1 (382)

Prévia do material em texto

Solutions for Aromatic Compounds
Br
Br
CH3
CH3 CH3
CH3
NO2
NO2
Br
Br
NO2
Br
Br
O2N
Br
Br
+ +
(c) The original compound had to have been meta-dibromobenzene as this is the only dibromo isomer that 
gives three mononitrated products.
only 1 isomer
(b)
37 continued
C
O
OO O
C
O
CH2
Cl
O
O
O
CH2CH3Br
+ CH2 + Cl
(a) The formula C8H7OCl has five elements of unsaturation, probably a benzene ring (4) plus either a double 
bond or a ring. The IR suggests a conjugated carbonyl at 1690 cm–1 and an aromatic ring at 1602 cm–1. The 
NMR shows a total of five aromatic protons, indicating a monosubstituted benzene. A 2H singlet at δ 4.7 is 
a deshielded methylene.
38
+
(b) The mass spectral evidence of molecular ion peaks of 1 : 1 intensity at 184 and 186 shows the presence of 
a bromine atom. The m/z 184 minus 79 for bromine gives a mass of 105 for the rest of the molecule, which is 
about a benzene ring plus two carbons and a few hydrogens. The NMR shows four aromatic hydrogens in a 
typical para pattern (two doublets), indicating a para-disubstituted benzene. The 2H quartet and 3H triplet 
are characteristic of an ethyl group.
+ Br + CH2CH3
39
like the ends of a 
conjugated diene
(a) (b)
Diels-Alder 
product
new sigma bonds shown in bold
377