The short answer
Turn a molecular formula plus two or three spectral clues into a structure you have never seen before.
Written and checked by GAMSAT tutors — not AI-generated.
Try the reasoning style
We treat forgetting as a failure — a lapse to be patched with reminders and records. Yet a mind that kept everything could not think; it would drown in the undifferentiated noise of every moment it had ever lived. To forget is not so much to lose information as to decide, mostly without our noticing, what was never worth keeping.
The author's argument relies most directly on which unstated assumption?
Pick an option to see how the tutor reasons to the answer — not just whether you were right.
Not quite — the answer is B.
Work backwards from the conclusion: a mind that ‘kept everything’ supposedly ‘could not think.’ That only follows if thinking means leaving most of experience out — so B is the premise the argument quietly rests on. A raises reliability, which the passage never weighs; C contradicts ‘mostly without our noticing’; D smuggles in a claim about intellect the passage never makes. The question rewards finding the hidden premise, not recalling a fact.
Spectroscopy is a logic puzzle. Each instrument answers one question — mass spec, how heavy and which halogen; IR, which functional group; ¹H NMR, how the hydrogens sit — and each answer deletes structures.
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Start with degrees of unsaturation
Take it off the formula
DoU = (2C + 2 + N − H − X)/2, X counting halogens, oxygen ignored.
Read what it licenses
Each degree buys one ring or one π bond; zero means a saturated open chain.
Spend it, then subtract
An IR band near 1700 cm⁻¹ spends one on C=O; if the formula bought only one, nothing is left — no ring, no alkene, whatever the options say.
What ¹H NMR gives you
Signals = environments
One per distinct hydrogen environment: the count measures symmetry.
Integration = ratio, never a headcount
3:2:3 is 3, 2, 3 in an eight-hydrogen molecule but 6, 4, 6 in a sixteen.
Splitting = n + 1 neighbours
n is the hydrogens on the adjacent carbons, never the signal's own.
| What you see | Technique | What it points to |
|---|---|---|
| Very broad band, 2500–3300 cm⁻¹ | IR | O–H of a carboxylic acid |
| Broad band, 3200–3600 cm⁻¹ | IR | O–H of an alcohol |
| Sharper band(s), 3300–3500 cm⁻¹ | IR | N–H (twin peaks = primary amine) |
| Strong band, 1670–1750 cm⁻¹ | IR | C=O, a carbonyl of some kind |
| δ 0.9–1.7 | ¹H NMR | H on a carbon with no electronegative neighbour |
| δ 2.0–2.6 | ¹H NMR | H on the carbon next to a C=O |
| δ 3.3–4.2 | ¹H NMR | H on a carbon bonded to O or a halogen |
| δ 6.5–8.0 | ¹H NMR | H on an aromatic ring |
| δ 9.5–10.1 | ¹H NMR | the H of an aldehyde, —CHO |
| δ 10–13, broad | ¹H NMR | the H of a carboxylic acid (position varies a lot) |
The deduction
An unknown liquid, C₄H₈O₂. IR: strong band at 1740 cm⁻¹, nothing broad from 2500 to 3600 cm⁻¹. ¹H NMR: δ 4.1 (2H, quartet), δ 2.0 (3H, singlet), δ 1.2 (3H, triplet).
The tallest peak is not the molecular ion
That is the base peak, usually a fragment (43 in Figure 1). The molecular ion is the heaviest peak that is not an isotope twin of a lighter one: 78, since 80 is its M+2.
Check yourself
An unknown liquid has molecular formula C₃H₆O. Its IR spectrum shows a strong band at 1715 cm⁻¹ and nothing at all in the 3200–3600 cm⁻¹ region. Its ¹H NMR spectrum is a single signal at δ 2.1. Which structure fits every one of those observations?
Key takeaways
- Take degrees of unsaturation off the formula first; the spectra spend them.
- Twin peaks two apart: ≈3:1 is one Cl, ≈1:1 is one Br.
- Signals = environments, integration = ratio, splitting = n + 1.
- You eliminate structures; you never build one.
Practise this with real GAMSAT-style questions
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