4.9.3Unit 4. Rates, Equilibria and Further Organic Chemistry
Low Resolution Proton NMR — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 4 parts Lesson 53 of 56 Unit quiz
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Section 1 of 4Basics
Hydrogen nuclei (protons) absorb energy in a magnetic field to give a spectrum.
Environments
- Environments Each set of chemically equivalent hydrogen atoms gives one peak, so the number of peaks equals the number of environments.
Chemical shift
- Chemical shift The position of each peak depends on nearby atoms;
- electronegative atoms or groups move the peak to a higher ppm value.
Integration
- Integration The area under each peak is proportional to the number of hydrogen atoms in that environment.
Solvent and reference
- Solvent and reference Solvents such as CDCl₃ or CCl₄ contain no hydrogen atoms, and TMS gives the 0 ppm reference.
Section 2 of 4Typical Shifts
The values are approximate and are given relative to TMS.
Typical Shifts
Alkyl groups
- Alkyl groups Hydrogens on saturated carbon, as in CH₃– and –CH₂–, appear at roughly 0.9 to 1.7 ppm.
Next to C=O or a ring
- Next to C=O or a ring Hydrogens on carbon next to a carbonyl group or a benzene ring appear at about 2.0 to 3.0 ppm.
Next to oxygen
- Next to oxygen Hydrogens on a carbon bonded to oxygen, as in alcohols and esters, appear at about 3.3 to 4.3 ppm.
Special groups
- Special groups Aromatic H appears at about 6.5 to 8.0 ppm, aldehyde H near 9 to 10 ppm and carboxylic acid H near 10 to 12 ppm.
Section 3 of 4Analysing a Spectrum
1
Count the peaks to find the number of different hydrogen environments.
2
Divide the integration values by the smallest to give the simplest whole number ratio of hydrogens.
3
Use the shift values to suggest the group that the hydrogens are attached to.
4
Add D₂O: peaks from –OH and –NH hydrogens disappear because the hydrogens exchange with deuterium.
5
Combine the information with the molecular formula to deduce a structure with the correct number of environments.
Section 4 of 4Exam Checklist & Tips
4.9.3 – Key Points for the Exam
- Proton environment Set of equivalent hydrogen atoms.
- Integration Relative area under a peak, indicating hydrogen count.
- Deuterium exchange Replacement of –OH hydrogen with deuterium.
- TMS Reference compound at 0 ppm.
Common Mistakes
- Treating the integration value as the actual number of hydrogens without simplifying the ratio.
- Assuming each CH₃ group gives a separate peak even when equivalent.
- Using CHCl₃ as the solvent in the sample.
- Forgetting that –OH peaks vanish with D₂O.
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