The boiling point of a compound depends on various factors such as molecular weight, type of bonding, and intermolecular forces. In this case, let’s analyze the given options:
Option (1) CH$_3$CH$_2$CH$_2$CH$_3$ (Butane): This is a hydrocarbon with weak Van der Waals forces. It has a relatively low boiling point due to the absence of hydrogen bonding.
Option (2) CH$_3$CH$_2$CH$_2$CH$_2$OH (Butanol): This compound contains an --OH (hydroxyl) group, which allows for hydrogen bonding between molecules. Hydrogen bonding significantly increases the boiling point compared to compounds with only Van der Waals interactions.
Option (3) CH$_3$CH$_2$CH$_2$CHO (Butanal): This compound has a carbonyl group (C=O), leading to dipole-dipole interactions. However, these interactions are weaker than the hydrogen bonding present in alcohols.
Option (4) C$_2$H$_5$OC$_2$H$_5$ (Diethyl ether): This compound contains an ether linkage, resulting in weak dipole-dipole interactions, but it lacks hydrogen bonding.
Conclusion: Among the given compounds, CH$_3$CH$_2$CH$_2$CH$_2$OH (butanol) has the highest boiling point due to the presence of strong intermolecular hydrogen bonding.
List-I Compound | List-II Product in Basic Medium (in NaOH + Heat) | ||
A | Ethanal | (I) | Benzoic acid + Phenyl methanol |
B | Methanal | (II) | 3-Hydroxybutanal + But-2-enal |
C | Benzenecarbaldehyde | (III) | 4-Hydroxy-4-methylpentan-2-one + 4-Methylpent-3-en-2-one |
D | Acetone | (IV) | Formic acid + Methanol |
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: