Question
Why is the boiling point of ethanol (78°C) much higher than that of dimethyl ether (−24°C), even though both have the same molecular formula ?
Solution — Step by Step
Both are 46 g/mol. So London dispersion forces are similar.
Ethanol has an O–H bond; dimethyl ether does not. O–H bonds allow hydrogen bonding between molecules.
Each ethanol molecule can donate 1 H and accept 2 lone pairs, forming an extended H-bond network. Dimethyl ether can only accept (it has no O–H to donate), so no network forms.
Breaking H-bonds requires extra energy (~20 kJ/mol per bond), raising the boiling point by about 100°C compared to ether.
Why This Works
Boiling point is controlled by the strength of intermolecular forces, not by molecular formula. Ethanol and dimethyl ether are isomers with identical mass, but functional group decides what forces operate. Hydrogen bonding is the decisive factor wherever O–H, N–H, or F–H is present.
Reading the question twice before touching the pen saves more time than any shortcut. Underline what is asked — students lose marks by solving the wrong quantity, not by arithmetic errors.
Alternative Method
Compare water (100°C) vs methane (−161°C) — same story. A 16 g/mol molecule with O–H boils 260°C higher than an 18 g/mol molecule without it. H-bonding dominates.
Common Mistake
Writing ‘ethanol has more polar bonds, so higher BP’ is vague and half-right. The precise reason is hydrogen bonding, which requires H on a highly electronegative atom (N, O, F). Polarity alone is not enough — acetone is very polar but boils at 56°C because it cannot donate H-bonds.
Exam Connection
This kind of question is a reliable marks-scorer in CBSE board papers and JEE Main. Examiners reuse the underlying concept with small cosmetic changes — different numbers, a renamed compound, a twist in the wording. If you understand the reasoning here, you will recognise the pattern in any future variant. That is the difference between solving one question and solving a question type.
Teachers often say, “practise till the method becomes boring.” What they mean is: work through enough variations that the decision tree inside your head becomes automatic. By the time you sit the exam, your pen should know the first three steps before your brain consciously thinks about them.
Practice Extension
Try these follow-up variations on your own and see if your method still works:
- Change the starting quantity (e.g. 4.6 g becomes 9.2 g) and confirm that the final answer scales linearly.
- Swap the reagent for a close analogue (e.g. instead of ) and check which steps stay the same and which change.
- Flip the question — given the product, work backwards to find the starting amount. This reverse drill tests whether you really understand the stoichiometry, or whether you were just plugging into a memorised chain.
- Add an impurity (e.g. “the sample is only 80% pure”) and adjust the mole calculation. Purity questions are a classic twist in JEE Main.
In JEE Main, the cut-off between a 95 percentile and a 99 percentile is often just 3 to 4 questions of this exact difficulty. Practising these variations matters more than learning new topics during the last month of preparation.
Quick Recap
- Identify the concept before touching numbers.
- Write units on every line — they catch errors automatically.
- Reach the final answer in a single bold or boxed expression so the examiner sees it instantly.
- When stuck, ask: “what quantity is conserved here?” Conservation of mass, charge, and energy solve most chemistry problems.