- 1(current)
Q. No.Match the bond line structures of hydrocarbons given in List-I with the corresponding boiling points given in List-II.
| List-I (Compound) | List-II (Boiling point in K) | ||
| (a) |  |
(i) | 300.9 |
| (b) |  |
(ii) | 282.5 |
| (c) |  |
(iii) | 309.1 |
| (d) |  |
(iv) | 341.9 |
Choose the correct answer from the options given below:
| (a) | (b) | (c) | (d) | |
| 1. | (i) | (iv) | (iii) | (ii) |
| 2. | (iii) | (i) | (iv) | (ii) |
| 3. | (iii) | (iv) | (i) | (ii) |
| 4. | (iv) | (i) | (ii) | (iii) |
Subtopic: Aliphatic Hydrocarbon- Physical Properties |
83%
Level 1: 80%+
NEET - 2022
Hints
2,2–dimethylpropane, C(CH3)4, is an isomer of pentane, CH3(CH2)3CH3. Pentane has a boiling point of 36 °C whilst the boiling point of 2,2–dimethylpropane is 10 °C.
The statement that explains the difference in the boiling points for these two substances is:
| 1. | The molecules have different relative molecular masses. |
| 2. | Isomers have different chemical properties. |
| 3. | Pentane has permanent dipoles. |
| 4. | Longer chain, lesser branched molecules have stronger spontaneous/induced dipoles. |
Subtopic: Aliphatic Hydrocarbon- Physical Properties |
85%
Level 1: 80%+
Please attempt this question first.
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Please attempt this question first.
- 1(current)
Q. No.
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