answer:Use the Data Booklet to help you answer this question - AQA - A-Level Chemistry - Question 6 - 2017 - Paper 2
Step 1
Draw the structure of the zwitterion formed by phenylalanine.
Answer
The zwitterion of phenylalanine features a positively charged ammonium group (–NH₃⁺) and a negatively charged carboxylate group (–COO⁻). The structure can be illustrated as follows:
HOOC
|
H₂N–C–C₆H₅
|
CH₂
In this structure, the phenyl group (C₆H₅) is connected to the alpha carbon (C), which is bonded to both the amino group and the carboxyl group.
Step 2
Step 3
Draw the structures of both dipeptides formed when phenylalanine reacts with serine.
Answer
When phenylalanine reacts with serine, two dipeptides can be formed based on the order of linkage:
- Phenylalanine-Serine:
HOOC
|
H₂N–C–C₆H₅
|
C–OH
|
CH₂
- Serine-Phenylalanine:
HOOC
|
H₂N–C–CH₂
|
C=C₆H₅
|
OH
In both structures, the amide link arises from the connection between the carbon of the phenylalanine and the nitrogen of the serine.
Step 4
Name and outline a mechanism for the reaction between CH₃CH₂COCl and CH₃CH₂NH₂.
Answer
When an acyl chloride (CH₃CH₂COCl) reacts with a primary amine (CH₃CH₂NH₂), an amide is formed through a nucleophilic addition-elimination mechanism. The steps are as follows:
- Nucleophilic attack: The amine attacks the carbonyl carbon of the acyl chloride.
- Tetrahedral intermediate formation: A tetrahedral intermediate is formed.
- Elimination of HCl: The intermediate collapses to eliminate HCl, forming the amide.
The overall reaction can be summarized as:
CH₃CH₂COCl + CH₃CH₂NH₂ → CH₃CH₂CONHCH₃CH₂ + HCl
IUPAC Name of Organic Product: Ethyl propanamide.