Identify the pair of reactants that upon reaction, with elimination of HCl will give rise to the dipeptide Gly-Ala.
Show Hint
- $\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}$ and $\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}$
- $\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}$ and $\mathrm{NH}_{3}-\mathrm{CH}-\mathrm{COCl}$
- $\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}$ and $\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COCl}$
- $\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}$ and $\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}$
The Correct Option is A
Approach Solution - 1
To determine the correct pair of reactants that can form the dipeptide Gly-Ala with the elimination of HCl, we need to understand a few concepts about peptide formation and reactivity of functional groups:
- A dipeptide is formed by the condensation reaction between two amino acids, typically involving the carboxylic acid group of one amino acid and the amine group of another.
- During peptide bond formation, a molecule of water is typically removed. However, when using acyl chlorides (like acid chlorides) for peptide bond formation, HCl is eliminated instead.
Let's analyze the given options one by one:
- The option \(NH_{2}-CH_{2}-COCl\) and \(NH_{2}-CH-COOH\) is the most promising candidate to form Gly-Ala because one reactant is an acid chloride (which is reactive) and the other is an amine with a free carboxylic acid intact for further reactions.
In the formation reaction:
- The acetyl end (acid chloride part) of Glycine will react with the amine group of Alanine. Formation proceeds with the loss of HCl.
- This matches the type of condensation reaction that connects amino acids to form peptides through peptide bonds.
Hence, the correct pair is: \(NH_{2}-CH_{2}-COCl\) and \(NH_{2}-CH-COOH\).
Let's summarize why other options do not work:
- \(NH_{2}-CH_{2}-COCl\) and \(NH_{3}-CH-COCl\): Both are acid chlorides, and no available amine group is free to bind resulting in no peptide bond formation.
- \(NH_{2}-CH_{2}-COOH\) and \(NH_{2}-CH-COCl\): Here, Glycine has no reactive site because it is a COOH end rather than a COCl; it won’t react spontaneously without activating the COOH end.
- \(NH_{2}-CH_{2}-COOH\) and \(NH_{2}-CH-COOH\): Both counterparts are in free acid form, making them less reactive and unable to spontaneously form a peptide linkage directly.
Therefore, the pair \(NH_{2}-CH_{2}-COCl\) and \(NH_{2}-CH-COOH\) is correct for forming the dipeptide Gly-Ala by losing HCl.
Approach Solution -2
1. Reactants: - $\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}$ (Glycine chloride) - $\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}$ (Alanine)
2. Reaction: - The reaction between these reactants with the elimination of HCl will produce the dipeptide Gly-Ala.
Therefore, the correct answer is (1) $\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}$ and $\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}$.
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