Chloroform and ethanolic KOH is used as a reagent in the following reaction:
(a) Hoffmann carbylamine reaction
(b) Hoffmann degradation reaction
(c) Reimer-Tiemann reaction
(d) Hoffmann mustard oil reaction
Code is:

Options

A. Only for (a)
B. Only for (a) and (b)
C. Only for (b) and (d)
D. Only for (a ) and (c)

Answer

To solve the question, we need to identify which reaction utilizes chloroform (CHCl3) and ethanolic KOH as reagents. Let's analyze each option step by step.

### Step 1: Analyze the Hoffmann Carbylamine Reaction
- In the Hoffmann carbylamine reaction, a primary amine reacts with chloroform (CHCl3) in the presence of alcoholic KOH.
- The product of this reaction is an isocyanide (R-N≡C).
- **Conclusion**: This reaction uses chloroform and ethanolic KOH.

### Step 2: Analyze the Hoffmann Degradation Reaction
- The Hoffmann degradation reaction involves the conversion of an amide (R-CO-NH2) to a primary amine (R-NH2) using bromine (Br2) and sodium hydroxide (NaOH).
- **Conclusion**: This reaction does not use chloroform or ethanolic KOH.

### Step 3: Analyze the Reimer-Tiemann Reaction
- The Reimer-Tiemann reaction involves the reaction of phenol with chloroform (CHCl3) in the presence of a strong base like NaOH.
- The product of this reaction is typically ortho-hydroxybenzaldehyde.
- **Conclusion**: This reaction uses chloroform but does not specifically use ethanolic KOH.

### Step 4: Analyze the Hoffmann Mustard Oil Reaction
- In the Hoffmann mustard oil reaction, a primary amine reacts with carbon disulfide (CS2) to produce thiourea derivatives, not involving chloroform or ethanolic KOH.
- **Conclusion**: This reaction does not use chloroform or ethanolic KOH.

### Final Conclusion
From the analysis:
- The Hoffmann carbylamine reaction (option a) is the only reaction that uses both chloroform and ethanolic KOH.
- The Reimer-Tiemann reaction (option c) uses chloroform but not ethanolic KOH.
- The Hoffmann degradation (option b) and Hoffmann mustard oil reactions (option d) do not use either reagent.

Thus, the correct answer is **(a) Hoffmann carbylamine reaction**.

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