JEE Main 2020 · 8 January · morning · Single correct · +4 / −1
Arrange the following compounds in increasing order of C–OH bond length :
methanol, phenol, p-ethoxyphenol
- (a) phenol < methanol < p-ethoxyphenol
- (b) methanol < p-ethoxyphenol < phenol
- (c) methanol < phenol < p-ethoxyphenol
- (d) phenol < p-ethoxyphenol < methanol
Answer and solution
Answer (d) phenol < p-ethoxyphenol < methanol
Solution
Chemistry ma’am · 4 steps
Ask how much double-bond character each C–O bond has: the more it has, the shorter it is.
- Methanol. The C–O bond in CH₃OH is a pure single bond to an sp³ carbon. There is no resonance, so it is the longest of the three, about 142 pm.
- Phenol. A lone pair on oxygen is delocalised into the benzene ring, so the C–O bond gets partial double-bond character. The carbon is also sp², which holds the oxygen closer. The bond shrinks to about 136 pm.
- p-Ethoxyphenol. The -OC₂H₅ group at the para position also donates electrons into the ring by resonance. With the ring already electron-rich, the -OH oxygen delocalises its lone pair less, so its C–O bond has less double-bond character than in phenol. It is still an sp² C–O bond with some resonance, so it stays shorter than in methanol.
- So. The increasing order of C–OH bond length is phenol < p-ethoxyphenol < methanol, which is option (d).
The idea to keep. Bond length falls as double-bond character rises. An electron-donating group on the ring competes with the –OH lone pair, so the C–OH bond gets less double-bond character and becomes longer.
Answer from the NTA final answer key. Solution by PrepWiser.