Organic Chemistry: Concepts and Applications - Headley Allan D. 2020
Acid—Base Reactions in Organic Chemistry
7.5 Factors That Affect Acid and Base Strengths
It is possible to predict the relative acidity of compounds without knowledge of the actual pKa values. For most of the reactions that will be examined in organic chemistry, knowledge of the acidity or basicity of the reactants will assist in the prediction of products and the type of reaction needed to produce desired products. There are various factors that affect the acidity of compounds and they do so in a predictable manner, and each is discussed in the next sections.
7.5.1 Electronegativity
By definition, electronegativity is the ability of an atom to attract electrons toward itself. Thus, the electronegativity of the atom, which is bonded to a hydrogen atom of an acid, plays an important role in the relative acidity of that compound. For example, the relative acidities of ammonia and methane can be determined from examination of the electronegativity of the atoms to which the acidic hydrogen is directly bonded. The dissociation of these compounds is shown in Reactions (7-12) and (7-13).
(7-12)−
(7-13)−
Since nitrogen is a more electronegative atom than carbon, the nitrogen can accommodate the negative charge of the conjugate base much better than the less electronegative carbon atom. Thus, the conjugate base, NH2−, is a weaker base compared to the conjugate base CH3−. As a result, methane is a weaker acid than ammonia. This information can be confirmed from the pKa table, which gives the pKa values of methane and ammonia as 50 and 38, respectively. The ability of the conjugate base to stabilize its negative charge is very important. As mentioned before, the most electronegative atoms are the most appropriate atoms to accommodate a negative charge.
Problem 7.5
For the Brønsted—Lowry pairs of acids shown below, give the structures of the conjugate bases? Without using the pKa table, determine which of the following pairs of acids is more acidic?i)
i. CH4 and CH3OH
ii. NH3 and CH3OH
iii. CH4 and NH3
7.5.2 Type of Hybridized Orbitals
As seen in Chapter 1, hybridized orbitals of carbon result from the mixture of s and p orbitals: sp3 orbitals are 75% p character and 25% s character; sp2 orbitals are 66% p character and 33% s character; and sp orbitals are 50% p character and 50% s character. Since the s orbitals are closer to the nucleus, compared to the more diffused p orbitals, hybridized orbitals that have more s character are closer to the nucleus than orbitals that have less s character. Thus, alkynes are more acidic than alkenes and alkanes as shown by Reactions.
(7-14)
(7-15)
(7-16)
Since the conjugate base of an alkyne has a negative charge on the more electronegative sp-hybridized carbon, it is a weaker base than that of the conjugate base of either an alkene or an alkane. Hence, alkynes are stronger acids than either alkenes or alkanes, and alkenes are stronger acids than alkanes. From the pKa table, the pKa values of most terminal alkynes are around 25, which make them acidic, even though weak acids and they can be involved in an acid—base reaction with an appropriate base, as shown in Reaction (7-17).
(7-17)≡−
As shown in the reaction in Reaction (7-17), in order to remove the proton from propyne, a strong enough base is needed to ensure that the equilibrium lies to the right. Sodium amide (NaNH2) is a strong enough base to remove the proton from an alkyne since the pKa of the conjugate acid of NaNH2 is 38.
Problem 7.6
Using the pKa table, predict appropriate bases that can be used to complete the following acid—base reactions.
7.5.3 Resonance
As mentioned in Chapter 1, resonance structures are equivalent Lewis dot structures and represent the movement of nonbonded electrons or pi (π) electrons across a number of adjacent atoms. Resonance structures for charged species also indicate the degree of possible charge delocalization within an ion, and hence the degree of stabilization of an ion. The greater the number of resonance structures there are for an ion, the greater is the stability of that ion. Hence, a conjugate base, which is typically negatively charged, that has a number of resonance structures is more stable than another similar conjugate base, which does not have as many resonance structures. The conjugate base of phenol has many more resonance structures, compared to the conjugate base of cyclohexanol, as shown in Reactions (7-18) and (7-19).
(7-18)
(7-19)
Since the conjugate base of phenol is resonance stabilized, it is more stable than the conjugate base of cyclohexanol, which is not resonance stabilized. Thus, phenol is a stronger acid (resulting in a weaker conjugate base) than cyclohexanol, which has a stronger conjugate base, as shown in the equilibrium reaction given in Reaction (7-20), which is to the right.
(7-20)
Problem 7.7
Of the pairs of molecules shown below, (a) remove the proton from the most acidic molecule and draw another resonance of the conjugate base; (b) which is the stronger acid?
As shown in Reaction (7-20) and Table 7.1, the pKa values of aliphatic alcohols are approximately 15—16, and as a result, very strong bases are required to abstract the acidic hydrogen bonded to the oxygen. One of the strongest bases in organic chemistry is sodium hydride, which can react with alcohols to form the conjugate alkoxide ion of the alcohol as shown in Reaction (7-21).
(7-21)
Compared to the hydrogens of alkanes, the hydrogens bonded to the α-carbon of carbonyl compounds are fairly acidic. The reason for the higher acidity (lower pKa values) of the hydrogens bonded to the α-carbon of carbonyl compounds is due to the stability of the conjugate base that results upon deprotonation as shown in Reaction (7-22).
(7-22)αα
As shown in the reaction in Reaction (7-22), the conjugate base is resonance stabilized and hence fairly stable resulting in pKa values of most ketones and aldehydes of approximately 20 as shown in Table 7.1. The conjugate bases of carbonyl compounds are called enolates since the protonation of one of the resonance structures results in enols. The removal of the α-hydrogen requires a strong enough base in which its conjugate acid’s pKa would have to be greater than 20. The conjugate base of water (the hydroxide anion) is often used to deprotonate the α-hydrogen of carbonyl compounds even though the pKa of the conjugate acid (water) is around 15. Since the pKa values are fairly close, an equilibrium exists and slightly favors the neutral carbonyl compound, but a mixture of the acids and conjugate bases will exist as shown in Reaction (7-23). The use of a stronger base would favor the enolate formation.
(7-23)
Problem 7.8
Give the structure of the conjugate bases that would result from the following acid—base reactions.
7.5.4 Polarizability/Atom Size
The size of the atom, which bears the negative charge of a conjugate base, is also an important factor in the determination of the stability of the conjugate base, and hence, the acidity of the conjugate acid. A large atom can best accommodate a negative charge compared to a smaller atom. This concept, also known as hard and soft bases, was discussed earlier in the chapter. For the two equilibria shown in Reaction (7-24) and (7-25), note that iodine is the larger atom, compared to chlorine, and as a result, the iodide anion can accommodate the negative charge of the conjugate base much better than the smaller chloride anion.
(7-24)
(7-25)
The ability of a large atom to accommodate a charge is based on the fact that it has more electrons that can be polarized (or moved about) to accommodate a negative charge.
Problem 7.9
Of the following pairs of acids, which is the stronger acid? Explain your answer.
1. HBr and HCI
2. HI and HCl
7.5.5 Inductive Effect
If the conjugate base of an acid contains an electronegative atom, then the negatively charged conjugate base can be stabilized by an electronegative atom that is close to the negative charge. From the periodic table, electronegativity of the atoms increases across a period from left to top right and decreases down a family or group. Thus, the carboxylate salt of a conjugate carboxylic acid that has a fluorine atom bonded in the α-position is more stable than the carboxylate salt that has a less electronegative chlorine atom bonded to the same α-carbon. As a result, a carboxylic acid that has a very electronegative atom bonded to the α-position is more acidic than another similar carboxylic acid that has a less electronegative atom bonded in the same α-position, this concept is illustrated below.
This concept applies to molecules that have electronegative atoms in other positions in relationship to the acidic proton of molecules. Thus, if a fluorine is in the β-position of a carboxylic acid, it will be more acidic than another similar carboxylic acid that has a less electronegative atom, such as chlorine, in the same β-position.
Problem 7.10
Of the following pairs of acids, which is the stronger acid?
Of course, if the same electronegative atom resides at different locations on different conjugate bases, the more stable (weaker) conjugate base is the one that has the electronegative atom closest to the negative charge of the conjugate base. Thus, 2-fluorobutanoic acid is a stronger acid than 4-fluorobutanoic acid.
The same is true for substituted benzoic acids, 2-fluorobenzoic acid is a stronger acid than 4-fluorobenzoic acid.
Problem 7.11
Of the two pairs of acids shown below, determine which is the stronger acid.
