With Chapter 7, you begin adding reactions to the framework of mechanism types that Chapter 6 introduces. This section describes how this and future chapters are set up. It also gives you some guidelines for studying. Early in each chapter, the fundamental mechanism discussed in that chapter is presented. Nucleophilic addition reactions take two mechanism modes. Figures 7.1, and 7.2 on page 000 show both mechanistic modes. Study both modes. Try to find
the underlying principles of acid-base chemistry that make the mechanism work.
Succeeding sections then present one or more specific reactions that follow the fundamental mechanism. Each section includes the generalized reaction and mechanism for that reaction type followed by one or more specific chemical examples. To study this material, strive to understand the generalized reaction and its mechanism. Consider how it relates to the fundamental mechanism given at the beginning of the chapter. Look at the examples and determine how they relate to the mechanism. Write a mechanism for one or more of the specific examples and compare it to the mechanism contained in the chapter.
Then work the exercises using that particular reaction type.
As you study each of the reaction types, learn the answers to all of the following four questions:
1. What mechanistic operations describe the electron flow from the substrate and reagents through any intermediates to form the product?
2. Given the substrate, reagent(s), and a set of reaction conditions, what is the expected product?
Substrate Reagents Conditions ?
3. Given both a substrate and a product, what reagents and reaction conditions favor the transformation?
Substrate ? Product
4. Given a product, reagent(s), and a set of reaction conditions, what is a probable substrate?
Product Reagents
? Conditions
The last three questions require that you thoroughly understand how a particular reaction type occurs. The fourth question focuses on the electron flow, or mechanism, of a substrate as it converts into a product. Reaction mechanisms are the key to your understanding of organic chemistry. Knowing the mechanism of a reaction allows you to generalize your knowledge about organic chemistry and to predict the outcome of a new reaction with confidence.
Being able to answer these four questions allows you to use the retrosynthetic analysis technique. This chapter and Chapter 15 discuss retrosynthetic analysis. To use retrosynthetic analysis, take the target molecule and search through the reactions you know to choose a reasonable precursor, which then becomes the new target molecule. Then search again through the reactions you know until you find a precursor for this new target molecule. Repeat this analysis until you find a readily available substrate. This approach allows you to logically plan the construction of a new molecule or to propose a new synthesis for existing molecules.
Retrosynthetic analysis literally means
backward synthetic analysis. As chemists propose a synthesis, they usually work backwards from the target to the substrate molecules.
7.4 The Cyanohydrin Reaction
One of the simplest nucleophilic additions to the carbonyl group is the cyanohydrin reaction. In this reaction hydrogen cyanide, HCN, adds to a carbonyl group to produce a 2-hydroxy nitrile.
Below is the generalized cyanohydrin reaction.
In the cyanohydrin reaction, HCN reacts with a carbonyl group to form an α hydroxy nitrile.
C CN OH HCN
C O +
The mechanism for a cyanohydrin reaction shows that a cyanide ion, which is the nucleophile, reacts with the carbonyl carbon in a 1,3-electron pair displacement operation to produce an unstable intermediate anion.
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••
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•• ••
CN H C O
R C O
CN H
R +
Because the intermediate is a stronger base than the c- CN ion, it abstracts a proton from the HCN molecule to form the cyanohydrin and another c- CN ion. This step is a 1,3-electron pair displacement operation.
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••
••
•••• + H CN ••
CN C O H
R
C OH CN H
R
+ CN
Exercise 7.3
The addition of HCN to a carbonyl group is a second order reversible reaction. The rate of a second order reaction depends on the concentrations of two reagents. The following equation expresses the rate for this reaction:
rate = k[RHC=O][c- CN]
a) What is the rate-limiting step for the reaction?
b) What does this tell you about the role of the proton in the reaction?
c) The equilibrium constant for the addition of HCN to 2- propanone is 32 at 27oC. At equilibrium, are the products or the reactants found in higher quantity?
d) What is the ∆Go of this reaction (see Section 6.2)?
HCN is a very weak acid (pKa=9.1), so it ionizes only slightly.
That is, at equilibrium only a few hydrogen and cyanide ions are present. Because of this low ionization, the reaction of the cyanide ion with the carbonyl group is relatively slow, but the reaction with the proton is relatively rapid. The rate law (see Exercise 7.3) for this reaction shows the concentrations of both the carbonyl group and the cyanide ion. Because the rate depends on the concentration of two reactants, the addition of HCN to an aldehyde or ketone is a second order reaction.
Exercise 7.4
Using the data, and your answers from Exercise 7.3, sketch a reaction profile for this reaction. Propose structures for all intermediates.
Because HCN is such a weak acid, it is a very poor source of cyanide ion nucleophiles. The rate of reaction with pure HCN is very low. However, the addition of a source of cyanide ions to the reaction flask causes a dramatic increase in the rate of reaction. Arthur Lapworth discovered this effect in 1903. Lapworth assumed that the reaction was ionic in nature and proposed a reaction mechanism remarkably similar to the above mechanism. His proposal was an extraordinary accomplishment considering that Lewis’ theory of bonding was more than ten years in the future. Lapworth was one of the originators of the mechanistic view of organic chemistry.
Exercise 7.5
Write a transformation for the conversion of each of the following compounds with HCN.
a)
C O
H
b)
O
c) 2-Butenal d) CH3CH2OCH2CH2CH2CHO e)
O
f)
CCH3
CH3OC O O
Sample solution
a)
C H OH
HCN CN C H
O
7.5 Addition of Water and Alcohols
The nucleophilic addition reaction of the carbonyl group in aldehydes and ketones with water is a rapidly reversible equilibrium reaction that produces a class of compounds known as hydrates.
A hydrate is a compound that contains two —OH groups bonded to the
same carbon. RCR'
O
RCR' OH
OH H2O
+
The acid-catalyzed reaction mechanism has three steps. The first step involves the protonation of the carbonyl oxygen in a 1,3-electron pair displacement operation. The next step is a nucleophilic attack by the water on the carbonyl carbon, and the last step is the loss of a proton to form the hydrate. The last two steps are also 1,3-electron pair displacement operations.
-
H2••O••
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••
••
•• ••
••
•• •• •• ••
•• ••
H2O C O
R R'
H OH2
C O R
R' H
C HO
R
R' OH H
C HO
R
R' OH
The position of equilibrium for the nucleophilic addition of water to aldehydes and ketones depends on the size and the number of groups attached to the carbonyl carbon. As the carbonyl group becomes increasingly hindered, the position of equilibrium favors the substrate. Generally, an aldehyde is more hydrated at equilibrium than a ketone. Table 7.1 lists the equilibrium constants for the hydration of some common aldehydes and ketones.
Carbonyl Compound
Ke (at 25oC)
% Reaction (at equilibrium) HCH
O
41 99.95
CH3CH O
1.8 x 10–2 50
(CH3)2CHCH O
1.2 x 10–3 32
(CH3)3CCH O
4.0 x 10–3 20
CH3CCH3
O
2.5 x 10–5 0.14
(CH3)3CCC(CH3)3
O
3.1 x 10–8 1.7 x 10–4
Table 7.1. Effect of group size on the equilibrium constant for the hydration of various carbonyl compounds.
The data in Table 7.1 show that Ke and the percentage of reaction to form the hydrate decreases as the size of the groups attached to the carbonyl carbon increases. This result is due mainly to steric hindrance. However, a second factor is the electronic effect of the alkyl groups attached to the carbonyl carbon. Because an alkyl group is weakly electron-donating, it tends to reduce the partial positive charge on the carbonyl carbon, thereby lowering its reactivity.
A strong electron-withdrawing substituent attached to the carbonyl group shifts the position of equilibrium to the product. For example, consider –CF3, a strong electron-withdrawing group. The negative end of the dipole moment for each of the three C—F bonds is the fluorine atom. Three such bonds create a large partial positive charge on the carbon atom, which, in turn, attempts to withdraw more electron density from the carbonyl group. Acetone has only a minute amount of hydration in contrast to hexafluoroacetone, which is almost completely hydrated.
Acetone is the common name for 2-propanone.
CF3CCF3
O
CF3CCF3
OH
OH H2O
+ Ke = 2.2 x 104
Most hydrates are so unstable that they cannot be isolated.
Two that are stable enough to isolate are the hydrate of methanal (formaldehyde) and the hydrate of trichloroethanal (chloral).
Formaldehyde is a gas at room temperature (b.p. –21oC), but chemists use it as a 37% aqueous solution of the hydrate. In aqueous solution they call it formalin. Formalin is often used to preserve biological specimens. The reaction of chloral with water has a large equilibrium constant of 3 x 104. When isolated as a stable solid, its melting point is 57oC. Chloral is a hypnotic or sedative—the "knockout drops" or
"Mickey Finn" of film and literature.
Chloral hydrate CCl3CH
OH
OH Formalin
HCH OH
OH Hemiacetals are
compounds containing one —OH group and one —OR group bonded to the same carbon.
A reaction equivalent to a hydration takes place between an aldehyde or ketone and an alcohol. This reaction involves one mole each of an alcohol and an aldehyde or ketone. The product is one of a class of compounds known as hemiacetals.
CH3CH2CH H O
CH3CH2CH OCH3
OH CH3OH
+
1-Hydroxypropyl methyl ether A Hemiacetal
(64%)
As with hydrate formation, hemiacetal formation is a reversible process. Hemiacetal formation follows a similar three-step reaction mechanism to a hydrate formation. The first step in an acid-catalyzed reaction is the protonation of the carbonyl oxygen in a heterogenic bond forming operation. The protonation of the carbonyl oxygen enhances the electrophilicity of the carbonyl carbon.
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••
••
••••
C O H
CH3CH2
C O H
H CH3CH2
C O H
H CH3CH2
+ H
Next, the nucleophilic oxygen of the alcohol adds to the carbonyl carbon of the resonance hybrid in a 1,3-electron pair displacement operation. The reaction consumes a proton to initiate the reaction, then releases a proton at the end of the sequence.
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••
••
••
••
••
••
C ••O CH3CH2
H
H
C O CH3CH2
H
H
CH3OH CH3CH2CH CH3O H