Question

The ¹H NMR spectra of three isomers with molecular formula C₄H₉Br are shown here. Which isomer produces which spectrum?

a)

b)

c)

Short answer

a)

b)

c)

Step-by-step solution

Step-by-step SolutionStep 1: Number of signals

NMR stands for Nuclear Magnetic Resonance Spectroscopy. This is the technique used for determining the structure of organic compounds.

The number of signals present in NMR reflects the number of magnetically different protons. Thus, the number of signal reflects the number of chemically different protons or sets of protons. For example, methanol has two different set of protons (methyl and hydroxyl).

Peak integration

The peak area variety of signals in NMR spectrum is measured by an electronic integrator, and the process is called integration. It is shown in the chart in the form of a stepped up curve called peak integration curve.

An NMR spectrum can not only identify different types of protons present in a molecule, but it can also identify the number of each type of proton in the molecule.

An NMR spectrometer is equipped with a computer that calculates the integrals electronically and then displays them as an integral trace superimposed on the original spectrum. The height of each step in the integral trace is proportional to the area under the corresponding signal, which, in turn, is proportional to the number of protons producing the signal.

Determining Double Bond equivalent (DBE)

Degree of unsaturation is also known as double bond equivalent (DBE). If the molecular formula is given, plug in the number into this formula. The DBE formula is given below.

$\mathrm{DBE}=\frac{2\mathrm{C}+2+\mathrm{N}-\mathrm{X}-\mathrm{H}}{2}$

Where, C is the number of Carbon.

N is the number of nitrogen.

H is the number of hydrogen.

X is the number of halogen.

Here, the given molecular formula of the compound is ${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Br}$.

For ${\mathrm{C}}_4{\mathrm{H}}_9{\mathrm{Br}}_{}$.

$\begin{array}{l}\mathrm{DBE}=\frac{2\left(4\right)+2+0-1-9}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=\frac{0}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=0\end{array}$

There is no unsaturation present in the given compound.

Identify the isomer producing a signal in the given spectra

As we know that the coupling constant is smaller in cis isomer than in trans-isomer, the NMR spectrum of the two isomers are showing different signals

(a) According to the data given in spectrum (a), the ¹H NMR spectrum of 2-bromobutane shows four signals. The structure of the isomer is shown below.

(b) According to the data given in spectrum (b), the ¹H NMR spectrum of 1-bromobutane shows four signals. The structure of the isomer is shown below.

(c) According to the data given in spectrum (c), the ¹H NMR spectrum of 1-bromo-2-methylpropane shows three signals.The structure of the isomer is shown below.