H3 Chemistry 9813 · Study focus: H3 Chemistry: Use peak integration to count protons

H3 Chemistry: Use peak integration to count protons

Start from the governing chemical model, test it against evidence, then transfer the reasoning to an unfamiliar case.

Your success criteria

  • Use peak integration to count protons
  • Use named chemical evidence.
  • Transfer the governing reason to an unfamiliar case.
Diagnose this objective

Area gives a ratio

A C4H8O spectrum has integral areas 30, 20 and 30 units. Divide by 10 to obtain 3:2:3, totalling eight protons.

Integration is area under a signal/multiplet, not its tallest line.

Explore this H3 topic and lesson sequence.

Integral and scaling

Integrated area is proportional to the number of equivalent protons contributing to a signal.

Convert areas to the simplest ratio, then scale to a known total or candidate molecular formula.

  1. For areas 1.5:1.0:1.5, divide by 0.5 to get 3:2:3. Confirm the sum equals the eight hydrogens available.

  2. Exchangeable OH/NH integrals can be unreliable; use supplied data and other evidence.

H3 Chemistry: Use peak integration to count protons: move from the evidence or givens, through the governing Chemistry idea, to a conclusion that stays inside the selected course boundary.
H3 Chemistry: Use peak integration to count protons evidence representation. Fixed raw-area and normalised-count columns make proportional scaling auditable.
Integration-area scaling
Caseδ / ppmMeasured areaCommon divisorNormalised ratioAssigned HChecksum
C₄H₈O2.13010333 + 2 + 3 = 8 H
C₄H₈O2.52010228 H total
C₄H₈O13010338 H total
area 1:2; total 61:2sum=3; scale ×21:22 H:4 H6 H
area 4.8:3.2:1.64.8:3.2:1.61.63:2:13 H:2 H:1 H6 H

Text alternative: Every bar length is duplicated by raw numeric area, divisor, normalised ratio and proton count.

Use totals as a checksum

For areas 1.5:1.0:1.5, divide by 0.5 to get 3:2:3. Confirm the sum equals the eight hydrogens available.

Exchangeable OH/NH integrals can be unreliable; use supplied data and other evidence.

Scale 1:2 to six H

Let counts be k and 2k. Since 3k=6, k=2.

The two environments contain 2H and 4H.

  • Scale an area ratio 1:2 to a six-proton total.
Open the feedback checkpoint after attempting
  • Credit algebra or proportional reasoning to 2H:4H.

Read an ethyl ester

Integrals 3:2:3 match acyl CH3, OCH2 and terminal CH3 in ethyl ethanoate.

Use chemical shift and splitting to decide which 3H is which.

  • Assign only the proton counts from 3:2:3.
Open the feedback checkpoint after attempting
  • Credit 3H, 2H and 3H; do not use area to claim neighbours.

Start the diagnostic and follow its feedback

Normalise decimal areas

Areas 4.8, 3.2 and 1.6 share factor 1.6.

The ratio is 3:2:1 and totals six protons.

  • Normalise 4.8:3.2:1.6 and check a C3H6O candidate.
Open the feedback checkpoint after attempting
  • Credit 3:2:1 and total six.

Height is not area

A narrow line can be taller than a broad line with the same or smaller area.

Use the integral trace/value supplied across the entire multiplet.

  • Repair: “The tallest signal contains the most protons.”
Open the feedback checkpoint after attempting
  • Replace height with integrated area and keep multiplicity separate.

Count, scale, then assign

Complete twelve integral checks, then unseen aromatic ketone assessment and different diol re-test.

Next objective: interpret spin–spin splitting at /learning/h3-nmr-spin-spin-splitting-lesson.html.

  • Show the scaling of 30:20:30 for C4H8O.
Open the feedback checkpoint after attempting
  • Credit simplest ratio 3:2:3, eight-proton checksum and bounded structural use.