Organic Chemistry

44 CHAPTER 1 Electronic Structure and Bonding • Acids and Bases

Protonated alcohols, protonated carboxylic acids, and protonated water have val-

ues less than 0, carboxylic acids have values of about 5, protonated amines have

values of about 10, and alcohols and water have values of about 15. These

values are also listed inside the back cover of this book for easy reference.

Now let’s see how we knew that water acts as a base in the first reaction in

Section 1.16 and as an acid in the second reaction. To determine which of the reactants

will be the acid, we need to compare their values: The of hydrogen chloride

is and the of water is 15.7. Because hydrogen chloride is the stronger acid, it

will donate a proton to water. Water, therefore, is a base in this reaction. When we

compare the values of the two reactants of the second reaction, we see that the

of ammonia is 36 and the of water is 15.7. In this case, water is the stronger

acid, so it donates a proton to ammonia. Water, therefore, is an acid in this reaction.

In determining the position of equilibrium for an acid–base reaction (i.e., whether

reactants or products are favored at equilibrium), remember that the equilibrium favors

reactionof the strong acid and strong base and formationof the weak acid and weak

base. In other words,strong reacts to give weak. Thus, the equilibrium lies away from

the stronger acid and toward the weaker acid.

PROBLEM 30

a. For each of the acid–base reactions in Section 1.17, compare the values of the acids

on either side of the equilibrium arrows and convince yourself that the position of equi-

librium is in the direction indicated. (The values you need can be found in

Section 1.17 or in Problem 31.

b. Do the same thing for the equilibria in Section 1.16. (The of is 9.4.)

The precise value of the equilibrium constant can be calculated by dividing the

of the reactant acid by the of the product acid.

Thus, the equilibrium constant for the reaction of acetic acid with ammonia is

and the equilibrium constant for the reaction of ethanol with methylamine

is The calculations are as follows:

reaction of acetic acid with ammonia:

reaction of ethanol with methylamine:

Keq=

10 - 15.9

10 - 10.7

= 10 - 5.2=6.3* 10 -^6

Keq=

10 - 4.8

10 - 9.4

= 10 4.6=4.0* 104

6.3* 10 -^6.

4.0* 104 ,

Keq=

Ka reactant acid

Ka product acid

Ka

Ka

+NH

pKa 4

pKa

pKa

stronger acid

pKa = 4.8

stronger base

+ NH 3

weaker base weaker acid

pKa = 9.4

+ NH 4

+

+

CH 3 CH 2 OH

weaker acid

pKa = 15.9

weaker base

+ CH 3 NH 2 CH 3 CH 2 O−

stronger base stronger acid

pKa = 10.7

+ CH 3 NH 3

CH 3

C

OH

O

CH 3

C

O–

O

pKa pKa

pKa

• 7 pKa

pKa pKa

pKa pKa

pKa

pKa

Strong reacts to give weak.

Be sure to learn the approximate
pKavalues given in Table 1.8.

Tutorial:

Acid–base reaction