Reactants of reaction -I are : `CH_(3)CONH_(2),KOH,Br_(2)`
Reactants of reaction-II are : `CH_(3)NH_(2), CHCl_(3), KOH`
The intermediate species of reaction-I and reaction-II are
respectively.

Options

A. Carbonium ion, carbene
B. Alkyl isocyanate, carbene
C. Carbene, nitrene
D. Carbocation, carbanion

Answer

To determine the intermediate species of the two reactions given, we will analyze each reaction step by step.

### Reaction I:
**Reactants:** CH3CONH2 (Acetamide), KOH, Br2

1. **Identify the Reaction Type:**
The reaction involves acetamide (CH3CONH2) and bromine (Br2) in the presence of a strong base (KOH). This is known as the Hofmann bromamide reaction.

2. **Mechanism Overview:**
- The hydroxide ion (OH-) from KOH deprotonates the amide nitrogen, forming an anion.
- This anion then attacks Br2, leading to the formation of an N-bromoamide intermediate.

3. **Formation of Intermediate:**
- The intermediate formed in this reaction is N-bromoacetamide (CH3C(O)NHBr).

### Reaction II:
**Reactants:** CH3NH2 (Methylamine), CHCl3 (Chloroform), KOH

1. **Identify the Reaction Type:**
This reaction involves methylamine and chloroform in the presence of KOH, which is known as the carbylamine reaction.

2. **Mechanism Overview:**
- KOH acts as a base, generating hydroxide ions (OH-).
- The hydroxide ion reacts with chloroform (CHCl3) to form a trichloromethyl anion (CCl3-).
- This anion then reacts with the methylamine to form an intermediate.

3. **Formation of Intermediate:**
- The intermediate formed in this reaction is methyl isocyanide (CH3N=C=O).

### Summary of Intermediates:
- **Intermediate of Reaction I:** N-bromoacetamide (CH3C(O)NHBr)
- **Intermediate of Reaction II:** Methyl isocyanide (CH3N=C=O)

### Final Answer:
The intermediate species of reaction-I and reaction-II are respectively:
1. N-bromoacetamide (CH3C(O)NHBr)
2. Methyl isocyanide (CH3N=C=O)

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