The predominant intermolecular forces present in ethyl acetate, a liquid, are:

We begin by recalling that in any molecular liquid three main types of intermolecular forces are possible—London dispersion forces, dipole–dipole interactions, and hydrogen bonding.

First, London dispersion forces arise from momentary fluctuations in electron distribution that create instantaneous dipoles. These forces act between all molecules, whether polar or non-polar. Hence every liquid, including ethyl acetate, necessarily possesses London dispersion forces.

Next, we must check for a permanent molecular dipole. Ethyl acetate has the molecular formula $$\mathrm{CH_3COOCH_2CH_3}$$. The carbonyl group $$\mathrm{C=O}$$ is strongly polarized because oxygen is more electronegative than carbon. Therefore the electron cloud is pulled toward the oxygen, creating a permanent partial negative charge ($$\delta^-$$) on the oxygen and a corresponding partial positive charge ($$\delta^+$$) on the carbon. This permanent charge separation means the molecule is polar. A polar molecule experiences dipole–dipole interactions with its neighbours. Thus dipole–dipole forces are also present in ethyl acetate.

Finally, we test the conditions for hydrogen bonding. The rule is that a hydrogen bond can form only when a hydrogen atom is covalently bonded to a highly electronegative atom, specifically $$\mathrm{F}$$, $$\mathrm{O}$$, or $$\mathrm{N}$$. In ethyl acetate every hydrogen is bonded to carbon, not to fluorine, oxygen, or nitrogen. Therefore ethyl acetate molecules cannot serve as hydrogen donors, and true intermolecular hydrogen bonding does not occur between them.

Collecting all our observations:

$$\text{Present forces} = \text{London dispersion} + \text{dipole–dipole}$$

$$\text{Absent force} = \text{hydrogen bonding}$$

Among the given choices, only Option A lists exactly London dispersion and dipole–dipole interactions.

Hence, the correct answer is Option A.