Essential Organic Chemistry
Bruice
Second Editon
Essential Organic Chemistry
Paula Y. Bruice
Second Edition
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ISBN 10: 1-292-02081-4
ISBN 13: 978-1-292-02081-5
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Printed in the United States of America
Delocalized Electrons and Their Effect on Stability, Reactivity, and pKa
Resonance contributors are shown with a double-headed arrow between them. The
double-headed arrow does not mean that the structures are in equilibrium with one another. Rather, it indicates that the actual structure lies somewhere between the structures of the resonance contributors. Resonance structures are merely a convenient way
to show the p electrons; they do not depict any real electron distribution.
The following analogy illustrates the difference between resonance contributors
and the resonance hybrid. Imagine that you are trying to describe to a friend what a
rhinoceros looks like. You might tell your friend that a rhinoceros looks like a cross
between a unicorn and a dragon. Like resonance contributors, the unicorn and the
dragon do not really exist. Furthermore, like resonance contributors, they are not in
equilibrium: a rhinoceros does not change back and forth between the two forms,
looking like a unicorn one instant and a dragon the next. The unicorn and the dragon
are simply ways to represent what the actual structure—the rhinoceros—looks like.
Resonance contributors, like unicorns and dragons, are imaginary, not real. Only the
resonance hybrid, like the rhinoceros, is real.
unicorn
resonance contributor
Electron delocalization is shown by
double-headed arrows ( ·) .
Equilibrium is shown by two arrows
pointing in opposite directions ( ∆ ).
dragon
resonance contributor
rhinoceros
resonance hybrid
Electron delocalization is most effective if all the atoms sharing the delocalized
electrons lie in or close to the same plane, so that their p orbitals can maximally overlap. For example, cyclooctatetraene is not planar; its sp2 carbons have bond angles of
120°, whereas a planar eight-member ring would have bond angles of 135°. Because
the ring is not planar, a p orbital can overlap with one adjacent p orbital, but it can have
little or no overlap with the other adjacent p orbital. Therefore, the eight p electrons
are not delocalized over the entire cyclooctatetraene ring, and its carbon–carbon bonds
do not all have the same length.
3-D Molecule:
Cyclooctatetraene
no overlap
cyclooctatetraene
183
Delocalized Electrons and Their Effect on Stability, Reactivity, and pKa
4 HOW TO DRAW RESONANCE CONTRIBUTORS
We have seen that an organic compound with delocalized electrons is generally represented as a structure with localized electrons to let us know how many p electrons are
present in the molecule. For example, nitroethane is represented as having a nitrogen–
oxygen double bond and a nitrogen–oxygen single bond.
O
+
CH3CH2
N
O
−
nitroethane
However, the two nitrogen–oxygen bonds in nitroethane are actually identical; they
each have the same bond length. A more accurate description of the molecule’s structure is obtained by drawing the two resonance contributors. Both resonance contributors show the compound with a nitrogen–oxygen double bond and a nitrogen–oxygen
single bond, but they indicate that the electrons are delocalized by depicting the double bond in one contributor as a single bond in the other.
−
O
O
+
CH3CH2
+
CH3CH2
N
−
O
O
resonance contributor
Delocalized electrons result from a p
orbital overlapping the p orbitals of
more than one adjacent atom.
N
resonance contributor
The resonance hybrid, in contrast, shows that the p orbital of nitrogen overlaps the
p orbital of each oxygen. In other words, it shows that the two p electrons are shared
by three atoms. The resonance hybrid also shows that the two nitrogen–oxygen bonds
are identical and that the negative charge is shared equally by both oxygen atoms.
Thus, we need to visualize and mentally average both resonance contributors to appreciate what the actual molecule—the resonance hybrid—looks like.
d−
O
+
CH3CH2
N
O
d−
resonance hybrid
Rules for Drawing Resonance Contributors
To draw a set of resonance contributors, we first draw a Lewis structure for the molecule—it becomes our first resonance contributor—and then we move the electrons,
following the rules listed below, to generate the next resonance contributor.
1. Only electrons move. Atoms never move.
2. Only p electrons (electrons in p bonds) and nonbonding electrons can move;
s electrons never move.
To draw resonance contributors, move only
P electrons or lone-pair electrons toward
an sp2 carbon.
3. The total number of electrons in the molecule does not change. Therefore, each
of the resonance contributors for a particular compound must have the same net
charge. If one has a net charge of zero, all the others must also have net charges
of zero. (A net charge of zero does not necessarily mean that there is no charge
on any of the atoms: a molecule with a positive charge on one atom and a negative charge on another atom has a net charge of zero.)
Notice, as you study the following resonance contributors and practice drawing
them yourself, that the electrons (p electrons or lone pairs) are always moved toward an
sp2 carbon. Remember that an sp2 carbon is either a positively charged carbon or a
184
Delocalized Electrons and Their Effect on Stability, Reactivity, and pKa
double-bonded carbon. Electrons cannot be moved toward an sp 3 carbon because
an sp 3 carbon has a complete octet so it cannot accommodate any more electrons.
The following carbocation has delocalized electrons. To draw its resonance contributor, we move the p electrons toward an sp2 carbon. A curved arrow can help you decide how to draw the second contributor. Remember that the tail of the curved arrow
shows where the electrons start from and the head of the arrow shows where the electrons are going. The resonance hybrid shows that the positive charge is shared by two
carbons.
Tutorial:
Drawing resonance contributors
an sp2 carbon
+
CH3CH
CH
+
CH3CH
CHCH3
CH
CHCH3
resonance contributors
d+
CH3CH
CH
d+
CHCH3
resonance hybrid
Let’s compare this carbocation with a similar compound in which all the electrons are
localized. The p electrons in the compound shown below cannot move because the
carbon they would move toward is an sp3 carbon; sp3 carbons cannot accept electrons.
Tutorial:
Localized and delocalized
electrons
an sp3 carbon cannot
accept electrons
+
CH2
CH
CH2CHCH3
localized electrons
In the next example, p electrons again move toward a positive charge. The resonance hybrid shows that the positive charge is shared by three carbons.
an sp2 carbon
+
CH3CH
CH
CH
CH
+
CH3CH
CH2
CH
CH
+
CH
CH3CH
CH2
CH
CH
CH
CH2
resonance contributors
d+
CH3CH
CH
d+
CH
CH
d+
CH2
resonance hybrid
The resonance contributor for the next compound is obtained by moving lone-pair
electrons toward an sp2 carbon. The sp2 carbon can accommodate the new electrons
by breaking a p bond. The lone-pair electrons in the example, on the right below, are
not delocalized because they would have to move toward an sp3 carbon.
O
lone-pair
electrons
O
C
R
an sp2 carbon
C
R
NH2
−
O
C
+
NH2
R
CH2
NH2
resonance contributors
an sp3 carbon
cannot accept
electrons
d−
O
C
R
d+
NH2
resonance hybrid
185
Delocalized Electrons and Their Effect on Stability, Reactivity, and pKa
The following resonance contributors are obtained by moving p electrons toward
an sp2 carbon. Notice that the electrons move toward (not away from) the most electronegative atom (the oxygen).
O
O
−
+
For additional practice drawing resonance
contributors, see Special Topics II in the
Study Guide and Solutions Manual.
CH3C
CH
CH3C
CH2
O
CH
CH2
d−
d+
CH3C
CH
CH2
«
resonance hybrid
The only time you move electrons away from the most electronegative atom in
order to arrive at a resonance contributor is when that is the only way electrons can be
moved. In other words, movement of electrons away from the most electronegative
atom is better than no movement at all, because electron delocalization makes a molecule more stable (as you will see in Section 5).
−
CH2
CH
OCH3
+
CH2
CH
OCH3
Radicals can also have delocalized electrons. The resonance contributors are obtained by moving single electrons toward sp2 carbons.
CH3
CH
CH
CH2
CH3
CH
CH
CH3
resonance contributors
d
d
CH3
CH
CH
CH2
resonance hybrid
PEPTIDE BONDS
Every third bond in a protein is a peptide bond. A
resonance contributor can be drawn for a peptide
bond by moving the lone pair on nitrogen toward
the sp2 carbon.
O
O−
R
CH
C
C
R
+
CH
N
CH
N
R
H
R
H
Because of the partial double bond character of the peptide
bond, the carbon and nitrogen atoms and the two atoms bonded to each are held rigidly in a plane, as represented below by
the blue and green boxes. This planarity affects the way
proteins fold, so it has important implications for the three-dimensional shape of these biological molecules.
CH
peptide bond
O
H
R
C
N
O
H
CH
C
N
CH
N
C
CH
R
H
O
R
C
N
CH
R
O
H
CH
C
N
N
C
CH
N
C
CH
H
O
R
H
O
R
a segment of a protein
186
O
H
R
Delocalized Electrons and Their Effect on Stability, Reactivity, and pKa
PROBLEM 1♦
Which of the following compounds have delocalized electrons?
a.
e.
NH2
N
H
⫹
b.
f. CH3CH “ CHCH “ CHCH2
CH2NH2
c.
g.
O
O
«
h. CH3CH2NHCH2CH
d. CH2 “ CHCH2CH “ CH2
CH2
PROBLEM 2
For the compounds in Problem 1 that have delocalized electrons, draw their resonance
contributors.
5 THE PREDICTED STABILITIES OF
RESONANCE CONTRIBUTORS
All resonance contributors do not necessarily contribute equally to the resonance hybrid. The degree to which each resonance contributor contributes depends on its predicted stability. Because resonance contributors are not real, their stabilities cannot be
measured. Therefore, the stabilities of resonance contributors have to be predicted
based on molecular features found in real molecules. The greater the predicted stability of the resonance contributor, the more it contributes to the structure of the resonance hybrid; and the more it contributes to the structure of the resonance hybrid, the
more similar the contributor is to the real molecule. The examples that follow illustrate these points.
The two resonance contributors for a carboxylic acid are shown below, labeled A
and B. Structure B has two features that make it less stable than structure A: one of its
oxygen atoms has a positive charge—not a comfortable situation for an electronegative atom—and the structure has separated charges. A molecule with separated
charges has a positive charge and a negative charge that can be neutralized by the
movement of electrons. Resonance contributors with separated charges are relatively
unstable (relatively high in energy) because energy is required to keep the opposite
charges separated. Structure A, therefore, is predicted to be more stable than structure
B. Consequently, A makes a greater contribution to the resonance hybrid, so the resonance hybrid looks more like A than like B.
O
O
C
R
−
C
OH
R
A
The greater the predicted stability of the
resonance contributor, the more it
contributes to the structure of the
resonance hybrid.
separated
charges
+
OH
B
a carboxylic acid
The two resonance contributors for a carboxylate ion are shown next.
O
O
C
R
O
−
−
C
R
C
O
D
a carboxylate ion
187