Sets of spectra are given for two compounds. For each

Chapter 18, Problem 32PS

(choose chapter or problem)

Sets of spectra are given for two compounds. For each set,

(1) Look at each spectrum individually, and list the structural characteristics you can determine from that spectrum.

(2) Look at the set of spectra as a group, and propose a tentative structure.

(3) Verify that your proposed structure accounts for the major features of each spectrum. The solution for compound 1 is given after the problem, but go as far as you can before looking at the solution.

Solution to Compound 1:

Mass spectrum: The MS shows an odd molecular weight at 121 and a large even-numbered fragment at 106. These features may indicate the presence of a nitrogen atom.

Infrared spectrum: The IR shows a sharp peak around \(3400 \mathrm{~cm}^{-1}\), possibly the \(\mathrm{N}-\mathrm{H}\) of an amine or the \(\equiv \mathrm{C}-\mathrm{H}\) of a terminal alkyne. Because the MS suggests a nitrogen atom, and there is no other evidence for an alkyne (no \(\mathrm{C} \equiv \mathrm{C}\) stretch around \(2200 \mathrm{~cm}^{-1}\)), the \(3400 \mathrm{~cm}^{-1}\) absorption is probably an \(\mathrm{N}-\mathrm{H}\) bond. The unsaturated \(=\mathrm{C}-\mathrm{H}\) absorptions above \(3000 \mathrm{~cm}^{-1}\), combined with an \(\mathrm{C}=\mathrm{C}\) aromatic stretch around \(1600 \mathrm{~cm}^{-1}\), indicate an aromatic ring.

NMR spectrum: The NMR shows complex splitting in the aromatic region, probably from a benzene ring: The total integral of 5 suggests the ring is monosubstituted. Part of the aromatic absorption is shifted upfield \(\delta 7.2\), of suggesting that the substituent on the benzene ring is a pi electron-donating group like an amine or an ether. An ethyl group (total area 5) is seen \(\delta 1.2\) at \(\delta 3.1\) and appropriate for protons on a carbon atom bonded to nitrogen. A broad singlet of area 1 appears at \(\delta 3.5\), probably resulting from the \(\mathrm{N}-\mathrm{H}\) seen in the IR spectrum. Combining this information, we propose a nitrogen atom bonded to a hydrogen atom, a benzene ring, and an ethyl group. The total molecular weight for this structure would be 121, in agreement with the molecular ion in the mass spectrum.

The proposed structure shows an aromatic ring with 5 protons, which explains the aromatic signals in the NMR and the \(\mathrm{C}=\mathrm{C}\) at \(1600 \mathrm{~cm}^{-1}\) and the \(= \mathrm {C-H}\) above \(3000 \mathrm{~cm}^{-1}\) in the IR. The aromatic ring is bonded to an electron-donating \(\mathrm {-NHR}\) group, which explains the odd molecular weight, the \(\mathrm {N-H}\) absorption in the IR, and the aromatic signals shifted above \(\delta 7.2\) in the NMR. The ethyl group bonded to nitrogen explains the ethyl signals in the NMR, deshielded to \(\delta 3.1\) by the nitrogen atom. The base peak in the MS \((\mathrm{M}-15=106)\) is explained by the loss of a methyl group to give a resonance-stabilized cation:

Equation Transcription:

Text Transcription:

121 M^+

m/z

({mu}m)

(cm^{-1})

delta(ppm)

M^+ 136

m/z

({mu}m)

(cm^{-1})

^{13}C NMR

CDCl_3

0Hz

50Hz

3.2{delta}

3.1{delta}

0Hz

50Hz

1.8{delta}

1.7{delta}

0Hz

50Hz

1.1{delta}

1.0{delta}

delta(ppm)

3400 cm^{-1}

N-H

{equiv}C-H

C{equiv}C

2200 cm^{-1}

3400 cm^{-1}

N-H

=C-H

3000 cm-1

C=C

1600 cm-1

{delta}7.2

{delta}1.2

{delta}3.1

{delta}3.5

N-H

C=C

C-ddot{N}-CH_{2}-CH_{3}

C=C

1600 cm^{-1}

=C-H

3000 cm-1

-NHR

N-H

{delta}7.2

NMR

NMR

{delta}3.1

(M-15=106)

Ph-ddot{N}-CH_{2}-CH+{3}

Ph-ddot{N}-C^+

Ph-N=C

CH_3