CBSE Class 12 Chemistry: Marking Scheme Solution

Question: Write equations involved in the following reactions: (i) Ethanamine with acetyl chloride, (ii) Aniline with bromine water, (iii) Aniline with chloroform and ethanolic KOH.

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Step 1: Reaction (i) - Acetylation of Ethanamine [1 Mark]

Ethanamine reacts with acetyl chloride in the presence of a base (like pyridine) to form N-ethylethanamide.

$$CH_3CH_2NH_2 + CH_3COCl \xrightarrow{Pyridine} CH_3CH_2NHCOCH_3 + HCl$$

Step 2: Reaction (ii) - Bromination of Aniline [1 Mark]

Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline due to the high activating effect of the $-NH_2$ group.

$$C_6H_5NH_2 + 3Br_2(aq) \rightarrow C_6H_2(Br)_3NH_2 + 3HBr$$

(Note: The product is 2,4,6-tribromoaniline)

Step 3: Reaction (iii) - Carbylamine Reaction [1 Mark]

Aniline reacts with chloroform ($CHCl_3$) and ethanolic potassium hydroxide ($KOH$) to form phenyl isocyanide (carbylamine), which has a foul smell.

$$C_6H_5NH_2 + CHCl_3 + 3KOH(alc.) \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O$$

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💡 EL10 Concept Insight:

• Acetylation: Think of this as "swapping" a hydrogen atom for an acetyl group to make a stable amide.
• Bromination: The $-NH_2$ group is like a "super-charger" that pushes bromine atoms into every available corner of the benzene ring instantly.
• Carbylamine: This is a "stinky test"—only primary amines can react with chloroform to create that unmistakable, awful smell of isocyanide!