The reaction of Propanamide with Bromine in aqueous NaOH is a Hoffmann Bromamide reaction. This reaction results in the formation of an amine. Here are the steps:

1. Propanamide (CH3CH2CONH2) reacts with bromine (Br2) in an aqueous solution of sodium hydroxide (NaOH).

2. The bromine is used to substitute the amide group (-CONH2) in the propanamide molecule.

3. The reaction undergoes rearrangement, leading to the loss of a molecule of carbon dioxide (CO2).

4. The final product is Propanamine (CH3CH2NH2).

So, Propanamide on reaction with bromine in aqueous NaOH gives Propanamine.

Question

The reaction of Propanamide with Bromine in aqueous NaOH is a Hoffmann Bromamide reaction. This reaction results in the formation of an amine. Here are the steps:

1. Propanamide (CH3CH2CONH2) reacts with bromine (Br2) in an aqueous solution of sodium hydroxide (NaOH).

2. The bromine is used to substitute the amide group (-CONH2) in the propanamide molecule.

3. The reaction undergoes rearrangement, leading to the loss of a molecule of carbon dioxide (CO2).

4. The final product is Propanamine (CH3CH2NH2).

So, Propanamide on reaction with bromine in aqueous NaOH gives Propanamine.

Knowee AI · Accepted Answer

The reaction of Propanamide with Bromine in aqueous NaOH is a Hoffmann Bromamide reaction. This reaction results in the formation of an amine. Here are the steps:

1. Propanamide (CH3CH2CONH2) reacts with bromine (Br2) in an aqueous solution of sodium hydroxide (NaOH).

2. The bromine is used to substitute the amide group (-CONH2) in the propanamide molecule.

3. The reaction undergoes rearrangement, leading to the loss of a molecule of carbon dioxide (CO2).

4. The final product is Propanamine (CH3CH2NH2).

So, Propanamide on reaction with bromine in aqueous NaOH gives Propanamine.

Propanamide on reaction with bromine in aqueous NaOH gives

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