Reaction of Propanamide with Bromine and Sodium Hydroxide

Propanamide is an organic compound with the chemical formula CH3CH2CONH2. When it is treated with bromine in an aqueous solution of sodium hydroxide, a substitution reaction takes place. The reaction can be represented as:

CH3CH2CONH2 + Br2 + NaOH → CH3CH2NH2 + NaBr + NaHCO3

Explanation of the Reaction and Products

The reaction involves the substitution of the amide group (-CONH2) with a bromine atom (-Br) in the presence of a strong base (NaOH) and water. The reaction proceeds through the following steps:

Step 1: Formation of a Bromoamide Intermediate

The bromine molecule (Br2) is polar and can interact with the electron-rich amide group of propanamide. As a result, the bromine molecule is polarized, and a bromoamide intermediate is formed.

CH3CH2CONH2 + Br2 → CH3CH2C(NHBr)Br

Step 2: Hydrolysis of the Bromoamide Intermediate

The bromoamide intermediate is then hydrolyzed by the aqueous sodium hydroxide solution. The strong base (NaOH) reacts with the bromoamide, causing the amide group to break down. The nitrogen atom of the amide group is protonated, and the resulting intermediate is attacked by hydroxide ion (OH-) from the NaOH solution. This results in the formation of an amine (CH3CH2NH2) and sodium bromide (NaBr).

CH3CH2C(NHBr)Br + NaOH → CH3CH2NH2 + NaBr + H2O

Step 3: Formation of Sodium Bicarbonate

The reaction between the strong base (NaOH) and the acidic hydrogen bromide (HBr) produced in step 2 results in the formation of sodium bicarbonate (NaHCO3).

NaOH + HBr → NaBr + H2O
NaHCO3 + NaOH → Na2CO3 + H2O

Conclusion

Therefore, the correct option is C, Ethanamine, as propanamide is converted into ethanamine upon treatment with bromine in an aqueous solution of sodium hydroxide.