Structure Determination: Worked Problem I

Introduction

This page works through a complete identification from start to finish. An unknown liquid is known to be a simple carbonyl compound. You are given its high-resolution mass spectrum, IR spectrum, ¹³C NMR spectrum and ¹H NMR spectrum. The aim is not just to reach the answer but to show how each piece of evidence narrows the possibilities, and how competing isomers are ruled out one by one.

Core explanation

The data for compound Z.

- High-resolution MS: M⁺ at m/z 86.0732; major fragments at 57 (base peak) and 29. - IR: strong band at 1715 cm⁻¹; no broad band above 3000 cm⁻¹; no bands near 2720 or 2820 cm⁻¹. - ¹³C NMR: three signals, near 212, 35 and 8 ppm. - ¹H NMR: 2.4 ppm (quartet, relative area 2) and 1.05 ppm (triplet, relative area 3).

Stage 1: molecular formula. Candidate formulae with nominal mass 86 include C₅H₁₀O (86.0732), C₄H₆O₂ (86.0368) and C₆H₁₄ (86.1096). Only C₅H₁₀O matches the measured exact mass.

Stage 2: degree of unsaturation. (2 × 5 + 2 − 10) ÷ 2 = 1. There is one ring or one π bond.

Stage 3: IR. The strong band at 1715 cm⁻¹ is a C=O stretch, which uses up the single degree of unsaturation — so there are no rings or C=C bonds. No O–H band rules out alcohols and acids; no bands at 2720 and 2820 cm⁻¹ argue against an aldehyde. Z is most likely a ketone .

Stage 4: listing candidates. The C₅H₁₀O ketones are pentan-2-one, pentan-3-one and 3-methylbutan-2-one. Pentanal and its branched isomers are aldehydes, already disfavoured.

Stage 5: ¹³C NMR. Only three carbon environments are seen for five carbon atoms, so the molecule must be symmetrical. Pentan-2-one has five different carbons, and 3-methylbutan-2-one has four. Pentan-3-one, CH₃CH₂COCH₂CH₃, has exactly three: the C=O (about 212 ppm), two equivalent CH₂ carbons and two equivalent CH₃ carbons.

Stage 6: ¹H NMR. The quartet and triplet in a 2:3 ratio are the signature of equivalent ethyl groups. The CH₂ quartet at 2.4 ppm is shifted downfield because it is next to C=O. There is no singlet near 2.1 ppm, which pentan-2-one and 3-methylbutan-2-one would both show for their CH₃CO group.

Stage 7: mass spectrum. α-Cleavage next to C=O loses an ethyl radical (29) to give the acylium ion C₂H₅CO⁺ at m/z 57, the base peak. C₂H₅⁺ appears at 29. Pentan-2-one would give its base peak at 43 (CH₃CO⁺).

Step-by-step reasoning

1. Exact mass → C₅H₁₀O. 2. DBE = 1. 3. IR → ketone C=O, no O–H, no aldehyde C–H. 4. ¹³C: three signals → symmetrical molecule. 5. ¹H: ethyl quartet/triplet, no methyl singlet. 6. MS: 57 = C₂H₅CO⁺ confirms ethyl groups beside C=O.

Visual explanation

Draw pentan-3-one with a mirror line through the C=O group. The ethyl group on the left is the mirror image of the ethyl group on the right, which is why the five carbons give only three ¹³C signals and the ten hydrogens give only two ¹H signals.

Real-world analogy

This process resembles a detective eliminating suspects. The exact mass lists who was present, IR narrows the suspects to those with a particular trait, and NMR and fragmentation each provide an alibi check that only one suspect fails to escape.

Real-world example

Pentan-3-one is used as a solvent and as a starting material in synthesis. Suppliers confirm its identity and purity by exactly this combination of techniques: a single sharp ¹H NMR pattern of quartet and triplet, the ketone C=O band and a clean GC–MS trace with the expected fragment at m/z 57.

Why?

Why does the ¹³C spectrum decide the problem so quickly? The three candidate ketones differ in symmetry, so they give three, four or five ¹³C signals. Counting signals is simple and unambiguous, making it one of the most powerful ways to separate isomers.

Common misconception

"A band near 1715 cm⁻¹ proves the compound is a ketone." Aldehydes, acids and some esters also absorb near this region. The ketone assignment relies on the absence of O–H and aldehyde C–H bands, and must be confirmed by NMR, where an aldehyde would show a signal near 9.5–10 ppm.

Worked example

Question: Explain how the ¹H NMR spectrum of 3-methylbutan-2-one would differ from that of Z.

Reasoning: 3-Methylbutan-2-one, CH₃COCH(CH₃)₂, has three proton environments: CH₃CO (3H, singlet near 2.1 ppm), CH (1H, septet near 2.6 ppm) and two equivalent CH₃ groups (6H, doublet near 1.1 ppm).

Answer: It would show three signals — a singlet, a septet and a doublet in a 3:1:6 ratio — instead of the single quartet and triplet of Z.

Quick check

1. Why is pentanal unlikely to be Z, based on the IR spectrum alone? Answer: Pentanal would show the two aldehyde C–H stretches near 2720 and 2820 cm⁻¹, which are absent from the IR spectrum of Z.

Exam focus

Structure-determination questions reward a clear audit trail. Label each deduction with the technique and the exact data used, and explicitly eliminate each alternative isomer. Always check that the final structure fits the molecular formula and every spectrum.

Advanced insight

The ¹³C chemical shift of the carbonyl carbon helps separate carbonyl classes: ketones usually appear near 205–220 ppm, aldehydes near 190–205 ppm, and acids, esters and amides at lower values, about 160–185 ppm, because the extra oxygen or nitrogen donates electron density to the carbonyl carbon.

Summary

Compound Z was identified as pentan-3-one. Exact mass fixed C₅H₁₀O, DBE = 1 matched a single C=O, and IR indicated a ketone. Three ¹³C signals showed symmetry, the ¹H quartet and triplet revealed equivalent ethyl groups, and the base peak at m/z 57 confirmed C₂H₅CO⁺. Each alternative ketone was ruled out by specific evidence.

Practice questions

1. How many ¹³C NMR signals would pentan-2-one give? Answer: Five, because all five carbon atoms are in different environments. 2. Which fragment ion would give the base peak for pentan-2-one, and at what m/z? Answer: CH₃CO⁺, the acylium ion, at m/z 43. 3. The exact mass of C₄H₆O₂ is 86.0368. Why can Z not have this formula? Answer: The measured mass of 86.0732 differs from 86.0368 by far more than the instrument's accuracy, so only C₅H₁₀O fits. 4. Explain why the CH₂ protons in pentan-3-one appear at 2.4 ppm rather than about 1.3 ppm. Answer: They are next to the electron-withdrawing C=O group, which deshields them and moves their signal downfield.