The increasing order of boiling points for the following compound is:
A. \[{{C}_{2}}{{H}_{5}}OH\]
B. \[{{C}_{2}}{{H}_{5}}Cl\]
C. \[{{C}_{2}}{{H}_{5}}C{{H}_{3}}\]
D. \[{{C}_{2}}{{H}_{5}}OC{{H}_{3}}\]

Hint:
Arranging the compound on the basis of their intermolecular bonding and molecular weight.

Complete step by step solution:
Ethyl Alcohol (\[{{C}_{2}}{{H}_{5}}OH\]) has the highest boiling point due to more extensive intermolecular H-Bonding. Ethyl alcohol because of the presence of polar hydroxyl group forms extensive hydrogen bonding with other ethyl alcohol molecules and forms a much greater bulk than ethyl chloride. So ethyl alcohol requires a higher boiling point in order to break the strong hydrogen bonds. C-O bond is polar & forms hydrogen bonding in \[{{H}_{2}}O\].

Ethyl chloride (\[{{C}_{2}}{{H}_{5}}Cl\]) has a higher molecular weight than ethyl alcohol so it should have had higher boiling point.C-Cl bond is polar but less than that of C-O bond so boiling point of \[{{C}_{2}}{{H}_{5}}OH\] less than that of \[{{C}_{2}}{{H}_{5}}OH\] . Chlorine is more polar than ether, so ethyl chloride has high boiling point than ethyl ether.
C-O bonds are polar but the weak polarity of ethers do not appreciably affect their boiling points and are lower than that of alkane.

The increasing order of order of boiling points:
\[{{C}_{2}}{{H}_{5}}OC{{H}_{3}}\] < \[{{C}_{2}}{{H}_{5}}C{{H}_{3}}\] < \[{{C}_{2}}{{H}_{5}}Cl\] < \[{{C}_{2}}{{H}_{5}}OH\]

Note:
Alkane has the lowest boiling point.ethyl alcohol requires a higher boiling point in order to break the strong hydrogen bonds.