Which of the following will produce isopropyl amine -
(I) `(CH_(3))_(2)CO overset(NH_(2)OH)rarr X overset(LiAlH_(4))rarr`
(II) `CH_(3)-CH_(2)-CHO overset(NH_(3))underset("heat")rarr X overset(LiAlH_(4))rarr`
(III) `(CH_(3))_(2)CH-OH+PCl_(5)rarr X overset(NH_(3))rarr`
(IV) `CH_(3)-CH_(2)-CH_(2)-NH_(2)overset("heat")rarr`

Options

A. I, II
B. II, III
C. I, III
D. IV only

Explanation

To determine which of the given reactions will produce isopropyl amine, let's analyze each option step by step.

### Step 1: Analyze Reaction (I)
The first reaction is:
\[ (CH_3)_2CO \overset{NH_2OH}{\rightarrow} X \overset{LiAlH_4}{\rightarrow} \]

1. **Formation of Hydroxylamine**:
- Acetone \((CH_3)_2CO\) reacts with hydroxylamine \((NH_2OH)\) to form an oxime. The reaction can be represented as:
\[ (CH_3)_2CO + NH_2OH \rightarrow (CH_3)_2C=NOH + H_2O \]
- This compound is called acetone oxime.

2. **Reduction with Lithium Aluminum Hydride**:
- Lithium aluminum hydride \((LiAlH_4)\) is a strong reducing agent that reduces the oxime to the corresponding amine:
\[ (CH_3)_2C=NOH + LiAlH_4 \rightarrow (CH_3)_2CH-NH_2 + H_2O \]
- The product is isopropyl amine \((C_3H_9N)\).

### Conclusion for Reaction (I):
- **Produces Isopropyl Amine**.

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### Step 2: Analyze Reaction (II)
The second reaction is:
\[ CH_3-CH_2-CHO \overset{NH_3}{\text{heat}} \rightarrow X \overset{LiAlH_4}{\rightarrow} \]

1. **Formation of Imine**:
- Propanal \((CH_3-CH_2-CHO)\) reacts with ammonia \((NH_3)\) upon heating to form an imine:
\[ CH_3-CH_2-CHO + NH_3 \rightarrow CH_3-CH_2-CH=NH + H_2O \]

2. **Reduction with Lithium Aluminum Hydride**:
- The imine can then be reduced by lithium aluminum hydride:
\[ CH_3-CH_2-CH=NH + LiAlH_4 \rightarrow CH_3-CH_2-CH_2-NH_2 \]
- The product is n-propyl amine \((C_3H_9N)\), not isopropyl amine.

### Conclusion for Reaction (II):
- **Does not produce Isopropyl Amine**.

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### Step 3: Analyze Reaction (III)
The third reaction is:
\[ (CH_3)_2CHOH + PCl_5 \rightarrow X \overset{NH_3}{\rightarrow} \]

1. **Formation of Alkyl Chloride**:
- The alcohol \((CH_3)_2CHOH\) reacts with phosphorus pentachloride \((PCl_5)\) to form an alkyl chloride:
\[ (CH_3)_2CHOH + PCl_5 \rightarrow (CH_3)_2CCl + POCl_3 + HCl \]

2. **Nucleophilic Attack by Ammonia**:
- The alkyl chloride then reacts with ammonia:
\[ (CH_3)_2CCl + NH_3 \rightarrow (CH_3)_2C-NH_2 + HCl \]
- The product is isopropyl amine \((C_3H_9N)\).

### Conclusion for Reaction (III):
- **Produces Isopropyl Amine**.

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### Step 4: Analyze Reaction (IV)
The fourth reaction is:
\[ CH_3-CH_2-CH_2-NH_2 \overset{\text{heat}}{\rightarrow} \]

1. **Heating Primary Amine**:
- Heating a primary amine like propyl amine does not lead to the formation of isopropyl amine. Instead, it may lead to the elimination of ammonia or other products but not isopropyl amine.

### Conclusion for Reaction (IV):
- **Does not produce Isopropyl Amine**.

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### Final Conclusion:
The reactions that produce isopropyl amine are:
- **(I)** and **(III)**.

### Answer:
**C. (I) and (III)**.

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