H NMR

A compound with the molecular formula C3H6Cl2 gives the following 1H NMR
spectrum: δ 2.20 ppm (quintet) and δ 3.70 ppm (triplet). Propose a
structure and assign the peak

Soln

I usually use the base values of 0.9 ppm for CH3, 1.3 for CH2, and 1.7
for CH. A triplet indicates that the number of numbers is two. A
quintet indicates that there are four neighboring pairs. 2.20 – 0.9 =
1.3 ppm, which suggest that there are CH2 groups. 3.70 – 0.9 = 2.8
ppm, which indicates the halides region (Cl, Br, and I). Since there
are four neighboring pairs, and the signal is in the CH2 region, the
H’s should be creating the signal, which is producing the quintet.
There should be a signal H proton, and two chlorine atoms on one carbon.
This should produce the triplet since two neighboring pairs are
required to do so.



A compound with the molecular formula C6H15N gives the following 1H NMR
spectrum: δ 0.90 ppm (triplet) and δ 2.40 ppm (quartet). Propose a
structure and assign the peaks.

soln

0.9 ppm indicates methyl groups in the structure

2.40 – 0.9 = 1.5 ppm suggest an aryl ring (Ar) or amine (NH2)

A triplet suggests two neighbors

A quartet suggests three neighbors.

b)

I’m not sure on this structure. I think it is option b.

H NMR spectrum: δ 1.10 ppm (doublet), δ 2.10 ppm (singlet), and δ
2.50 ppm (septet). In its IR spectrum a strong peak is observed at 1720
cm-1. Propose a structure and assign the peaks of both spectra.

Ans.

Doublet suggests one neighboring atoms.

A singlet doesn’t have any neighboring atoms.

A septet has six neighboring atoms.

1.10 ppm suggest the presence of methyl group

2.50 and 2.10 suggest a carbonyl group

An IR spectra with an observed strong peak of 1720 cm-1 confirms the
presence of a carbonyl group within the structure.