The solubility in carbonyl compound with water is due to hydrogenbonding.

Given below are two statements:
Statement $\mathrm{I}$: The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole - dipole interactions.
Statement $\mathrm{II}$: The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of $\mathrm{H}$-bonding.

In the light of the above statements, choose the most appropriate answer from the options given below :

How would you account for:

Boiling points of aldehydes are lower than alcohols.

Arrange the following compounds in increasing order of their boiling points: ${\mathrm{CH}}_3-\mathrm{CHO}, {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}, {\mathrm{CH}}_3-\mathrm{O}-{\mathrm{CH}}_3, {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3$

Which of the compound is soluble in ${\mathrm{H}}_{\mathrm{2}}\mathrm{O}$?

The odour from vanilla extract comes from the molecule vanillin.

Decreasing order of boiling point of $\mathrm{I}$ to $\mathrm{IV}$ follow:

Arrange the following compounds in the increasing order of their boiling points.

Acetone, n-propyl alcohol, ethyl methyl ether and n-butane.

Decreasing order of boiling point of $\mathrm{I}$ to $\mathrm{IV}$ follows:

At room temperature formaldehyde is