Answer :
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.
### 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.
