The order of increasing boiling point for the following compounds is
(A) \( CH_3CH_2CH_2CH_3 \)
(B) \( CH_3CH_2CH_2 CH_2OH \)
(C) \( C_2H_5OC_2H_5 \)
(D) \( CH_3CH_2CH_2CHO \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C), (D), (B)
The four compounds have similar molar masses (58–74 g mol−1), so the boiling point is decided by the strength of the intermolecular forces:
Hydrogen bonding (O–H group) is the strongest, so alcohols boil highest.
Dipole–dipole attraction is stronger for the highly polar C=O group of an aldehyde than for the weakly polar C–O–C of an ether.
A hydrocarbon has only weak van der Waals forces.
The compounds:
(A) \( \ce{CH3CH2CH2CH3} \): butane, non-polar, b.p. about 273 K.
(B) \( \ce{CH3CH2CH2CH2OH} \): butan-1-ol, hydrogen bonded, b.p. about 391 K.
(C) \( \ce{C2H5OC2H5} \): ethoxyethane, weakly polar, no O–H, b.p. about 308 K.
(D) \( \ce{CH3CH2CH2CHO} \): butanal, strong dipole–dipole attraction, b.p. about 348 K.
In increasing order of boiling point:
\( \displaystyle \text{butane} \lt \text{ethoxyethane} \lt \text{butanal} \lt \text{butan-1-ol} \)
Shortcut: The hydrocarbon must come first and the alcohol last. Only Option (3) starts with (A) and ends with (B).
NCERT (2026-27 edition):
Physical Properties – “… The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular masses. It is due to weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions. Also, their boiling points are lower than those of alcohols of similar molecular masses due to absence of intermolecular hydrogen bonding.”