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1. Give the structures of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] in the following sequence of reactions:

(a) [tex]\(CH_3COOH \xrightarrow[\Delta]{NH_3} A \xrightarrow{NaOBr} B\)[/tex]

(b)

(c)

Sagot :

GinnyAnswer
Certainly! Let's break down the reaction sequence step by step to determine the structures of compounds [tex]\( A \)[/tex] and [tex]\( B \)[/tex].

### Reaction Sequence
[tex]\[ CH_3COOH \xrightarrow[\Delta]{NH_3} A \xrightarrow{NaOBr} B \][/tex]

1. Step 1: Reaction of acetic acid with ammonia (NH[tex]\(_3\)[/tex])

- Reactant: Acetic acid ([tex]\(CH_3COOH\)[/tex])

When acetic acid is treated with ammonia ([tex]\(NH_3\)[/tex]) under heat ([tex]\(\Delta\)[/tex]), it undergoes an amide formation reaction to yield acetamide ([tex]\(CH_3CONH_2\)[/tex]).

[tex]\[ CH_3COOH + NH_3 \xrightarrow[\Delta]{} CH_3CONH_2 \][/tex]

So, compound [tex]\(A\)[/tex] is acetamide:
[tex]\[ A = CH_3CONH_2 \][/tex]

2. Step 2: Reaction of acetamide with sodium hypobromite (NaOBr)

- Reactant: Acetamide ([tex]\(CH_3CONH_2\)[/tex])

When acetamide ([tex]\(CH_3CONH_2\)[/tex]) is treated with sodium hypobromite (NaOBr), it undergoes the Hofmann rearrangement reaction. This reaction results in the formation of a primary amine. In this case, the product is methylamine ([tex]\(CH_3NH_2\)[/tex]).

[tex]\[ CH_3CONH_2 + NaOBr \xrightarrow{} CH_3NH_2 \][/tex]

So, compound [tex]\(B\)[/tex] is methylamine:
[tex]\[ B = CH_3NH_2 \][/tex]

### Summary of Structures
- Compound [tex]\(A\)[/tex]: [tex]\(CH_3CONH_2\)[/tex] (acetamide)
- Compound [tex]\(B\)[/tex]: [tex]\(CH_3NH_2\)[/tex] (methylamine)

So, the final structures of compounds [tex]\(A\)[/tex] and [tex]\(B\)[/tex] based on the given reactions are:
- [tex]\(A = CH_3CONH_2\)[/tex]
- [tex]\(B = CH_3NH_2\)[/tex]

If there are additional parts (b) and (c) to the question, please provide the rest of the reactions for further assistance.
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