Matching peaks to groups

Each spectral observation supports the feature it measures — IR peaks bonds, the molecular ion the mass, M+2 patterns halogens, fragment losses pieces — and a proposed structure must explain every one, so a single contradiction rules it out.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

You will match observations to structural features, tell propanone from propanal, pick out evidence specific to aldehydes, name the technique for each question, weigh a contradiction, and check a formula against a molecular ion.

2. What you already have

You can read functional groups from infrared and masses and fragments from a mass spectrum. This lesson puts each observation to work on the question it can answer.

3. Words for this lesson

An observation is one feature of a spectrum. It supports a structural feature when that feature explains it, and contradicts a structure when the structure cannot explain it.

4. Each observation answers its own question

Observation	Technique	Supports
broad band 3200–3550 cm−1	IR	an O–H
strong, sharp 1680–1750 cm−1	IR	a C=O
weak 2720 and 2820 cm−1	IR	an aldehyde C–H
heaviest peak	MS	the molecular mass
M+2 at a third of M	MS	one chlorine
M+2 equal to M	MS	one bromine
M − 15	MS	a CH3 that can break off

Propanone and propanal are both C3H6O, mass 58, with a C=O. The observations that separate them are the ones only one explains: the aldehyde's C–H peaks near 2720 and 2820 cm−1 and its HCO+ fragment at m/z 29.

Another way: steps

To match evidence to a structure:

Check the molecular ion against the formula's mass.
Match each IR peak to its bond.
Match each large fragment loss to a piece.
Look for the observation that separates the remaining candidates.

5. Contradictions count most

A proposed structure must explain every observation. Three observations that fit butan-1-ol do not outweigh the one that does not: an alcohol must show an O–H band, and if the spectrum has none, the compound is not an alcohol. That is the logic of the next lesson, where several constraints are combined to rule structures out until one remains.

6. Where this goes wrong

A peak supports every functional group equally. Each supports the feature it measures.

More agreeing observations prove a structure. One contradiction rules it out.

A C=O peak separates ketones from aldehydes. Both have one; look for the aldehyde C–H.

Any technique answers any question. IR sees bonds; MS sees mass.

7. Checking ethanenitrile, CH3CN

Formula mass: 24 + 3 + 14 = 41; the molecular ion is at m/z 41.
Mass fits.
A sharp IR peak at 2250 cm−1: the C≡N.
Bond fits.
No O–H and no C=O peaks, as expected.
No contradiction: consistent.

8. Your turn: a proposed C4H8O2 ester

What m/z should the molecular ion have?
48 + 8 + 32 = 88.
Which IR peak should appear?
Your turn: work this step out. Its working is at the end of the packet.
A strong C=O near 1740 cm−1, with no broad O–H.

9. Guided practice

Match each observation to the structural feature it supports.

	an O–H group	a C=O group	one bromine atom	a molecular mass of 74
a broad IR band at 3300 cm−1
a strong, sharp IR peak at 1715 cm−1
M and M+2 peaks of equal height
a molecular ion at m/z 74

10. Guided practice

A compound has its molecular ion at m/z 58 and a strong IR peak near 1720 cm−1, so it is propanone or propanal, both C3H6O. It also shows no peaks near 2720 or 2820 cm−1, and a strong fragment at m/z 43 (loss of CH3). Which is it?

Propanone
Propanal
Either: a C=O peak supports every carbonyl compound equally
Neither: the molecular mass must be wrong

11. Practice

A spectrum set is being used to decide whether a compound is an aldehyde. Mark every observation that points to an aldehyde specifically.

This task has no paper form; do it on a device.

12. Practice

For each question about an unknown compound, name the technique that answers it: infrared or mass spectrometry.

	technique
Is there a C=O?
What is the molecular mass?
Is there an O–H?
Does it contain chlorine?

13. Somewhere new

A student proposes that an unknown is butan-1-ol, C4H10O, molecular mass 74. The spectra show a molecular ion at m/z 74, C–H peaks near 2950 cm−1, a C–O peak near 1100 cm−1, and no band at all above 3000 cm−1 apart from the C–H. What should the student conclude?

It is not butan-1-ol: the missing O–H band rules it out; an ether fits
It is butan-1-ol, since three observations agree
It is butan-1-ol: one peak matching proves a structure
The mass spectrum must be wrong

14. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

15. Test question

A structure is proposed with formula C3H6O. At what m/z should its molecular ion appear? (C = 12, H = 1, O = 16, N = 14.)

The molecular ion should appear at m/z _a.

16. What you can do now

You can match spectral evidence to the structure it supports. Tell someone why one missing peak can rule out a structure. Next: combining every constraint to settle a structure.

Working for the steps left to you

8. Your turn: a proposed C4H8O2 ester, step 3