There are many millions of organic compounds. If you had to memorize how each of them reacts, studying organic chemistry would not be a very pleasant experience. For- tunately, organic compounds can be divided into families, and all the members of a family react in similar ways. What makes learning organic chemistry even easier is that there are only a few rules that govern the reactivity of each family.
The family that an organic compound belongs to is determined by its functional group. The functional group is the center of reactivity of a molecule. You will find a table of common functional groups inside the back cover of this book. You are already familiar with the functional group of an alkene: the carbon–carbon double bond. All compounds with a carbon–carbon double bond react in similar ways, whether the compound is a small molecule like ethene or a large molecule like cholesterol.
Electron-rich atoms or molecules are attracted to electron-deficient atoms or molecules.
It is not sufficient to know that a compound with a carbon–carbon double bond re- acts with HBr to form a product in which the H and Br atoms have taken the place of the bond; we need to understand why the compound reacts with HBr. In each chap- ter that discusses the reactivity of a particular functional group, we will see how the nature of the functional group allows us to predict the kind of reactions it will under- go. Then, when you are confronted with a reaction you have never seen before, know- ing how the structure of the molecule affects its reactivity will help you predict the products of the reaction.
In essence, organic chemistry is all about the interaction between electron-rich atoms or molecules and electron-deficient atoms or molecules. These are the forces that make chemical reactions happen. From this observation follows a very important rule for predicting the reactivity of organic compounds: electron-rich atoms or mole- cules are attracted to electron-deficient atoms or molecules. Each time you study a new functional group, remember that the reactions it undergoes can be explained by this very simple rule.
Therefore, to understand how a functional group reacts, you must first learn to rec- ognize electron-deficient and electron-rich atoms and molecules. An electron- deficient atom or molecule is called an electrophile. Literally, “electrophile” means
“electron loving” (phile is the Greek suffix for “loving”). An electrophile looks for electrons.
p
CH3CH CHCH3 + H Br CH3CH CHCH3 CH3CH CHCH3
Br H
H
a carbocation 2-bromobutane
an alkyl halide Br−
d+ d−
+ +
these are nucleophiles because they have a pair of electrons to share
Cl − H2O
HO − CH3NH2
AlCl3 + NH3
H2O NH3
Cl3Al +
H Br + +
a. −
b. HO − Br−
CH3CH CHCH3 H Br CH3CH CHCH3 Br− H
+ d+ d− + +
p bond has broken
new s bond has formed
A nucleophile reacts with an electrophile.
An electron-rich atom or molecule is called a nucleophile. A nucleophile has a pair of electrons it can share. Because a nucleophile has electrons to share and an elec- trophile is seeking electrons, it should not be surprising that they attract each other.
Thus, the preceding rule can be restated as a nucleophile reacts with an electrophile.
Curved arrows are used to show the bonds that break and the bonds that form; they are drawn from an electron-rich center to an electron-deficient center.
PROBLEM 13♦
Identify the electrophile and the nucleophile in each of the following acid–base reactions:
We have seen that the bond of an alkene consists of a cloud of electrons above and below the plane defined by the sp2carbons and the four atoms bonded to them. As a result of this cloud of electrons, an alkene is an electron-rich molecule—it is a nu- cleophile. (Notice the relatively electron-rich pale orange area in the electrostatic po- tential maps for cis- and trans-2-butene in Section 4.) We have also seen that a bond is weaker than a bond. The bond, therefore, is the bond that is most easily broken when an alkene undergoes a reaction. For these reasons, we can predict that an alkene will react with an electrophile, and, in the process, the bond will break.
So if a reagent such as hydrogen bromide is added to an alkene, the alkene (a nucle- ophile) will react with the partially positively charged hydrogen (an electrophile) of hydrogen bromide and a carbocation will be formed. In the second step of the reac- tion, the positively charged carbocation (an electrophile) will react with the negative- ly charged bromide ion (a nucleophile) to form an alkyl halide.
p p
s
p p
This step-by-step description of the process by which reactants (alkene HBr) are changed into products (alkyl halide) is called the mechanism of the reaction. To help us understand a mechanism, curved arrows are drawn to show how the electrons move as new covalent bonds are formed and existing covalent bonds are broken. These are called “curved” arrows to distinguish them from the “straight” arrow used to link reac- tants with products in a chemical reaction. Each curved arrow represents the simultane- ous movement of two electrons from an electron-rich center (at the tail of the arrow) and toward an electron-deficient center (at the point of the arrow). In this way, the curved arrows show which bonds are broken and which bonds are formed.
+
For the reaction of 2-butene with HBr, an arrow is drawn to show that the two elec- trons of the bond of the alkene are attracted to the partially positively charged hydrogen of HBr. The hydrogen is not immediately free to accept this pair of electrons because it is already bonded to a bromine, and hydrogen can be bonded to only one
p
Mechanistic Tutorial:
Addition of HBr to an alkene
A FEW WORDS ABOUT CURVED ARROWS
1. Draw a curved arrow so that it points in the direction of electron flow and never away from the flow. This means that an arrow will always be drawn away from a negative charge and/or toward a positive charge. An arrow is used to show both the bond that forms and the bond that breaks.
CH3 C CH3
+ O−
Br CH3 C
CH3
O
Br−
CH3 +O H
+ O
H CH3 H H+
correct
correct
CH3 C CH3
+ O−
Br CH3 C
CH3
O
Br−
CH3 +O H
+ O
H CH3 H H+
incorrect
incorrect
atom at a time. However, as the electrons of the alkene move toward the hydrogen, the bond breaks, with bromine keeping the bonding electrons. Notice that the electrons are pulled away from one carbon but remain attached to the other. Thus, the two electrons that formerly formed the bond now form a bond between carbon and the hydrogen from HBr. The product of this first step in the reaction is a carboca- tion because the sp2carbon that did not form the new bond with hydrogen has lost a share in an electron pair (the electrons of the bond) and is, therefore, positively charged.
In the second step of the reaction, a lone pair on the negatively charged bromide ion forms a bond with the positively charged carbon of the carbocation. Notice that in both steps of the reaction an electrophile reacts with a nucleophile.
p
s p
p
HơBr
p
It will be helpful to do the exercise on drawing curved arrows in the Study Guide/Solution Manual (Special Topic III).
H− CH3O− CH3C CH CH3CHCH+ 3 NH3
CH3CH CHCH3
H Br CH3CH CHCH3 Br−
H
+ +
new s bond
Solely from the knowledge that an electrophile reacts with a nucleophile and a bond is the weakest bond in an alkene, we have been able to predict that the product of the reaction of 2-butene and HBr is 2-bromobutane. The overall reaction involves the addition of 1 mole of HBr to 1 mole of the alkene. The reaction, therefore, is called an addition reaction. Because the first step of the reaction is the addition of an elec- trophile (H ) to the alkene, the reaction is more precisely called an electrophilic addi- tion reaction. Electrophilic addition reactions are the characteristic reactions of alkenes.
At this point, you may think that it would be easier just to memorize the fact that 2- bromobutane is the product of the reaction, without trying to understand the mecha- nism that explains why 2-bromobutane is the product. Keep in mind, however, that you will soon be encountering a great number of reactions, and you will not be able to memorize them all. If you strive to understand the mechanism of each reaction, how- ever, the unifying principles of organic chemistry will soon be clear to you, making mastery of the material much easier and a lot more fun.
PROBLEM 14♦
Which of the following are electrophiles, and which are nucleophiles?
+
p
2. Curved arrows are meant to indicate the movement of electrons. Never use a curved arrow to indicate the movement of an atom. For example, do not use an arrow as a lasso to remove the proton, as shown here:
3. A head of a curved arrow always points at an atom or at a bond. Never draw the arrow head pointing out into space.
4. The arrow always starts at the electron source. In the following example, the arrow starts at the electron-rich bond, not at a carbon atom:
p O
CH3CCH3
H
+ H+
+ O
CH3CCH3
O
CH3CCH3 CH3CCH3
H
+ H+
+ O
incorrect correct
HO − HO −
O O
CH3COCH3 + CH3COCH3
O − CH3COCH3
OH
O − CH3COCH3
OH +
incorrect correct
PROBLEM 16
For reactions a-c in Problem 15, indicate which reactant is the nucleophile and which is the electrophile.
+ + Br−
CHCH3
CH3CH + H Br CH3CH CHCH3
H
+ + Br−
CHCH3
CH3CH + H Br CH3CH CHCH3
H incorrect
correct
PROBLEM 15
Use curved arrows to show the movement of electrons in each of the following reaction steps. (Hint: Look at the starting material and look at the products, then draw the arrows.) a.
b.
c.
d.
CH3
CH3
CH3
CH3
Cl C+
C
CH3 CH3 + Cl−
H
H O H
CH3COH O
H2O +
+ + CH3COH
+OH H Br
H
+ + Br−
+
O H HO−
CH3C O
+ CH3C O− H2O O
+
+ + +
CH3CH CH3CH CHCH3
‡
CH3CHCH2CH3
H Br transition state
CHCH3 H Br Br−
‡
transition state
Br Br
Br− CH3CHCH2CH3 CH3CHCH2CH3 CH3CHCH+ 2CH3 +
d+
d+
d−
d−
Progress of the reaction
a b
Free energy Free energy
Progress of the reaction ΔG‡
reactants
transition state transition state
product carbocation
carbocation ΔG‡
왘Figure 2
The reaction coordinate diagrams for the two steps in the addition of HBr to 2-butene:
(a) the first step;
(b) the second step.