Solution - 19.11 Nucleophilic Addition of Phosphorus Ylides: The Wittig Reaction
Aldehydes and ketones are converted into alkenes by means of a nucleophilic addition called the Wittig reaction, after the German chemist Georg Wittig who won the 1979 Nobel Prize in Chemistry. The reaction has no direct biological counterpart but is important both because of its wide use in drug manufacture and because of its mechanistic similarity to reactions of the coenzyme thiamin diphosphate, which we'll see in Section 29.6.

In the Wittig reaction, a triphenylphosphorus ylide, $\mathrm{R}_{2} \stackrel{-}{\mathrm{C}}-\stackrel{+}{\mathrm{P}} \mathrm{Ph}_{3}$ also called a phosphorane and sometimes written in the resonance form $\mathrm{R}_{2} \mathrm{C}=\mathrm{P}(\mathrm{Ph})_{3}$, adds to an aldehyde or ketone to yield a four-membered cyclic intermediate called an oxaphosphetane. The oxaphosphetane is not isolated, but instead spontaneously decomposes to give an alkene plus triphenylphosphine oxide, $\mathrm{O}=\mathrm{PPh}_{3}$. In effect, the oxygen atom of the aldehyde or ketone and the $\mathrm{R}_{2} \mathrm{C}=$ bonded to phosphorus exchange places. (An ylide-pronounced ill-id-is a neutral, dipolar compound with adjacent positive and negative charges.)

The initial addition takes place by different pathways depending on the structure of the reactants and the exact experimental conditions. One pathway involves a one-step cycloaddition process analogous to the Diels-Alder cycloaddition reaction (Section 14.4). The other pathway involves a nucleophilic addition reaction to give a dipolar intermediate called a betaine (bay-ta-een), which undergoes ring closure.


Oxaphosphetane -
The phosphorus ylides necessary for Wittig reaction are easily prepared by $\mathrm{S}_{\mathrm{N}} 2$ reaction of primary (and some secondary) alkyl halides with triphenylphosphine, ( Ph$)_{3} \mathrm{P}$, followed by treatment with base. Triphenylphosphine is a good nucleophile in $\mathrm{S}_{\mathrm{N}} 2$ reactions, and yields of the resultant alkyltriphenylphosphonium salts are high. Because of the positive charge on phosphorus, the hydrogen on the neighboring carbon is weakly acidic and can be removed by a strong base such as butyllithium (BuLi) to generate the neutral ylide. For example:

Triphenylphosphine

Methyltriphenylphosphonium bromide

Methylenetriphenylphosphorane

The Wittig reaction is extremely general, and a great many monosubstituted, disubstituted, and trisubstituted alkenes can be prepared from the appropriate combination of phosphorane and aldehyde or ketone. Tetrasubstituted alkenes, however, can't be prepared this way because of steric hindrance.

The real value of the Wittig reaction is that it yields a pure alkene of predictable structure. The $\mathrm{C}=\mathrm{C}$ bond in the product is always exactly where the $\mathrm{C}=\mathrm{O}$ group was in the reactant, and no alkene isomers (except $E, Z$ isomers) are formed. For example, Wittig reaction of cyclohexanone with methylenetriphenylphosphorane yields only the single alkene product methylenecyclohexane. By contrast, addition of methylmagnesium bromide to cyclohexanone, followed by dehydration with $\mathrm{POCl}_{3}$, yields a roughly $9: 1$ mixture of two alkenes.

Wittig reactions are used commercially in the synthesis of numerous pharmaceutical products. For example, the German chemical company BASF prepares vitamin A by using a Wittig reaction between a 15 -carbon ylide and a 5-carbon aldehyde.


Oxaphosphetane -
$$
\downarrow \begin{aligned}
& \mathrm{Na}^{+-} \mathrm{OCH}_{3} \\
& \mathrm{CH}_{3} \mathrm{OH}
\end{aligned}
$$

Vitamin A acetate


Synthesizing an Alkene Using a Wittig Reaction -
What carbonyl compound and what phosphorus ylide might you use to prepare 3-ethyl-2-pentene?


Strategy -
An aldehyde or ketone reacts with a phosphorus ylide to yield an alkene in which the oxygen atom of the carbonyl reactant is replaced by the $=\mathrm{CR}_{2}$ of the ylide. Preparation of the phosphorus ylide itself usually involves $\mathrm{S}_{\mathrm{N}} 2$ reaction of a primary alkyl halide with triphenylphosphine, so the ylide is typically primary, $\mathrm{RCH}=\mathrm{P}(\mathrm{Ph})_{3}$. This means that the disubstituted alkene carbon in the product comes from the carbonyl reactant, while the monosubstituted alkene carbon comes from the ylide.

Solution

PROBLEM What carbonyl compound and what phosphorus ylide might you use to prepare each of the 19-16 following compounds?
(a)

(b)

(c)

(d)

(e)

(f)

PROBLEM $\beta$-Carotene, a yellow food-coloring agent and dietary source of vitamin A can be prepared by a
19-17 double Wittig reaction between 2 equivalents of $\beta$-ionylideneacetaldehyde and a diylide. Show the structure of the $\beta$-carotene product.


Strategy - 19.12 Biological Reductions
As a general rule, nucleophilic addition reactions are characteristic only of aldehydes and ketones, not of carboxylic acid derivatives. The reason for the difference is structural. As discussed previously in the Preview of Carbonyl Chemistry, and shown in FIGURE 19.12, the tetrahedral intermediate produced by addition of a nucleophile to a carboxylic acid derivative can eliminate a leaving group, leading to a net nucleophilic acyl substitution reaction. The tetrahedral intermediate produced by addition of a nucleophile to an aldehyde or ketone, however, has only alkyl or hydrogen substituents and thus can't usually expel a leaving group. One exception to this rule is the Cannizzaro reaction, discovered in 1853.

$$
\begin{array}{ll}
\text { Reaction occurs when: } & \mathrm{Y}=-\mathrm{Br},-\mathrm{Cl},-\mathrm{OR},-\mathrm{NR}_{2} \\
\text { Reaction does NOT occur when: } & \mathrm{Y}=-\mathrm{H},-\mathrm{R}
\end{array}
$$

FIGURE 19.12 Carboxylic acid derivatives have an electronegative substituent $\mathbf{Y}=-\mathrm{Br},-\mathrm{Cl},-\mathrm{OR},-\mathrm{NR}_{\mathbf{2}}$ that can be expelled as a leaving group from the tetrahedral intermediate formed by nucleophilic addition. Aldehydes and ketones have no such leaving group and thus do not usually undergo this reaction.

The Cannizzaro reaction takes place by nucleophilic addition of $\mathrm{OH}^{-}$to an aldehyde to give a tetrahedral intermediate, which expels hydride ion as a leaving group and is thereby oxidized. A second aldehyde molecule accepts the hydride ion in another nucleophilic addition step and is thereby reduced. Benzaldehyde, for instance, yields benzyl alcohol plus benzoic acid when heated with aqueous NaOH .


Strategy - 19.12 Biological Reductions
The Cannizzaro reaction is rarely used today but is interesting mechanistically because it is a simple laboratory analogy for the primary biological pathway by which carbonyl reductions occur in living organisms. In nature, as we saw in Section 17.4, one of the most important reducing agents is NADH, reduced nicotinamide adenine dinucleotide. NADH donates $\mathrm{H}^{-}$to aldehydes and ketones, thereby reducing them, in much the same way that the tetrahedral alkoxide intermediate in a Cannizzaro reaction does. The electron lone pair on a nitrogen atom of NADH expels $\mathrm{H}^{-}$as leaving group, which adds to a carbonyl group in another molecule (FIGURE 19.13). As an example, pyruvate is converted during intense muscle activity to (S)-lactate, a reaction catalyzed by lactate dehydrogenase.

NADH

FIGURE 19.13 Mechanism of biological aldehyde and ketone reductions by the coenzyme NADH. The key step is an expulsion of hydride ion from NADH and donation to the carbonyl group.

PROBLEM When $o$-phthalaldehyde is treated with base, $o$-(hydroxymethyl)benzoic acid is formed. Show the
19-18 mechanism of this reaction.

PROBLEM What is the stereochemistry of the pyruvate reduction shown in Figure 19.13? Does NADH lose its
19-19 pro- $R$ or pro- $S$ hydrogen? Does addition occur to the Si face or Re face of pyruvate? (Review Section 5.11.)


Strategy - 19.13 Conjugate Nucleophilic Addition to $\boldsymbol{\alpha}, \boldsymbol{\beta}$-Unsaturated Aldehydes and Ketones
All the reactions we've been discussing to this point have involved the addition of a nucleophile directly to the carbonyl group, a so-called $\mathbf{1 , 2}$-addition. Closely related to this direct addition is the conjugate addition, or 1,4-addition, of a nucleophile to the $\mathrm{C}=\mathrm{C}$ bond of an $\alpha, \beta$-unsaturated aldehyde or ketone. (The carbon atom next to a carbonyl group is often called the $\alpha$ carbon, the next carbon is the $\beta$ carbon, and so on. Thus, an $\alpha, \beta$-unsaturated aldehyde or ketone has a double bond conjugated with the carbonyl group.) The initial product of conjugate addition is a resonance-stabilized enolate ion, which typically undergoes protonation on the $\alpha$ carbon to give a saturated aldehyde or ketone product (FIGURE 19.14).


Conjugate (1,4) addition -
FIGURE 19.14 A comparison of direct $(1,2)$ and conjugate $(1,4)$ nucleophilic addition reactions. In conjugate addition, a nucleophile adds to the $\beta$ carbon of an $\alpha, \beta$-unsaturated aldehyde or ketone and protonation occurs on the $\alpha$ carbon.

The conjugate addition of a nucleophile to an $\alpha, \beta$-unsaturated aldehyde or ketone is caused by the same electronic factors that are responsible for direct addition. The electronegative oxygen atom of the $\alpha, \beta$-unsaturated carbonyl compound withdraws electrons from the $\beta$ carbon, thereby making it electron-poor and more electrophilic than a typical alkene carbon.

As noted, conjugate addition of a nucleophile to the $\beta$ carbon of an $\alpha, \beta$-unsaturated aldehyde or ketone leads to an enolate ion intermediate, which is protonated on the $\alpha$ carbon to give the saturated product (FIGURE 19.14). The net effect is addition of the nucleophile to the $\mathrm{C}=\mathrm{C}$ bond, with the carbonyl group itself unchanged. In fact, of course, the carbonyl group is crucial to the success of the reaction. Without the carbonyl group, the $\mathrm{C}=\mathrm{C}$ bond would not be activated for addition, and no reaction would occur.

Activated double bond

Unactivated double bond


Conjugate Addition of Amines -
Both primary and secondary amines add to $\alpha, \beta$-unsaturated aldehydes and ketones to yield $\beta$-amino aldehydes and ketones rather than the alternative imines. Under typical reaction conditions, both modes of addition occur
rapidly. But because the reactions are reversible, they generally proceed with thermodynamic control rather than kinetic control (Section 14.3), so the more stable conjugate addition product is often obtained with the complete exclusion of the less stable direct addition product.


Conjugate Addition of Water -
Water can add reversibly to $\alpha, \beta$-unsaturated aldehydes and ketones to yield $\beta$-hydroxy aldehydes and ketones, although the position of the equilibrium generally favors unsaturated reactant rather than saturated adduct. Related additions to $\alpha, \beta$-unsaturated carboxylic acids occur in numerous biological pathways, such as the citric acid cycle of food metabolism in which cis-aconitate is converted into isocitrate by conjugate addition of water to a double bond.

PROBLEM Assign $R$ or $S$ stereochemistry to the two chirality centers in isocitrate, and tell whether OH and H 19-20 add to the Si face or the Re face of the double bond.


Conjugate Addition of Alkyl Groups: Organocopper Reactions -
The conjugate addition of an alkyl or other organic group to an $\alpha, \beta$-unsaturated ketone (but not aldehyde) is one of the more useful 1,4-addition reactions, just as direct addition of a Grignard reagent is one of the more useful 1,2-additions.

$\alpha, \beta$-Unsaturated ketone
Conjugate addition of an organic group is carried out by treating the $\alpha, \beta$-unsaturated ketone with a lithium diorganocopper reagent, $\mathrm{R}_{2} \mathrm{CuLi}$. As we saw in Section 10.7, lithium diorganocopper (Gilman) reagents are prepared by reaction between 1 equivalent of copper(I) iodide and 2 equivalents of an organolithium regent, RLi. The organolithium reagent, in turn, is formed by reaction of lithium metal with an organohalide in the same way that a Grignard reagent is prepared by reaction of magnesium metal with an organohalide.

# $\mathrm{RX} \xrightarrow[\text { Pentane }]{2 \mathrm{Li}} \mathrm{RLi}+\mathrm{Li}^{+} \mathrm{X}^{-}$ <br> $2 \mathrm{RLi} \xrightarrow[\text { Ether }]{\mathrm{CuI}} \mathrm{Li}^{+}(\mathrm{RCuR})+\mathrm{Li}^{-} \mathrm{I}^{-}$


A lithium
 diorganocopper
 (Gilman reagent) -
Primary, secondary, and even tertiary alkyl groups undergo the conjugate addition reaction, as do aryl and alkenyl groups. Alkynyl groups, however, react poorly in the conjugate addition process. Diorganocopper reagents are unique in their ability to give conjugate addition products. Other organometallic reagents, such as Grignard reagents and organolithiums, normally result in direct carbonyl addition on reaction with $\alpha, \beta$-unsaturated ketones.

The mechanism of this reaction is thought to involve conjugate nucleophilic addition of the diorganocopper anion, $\mathrm{R}_{2} \mathrm{Cu}^{-}$, to the unsaturated ketone to give a copper-containing intermediate. Transfer of an R group from copper to carbon, followed by elimination of a neutral organocopper species, RCu, gives the final product.


Using a Conjugate Addition Reaction -
How might you use a conjugate addition reaction to prepare 2-methyl-3-propylcyclopentanone?

2-Methyl-3-propylcyclopentanone


Strategy -
A ketone with a substituent group in its $\beta$ position might be prepared by a conjugate addition of that group to an $\alpha, \beta$-unsaturated ketone. In the present instance, the target molecule has a propyl substituent on the $\beta$ carbon and might therefore be prepared from 2-methyl-2-cyclopentenone by reaction with lithium dipropylcopper.


Solution -
PROBLEM Treatment of 2-cyclohexenone with HCN/KCN yields a saturated keto nitrile rather than an
19-21 unsaturated cyanohydrin. Show the structure of the product, and propose a mechanism for the reaction.

PROBLEM How might conjugate addition reactions of lithium diorganocopper reagents be used to synthesize 19-22 the following compounds?
(a)
(b)

(c)

(d)


Infrared Spectroscopy -
Aldehydes and ketones show a strong $\mathrm{C}=\mathrm{O}$ bond absorption in the IR region from 1660 to $1770 \mathrm{~cm}^{-1}$, as the spectra of benzaldehyde and cyclohexanone demonstrate (FIGURE 19.15). In addition, aldehydes show two characteristic C-H absorptions between $2700-2760$ and $2800-2860 \mathrm{~cm}^{-1}$. These absorbances are important for distinguishing between aldehydes and ketones. The higher-frequency absorbance is sometimes obscured in cases where the compound has numerous saturated $\mathrm{C}-\mathrm{H}$ groups, but the lower-frequency peak is almost always visible.

FIGURE 19.15 Infrared spectra of (a) benzaldehyde and (b) cyclohexanone.
The exact position of the $\mathrm{C}=\mathrm{O}$ absorption is diagnostic of the nature of the carbonyl group. As the data in TABLE 19.2 indicate, saturated aldehydes usually show carbonyl absorptions near $1730 \mathrm{~cm}^{-1}$ in the IR spectrum, but conjugation of the aldehyde to an aromatic ring or a double bond lowers the absorption by $25 \mathrm{~cm}^{-1}$ to near $1705 \mathrm{~cm}^{-1}$. Saturated aliphatic ketones and cyclohexanones both absorb near $1715 \mathrm{~cm}^{-1}$, and conjugation with a double bond or an aromatic ring again lowers the absorption by $30 \mathrm{~cm}^{-1}$ to $1685-1690 \mathrm{~cm}^{-1}$. Angle strain in the carbonyl group, caused by reducing the ring size of cyclic ketones to four or five, raises the absorption position. Cyclohexanone has its $\mathrm{C}=\mathrm{O}$ stretch absorbance at $1715 \mathrm{~cm}^{-1}$ while cyclopentanone's carbonyl stretch is at $1750 \mathrm{~cm}^{-1}$ and cyclobutanone's is at $1785 \mathrm{~cm}^{-1}$.

TABLE 19.2 Infrared Absorptions of Some Aldehydes and Ketones


Infrared Spectroscopy -
TABLE 19.2 Infrared Absorptions of Some Aldehydes and Ketones

| Carbonyl type | Example | Absorption (cm $\left.{ }^{\mathbf{- 1}}\right)$ |
| :--- | :--- | :---: |
| Saturated aldehyde | $\mathrm{CH}_{3} \mathrm{CHO}$ | 1730 |
| Aromatic aldehyde | PhCHO | 1705 |
| $\alpha, \beta$-Unsaturated aldehyde | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCHO}$ | 1705 |
| Saturated ketone | $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ | 1715 |
| Cyclohexanone |  | 1715 |
| Cyclopentanone |  | 1750 |
| Cyclobutanone | $\mathrm{PhCOCH}_{3}$ | 1785 |
| Aromatic ketone | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCOCH}_{3}$ | 1685 |
| $\alpha, \beta$-Unsaturated ketone |  |  |

The values given in TABLE 19.2 are remarkably constant from one aldehyde or ketone to another. As a result, IR spectroscopy is a powerful tool for identifying the kind of a carbonyl group in a molecule of unknown structure. An unknown that shows an IR absorption at $1730 \mathrm{~cm}^{-1}$ is almost certainly an aldehyde rather than a ketone; an unknown that shows an IR absorption at $1750 \mathrm{~cm}^{-1}$ is almost certainly a cyclopentanone, and so on.

PROBLEM How might you use IR spectroscopy to determine whether reaction between 2-cyclohexenone and 19-23 lithium dimethylcopper gives the direct addition product or the conjugate addition product?


Infrared Spectroscopy -
PROBLEM How might you use IR spectroscopy to determine whether reaction between 2-cyclohexenone and 19-23 lithium dimethylcopper gives the direct addition product or the conjugate addition product?

PROBLEM Where would you expect each of the following compounds to absorb in the IR spectrum?
19-24 (a) 4-Penten-2-one
(b) 3-Penten-2-one
(c) 2,2-Dimethylcyclopentanone
(d) m-Chlorobenzaldehyde
(e) 3-Cyclohexenone
(f) 2-Hexenal


Nuclear Magnetic Resonance Spectroscopy -
Aldehyde protons (RCHO) absorb near $10 \delta$ in the ${ }^{1} \mathrm{H}$ NMR spectrum and are very distinctive because no other absorptions occur in this region. The aldehyde proton shows spin-spin coupling with protons on the neighboring carbon, with coupling constant $J \approx 3 \mathrm{~Hz}$. Acetaldehyde, for example, shows a quartet at $9.79 \delta$ for the aldehyde proton, indicating that there are three protons neighboring the - CHO group (FIGURE 19.16).

FIGURE $19.16{ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of acetaldehyde. The absorption of the aldehyde proton appears at $9.79 \delta$ and is split into a quartet.
Hydrogens on the carbon next to a carbonyl group are slightly deshielded and usually absorb near 2.0 to $2.3 \delta$. The acetaldehyde methyl group in FIGURE 19.16, for instance, absorbs at $2.20 \delta$. Methyl ketones are particularly distinctive because they always show a sharp three-proton singlet near 2.1 $\delta$.

The carbonyl-group carbon atoms of aldehydes and ketones have characteristic ${ }^{13} \mathrm{C}$ NMR resonances in the range 190 to $215 \delta$. Since no other kinds of carbons absorb in this range, the presence of an NMR absorption near $200 \delta$ is clear evidence for a carbonyl group. Saturated aldehyde or ketone carbons usually absorb in the region from 200 to $215 \delta$, while aromatic and $\alpha, \beta$-unsaturated carbonyl carbons absorb in the 190 to $200 \delta$ region.


Mass Spectrometry -
As discussed in Section 12.3, aliphatic aldehydes and ketones that have hydrogens on their gamma ( $\gamma$ ) carbon atoms undergo a characteristic mass spectral cleavage called the McLafferty rearrangement. A hydrogen atom is transferred from the $\gamma$ carbon to the carbonyl oxygen, the bond between the $\alpha$ and $\beta$ carbons is broken, and a neutral alkene fragment is produced. The charge remains with the oxygen-containing fragment.

In addition to fragmentation by the McLafferty rearrangement, aldehydes and ketones also undergo cleavage of the bond between the carbonyl group and the $\alpha$ carbon, called an $\alpha$ cleavage. Alpha cleavage yields a neutral radical and a resonance-stabilized acyl cation.

Fragment ions from both McLafferty rearrangement and $\alpha$ cleavage are visible in the mass spectrum of 5-methyl-2-hexanone shown in FIGURE 19.17. McLafferty rearrangement with loss of 2-methylpropene yields a fragment with $m / z=58$. Alpha cleavage occurs primarily at the more substituted side of the carbonyl group, leading to a $\left[\mathrm{CH}_{3} \mathrm{CO}\right]^{+}$fragment with $m / z=43$.

FIGURE 19.17 Mass spectrum and the related reactions of 5-methyl-2-hexanone. The peak at $m / z=58$ is due to McLafferty rearrangement. The abundant peak at $m / z=43$ is due to $\alpha$ cleavage at the more highly substituted side of the carbonyl group. Note that the peak due to the molecular ion is very small.

PROBLEM How might you use mass spectrometry to distinguish between the following pairs of isomers?
19-25 (a) 3-Methyl-2-hexanone and 4-methyl-2-hexanone (b) 3-Heptanone and 4-heptanone
(c) 2-Methylpentanal and 3-methylpentanal

PROBLEM Describe the prominent IR absorptions and mass spectral peaks you would expect for the following 19-26 compound:


Enantioselective Synthesis -
Whenever a chiral product is formed by reaction between achiral reagents, the product is racemic; that is, both enantiomers of the product are formed in equal amounts. The epoxidation reaction of geraniol with $m$-chloroperoxybenzoic acid, for instance, gives a racemic mixture of $(2 S, 3 S)$ and $(2 R, 3 R)$ epoxides.

Unfortunately, it's usually the case that only one enantiomer of a given drug or other important substance has the desired biological properties. The other enantiomer might be inactive or even dangerous. Thus, much work is currently being done on developing enantioselective methods of synthesis, which yield only one of the two possible enantiomers. So important has enantioselective synthesis become that the 2001 Nobel Prize in Chemistry was awarded to three pioneers in the field: William S. Knowles of the Monsanto Co., K. Barry Sharpless of the Scripps Research Institute, and Ryoji Noyori of Nagoya University.
Several approaches to enantioselective synthesis have been developed, but the most efficient are those that use chiral catalysts to temporarily hold a substrate molecule in an unsymmetrical environment-the same strategy that nature uses when catalyzing reactions with chiral enzymes. While in that unsymmetrical environment, the substrate may be more open to reaction on one side than on another, leading to an excess of one enantiomeric product over another. As an analogy, think about picking up a coffee mug in your right hand to take a drink. The mug by itself is achiral, but as soon as you pick it up by the handle, it becomes chiral. One side of the mug now faces toward you so you can drink from it, but the other side faces away. The two sides are different, with one side much more accessible to you than the other.

FIGURE 19.18 A substance made from the tartaric acid found at the bottom of this wine vat catalyzes enantioselective reactions. (credit: "Pinot Noir fermentation in open vat" by Pipers Brook Vineyard Media/Flickr, CC BY 2.0)


Enantioselective Synthesis -
Among the thousands of enantioselective reactions now known, one of the most useful is the so-called Sharpless epoxidation, in which an allylic alcohol, such as geraniol, is treated with tert-butyl hydroperoxide,
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OOH}$, in the presence of titanium tetraisopropoxide and diethyl tartrate (DET) as a chiral auxiliary reagent. When $(R, R)$ tartrate is used, geraniol is converted into its $2 S, 3 S$ epoxide with $98 \%$ selectivity, whereas the use of $(S, S)$ tartrate gives the $2 R, 3 R$ epoxide enantiomer. We say that the major product in each case is formed with an enantiomeric excess of $96 \%$, meaning that $4 \%$ of the product is racemic ( $2 \% 2 S, 3 S$ plus $2 \%$ $2 R, 3 R$ ) and an extra $96 \%$ of a single enantiomer is formed. The mechanistic details by which the chiral catalyst works are a bit complex, although it appears that a chiral complex of two tartrate molecules with one titanium is involved.

2S,3S isomer-98\%

Geraniol

$2 R, 3 R$ isomer-98\%


Key Terms -
```
- acetal, R2C(OR')2
- imine ( }\mp@subsup{\textrm{R}}{2}{}\textrm{C}=\textrm{NR}
- acyl group
- ketal
- 1,2-addition
- ketone (R2CO)
- 1,4-addition
- McLafferty rearrangement
- aldehyde (RCHO)
- nucleophilic addition reaction
- Cannizzaro reaction
- phosphorane
- carbinolamine
- Schiff bases
- conjugate addition
- Wittig reaction
- cyanohydrin
- Wolff-Kishner reaction
- enamine ( }\mp@subsup{\textrm{R}}{2}{}\textrm{N}-\textrm{CR}=\mp@subsup{\textrm{CR}}{2}{}
- ylide
- hemiacetal
```


Summary -
Aldehydes and ketones are among the most important of all functional groups, both in the chemical industry and in biological pathways. In this chapter, we've looked at some of their typical reactions. Aldehydes are normally prepared in the laboratory by oxidation of primary alcohols or by partial reduction of esters. Ketones are similarly prepared by oxidation of secondary alcohols.

The nucleophilic addition reaction is the most common general reaction type for aldehydes and ketones. Many different kinds of products can be prepared by nucleophilic additions. Aldehydes and ketones are reduced by $\mathrm{NaBH}_{4}$ or $\mathrm{LiAlH}_{4}$ to yield primary and secondary alcohols, respectively. Addition of Grignard reagents to aldehydes and ketones also gives alcohols (secondary and tertiary, respectively), and addition of HCN yields cyanohydrins. Primary amines add to carbonyl compounds yielding imines, or Schiff bases, and secondary
amines yield enamines. Reaction of an aldehyde or ketone with hydrazine and base gives an alkane (the Wolff-Kishner reaction). Alcohols add to carbonyl groups to yield acetals, which are valuable as protecting groups. Phosphorus ylides add to aldehydes and ketones in the Wittig reaction to give alkenes.
$\alpha, \beta$-Unsaturated aldehydes and ketones often react with nucleophiles to give the product of conjugate addition, or 1,4-addition. Particularly useful are the conjugate addition of an amine and the conjugate addition of an organic group by reaction with a diorganocopper reagent.


Summary -
IR spectroscopy is helpful for identifying aldehydes and ketones. Carbonyl groups absorb in the IR range 1660 to $1770 \mathrm{~cm}^{-1}$, with the exact position highly diagnostic of the kind of carbonyl group present in the molecule. ${ }^{13} \mathrm{C}$ NMR spectroscopy is also useful for aldehydes and ketones because their carbonyl carbons show resonances in the 190 to $215 \delta$ range. ${ }^{1} \mathrm{H}$ NMR is useful for aldehyde - CHO protons, which absorb near $10 \delta$. Aldehydes and ketones undergo two characteristic kinds of fragmentation in the mass spectrometer: $\alpha$ cleavage and McLafferty rearrangement.


Summary of Reactions -
1. Preparation of aldehydes (Section 19.2)
a. Oxidation of primary alcohols (Section 17.7)

b. Partial reduction of esters (Section 19.2)

2. Preparation of ketones
a. Oxidation of secondary alcohols (Section 17.7)

b. Diorganocopper reaction with acid chlorides (Section 19.2)

3. Oxidation of aldehydes (Section 19.3)

4. Nucleophilic addition reactions of aldehydes and ketones
a. Addition of hydride to give alcohols: reduction (Section 19.7)

b. Addition of Grignard reagents to give alcohols (Section 19.7)

c. Addition of HCN to give cyanohydrins (Section 19.6)

d. Addition of primary amines to give imines (Section 19.8)

e. Addition of secondary amines to give enamines (Section 19.8)

f. Wolff-Kishner reaction to give alkanes (Section 19.9)

g. Addition of alcohols to give acetals (Section 19.10)

h. Addition of phosphorus ylides to give alkenes: Wittig reaction (Section 19.11)

5. Conjugate additions to $\alpha, \beta$-unsaturated aldehydes and ketones (Section 19.13)
a. Conjugate addition of amines

b. Conjugate addition of water

c. Conjugate addition of alkyl groups by diorganocopper reaction


Visualizing Chemistry -
PROBLEM Each of the following substances can be prepared by a nucleophilic addition reaction between an
19-27 aldehyde or ketone and a nucleophile. Identify the reactants from which each was prepared. If the substance is an acetal, identify the carbonyl compound and the alcohol; if it is an imine, identify the carbonyl compound and the amine; and so forth.
(a)

(b)

(c)

(d)

PROBLEM The following molecular model represents a tetrahedral intermediate resulting from addition of a
19-28 nucleophile to an aldehyde or ketone. Identify the reactants, and write the structure of the final product when the nucleophilic addition reaction is complete.

PROBLEM The enamine prepared from acetone and dimethylamine is shown in its lowest-energy form.
19-29 (a) What is the geometry and hybridization of the nitrogen atom?
(b) What orbital on nitrogen holds the lone pair of electrons?
(c) What is the geometric relationship between the $p$ orbitals of the double bond and the nitrogen orbital that holds the lone pair? Why do you think this geometry represents the minimum energy?


Mechanism Problems -
PROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:
19-30 (a)

(b)

$+$

PROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:
19-31 (a)

(b)

PROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:
19-32

b)

PROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:
19-33 (a)

(b)

PROBLEM Predict the product(s) and propose mechanisms for the following reactions:

19-34 (a)

(b)

PROBLEM The following reaction begins with an acetal and converts it into a different acetal. Predict the
19-35 product(s) and propose a mechanism.
(a)

(b)

PROBLEM When $\alpha$-glucose is treated with an acid catalyst in the presence of an alcohol, an acetal is formed.
19-36 Propose a mechanism for this process and give the structure of the stereoisomeric acetal that you would also expect as a product.

PROBLEM Predict the products of the following Wolff-Kishner reactions. Write the mechanism for each,
19-37 beginning from the hydrazone intermediate.
(a)

(b)

PROBLEM Aldehydes can be prepared by the Wittig reaction using (methoxymethylene)triphenylphosphorane 19-38 as the Wittig reagent and then hydrolyzing the product with acid. For example,

(a) How would you prepare the necessary phosphorane?
(b) Propose a mechanism for the hydrolysis step.

PROBLEM One of the steps in the metabolism of fats is the reaction of an unsaturated acyl CoA with water to 19-39 give a $\beta$-hydroxyacyl CoA. Propose a mechanism.

Unsaturated acyl CoA
$\beta$-Hydroxyacyl CoA
PROBLEM Aldehydes and ketones react with thiols to yield thioacetals just as they react with alcohols to yield 19-40 acetals. Predict the product of the following reaction, and propose a mechanism:

PROBLEM Ketones react with dimethylsulfonium methylide to yield epoxides. Suggest a mechanism for the
19-41 reaction.


Mechanism Problems -
PROBLEM Ketones react with dimethylsulfonium methylide to yield epoxides. Suggest a mechanism for the
19-41 reaction.

PROBLEM Propose a mechanism for the following reaction.
19-42

PROBLEM Paraldehyde, a sedative and hypnotic agent, is prepared by treatment of acetaldehyde with an acidic
19-43 catalyst. Propose a mechanism for the reaction.


Paraldehyde -
PROBLEM The Meerwein-Ponndorf-Verley reaction involves reduction of a ketone by treatment with an
19-44 excess of aluminum triisopropoxide, $\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHO}\right]_{3} \mathrm{Al}$. The mechanism of the process is closely related to the Cannizzaro reaction in that a hydride ion acts as a leaving group. Propose a mechanism.

PROBLEM Propose a mechanism to account for the formation of 3,5-dimethylpyrazole from hydrazine and
19-45 2,4-pentanedione. What has happened to each carbonyl carbon in going from starting material to product.

PROBLEM In light of your answer to Problem 19-45, propose a mechanism for the formation of 19-46 3,5-dimethylisoxazole from hydroxylamine and 2,4-pentanedione.

3,5-Dimethylisoxazole

PROBLEM Trans alkenes are converted into their cis isomers and vice versa on epoxidation followed by 19-47 treatment of the epoxide with triphenylphosphine. Propose a mechanism for the reaction.

PROBLEM Treatment of an $\alpha, \beta$-unsaturated ketone with basic aqueous hydrogen peroxide yields an epoxy
19-48 ketone. The reaction is specific to unsaturated ketones; isolated alkene double bonds do not react. Propose a mechanism.

PROBLEM One of the biological pathways by which an amine is converted to a ketone involves two steps: (1)
19-49 oxidation of the amine by $\mathrm{NAD}^{+}$to give an imine and (2) hydrolysis of the imine to give a ketone plus ammonia. Glutamate, for instance, is converted by this process into $\alpha$-ketoglutarate. Show the structure of the imine intermediate, and propose mechanisms for both steps.


Paraldehyde -
PROBLEM Primary amines react with esters to yield amides: $\mathrm{RCO}_{2} \mathrm{R}^{\prime}+\mathrm{R}^{\prime \prime} \mathrm{NH}_{2} \rightarrow \mathrm{RCONHR}{ }^{\prime \prime}+\mathrm{R}^{\prime} \mathrm{OH}$. Propose a 19-50 mechanism for the following reaction of an $\alpha, \beta$-unsaturated ester.

PROBLEM When crystals of pure $\alpha$-glucose are dissolved in water, isomerization occurs slowly to produce 19-51 $\beta$-glucose. Propose a mechanism for the isomerization.

PROBLEM The Wharton reaction converts an epoxy ketone to an allylic alcohol by reaction with hydrazine.
19-52 Review the Wolff-Kishner reaction in Section 19.9 and then propose a mechanism.

Naming Aldehydes and Ketones
PROBLEM Draw structures corresponding to the following names:
19-53
(a) Bromoacetone
(b) (S)-2-Hydroxypropanal
(c) 2-Methyl-3-heptanone
(d) $(2 S, 3 R)-2,3,4$-Trihydroxybutanal (e) 2,2,4,4-Tetramethyl-3-pentanone
(f) 4-Methyl-3-penten-2-one (g) Butanedial (h) 3-Phenyl-2-propenal
(i) 6,6-Dimethyl-2,4-cyclohexadienone (j) p-Nitroacetophenone

PROBLEM Draw and name the seven aldehydes and ketones with the formula $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$. Which are chiral? 19-54

PROBLEM Give IUPAC names for the following compounds:
19-55 (a)

(b)

(c)

(d)

(e)

(f)


Paraldehyde -
PROBLEM Give IUPAC names for the following compounds:
19-55 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Draw structures of compounds that fit the following descriptions:
19-56 (a) An $\alpha, \beta$-unsaturated ketone, $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}$
(b) An $\alpha$-diketone (c) An aromatic ketone, $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}$
(d) A diene aldehyde, $\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}$

Reactions of Aldehydes and Ketones
PROBLEM Predict the products of the reaction of (1) phenylacetaldehyde and (2) acetophenone with the
19-57 following reagents:
(a) $\mathrm{NaBH}_{4}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$
(b) Dess-Martin reagent
(c) $\mathrm{NH}_{2} \mathrm{OH}, \mathrm{HCl}$ catalyst
(d) $\mathrm{CH}_{3} \mathrm{MgBr}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$
(e) $2 \mathrm{CH}_{3} \mathrm{OH}, \mathrm{HCl}$ catalyst (f) $\mathrm{H}_{2} \mathrm{NNH}_{2}, \mathrm{KOH}$
(g) $\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{P}=\mathrm{CH}_{2}$
(h) $\mathrm{HCN}, \mathrm{KCN}$

PROBLEM Show how you might use a Wittig reaction to prepare the following alkenes. Identify the alkyl halide 19-58 and the carbonyl components.
(a)

(b)

PROBLEM How would you use a Grignard reaction on an aldehyde or ketone to synthesize the following 19-59 compounds?
(a) 2-Pentanol
(b) 1-Butanol
(c) 1-Phenylcyclohexanol
(d) Diphenylmethanol


Paraldehyde -
PROBLEM How might you carry out the following selective transformations? One of the two schemes requires
19-60 a protection step. (Recall from Section 19.4 that aldehydes are more reactive than ketones toward nucleophilic addition.)
(a)

(b)

PROBLEM How would you prepare the following substances from 2-cyclohexenone? More than one step may
19-61 be needed.
(a)

(b)

(c)

(d)

(Two ways)

PROBLEM How would you synthesize the following substances from benzaldehyde and any other reagents 19-62 needed?
(a)

(b)

(c)

PROBLEM Carvone is the major constituent of spearmint oil. What products would you expect from reaction 19-63 of carvone with the following reagents?

Carvone
(a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Cu}^{-} \mathrm{Li}^{+}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$
(b) $\mathrm{LiAlH}_{4}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$
(c) $\mathrm{CH}_{3} \mathrm{NH}_{2}$
(d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$
(e) $\mathrm{H}_{2} / \mathrm{Pd}$
(f) $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{HCl}$
(g) $\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \stackrel{+-}{\mathrm{P}} \mathrm{HCH}_{3}$

PROBLEM How would you synthesize the following compounds from cyclohexanone?
19-64 (a) 1-Methylcyclohexene (b) 2-Phenylcyclohexanone (c) cis-1,2-Cyclohexanediol
(d) 1-Cyclohexylcyclohexanol


Spectroscopy -
PROBLEM At what position would you expect to observe IR absorptions for the following molecules?


19-65 (a) -
(b)

1-Indanone
(c)

PROBLEM Acid-catalyzed dehydration of 3-hydroxy-3-phenylcyclohexanone leads to an unsaturated ketone.
19-66 What possible structures are there for the product? At what position in the IR spectrum would you expect each to absorb? If the actual product has an absorption at $1670 \mathrm{~cm}^{-1}$, what is its structure?

PROBLEM Choose the structure that best fits the IR spectrum shown.
19-67

(a)

(b)

(c)

(d)


19-65 (a) -
PROBLEM Choose the structure that best fits the IR spectrum shown.
19-67

(a)

(b)

(c)

(d)

PROBLEM Propose structures for molecules that meet the following descriptions. Assume that the kinds of 19-68 carbons ( $1^{\circ}, 2^{\circ}, 3^{\circ}$, or $4^{\circ}$ ) have been assigned by DEPT-NMR.
(a) $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}$; IR: $1715 \mathrm{~cm}^{-1}$; ${ }^{13} \mathrm{C}$ NMR: $8.0 \delta\left(1^{\circ}\right)$, $18.5 \delta\left(1^{\circ}\right), 33.5 \delta\left(2^{\circ}\right), 40.6 \delta\left(3^{\circ}\right), 214.0 \delta\left(4^{\circ}\right)$
(b) $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$; IR: $1730 \mathrm{~cm}^{-1}$; ${ }^{13} \mathrm{C}$ NMR: $22.6 \delta\left(1^{\circ}\right), 23.6 \delta\left(3^{\circ}\right), 52.8 \delta\left(2^{\circ}\right), 202.4 \delta\left(3^{\circ}\right)$
(c) $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}$; IR: $1680 \mathrm{~cm}^{-1}$; ${ }^{13} \mathrm{C}$ NMR: $22.9 \delta\left(2^{\circ}\right), 25.8 \delta\left(2^{\circ}\right), 38.2 \delta\left(2^{\circ}\right), 129.8 \delta\left(3^{\circ}\right), 150.6 \delta\left(3^{\circ}\right)$, 198.7 ( $4^{\circ}$ )


19-65 (a) -
PROBLEM Compound $\mathbf{A}, \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2}$, has an intense IR absorption at $1750 \mathrm{~cm}^{-1}$ and gives the ${ }^{13} \mathrm{C}$ NMR spectrum 19-69 shown. Propose a structure for A.

PROBLEM Propose structures for ketones or aldehydes that have the following ${ }^{1} \mathrm{H}$ NMR spectra:
19-70 (a) $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{ClO}$
IR: $1715 \mathrm{~cm}^{-1}$

(b) $\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}$

IR: $1710 \mathrm{~cm}^{-1}$


General Problems -
PROBLEM When 4-hydroxybutanal is treated with methanol in the presence of an acid catalyst, 19-71 2-methoxytetrahydrofuran is formed. Explain.

PROBLEM The $\mathrm{S}_{\mathrm{N}} 2$ reaction of (dibromomethyl)benzene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHBr}_{2}$, with NaOH yields benzaldehyde rather 19-72 than (dihydroxymethyl)benzene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH})_{2}$. Explain.

PROBLEM Reaction of 2-butanone with HCN yields a chiral product. What stereochemistry does the product 19-73 have? Is it optically active?

PROBLEM The amino acid methionine is biosynthesized by a multistep route that includes reaction of an 19-74 imine of pyridoxal phosphate (PLP) to give an unsaturated imine, which then reacts with cysteine. What kinds of reactions are occurring in the two steps?


O-SuccinylhomoserinePLP imine <br> Unsaturated imine -
PROBLEM Each of the following reaction schemes has one or more flaws. What is wrong in each case? How 19-75 would you correct each scheme?
(a)

(b)

PROBLEM 6-Methyl-5-hepten-2-one is a constituent of lemongrass oil. How could you synthesize this 19-76 substance from methyl 4-oxopentanoate?
Methyl 4-oxopentanoate

PROBLEM Tamoxifen is a drug used in the treatment of breast cancer. How would you prepare tamoxifen from 19-77 benzene, the following ketone, and any other reagents needed?

PROBLEM Compound $\mathbf{A}, \mathrm{MW}=86$, shows an IR absorption at $1730 \mathrm{~cm}^{-1}$ and a very simple ${ }^{1} \mathrm{H}$ NMR spectrum 19-78 with peaks at $9.7 \delta(1 \mathrm{H}$, singlet) and $1.2 \delta(9 \mathrm{H}$, singlet). Propose a structure for A.
PROBLEM Compound $\mathbf{B}$ is isomeric with $\mathbf{A}$ (Problem 19-78) and shows an IR peak at $1715 \mathrm{~cm}^{-1}$. The ${ }^{1} \mathrm{H}$ NMR 19-79 spectrum of $\mathbf{B}$ has peaks at $2.4 \delta(1 \mathrm{H}$, septet, $J=7 \mathrm{~Hz}), 2.1 \delta(3 \mathrm{H}$, singlet), and $1.2 \delta(6 \mathrm{H}$, doublet, $J$
$=7 \mathrm{~Hz})$. What is the structure of $\mathbf{B}$ ?
PROBLEM The ${ }^{1} \mathrm{H}$ NMR spectrum shown is that of a compound with the formula $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}$. How many double 19-80 bonds and/or rings does this compound contain? If the unknown compound has an IR absorption at $1690 \mathrm{~cm}^{-1}$, what is a likely structure?


O-SuccinylhomoserinePLP imine <br> Unsaturated imine -
PROBLEM The ${ }^{1} \mathrm{H}$ NMR spectrum shown is that of a compound isomeric with the one in Problem 19-80. This 19-81 isomer has an IR absorption at $1730 \mathrm{~cm}^{-1}$. Propose a structure. [Note: Aldehyde protons (CHO) often show low coupling constants to adjacent hydrogens, so the splitting of aldehyde signals is not always apparent.]

PROBLEM Propose structures for ketones or aldehydes that have the following ${ }^{1} \mathrm{H}$ NMR spectra:
19-82 (a) $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}_{2}$ : IR: $1695 \mathrm{~cm}^{-1}$

(b) $\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}:$ IR: $1690 \mathrm{~cm}^{-1}$

PROBLEM Propose structures for ketones or aldehydes that have the following ${ }^{1} \mathrm{H}$ NMR spectra.
19-83 (a) $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}$ : IR: $1710 \mathrm{~cm}^{-1}$

(b) $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}$ : IR: $1715 \mathrm{~cm}^{-1}$

PROBLEM When glucose (Problem 19-51) is treated with $\mathrm{NaBH}_{4}$, reaction occurs to yield sorbitol, a 19-84 polyalcohol commonly used as a food additive. Show how this reduction occurs.

Glucose
Sorbitol

PROBLEM The proton and carbon NMR spectra for each of three isomeric ketones with the formula $\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}$ are
19-85 shown. Assign a structure to each pair of spectra.


O-SuccinylhomoserinePLP imine <br> Unsaturated imine -
PROBLEM The proton NMR spectrum for a compound with formula $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ is shown below. The infrared 19-86 spectrum has a strong band at $1711 \mathrm{~cm}^{-1}$. The broadband-decoupled 13C NMR spectral results are tabulated along with the DEPT-135 and DEPT-90 information. Draw the structure of this compound.


CHAPTER 20 -
Carboxylic Acids and Nitriles

FIGURE 20.1 The burning sensation produced by touching or eating chili peppers is due to capsaicin, a carboxylic acid derivative called an amide. (credit: modification of "Chillis" by Lucas Cobb/Flickr, CC BY 2.0)


CHAPTER CONTENTS -
20.1 Naming Carboxylic Acids and Nitriles
20.2 Structure and Properties of Carboxylic Acids
20.3 Biological Acids and the Henderson-Hasselbalch Equation
20.4 Substituent Effects on Acidity
20.5 Preparing Carboxylic Acids
20.6 Reactions of Carboxylic Acids: An Overview
20.7 Chemistry of Nitriles
20.8 Spectroscopy of Carboxylic Acids and Nitriles

WHY THIS CHAPTER? Carboxylic acids are present in many industrial processes and most biological pathways and are the starting materials from which other acyl derivatives are made. Thus, an understanding of their properties and reactions is fundamental to understanding organic chemistry. We'll look both at acids and at their close relatives, nitriles ( $\mathrm{RC} \equiv \mathrm{N}$ ), in this chapter and at carboxylic acid derivatives in the next chapter.

Carboxylic acids, $\mathbf{R C O}_{\mathbf{2}} \mathbf{H}$, occupy a central place among carbonyl compounds. Not only are they valuable in themselves, they also serve as starting materials for preparing numerous carboxylic acid derivatives such as acid chlorides, esters, amides, and thioesters. In addition, carboxylic acids are present in the majority of biological pathways.

An acid chloride $\quad$ An ester


CHAPTER CONTENTS -
An acid chloride $\quad$ An ester

A great many carboxylic acids are found in nature: acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, is the chief organic component of vinegar; butanoic acid, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$, is responsible for the rancid odor of sour butter; and hexanoic acid (caproic acid), $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CO}_{2} \mathrm{H}$, is responsible for the unmistakable aroma of goats and dirty gym socks (the name comes from the Latin caper, meaning "goat"). Other examples are cholic acid, a major component of human bile, and long-chain aliphatic acids such as palmitic acid, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{14} \mathrm{CO}_{2} \mathrm{H}$, a biological precursor of fats and vegetable oils.

Cholic acid
Approximately 20 million tons of acetic acid is produced worldwide each year for a variety of purposes, including preparation of the vinyl acetate polymer used in paints and adhesives. About $20 \%$ of the acetic acid synthesized industrially is obtained by oxidation of acetaldehyde. Much of the remaining $80 \%$ is prepared by the rhodium-catalyzed reaction of methanol with carbon monoxide.


Carboxylic Acids, $\mathrm{RCO}_{2} \mathrm{H}$ -
Simple carboxylic acids derived from open-chain alkanes are systematically named by replacing the terminal $-e$ of the corresponding alkane name with -oic acid. The $-\mathrm{CO}_{2} \mathrm{H}$ carbon atom is numbered C 1 .

Propanoic acid

4-Methylpentanoic acid

3-Ethyl-6-methyloctanedioic acid

Compounds that have $\mathrm{a}-\mathrm{CO}_{2} \mathrm{H}$ group bonded to a ring are named using the suffix -carboxylic acid. The $\mathrm{CO}_{2} \mathrm{H}$ carbon is attached to C 1 in this system and is not itself numbered. As a substituent, the $\mathrm{CO}_{2} \mathrm{H}$ group is called a carboxyl group.

trans-4-Hydroxycyclohexanecarboxylic acid

1-Cyclopentenecarboxylic acid

Because many carboxylic acids were among the first organic compounds to be isolated and purified, quite a few common names exist (TABLE 20.1). Biological chemists, in particular, make frequent use of these names, so you may find yourself referring back to this list on occasion. We'll use systematic names in this book, with a few exceptions such as formic (methanoic) acid and acetic (ethanoic) acid, whose names are accepted by IUPAC and are so well known that it makes little sense to refer to them any other way.
Also listed in TABLE 20.1 are the names of acyl groups $\binom{\mathrm{O}}{\mathrm{R}-\mathrm{C}-}$ derived from the parent acids. Except for the eight entries at the top of TABLE 20.1, whose names have a -yl ending, all other acyl groups are named using an -oyl ending.

TABLE 20.1 Common Names of Some Carboxylic Acids and Acyl Groups


Carboxylic Acids, $\mathrm{RCO}_{2} \mathrm{H}$ -
| Structure | Name | Acyl group |
| :---: | :---: | :---: |
| $\mathrm{HCO}_{2} \mathrm{H}$ | Formic | Formyl |
| $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ | Acetic | Acetyl |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Propionic | Propionyl |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Butyric | Butyryl |
| $\mathrm{HO}_{2} \mathrm{CCO}_{2} \mathrm{H}$ | Oxalic | Oxalyl |
| $\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Malonic | Malonyl |
| $\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Succinic | Succinyl |
| $\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Glutaric | Glutaryl |
| $\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Adipic | Adipoyl |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCO}_{2} \mathrm{H}$ | Acrylic | Acryloyl |
| $\mathrm{HO}_{2} \mathrm{CCH}=\mathrm{CHCO}_{2} \mathrm{H}$ | Maleic (cis) Fumaric (trans) | Maleoyl Fumaroyl |
| $\mathrm{HOCH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Glycolic | Glycoloyl |


Carboxylic Acids, $\mathrm{RCO}_{2} \mathrm{H}$ -
| $\mathrm{HOCH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | Glycolic | Glycoloyl |
| <smiles>CC(O)C(=O)O</smiles> | Lactic | Lactoyl |
| <smiles>CC(=O)O</smiles> | Pyruvic | Pyruvoyl |
| <smiles>O=C(O)C(O)CO</smiles> | Glyceric | Glyceroyl |


Carboxylic Acids, $\mathrm{RCO}_{2} \mathrm{H}$ -
TABLE 20.1 Common Names of Some Carboxylic Acids and Acyl Groups

| Structure | Name | Acyl group |
| :---: | :---: | :---: |
| <smiles>O=C(O)CC(O)C(=O)O</smiles> | Malic | Maloyl |
| <smiles>O=C(O)CC(=O)O</smiles> | Oxaloacetic | Oxaloacetyl |
| <smiles>O=C(O)c1ccccc1</smiles> | Benzoic | Benzoyl |
| <smiles>O=C(O)c1ccccc1C(=O)O</smiles> | Phthalic | Phthaloyl |


Nitriles, $\mathrm{RC} \equiv \mathrm{N}$ -
Compounds containing the $-\mathrm{C} \equiv \mathrm{N}$ functional group are called nitriles and can undergo some chemistry similar to that of carboxylic acids. Simple open-chain nitriles are named by adding -nitrile as a suffix to the alkane name, with the nitrile carbon numbered C 1 .

$$
\underset{5}{\mathrm{CH}_{3}} \stackrel{\mathrm{C}_{4}^{\mathrm{CH}} \mathrm{CH}_{3}}{\mathrm{CH}_{3}}
$$

Nitriles can also be named as derivatives of carboxylic acids by replacing the -ic acid or -oic acid ending with -onitrile, or by replacing the -carboxylic acid ending with -carbonitrile. The nitrile carbon atom is attached to C 1 but is not itself numbered.

$$
\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{~N}
$$


Acetonitrile (from acetic acid)
 Benzonitrile
 (from benzoic acid) -
2,2-Dimethylcyclohexanecarbonitrile (from 2,2-dimethylcyclohexanecarboxylic acid)

If another carboxylic acid derivative is present in the same molecule, the prefix cyano- is used for the $\mathrm{C} \equiv \mathrm{N}$ group.


Methyl 4-cyanopentanoate -
PROBLEM Give IUPAC names for the following compounds:
20-1 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Draw structures corresponding to the following IUPAC names:
20-2 (a) 2,3-Dimethylhexanoic acid (b) 4-Methylpentanoic acid
(c) trans-1,2-Cyclobutanedicarboxylic acid (d) o-Hydroxybenzoic acid
(e) (9Z,12Z)-9,12-Octadecadienoic acid (f) 2-Pentenenitrile


Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids
Carboxylic acids are similar in some respects to both ketones and alcohols. Like ketones, the carboxyl carbon is $s p^{2}$-hybridized, and carboxylic acid groups are therefore planar with $\mathrm{C}-\mathrm{C}=\mathrm{O}$ and $\mathrm{O}=\mathrm{C}-\mathrm{O}$ bond angles of approximately $120^{\circ}$ (TABLE 20.2).

| TABLE 20.2 Physical Parameters for Acetic Acid |  |  |  |  |
| :--- | :--- | :--- | :--- | :---: |
|  |  |  |  |  |
|  |  |  |  |  |
| Bond angle | (degrees) | Bond length | (pm) |  |
| $\mathrm{C}-\mathrm{C}=\mathrm{O}$ | 119 | $\mathrm{C}-\mathrm{C}$ | 152 |  |
| $\mathrm{C}-\mathrm{C}-\mathrm{OH}$ | 119 | $\mathrm{C}=\mathrm{O}$ | 125 |  |
| $\mathrm{O}=\mathrm{C}-\mathrm{OH}$ | 122 | $\mathrm{C}-\mathrm{OH}$ | 131 |  |

Like alcohols, carboxylic acids are strongly associated because of hydrogen-bonding. Most carboxylic acids exist as cyclic dimers held together by two hydrogen bonds. This strong hydrogen-bonding has a noticeable effect on boiling points, making carboxylic acids boil far less easily than their corresponding alcohols. Acetic acid, for instance, has a boiling point of $117.9^{\circ} \mathrm{C}$, versus $78.3^{\circ} \mathrm{C}$ for ethanol, even though both compounds have two carbons.

Acetic acid dimer


Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids
Acetic acid dimer

The most obvious property of carboxylic acids is implied by their name: carboxylic acids are acidic. They therefore react with bases such as NaOH and $\mathrm{NaHCO}_{3}$ to give metal carboxylate salts, $\mathrm{RCO}_{2}{ }^{-} \mathrm{M}^{+}$. Carboxylic acids with more than six carbons are only slightly soluble in water, but the alkali metal salts of carboxylic acids are often highly water-soluble. In fact, it's often possible to purify an acid by extracting its salt into aqueous base, then reacidifying and extracting the pure acid back into an organic solvent.

Like other Brønsted-Lowry acids discussed in Section 2.7, carboxylic acids dissociate slightly in dilute aqueous solution to give $\mathrm{H}_{3} \mathrm{O}^{+}$and the corresponding carboxylate anions, $\mathrm{RCO}_{2}{ }^{-}$. The extent of dissociation is given by an acidity constant, $K_{\mathrm{a}}$.

A list of $K_{\mathrm{a}}$ values for various carboxylic acids is given in TABLE 20.3. For most, $K_{\mathrm{a}}$ is approximately $10^{-4}$ to $10^{-5}$. Acetic acid, for instance, has $K_{\mathrm{a}}=1.75 \times 10^{-5}$ at $25^{\circ} \mathrm{C}$, which corresponds to a $\mathrm{p} K_{\mathrm{a}}$ of 4.76 . In practical terms, a $K_{\mathrm{a}}$ value near $10^{-5}$ means that only about $0.1 \%$ of the molecules in a 0.1 M solution are dissociated, as opposed to the $100 \%$ dissociation found with strong mineral acids like HCl .


Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids
| Structure | $K_{\text {a }}$ | $\mathrm{p} K_{\mathrm{a}}$ |  |
| :---: | :---: | :---: | :---: |
| $\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}$ | 0.59 | 0.23 | Stronger acid |
| $\mathrm{HCO}_{2} \mathrm{H}$ | $1.77 \times 10^{-4}$ | 3.75 |  |
| $\mathrm{HOCH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | $1.5 \times 10^{-4}$ | 3.84 |  |
| $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}$ | $6.46 \times 10^{-5}$ | 4.19 |  |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCO}_{2} \mathrm{H}$ | $5.6 \times 10^{-5}$ | 4.25 |  |
| $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ | $1.75 \times 10^{-5}$ | 4.76 |  |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | $1.34 \times 10^{-5}$ | 4.87 | acid |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ (ethanol) | $\left(1 \times 10^{-16}\right)$ | (16) |  |

Although much weaker than mineral acids, carboxylic acids are nevertheless much stronger acids than alcohols and phenols. The $K_{\mathrm{a}}$ of ethanol, for example, is approximately $10^{-16}$, making it a weaker acid than acetic acid by a factor of $10^{11}$.


Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids
Why are carboxylic acids so much more acidic than alcohols, even though both contain - OH groups? An alcohol dissociates to give an alkoxide ion, in which the negative charge is localized on a single electronegative atom. A carboxylic acid, however, gives a carboxylate ion, in which the negative charge is delocalized over two equivalent oxygen atoms (FIGURE 20.2). In resonance terms (Section 2.4), a carboxylate ion is a stabilized resonance hybrid of two equivalent structures. Since a carboxylate ion is more stable than an alkoxide ion, it is lower in energy and more favored in the dissociation equilibrium.

Ethanol

$+$

Ethoxide ion (localized charge)

Acetate ion
(delocalized charge)

FIGURE 20.2 An alkoxide ion has its charge localized on one oxygen atom and is less stable, while a carboxylate ion has the charge spread equally over both oxygens and is therefore more stable.

Experimental evidence for the equivalence of the two carboxylate oxygens comes from X-ray crystallographic studies on sodium formate. Both carbon-oxygen bonds are 127 pm in length, midway between the $\mathrm{C}=\mathrm{O}$ double bond ( 120 pm ) and the $\mathrm{C}-\mathrm{O}$ single bond ( 134 pm ) of formic acid. An electrostatic potential map of the formate ion also shows how the negative charge (red) is spread equally over both oxygens.

PROBLEM Assume you have a mixture of naphthalene and benzoic acid that you want to separate. How might 20-3 you take advantage of the acidity of one component in the mixture to effect a separation?

PROBLEM The $K_{\mathrm{a}}$ for dichloroacetic acid is $3.32 \times 10^{-2}$. Approximately what percentage of the acid is 20-4 dissociated in a 0.10 M aqueous solution?


Methyl 4-cyanopentanoate - 20.3 Biological Acids and the Henderson-Hasselbalch Equation
In acidic solution, at low pH , a carboxylic acid is completely undissociated and exists entirely as RCOH. In basic solution, at high pH , a carboxylic acid is completely dissociated and exists entirely as $\mathrm{RCO}_{2}{ }^{-}$. Inside living cells, however, the pH is neither acidic nor basic but is instead buffered to a nearly neutral pH of $3.5-4.5$ in humans, a value often referred to as physiological $p H$. In what form, then, do carboxylic acids exist inside cells? The question is an important one for understanding the acid catalysts so often found in biological reactions.

If the $\mathrm{p} K_{\mathrm{a}}$ value of a given acid and the pH of the medium are known, the percentages of dissociated and
undissociated forms can be calculated using the Henderson-Hasselbalch equation.
For any acid HA, we have

$$
\begin{aligned}
\mathrm{p} K_{\mathrm{a}} & =-\log \frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \\
& =\mathrm{pH}-\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}
\end{aligned}
$$

which can be rearranged to give

$$
\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \quad \text { Henderson-Hasselbalch equation }
$$

so

$$
\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}
$$


Methyl 4-cyanopentanoate - 20.3 Biological Acids and the Henderson-Hasselbalch Equation
so

$$
\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}
$$

This equation says that the logarithm of the concentration of dissociated acid [ $\mathrm{A}^{-}$] divided by the concentration of undissociated acid [HA] is equal to the pH of the solution minus the $\mathrm{p} K_{\mathrm{a}}$ of the acid. Thus, if we know both the pH of the solution and the $\mathrm{p} K_{\mathrm{a}}$ of the acid, we can calculate the ratio of [ $\mathrm{A}^{-}$] to [HA]. Furthermore, when $\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}$, then HA and $\mathrm{A}^{-}$are present in equal amounts because $\log 1=0$.

As an example of how to use the Henderson-Hasselbalch equation, let's find out what species are present in a 0.0010 M solution of acetic acid at $\mathrm{pH}=7.3$. According to Table $20-3$, the $\mathrm{p} K_{\mathrm{a}}$ of acetic acid is 4.76 . From the Henderson-Hasselbalch equation, we have

$$
\begin{aligned}
\log \frac{\left[\mathrm{A}^{-}\right]}{\mathrm{HA}} & =\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}=7.3-4.76=2.54 \\
\frac{\left[\mathrm{~A}^{-}\right]}{\mathrm{HA}} & =\operatorname{antilog}(2.54)=3.5 \times 10^{2} \quad \text { so }\left[\mathrm{A}^{-}\right]=\left(3.5 \times 10^{2}\right)[\mathrm{HA}]
\end{aligned}
$$

In addition, we know that

$$
\left[\mathrm{A}^{-}\right]+[\mathrm{HA}]=0.0010 \mathrm{M}
$$


Methyl 4-cyanopentanoate - 20.3 Biological Acids and the Henderson-Hasselbalch Equation
In addition, we know that

$$
\left[\mathrm{A}^{-}\right]+[\mathrm{HA}]=0.0010 \mathrm{M}
$$

Solving the two simultaneous equations gives $\left[\mathrm{A}^{-}\right]=0.0010 \mathrm{M}$ and $[\mathrm{HA}]=3 \times 10^{-6} \mathrm{M}$. In other words, at a physiological pH of 7.3 , essentially $100 \%$ of acetic acid molecules in a 0.0010 M solution are dissociated to the acetate ion.

What is true for acetic acid is also true for other carboxylic acids: At the physiological pH that occurs inside cells, carboxylic acids are almost entirely dissociated. To reflect this fact, we always refer to cellular carboxylic acids by the name of their anion-acetate, lactate, citrate, and so forth, rather than acetic acid, lactic acid, and citric acid.

PROBLEM Calculate the percentages of dissociated and undissociated forms present in the following 20-5 solutions:
(a) 0.0010 M glycolic acid $\left(\mathrm{HOCH}_{2} \mathrm{CO}_{2} \mathrm{H} ; \mathrm{p} K_{\mathrm{a}}=3.83\right)$ at $\mathrm{pH}=4.50$
(b) 0.0020 M propanoic $\operatorname{acid}\left(\mathrm{p} K_{\mathrm{a}}=4.87\right)$ at $\mathrm{pH}=5.30$


Methyl 4-cyanopentanoate - 20.4 Substituent Effects on Acidity
The listing of $\mathrm{p} K_{\mathrm{a}}$ values shown previously in TABLE 20.3 indicates that there are substantial differences in acidity from one carboxylic acid to another. For example, trifluoroacetic acid ( $K_{\mathrm{a}}=0.59$ ) is 33,000 times as strong as acetic acid $\left(K_{\mathrm{a}}=1.75 \times 10^{-5}\right)$. How can we account for such differences?

Because the dissociation of a carboxylic acid is an equilibrium process, any factor that stabilizes the carboxylate anion relative to undissociated carboxylic acid will drive the equilibrium toward increased dissociation and result in increased acidity. For instance, three electron-withdrawing fluorine atoms delocalize the negative charge in the trifluoroacetate anion, thereby stabilizing the ion and increasing the acidity of $\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}$. In the same way, glycolic acid $\left(\mathrm{HOCH}_{2} \mathrm{CO}_{2} \mathrm{H} ; \mathrm{p} K_{\mathrm{a}}=3.83\right)$ is stronger than acetic acid because of the electronwithdrawing effect of the electronegative oxygen atom.

Because inductive effects operate through $\sigma$ bonds and are dependent on distance, the effect of halogen substitution decreases as the substituent moves farther from the carboxyl. Thus, 2-chlorobutanoic acid has $\mathrm{p} K_{\mathrm{a}}=2.86,3$-chlorobutanoic acid has $\mathrm{p} K_{\mathrm{a}}=4.05$, and 4-chlorobutanoic acid has $\mathrm{p} K_{\mathrm{a}}=4.52$, similar to that of butanoic acid itself.

$\mathrm{p} K_{\mathrm{a}}=4.52$

$\mathrm{p} K_{\mathrm{a}}=4.05$
Acidity


Methyl 4-cyanopentanoate - 20.4 Substituent Effects on Acidity
$\mathrm{p} K_{\mathrm{a}}=4.52$

$\mathrm{p} K_{\mathrm{a}}=4.05$
Acidity

Substituent effects on acidity are also found in substituted benzoic acids. We said during the discussion of electrophilic aromatic substitution in Section 16.4 that substituents on the aromatic ring strongly affect reactivity. Aromatic rings with electron-donating groups are activated toward further electrophilic substitution, and aromatic rings with electron-withdrawing groups are deactivated. Exactly the same effects can be observed on the acidity of substituted benzoic acids (TABLE 20.4).

|  | Y | <smiles>[Y]c1ccc(C(=O)O)cc1</smiles> |  |  |
| :---: | :---: | :---: | :---: | :---: |
|  |  | $K_{\mathrm{a}} \times 1$ | $\mathrm{p} K_{\mathrm{a}}$ |  |
| Stronger acid | $-\mathrm{NO}_{2}$ | 39 | 3.41 | Deactivating groups |
|  | -CN | 28 | 3.55 |  |
|  | - CHO | 18 | 3.75 |  |
|  | -Br | 11 | 3.96 |  |
|  | -Cl | 10 | 4.0 |  |
|  | -H | 6.46 | 4.19 |  |
| Weaker | $-\mathrm{CH}_{3}$ | 4.3 | 4.34 | Activating groups |
|  | $-\mathrm{OCH}_{3}$ | 3.5 | 4.46 |  |
|  | -OH | 3.3 | 4.48 |  |

As TABLE 20.4 shows, an electron-donating (activating) group such as methoxy decreases acidity by destabilizing the carboxylate anion, and an electron-withdrawing (deactivating) group such as nitro increases acidity by stabilizing the carboxylate anion.


Methyl 4-cyanopentanoate - 20.4 Substituent Effects on Acidity
$p$-Methoxybenzoic acid ( $\mathrm{p} K_{\mathrm{a}}=4.46$ )

Benzoic acid
( $\mathrm{p} K_{\mathrm{a}}=4.19$ )

$p$-Nitrobenzoic acid $\left(\mathrm{p} K_{\mathrm{a}}=3.41\right)$


Acidity -
Because it's much easier to measure the acidity of a substituted benzoic acid than it is to determine the relative reactivity of an aromatic ring toward electrophilic substitution, the correlation between the two effects is useful for predicting reactivity. If we want to know the effect of a certain substituent on electrophilic reactivity, we can simply find the acidity of the corresponding benzoic acid. Worked Example 20.1 gives an illustration.

... lets us predict the reactivity of this substituted benzene to electrophilic attack.


Predicting the Effect of a Substituent on the Reactivity of an Aromatic Ring toward Electrophilic Substitution -
The $\mathrm{p} K_{\mathrm{a}}$ of $p$-(trifluoromethyl)benzoic acid is 3.6 . Is the trifluoromethyl substituent an activating or deactivating group in electrophilic aromatic substitution?


Strategy -
Decide whether $p$-(trifluoromethyl)benzoic acid is stronger or weaker than benzoic acid. A substituent that strengthens the acid is a deactivating group because it withdraws electrons, and a substituent that weakens the acid is an activating group because it donates electrons.


Solution -
A p $K_{\mathrm{a}}$ of 3.6 means that $p$-(trifluoromethyl)benzoic acid is stronger than benzoic acid, whose $\mathrm{p} K_{\mathrm{a}}$ is 4.19. Thus, the trifluoromethyl substituent favors dissociation by helping stabilize the negative charge. Trifluoromethyl must therefore be an electron-withdrawing, deactivating group.

PROBLEM Which would you expect to be a stronger acid, the lactic acid found in tired muscles or acetic acid? 20-6 Explain.

Lactic acid

PROBLEM Dicarboxylic acids have two dissociation constants, one for the initial dissociation into a monoanion 20-7 and one for the second dissociation into a dianion. For oxalic acid, $\mathrm{HO}_{2} \mathrm{C}-\mathrm{CO}_{2} \mathrm{H}$, the first ionization constant is $\mathrm{p} K_{\mathrm{a} 1}=1.2$ and the second ionization constant is $\mathrm{p} K_{\mathrm{a} 2}=4.2$. Why is the second carboxyl group far less acidic than the first?

PROBLEM The $\mathrm{p} K_{\mathrm{a}}$ of $p$-cyclopropylbenzoic acid is 4.45. Is cyclopropylbenzene likely to be more reactive or 20-8 less reactive than benzene toward electrophilic bromination? Explain.

PROBLEM Rank the following compounds in order of increasing acidity. Don't look at a table of $\mathrm{p} K_{\mathrm{a}}$ data to help 20-9 with your answer.
(a) Benzoic acid, $p$-methylbenzoic acid, $p$-chlorobenzoic acid
(b) $p$-Nitrobenzoic acid, acetic acid, benzoic acid


Solution - 20.5 Preparing Carboxylic Acids
Let's review briefly some of the methods for preparing carboxylic acids that we've seen in previous chapters.

- Oxidation of a substituted alkylbenzene with $\mathrm{KMnO}_{4}$ gives a substituted benzoic acid (Section 16.8). Both primary and secondary alkyl groups can be oxidized, but tertiary groups are not affected.

$p-$ Nitrotoluene
$p$-Nitrobenzoic acid (88\%)
- Oxidation of a primary alcohol or an aldehyde yields a carboxylic acid (Section 17.7 and Section 19.3). Primary alcohols are often oxidized with the Dess Martin periodinane, and aldehydes are similarly oxidized with alkaline $\mathrm{KMnO}_{4}$.

4-Methyl-1-pentanol
4-Methylpentanoic acid

Hexanal
Hexanoic acid


Hydrolysis of Nitriles -
Carboxylic acids can be prepared from nitriles on heating with aqueous acid or base by a mechanism that we'll discuss in Section 20.7. Since nitriles themselves are usually made by $\mathrm{S}_{\mathrm{N}} 2$ reaction of a primary or secondary alkyl halide with $\mathrm{CN}^{-}$, the two-step sequence of cyanide displacement followed by nitrile hydrolysis is a good way to make a carboxylic acid from an alkyl halide ( $\mathrm{RBr} \rightarrow \mathrm{RC} \equiv \mathrm{N} \rightarrow \mathrm{RCO}_{2} \mathrm{H}$ ). Note that the product acid has one more carbon than the starting alkyl halide. One example occurs in a commercial route for the synthesis of the nonsteroidal anti-inflammatory drug ibuprofen. (See Chapter 15 Chemistry Matters.)

$\downarrow$ 1. $\mathrm{NaOH}, \mathrm{H}_{2} \mathrm{O}$

Ibuprofen


Carboxylation of Grignard Reagents -
Another method for preparing carboxylic acids is by reaction of a Grignard reagent with $\mathrm{CO}_{2}$ to yield a metal carboxylate, followed by protonation to give a carboxylic acid. This carboxylation reaction is usually carried out by bubbling a stream of dry $\mathrm{CO}_{2}$ gas through a solution of the Grignard reagent. The organomagnesium halide adds to a $\mathrm{C}=\mathrm{O}$ bond of carbon dioxide in a typical nucleophilic carbonyl addition reaction, and protonation of
the carboxylate by addition of aqueous HCl in a separate step then gives the free carboxylic acid. For example:

As noted previously, there are no Grignard reagents inside living cells, but there are other types of stabilized carbanions that are often carboxylated. One of the initial steps in fatty-acid biosynthesis, for instance, involves the formation of a carbanion from acetyl CoA, followed by carboxylation to yield malonyl CoA.


Devising a Synthesis Route for a Carboxylic Acid -
How would you prepare phenylacetic acid $\left(\mathrm{PhCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)$ from benzyl bromide $\left(\mathrm{PhCH}_{2} \mathrm{Br}\right)$ ?


Strategy -
We've seen two methods for preparing carboxylic acids from alkyl halides: (1) cyanide ion displacement followed by hydrolysis and (2) formation of a Grignard reagent followed by carboxylation. The first method involves an $\mathrm{S}_{\mathrm{N}} 2$ reaction and is therefore limited to use with primary and some secondary alkyl halides. The second method involves formation of a Grignard reagent and is therefore limited to use with organic halides that have no acidic hydrogens or reactive functional groups elsewhere in the molecule. In the present instance, either method would work well.

Solution

PROBLEM How would you prepare the following carboxylic acids?
20-10 (a) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCO}_{2} \mathrm{H}$ from $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ from $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$


Strategy - 20.6 Reactions of Carboxylic Acids: An Overview
We commented earlier in this chapter that carboxylic acids are similar in some respects to both alcohols and
ketones. Like alcohols, carboxylic acids can be deprotonated to give anions, which are good nucleophiles in $\mathrm{S}_{\mathrm{N}} 2$ reactions. Like ketones, carboxylic acids undergo addition of nucleophiles to the carbonyl group. However, carboxylic acids also undergo other reactions characteristic of neither alcohols nor ketones. FIGURE 20.3 shows some of the general reactions of carboxylic acids.

Deprotonation

Reduction

FIGURE 20.3 Some general reactions of carboxylic acids.
Reactions of carboxylic acids can be grouped into the four categories indicated in FIGURE 20.3. Of the four, we've already discussed the acidic behavior of carboxylic acids in Section 20.2 through Section 20.4, and we mentioned reduction by treatment with $\mathrm{LiAlH}_{4}$ in Section 17.4. The remaining two categories are examples of fundamental carbonyl-group reaction mechanisms-nucleophilic acyl substitution and $\alpha$ substitution-that will be discussed in detail in Chapters 21 and 22.

PROBLEM How might you prepare 2-phenylethanol from benzyl bromide? More than one step is needed.
20-11

PROBLEM How might you carry out the following transformation? More than one step is needed.
20-12


Strategy - 20.7 Chemistry of Nitriles
Nitriles are analogous to carboxylic acids in that both have a carbon atom with three bonds to an electronegative atom and both contain a $\pi$ bond. Thus, some reactions of nitriles and carboxylic acids are similar. Both kinds of compounds are electrophiles, for instance, and both undergo nucleophilic addition reactions.

$$
\mathrm{R}-\mathrm{C} \equiv \mathrm{~N}
$$


A nitrile-three bonds to nitrogen -
An acid-three bonds to two oxygens

Nitriles occur infrequently in living organisms, although several hundred examples are known. Cyanocycline A, for instance, has been isolated from the bacterium Streptomyces lavendulae and was found to have both antimicrobial and antitumor activity. In addition, more than 1000 compounds called cyanogenic glycosides are known. Derived primarily from plants, cyanogenic glycosides contain a sugar with an acetal carbon, one oxygen of which is bonded to a nitrile-bearing carbon (sugar-O-C-CN). On hydrolysis with aqueous acid, the acetal is cleaved (Section 19.10), generating a cyanohydrin (HO-C-CN), which releases hydrogen cyanide. It's thought that the primary function of cyanogenic glycosides is to protect the plant by poisoning any animal foolish enough to eat it. Lotaustralin from the cassava plant is an example.

Cyanocycline A

Lotaustralin
(a cyanogenic glycoside)


Preparation of Nitriles -
The simplest method of nitrile preparation is the $\mathrm{S}_{\mathrm{N}} 2$ reaction of $\mathrm{CN}^{-}$with a primary or secondary alkyl halide, as discussed in Section 20.5. Another method for preparing nitriles is by dehydration of a primary amide, $\mathrm{RCONH}_{2}$. Thionyl chloride is often used for this reaction, although other dehydrating agents such as $\mathrm{POCl}_{3}$ also work.

The dehydration occurs by initial reaction of $\mathrm{SOCl}_{2}$ on the nucleophilic amide oxygen atom, followed by deprotonation and a subsequent E2-like elimination reaction.

Both methods of nitrile synthesis- $\mathrm{S}_{\mathrm{N}} 2$ displacement by $\mathrm{CN}^{-}$on an alkyl halide and amide dehydration-are useful, but the synthesis from amides is more general because it is not limited by steric hindrance.


Reactions of Nitriles -
Like a carbonyl group, a nitrile group is strongly polarized and has an electrophilic carbon atom. Nitriles therefore react with nucleophiles to yield $s p^{2}$-hybridized imine anions in a reaction analogous to the formation of a $s p^{3}$-hybridized alkoxide ion by nucleophilic addition to a carbonyl group.


Carbonyl compound -
Nitrile

Some general reactions of nitriles are shown in FIGURE 20.4.

Hydrolysis: Conversion of Nitriles into Carboxylic Acids
Among the most useful reactions of nitriles is their hydrolysis to yield first an amide and then a carboxylic acid plus ammonia or an amine. The reaction occurs in either basic or acidic aqueous solution:

As shown in FIGURE 20.5, base-catalyzed nitrile hydrolysis involves nucleophilic addition of hydroxide ion to the polar $\mathrm{C} \equiv \mathrm{N}$ bond to give an imine anion in a process similar to the nucleophilic addition to a polar $\mathrm{C}=\mathrm{O}$ bond to give an alkoxide anion. Protonation then gives a hydroxy imine, which tautomerizes (Section 9.4) to an amide in a step similar to the tautomerization of an enol to a ketone. Further hydrolysis gives a carboxylate ion.


FIGURE 20.5 MECHANISM -
Mechanism for the basic hydrolysis of a nitrile to yield an amide, which is then hydrolyzed further to a carboxylic acid anion.
(1) Nucleophilic addition of hydroxide ion to the CN triple bond gives an imine anion addition product.
(2) Protonation of the imine anion by water yields a hydroxyimine and regenerates the base catalyst.

(3) Tautomerization of the hydroxyimine yields an amide in a reaction analogous to the tautomerization of an enol to give a ketone.

(4) Further hydrolysis of the amide gives the anion of a carboxylic acid by a mechanism we'll discuss in Section 21.7.

Carboxylate ion

The further hydrolysis of the amide intermediate takes place by a nucleophilic addition of hydroxide ion to the amide carbonyl group, which yields a tetrahedral alkoxide ion. Expulsion of amide ion, $\mathrm{NH}_{2}{ }^{-}$, as leaving group gives the carboxylate ion, thereby driving the reaction toward the products. Subsequent acidification in a separate step yields the carboxylic acid. We'll look at this process in more detail in Section 21.7.


Reduction: Conversion of Nitriles into Amines -
Reduction of a nitrile with $\mathrm{LiAlH}_{4}$ gives a primary amine, $\mathrm{RNH}_{2}$. The reaction occurs by nucleophilic addition of hydride ion to the polar $\mathrm{C} \equiv \mathrm{N}$ bond, yielding an imine anion, which still contains a $\mathrm{C}=\mathrm{N}$ bond and therefore undergoes a second nucleophilic addition of hydride to give a dianion. Both monoanion and dianion intermediates are undoubtedly stabilized by Lewis acid-base complexation to an aluminum species, facilitating the second addition that would otherwise be difficult. Protonation of the dianion by addition of water in a subsequent step gives the amine.

Reaction of Nitriles with Grignard Reagents
Grignard reagents add to a nitrile to give an intermediate imine anion that is hydrolyzed by addition of water to yield a ketone. The mechanism of the hydrolysis is the exact reverse of imine formation (see FIGURE 19.7).

This reaction is similar to the reduction of a nitrile to an amine, except that only one nucleophilic addition occurs rather than two and the attacking nucleophile is a carbanion ( $\mathrm{R}:^{-}$) rather than a hydride ion. For example:


Synthesizing a Ketone from a Nitrile -
How would you prepare 2-methyl-3-pentanone from a nitrile?

2-Methyl-3-pentanone


Strategy -
A ketone results from the reaction between a Grignard reagent and a nitrile, with the $\mathrm{C} \equiv \mathrm{N}$ carbon of the nitrile becoming the carbonyl carbon. Identify the two groups attached to the carbonyl carbon atom in the product. One will come from the Grignard reagent and the other will come from the nitrile.


Solution -
There are two possibilities.

PROBLEM How would you prepare the following carbonyl compounds from a nitrile?
20-13
(a)

(b)

PROBLEM How would you prepare 1-phenyl-2-butanone, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}$, from benzyl bromide,
20-14 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}$ ? More than one step is needed.


Infrared Spectroscopy -
Carboxylic acids have two characteristic IR absorptions that make the $-\mathrm{CO}_{2} \mathrm{H}$ group easily identifiable. The $\mathrm{O}-\mathrm{H}$ bond of the carboxyl group gives rise to a very broad absorption over the range 2500 to $3300 \mathrm{~cm}^{-1}$. The $\mathrm{C}=\mathrm{O}$ bond shows an absorption between 1710 and $1760 \mathrm{~cm}^{-1}$. The exact position of $\mathrm{C}=\mathrm{O}$ absorption depends both on the structure of the molecule and on whether the acid is free (monomeric) or hydrogen-bonded (dimeric). Free carboxyl groups absorb at $1760 \mathrm{~cm}^{-1}$, but the more commonly encountered dimeric carboxyl groups absorb in a broad band centered around $1710 \mathrm{~cm}^{-1}$. As with other carbonyl-containing functional groups, conjugation with an alkene or benzene ring lowers the frequency of the $\mathrm{C}=\mathrm{O}$ stretch by 20 to $30 \mathrm{~cm}^{-1}$.

Associated carboxyl (usual case), $1710 \mathrm{~cm}^{-1}$

Both the broad $\mathrm{O}-\mathrm{H}$ absorption and the $\mathrm{C}=\mathrm{O}$ absorption at $1710 \mathrm{~cm}^{-1}$ (dimeric) are identified in the IR spectrum of butanoic acid shown in FIGURE 20.6.

FIGURE 20.6 IR spectrum of butanoic acid, $\mathbf{C H}_{3} \mathbf{C H}_{2} \mathbf{C H}_{2} \mathbf{C O}_{2} \mathbf{H}$.
Nitriles show an intense and easily recognizable $\mathrm{C} \equiv \mathrm{N}$ bond absorption near $2250 \mathrm{~cm}^{-1}$ for saturated compounds and $2230 \mathrm{~cm}^{-1}$ for aromatic and conjugated molecules. Few other functional groups absorb in this region, so IR spectroscopy is highly diagnostic for nitriles.


Infrared Spectroscopy -
PROBLEM Cyclopentanecarboxylic acid and 4-hydroxycyclohexanone have the same formula $\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{2}\right)$, and
20-15 both contain an -OH and a $\mathrm{C}=\mathrm{O}$ group. How could you distinguish between them using IR spectroscopy?


Nuclear Magnetic Resonance Spectroscopy -
Carboxyl carbon atoms absorb in the range 165 to $185 \delta$ in the ${ }^{13} \mathrm{C}$ NMR spectrum, with aromatic and $\alpha, \beta$-unsaturated acids near the upfield end of the range ( $\sim 165 \delta$ ) and saturated aliphatic acids near the downfield end ( $\sim 185 \delta$ ). Nitrile carbons absorb in the range 115 to $130 \delta$.

In the ${ }^{1} \mathrm{H}$ NMR spectrum, the acidic $-\mathrm{CO}_{2} \mathrm{H}$ proton normally absorbs as a singlet near $12 \delta$. The chemical shift of the carboxyl proton is concentration and solvent dependent, as these variables can change the extent of hydrogen-bonding in the sample. In some cases, the carboxyl-proton resonance is broadened to the point of being nearly undetectable. Traces of water in the sample can exacerbate the situation. As with alcohols (Section 17.11), the $-\mathrm{CO}_{2} \mathrm{H}$ proton can be replaced by deuterium when $\mathrm{D}_{2} \mathrm{O}$ is added to the sample tube, causing the absorption to disappear from the NMR spectrum. FIGURE 20.7 shows the ${ }^{1} \mathrm{H}$ NMR spectrum of phenylacetic acid. Note that the $-\mathrm{CO}_{2} \mathrm{H}$ absorption occurs at $12.0 \delta$.

FIGURE 20.7 Proton NMR spectrum of phenylacetic acid, $\mathbf{C}_{\mathbf{6}} \mathbf{H}_{\mathbf{5}} \mathbf{C H}_{\mathbf{2}} \mathbf{C O}_{\mathbf{2}} \mathbf{H}$.
PROBLEM How could you distinguish between the isomers cyclopentanecarboxylic acid and 20-16 4-hydroxycyclohexanone by ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ NMR spectroscopy? (See Problem 20-15.)


Vitamin C -
The word vitamin, despite its common usage, is actually an imprecise term. Generally speaking, a vitamin is an organic substance that a given organism requires in small amounts to live and grow but is unable to synthesize and must obtain in its diet. Thus, to be considered a vitamin, only a small amount of the substance is needed-anywhere from a few micrograms to 100 mg or so per day. Dietary substances needed in larger amounts, such as some amino acids and unsaturated fats, are not considered vitamins.

FIGURE 20.8 In addition to the hazards of weather, participants in early polar expeditions often suffered from scurvy, caused by a dietary vitamin C deficiency. (credit: "Shackleton's expedition to the Antarctic last moments of the Endurance" by Underwood \& Underwood/ Wikimedia Commons, Public Domain)

Furthermore, different organisms need different vitamins. More than 4000 species of mammals can synthesize ascorbic acid in their bodies, for instance, but humans are not among them. Ascorbic acid is therefore a human vitamin-what we all know as vitamin C -and must be provided by our diet. Small amounts of more than a dozen other substances are similarly required by humans: retinol (vitamin A), thiamine (vitamin $B_{1}$ ), and tocopherol (vitamin E), for instance.

Vitamin C is surely the best known of all human vitamins. It was the first to be discovered (1912), the first to be isolated (1928), the first to be structurally characterized (1933), and the first to be synthesized in the laboratory (1933). Over 121,000 tons of vitamin C is synthesized worldwide each year, more than the total amount of all other vitamins combined. In addition to its use as a supplement, vitamin C is used as a food preservative, a "flour improver" in bakeries, and an animal-food additive.

Vitamin C (ascorbic acid)

Vitamin C is perhaps most well-known for its antiscorbutic properties, meaning that it prevents the onset of scurvy, a bleeding disease affecting those with a deficiency of fresh vegetables and citrus fruits in their diet. Sailors in the Age of Exploration were particularly susceptible to scurvy, and the death toll was high. The


Vitamin C -
Portuguese explorer Vasco da Gama lost more than half his crew to scurvy during his two-year voyage around the Cape of Good Hope in 1497-1499.

More recently, large doses of vitamin $C$ have been claimed to prevent the common cold, cure infertility, delay the onset of symptoms in acquired immunodeficiency syndrome (AIDS), and inhibit the development of gastric and cervical cancers. None of these claims have been backed by medical evidence, however. In the largest study yet of the effect of vitamin $C$ on the common cold, a meta-analysis of more than 100 separate trials covering 40,000 people found no difference in the incidence of colds between those who took supplemental vitamin C regularly and those who did not. When taken during a cold, however, vitamin $C$ does appear to decrease the cold's duration by perhaps a day.

The industrial preparation of vitamin $C$ involves an unusual blend of biological and laboratory organic chemistry, beginning with glucose and following the five-step route shown in FIGURE 20.9. Glucose, a pentahydroxy aldehyde, is first reduced to sorbitol, which is then oxidized by the microorganism Acetobacter suboxydans. No chemical reagent is known that is selective enough to oxidize only one of the six alcohol groups in sorbitol, so an enzymatic reaction is used. Treatment with acetone and an acid catalyst then converts four of the other hydroxyl groups into acetal linkages, and the remaining hydroxyl group is chemically oxidized to a carboxylic acid by reaction with aqueous NaOCl (household bleach). Hydrolysis with acid then removes the two acetal groups and causes an internal ester-forming reaction to give ascorbic acid. Each of the five steps has a yield better than $90 \%$.

FIGURE 20.9 The industrial synthesis of ascorbic acid from glucose.


Key Terms -
- carboxyl group - Henderson-Hasselbalch equation
- carboxylation
- nitrile
- carboxylic acid, $\mathrm{RCO}_{2} \mathrm{H}$


Summary -
Carboxylic acids are among the most useful building blocks for synthesizing other molecules, both in nature and in the laboratory. Thus, an understanding of their properties and reactions is fundamental to understanding biological chemistry. In this chapter, we've looked both at acids and at their close relatives, nitriles ( $\mathrm{RC} \equiv \mathrm{N}$ ).

Carboxylic acids are named systematically by replacing the terminal -e of the corresponding alkane name with -oic acid. Like aldehydes and ketones, the carbonyl carbon atom is $s p^{2}$-hybridized; like alcohols, carboxylic acids are associated through hydrogen-bonding and therefore have high boiling points.

The distinguishing characteristic of carboxylic acids is their acidity. Although weaker than mineral acids such as HCl , carboxylic acids dissociate much more readily than alcohols because the resultant carboxylate ions are stabilized by resonance between two equivalent forms.

Most carboxylic acids have $\mathrm{p} K_{\mathrm{a}}$ values near 5, but the exact $\mathrm{p} K_{\mathrm{a}}$ of a given acid depends on structure. Carboxylic acids substituted by electron-withdrawing groups are more acidic (have a lower $\mathrm{p} K_{\mathrm{a}}$ ) because their carboxylate ions are stabilized. Carboxylic acids substituted by electron-donating groups are less acidic (have a higher $\mathrm{p} K_{\mathrm{a}}$ ) because their carboxylate ions are destabilized. The extent of dissociation of a carboxylic acid in a buffered solution of a given pH can be calculated with the Henderson-Hasselbalch equation. Inside living cells, where the physiological $\mathrm{pH}=7.3$, carboxylic acids are entirely dissociated and exist as their carboxylate anions.


Summary -
Methods of synthesis for carboxylic acids include (1) oxidation of alkylbenzenes, (2) oxidation of primary alcohols or aldehydes, (3) reaction of Grignard reagents with $\mathrm{CO}_{2}$ (carboxylation), and (4) hydrolysis of nitriles. General reactions of carboxylic acids include (1) loss of the acidic proton, (2) nucleophilic acyl substitution at the carbonyl group, (3) substitution on the $\alpha$ carbon, and (4) reduction.

Nitriles are similar in some respects to carboxylic acids and are prepared either by $\mathrm{S}_{\mathrm{N}} 2$ reaction of an alkyl halide with cyanide ion or by dehydration of an amide. Nitriles undergo nucleophilic addition to the polar $\mathrm{C} \equiv \mathrm{N}$ bond in the same way that carbonyl compounds do. The most important reactions of nitriles are their hydrolysis to carboxylic acids, reduction to primary amines, and reaction with Grignard reagents to yield ketones.

Carboxylic acids and nitriles are easily distinguished spectroscopically. Acids show a characteristic IR absorption at 2500 to $3300 \mathrm{~cm}^{-1}$ due to the $\mathrm{O}-\mathrm{H}$ bond and another at 1710 to $1760 \mathrm{~cm}^{-1}$ due to the $\mathrm{C}=\mathrm{O}$ bond; nitriles have an absorption at $2250 \mathrm{~cm}^{-1}$. Acids also show ${ }^{13} \mathrm{C}$ NMR absorptions at 165 to $185 \delta$ and ${ }^{1} \mathrm{H}$ NMR absorptions near $12 \delta$. Nitriles have a ${ }^{13} \mathrm{C}$ NMR absorption in the range 115 to $130 \delta$.


Summary of Reactions -
1. Preparation of carboxylic acids (Section 20.5)
a. Carboxylation of Grignard reagents

$$
\mathrm{R}-\mathrm{MgX} \stackrel{\text { 1. } \mathrm{CO}_{2}}{\text { 2. } \mathrm{H}_{3} \mathrm{O}^{+}} \quad \stackrel{\stackrel{\mathrm{O}}{\|}}{\mathrm{R}^{-}}{ }_{\mathrm{OH}}
$$

b. Hydrolysis of nitriles

$$
\mathrm{R}-\mathrm{C} \equiv \mathrm{~N} \stackrel{\mathrm{or}^{\mathrm{NaOH}, \mathrm{H}_{2} \mathrm{O}}}{\mathrm{R}^{-} \stackrel{\mathrm{O}}{\mathrm{C}} \mathrm{C}_{\mathrm{OH}}^{+}}
$$

2. Preparation of nitriles (Section 20.7)
a. $\mathrm{S}_{\mathrm{N}} 2$ reaction of alkyl halides

$$
\mathrm{RCH}_{2} \mathrm{Br} \quad \xrightarrow{\mathrm{NaCN}} \mathrm{RCH}_{2} \mathrm{C} \equiv \mathrm{~N}
$$

b. Dehydration of amides

3. Reactions of nitriles (Section 20.7)
a. Hydrolysis to yield carboxylic acids

$$
\mathrm{R}-\mathrm{C} \equiv \mathrm{~N} \xrightarrow[\text { 2. } \mathrm{H}_{3} \mathrm{O}^{+}]{\stackrel{\text { 1. } \mathrm{NaOH}, \mathrm{H}_{2} \mathrm{O}}{\mathrm{R}^{-\quad}} \stackrel{\mathrm{O}}{\mathrm{C}} \mathrm{OH}_{\mathrm{OH}}}+\mathrm{NH}_{3}
$$

b. Reduction to yield primary amines

c. Reaction with Grignard reagents to yield ketones


Summary of Reactions -
b. Reduction to yield primary amines

c. Reaction with Grignard reagents to yield ketones

$$
\mathrm{R}-\mathrm{C} \equiv \mathrm{~N} \quad \stackrel{\text { 1. } \mathrm{R}^{\prime} \mathrm{MgX} \text {, ether }}{\text { 2. } \mathrm{H}_{3} \mathrm{O}^{+}} \quad \stackrel{\text { R }}{\stackrel{\mathrm{O}}{\mathrm{C}}>_{\mathrm{R}^{\prime}}}+\mathrm{NH}_{3}
$$


Visualizing Chemistry -
PROBLEM Give IUPAC names for the following carboxylic acids (reddish brown $=\mathrm{Br}$ ):

20-17 (a)

(c)

(b)

(d)

PROBLEM Would you expect the following carboxylic acids to be more acidic or less acidic than benzoic acid?
20-18 Explain. (Reddish brown $=\mathrm{Br}$.)
(a)

(b)

PROBLEM The following carboxylic acid can't be prepared from an alkyl halide by either the nitrile hydrolysis
20-19 route or the Grignard carboxylation route. Explain.

PROBLEM Electrostatic potential maps of anisole and thioanisole are shown. Which do you think is the 20-20 stronger acid, $p$-methoxybenzoic acid or $p$-(methylthio) benzoic acid? Explain.

Anisole $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OCH}_{3}\right)$

Thioanisole $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SCH}_{3}\right)$


Mechanism Problems -
PROBLEM Predict the product(s) and write the mechanism of each of the following reactions:

20-21 (a)

1. Mg
$\xrightarrow[\text { 3. } \mathrm{H}_{3} \mathrm{O}^{+}]{\text {2. } \mathrm{CO}_{2}}$ ?
(b)

PROBLEM Predict the product(s) and write the mechanism of each of the following reactions:


20-22 (a) -
(b)

PROBLEM Predict the product(s) and write the mechanism of each of the following reactions:

20-23 (a)

(b)

PROBLEM Predict the product(s) and write the complete mechanism of each of the following reactions:
20-24 (a)

(b)

PROBLEM Acid-catalyzed hydrolysis of a nitrile to give a carboxylic acid occurs by initial protonation of the
20-25 nitrogen atom, followed by nucleophilic addition of water. Review the mechanism of base-catalyzed nitrile hydrolysis in Section 20.7 and then predict the products for the following reactions. Write the steps involved in the acid-catalyzed reaction, using curved arrows.
(a)

(b)

PROBLEM Nitriles can be converted directly to esters by the Pinner reaction, which first produces an
$\mathbf{2 0 - 2 6}$ iminoester salt that is isolated and then treated with water to give the final product. Propose a mechanism for the Pinner reaction using curved arrows to show the flow of electrons at each step.

PROBLEM Naturally occurring compounds called cyanogenic glycosides, such as lotaustralin, release
20-27 hydrogen cyanide, HCN, when treated with aqueous acid. The reaction occurs by hydrolysis of the acetal linkage to form a cyanohydrin, which then expels HCN and gives a carbonyl compound.
(a) Show the mechanism of the acetal hydrolysis and the structure of the cyanohydrin that results.
(b) Propose a mechanism for the loss of HCN, and show the structure of the carbonyl compound that forms.


Lotaustralin -
PROBLEM 2-Bromo-6,6-dimethylcyclohexanone gives 2,2-dimethylcyclopentanecarboxylic acid on treatment
20-28 with aqueous NaOH followed by acidification, a process called the Favorskii reaction. Propose a mechanism.

PROBLEM Naturally occurring compounds called terpenoids, which we'll discuss in Section 27.5, are
20-29 biosynthesized by a pathway that involves loss of $\mathrm{CO}_{2}$ from 3-phosphomevalonate 5-diphosphate to yield isopentenyl diphosphate. Use curved arrows to show the mechanism of this reaction.

PROBLEM In the Ritter reaction, an alkene reacts with a nitrile in the presence of strong aqueous sulfuric acid 20-30 to yield an amide. Propose a mechanism.


Naming Carboxylic Acids and Nitriles -
PROBLEM Give IUPAC names for the following compounds:
20-31 (a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

PROBLEM Draw structures corresponding to the following IUPAC names:
20-32 (a) cis-1,2-Cyclohexanedicarboxylic acid (b) Heptanedioic acid (c) 2-Hexen-4-ynoic acid
(d) 4-Ethyl-2-propyloctanoic acid (e) 3-Chlorophthalic acid (f) Triphenylacetic acid
(g) 2-Cyclobutenecarbonitrile (h) m-Benzoylbenzonitrile

PROBLEM Draw and name the following compounds:
20-33 (a) The eight carboxylic acids with the formula $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}$
(b) Three nitriles with the formula $\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{~N}$

PROBLEM Pregabalin, marketed as Lyrica, is an anticonvulsant drug that is also effective in treating chronic
20-34 pain. The IUPAC name of pregabalin is (S)-3-(aminomethyl)-5-methylhexanoic acid. (An aminomethyl group is $-\mathrm{CH}_{2} \mathrm{NH}_{2}$.) Draw the structure of pregabalin.

PROBLEM Isocitric acid, an intermediate in the citric acid cycle of food metabolism, has the systematic name
20-35 ( $2 R, 3 S$ )-3-carboxy-2-hydroxypentanedioic acid. Draw the structure.


Acidity of Carboxylic Acids -
PROBLEM Order the compounds in each of the following sets with respect to increasing acidity:
20-36 (a) Acetic acid, oxalic acid, formic acid
(b) $p$-Bromobenzoic acid, $p$-nitrobenzoic acid, 2,4-dinitrobenzoic acid
(c) Fluoroacetic acid, 3-fluoropropanoic acid, iodoacetic acid

PROBLEM Arrange the compounds in each of the following sets in order of increasing basicity:
20-37 (a) Magnesium acetate, magnesium hydroxide, methylmagnesium bromide
(b) Sodium benzoate, sodium $p$-nitrobenzoate, sodium acetylide
(c) Lithium hydroxide, lithium ethoxide, lithium formate


Acidity of Carboxylic Acids -
PROBLEM Calculate the $\mathrm{p} K_{\mathrm{a}}$ 's of the following acids:
20-38 (a) Lactic acid, $K_{\mathrm{a}}=8.4 \times 10^{-4}$ (b) Acrylic acid, $K_{\mathrm{a}}=5.6 \times 10^{-6}$
PROBLEM Calculate the $K_{\mathrm{a}}$ 's of the following acids:
20-39 (a) Citric acid, $\mathrm{p} K_{\mathrm{a}}=3.14$ (b) Tartaric acid, $\mathrm{p} K_{\mathrm{a}}=2.98$
PROBLEM Thioglycolic acid, $\mathrm{HSCH}_{2} \mathrm{CO}_{2} \mathrm{H}$, a substance used in depilatory agents (hair removers) has $\mathrm{p} K_{\mathrm{a}}=$
20-40 3.42. What is the percent dissociation of thioglycolic acid in a buffer solution at $\mathrm{pH}=3.0$ ?
PROBLEM In humans, the final product of purine degradation from DNA is uric acid, $\mathrm{p} K_{\mathrm{a}}=5.61$, which is 20-41 excreted in the urine. What is the percent dissociation of uric acid in urine at a typical $\mathrm{pH}=6.0$ ? Why do you think uric acid is acidic even though it does not have a $\mathrm{CO}_{2} \mathrm{H}$ group?

Uric acid

PROBLEM Some $\mathrm{p} K_{\mathrm{a}}$ data for simple dibasic acids is shown. How can you account for the fact that the difference
20-42 between the first and second ionization constants decreases with increasing distance between the carboxyl groups?


Acidity of Carboxylic Acids -
| Name | Structure | $\mathbf{p K}$ | $\mathbf{p} \mathbf{K}_{\mathbf{2}}$ |
| :---: | :---: | :---: | :---: |
| Oxalic | $\mathrm{HO}_{2} \mathrm{CCO}_{2} \mathrm{H}$ | 1.2 | 4.2 |

| Succinic | $\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ | 4.2 | 5.6 |
| :---: | :---: | :---: | :---: |
| Adipic | $\mathrm{HO}_{2} \mathrm{C}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CO}_{2} \mathrm{H}$ | 4.4 | 5.4 |

Reactions of Carboxylic Acids and Nitriles
PROBLEM How could you convert butanoic acid into the following compounds? Write each step showing the 20-43 reagents needed.
(a) 1-Butanol
(b) 1-Bromobutane
(c) Pentanoic acid
(d) 1-Butene
(e) Octane

PROBLEM How could you convert each of the following compounds into butanoic acid? Write each step 20-44 showing all reagents.
(a) 1-Butanol
(b) 1-Bromobutane
(c) 1-Butene
(d) 1-Bromopropane
(e) 4-Octene

PROBLEM How could you convert butanenitrile into the following compounds? Write each step showing the 20-45 reagents needed.
(a) 1-Butanol
(b) Butylamine
(c) 2-Methyl-3-hexanone