Treating an amide with excess LiAlH4, followed by water, will produce a(n) ___________________.Select answer from the options belowalcoholaminecarboxylic acidesterenaminecarbinolamineSave for LaterSubmit Answer
Question
Treating an amide with excess LiAlH4, followed by water, will produce a(n) ___________________.Select answer from the options belowalcoholaminecarboxylic acidesterenaminecarbinolamineSave for LaterSubmit Answer
Solution
The reaction of an amide with excess Lithium aluminium hydride (LiAlH4), followed by water, will produce an amine.
Here's the step-by-step process:
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LiAlH4 is a strong reducing agent. When an amide comes into contact with LiAlH4, the LiAlH4 will donate a hydride (H-) ion to the carbonyl carbon of the amide.
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This reduces the carbonyl group to an alcohol group, and breaks the pi bond, leaving a negatively charged oxygen atom and a positively charged nitrogen atom.
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The nitrogen atom then donates a pair of electrons to form a bond with a hydrogen atom, which is donated by water in the second step of the reaction.
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The oxygen atom also bonds with a hydrogen atom from water, forming an alcohol group.
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However, the alcohol group is not stable in this situation and leaves as a water molecule, resulting in the formation of an amine.
So, the answer is "amine".
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