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.
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Answer
The zwitterion of phenylalanine includes both a positively charged amino group and a negatively charged carboxylate group. The structure can be represented as follows:
H
|
H2N-C-COO⁻
|
C₆H₅
This structure illustrates that the amino group is protonated while the carboxylic group is deprotonated.
Step 2
Draw the structure of serine at high pH.
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Answer
At high pH, serine possesses a carboxylate ion and a neutral amino group. Its structure can be depicted as:
H
|
H2N-C-COO⁻
|
CH₂
|
OH
This representation shows that the amino group remains protonated while the carboxyl group is deprotonated.
Step 3
Draw the structures of both dipeptides formed when phenylalanine reacts with serine.
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Answer
The two dipeptides formed when phenylalanine reacts with serine are:
Phenylalanine-Serine (Phe-Ser):
H2N-C-CO-NH-C-COO⁻
| |
C₆H₅ CH₂
Serine-Phenylalanine (Ser-Phe):
H
|
H2N-C-CO-NH-C-COO⁻
| |
CH₂ C₆H₅
In both structures, it is essential to show the amide bond (–CO-NH–).
Step 4
Name and outline a mechanism for the reaction between CH₃CH₂COCl and CH₃CH₂NH₂.
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Answer
Name of Mechanism
Nucleophilic addition-elimination.
Mechanism Steps:
Nucleophilic Attack:
The lone pair on the nitrogen atom of the amine attacks the carbonyl carbon of the acyl chloride, resulting in a tetrahedral intermediate.
Formation of the Tetrahedral Intermediate:
This temporary structure has a negative charge on the oxygen atom that can expel the chloride ion.
Elimination of Chloride Ion:
The intermediate collapses to form the amide bond while releasing Cl⁻, leading to the formation of the product.
The resulting product can be named as N-ethylpropanamide.