JEE Main 2020 · 5 September · evening · Single correct · +4 / −1
The increasing order of boiling points of the following compounds is :
- (a) I < III < IV < II
- (b) IV < I < II < III
- (c) I < IV < III < II
- (d) III < I < II < IV
Answer and solution
Answer (c) I < IV < III < II
Solution
Chemistry ma’am · 3 steps
All four have the same phenolic OH, so let the para group decide how strongly the molecules hold together.
- Same OH everywhere. Each compound is a para-substituted phenol, so each has one O–H that can hydrogen bond. The boiling point order therefore depends on what the para group adds: polarity, extra hydrogen bonding and mass.
Compound Para group What it adds Rank I -CH₃ Nonpolar, no hydrogen bonding, lightest (M = 108) lowest IV -OCH₃ Polar C–O bond and more mass (M = 124), but no extra N–H or O–H donor second III -NH₂ Two extra N–H donors plus a lone pair, so many more hydrogen bonds third II -NO₂ Very polar group with good O acceptors. It also makes the O–H more acidic, so the O–H forms stronger hydrogen bonds (M = 139) highest - Why nitro wins. In p-nitrophenol the electron-withdrawing -NO₂ pulls charge away from the O–H, which makes it a strong hydrogen bond donor. The nitro oxygens of the next molecule are good acceptors, so the molecules link into strong intermolecular chains, and the large dipole adds to this.
- So. The increasing order is I < IV < III < II, which is option (c).
The idea to keep. When the parent group is the same, rank boiling points by what the substituent adds: extra hydrogen bonding and strong polarity count far more than a little extra mass.
Answer from the NTA final answer key. Solution by PrepWiser.