Arrange the following in order of increase//decrease in boiling
point.
`CH_(3)CH_(2)underset(I)(CH_(2))CH_(2)CH_(3),` `(CH_(3))_(2)CHunderset(II)(CH_(2))CH_(3),` `(CH_(3))_(4)underset(III)C`

Options

A. `IgtIIgtIII`
B. `IIgtIgtIII`
C. `IIIgtIgtII`
D. `IIIltIIltI`

Answer

To solve the problem of arranging the given alkanes in order of increasing or decreasing boiling points, we will follow these steps:

### Step 1: Identify the Structures
We have three compounds to analyze:
1. **(I)** n-Pentane: CH₃CH₂CH₂CH₂CH₃
2. **(II)** Isopentane: (CH₃)₂CH(CH₂)CH₃
3. **(III)** Neopentane: (CH₃)₄C

### Step 2: Determine the Type of Isomer
- **n-Pentane** is a straight-chain alkane.
- **Isopentane** is a branched-chain alkane with one branch.
- **Neopentane** is a more highly branched alkane with four methyl groups attached to a central carbon.

### Step 3: Analyze the Boiling Points
Boiling points in alkanes are influenced by molecular structure:
- Straight-chain alkanes generally have higher boiling points than their branched isomers due to greater surface area and stronger Van der Waals forces.
- As branching increases, the boiling point decreases because the surface area available for intermolecular interactions decreases.

### Step 4: Compare the Boiling Points
- **n-Pentane (I)**: Highest boiling point due to its straight-chain structure.
- **Isopentane (II)**: Lower boiling point than n-pentane due to one branching.
- **Neopentane (III)**: Lowest boiling point due to the highest degree of branching.

### Step 5: Arrange in Order
- **Decreasing order of boiling points**:
1. n-Pentane (I) > Isopentane (II) > Neopentane (III)

- **Increasing order of boiling points**:
1. Neopentane (III) < Isopentane (II) < n-Pentane (I)

### Final Answer
- **Decreasing order**: n-Pentane (I) > Isopentane (II) > Neopentane (III)
- **Increasing order**: Neopentane (III) < Isopentane (II) < n-Pentane (I)

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