H3 Chemistry 9813 · Study focus: H3 Chemistry: Interpret first-order spin–spin splitting and multiplicity
H3 Chemistry: Interpret first-order spin–spin splitting and multiplicity
Start from the governing chemical model, test it against evidence, then transfer the reasoning to an unfamiliar case.
Your success criteria
- Interpret first-order spin–spin splitting and multiplicity
- Use named chemical evidence.
- Transfer the governing reason to an unfamiliar case.
Neighbours split one environment
In CH3CH2Br, CH3 is split by two equivalent CH2 protons into a triplet; CH2 is split by three CH3 protons into a quartet.
Use n+1 only for first-order coupling to n equivalent neighbouring protons under the supplied simple conditions.
Multiplicity and coupling
Spin–spin coupling splits a signal according to neighbouring nuclear spin states.
Singlet, doublet, triplet, quartet and septet describe line count; they do not state integration.
A 3H triplet plus 2H quartet with matching coupling supports CH3CH2. Pascal intensities are 1:2:1 for a triplet and 1:3:3:1 for a quartet.
Equivalent protons do not split one another in this first-order treatment; labile-proton coupling may be absent through exchange.
Text alternative: Text lists each signal centre, line count, relative line intensities, integral and exact neighbouring group.
Read an ethyl pattern
A 3H triplet plus 2H quartet with matching coupling supports CH3CH2. Pascal intensities are 1:2:1 for a triplet and 1:3:3:1 for a quartet.
Equivalent protons do not split one another in this first-order treatment; labile-proton coupling may be absent through exchange.
Assign bromoethane
CH3 has n=2 neighbours, so n+1=3; CH2 has n=3, so four lines.
Combine with integrals 3H and 2H rather than assigning from multiplicity alone.
- Predict both multiplicities in CH3CH2Br.
Open the feedback checkpoint after attempting
- Credit CH3 triplet, CH2 quartet and correct neighbour counts.
Recognise isopropyl
A CH next to six equivalent methyl protons gives a septet; two equivalent CH3 groups next to one CH give a 6H doublet.
Together these form the characteristic isopropyl pattern.
- Assign a 1H septet and 6H doublet.
Open the feedback checkpoint after attempting
- Credit central CH septet and two equivalent methyl groups doublet.
Transfer to propan-2-ol
Ignoring exchange coupling, the two equivalent CH3 groups are split by CH into a doublet; CH is split by six methyl protons into a septet.
OH may appear broad and unsplit under exchange.
- Predict multiplicity and integration for the carbon-bound signals.
Open the feedback checkpoint after attempting
- Credit 6H doublet and 1H septet.
Do not split a group by itself
The three equivalent protons within one CH3 environment do not split one another.
Count equivalent neighbours on adjacent atoms in the stated first-order system.
- Repair: “A CH3 group is always a quartet because it contains three protons.”
Open the feedback checkpoint after attempting
- Multiplicity depends on neighbouring protons; an isolated CH3 can be a singlet.
Combine n+1 with shift and area
Complete twelve fixed splitting checks, then unseen ether assessment and different isopropyl ketone re-test.
Next objective: use TMS and δ at /learning/h3-nmr-tms-delta-scale-lesson.html.
- Annotate the fixed ethyl spectrum with n, n+1, integral and Pascal ratio.
Open the feedback checkpoint after attempting
- Credit all four features separately for triplet and quartet.