Consider the following substances
1. Propan -2-ol
2. Propanone
3. 2 - methyl propene
The correct sequence of increasing order of boiling point is

Options

A. `1 gt 2 gt 3`
B. `1 gt 3 gt 2`
C. `2 gt 1 gt 3`
D. `3 gt 2 gt 1`

Answer

To determine the correct sequence of increasing order of boiling points for the substances: Propan-2-ol, Propanone, and 2-methylpropene, we need to analyze the structures and the types of intermolecular forces present in each compound.

### Step-by-step Solution:

1. **Identify the Structures:**
- **Propan-2-ol (C3H8O)**: This is an alcohol with the hydroxyl (–OH) group attached to the second carbon. It has hydrogen bonding due to the presence of the –OH group.
- **Propanone (C3H6O)**: This is a ketone (commonly known as acetone) with a carbonyl (C=O) group. It can also engage in dipole-dipole interactions.
- **2-methylpropene (C4H8)**: This is an alkene with a double bond between two carbons. It is non-polar and primarily experiences Van der Waals forces.

2. **Analyze Intermolecular Forces:**
- **Propan-2-ol**: Exhibits strong hydrogen bonding due to the –OH group, which significantly increases its boiling point.
- **Propanone**: Has dipole-dipole interactions due to the carbonyl group. While it has a higher boiling point than alkenes, it is lower than that of alcohols.
- **2-methylpropene**: Being an alkene, it primarily has weak Van der Waals forces, leading to the lowest boiling point among the three.

3. **Rank the Boiling Points:**
- Since hydrogen bonding is the strongest intermolecular force, Propan-2-ol will have the highest boiling point.
- Propanone, with dipole-dipole interactions, will have a higher boiling point than 2-methylpropene, which only has Van der Waals forces.
- Therefore, the order from lowest to highest boiling point is:
- 2-methylpropene < Propanone < Propan-2-ol

### Conclusion:
The correct sequence of increasing order of boiling points is:
**2-methylpropene < Propanone < Propan-2-ol**