Question

The proton NMR spectrum of a compound with formula ${\mathbf{C}}_6{\mathbf{H}}_5{\mathbf{NCl}}_2$is shown. The normal carbon-13 and DEPT experimental results are tabulated. The infrared spectrum shows peaks at 3432 and $\mathbf{3313}{\mathbf{cm}}^{-1}$and a series of medium-sized peaks between 1618 and $\mathbf{1466}{\mathbf{cm}}^{-1}$. Draw the structure of this compound.

Short answer

${}^1\mathrm{H}\mathrm{NMR}\mathrm{spectrum}$

${}^{13}\mathrm{C}\mathrm{NMR}\mathrm{spectrum}$

Step-by-step solution

Determination of the structure

By observing the tabular form for C-13 NMR DEPT value the chemical shift values for the carbon are assigned and by observing the NMR spectrum, the values for hydrogens are assigned.

Proton NMR and C-13 Spectrum

The assignment of ${}^1\mathrm{H}$NMR spectrum of the compound with molecular formula ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NCl}}_2$is as follows:

The key features of this${}^1\mathrm{H}$NMR spectrum are:

1. The amine protons at $\mathrm{\delta }4.35$as a 2H, broad singlet as the protons are attached to the electronegative nitrogen atom and as these protons are exchangeable, the signal is broad.
2. The protons give a doublet at $\mathrm{\delta }7.25$as these are close to electronegative chlorine atoms and there is one adjacent proton.
3. The aromatic protons give a peak around $\mathrm{\delta }6.70$as this proton is ortho to the amine group.

${}^1\mathrm{H}$NMR spectrum

The IR peaks at 3432 and $3313{\mathrm{cm}}^{-1}$corresponds to the N-H stretching frequency. The peaks in the range 1466-1618 ${\mathrm{cm}}^{-1}$corresponds to the aromatic ring.

For ${}^{13}C$NMR, ${\mathrm{CH}}_2$protons give the negative signal at DEPT-135 whereas ${\mathrm{CH}}_3$protons give the positive signal at DEPT-135. The assignment of ${}^{13}\mathrm{C}$peaks is as shown:

${}^{13}\mathrm{C}$NMR spectrum