Test: NMR Spectroscopy - Chemistry MCQ

In compound a;

Here only 2 types of Hydrogen are present (two Ha & twelve Hb). So, 2 singlets only. There is no need to check ¹³C NMR. Still, if you check three types carbon are present (two C₁, four C₂ and four C₂ carbon). Hence, there peak for ¹³C NMR.

Geminal proton 2J coupling i.e. coupling of H & H on same carbon.

as ring size increases, bond angle outside decreases i.e. α1 > α2 > α3 > α4 (according % s character). If B.A. decreases, 2J coupling increases.

Different types of proton give the different type of signals, so the different hydrogen will give different numbers of signals.
As we can see first compound has two type of hydrogen Ha and Hb. Compound second has six type of hydrogen Ha, Hb, Hc, Hd, He and H. Compound third has four type of hydrogen Ha, Hb, Hc, Hd and l forth compound has two type of hydrogen Ha and Hb.

By using the below formula we can calculate NMR frequency, where h is Plank’s constant.

C₈H₁₀O

3Ha → 3.8, singlet (deshielded because of –I of oxygen)
3Hb → 2.5, singlet (No such –I)
2Hc → 7.2, doublet (deshielded because of anisotropy effect of benzene as well as because of –I of oxygen)
2Hd → 6.8, doublet (deshielded because of anisotropy of benzene).

In NMR spectroscopy. the product of Nuclear ‘g’ factor (g_N), the nuclear magneton (β_N) and the magnetic field strength (B₀) gives the energy of transition from alpha to beta state. As shown below:

If intensity is 1:1, the ratio must be 1:3

Where, I: nucleus spin = 1/2
μn: nuclear magneton = 5.05 × 10^-27
μ: magnetic moment by protons = 2.79
h: Plank constant = 6.63 x 10^-34 joule-seconds.

Vinyl chloride compound has three types of proton as shown below and ethyl cyclopropane has five types of protons so five peaks will be observed.

As shown below; three signals are observed. 3Ha will experience shielding due to O^–, so They are different from 3Hb.