CLAYDEN. Organic Chemistry. 2ª edição. Oxford. 2012.
1261 pág.

CLAYDEN. Organic Chemistry. 2ª edição. Oxford. 2012.


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alone have two identical carbon 
atoms (the two methyl groups in each case); all the other structures would have four separate 
signals in the NMR.
So, how can we choose between these? The solution is in the 1H NMR spectrum, which is 
shown below. There are only two peaks visible: one at 3.3 and one at 1.1 ppm. It\u2019s quite com-
mon in 1H NMR spectra not to see signals for protons attached to O or N (you will see why in 
Chapter 13) so we can again rule out all structures with more than two different types of H 
attached to C. Again, we are left with A and B, con\ufb01 rming our earlier deductions. But the 
chemical shift of the signal at \u3b4 3.3 tells us more: it has to be due to H atoms next to an oxygen 
atom because it is deshielded. The industrial emulsi\ufb01 er must therefore be A: 2-amino-2-
methylpropan-1-ol.
HO
NH2
2-amino-2-methylpropan-1-ol
1H NMR spectrum
8 7 6 5 4
ppm
3 2 1 0
Double bond equivalents help in the search for a structure
The last example was fully saturated but it is usually a help in deducing the structure of an 
unknown compound if, once you know the atomic composition, you immediately work out 
how much unsaturation there is. It may seem obvious to you that, as C4H11NO has no double 
bonds, then C4H9NO (losing two hydrogen atoms) must have one double bond, C4H7NO two 
double bonds, and so on. Well, it\u2019s not quite as simple as that. Some possible structures for 
these formulae are shown below.
HO
NH2
2-amino-2-methylpropan-1-ol
CHAPTER 3   DETERMINING ORGANIC STRUCTURES74
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HO
NH2
some structures for C4H9NO
O NH2
O
NH2
O
NH2 HO
NH2
some structures for C4H7NO
N
H
O
N
H
O
Some of these structures have the right number of double bonds (C\ufffdC and C\ufffdO), one 
has a triple bond, and three compounds use rings as an alternative way of \u2018losing\u2019 some 
hydrogen atoms. Each time you make a ring or a double bond, you have to lose two more 
hydrogen atoms. So double bonds (of all kinds) and rings are called double bond equiva-
lents (DBEs).
You can work out how many DBEs there are in a given atomic composition just by making 
a drawing of one possible structure for the formula (all possible structures for the same for-
mula have the same number of DBEs). Alternatively, you can calculate the DBEs if you wish. 
A saturated hydrocarbon with n carbon atoms has (2n + 2) hydrogens. Oxygen doesn\u2019t make 
any difference to this: there are the same number of Hs in a saturated ether or alcohol as in a 
saturated hydrocarbon.
O
OH
saturated hydrocarbon C7H16 saturated alcohol C7H16O saturated ether C7H16O
All have
(2n + 2)
H atoms
So, for a compound containing C, H, and O only, take the actual number of hydrogen atoms 
away from (2n + 2) and divide by two. Just to check that it works, for the unsaturated ketone 
C7H12O the calculation becomes:
 1. Maximum number of H atoms for 7Cs: 2n + 2 = 16
 2. Subtract the actual number of H atoms (12): 16 \u2013 12 = 4
 3. Divide by 2 to give the DBEs: 4/2 = 2
Here are two more examples to illustrate the method. This unsaturated cyclic acid has: 
16 \u2013 10 = 6 divided by 2 = 3 DBEs and it has one alkene, one C\ufffdO, and one ring. Correct.
The aromatic ether has 16 \u2013 8 = 8 divided by 2 gives 4 DBEs and it has three double bonds in 
the ring and the ring itself. Correct again. A benzene ring always gives four DBEs: three for the 
double bonds and one for the ring.
Nitrogen makes a difference. Every nitrogen adds one extra hydrogen atom because nitrogen 
can make three bonds. This means that the formula becomes: subtract actual number of 
hydrogens from (2n + 2), add one for each nitrogen atom, and divide by two. We can try this out 
too. Here are some example structures of compounds with seven C atoms, one N and an 
assortment of unsaturation and rings.
NH2 NO2 NMe
O
NH2
N
NMe2
saturated C7 compound with nitrogen
C7H17N = (2n + 3) H atoms C7H15NO2 = one DBE C7H13NO = two DBE C7H9N = four DBE C7H10N2 = four DBE
The saturated compound has (2n + 3) Hs instead of (2n + 2). The saturated nitro compound 
has (2n + 2) = 16, less 15 (the actual number of Hs) plus one (the number of nitrogen atoms) = 2. 
O C7H12O
= two DBE
CO2H
C7H10O2
= three DBE
OMe
C7H8O
= four DBE
DOUBLE BOND EQUIVALENTS HELP IN THE SEARCH FOR A STRUCTURE 75
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Divide this by 2 and you get 1 DBE, which is the N\ufffdO bond. We leave the third and fourth 
examples for you to work out, but the last compound (we shall meet this later as DMAP) has:
 1. Maximum number of H atoms for 7Cs: 2n + 2 = 16
 2. Subtract the actual number of H atoms (10): 16 \u2013 10 = 6
 3. Add number of nitrogens: 6 + 2 = 8
 4. Divide by 2 to give the DBEs: 8/2 = 4
There are indeed three double bonds and a ring, making four in all. Make sure that you can 
do these calculations without much trouble.
If you have other elements too it is simpler just to draw a trial structure and \ufb01 nd out how 
many DBEs there are. You may prefer this method for all compounds as it has the advantage 
of giving you one possible structure before you really start. One good tip is that if you have 
few hydrogens relative to the number of carbon atoms (at least four DBEs) then there is prob-
ably an aromatic ring in the compound.
Knowing the number of double bond equivalents for a formula derived by high-resolution 
mass spectrometry is a quick short cut to generating some plausible structures. You can then 
rule them in or rule them out by comparing with IR and NMR data.
 \u25cf Working out the DBEs for an unknown compound
1 Calculate the expected number of Hs in the saturated structure
(a) For Cn there would be 2n + 2H atoms if C, H, O only.
(b) For CnNm there would be 2n + 2 + mH atoms.
2 Subtract the actual number of Hs and divide by 2. This gives the DBEs.
3 If there are other atoms (Cl, B, P, etc.) it is best to draw a trial structure.
4 A DBE indicates either a ring or a double bond (a triple bond is two DBEs).
5 A benzene ring has four DBEs (three for the double bonds and one for the ring).
6 If there are few Hs, e.g. less than the number of Cs, suspect a benzene ring.
7 A nitro group has one DBE only.
An unknown compound from a chemical reaction
Our last example addresses a situation very common in chemistry\u2014working out the structure 
of a product of a reaction. The situation is this: you have treated propenal (acrolein) with HBr 
in ethane-1,2-diol (or glycol) as solvent for 1 hour at room temperature. Distillation of the 
reaction mixture gives a colourless liquid, compound X. What is it?
100
80
60
40
20
0
R
el
at
iv
e 
ab
un
da
nc
e
150
Mass spectrum of compound X
100
m/z
163.0
152.0
136.9
101.0
122.9
108.9
80.9
73.0
90
70
50
30
10
181.0
179.0
 \u25a0 Do not confuse this 
calculation with the observation 
we made about mass spectra 
that the molecular weight of a 
compound containing one 
nitrogen atom must be odd. This 
observation and the number of 
DBEs are, of course, related but 
they are different calculations 
made for different purposes.
O
H HO
OH
acrolein
(propenal) ethylene glycol(ethane-1,2-diol)
HBr
X
CHAPTER 3   DETERMINING ORGANIC STRUCTURES76
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The mass spectrum shows a molecular ion (181) much heavier than that of the starting 
material, C3H4O = 56. Indeed it shows two molecular ions at 181 and 179, typical of a bromo 
compound, so it looks as if HBr has added to the aldehyde somehow. High resolution mass 
spectrometry reveals a formula of C5H9BrO2, and the \ufb01 ve carbon atoms make it look as though 
the glycol has added in too. If we add everything together we \ufb01 nd that the unknown com-
pound is the result of the three reagents added together less