In 1923, G. N. Lewis offered new definitions for the terms acid and base. He defined an acid as a species that accepts a share in an electron pair and a base as a species that donates a share in an electron pair. All proton-donating acids fit the Lewis definition because all proton-donating acids lose a proton and the proton accepts a share in an electron pair.
Lewis acids, however, are not limited to compounds that donate protons. Accord- ing to the Lewis definition, compounds such as aluminum trichloride ( ) and borane ( ) are acids because they have unfilled valence orbitals and thus can ac- cept a share in an electron pair. These compounds react with a compound that has a lone pair, just as a proton reacts with ammonia. Thus, the Lewis definition of an acid includes all proton-donating compounds and some additional compounds that do not have protons. Throughout this text, the term acid is used to mean a proton-donating acid, and the term Lewis acid is used to refer to a non-proton-donating acid such as
or .BH3 AlCl3
BH3
AlCl3
H+ NH3 NH3
acid base + H
accepts a share in an electron pair
donates a share in an electron pair
the curved arrow indicates where the pair of electrons starts from and where it ends up
+
HO- H+ CH3COO-Na+
CH3COOH
Lewis acid: need two from you.
Lewis base: have pair, will share.
All bases are Lewis bases because they all have a pair of electrons that they can share, either with an atom such as aluminum or boron or with a proton.
PROBLEM 19
Write the products of the following reactions using arrows to show where the pair of electrons starts and where it ends up:
PROBLEM 20
Write the products formed from the reaction of each of the following species with :HO-
ZnCl2
a.
FeBr3
b.
AlCl3
c.
CH3OH +
+ +
Br− Cl−
Cl CH3OCH3 CH3
CH3
Cl Cl
− +
Al Cl
Cl Cl Al O +
aluminum trichloride a Lewis acid
dimethyl ether a Lewis base
H H H
H H
H
H H
+
B N H −
H
H H
B N
+
borane a Lewis acid
ammonia a Lewis base
SUMMARY
An acid is a species that donates a proton; a base is a species that accepts a proton. A Lewis acid is a species that accepts a share in an electron pair; a Lewis base is a species that donates a share in an electron pair.
Acidity is a measure of the tendency of a compound to give up a proton. Basicity is a measure of a compound’s affinity for a proton. The stronger the acid, the weaker is its conjugate base. The strength of an acid is given by the acid dissociation constant ( ). Approximate values are as follows: protonated alcohols, protonated carboxylic acids, protonated water ; carboxylic acids ; protonated amines ; alcohols and water . The pH of a solu- tion indicates the concentration of positively charged hy- drogen ions in the solution. In acid–base reactions, the
'15 '10
'5 6 0
pKa
Ka
equilibrium favors reaction of the stronger acid and forma- tion of the weaker acid.
The strength of an acid is determined by the stability of its conjugate base: the more stable the base, the stronger is its conjugate acid. When atoms are similar in size, the strongest acid will have its hydrogen attached to the most electronegative atom. When atoms are very different in size, the strongest acid will have its hydrogen attached to the largest atom.
A compound exists primarily in its acidic form in solu- tions more acidic than its value and primarily in its basic form in solutions more basic than its value. A buffer solution contains both a weak acid and its conjugate base.
pKa
pKa a.
b.
c.
d.
e.
f.
g.
h. CH3COOH AlCl3
FeBr3
+CH3 BF3
CH3N
+
H3 +NH4
CH3OH
PROBLEMS
21. For each of the following compounds, draw the form in which it will predominate at pH = 3, pH = 6, pH = 10,and pH = 14:
a. b. c.
pKa=12.4 CF3CH2OH pKa=11.0
CH3CH2N
+
H3 pKa=4.8
CH3COOH
22. Write the products of the following acid–base reactions, and indicate whether reactants or products are favored at equilibrium (use the pKavalues that are given in Section 3):
a.
b.
c.
d. CH3CH2OH + HCl O
+
CH3COH CH3NH2 +
CH3CH2OH −NH2 O
+
CH3COH CH3O−
23. a. Which of the following is the strongest acid?
b. Which is the weakest acid?
c. Which acid has the strongest conjugate base?
1. nitrous acid 2. nitric acid 3. bicarbonate
4. hydrogen cyanide (HCN),
5. formic acid (HCOOH), Ka = 2.0 * 10-4 Ka = 7.9 * 10-10
(HCO3-), Ka = 6.3 * 10-11 (HNO3), Ka = 22
(HNO2), Ka = 4.0 * 10-4
24. Which is the stronger base?
25. Locate the three nitrogen atoms in the electrostatic potential map of histamine, the compound that causes the symptoms associated with the common cold and allergic responses. Which of the two nitrogen atoms in the ring is the most basic?
:
This question has been intentionally removed
27. As long as the pH is greater than __________, more than 50% of a protonated amine with a of 10.4 will be in its neutral, nonprotonated form.
28. a. List the following carboxylic acids in order of decreasing acidity:
1. 2. 3. 4.
b. How does the presence of an electronegative substituent such as Cl affect the acidity of a carboxylic acid?
c. How does the location of the substituent affect the acidity of a carboxylic acid?
d. Why does the electronegative substituent affect the acidity of the carboxylic acid?
CH3CHCH2COOH Ka = 8.9 ⴛ 10−5
Cl ClCH2CH2CH2COOH
Ka = 2.96 ⴛ 10−5 CH3CH2CHCOOH
Ka = 1.39 ⴛ 10−3 Cl CH3CH2CH2COOH
Ka = 1.52 ⴛ 10−5
pKa
pKa
CH2CH2NH3
N H N
H H
+
histamine
a. HS– or HO– b. CH3O– or CH3NH– c. CH3OH or CH3O– d. Cl– or Br–
29. Explain the difference in the values of the following compounds:
30. a. List the following alcohols in order of decreasing acidity:
b. Explain their relative acidities.
31. Ethyne has a value of 25, water has a value of 15.7, and ammonia ( ) has a value of 36. Draw the equation, showing equilibrium arrows that indicate whether reactants or products are favored, for the reaction of ethyne with
a.
b.
c. Which would be a better base to use if you wanted to remove a proton from ethyne, or ?
32. For each of the following pairs of reactions, indicate which one has the more favorable equilibrium constant (that is, which one most favors products):
a.
or
b.
or
33. Carbonic acid has a of 6.1 at physiological temperature. Is the carbonic acid/bicarbonate buffer system that maintains the pH of the blood at 7.3 better at neutralizing excess acid or excess base?
34. Water and diethyl ether are immiscible liquids. Charged compounds dissolve in water, and uncharged compounds dissolve in ether. has a of 4.8 and
.
a. What pH would you make the water layer in order to cause both compounds to dissolve in it?
b. What pH would you make the water layer in order to cause the acid to dissolve in the water layer and the amine to dissolve in the ether layer?
c. What pH would you make the water layer in order to cause the acid to dissolve in the ether layer and the amine to dissolve in the water layer?
35. How could you separate a mixture of the following compounds? The reagents available to you are water, ether, 1.0 M HCl, and 1.0 M NaOH. (Hint: See Problem 34.)
+NH3Cl−
pKa = 10.66 Cl
OH
pKa = 9.95 +NH3Cl−
pKa = 4.60 COOH
pKa = 4.17 C6H11N
+
H3 has a pKa of 10.7
pKa
C6H11COOH pKa
CH3CH2OH+CH3NH2 Δ CH3CH2O- + CH3N
+
H3 CH3CH2OH+NH3 Δ CH3CH2O- + +NH4 CH3OH+NH3 Δ CH3O- + +NH4 CH3CH2OH+NH3 Δ CH3CH2O- ++NH4
-NH2 HO-
-NH2 HO-
pKa NH3
pKa pKa
CCl3CH2OH
Ka ⴝ 5.75 ⴛ 10ⴚ13 Ka ⴝ 1.29 ⴛ 10ⴚ13 Kaⴝ 4.90 ⴛ 10ⴚ13 CH2ClCH2OH CHCl2CH2OH
pKa = 4.76 ICHpK2a = 3.15 BrCHpK2a = 2.86 ClCHpK2a = 2.81 FCHpK2a = 2.66 C
OH O
C OH O
C OH O
C OH O CH3
C OH O
pKa
ether water
PHOTO CREDITS
Credits listed in order of appearance
Edgar Fahs Smith Memorial Collection; (left) © 1994 NYC Parks Photo Archive, Fundamental Photographs, NYC; (right) © 1994 Kristen Brochmann, Fundamental Photographs, NYC; Richard Megna\Fundamental Photographs, NYC.
ANSWERS TO SELECTED PROBLEMS
1. a. 1. 2. HCl 3. H2O 4. b. 1. 2. 3. 4.
3. a. compound with pKa= 5.2 b. compound with dissociation
4. ; weaker 6. a. basic b. acidic
c. basic 7. a. b. c.
8.
9.
11. a. HBr b. c. CH3CH2CH2O structure on the right 12. a. F- b. I- +
H2 CH3O− + CH3NH+ 3
CH3NH− > CH3O− > CH3NH2 > CH3CO− > CH3OH O
H2O -NH2 CH3COO-
Ka=1.51 10-5 constant= 3.4*10-3
HO- NO3- Br- -NH2 H3O+
+NH4 13. a. oxygen b. H2S c. CH3SH d. 14. a. b. NH3
c. d. 15. a. b. c. H2O d.
e. f. g. h. 16. a. 1. neutral 2. neutral 3. charged 4. charged 5. charged 6. charged b. 1. charged 2. charged 3. charged 4. charged 5. neutral 6. neutral c. 1. neutral 2. neutral 3. neutral 4. neutral 5. neutral 6. neutral
17. a. CH3CHCO− b. no O
+NH3
NO3- NO2- HC‚N +NH4
Br- CH3CH2N
+ H3 CH3COO-
CH3O- CH3O-
HO- CH3C(“O)SH
If we are going to talk about organic compounds, we need to know how to name them.
First, we will learn how alkanes are named because their names form the basis for the names of almost all organic compounds. Alkanes are composed of only carbon atoms and hydrogen atoms and contain only single bonds. Compounds that contain only carbon and hydrogen are called hydrocarbons. Thus, alkanes are hydrocarbons.
Alkanes in which the carbons form a continuous chain with no branches are called straight-chain alkanes. The names of several straight-chain alkanes are given in Table 1.
An Introduction to Organic Compounds
Nomenclature, Physical Properties, and Representation of Structure
CH3CH2NH2 CH3CH2Br
CH3OCH3
CH3CH2Cl CH3CH2OH
Table 1 Nomenclature and Physical Properties of Some Straight-Chain Alkanes Number
of carbons
Molecular
formula Name
Condensed structure
Boiling point (°C)
Melting point (°C)
Densitya (g/mL)
1 CH4 methane CH4 -167.7 -182.5
2 C2H6 ethane CH3CH3 -88.6 -183.3
3 C3H8 propane CH3CH2CH3 -42.1 -187.7 0.5005
4 C4H10 butane CH3CH2CH2CH3 -0.5 -138.3 0.5787 5 C5H12 pentane CH3(CH2)3CH3 36.1 -129.8 0.5572 6 C6H14 hexane CH3(CH2)4CH3 68.7 -95.3 0.6603 7 C7H16 heptane CH3(CH2)5CH3 98.4 -90.6 0.6837 8 C8H18 octane CH3(CH2)6CH3 125.7 -56.8 0.7026 9 C9H20 nonane CH3(CH2)7CH3 150.8 -53.5 0.7177 10 C10H22 decane CH3(CH2)8CH3 174.0 -29.7 0.7299
aDensity is temperature dependent. The densities given are those determined at 20 °C.
From Chapter 3 of Essential Organic Chemistry, Second Edition. Paula Y. Bruice.
Copyright © 2010 by Pearson Education, Inc. All rights reserved.
If you look at the relative numbers of carbon and hydrogen atoms in the alkanes listed in Table 1, you will see that the general molecular formula for an alkane is where n is any integer. So, if an alkane has one carbon atom, it must have four hydrogen atoms; if it has two carbon atoms, it must have six hydrogen atoms.
Carbon forms four covalent bonds and hydrogen forms only one covalent bond.
This means that there is only one possible structure for an alkane with molecular for- mula CH4(methane) and only one structure for an alkane with molecular formula C2H6(ethane). There is also only one possible structure for an alkane with molecular formula C3H8(propane).
C H
H Kekulé structure
name condensed structure ball-and-stick model
H CH4
H methane
C H
H C H
H CH3CH3
H H ethane
C H
H C H
C H
H
H CH3CH2CH3 H
H propane
C H
H C H
C H
H C H
H
H CH3CH2CH2CH3 H
H butane
CnH2n⫹2,
As the number of carbons in an alkane increases beyond three, the number of possible structures increases. There are two possible structures for an alkane with molecular formula C4H10. In addition to butane—a straight-chain alkane—there is a branched butane called isobutane. Both of these structures fulfill the requirement that each carbon forms four bonds and each hydrogen forms only one bond.
Compounds such as butane and isobutane that have the same molecular formula but differ in the order in which the atoms are connected are called constitutional isomers—their molecules have different constitutions. In fact, isobutane got its name because it is an “iso”mer of butane. The structural unit consisting of a carbon bonded to a hydrogen and two CH3groups—that occurs in isobutane—has come to be called
“iso.” Thus, the name isobutane tells you that the compound is a four-carbon alkane with an iso structural unit.
CH3CH2CH2CH3 butane
CH3CHCH3 isobutane
CH3
CH3CH an “iso”
structural unit CH3
3-D Molecules:
Methane; Ethane;
Propane; Butane
There are three alkanes with molecular formula C5H12. You have already learned to name two of them. Pentane is the straight-chain alkane. Isopentane, as its name indi- cates, has an iso structural unit and five carbon atoms. We cannot name the other branched-chain alkane without defining a name for a new structural unit. (For now, ignore the names written in blue.)
There are five constitutional isomers with molecular formula C6H14. Again, we are able to name only two of them, unless we define new structural units.
The number of constitutional isomers increases rapidly as the number of carbons in an alkane increases. For example, there are 75 alkanes with molecular formula C10H22 and 4347 alkanes with molecular formula C15H32. To avoid having to memorize the names of thousands of structural units, chemists have devised rules for creating systematic names that describe the compound’s structure. That way, only the rules have to be learned. Because the name describes the structure, these rules also make it possible to deduce the structure of a compound from its name.
CH3CH2CH2CH2CH2CH3 hexane hexane
CH3CHCH2CH2CH3 isohexane 2-methylpentane
CH3
CH3CH2CHCH2CH3 3-methylpentane
CH3
CH3CH CHCH3 2,3-dimethylbutane
CH3 CH3
CH3CCH2CH3 2,2-dimethylbutane
CH3 CH3 common name:
systematic name:
CH3CH2CH2CH2CH3 pentane
CH3CHCH2CH3 isopentane
CH3
CH3CCH3 2,2-dimethylpropane
CH3
CH3
FOSSIL FUELS
Alkanes are widespread both on Earth and on other planets. The atmospheres of Jupiter, Saturn, Uranus, and Neptune contain large quantities of methane (CH4), the smallest alkane, an odorless and flamma- ble gas. In fact, the blue colors of Uranus and Neptune are the result of methane in their atmospheres. Alkanes on Earth are found in natural gas and petroleum, which are formed by the decomposition of plant and animal material that has been buried for long periods in the Earth’s crust, where oxygen is scarce. Natural gas and petroleum, therefore, are known as fossil fuels.
Natural gas is approximately 75% methane. The remaining 25% is composed of other small alkanes such as ethane, propane, and butane. In the 1950s, natural gas replaced coal as the main energy source for domestic and industrial heating in the United States.
Petroleum is a complex mixture of alkanes that can be sepa- rated into fractions by distillation. The fraction that boils off at the lowest temperature (hydrocarbons containing three and four
carbons) is a gas that can be liquefied under pressure.
This gas is used as a fuel for cigarette lighters, camp stoves, and barbecues. The fraction that boils at some- what higher temperatures (hydrocarbons containing 5 to 11 carbons) is gasoline;
the next fraction (9 to 16 carbons) includes kerosene and jet fuel. The fraction with 15 to 25 carbons is used for heating oil and diesel oil, and the highest boiling fraction is used for lubricants and greases.
After distillation, a non-
volatile residue called asphalt or tar is left behind.
gasoline kerosene, jet fuel heating oil, diesel oil lubricants, greases
heating element natural gas
asphalt, tar
A PROBLEMATIC ENERGY SOURCE
Modern society faces three major problems as a consequence of our dependence on fossil fuels for energy. First, these fuels are a nonrenewable resource and the world’s supply is continually decreasing.
Second, a group of Middle Eastern and South American countries controls a large portion of the world’s supply of petroleum. These countries have formed a cartel known as the Organization of Petroleum Exporting Countries (OPEC), which controls both the supply and the price of crude oil. Political instability in any OPEC country can seri- ously affect the world oil supply. Third, burning fossil fuels increases the concentrations of CO2 in the atmosphere;
burning coal increases the concentration of both CO2 and SO2. Scientists have established experimentally that atmo- spheric SO2causes “acid rain,” which represents a threat to
the Earth’s plants and, therefore, to our food and oxygen supplies.
Since 1958, the concentration of atmospheric CO2 at Mauna Loa, Hawaii has been measured periodically. The con- centration has increased 20% since the first measurements were taken, causing scientists to predict an increase in the Earth’s temperature as a result of the absorption of infrared ra- diation by CO2(the greenhouse effect). A steady increase in the temperature of the Earth would have devastating conse- quences, including the formation of new deserts, massive crop failure, decreasing polar ice sheets, and the melting of glaciers with a concomitant rise in sea level. Clearly, what we need is a renewable, nonpolitical, nonpolluting, and economically affordable source of energy.
This method of nomenclature is called systematic nomenclature. It is also called IUPAC nomenclature because it was designed by a commission of the International Union of Pure and Applied Chemistry (abbreviated IUPAC and pronounced “eye-you- pack”). Names such as isobutane—nonsystematic names—are called common names and are shown in red in this text. The systematic (IUPAC) names are shown in blue.
Before we can understand how a systematic name for an alkane is constructed, we must learn how to name alkyl substituents.
methyl alcohol
methyl chloride
methylamine