Chemical shift depends on the molecular environment.

The idea

A nucleus never feels the whole applied field, because the electrons around it circulate and shield it. Anything that pulls electron density away, an oxygen, a halogen, a carbonyl group or a benzene ring, leaves the nucleus less shielded, so it resonates further left at a higher delta value. That is why a methyl group bonded only to carbon appears near 0.7 to 1.2 ppm, a CH₂ group attached to the single-bonded oxygen of an ester appears near 4.1 ppm, a methyl next to a carbonyl near 2.1 to 2.6 ppm, an aldehyde proton near 9.5 ppm and a carboxylic acid O-H near 11 ppm. Protons in identical surroundings give a single peak, so symmetry cuts the number of peaks you actually see.

What earns the marks

• Shielding by surrounding electrons as the reason a nucleus resonates where it does.
• Stating that an electron-withdrawing neighbour deshields the nucleus and raises the shift.
• Using a quoted shift to name a specific environment, such as a proton on a carbon bonded to the single-bonded oxygen of an ester at about 4.1 ppm.
• Recognising equivalent protons as giving one peak rather than several.

Common mistakes and how to put them right

• Assuming a higher shift means more hydrogens, which is what integration tells you and shift does not.

• Expecting a proton three or four bonds from an oxygen to be shifted as far downfield as one on the carbon bonded to it.

• Treating an O-H or N-H peak as having a fixed shift, when it is broad and variable.

• Counting every hydrogen as its own environment and predicting far too many peaks.

Exam tips

• Name the environment in the Data Booklet's own wording when you justify a shift; the mark scheme is written from that table.
• Read shift and splitting together, because each narrows the possibilities the other leaves open.
• Check for symmetry first: two identical methyl groups are one environment and one peak.