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Treatment of amide A with LiAlH \(_{4}\), followed by acidic aqueous work-up, gave compound \(\mathrm{B}\). Explain.

Short Answer

Expert verified
Amide A is reduced by LiAlH鈧 to amine B (\text{鈥揅H}_2\text{NH}_2\text{)}.

Step by step solution

01

Understand the Reactants

Identify the compounds involved. Amide A is a compound with the functional group \(\text{鈥揅ONH}_2\). LiAlH鈧 (Lithium Aluminum Hydride) is a strong reducing agent.
02

Reduction by LiAlH鈧

LiAlH鈧 reduces amides to amines. The reaction of LiAlH鈧 with amide A will replace the \(\text{C=O}\) group of the amide with two hydrogen atoms, turning the \(\text{鈥揅ONH}_2\) group into an \(\text{鈥揅H}_2\text{NH}_2\) group.
03

Acidic Work-Up

After the reduction, an acidic aqueous work-up is performed to neutralize the reaction mixture and to protonate any intermediate species, resulting in amine B.
04

Identify Compound B

Based on these steps, compound B is the amine with the structure corresponding to amide A but with the \(\text{鈥揅ONH}_2\) group converted to an \(\text{鈥揅H}_2\text{NH}_2\) group.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amide Reduction
To understand amide reduction, let's start by identifying the functional group. An amide has the structure \(\text{鈥揅ONH}_2\) and is essentially a carbonyl group bonded to a nitrogen atom.
Reduction of an amide alters this structure substantially.
Instead of having a \(\text{C=O}\) double bond, you're left with a simple \(\text{鈥揅H}_2\) group attached to the nitrogen, transforming the amide into an amine.
Amides are relatively stable, making this transformation an interesting and powerful chemical reaction.
One of the most effective ways to achieve this reduction is by using a special reagent called Lithium Aluminum Hydride, or LiAlH鈧.
LiAlH鈧 Reduction
LiAlH鈧 is a potent reducing agent, often used in organic chemistry for reducing various functional groups.
When LiAlH鈧 interacts with an amide, it donates hydride ions (H鈦), which rapidly attack the carbonyl carbon (C=O).
This leads to the breakdown and replacement of the double bond, eventually resulting in the formation of an amine group.
Consider our exercise: Amide A undergoes reduction by LiAlH鈧.
The amide's \(\text{鈥揅ONH}_2\) group is broken down and converted to a \(\text{鈥揅H}_2\text{NH}_2\) group.
We see hydrogen atoms adding where the oxygen used to be, effectively removing the \(\text{C=O}\) double bond and simplifying the overall molecule structure.
This key step transforms the amide into an amine.
Acidic Aqueous Work-Up
Following the reduction stage with LiAlH鈧, an acidic aqueous work-up is essential.
During this step, an acid (like HCl or H鈧係O鈧) is added to the reaction mixture.
The purpose here is to neutralize the chemical environment and finalize the reduction process.
By adding acid, any remaining intermediate compounds are protonated, ensuring that we end up with a stable product.
This step is crucial as it converts any remaining intermediate species into the desired amine form.
So, after the reduction of amide A by LiAlH鈧, the acidic aqueous work-up ensures the correct formation of compound B, a stable amine with the \(\text{鈥揅H}_2\text{NH}_2\) group in place of the original \(\text{鈥揅ONH}_2\).

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Most popular questions from this chapter

Reaction review. Suggest reagents to convert each of the following starting materials into the indicated product: (a) hexanoyl chloride into acetic hexanoic anhydride; (b) methyl hexanoate into \(N\) -methylhexanamide; (c) hexanoyl chloride into hexanal; (d) hexanenitrile into hexanoic acid; (e) hexanamide into hexanamine; (f) hexanamide into pentanamine; (g) ethyl hexanoate into 3-ethyl-3-octanol; (h) hexanenitrile into 1-phenyl-1-hexanone \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}\right]\).

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