There are 6 amide isomers that exist for C4H9ON without forming amine on treatment with Br2-NaOH.

Explanation:

To understand this, let's break down the question into smaller parts:

- What is an amide?
An amide is a compound that contains a carbonyl group (C=O) bonded to a nitrogen atom (N).

- What is an isomer?
Isomers are compounds that have the same molecular formula but different arrangements of atoms.

- What is Br2-NaOH?
Br2-NaOH is a reagent used for the detection of primary and secondary amines. It reacts with amines to form a yellow precipitate.

Now, let's apply this information to the question.

- C4H9ON is a compound that contains a carbonyl group (C=O) bonded to a nitrogen atom (N).

- The question states that the compound should not form an amine on treatment with Br2-NaOH. This means that the compound should not contain any primary or secondary amines.

- To find the number of isomers that meet this criteria, we need to consider all possible arrangements of atoms for C4H9ON that do not contain primary or secondary amines.

- There are 6 such isomers that exist for C4H9ON:

1. N-methylacetamide
2. N,N-dimethylformamide
3. N-ethylacetamide
4. N,N-diethylformamide
5. N-methyl-N-propionamide
6. N-ethyl-N-propionamide

- These isomers do not contain primary or secondary amines and, therefore, will not react with Br2-NaOH to form a yellow precipitate.

In conclusion, there are 6 amide isomers that exist for C4H9ON that do not form amine on treatment with Br2-NaOH.

U need to see that only 1 degree amines with c=0 group will effect with x2+naoh
so none of them reacts