 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/1/c/ii/solution

300 MHz

J / Hz = 300 Δ δ / ppm

^{3} J_{b c}^{3} J_{ab}^{4} J_{a c}^{3} J_{d e}^{3} J_{df}^{2} J_{e f} = (5.893 - 5.842) 300 = 15.3 Hz, = (7.071 - 7.048) 300 = 6.9 Hz, = (5.848 - 5.842) 300 = 1.8 Hz, = (4.932 - 4.885) 300 = 14.1 Hz, = (4.589 - 4.568) 300 = 6.3 Hz, = (4.932 - 4.927) 300 = 1.5 Hz .

For example, the two quartets of

c

are

{5.842, 5.848, 5.854, 5.860}

and

{5.893, 5.899, 5.905, 5.911}

ppm. Their origins differ by

0.051

ppm, while the spacing within either quartet is

0.006

ppm. This separates the trans alkene interaction from the smaller allylic methyl interaction.

The four

d

lines independently give

(7.377 - 7.330) 300 = 14.1

and

(7.377 - 7.356) 300 = 6.3 Hz

. The large interaction identifies

e

as trans to

d

;

f

has the smaller cis interaction. The double-headed arrows in the structural map identify the actual interacting nuclei. Coupling to

a

means the same coupling to each of its three equivalent hydrogens.

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