Answer the question on basis of passage given below:

A ketone [A] which undergoes haloform reaction gives compound [B] on reduction. Compound [B] on heating with sulphuric acid gives compound [C] which forms monozonide [D]. The compound [D] on hydrolysis in the presence of zinc dust gives compound [E]. Compound [E] on oxidation gives \(C_2H_4O_2\)[F]

The boiling point of [A] is observed exceptionally high because of:

Dipole-Dipole interaction

The correct answer is option 4. Dipole-Dipole interaction.

From the given data in the passage the various reactions and be depicted as:

The compound A is butan-2-one.

Butan-2-one is a ketone, and its structure consists of a carbonyl group (\(\text{C=O}\)) where oxygen is more electronegative than carbon. This means that oxygen pulls the electron density towards itself, making the oxygen partially negative (\(\delta^-\)) and the carbon partially positive (\(\delta^+\)).

The partial charges create a dipole moment in the molecule, with the oxygen having a slight negative charge and the carbonyl carbon a slight positive charge.

Because of the polarity of the carbonyl group, each butan-2-one molecule can attract neighboring molecules through dipole-dipole interactions. These interactions occur when the positive end of the dipole in one molecule (the carbonyl carbon) is attracted to the negative end of a neighboring molecule (the oxygen of another molecule).

This attraction is stronger than the London dispersion forces (which are present in all molecules), but weaker than hydrogen bonding.

Boiling point is influenced by the strength of intermolecular forces—the stronger these forces, the more energy (in the form of heat) is required to break them and convert the substance from liquid to gas.

In butan-2-one:

The dipole-dipole forces between the polar carbonyl groups require significant energy to overcome. As a result, butan-2-one has a higher boiling point compared to non-polar molecules of similar molecular weight, which only experience weak London dispersion forces

Why Other Forces Are Not Involved:

Hydrogen Bonding (Intermolecular or Intramolecular):  

Butan-2-one cannot form hydrogen bonds because it lacks a hydrogen atom attached to a highly electronegative atom like oxygen or nitrogen, which is necessary for hydrogen bonding. So, options 1 and 2 (Intermolecular or Intramolecular H-bonds) are not responsible for the higher boiling point.

Like Interaction:  

This option is vague. While like interactions could refer to interactions between similar molecules, it doesn't specify a type of force. The primary force here is the dipole-dipole interaction, not just any generic interaction.

Conclusion

The dipole-dipole interactions between the molecules of butan-2-one are responsible for its higher boiling point compared to non-polar molecules of similar size. This is because the partial positive and partial negative charges on the molecules create attractions that need more energy to break during boiling.

Thus, the correct answer is: 4. Dipole-Dipole interaction.