292 Section F – Combined techniques
Example 3
The spectra shown in Figures 3(a)–(d)are for a compound boiling at 205°C and
insoluble in water. The composition is C 52.2%, H 3.7%, Cl 44.1%.(i) Empirical formula: C 7 H 6 Cl 2 ; RMM = 161(ii) DBE = 4(iii) 3a: IR (liquid film)
3000–3050 cm−^1 H−C aromatic
2000–1600 Monosubstituted aromatic bands
1700 not a carbonyl, since no oxygen: probably aromatic
1500, 1450 aromatic ring vibrations
696 C−Cl stretch
(iv) UV: The weak absorbance at 270 nm may indicate an aromatic compound.(v) 3b: MS (EI). The multiple molecular ion at m/z 160, 162 and 164 and the
doubled fragment ions at 125 and 127 would strongly suggest 2 Cl atoms,
even without the analytical information.
m/z Relative abundance
164 M+• C 7 H 637 Cl 2 9
162 M+• C 7 H 6 35 Cl^37 Cl 6
160 M+• C 7 H 635 Cl 2 1
127 M−Cl C 7 H 637 Cl 33
125 M−Cl C 7 H 635 Cl 100
Other peaks suggest aromatic ring residues.(vi) 3c: 1-H NMR(200 MHz, CDCl 3 solution)
d/ppm Relative integral Multiplicity Assignment
6.7 1 1 H-C(Cl)−Ar
7.3 3 m m-, p-ArH
7.5 2 m o-ArH
(m = multiplet)
This suggests a monosubstituted aromatic compound.3d:13-C NMR (50.0 MHz, CDCl 3 solution)
The presence of 5 carbon resonances when 7 carbons are present means
that at least two pairs are equivalent, which would correspond to a mono-
substituted aromatic compound.
d/ppm Multiplicity Assign
73.0 2 H−C(Cl)−Ar
128.0 2 ArC−H
130.5 2 ArC−H
132.0 2 ArC−H
142.0 1 ArC−CH(Cl)
(vii) The deductions above indicate that this is an aromatic compound, which is
consistent with a DBE of 4. Monosubstitution is confirmed by the IR spec-
trum and by both NMR spectra. This means that the two chlorines cannot
be substituted on the ring. Since the carbon external to the ring has
only a single hydrogen, both the chlorines must also be attached to
it. The compound is therefore dichloromethyl benzene, Cl 2 CH-C 6 H 5
(benzylidene chloride).