 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/1/c/i/solution

C is 2-Chloroethyl vinyl ether, $C H_{2} = CH - O - C H_{2} - C H_{2} Cl$ . There is one alkene and no carbonyl. Assign the carbon-13 NMR resonances as

site C H_{2} = = CH - O OC H_{2} C H_{2} Cl δ_{C} / ppm 87.4 151.1 67.8 41.9

The strongly differentiated pair of alkene shifts is characteristic of a vinyl ether; interpreting

151.1

ppm as a carbonyl would leave the unusually shielded unsaturated

87.4

ppm carbon and the three distinct vinylic protons unexplained.

 New to topics? Read the docs here!