17.1.1. Arrhenius concept

Weak acids are those that do not ionize completely and the dissociation of these acids provides us with an equilibrium reaction. A couple of examples are HC₂H₃O₂ and HNO₂.

An equilibrium constant can be calculated. Since the substance on the left of the written reaction is an acid and is behaving as one by releasing H⁺, the equilibrium constant is given a special symbol, K_a. For acetic acid, the K_a is calculated by

(17-1)

Note that the K_a expression is set up in the same way as the K's in Chapter 16: K_a is the product of the products divided by the product of the reactants, as in (17-1) above. As with any K, pure solids and liquids (including liquid water, the solvent, in these solutions) are not included in the K expression.

Bases are those substances that ionize in water to release OH⁻ ions. NaOH is a strong base, ionizing completely in water to yield Na⁺ and OH⁻ ions. However, even those hydroxide bases that do not dissolve well, such as Ca(OH)₂, do ionize completely to the extent they dissolve. A weak base is one that does not ionize completely. As with the acids, a K expression can be calculated, and that expression is referred to as K_b (the K for a base, as in (17-4) below).

An interesting situation that may occur when ammonia, NH₃, dissolves in water is that hydroxide ions, OH⁻, appear in the solution and the concentration is measurable. The implication is that NH₄OH forms, then ionizes to yield ammonium and hydroxide ions. This is logical, but NH₄OH has not been detected in aqueous solution. Since the OH⁻ concentration is only a few percent of the ammonia concentration, NH₃ is considered a weak base.

17.1.2. Brönsted-Lowry concept

A point of this reaction is the release of H⁺ ions—the Brönsted-Lowry approach considers this the appearance of a proton from the acid. A Brönsted-Lowry acid is a proton donor. Note that the hydrogen ion is a proton, hydrogen's nucleus without the electron found in the atom. Then, a Brönsted-Lowry acid must contain a hydrogen. Of course, if the solvent were not to be water, this statement may not work because the cation released could be other than the hydrogen ion, but there might be other ions performing the same service (liquid ammonia autoionizes, Problem 17.3).

The base in the Brönsted-Lowry concept is any substance that can accept the proton; it can even be the solvent. A Brönsted-Lowry base has an electron pair (a lone pair) that will accept the proton. The point that needs to be stressed here is that the proton is involved in both the definition of an acid (donates a proton) and a base (accepts a proton).

The Brönsted-Lowry concept looks at the equilibrium reaction and ties the acid on the left to a base on the right, called a conjugate acid-base pair or, more simply, a conjugate pair. Suppose we were to consider an acid reacting with a compound in equilibrium with the acid's anion and the products, as

(17-2)

The conjugate pair would contain the acid, HA, on the left and A⁻ on the right (bold participants). A⁻ is the result of acid's loss of its proton (H⁺). This relationship can be read as “A⁻ is the conjugate base of the acid, HA.” A⁻ is a base, because in the reverse reaction (reading from right to left), it will accept a proton (H⁺) and become the starting compound, HA.

B is a base, because it will accept the proton, whereas BH⁺ is an acid. BH⁺ is an acid because, if the reaction is read from right to left, it will give up the proton (H⁺). So, the relationship can be read as “B is the conjugate base of the acid, BH⁺.”

Note that HA and B are not necessarily neutral. They could be ions that are capable of acting as an acid or a base. This is one of the features of the Brönsted-Lowry concept that broadens the definitions of acids and bases over the Arrhenius concept—there are many more substances that can behave as acids or bases. Further, we can write the reaction including the solvent, water in this case, and the associated K_a.

(17-3)

Notice that water is not included in the K_a; it is a pure liquid and is omitted, as discussed in Chapter 16.

In a way similar to (17-3), we can write the ionization of the weak base, ammonia, and its K_b.

(17-4)

Water can act as either an acid or a base, depending on the circumstances. This ability to act as either an acid or a base is referred to by stating that water is amphoteric. Water serves as a base in (17-3) and as an acid in (17-4). Note that the bare H⁺ (a proton) becomes the hydronium ion, H₃O⁺, which is a hydrated proton (H₃O⁺ is H⁺ + H₂O) because the bare proton does not really exist in solution. When we write the equilibrium constant expression for an aqueous equilibrium, we can use either the hydrogen ion, H⁺, or the hydrated form, H₃O⁺. Although the proton is hydrated in aqueous solution (as is the hydroxide), the use of H⁺ and H₃O⁺ is up to the style of the person working the problem and the problem itself. More often than not, leaving out water on both sides of the equation is used to keep the solutions to the problems visually simple. So long as water is in its standard state (liquid), it is not included in the K expression and, therefore, not necessary in the chemical equation.

The strengths of acids can be compared in terms of their K_a's; the stronger the acid, the larger its K_a. This statement also applies to bases and their K_b's—the larger the K_b, the stronger the base. Also, the strengths of acids and bases are due to the magnitude of their respective K's in solvents other than water. For instance, HNO₃ is a strong acid in water and it is a weak acid in ethanol solvent—a K_a can be determined that is much smaller than that in water.

17.1.3. Lewis concept

Some other possible Lewis acids can be transition-metal ions, which can react with ligands (bases) to form complexes. Other substances that are short electrons are like BF₃, which can react with a base like NH₃ to form a compound, as below:

Nomenclature and Structure

Acetylene, C₂H₂, is a linear molecule in which each C uses two sp HO's to form two σ bonds with a 180° angle. The unhybridized p orbitals form two π bonds.

Problem 8.1 Name the structures below by the IUPAC system:

1. 2-Butyne

2. 2-Pentyne

3. 2,2,5-Trimethyl-3-hexyne

4. 1-Penten-4-yne

   and gets the smaller number.

5. 4-Chloro-1-butyne

6. 5-Hepten-1,3-diyne

Problem 8.2 Supply structural formulas and IUPAC names for all alkynes with the molecular formula (a) C₅H₈, (b) C₆H₁₀.

1. Insert a triple bond where possible in n-pentane, isopentane, and neopentane. Placing a triple bond in an n-pentane chain gives (1-pentyne) and (2-pentyne). Isopentane gives one compound,

   

   because a triple bond cannot be placed on a 3° C. No alkyne is obtainable from neopentane, (CH₃)₂C(CH₃)₂.

2. Inserting a triple bond in n-hexane gives

   

   Isohexane yields two alkynes, and 3-methylpentane and 2,2-dimethylbutane one alkyne each.

   

Problem 8.3 Draw models of (a) sp-hybridized C and (b) C₂H₂ to show bonds formed by orbital overlap.

1. See Fig. 8.1(a). Only one of three p orbitals of C is hybridized. The two unhybridized p orbitals (p_z and p_y) are at right angles to each other and also to the axis of the sp hybrid orbitals.

2. See Fig. 8.1(b). Sidewise overlap of the p_y and p_z orbitals on each C forms the π_y and π_z bonds, respectively.

Figure 8.1  

Problem 8.4 Why is the distance (0.120 nm) shorter than the (0.133 nm) and C—C (0.154 nm)?

The carbon nuclei in are shielded by six electrons (from three bonds) rather than by four or two electrons as in or C—C, respectively. With more shielding electrons present, the C's of can get closer, thereby affording more orbital overlap and stronger bonds.

Problem 8.5 Explain how the orbital picture of accounts for (a) the absence of geometric isomers in ; (b) the acidity of an acetylenic H, for example:

1. The sp-hybridized bonds are linear, ruling out cis-trans isomers in which substituents must be on different sides of the multiple bond.

2. We apply the principle: "The more s character in the orbital used by the C of the C—H bond, the more acidic is the H." Therefore, the order of acidity of hydrocarbons is

   

Problem 8.6 (a) Relate the observed C—H and C—C bond lengths and bond energies given in Table 8.1 in terms of the hybrid orbitals used by the C's involved. (b) Predict the relative C—C bond lengths in CH₃CH₃, , and .

Table 8.1  

COMPOUND	BOND	BOND LENGTH, nm	BOND ENERGY, kJ/mol
(1) CH3—CH3	—C—H	0.110	410
		0.108	423
		0.106	460
(4) CH3—CH3	C—C—	0.154	356
		0.151	377
		0.146	423

Bond energy increases as bond length decreases; the shorter bond length makes for greater orbital overlap and a stronger bond.

1. The hybrid nature of C is: (1) Csp3-Hs, (2) Csp2-Hs, (3) Csp-Hs, (4) Csp3-Csp3, (5) Csp3-Csp2, and (6) Csp3-Csp. In going from (1) to (3), the C—H bond length decreases as the s character of the hybrid orbital used by C increases. The same situation prevails for the C—C bond in going from (4) to (6). Bonds to C therefore become shorter as the s character of the hybridized orbital used by C increases.

2. The hybrid character of the C's in the C—C bond is: for CH₃—CH₃, Csp3-Csp3; , Csp2-Csp2; and , C_sp-C_sp. Bond length becomes shorter as s character increases and hence relative C—C bond lengths should decrease in the order

   

The observed bond lengths are, respectively, 0.154 nm, 0.149 nm, and 0.138 nm.

Laboratory Methods of Preparation

1. Dehydrohalogenation of vic-Dihalides or gem-Dihalides

   The vinyl (alkenyl) halide requires the stronger base sodamide (NaNH₂).

   

2. Primary Alkyl Substitution in Acetylene; Acidity of [see Problem 8.5(b)]

   

In E2 eliminations, the more acidic H is removed preferably. The inductive effect of the Br's increases the acidities of the H's on the C's to which the Br's are bonded. To get this product, the less acidic H (one of the CH₃ group) must be removed.

Problem 8.8 Outline a synthesis of propyne from isopropyl or propyl bromide.

The needed vic-dihalide is formed from propene, which is prepared from either of the alkyl halides.

Problem 8.9 Synthesize the following compounds from and any other organic and inorganic reagents (do not repeat steps): (a) 1-pentyne, (b) 2-hexyne.

Problem 8.10 Industrially, acetylene is made from calcium carbide, . Formulate the reaction as a Brönsted acid-base reaction.

The carbide anion C22− is the base formed when loses two H⁺'s.

Nomenclature; Aromaticity

The three most common five-membered aromatic rings are furan, with an O atom; pyrrole, with an N atom; and thiophene, with an S atom.

Problem 20.1 Name the following compounds, using (i) numbers and (ii) Greek letters.

(a) 2-methylthiophene (2-methylthiole) or α-methylthiophene, (b) 2,5-dimethylfuran (2,5-dimethyloxole) or α,α′-dimethylfuran, (c) 2,4-dimethylfuran or α,β′-dimethylfuran (2,4-dimethyloxole), (d) 1-ethyl-5-bromo-2-pyrrolecarboxylic acid or N-ethyl-α-bromo-α′-pyrrolecarboxylic acid (N-ethyl-5-bromazole-2-carboxylic acid).

Problem 20.2 Write structures for (a) 2-benzoylthiophene, (b) 3-furansulfonic acid, (c) α,β′-dichloropyrrole.

Problem 20.3 Account for the aromaticity of furan, pyrrole, and thiophene, which are planar molecules with bond angles of 120°.

See Fig. 20.1. The four C's and the heteroatom Z use sp²-hybridized atomic orbitals to form the σ bonds. When Z is O or S, one of the unshared pairs of e⁻'s is in an sp² HO. Each C has a p orbital with one electron, and the heteroatom Z has a p orbital with two electrons. These five p orbitals are parallel to each other and overlap side by side to give a cyclic π system with six p electrons. These compounds are aromatic because six electrons fit Hückel's 4n + 2 rule, which is extended to include heteroatoms.

Figure 20.1  

It is noteworthy that typically these heteroatoms would use sp³ HO's for bonding. The exceptional sp² HO's lead to a p AO for the cyclic aromatic π system.

Problem 20.4 Account for the following dipole moments: furan, 0.7D (away from O); tetrahydrofuran, 1.7 D (toward O).

In tetrahydrofuran, the greater electronegativity of O directs the moment of the C—O bond toward O. In furan, delocalization of an electron pair from O makes the ring C's negative and O positive; the moment is away from O. See Fig. 20.1.

Preparation

The carbonyl C's become the α C's in the heterocyclic compound.

Problem 20.6 Prepare pyrrole from succinic anhydride.

Problem 20.7 Identify compounds (A) through (D).

1. PhCOOEt

2. 

3. 

4. 

Problem 20.8 Dilantin (5,5-diphenylhydantoin), an anticonvulsant drug used in the treatment of epileptic seizures, is a pyrrole with the molecular formula C₁₅H₁₂N₂O₂. What is the structural formula for Dilantin?

Chemical Properties

1. The transition state and the intermediate R⁺ formed by α-attack is a hybrid of three resonance structures which possess less energy; the intermediate from β-attack is less stable and has more energy because it is a hybrid of only two resonance structures. I and II are also more stable allylic carbocations; V is not allylic.

   

2. This is ascribed to resonance structure III, in which Z has + charge and in which all ring atoms have an octet of electrons. These heterocyclics are as reactive as PhOH and PhNH₂.

Problem 20.10 Explain why pyrrole is not basic.

The unshared pair of electrons on N is delocalized and an "aromatic sextet." Adding an acid to N could prevent delocalization and destroy the aromaticity.

Problem 20.11 Give the type of reaction and the structures and names of the products obtained from: (a) furfural,

and concentrated aq. KOH; (b) furan with (i) CH₃CO—ONO₂ (acetyl nitrate), (ii) (CH₃CO)₂O and BF₃ and then H₂O; (c) pyrrole with (i) SO₃ and pyridine, (ii) CHCl₃ and KOH, (iii) PhN2+Cl−, (iv) Br₂ and C₂H₅OH; (d) thiophene and (i) H₂SO₄, (ii) (CH₃CO)₂O and CH₃COONO₂, (iii) Br₂ in benzene.

1. Cannizzaro reaction:

   

2. 1. Nitration; 2-nitrofuran:

      

   2. Acetylation; 2-acetylfuran:

      

3. 1. Sulfonation; 2-pyrrolesulfonic acid:

      

   2. Reimer-Tiemann formylation; 2-pyrrolecarboxaldehyde (2-formylpyrrole),

      

   3. Coupling; 2-phenylazopyrrole:

      

   4. Bromination; 2,3,4,5-tetrabromopyrrole.

4. 1. Sulfonation; thiophene-2-sulfonic acid:

      

   2. Nitration; 2-nitrothiophene:

      

   3. Bromination; 2,5-dibromothiophene. (Thiophene is less reactive than pyrrole and furan.)

Problem 20.12 Write structures for the mononitration products of the following compounds, and explain their formation: (a) 3-nitropyrrole, (b) 3-methoxythiophene, (c) 2-acetylthiophene, (d) 5-methyl-2-methoxythiophene, (e) 5-methylfuran-2-carboxylic acid.

1. Nitration at C⁵ to form 2,4-dinitropyrrole. After nitration, C⁵ (i) becomes C², and C³ becomes C⁴. Nitration at C² (ii) would form an intermediate with a + on C³, which has the electron-attracting —NO₂ group.

   

2. 

3. 

4. 

5. 

   Attack of NO⁺₂ at C², followed by elimination of CO₂ and H⁺.

Problem 20.13 Give the Diels-Alder product for the reaction of furan and maleic anhydride.

Furan is the least aromatic of the five-membered ring heterocyclics and acts as a diene toward strong dienophiles.

Problem 20.14 Give the products of reaction of pyrrole with (a) I₂ in aqueous KI; (b) CH₃CN + HCl, followed by hydrolysis; (c) CH₃MgI.

1. 2,3,4,5-Tetraiodopyrrole

2. α-Acetylpyrrole [see Problem 19.21 (c)]

3.