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Question 57

The major product of the following reaction is:

Solution

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  1. The Starting Materials
  • Substrate: 4-aminobutan-2-ol. This molecule contains both a secondary alcohol (OH group) and a primary amine (NH2 group).
  • Reagent: Ethyl formate, an ester used here as a formylating agent.
  • Base: Triethylamine, which serves as a mild base to help the reaction along.
  1. Why the Amine Reacts First (Chemoselectivity) When a molecule has both an alcohol and an amine, they compete to attack the reagent. Nitrogen is less electronegative than oxygen, meaning it shares its electrons much more easily. Because of this, the amine is a significantly stronger nucleophile than the alcohol. Since exactly 1 equivalent of ethyl formate is used, the reaction happens exclusively at the amine group, leaving the alcohol completely untouched.
  2. How the Reaction Works First, the lone pair on the primary amine attacks the carbon atom in the carbonyl group (the C=O bond) of the ethyl formate. Next, the ester kicks off its ethoxide leaving group. Finally, that leaving group picks up a spare proton to form ethanol as a byproduct. This specific process replaces a hydrogen on the amine with a formyl group.
  3. Final Product The reaction selectively converts the primary amine into a formamide group, while the alcohol stays exactly the same.
  • Chemical Formula: CH3CH(OH)CH2CH2NHCHO
  • IUPAC Name: N-(3-hydroxybutyl)formamide

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