Structure Determination: Worked Problem II

Introduction

Isomers are the ultimate test of spectral interpretation, because they share a molecular formula and molecular mass. A molecular ion alone cannot tell them apart. This page takes the three isomers of C₃H₈O — propan-1-ol, propan-2-ol and methoxyethane — predicts how each technique responds to them, and then uses a set of data to identify an unknown. The aim is to learn which technique answers which question most efficiently.

Core explanation

The three isomers.

- Propan-1-ol, CH₃CH₂CH₂OH (primary alcohol) - Propan-2-ol, (CH₃)₂CHOH (secondary alcohol) - Methoxyethane, CH₃OCH₂CH₃ (ether)

All have Mᵣ 60 and DBE 0, so there are no rings or multiple bonds in any of them.

IR: alcohol or ether? Both alcohols show a broad O–H stretch at about 3200–3550 cm⁻¹, broadened by hydrogen bonding. Methoxyethane has no O–H band; it shows only C–H stretches just below 3000 cm⁻¹ and a strong C–O stretch near 1100 cm⁻¹. IR therefore separates the ether from the alcohols at once, but it cannot reliably distinguish the two alcohols.

Mass spectrometry: primary or secondary? α-Cleavage gives different oxonium ions. Propan-1-ol loses C₂H₅· to give CH₂=OH⁺ at m/z 31 (base peak). Propan-2-ol loses CH₃· to give CH₃CH=OH⁺ at m/z 45 (base peak). Methoxyethane loses CH₃· from its ethyl group to give CH₃O=CH₂⁺, also at m/z 45, so mass spectrometry alone can confuse the ether with propan-2-ol.

¹³C NMR: symmetry. Propan-1-ol has three carbon environments and methoxyethane also has three, but propan-2-ol has only two , because its two methyl groups are equivalent.

¹H NMR: the most complete picture.

Compound Predicted signals --- --- Propan-1-ol 0.9 (t, 3H), 1.6 (sextet, 2H), 3.6 (t, 2H), OH (broad s, 1H) Propan-2-ol 1.2 (d, 6H), 4.0 (septet, 1H), OH (broad s, 1H) Methoxyethane 1.2 (t, 3H), 3.3 (s, 3H), 3.4 (q, 2H)

The O–H signal has a variable position and normally appears as a broad singlet that disappears on a D₂O shake. Methoxyethane has no such signal, and its singlet at 3.3 ppm (OCH₃) is highly distinctive.

Step-by-step reasoning

A quick decision route:

1. IR: broad O–H? If no, the compound is methoxyethane. 2. If yes, count ¹³C signals: two means propan-2-ol, three means propan-1-ol. 3. Confirm with the mass spectrum base peak: 31 for propan-1-ol, 45 for propan-2-ol. 4. Check the ¹H NMR splitting pattern against the table.

Visual explanation

Draw the three isomers side by side and circle equivalent groups in the same colour. Propan-2-ol has two methyl groups in one colour and the CH in another — only two carbon colours in total. Propan-1-ol and methoxyethane each need three colours, which is why they give three ¹³C signals.

Real-world analogy

Distinguishing isomers is like identifying three people wearing identical uniforms. Height (Mᵣ) is the same for all of them, so you need other features — voice, walk, handwriting — each of which separates some but not all. Using several features together identifies everyone.

Real-world example

Propan-2-ol is widely used as a cleaning solvent and disinfectant, while propan-1-ol is used in some hand rubs and printing inks. Manufacturers must confirm which isomer is present because their properties and regulatory limits differ; a quick ¹H NMR or GC–MS check separates them reliably.

Why?

Why do the two alcohols give different base peaks? α-Cleavage breaks the C–C bond next to the C–OH carbon. In propan-1-ol, that carbon carries two hydrogens, so the ion formed is CH₂=OH⁺ (31). In propan-2-ol, it carries a methyl group and a hydrogen, so the ion is CH₃CH=OH⁺ (45).

Common misconception

"The OH proton in propan-1-ol will be a triplet because it is next to a CH₂." In most samples the OH proton exchanges rapidly between molecules, so coupling is averaged out and it appears as a broad singlet. Likewise, the neighbouring CH₂ is not split by the OH proton.

Worked example

Question: Unknown W, C₃H₈O, shows a broad IR band at 3350 cm⁻¹. Its ¹³C NMR has three signals, and its mass spectrum has its base peak at m/z 31. The ¹H NMR shows a triplet (3H), a sextet (2H), a triplet (2H) at 3.6 ppm and a broad singlet (1H) that disappears with D₂O. Identify W.

Reasoning: The broad O–H band rules out methoxyethane. Three ¹³C signals rule out propan-2-ol. The base peak at 31 (CH₂=OH⁺) indicates a primary alcohol. The ¹H pattern matches CH₃CH₂CH₂–, with the CH₂ next to oxygen at 3.6 ppm.

Answer: Propan-1-ol, CH₃CH₂CH₂OH.

Quick check

1. Which single technique most quickly distinguishes methoxyethane from both alcohols, and why? Answer: IR spectroscopy, because only the alcohols show the broad O–H stretching band near 3200–3550 cm⁻¹.

Exam focus

When asked to distinguish isomers, give a specific difference in the data for each technique you use — for example "propan-2-ol gives two ¹³C signals, propan-1-ol gives three". Vague statements such as "the spectra will be different" earn no credit.

Advanced insight

In very pure, dry samples or in solvents such as DMSO-d₆, OH exchange is slowed and coupling to the OH proton can be observed: the OH of propan-1-ol then appears as a triplet. This is why the appearance of exchangeable protons depends strongly on the sample conditions and should be interpreted with care.

Summary

The isomers of C₃H₈O share Mᵣ 60 and DBE 0. IR separates the ether from the alcohols using the O–H band; ¹³C NMR separates propan-2-ol (two signals) from propan-1-ol (three); mass spectrometry gives base peaks at 31 (propan-1-ol) and 45 (propan-2-ol and the ether); ¹H NMR splitting patterns confirm each structure.

Practice questions

1. How many ¹H NMR signals would methoxyethane show, and what are their splitting patterns? Answer: Three: a singlet (OCH₃), a quartet (OCH₂) and a triplet (CH₃). 2. Explain why propan-2-ol gives a doublet with an integration of 6. Answer: The two equivalent CH₃ groups contain six protons, each set coupled to the single CH proton, so n+1 = 2 gives a doublet. 3. Why can the mass spectrum alone fail to distinguish methoxyethane from propan-2-ol? Answer: Both lose a CH₃ radical to form an ion at m/z 45, so their base peaks can coincide. 4. What is observed in the ¹H NMR spectrum of propan-2-ol after a D₂O shake? Answer: The broad O–H signal disappears because the OH hydrogen is replaced by deuterium, which does not give a ¹H signal.