Question

Draw a structure for the compound, C₄H₇BrO, that fits the following ¹H NMR data:

δ 2.11 (3H, singlet)

δ 3.52 (2H, triplet, J = 6 Hz)

δ 4.40 (2H, triplet, J = 6 Hz)

You do not have to consider stereochemistry.

You do not have to explicitly draw H atoms.

In cases where there is more than one answer, just draw one.

Answer 1

Adjacent carbon atom(C2) to C1 is not linked to any H atoms, so no spin coupling occurs for the H atoms at C1; these 3 hydrogen atoms appear as a 3 hydrogen singlet in the spectrum. C3-H atoms are spin coupled by 2 adjacent C4 hydrogen, similarly C4-H atoms are spin coupled by 2 adjacent C3 hydrogen. H atoms at C3 and C4 thus appear as two set of triplets in the spectrum. This is followed by the (n+1) rule[here n=2].

[ Note: ?-Hydrogens in ketones are deshilded by the anisotropy of the adjacent C=O group. Chemical shift of these hydrogens remains between 2.1-2.4 ppm.

The chemical shift of a hydrogen atom attached to a carbon atom same as the halide atom decreases due to the electronegativity of the halide atom. Chemical shift of these -CH-Br remains between 2.7-4.1 ppm.]