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4.6 • Axial and Equatorial Bonds in Cyclohexane
The chair conformation of cyclohexane has many consequences. We’ll see in Section 11.9, for instance, that the chemical behavior of many substituted cyclohexanes is influenced by their conformation. In addition, we’ll see in Section 25.5 that simple carbohydrates, such as glucose, adopt a conformation based on the cyclohexane chair and that their chemistry is directly affected as a result.
Another trait of the chair conformation is that there are two kinds of positions for substituents on the cyclohexane ring: axial positions and equatorial positions (as shown in Figure 4.9). The six axial positions are parallel to the ring axis, while the six equatorial positions are in the rough plane of the ring, around the ring equator.
and equatorial positions in chair cyclohexane. The six axial hydrogens are parallel to the ring axis, and the six equatorial hydrogens are in a band around the ring equator.
As shown in Figure 4.9, each carbon atom in chair cyclohexane has one axial and one equatorial hydrogen. Furthermore, each side of the ring has three axial and three equatorial hydrogens in an alternating arrangement. For example, if the top side of the ring has axial hydrogens on carbons 1, 3, and 5, then it has equatorial hydrogens on carbons 2, 4, and 6. The reverse is true for the bottom side: carbons 1, 3, and 5 have equatorial hydrogens, but carbons 2, 4, and 6 have axial hydrogens (Figure 4.10).
and equatorial positions in chair cyclohexane, looking directly down the ring axis. Each carbon atom has one axial and one equatorial position, and each face has alternating axial and equatorial positions.
Note that we haven’t used the words cis and trans in this discussion of cyclohexane conformation. Two hydrogens on the same side of the ring are always cis, regardless of whether they’re axial or equatorial and regardless of whether they’re adjacent. Similarly, two hydrogens on opposite sides of the ring are always trans.
Axial and equatorial bonds can be drawn following the procedure shown in Figure 4.11. If possible, look at a molecular model as you practice.
Because chair cyclohexane has two kinds of positions—axial and equatorial—we might expect to find two isomeric forms of a monosubstituted cyclohexane. In fact, we don’t. There is only one methylcyclohexane, one bromocyclohexane, one cyclohexanol (hydroxycyclohexane), and so on, because cyclohexane rings are conformationally mobile at room temperature. Different chair conformations readily interconvert, exchanging axial and equatorial positions. This interconversion, called a ring-flip, is shown in Figure 4.12.
in the starting structure becomes equatorial in the ring-flipped structure, and what is equatorial in the starting structure is axial after ring-flip.
As shown in Figure 4.12, a chair cyclohexane can be ring-flipped by keeping the middle four carbon atoms in place while folding the two end carbons in opposite directions. In so doing, an axial substituent in one chair form becomes an equatorial substituent in the ring-flipped chair form and vice versa. For example, axial bromocyclohexane becomes equatorial bromocyclohexane after a ring-flip. Since the energy barrier to chair–chair interconversion is only about 45 kJ/mol (10.8 kcal/mol), the process is rapid at room temperature and we see what appears to be a single structure rather than distinct axial and equatorial isomers.
Worked Example 4.2
Drawing the Chair Conformation of a Substituted Cyclohexane
Draw 1,1-dimethylcyclohexane in a chair conformation, indicating which methyl group in your drawing is axial and which is equatorial.
Strategy
Draw a chair cyclohexane ring using the procedure in Figure 4.11, and then put two methyl groups on the same carbon. The methyl group in the rough plane of the ring is equatorial, and the one directly above or below the ring is axial.
Solution
Problem 4-12
Draw two different chair conformations of cyclohexanol (hydroxycyclohexane), showing all hydrogen atoms. Identify each position as axial or equatorial.
Problem 4-13
Draw two different chair conformations of trans-1,4-dimethylcyclohexane, and label all positions as axial or equatorial.
Problem 4-14
Identify each of the colored positions—red, blue, and green—as axial or equatorial. Then carry out a ring-flip, and show the new positions occupied by each color. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.07%3A_Axial_and_Equatorial_Bonds_in_Cyclohexane.txt |
Even though cyclohexane rings flip rapidly between chair conformations at room temperature, the two conformations of a monosubstituted cyclohexane aren’t equally stable. In methylcyclohexane, for instance, the equatorial conformation is more stable than the axial conformation by 7.6 kJ/mol (1.8 kcal/mol). The same is true of other monosubstituted cyclohexanes: a substituent is almost always more stable in an equatorial position than in an axial position.
You might recall from your general chemistry course that it’s possible to calculate the percentages of two isomers at equilibrium using the equation ΔE = –RT ln K, where ΔE is the energy difference between isomers, R is the gas constant [8.315 J/(K·mol)], T is the Kelvin temperature, and K is the equilibrium constant between isomers. For example, an energy difference of 7.6 kJ/mol means that about 95% of methylcyclohexane molecules have an equatorial methyl group at any given instant while only 5% have an axial methyl group. Figure 4.13 plots the relationship between energy and isomer percentages.
The energy difference between axial and equatorial conformations is due to steric strain caused by 1,3-diaxial interactions. The axial methyl group on C1 is too close to the axial hydrogens three carbons away on C3 and C5, resulting in 7.6 kJ/mol of steric strain (Figure 4.14).
The 1,3-diaxial steric strain in substituted methylcyclohexane is already familiar—we saw it previously as the steric strain between methyl groups in gauche in Section 3.7). Gauche butane is less stable than anti butane by 3.8 kJ/mol (0.9 kcal/mol) because of steric interference between hydrogen atoms on the two methyl groups. Comparing a four-carbon fragment of axial methylcyclohexane with gauche butane shows that the steric interaction is the same in both (Figure 4.15). Because axial methylcyclohexane has two such interactions, it has 2 × 3.8 = 7.6 kJ/mol of steric strain. Equatorial methylcyclohexane has no such interactions and is therefore more stable.
and an axial hydrogen atom three carbons away is identical to the steric strain in gauche butane. (To display clearly the diaxial interactions in methylcyclohexane, two of the equatorial hydrogens are not shown.)
The exact amount of 1,3-diaxial steric strain in a given substituted cyclohexane depends on the nature and size of the substituent, as indicated in Table 4.1. Not surprisingly, the amount of steric strain increases through the series H3C− < CH3CH2− < (CH3)2CH–<< (CH3)3C−, paralleling the increasing size of the alkyl groups. Note that the values given in Table 4.1 refer to 1,3-diaxial interactions of the substituent with a single hydrogen atom. These values must be doubled to arrive at the amount of strain in a monosubstituted cyclohexane.
Table 4.1 Steric Strain in Monosubstituted Cyclohexanes
a 1,3-Diaxial strain
Y (kJ/mol) (kcal/mol)
F 0.5 0.12
Cl, Br 1.0 0.25
OH 2.1 0.5
CH3 3.8 0.9
CH2CH3 4.0 0.95
CH(CH3)2 4.6 1.1
C(CH3)3 11.4 2.7
C6H5 6.3 1.5
CO2H 2.9 0.7
CN 0.4 0.1
Problem 4-15
What is the energy difference between the axial and equatorial conformations of cyclohexanol (hydroxycyclohexane)?
Problem 4-16
Why do you suppose an axial cyano (–CN) substituent causes practically no 1,3-diaxial steric strain (0.4 kJ/mol)?
Problem 4-17
Look back at Figure 4.13 and estimate the percentages of axial and equatorial conformations present at equilibrium in bromocyclohexane. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.08%3A_Conformations_of_Monosubstituted_Cyclohexanes.txt |
Monosubstituted cyclohexanes are always more stable with their substituent in an equatorial position, but the situation with disubstituted cyclohexanes is more complex because the steric effects of both substituents must be taken into account. All steric interactions for both possible chair conformations must be analyzed before deciding which conformation is favored.
Let’s look at 1,2-dimethylcyclohexane as an example. There are two isomers, cis-1,2-dimethylcyclohexane and trans-1,2-dimethylcyclohexane, which must be considered separately. In the cis isomer, both methyl groups are on the same face of the ring and the compound can exist in either of the two chair conformations shown in Figure 4.16. (It may be easier for you to see whether a compound is cis- or trans-disubstituted by first drawing the ring as a flat representation and then converting it to a chair conformation.)
Both chair conformations of cis-1,2-dimethylcyclohexane have one axial methyl group and one equatorial methyl group. The top conformation in Figure 4.16 has an axial methyl group at C2, which has 1,3-diaxial interactions with hydrogens on C4 and C6. The ring-flipped conformation has an axial methyl group at C1, which has 1,3-diaxial interactions with hydrogens on C3 and C5. In addition, both conformations have gauche butane interactions between the two methyl groups. The two conformations are equal in energy, with a total steric strain of 3 × 3.8 kJ/mol = 11.4 kJ/mol (2.7 kcal/mol).
In trans-1,2-dimethylcyclohexane, the two methyl groups are on opposite sides of the ring and the compound can exist in either of the two chair conformations shown in Figure 4.17. The situation here is quite different from that of the cis isomer. The top conformation in Figure 4.17 has both methyl groups equatorial with only a gauche butane interaction between them (3.8 kJ/mol) but no 1,3-diaxial interactions. The ring-flipped conformation, however, has both methyl groups axial. The axial methyl group at C1 interacts with axial hydrogens at C3 and C5, and the axial methyl group at C2 interacts with axial hydrogens at C4 and C6. These four 1,3-diaxial interactions produce a steric strain of 4 × 3.8 kJ/mol = 15.2 kJ/mol and make the diaxial conformation 15.2 − 3.8 = 11.4 kJ/mol less favorable than the diequatorial conformation. We therefore predict that trans-1,2-dimethylcyclohexane will exist almost exclusively in the diequatorial conformation.
The same kind of conformational analysis just carried out for cis- and trans-1,2-dimethylcyclohexane can be done for any substituted cyclohexane, such as cis-1-tert-butyl-4-chlorocyclohexane (see Worked Example 4.3). As you might imagine, though, the situation becomes more complex as the number of substituents increases. For instance, compare glucose with mannose, a carbohydrate present in seaweed. Which do you think is more strained? In glucose, all substituents on the six-membered ring are equatorial, while in mannose, one of the −OH groups is axial, making it more strained.
A summary of the various axial and equatorial relationships among substituent groups in the different possible cis and trans substitution patterns for disubstituted cyclohexanes is given in Table 4.2.
Table 4.2 Axial and Equatorial Relationships in Cis- and Trans-Disubstituted Cyclohexanes
Cis/trans substitution pattern Axial/equatorial relationships
1,2-Cis disubstituted a,e or e,a
1,2-Trans disubstituted a,a or e,e
1,3-Cis disubstituted a,a or e,e
1,3-Trans disubstituted a,e or e,a
1,4-Cis disubstituted a,e or e,a
1,4-Trans disubstituted a,a or e,e
Worked Example 4.3
Drawing the Most Stable Conformation of a Substituted Cyclohexane
Draw the more stable chair conformation of cis-1-tert-butyl-4-chlorocyclohexane. By how much is it favored?
Strategy
Draw the two possible chair conformations, and calculate the strain energy in each. Remember that equatorial substituents cause less strain than axial substituents.
Solution
First draw the two chair conformations of the molecule:
In the conformation on the left, the tert-butyl group is equatorial and the chlorine is axial. In the conformation on the right, the tert-butyl group is axial and the chlorine is equatorial. These conformations aren’t of equal energy because an axial tert-butyl substituent and an axial chloro substituent produce different amounts of steric strain. Table 4.1 shows that the 1,3-diaxial interaction between a hydrogen and a tert-butyl group costs 11.4 kJ/mol (2.7 kcal/mol), whereas the interaction between a hydrogen and a chlorine costs only 1.0 kJ/mol (0.25 kcal/mol). An axial tert-butyl group therefore produces (2 × 11.4 kJ/mol) − (2 × 1.0 kJ/mol) = 20.8 kJ/mol (4.9 kcal/mol) more steric strain than an axial chlorine, and the compound preferentially adopts the conformation with the chlorine axial and the tert-butyl equatorial.
Problem 4-18 Draw the more stable chair conformation of the following molecules, and estimate the amount of strain in each: (a)
trans-1-Chloro-3-methylcyclohexane
(b) cis-1-Ethyl-2-methylcyclohexane (c) cis-1-Bromo-4-ethylcyclohexane (d) cis-1-tert-Butyl-4-ethylcyclohexane
Problem 4-19
Identify each substituent in the following compound as axial or equatorial, and tell whether the conformation shown is the more stable or less stable chair form (green = Cl): | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.09%3A_Conformations_of_Disubstituted_Cyclohexanes.txt |
The final point we’ll consider about cycloalkane stereochemistry is to see what happens when two or more cycloalkane rings are fused together along a common bond to construct a polycyclic molecule—for example, decalin.
Decalin consists of two cyclohexane rings joined to share two carbon atoms (the bridgehead carbons, C1 and C6) and a common bond. Decalin can exist in either of two isomeric forms, depending on whether the rings are trans fused or cis fused. In cis-decalin, the hydrogen atoms at the bridgehead carbons are on the same side of the rings; in trans-decalin, the bridgehead hydrogens are on opposite sides. Figure 4.18 shows how both compounds can be represented using chair cyclohexane conformations. Note that the two decalin isomers are not interconvertible by ring-flips or other rotations. They are cis–trans stereoisomers and have the same relationship to each other that cis- and trans-1,2-dimethylcyclohexane have.
at the bridgehead carbons are on the same face of the rings in the cis isomer but on opposite faces in the trans isomer.
Polycyclic compounds are common in nature, and many valuable substances have fused-ring structures. For example, steroids, such as testosterone, the primary sex hormone in males, have three six-membered rings and one five-membered ring fused together. Although steroids look complicated compared with cyclohexane or decalin, the same principles that apply to the conformational analysis of simple cyclohexane rings apply equally well (and often better) to steroids.
Another common ring system is the norbornane, or bicyclo[2.2.1]heptane, structure. Like decalin, norbornane is a bicycloalkane, so called because two rings would have to be broken open to generate an acyclic structure. Its systematic name, bicyclo[2.2.1]heptane, reflects the fact that the molecule has seven carbons, is bicyclic, and has three “bridges” of 2, 2, and 1 carbon atoms connecting the two bridgehead carbons.
Norbornane has a conformationally locked boat cyclohexane ring (Section 4.5) in which carbons 1 and 4 are joined by an additional CH2 group. In drawing this structure, a break in the rear bond indicates that the vertical bond crosses in front of it. Making a molecular model is particularly helpful when trying to see the three-dimensionality of norbornane.
Substituted norbornanes, such as camphor, are found widely in nature, and many have been important historically in developing organic structural theories.
Problem 4-20
Which isomer is more stable, cis-decalin or trans-decalin (Figure 4.18)? Explain.
Problem 4-21
Look at the following structure of estrone, the primary sex hormone in females, and tell whether each of the two indicated (red) ring-fusions is cis or trans.
4.11: Chemistry MattersMolecular Mechanics
4 • Chemistry Matters 4 • Chemistry Matters
All the structural models in this book are computer-drawn. To make sure they accurately represent bond angles, bond lengths, torsional interactions, and steric interactions, the most stable geometry of each molecule has been calculated on a desktop computer using a commercially available molecular mechanics program based on work by Norman Allinger at the University of Georgia.
The idea behind molecular mechanics is to begin with a rough geometry for a molecule and then calculate a total strain energy for that starting geometry, using mathematical equations that assign values to specific kinds of molecular interactions. Bond angles that are too large or too small cause angle strain; bond lengths that are too short or too long cause stretching or compressing strain; unfavorable eclipsing interactions around single bonds cause torsional strain; and nonbonded atoms that approach each other too closely cause steric, or van der Waals, strain.
$Etotal= Ebond stretching+ Eangle strain+ Etorsional strain + Evan der WaalsEtotal=Ebond stretching+Eangle strain+Etorsional strain+Evan der Waals$
After calculating a total strain energy for the starting geometry, the program automatically changes the geometry slightly in an attempt to lower strain—perhaps by lengthening a bond that is too short or decreasing an angle that is too large. Strain is recalculated for the new geometry, more changes are made, and more calculations are done. After dozens or hundreds of iterations, the calculation ultimately converges on a minimum energy that corresponds to the most favorable, least strained conformation of the molecule.
Similar calculations have proven to be particularly useful in pharmaceutical research, where a complementary fit between a drug molecule and a receptor molecule in the body is often the key to designing new pharmaceutical agents (Figure 4.20).
4.12: Key Terms
4 • Key Terms 4 • Key Terms
• alicyclic
• angle strain
• axial position (cyclohexane)
• boat cyclohexane
• Bridgehead atom
• chair conformation
• cis–trans isomers
• conformational analysis
• cycloalkanes
• 1,3-diaxial interaction
• Equatorial position (cyclohexane)
• polycyclic molecule
• ring-flip (cyclohexane)
• stereoisomer
• steric strain
• torsional strain
• twist-boat conformation
4.13: Summary
4 • Summary 4 • Summary
Cyclic molecules are so commonly encountered throughout organic and biological chemistry that it’s important to understand the consequences of their cyclic structures. Thus, we’ve taken a close look at cyclic structures in this chapter.
Cycloalkanes are saturated cyclic hydrocarbons with the general formula CnH2n. In contrast to open-chain alkanes, where nearly free rotation occurs around C−C bonds, rotation is greatly reduced in cycloalkanes. Disubstituted cycloalkanes can therefore exist as cis–trans isomers. The cis isomer has both substituents on the same side of the ring; the trans isomer has substituents on opposite sides. Cis–trans isomers are just one kind of stereoisomer—compounds that have the same connections between atoms but different three-dimensional arrangements.
Not all cycloalkanes are equally stable. Three kinds of strain contribute to the overall energy of a cycloalkane: (1) angle strain is the resistance of a bond angle to compression or expansion from the normal 109° tetrahedral value, (2) torsional strain is the energy cost of having neighboring C−H bonds eclipsed rather than staggered, and (3) steric strain is the repulsive interaction that arises when two groups attempt to occupy the same space.
Cyclopropane (115 kJ/mol strain) and cyclobutane (110.4 kJ/mol strain) have both angle strain and torsional strain. Cyclopentane is free of angle strain but has a substantial torsional strain due to its large number of eclipsing interactions. Both cyclobutane and cyclopentane pucker slightly away from planarity to relieve torsional strain.
Cyclohexane is strain-free because it adopts a puckered chair conformation, in which all bond angles are near 109° and all neighboring C–H bonds are staggered. Chair cyclohexane has two kinds of positions: axial and equatorial. Axial positions are oriented up and down, parallel to the ring axis, while equatorial positions lie in a belt around the equator of the ring. Each carbon atom has one axial and one equatorial position.
Chair cyclohexanes are conformationally mobile and can undergo a ring-flip, which interconverts axial and equatorial positions. Substituents on the ring are more stable in the equatorial position because axial substituents cause 1,3-diaxial interactions. The amount of 1,3-diaxial steric strain caused by an axial substituent depends on its size. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.10%3A_Conformations_of_Polycyclic_Molecules.txt |
4 • Additional Problems 4 • Additional Problems
Visualizing Chemistry
Problem 4-22
Name the following cycloalkanes: (a)
(b)
Problem 4-23 Name the following compound, identify each substituent as axial or equatorial, and tell whether the conformation shown is the more stable or less stable chair form (green = Cl): Problem 4-24 A trisubstituted cyclohexane with three substituents—red, green, and blue—undergoes a ring-flip to its alternate chair conformation. Identify each substituent as axial or equatorial, and show the positions occupied by the three substituents in the ring-flipped form. Problem 4-25 The following cyclohexane derivative has three substituents—red, green, and blue. Identify each substituent as axial or equatorial, and identify each pair of relationships (red–blue, red–green, and blue–green) as cis or trans. Problem 4-26 Glucose exists in two forms having a 36:64 ratio at equilibrium. Draw a skeletal structure of each, describe the difference between them, and tell which of the two you think is more stable (red = O). Cycloalkane Isomers Problem 4-27 Draw the five cycloalkanes with the formula C5H10. Problem 4-28 Give IUPAC names for the following compounds. (a) (b) (c) (d) (e) Problem 4-29 Draw a stereoisomer of trans-1,3-dimethylcyclobutane. Problem 4-30 Tell whether the following pairs of compounds are identical, constitutional isomers, stereoisomers, or unrelated. (a) cis-1,3-Dibromocyclohexane and trans-1,4-dibromocyclohexane (b)
2,3-Dimethylhexane and 2,3,3-trimethylpentane
(c)
Problem 4-31
Draw three isomers of trans-1,2-dichlorocyclobutane, and label them as either constitutional isomers or stereoisomers.
Problem 4-32
Identify each pair of relationships among the –OH groups in glucose (red–blue, red–green, red–black, blue–green, blue–black, green–black) as cis or trans.
Problem 4-33
Draw 1,3,5-trimethylcyclohexane using a hexagon to represent the ring. How many cis–trans stereoisomers are possible?
Cycloalkane Conformation and Stability
Problem 4-34
Hydrocortisone, a naturally occurring hormone produced in the adrenal glands, is often used to treat inflammation, severe allergies, and numerous other conditions. Is the indicated −OH group axial or equatorial?
Problem 4-35
A 1,2-cis disubstituted cyclohexane, such as cis-1,2-dichlorocyclohexane, must have one group axial and one group equatorial. Explain.
Problem 4-36
A 1,2-trans disubstituted cyclohexane must have either both groups axial or both groups equatorial. Explain.
Problem 4-37
Why is a 1,3-cis disubstituted cyclohexane more stable than its trans isomer?
Problem 4-38
Which is more stable, a 1,4-trans disubstituted cyclohexane or its cis isomer?
Problem 4-39
cis-1,2-Dimethylcyclobutane is less stable than its trans isomer, but cis-1,3-dimethylcyclobutane is more stable than its trans isomer. Draw the most stable conformations of both, and explain.
Problem 4-40 From the data in Figure 4.13 and Table 4.1, estimate the percentages of molecules that have their substituents in an axial orientation for the following compounds: (a)
Isopropylcyclohexane
(b) Fluorocyclohexane (c) Cyclohexanecarbonitrile, C6H11CN
Problem 4-41 Assume that you have a variety of cyclohexanes substituted in the positions indicated. Identify the substituents as either axial or equatorial. For example, a 1,2-cis relationship means that one substituent must be axial and one equatorial, whereas a 1,2-trans relationship means that both substituents are axial or both are equatorial. (a)
1,3-Trans disubstituted
(b) 1,4-Cis disubstituted (c) 1,3-Cis disubstituted (d) 1,5-Trans disubstituted (e) 1,5-Cis disubstituted (f) 1,6-Trans disubstituted
Cyclohexane Conformational Analysis
Problem 4-42
Draw the two chair conformations of cis-1-chloro-2-methylcyclohexane. Which is more stable, and by how much?
Problem 4-43
Draw the two chair conformations of trans-1-chloro-2-methylcyclohexane. Which is more stable?
Problem 4-44
Galactose, a sugar related to glucose, contains a six-membered ring in which all the substituents except the –OH group, indicated below in red, are equatorial. Draw galactose in its more stable chair conformation.
Problem 4-45
Draw the two chair conformations of menthol, and tell which is more stable.
Problem 4-46
There are four cis–trans isomers of menthol (Problem 4-45), including the one shown. Draw the other three.
Problem 4-47
The diaxial conformation of cis-1,3-dimethylcyclohexane is approximately 23 kJ/mol (5.4 kcal/mol) less stable than the diequatorial conformation. Draw the two possible chair conformations, and suggest a reason for the large energy difference.
Problem 4-48
Approximately how much steric strain does the 1,3-diaxial interaction between the two methyl groups introduce into the diaxial conformation of cis-1,3-dimethylcyclohexane? (See Problem 4-47.)
Problem 4-49
In light of your answer to Problem 4-48, draw the two chair conformations of 1,1,3-trimethylcyclohexane and estimate the amount of strain energy in each. Which conformation is favored?
Problem 4-50
One of the two chair structures of cis-1-chloro-3-methylcyclohexane is more stable than the other by 15.5 kJ/mol (3.7 kcal/mol). Which is it? What is the energy cost of a 1,3-diaxial interaction between a chlorine and a methyl group?
General Problems
Problem 4-51
We saw in Problem 4-20 that cis-decalin is less stable than trans-decalin. Assume that the 1,3-diaxial interactions in cis-decalin are similar to those in axial methylcyclohexane [that is, one CH2↔H interaction costs 3.8 kJ/mol (0.9 kcal/mol)], and calculate the magnitude of the energy difference between cis- and trans-decalin.
Problem 4-52
Using molecular models as well as structural drawings, explain why trans-decalin is rigid and cannot ring-flip whereas cis-decalin can easily ring-flip.
Problem 4-53
trans-Decalin is more stable than its cis isomer, but cis-bicyclo[4.1.0]heptane is more stable than its trans isomer. Explain.
Problem 4-54 As mentioned in Problem 3-53, the statin drugs, such as simvastatin (Zocor), pravastatin (Pravachol), and atorvastatin (Lipitor) are the most widely prescribed drugs in the world. (a)
Are the two indicated bonds on simvastatin cis or trans?
(b) What are the cis/trans relationships among the three indicated bonds on pravastatin? (c) Why can’t the three indicated bonds on atorvastatin be identified as cis or trans?
Problem 4-55
myo-Inositol, one of the isomers of 1,2,3,4,5,6-hexahydroxycyclohexane, acts as a growth factor in both animals and microorganisms. Draw the most stable chair conformation of myo-inositol.
Problem 4-56
How many cis–trans stereoisomers of myo-inositol (Problem 4-55) are there? Draw the structure of the most stable isomer.
Problem 4-57
Julius Bredt, discoverer of the structure of camphor, proposed in 1935 that bicycloalkenes such as 1-norbornene, which have a double bond to a bridgehead carbon, are too strained to exist. Explain. (Making a molecular model will be helpful.
Problem 4-58 Tell whether each of the following substituents on a steroid is axial or equatorial. (A substituent that is “up” is on the top side of the molecule as drawn, and a substituent that is “down” is on the bottom side. (a)
Substituent up at C3
(b) Substituent down at C7 (c)
Substituent down at C11
Problem 4-59
Amantadine is an antiviral agent that is active against influenza type A infection. Draw a three-dimensional representation of amantadine, showing the chair cyclohexane rings.
Problem 4-60
There are two different isomers named trans-1,2-dimethylcyclopentane. Similarly, you have two different appendages called hands. What is the relationship between them? (We’ll explore this kind of isomerism in the next chapter.)
Problem 4-61
Ketones react with alcohols to yield products called ketals. Why does the all-cis isomer of 4-tert-butyl-1,3-cyclohexanediol react readily with acetone and an acid catalyst to form a ketal, but other stereoisomers do not react? In formulating your answer, draw the more stable chair conformations of all four stereoisomers and the product ketal for each one.
Problem 4-62
Alcohols undergo an oxidation reaction to yield carbonyl compounds when treatment with CrO3. For example, 2-tert-butylcyclohexanol gives 2-tert-butylcyclohexanone. If axial −OH groups are generally more reactive than their equatorial isomers, which do you think reacts faster, the cis isomer of 2-tert-butylcyclohexanol or the trans isomer? Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.14%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 5, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• use molecular models in solving problems on stereochemistry.
• solve road-map problems that include stereochemical information.
• define, and use in context, the new key terms.
This chapter introduces the concept of chirality, and discusses the structure of compounds containing one or two chiral centers. A convenient method of representing the three-dimensional arrangement of the atoms in chiral compounds is explained; furthermore, throughout the chapter , considerable emphasis is placed on the use of molecular models to assist in the understanding of the phenomenon of chirality. The chapter continues with an examination of stereochemistry—the three-dimensional nature of molecules. The subject is introduced using the experimental observation that certain substances have the ability to rotate plane-polarized light. Finally, certain reactions of alkenes are re-examined in the light of the new material encountered in this chapter.
• 5.1: Why This Chapter?
Understanding the causes and consequences of molecular handedness is crucial to understanding organic and biological chemistry. The subject can be a bit complex at first, but the material covered in this chapter nevertheless forms the basis for much of the remainder of the book.
• 5.2: Enantiomers and the Tetrahedral Carbon
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of the most interesting types of isomer is the mirror-image stereoisomer, a non-superimposable set of two molecules that are mirror images of one another. The existence of these molecules are determined by a a concept known as chirality.
• 5.3: The Reason for Handedness in Molecules - Chirality
• 5.4: Optical Activity
Identifying and distinguishing enantiomers is inherently difficult, since their physical and chemical properties are largely identical. Fortunately, a nearly two hundred year old discovery by the French physicist Jean-Baptiste Biot has made this task much easier. This discovery disclosed that the right- and left-handed enantiomers of a chiral compound perturb plane-polarized light in opposite ways. This perturbation is unique to chiral molecules, and has been termed optical activity.
• 5.5: Pasteur's Discovery of Enantiomers
Because enantiomers have identical physical and chemical properties in achiral environments, separation of the stereoisomeric components of a racemic mixture or racemate is normally not possible by the conventional techniques of distillation and crystallization. In some cases, however, the crystal habits of solid enantiomers and racemates permit the chemist (acting as a chiral resolving agent) to discriminate enantiomeric components of a mixture
• 5.6: Sequence Rules for Specifying Configuration
The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and, as such, is also often called the Cahn-Ingold-Prelog rules.
• 5.7: Diastereomers
• 5.8: Meso Compounds
A meso compound is an achiral compound that has chiral centers. A meso compound contains an internal plane of symmetry which makes it superimposable on its mirror image and is optically inactive although it contains two or more stereocenters. Remember, an internal plane of symmetry was shown to make a molecule achiral.
• 5.9: Racemic Mixtures and the Resolution of Enantiomers
A racemic mixture is a 50:50 mixture of two enantiomers. Because they are mirror images, each enantiomer rotates plane-polarized light in an equal but opposite direction and is optically inactive. If the enantiomers are separated, the mixture is said to have been resolved. A common experiment in the laboratory component of introductory organic chemistry involves the resolution of a racemic mixture.
• 5.10: A Review of Isomerism
• 5.11: Chirality at Nitrogen, Phosphorus, and Sulfur
• 5.12: Prochirality
When a tetrahedral carbon can be converted to a chiral center by changing only one of its attached groups, it is referred to as a ‘prochiral' center.
• 5.13: Chirality in Nature and Chiral Environments
• 5.14: Chemistry Matters—Chiral Drugs
• 5.15: Key Terms
• 5.16: Summary
• 5.17: Additional Problems
Thumbnail: Two enantiomers of a generic amino acid that are chiral. (Public Domain; unknonw author via Wikipedia)
05: Stereochemistry at Tetrahedral Centers
Understanding the causes and consequences of molecular handedness is crucial to understanding organic and biological chemistry. The subject can be a bit complex at first, but the material covered in this chapter nevertheless forms the basis for much of the remainder of the book.
Are you right-handed or left-handed? You may not spend much time thinking about it, but handedness plays a surprisingly large role in your daily activities. Many musical instruments, such as oboes and clarinets, have a handedness to them; the last available softball glove always fits the wrong hand. The reason for these difficulties is that our hands aren’t identical; rather, they’re mirror images. When you hold a left hand up to a mirror, the image you see looks like a right hand. Try it.
Handedness is also important in organic and biological chemistry, where it arises primarily as a consequence of the tetrahedral stereochemistry of sp3-hybridized carbon atoms. Many drugs and almost all the molecules in our bodies—amino acids, carbohydrates, nucleic acids, and many more—have a handedness. Furthermore, molecular handedness enables the precise interactions between enzymes and their substrates that are involved in the hundreds of thousands of chemical reactions on which life is based. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.01%3A_Why_This_Chapter.txt |
What causes molecular handedness? Look at generalized molecules of the type CH3X, CH2XY, and CHXYZ shown in Figure 5.2. On the left are three molecules, and on the right are their images reflected in a mirror. The CH3X and CH2XY molecules are identical to their mirror images and thus are not handed. If you make a molecular model of each molecule and its mirror image, you find that you can superimpose one on the other so that all atoms coincide. The CHXYZ molecule, by contrast, is not identical to its mirror image. You can’t superimpose a model of this molecule on a model of its mirror image for the same reason that you can’t superimpose a left hand on a right hand: they simply aren’t the same.
A molecule that is not identical to its mirror image is a kind of stereoisomer (Section 4.2) called an enantiomer (e-nan-tee-oh-mer, from the Greek enantio, meaning “opposite”). Enantiomers are related to each other as a right hand is related to a left hand and result whenever a tetrahedral carbon is bonded to four different substituents (one need not be H). For example, lactic acid (2-hydroxypropanoic acid) exists as a pair of enantiomers because there are four different groups (−H, −OH, −CH3, −CO2H) bonded to the central carbon atom. The enantiomers are called (+)-lactic acid and (−)-lactic acid. Both are found in sour milk, but only the (+) enantiomer occurs in muscle tissue.
No matter how hard you try, you can’t superimpose a molecule of (+)-lactic acid on a molecule of (−)-lactic acid. If any two groups match up, say −H and −CO2H, the remaining two groups don’t match (Figure 5.3).
5.03: The Reason for Handedness in Molecules - Chirality
A molecule that is not identical to its mirror image is said to be chiral (ky-ral, from the Greek cheir, meaning “hand”). You can’t take a chiral molecule and its enantiomer and place one on the other so that all atoms coincide.
How can you predict whether a given molecule is or is not chiral? A molecule is not chiral if it has a plane of symmetry. A plane of symmetry is a plane that cuts through the middle of a molecule (or any object) in such a way that one half of the molecule or object is a mirror image of the other half. A coffee mug, for example, has a plane of symmetry. If you were to cut the coffee mug in half from top to bottom, one half would be a mirror image of the other half. A hand, however, does not have a plane of symmetry. One “half” of a hand is not a mirror image of the other half (Figure 5.4).
A molecule that has a plane of symmetry in any conformation must be identical to its mirror image and must be nonchiral, or achiral. Thus, propanoic acid, CH3CH2CO2H, has a plane of symmetry when lined up as shown in Figure 5.5 and is achiral, while lactic acid, CH3CH(OH)CO2H, has no plane of symmetry in any conformation and is chiral.
The most common, although not the only, cause of chirality in organic molecules is the presence of a tetrahedral carbon atom bonded to four different groups—for example, the central carbon atom in lactic acid. Such carbons are referred to as chirality centers, although other terms such as stereocenter, asymmetric center, and stereogenic center have also been used. Note that chirality is a property of the entire molecule, whereas a chirality center is the cause of chirality.
Detecting a chirality center in a complex molecule takes practice because it’s not always immediately apparent whether four different groups are bonded to a given carbon. The differences don’t necessarily appear right next to the chirality center. For example, 5-bromodecane is a chiral molecule because four different groups are bonded to C5, the chirality center (marked with an asterisk). A butyl substituent is similar to a pentyl substituent, but it isn’t identical. The difference isn’t apparent until looking four carbon atoms away from the chirality center, but there’s still a difference.
As other possible examples, look at methylcyclohexane and 2-methylcyclohexanone. Methylcyclohexane is achiral because no carbon atom in the molecule is bonded to four different groups. You can immediately eliminate all −CH2− carbons and the −CH3 carbon from consideration, but what about C1 on the ring? The C1 carbon atom is bonded to a −CH3 group, to an −H atom, and to C2 and C6 of the ring. Carbons 2 and 6 are equivalent, however, as are carbons 3 and 5. Thus, the C6–C5–C4 “substituent” is equivalent to the C2–C3–C4 substituent, and methylcyclohexane is achiral. Another way of reaching the same conclusion is to realize that methylcyclohexane has a symmetry plane, which passes through the methyl group and through C1 and C4 of the ring.
The situation is different for 2-methylcyclohexanone. 2-Methylcyclohexanone has no symmetry plane and is chiral because its C2 is bonded to four different groups: a −CH3 group, an −H atom, a −COCH2− ring bond (C1), and a −CH2CH2− ring bond (C3).
Several more examples of chiral molecules are shown below. Check for yourself that the labeled carbons are chirality centers. You might note that carbons in −CH2−, −CH3, $C═OC═O$, $C═CC═C$, and $C≡CC≡C$ groups can’t be chirality centers. (Why not?)
Worked Example 5.1
Drawing the Three-Dimensional Structure of a Chiral Molecule
Draw the structure of a chiral alcohol.
Strategy
An alcohol is a compound that contains the −OH functional group. To make an alcohol chiral, we need to have four different groups bonded to a single carbon atom, say −H, −OH, −CH3, and −CH2CH3.
Solution
Problem 5-1
Which of the following objects are chiral?
(a) Soda can (b) Screwdriver (c) Screw (d) Shoe
Problem 5-2 Which of the following molecules are chiral? Identify the chirality center(s) in each. (a) (b) (c) Problem 5-3
Alanine, an amino acid found in proteins, is chiral. Draw the two enantiomers of alanine using the standard convention of solid, wedged, and dashed lines.
Problem 5-4 Identify the chirality centers in the following molecules (gray = H, black = C, red = O, green = Cl, yellow-green = F): (a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.02%3A_Enantiomers_and_the_Tetrahedral_Carbon.txt |
5.3 • Optical Activity
The study of chirality originated in the early 19th century during investigations by the French physicist Jean-Baptiste Biot into the nature of plane-polarized light. A beam of ordinary light consists of electromagnetic waves that oscillate in an infinite number of planes at right angles to its direction of travel. When a beam of ordinary light passes through a device called a polarizer, however, only the light waves oscillating in a single plane pass through and the light is said to be plane-polarized. Light waves in all other planes are blocked out.
Biot made the remarkable observation that when a beam of plane-polarized light passes through a solution of certain organic molecules, such as sugar or camphor, the plane of polarization is rotated through an angle, α. Not all organic substances exhibit this property, but those that do are said to be optically active.
The angle of rotation can be measured with an instrument called a polarimeter, represented in Figure 5.6. A solution of optically active organic molecules is placed in a sample tube, plane-polarized light is passed through the tube, and rotation of the polarization plane occurs. The light then goes through a second polarizer called the analyzer. By rotating the analyzer until the light passes through it, we can find the new plane of polarization and can tell to what extent rotation has occurred.
In addition to determining the extent of rotation, we can also find the direction. From the vantage point of the observer looking directly at the analyzer, some optically active molecules rotate polarized light to the left (counterclockwise) and are said to be levorotatory, whereas others rotate polarized light to the right (clockwise) and are said to be dextrorotatory. By convention, rotation to the left is given a minus sign (−) and rotation to the right is given a plus sign (+). (−)-Morphine, for example, is levorotatory, and (+)-sucrose is dextrorotatory.
The extent of rotation observed in a polarimetry experiment depends on the number of optically active molecules encountered by the light beam. This number, in turn, depends on sample concentration and sample pathlength. If the concentration of the sample is doubled, the observed rotation doubles. If the concentration is kept constant but the length of the sample tube is doubled, the observed rotation doubles. It also happens that the angle of rotation depends on the wavelength of the light used.
To express optical rotations in a meaningful way so that comparisons can be made, we have to choose standard conditions. The specific rotation, [α]D, of a compound is defined as the observed rotation when light of 589.6 nanometer (nm; 1 nm = 10−9 m) wavelength is used with a sample pathlength l of 1 decimeter (dm; 1 dm = 10 cm) and a sample concentration c of 1 g/cm3. (Light of 589.6 nm, the so-called sodium D line, is the yellow light emitted from common sodium street lamps.)
$[α]D=Observed rotation (degrees)Pathlength,l(dm)×Concentration,c(g/cm3)=αl×c[α]D=Observed rotation (degrees)Pathlength,l(dm)×Concentration,c(g/cm3)=αl×c$
When optical rotations are expressed in this standard way, the specific rotation, [α]D, is a physical constant characteristic of a given optically active compound. For example, (+)-lactic acid has [α]D = +3.82, and (−)-lactic acid has [α]D = −3.82. That is, the two enantiomers rotate plane-polarized light to exactly the same extent but in opposite directions. Note that the units of specific rotation are [(deg · cm2)/g] but the values are usually expressed without units. Some additional examples are listed in Table 5.1.
Table 5.1 Specific Rotations of Some Organic Molecules
Compound [α]D Compound [α]D
Penicillin V +233 Cholesterol −31.5
Sucrose +66.47 Morphine −132
Camphor +44.26 Cocaine −16
Chloroform 0 Acetic acid 0
Worked Example 5.2
Calculating an Optical Rotation
A 1.20 g sample of cocaine, [α]D = −16, was dissolved in 7.50 mL of chloroform and placed in a sample tube having a pathlength of 5.00 cm. What was the observed rotation?
Strategy
Since $[α]D=αl×c[α]D=αl×c$
Then $α=l×c×[α]Dα=l×c×[α]D$
where [α]D = −16; l = 5.00 cm = 0.500 dm; c = 1.20 g/7.50 cm3 = 0.160 g/cm3
Solution
α = (−16) (0.500) (0.160) = −1.3°.
Problem 5-5
Is cocaine (Worked Example 5.2) dextrorotatory or levorotatory?
Problem 5-6
A 1.50 g sample of coniine, the toxic extract of poison hemlock, was dissolved in 10.0 mL of ethanol and placed in a sample cell with a 5.00 cm pathlength. The observed rotation at the sodium D line was +1.21°. Calculate [α]D for coniine.
5.05: Pasteur's Discovery of Enantiomers
Little was done to build on Biot’s discovery of optical activity until 1848, when Louis Pasteur began work on a study of crystalline tartaric acid salts derived from wine. On crystallizing a concentrated solution of sodium ammonium tartrate below 28 °C, Pasteur made the surprising observation that two distinct kinds of crystals were obtained. Furthermore, the two kinds of crystals were nonsuperimposable mirror images and were related in the same way that a right hand is related to a left hand.
Working carefully with tweezers, Pasteur was able to separate the crystals into two piles, one of “right-handed” crystals and one of “left-handed” crystals, like those shown in Figure 5.7. Although the original sample, a 50 : 50 mixture of right and left, was optically inactive, solutions of the crystals from each of the sorted piles were optically active and their specific rotations were equal in magnitude but opposite in sign.
Pasteur was far ahead of his time. Although the structural theory of Kekulé had not yet been proposed, Pasteur explained his results by speaking of the molecules themselves, saying, “There is no doubt that [in the dextro tartaric acid] there exists an asymmetric arrangement having a nonsuperimposable image. It is no less certain that the atoms of the levo acid have precisely the inverse asymmetric arrangement.” Pasteur’s vision was extraordinary, for it was not until 25 years later that his ideas regarding asymmetric carbon atoms were confirmed.
Today, we would describe Pasteur’s work by saying that he had discovered enantiomers. Enantiomers, also called optical isomers, have identical physical properties, such as melting point and boiling point, but differ in the direction in which their solutions rotate plane-polarized light. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.04%3A_Optical_Activity.txt |
Structural drawings provide a visual representation of stereochemistry, but a written method for indicating the three-dimensional arrangement, or configuration, of substituents at a chirality center is also needed. The method used a set of sequence rules to rank the four groups attached to the chirality center and then looks at the handedness with which those groups are attached. Called the Cahn–Ingold–Prelog rules after the chemists who proposed them, the sequence rules are as follows:
RULE 1
Look at the four atoms directly attached to the chirality center, and rank them according to atomic number. The atom with the highest atomic number has the highest ranking (first), and the atom with the lowest atomic number (usually hydrogen) has the lowest ranking (fourth). When different isotopes of the same element are compared, such as deuterium (2H) and protium (1H), the heavier isotope ranks higher than the lighter isotope. Thus, atoms commonly found in organic compounds have the following order.
RULE 2
If a decision can’t be reached by ranking the first atoms in the substituent, look at the second, third, or fourth atoms away from the chirality center until the first difference is found. A −CH2CH3 substituent and a −CH3 substituent are equivalent by rule 1 because both have carbon as the first atom. By rule 2, however, ethyl ranks higher than methyl because ethyl has a carbon as its highest second atom, while methyl has only hydrogen as its second atom. Look at the following pairs of examples to see how the rule works:
RULE 3
Multiple-bonded atoms are equivalent to the same number of single-bonded atoms. For example, an aldehyde substituent ($–CH═O–CH═O$), which has a carbon atom doubly bonded to one oxygen, is equivalent to a substituent having a carbon atom singly bonded to two oxygens:
As further examples, the following pairs are equivalent:
Having ranked the four groups attached to a chiral carbon, we describe the stereochemical configuration around the carbon by orienting the molecule so that the group with the lowest ranking (4) points directly away from us. We then look at the three remaining substituents, which now appear to radiate toward us like the spokes on a steering wheel (Figure 5.8). If a curved arrow drawn from the highest to second-highest to third-highest ranked substituent (1 → 2 → 3) is clockwise, we say that the chirality center has the R configuration (S for the Latin rectus, meaning “right”). If an arrow from 1 → 2 → 3 is counterclockwise, the chirality center has the S configuration (Latin sinister, meaning “left”). To remember these assignments, think of a car’s steering wheel when making a Right (clockwise) turn.
23 is clockwise (right turn), the center has the R configuration. If the direction of travel 123 is counterclockwise (left turn), the center is S.
Look at (−)-lactic acid in Figure 5.9 for an example of how to assign configuration. Sequence rule 1 says that −OH is ranked 1 and −H is ranked 4, but it doesn’t allow us to distinguish between −CH3 and −CO2H because both groups have carbon as their first atom. Sequence rule 2, however, says that −CO2H ranks higher than −CH3 because O (the highest second atom in −CO2H) outranks H (the highest second atom in −CH3). Now, turn the molecule so that the fourth-ranked group (−H) is oriented toward the rear, away from the observer. Since a curved arrow from 1 (−OH) to 2 (−CO2H) to 3 (−CH3) is clockwise (right turn of the steering wheel), (−)-lactic acid has the R configuration. Applying the same procedure to (+)-lactic acid leads to the opposite assignment.
Further examples are provided by naturally occurring (−)-glyceraldehyde and (+)-alanine, which both have the S configuration as shown in Figure 5.10. Note that the sign of optical rotation, (+) or (−), is not related to the R,S designation. (S)-Glyceraldehyde happens to be levorotatory (−), and (S)-alanine happens to be dextrorotatory (+). There is no simple correlation between R,S configuration and direction or magnitude of optical rotation.
One additional point needs to be mentioned—the matter of absolute configuration. How do we know that the assignments of R and S configuration are correct in an absolute sense, rather than a relative, sense? Since there is no correlation between the R,S configuration and the direction or magnitude of optical rotation, how do we know that the R configuration belongs to the levorotatory enantiomer of lactic acid? This difficult question was finally solved in 1951, when an X-ray diffraction method was found for determining the absolute spatial arrangement of atoms in a molecule. Based on those results, we can say with certainty that the R,S conventions are correct.
Worked Example 5.3
Assigning Configuration to Chirality Centers
Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then assign R or S configuration to each:
Strategy
It takes practice to be able to visualize and orient a chirality center in three dimensions. You might start by indicating where the observer must be located—180° opposite the lowest-ranked group. Then imagine yourself in the position of the observer, and redraw what you would see.
Solution
In (a), you would be located in front of the page toward the top right of the molecule, and you would see group 2 to your left, group 3 to your right, and group 1 below you. This corresponds to an R configuration.
In (b), you would be located behind the page toward the top left of the molecule from your point of view, and you would see group 3 to your left, group 1 to your right, and group 2 below you. This also corresponds to an R configuration.
Worked Example 5.4
Drawing the Three-Dimensional Structure of a Specific Enantiomer
Draw a tetrahedral representation of (R)-2-chlorobutane.
Strategy
Begin by ranking the four substituents bonded to the chirality center: (1) −Cl, (2) −CH2CH3, (3) −CH3, (4) −H. To draw a tetrahedral representation of the molecule, orient the lowest-ranked group (−H) away from you and imagine that the other three groups are coming out of the page toward you. Then, place the remaining three substituents such that the direction of travel 1 → 2 → 3 is clockwise (right turn), and tilt the molecule toward you to bring the rear hydrogen into view. Using molecular models is a real help in working problems of this sort.
Solution
Problem 5-7 Which member in each of the following sets ranks higher? (a)
−H or −Br
(b) −Cl or −Br (c) −CH3 or −CH2CH3 (d) −NH2 or −OH (e) −CH2OH or −CH3 (f) −CH2OH or −CH$\text{=}$O
Problem 5-8 Rank each of the following sets of substituents: (a)
−H, −OH, −CH2CH3, −CH2CH2OH
(b) −CO2H, −CO2CH3, −CH2OH, −OH (c) −CN, −CH2NH2, −CH2NHCH3, −NH2 (d) −SH, −CH2SCH3, −CH3, −SSCH3
Problem 5-9 Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then assign R or S configuration: (a) (b) (c) Problem 5-10 Assign R or S configuration to the chirality center in each of the following molecules: (a) (b) (c) Problem 5-11
Draw a tetrahedral representation of (S)-2-pentanol (2-hydroxypentane).
Problem 5-12
Assign R or S configuration to the chirality center in the following molecular model of the amino acid methionine (blue = N, yellow = S): | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.06%3A_Sequence_Rules_for_Specifying_Configuration.txt |
Molecules like lactic acid, alanine, and glyceraldehyde are relatively simple because each has only one chirality center and thus only two stereoisomers. The situation becomes more complex, however, with molecules that have more than one chirality center. As a general rule, a molecule with n chirality centers can have up to 2n stereoisomers (although it may have fewer, as we’ll see below). Take the amino acid threonine (2-amino-3-hydroxybutanoic acid), for example. Since threonine has two chirality centers (C2 and C3), there are four possible stereoisomers, as shown in Figure 5.11. Check for yourself that the R,S configurations of all stereoisomers are correct.
The four stereoisomers of 2-amino-3-hydroxybutanoic acid can be grouped into two pairs of enantiomers. The 2R,3R stereoisomer is the mirror image of 2S,3S, and the 2R,3S stereoisomer is the mirror image of 2S,3R. But what is the relationship between any two molecules that are not mirror images? What, for instance, is the relationship between the 2R,3R isomer and the 2R,3S isomer? They are stereoisomers, yet they aren’t enantiomers. To describe such a relationship, we need a new term—diastereomer.
Diastereomers (dia-stair-e-oh-mers) are stereoisomers that are not mirror images. Since we used the right-hand/left-hand analogy to describe the relationship between two enantiomers, we might extend the analogy by saying that the relationship between diastereomers is like that of hands from different people. Your hand and your friend’s hand look similar, but they aren’t identical and they aren’t mirror images. The same is true of diastereomers: they’re similar, but they aren’t identical and they aren’t mirror images.
Note carefully the difference between enantiomers and diastereomers: enantiomers have opposite configurations at all chirality centers, whereas diastereomers have opposite configurations at some (one or more) chirality centers but the same configuration at others. A full description of the four stereoisomers of threonine is given in Table 5.2. Of the four, only the 2S,3R isomer, [α]D = −28.3, occurs naturally in plants and animals and is an essential nutrient for humans. This result is typical: most biological molecules are chiral, and usually only one stereoisomer is found in nature.
Table 5.2 Relationships among the Four Stereoisomers of Threonine
Stereoisomer Enantiomer Diastereomer
2R,3R 2S,3S 2R,3S and 2S,3R
2S,3S 2R,3R 2R,3S and 2S,3R
2R,3S 2S,3R 2R,3R and 2S,3S
2S,3R 2R,3S 2R,3R and 2S,3S
In the special case where two diastereomers differ at only one chirality center but are the same at all others, we say that the compounds are epimers. Cholestanol and coprostanol, for instance, are both found in human feces, and both have nine chirality centers. Eight of the nine are identical, but the one at C5 is different. Thus, cholestanol and coprostanol are epimeric at C5.
Problem 5-13
One of the following molecules (a)(d) is D-erythrose 4-phosphate, an intermediate in the Calvin photosynthetic cycle by which plants incorporate CO2 into carbohydrates. If D-erythrose 4-phosphate has R stereochemistry at both chirality centers, which of the structures is it? Which of the remaining three structures is the enantiomer of D-erythrose 4-phosphate, and which are diastereomers?
(a)(b)(c)(d)
Problem 5-14
How many chirality centers does morphine have? How many stereoisomers of morphine are possible in principle?
Problem 5-15
Assign R or S configuration to each chirality center in the following molecular model of the amino acid isoleucine (blue = N):
5.08: Meso Compounds
Let’s look at another example (Section 5.4) of a compound with more than one chirality center: the tartaric acid used by Pasteur. The four stereoisomers can be drawn as follows:
The 2R,3R and 2S,3S structures are nonsuperimposable mirror images and therefore represent a pair of enantiomers. A close look at the 2R,3S and 2S,3R structures, however, shows that they are superimposable, and thus identical, as can be seen by rotating one structure 180°.
The 2R,3S and 2S,3R structures are identical because the molecule has a plane of symmetry and is therefore achiral. The symmetry plane cuts through the C2–C3 bond, making one half of the molecule a mirror image of the other half (Figure 5.12). Because of the plane of symmetry, the molecule is achiral, despite the fact that it has two chirality centers. Such compounds, which are achiral, yet contain chirality centers, are called meso compounds (me-zo). Thus, tartaric acid exists in three stereoisomeric forms: two enantiomers and one meso form.
Some physical properties of the three stereoisomers are listed in Table 5.3. The (+)- and (−)-tartaric acids have identical melting points, solubilities, and densities, but they differ in the sign of their rotation of plane-polarized light. The meso isomer, by contrast, is diastereomeric with the (+) and (−) forms. It has no mirror-image relationship to (+)- and (−)-tartaric acids, is a different compound altogether, and thus has different physical properties.
Table 5.3 Some Properties of the Stereoisomers of Tartaric Acid Some Properties of the Stereoisomers of Tartaric Acid
Stereoisomer Melting point (°C) [α]D Density (g/cm3) Solubility at 20 °C
(g/100 mL H2O)
(+) 168–170 +12 1.7598 139.0
(−) 168–170 −12 1.7598 139.0
Meso 146–148 0 1.6660 125.0
Worked Example 5.5
Distinguishing Chiral Compounds from Meso Compounds
Does cis-1,2-dimethylcyclobutane have any chirality centers? Is it chiral?
Strategy
To see whether a chirality center is present, look for a carbon atom bonded to four different groups. To see whether the molecule is chiral, look for the presence or absence of a symmetry plane. Not all molecules with chirality centers are chiral overall—meso compounds are an exception.
Solution
A look at the structure of cis-1,2-dimethylcyclobutane shows that both methyl-bearing ring carbons (C1 and C2) are chirality centers. Overall, though, the compound is achiral because there is a symmetry plane bisecting the ring between C1 and C2. Thus, the molecule is a meso compound.
Problem 5-16
Which of the following structures represent meso compounds?
(a) (b) (c) (d)
Problem 5-17
Which of the following have a meso form? (Recall that the -ol suffix refers to an alcohol, ROH.)
(a) 2,3-Butanediol (b) 2,3-Pentanediol (c) 2,4-Pentanediol
Problem 5-18
Does the following structure represent a meso compound? If so, indicate the symmetry plane. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.07%3A_Diastereomers.txt |
To end this discussion of stereoisomerism, let’s return for a last look at Pasteur’s pioneering work, described in Section 5.4. Pasteur took an optically inactive tartaric acid salt and found that he could crystallize from it two optically active forms having what we would now call 2R,3R and 2S,3S configurations. But what was the optically inactive form he started with? It couldn’t have been meso-tartaric acid, because meso-tartaric acid is a different chemical compound and can’t interconvert with the two chiral enantiomers without breaking and re-forming chemical bonds.
The answer is that Pasteur started with a 50 : 50 mixture of the two chiral tartaric acid enantiomers. Such a mixture is called a racemate (rass-uh-mate), or racemic mixture, and is denoted by either the symbol (±) or the prefix d,l to indicate an equal mixture of dextrorotatory and levorotatory forms. Racemates show no optical rotation because the (+) rotation from one enantiomer exactly cancels the (−) rotation from the other. Through good luck, Pasteur was able to separate, or resolve, racemic tartaric acid into its (+) and (−) enantiomers. Unfortunately, the fractional crystallization technique he used doesn’t work for most racemates, so other methods are needed.
The most common method for resolving the racemate of a chiral carboxylic acid (RCO2H) is to carry out an acid-base reaction between the acid and an amine base (RNH2) to yield an ammonium salt:
To understand how this method of resolution works, let’s see what happens when a racemic mixture of chiral acids, such as (+)- and (−)-lactic acids, reacts with an achiral amine base, such as methylamine, CH3NH2. The situation is analogous to what happens when left and right hands (chiral) pick up a ball (achiral). Both left and right hands pick up the ball equally well, and the products—ball in right hand versus ball in left hand—are mirror images. In the same way, both (+)- and (−)-lactic acid react with methylamine equally well, and the product is a racemic mixture of the two enantiomers methylammonium (+)-lactate and methylammonium (−)-lactate (Figure 5.13).
Now let’s see what happens when the racemic mixture of (+)- and (−)-lactic acids reacts with a single enantiomer of a chiral amine base, such as (R)-1-phenylethylamine. The situation is analogous to what happens when left and right hands (chiral) put on a right-handed glove (also chiral). Left and right hands don’t put on the right-handed glove in the same way, so the products—right hand in right glove versus left hand in right glove—are not mirror images; they’re similar but different.
In the same way, (+)- and (−)-lactic acids react with (R)-1-phenylethylamine to give two different products (Figure 5.14). (R)-Lactic acid reacts with (R)-1-phenylethylamine to give the R,R salt, and (S)-lactic acid reacts with the R amine to give the S,R salt. The two salts are diastereomers, not enantiomers. They have different chemical and physical properties, and it may therefore be possible to separate them by crystallization or some other means. Once separated, acidification of the two diastereomeric salts with a strong acid makes it possible to isolate the two pure enantiomers of lactic acid and to recover the chiral amine for reuse.
Worked Example 5.6
Predicting the Chirality of a Reaction Product
We’ll see in Section 21.3 that carboxylic acids (RCO2H) react with alcohols (R′OH) to form esters (RCO2R′). Suppose that (±)-lactic acid reacts with CH3OH to form the ester, methyl lactate. What stereochemistry would you expect the product(s) to have? What is the relationship of the products?
Solution
Reaction of a racemic acid with an achiral alcohol such as methanol yields a racemic mixture of mirror-image (enantiomeric) products.
Problem 5-19
Suppose that acetic acid (CH3CO2H) reacts with (S)-2-butanol to form an ester (see Worked Example 5.6). What stereochemistry would you expect the product(s) to have? What is the relationship of the products?
Problem 5-20
What stereoisomers would result from reaction of (±)-lactic acid with (S)-1-phenylethylamine, and what is the relationship between them? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.09%3A_Racemic_Mixtures_and_the_Resolution_of_Enantiomers.txt |
As noted on several previous occasions, isomers are compounds with the same chemical formula but different structures. We’ve seen several kinds of isomers in the past few chapters, and it’s a good idea at this point to see how they relate to one another (Figure 5.15).
There are two fundamental types of isomers, both of which we’ve now encountered: constitutional isomers and stereoisomers.
Constitutional isomers (Section 3.2) are compounds whose atoms are connected differently. Among the kinds of constitutional isomers we’ve seen are skeletal, functional, and positional isomers.
Stereoisomers (Section 4.2) are compounds whose atoms are connected in the same order but with a different spatial arrangement. Among the kinds of stereoisomers we’ve seen are enantiomers, diastereomers, and cis–trans isomers of cycloalkanes. Actually, cis–trans isomers are just a subclass of diastereomers because they are non–mirror-image stereoisomers:
Problem 5-21 What kinds of isomers are the following pairs? (a)
(S)-5-Chloro-2-hexene and chlorocyclohexane
(b) (2R,3R)-Dibromopentane and (2S,3R)-dibromopentane
5.11: Chirality at Nitrogen Phosphorus and Sulfur
As noted previously, the most common cause of chirality in a molecule is the presence of four different substituents bonded to a tetrahedral atom. Although that atom is usually carbon, it doesn’t necessarily have to be. Nitrogen, phosphorus, and sulfur atoms are all commonly encountered in organic molecules, and can all be chirality centers. We know, for instance, that trivalent nitrogen is tetrahedral, with its lone pair of electrons acting as the fourth “substituent” (Section 1.10). Is trivalent nitrogen chiral? Does a compound such as ethylmethylamine exist as a pair of enantiomers?
The answer is both yes and no. Yes in principle, but no in practice. It turns out that most trivalent nitrogen compounds undergo a rapid umbrella-like inversion that interconverts enantiomers, so we can’t isolate individual enantiomers except in special cases.
A similar situation occurs in trivalent phosphorus compounds, called phosphines, but the inversion at phosphorus is substantially slower than inversion at nitrogen, so stable chiral phosphines can be isolated. (R)- and (S)-methylpropylphenylphosphine, for instance, are configurationally stable for several hours at 100 °C. We’ll see the importance of phosphine chirality in Section 26.7 in connection with the synthesis of chiral amino acids.
Divalent sulfur compounds are achiral, but trivalent sulfur compounds called sulfonium salts (R3S+) can be chiral. Like phosphines, sulfonium salts undergo relatively slow inversion, so chiral sulfonium salts are configurationally stable and can be isolated. Perhaps the best known example is the coenzyme S-adenosylmethionine, the so-called biological methyl donor, which is involved in many metabolic pathways as a source of CH3 groups. (The “S” in the name S-adenosylmethionine stands for sulfur and means that the adenosyl group is attached to the sulfur atom of the amino acid methionine.) The molecule has S stereochemistry at sulfur and is configurationally stable for several days at room temperature. Its R enantiomer is also known but is not biologically active.
5.12: Prochirality
Closely related to the concept of chirality, and particularly important in biological chemistry, is the notion of prochirality. A molecule is said to be prochiral if it can be converted from achiral to chiral in a single chemical step. For instance, an unsymmetrical ketone like 2-butanone is prochiral because it can be converted to the chiral alcohol 2-butanol by the addition of hydrogen, as we’ll see in Section 17.4.
Which enantiomer of 2-butanol is produced depends on which face of the planar carbonyl group undergoes reaction. To distinguish between the possibilities, we use the stereochemical descriptors Re and Si. Rank the three groups attached to the trigonal, sp2-hybridized carbon, and imagine curved arrows from the highest to second-highest to third-highest ranked substituents. The face on which the arrows curve clockwise is designated the Re face (similar to R), and the face on which the arrows curve counterclockwise is designated the Si face (similar to S). In this example, addition of hydrogen from the Re face gives (S)-2-butane, and addition from the Si face gives (R)-2-butane.
In addition to compounds with planar, sp2-hybridized atoms, compounds with tetrahedral, sp3-hybridized atoms can also be prochiral. An sp3-hybridized atom is said to be a prochirality center if, by changing one of its attached groups, it becomes a chirality center. The −CH2OH carbon atom of ethanol, for instance, is a prochirality center because changing one of its attached −H atoms converts it into a chirality center.
To distinguish between the two identical atoms (or groups of atoms) on a prochirality center, we imagine a change that will raise the ranking of one atom over the other without affecting its rank with respect to other attached groups. On the −CH2OH carbon of ethanol, for instance, we might imagine replacing one of the 1H atoms (protium) by 2H (deuterium). The newly introduced 2H atom ranks higher than the remaining 1H atom, but it remains lower than other groups attached to the carbon. Of the two identical atoms in the original compound, the atom whose replacement leads to an R chirality center is said to be pro-R and the atom whose replacement leads to an S chirality center is pro-S.
A large number of biological reactions involve prochiral compounds. One of the steps in the citric acid cycle by which food is metabolized, for instance, is the addition of H2O to fumarate to give malate. Addition of −OH occurs on the Si face of a fumarate carbon and gives (S)-malate as product.
As another example, studies with deuterium-labeled substrates have shown that the reaction of ethanol with the coenzyme nicotinamide adenine dinucleotide (NAD+), catalyzed by yeast alcohol dehydrogenase, occurs with exclusive removal of the pro-R hydrogen from ethanol and with addition only to the Re face of NAD+.
Determining the stereochemistry of reactions at prochirality centers is a powerful method for studying detailed mechanisms in biochemical reactions. As just one example, the conversion of citrate to cis-aconitate in the citric acid cycle has been shown to occur with loss of a pro-R hydrogen, implying that the OH and H groups leave from opposite sides of the molecule.
Problem 5-22 Identify the indicated hydrogens in the following molecules as pro-R or pro-S: (a)
(b)
Problem 5-23 Identify the indicated faces of carbon atoms in the following molecules as Re or Si: (a)
(b)
Problem 5-24
The lactic acid that builds up in tired muscles is formed from pyruvate. If the reaction occurs with addition of hydrogen to the Re face of pyruvate, what is the stereochemistry of the product?
Problem 5-25
The aconitase-catalyzed addition of water to cis-aconitate in the citric acid cycle occurs with the following stereochemistry. Does the addition of the OH group occur on the Re or Si face of the substrate? What about the addition of the H? Do the H and OH groups add from the same side of the double bond or from opposite sides? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.10%3A_A_Review_of_Isomerism.txt |
Although the different enantiomers of a chiral molecule have the same physical properties, they usually have different biological properties. For example, a change in chirality can affect the biological properties of many drugs, such as fluoxetine, a heavily prescribed medication sold under the trade name Prozac. Racemic fluoxetine is an effective antidepressant but has no activity against migraine. The pure S enantiomer, however, works remarkably well in preventing migraine. Other examples of how chirality affects biological properties are given in the Chapter 5 Chemistry Matters at the end of this chapter.
Why do different enantiomers have different biological properties? To have a biological effect, a substance typically must fit into an appropriate receptor that has a complementary shape. But because biological receptors are chiral, only one enantiomer of a chiral substrate can fit, just as only a right hand can fit into a right-handed glove. The mirror-image enantiomer will be a misfit, like a left hand in a right-handed glove. A representation of the interaction between a chiral molecule and a chiral biological receptor is shown in Figure 5.16: one enantiomer fits the receptor perfectly, but the other does not.
, red palm, and gray pinkie finger, with the blue substituent exposed. (b) The other enantiomer, however, can’t fit into the hand. When the green thumb and gray pinkie finger interact appropriately, the palm holds a blue substituent rather than a red one, with the red substituent exposed.
The hand-in-glove fit of a chiral substrate into a chiral receptor is relatively straightforward, but it’s less obvious how a prochiral substrate can undergo a selective reaction. Take the reaction of ethanol with NAD+ catalyzed by yeast alcohol dehydrogenase. As we saw at the end of Section 5.11, this reaction occurs with exclusive removal of the pro-R hydrogen from ethanol and with addition only to the Re face of the NAD+ carbon.
We can understand this result by imagining that the chiral enzyme receptor again has three binding sites, as in Figure 5.16. When green and gray substituents of a prochiral substrate are held appropriately, however, only one of the two red substituents—say, the pro-S one—is also held while the other, pro-R, substituent is exposed for reaction.
We describe the situation by saying that the receptor provides a chiral environment for the substrate. In the absence of a chiral environment, the two red substituents are chemically identical, but in the presence of a chiral environment, they are chemically distinctive (Figure 5.17a). The situation is similar to what happens when you pick up a coffee mug. By itself, the mug has a plane of symmetry and is achiral. When you pick up the mug, however, your hand provides a chiral environment so that one side becomes much more accessible and easier to drink from than the other (Figure 5.17b).
5.14: Chemistry MattersChiral Drugs
5 • Chemistry Matters 5 • Chemistry Matters
The many hundreds of different pharmaceutical agents approved for use by the U.S. Food and Drug Administration come from many sources. Many drugs are isolated directly from plants or bacteria, and others are made by chemical modification of naturally occurring compounds. An estimated 33%, however, are made entirely in the laboratory and have no relatives in nature.
Those drugs that come from natural sources, either directly or after laboratory modification, are usually chiral and are generally found only as a single enantiomer rather than as a racemate. Penicillin V, for example, an antibiotic isolated from the Penicillium mold, has the 2S,5R,6R configuration. Its enantiomer, which does not occur naturally but can be made in the laboratory, has no antibiotic activity.
In contrast to drugs from natural sources, drugs that are made entirely in the laboratory are either achiral or, if chiral, are often produced and sold as racemates. Ibuprofen, for example, has one chirality center and is sold commercially under such trade names as Advil, Nuprin, and Motrin as a 50 : 50 mixture of R and S. It turns out, however, that only the S enantiomer is active as an analgesic and anti-inflammatory agent. The R enantiomer of ibuprofen is inactive, although it is slowly converted in the body to the active S form.
Not only is it chemically wasteful to synthesize and administer an enantiomer that does not serve the intended purpose, many instances are now known where the presence of the “wrong” enantiomer in a racemic mixture either affects the body’s ability to utilize the “right” enantiomer or has unintended pharmacological effects of its own. The presence of (R)-ibuprofen in the racemic mixture, for instance, slows the rate at which the S enantiomer takes effect in the body, from 12 minutes to 38 minutes.
To get around this problem, pharmaceutical companies attempt to devise methods of enantioselective synthesis, which allow them to prepare only a single enantiomer rather than a racemic mixture. Viable methods have been developed for the preparation of (S)-ibuprofen, which is marketed in Europe. We’ll look further into enantioselective synthesis in the Chapter 19 Chemistry Matters.
5.15: Key Terms
5 • Key Terms 5 • Key Terms
• absolute configuration
• achiral
• Cahn–Ingold–Prelog rules
• chiral
• chiral environment
• chirality center
• configuration
• dextrorotatory
• diastereomer
• enantiomer
• epimer
• levorotatory
• meso compound
• optically active
• prochiral
• prochirality center
• pro-R configuration
• pro-S configuration
• R configuration
• racemate
• Re face
• resolution
• S configuration
• Si face
• specific rotation, [α]D
5.16: Summary
5 • Summary 5 • Summary
In this chapter, we’ve looked at some of the causes and consequences of molecular handedness—a topic of particular importance in understanding biological chemistry. The subject can be a bit complex but is so important that it’s worthwhile spending time to become familiar with it.
An object or molecule that is not superimposable on its mirror image is said to be chiral, meaning “handed.” A chiral molecule is one that does not have a plane of symmetry cutting through it so that one half is a mirror image of the other half. The most common cause of chirality in organic molecules is the presence of a tetrahedral, sp3-hybridized carbon atom bonded to four different groups—a so-called chirality center. Chiral compounds can exist as a pair of nonsuperimposable mirror-image stereoisomers called enantiomers. Enantiomers are identical in all physical properties except for the direction in which they rotate plane-polarized light.
The stereochemical configuration of a chirality center can be specified as either R (rectus) or S (sinister) by using the Cahn–Ingold–Prelog rules. First rank the four substituents on the chiral carbon atom, and then orient the molecule so that the lowest-ranked group points directly back. If a curved arrow drawn in the direction of decreasing rank (1 → 2 → 3) for the remaining three groups is clockwise, the chirality center has the R configuration. If the direction is counterclockwise, the chirality center has the S configuration.
Some molecules have more than one chirality center. Enantiomers have opposite configuration at all chirality centers, whereas diastereomers have the same configuration in at least one center but opposite configurations at the others. Epimers are diastereomers that differ in configuration at only one chirality center. A compound with n chirality centers can have a maximum of 2n stereoisomers.
A meso compound contains a chirality center but is achiral overall because it has a plane of symmetry. Racemic mixtures, or racemates, are 50 : 50 mixtures of (+) and (−) enantiomers. Racemates and individual diastereomers differ in their physical properties, such as solubility, melting point, and boiling point.
A molecule is prochiral if it can be converted from achiral to chiral in a single chemical step. A prochiral sp2-hybridized atom has two faces, described as either Re or Si. An sp3-hybridized atom is a prochirality center if, by changing one of its attached atoms, a chirality center results. The atom whose replacement leads to an R chirality center is pro-R, and the atom whose replacement leads to an S chirality center is pro-S. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.13%3A_Chirality_in_Nature_and_Chiral_Environments.txt |
5 • Additional Problems 5 • Additional Problems
Visualizing Chemistry
Problem 5-26
Which of the following structures are identical? (Green = Cl.)
(a) (b) (c)
(d)
Problem 5-27 Assign R or S configurations to the chirality centers in the following molecules (blue = N): (a) (b) Problem 5-28 Which, if any, of the following structures represent meso compounds? (Blue = N, green = Cl.) (a) (b) (c) Problem 5-29 Assign R or S configuration to each chirality center in pseudoephedrine, an over-the-counter decongestant found in cold remedies (blue = N). Problem 5-30 Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then assign R or S configuration: (a) (b) (c) Chirality and Optical Activity Problem 5-31 Which of the following objects are chiral? (a) A basketball (b) A fork (c) A wine glass (d) A golf club (e) A spiral staircase (f) A snowflake Problem 5-32 Which of the following compounds are chiral? Draw them, and label the chirality centers. (a) 2,4-Dimethylheptane (b)
5-Ethyl-3,3-dimethylheptane
(c) cis-1,4-Dichlorocyclohexane
Problem 5-33 Draw chiral molecules that meet the following descriptions: (a)
A chloroalkane, C5H11Cl
(b) An alcohol, C6H14O (c) An alkene, C6H12 (d) An alkane, C8H18
Problem 5-34
Eight alcohols have the formula C5H12O. Draw them. Which are chiral?
Problem 5-35 Draw compounds that fit the following descriptions: (a)
A chiral alcohol with four carbons
(b) A chiral carboxylic acid with the formula C5H10O2 (c) A compound with two chirality centers (d) A chiral aldehyde with the formula C3H5BrO
Problem 5-36
Erythronolide B is the biological precursor of erythromycin, a broad-spectrum antibiotic. How many chirality centers does erythronolide B have? Identify them.
Assigning Configuration to Chirality Centers
Problem 5-37 Which of the following pairs of structures represent the same enantiomer, and which represent different enantiomers? (a) (b) (c) (d) Problem 5-38 What is the relationship between the specific rotations of (2R,3R)-dichloropentane and (2S,3S)-dichloropentane? Between (2R,3S)-dichloropentane and (2R,3R)-dichloropentane? Problem 5-39 What is the stereochemical configuration of the enantiomer of (2S,4R)-2,4-octanediol? Problem 5-40 What are the stereochemical configurations of the two diastereomers of (2S,4R)-2,4-octanediol? (A diol is a compound with two – OH groups.) Problem 5-41 Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then assign R or S configuration: (a) (b) (c) Problem 5-42 Assign Cahn–Ingold–Prelog rankings to the following sets of substituents: (a) (b) (c) (d) Problem 5-43 Assign R or S configurations to each chirality center in the following molecules: (a) (b) (c) Problem 5-44 Assign R or S configuration to each chirality center in the following molecules: (a) (b) (c) Problem 5-45 Assign R or S configuration to each chirality center in the following biological molecules: (a) (b) Problem 5-46 Draw tetrahedral representations of the following molecules: (a) (S)-2-Chlorobutane (b)
(R)-3-Chloro-1-pentene [H2C$\text{=}$CHCH(Cl)CH2CH3]
Problem 5-47 Assign R or S configuration to each chirality center in the following molecules: (a) (b) Problem 5-48 Assign R or S configurations to the chirality centers in ascorbic acid (vitamin C). Problem 5-49 Assign R or S stereochemistry to the chirality centers in the following Newman projections: (a) (b) Problem 5-50 Xylose is a common sugar found in many types of wood, including maple and cherry. Because it is much less prone to cause tooth decay than sucrose, xylose has been used in candy and chewing gum. Assign R or S configurations to the chirality centers in xylose. Meso Compounds Problem 5-51 Draw examples of the following: (a) A meso compound with the formula C8H18 (b)
A meso compound with the formula C9H20
(c) A compound with two chirality centers, one R and the other S
Problem 5-52 Draw the meso form of each of the following molecules, and indicate the plane of symmetry in each: (a) (b) (c) Problem 5-53 Draw the structure of a meso compound that has five carbons and three chirality centers. Problem 5-54 Ribose, an essential part of ribonucleic acid (RNA), has the following structure: (a) How many chirality centers does ribose have? Identify them. (b)
How many stereoisomers of ribose are there?
(c) Draw the structure of the enantiomer of ribose. (d) Draw the structure of a diastereomer of ribose.
Problem 5-55
On reaction with hydrogen gas in the presence of a platinum catalyst, ribose (Problem 5-54) is converted into ribitol. Is ribitol optically active or inactive? Explain.
Prochirality
Problem 5-56 Identify the indicated hydrogens in the following molecules as pro-R or pro-S: (a) (b) (c) Problem 5-57 Identify the indicated faces in the following molecules as Re or Si: (a) (b) Problem 5-58 One of the steps in fat metabolism is the hydration of crotonate to yield 3-hydroxybutyrate. The reaction occurs by addition of – OH to the Si face at C3, followed by protonation at C2, also from the Si face. Draw the product of the reaction, showing the stereochemistry of each step. Problem 5-59 The dehydration of citrate to yield cis-aconitate, a step in the citric acid cycle, involves the pro-R “arm” of citrate rather than the pro-S arm. Which of the following two products is formed? Problem 5-60 The first step in the metabolism of glycerol, formed by digestion of fats, is phosphorylation of the pro-R – CH2OH group by reaction with adenosine triphosphate (ATP) to give the corresponding glycerol phosphate plus adenosine diphosphate (ADP). Show the stereochemistry of the product. Problem 5-61 One of the steps in fatty-acid biosynthesis is the dehydration of (R)-3-hydroxybutyryl ACP to give trans-crotonyl ACP. Does the reaction remove the pro-R or the pro-S hydrogen from C2? General Problems Problem 5-62 Draw all possible stereoisomers of 1,2-cyclobutanedicarboxylic acid, and indicate the interrelationships. Which, if any, are optically active? Do the same for 1,3-cyclobutanedicarboxylic acid. Problem 5-63 Draw tetrahedral representations of the two enantiomers of the amino acid cysteine, HSCH2CH(NH2)CO2H, and identify each as R or S. Problem 5-64 The naturally occurring form of the amino acid cysteine (Problem 5-63) has the R configuration at its chirality center. On treatment with a mild oxidizing agent, two cysteines join to give cystine, a disulfide. Assuming that the chirality center is not affected by the reaction, is cystine optically active? Explain. Problem 5-65 Draw tetrahedral representations of the following molecules: (a) The 2S,3R enantiomer of 2,3-dibromopentane (b)
The meso form of 3,5-heptanediol
Problem 5-66
Assign R or S configurations to the chiral centers in cephalexin, trade-named Keflex, the most widely prescribed antibiotic in the United States.
Problem 5-67
Chloramphenicol, a powerful antibiotic isolated in 1947 from the Streptomyces venezuelae bacterium, is active against a broad spectrum of bacterial infections and is particularly valuable against typhoid fever. Assign R or S configurations to the chirality centers in chloramphenicol.
Problem 5-68
Allenes are compounds with adjacent carbon–carbon double bonds. Many allenes are chiral, even though they don’t contain chirality centers. Mycomycin, for example, a naturally occurring antibiotic isolated from the bacterium Nocardia acidophilus, is chiral and has [α]D = −130. Explain why mycomycin is chiral.
Problem 5-69
Long before chiral allenes were known (Problem 5-68), the resolution of 4-methylcyclohexylideneacetic acid into two enantiomers had been carried out. Why is it chiral? What geometric similarity does it have to allenes?
Problem 5-70 (S)-1-Chloro-2-methylbutane undergoes light-induced reaction with Cl2 to yield a mixture of products, among which are 1,4-dichloro-2-methylbutane and 1,2-dichloro-2-methylbutane. (a)
Write the reaction, showing the correct stereochemistry of the reactant.
(b) One of the two products is optically active, but the other is optically inactive. Which is which?
Problem 5-71
How many stereoisomers of 2,4-dibromo-3-chloropentane are there? Draw them, and indicate which are optically active.
Problem 5-72 Draw both cis- and trans-1,4-dimethylcyclohexane in their more stable chair conformations. (a)
How many stereoisomers are there of cis-1,4-dimethylcyclohexane, and how many of trans-1,4-dimethylcyclohexane?
(b) Are any of the structures chiral? (c) What are the stereochemical relationships among the various stereoisomers of 1,4-dimethylcyclohexane?
Problem 5-73 Draw both cis- and trans-1,3-dimethylcyclohexane in their more stable chair conformations. (a)
How many stereoisomers are there of cis-1,3-dimethylcyclohexane, and how many of trans-1,3-dimethylcyclohexane?
(b) Are any of the structures chiral? (c) What are the stereochemical relationships among the various stereoisomers of 1,3-dimethylcyclohexane?
Problem 5-74
cis-1,2-Dimethylcyclohexane is optically inactive even though it has two chirality centers. Explain.
Problem 5-75
We’ll see in Chapter 11 that alkyl halides react with hydrosulfide ion (HS) to give a product whose stereochemistry is inverted from that of the reactant.
Draw the reaction of (S)-2-bromobutane with HS ion to yield 2-butanethiol, CH3CH2CH(SH)CH3. Is the stereochemistry of the product R or S?
Problem 5-76 Ketones react with sodium acetylide (the sodium salt of acetylene, Na+– : C$\text{≡}$CH) to give alcohols. For example, the reaction of sodium acetylide with 2-butanone yields 3-methyl-1-pentyn-3-ol: (a)
Is the product chiral?
(b) Assuming that the reaction takes place with equal likelihood from both Re and Si faces of the carbonyl group, is the product optically active? Explain.
Problem 5-77 Imagine that a reaction similar to that in Problem 5-76 is carried out between sodium acetylide and (R)-2-phenylpropanal to yield 4-phenyl-1-pentyn-3-ol: (a)
Is the product chiral?
(b) Draw both major and minor reaction products, assuming that the reaction takes place preferentially from the Re face of the carbonyl group. Is the product mixture optically active? Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.17%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 6, you should be able to
1. fulfill the detailed objectives listed under each individual section.
2. identify the polarity pattern in the common functional groups, and explain the importance of being able to do so.
3. describe the essential differences between polar and radical reactions, and assign a given reaction to one of these two categories.
4. discuss how kinetic and thermodynamic factors determine the rate and extent of a chemical reaction.
5. use bond dissociation energies to calculate the ΔH° of simple reactions, and vice versa.
6. draw and interpret reaction energy diagrams.
7. define, and use in context, the new key terms.
This chapter is designed to provide a gentle introduction to the subject of reaction mechanisms. Two types of reactions are introduced—polar reactions and radical reactions. The chapter briefly reviews a number of topics you should be familiar with, including rates and equilibria, elementary thermodynamics and bond dissociation energies. You must have a working knowledge of these topics to obtain a thorough understanding of organic reaction mechanisms. Reaction energy diagrams are used to illustrate the energy changes that take place during chemical reactions, and to emphasize the difference between a reaction intermediate and a transition state.
06: An Overview of Organic Reactions
All chemical reactions, whether they take place in the laboratory or in living organisms, follow the same “rules.” Reactions in living organisms often look more complex than laboratory reactions because of the size of the biomolecules and the involvement of biological catalysts called enzymes, but the principles governing all chemical reactions are the same.
To understand both organic and biological chemistry, it’s necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we’ll start with an overview of the fundamental kinds of organic reactions, we’ll see why reactions occur, and we’ll see how reactions can be described. Once this background is out of the way, we’ll then be ready to begin studying the details of organic chemistry in future chapters.
When first approached, organic chemistry might seem overwhelming. It’s not so much that any one part is difficult to understand, it’s that there are so many parts: tens of millions of compounds, dozens of functional groups, and an apparently endless number of reactions. With study, though, it becomes evident that there are only a few fundamental ideas that underlie all organic reactions. Far from being a collection of isolated facts, organic chemistry is a beautifully logical subject that is unified by a few broad themes. When these themes are understood, learning organic chemistry becomes much easier and memorization is minimized. The aim of this book is to describe the themes and clarify the patterns that unify organic chemistry in future chapters.
6.02: Kinds of Organic Reactions
Organic chemical reactions can be organized broadly in two ways—by what kinds of reactions occur and by how those reactions occur. Let’s look first at the kinds of reactions that take place. There are four general types of organic reactions: additions, eliminations, substitutions, and rearrangements.
• Addition reactions occur when two reactants add together to form a single product with no atoms “left over.” An example that we’ll be studying soon is the reaction of an alkene, such as ethylene, with HBr to yield an alkyl bromide.
• Elimination reactions are, in a sense, the opposite of addition reactions. They occur when a single reactant splits into two products, often with the formation of a small molecule such as water or HBr. An example is the acid-catalyzed reaction of an alcohol to yield water and an alkene.
• Substitution reactions occur when two reactants exchange parts to give two new products. An example is the reaction of an ester such as methyl acetate with water to yield a carboxylic acid plus an alcohol. Similar reactions occur in many biological pathways, including the metabolism of dietary fats.
• Rearrangement reactions occur when a single reactant undergoes a reorganization of bonds and atoms to yield an isomeric product. An example is the conversion of dihydroxyacetone phosphate into its constitutional isomer glyceraldehyde 3-phosphate, a step in the glycolysis pathway by which carbohydrates are metabolized.
Problem 6-1 Classify each of the following reactions as an addition, elimination, substitution, or rearrangement: (a)
CH3Br + KOH $\text{⟶}$ CH3OH + KBr
(b) CH3CH2Br $\text{⟶}$ H2C$\text{=}$CH2 + HBr (c) H2C$\text{=}$CH2 + H2 $\text{⟶}$ CH3CH3
6.03: How Organic Reactions Occur - Mechanisms
Having looked at the kinds of reactions that take place, let’s now see how they occur. An overall description of how a reaction occurs is called a reaction mechanism. A mechanism describes in detail exactly what takes place at each stage of a chemical transformation—which bonds are broken and in what order, which bonds are formed and in what order, and what the relative rates are for each step. A complete mechanism must also account for all reactants used and all products formed.
All chemical reactions involve bond-breaking and bond-making. When two molecules come together, react, and yield products, specific bonds in the reactant molecules are broken and specific bonds in the product molecules are formed. Fundamentally, there are two ways in which a covalent two-electron bond can break. A bond can break in an electronically unsymmetrical way so that both bonding electrons remain with one product fragment, leaving the other with a vacant orbital, or a bond can break in an electronically symmetrical way so that one electron remains with each product fragment. The unsymmetrical cleavage is said to be heterolytic, and the symmetrical cleavage is said to be homolytic.
We’ll develop this point in more detail later, but note for now that the movement of two electrons in the unsymmetrical process is indicated using a full-headed curved arrow (), whereas the movement of one electron in the symmetrical process is indicated using a half-headed, or “fishhook,” arrow ().
Just as there are two ways in which a bond can break, there are two ways in which a covalent two-electron bond can form. A bond can form in an electronically unsymmetrical way if both bonding electrons are donated to the new bond by one reactant, or in a symmetrical way if one electron is donated by each reactant.
Processes that involve unsymmetrical bond-breaking and bond-making are called polar reactions. Polar reactions involve species that have an even number of electrons and thus have only electron pairs in their orbitals. Polar processes are by far the more common reaction type in both organic and biological chemistry, and a large part of this book is devoted to their description.
Processes that involve symmetrical bond-breaking and bond-making are called radical reactions. A radical, often called a free radical, is a neutral chemical species that contains an odd number of electrons and thus has a single, unpaired electron in one of its orbitals.
In addition to polar and radical reactions, there is a third, less commonly encountered process called a pericyclic reaction. Rather than explain pericyclic reactions now, though, we’ll look at them more carefully in Chapter 30. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.01%3A_Why_This_Chapter.txt |
Polar reactions occur because of the electrical attraction between positively polarized and negatively polarized centers on functional groups in molecules. To see how these reactions take place, let’s first recall the discussion of polar covalent bonds in Section 2.1 and then look more deeply into the effects of bond polarity on organic molecules.
Most organic compounds are electrically neutral; they have no net charge, either positive or negative. We saw in Section 2.1, however, that certain bonds within a molecule, particularly the bonds in functional groups, are polar. Bond polarity is a consequence of an unsymmetrical electron distribution in a bond and is due to the difference in electronegativity of the bonded atoms.
Elements such as oxygen, nitrogen, fluorine, and chlorine are more electronegative than carbon, so a carbon atom bonded to one of these atoms has a partial positive charge (δ+). Metals are less electronegative than carbon, so a carbon atom bonded to a metal has a partial negative charge (δ−). Electrostatic potential maps of chloromethane and methyllithium illustrate these charge distributions, showing that the carbon atom in chloromethane is electron-poor (blue) while the carbon in methyllithium is electron-rich (red).
The polarity patterns of some common functional groups are shown in Table 6.1. Note that carbon is always positively polarized except when bonded to a metal.
Table 6.1 Polarity Patterns in Some Common Functional Groups
Compound type Functional group structure
This discussion of bond polarity is oversimplified in that we’ve considered only bonds that are inherently polar due to differences in electronegativity. Polar bonds can also result from the interaction of functional groups with acids or bases. Take an alcohol such as methanol, for example. In neutral methanol, the carbon atom is somewhat electron-poor because the electronegative oxygen attracts the electrons in the C−O bond. On protonation of the methanol oxygen by an acid, however, a full positive charge on oxygen attracts the electrons in the C−O bond much more strongly and makes the carbon much more electron-poor. We’ll see numerous examples throughout this book of reactions that are catalyzed by acids because of the resultant increase in bond polarity upon protonation.
Yet a further consideration is the polarizability (as opposed to polarity) of atoms in a molecule. As the electric field around a given atom changes because of changing interactions with solvent or other polar molecules nearby, the electron distribution around that atom also changes. The measure of this response to an external electrical influence is called the polarizability of the atom. Larger atoms with more loosely held electrons are more polarizable, and smaller atoms with fewer, tightly held electrons are less polarizable. Thus, sulfur is more polarizable than oxygen, and iodine is more polarizable than chlorine. The effect of this higher polarizability of sulfur and iodine is that carbon–sulfur and carbon–iodine bonds, although nonpolar according to electronegativity values (Figure 2.3), nevertheless usually react as if they were polar.
What does functional-group polarity mean with respect to chemical reactivity? Because unlike charges attract, the fundamental characteristic of all polar organic reactions is that electron-rich sites react with electron-poor sites. Bonds are made when an electron-rich atom donates a pair of electrons to an electron-poor atom, and bonds are broken when one atom leaves with both electrons from the former bond.
As we saw in Section 2.11, the movement of an electron pair during a polar reaction is indicated using a curved, full-headed arrow to show where electrons move when reactant bonds are broken and product bonds are formed during the reaction.
In referring to the electron-rich and electron-poor species involved in polar reactions, chemists use the words nucleophile and electrophile. A nucleophile is a substance that is “nucleus-loving.” (Remember that a nucleus is positively charged.) A nucleophile has a negatively polarized, electron-rich atom and can form a bond by donating a pair of electrons to a positively polarized, electron-poor atom. Nucleophiles can be either neutral or negatively charged; ammonia, water, hydroxide ion, and chloride ion are examples. An electrophile, by contrast, is “electron-loving.” An electrophile has a positively polarized, electron-poor atom and can form a bond by accepting a pair of electrons from a nucleophile. Electrophiles can be either neutral or positively charged. Acids (H+ donors), alkyl halides, and carbonyl compounds are examples (Figure 6.2).
and electrophilic (positive) atoms.
Note that neutral compounds can often react either as nucleophiles or as electrophiles, depending on the circumstances. After all, if a compound is neutral yet has an electron-rich nucleophilic site, it must also have a corresponding electron-poor electrophilic site. Water, for instance, acts as an electrophile when it donates H+ but acts as a nucleophile when it donates a nonbonding pair of electrons. Similarly, a carbonyl compound acts as an electrophile when it reacts at its positively polarized carbon atom, yet acts as a nucleophile when it reacts at its negatively polarized oxygen atom.
If the definitions of nucleophiles and electrophiles sound similar to those given in Section 2.11 for Lewis acids and Lewis bases, that’s because there is indeed a correlation. Lewis bases are electron donors and behave as nucleophiles, whereas Lewis acids are electron acceptors and behave as electrophiles. Thus, much of organic chemistry is explainable in terms of acid–base reactions. The main difference is that the words acid and base are used broadly in all fields of chemistry, while the words nucleophile and electrophile are used primarily in organic chemistry when carbon bonding is involved.
Worked Example 6.1
Identifying Electrophiles and Nucleophiles
Which of the following species is likely to behave as a nucleophile and which as an electrophile?
(a) NO2+(b) CN(c) CH3NH2(d) (CH3)3S+
Strategy
A nucleophile has an electron-rich site, either because it is negatively charged or because it has a functional group containing an atom that has a lone pair of electrons. An electrophile has an electron-poor site, either because it is positively charged or because it has a functional group containing an atom that is positively polarized.
Solution
(a) NO2+ (nitronium ion) is likely to be an electrophile because it is positively charged.
(b)$:C≡N−:C≡N−$ (cyanide ion) is likely to be a nucleophile because it is negatively charged.
(c) CH3NH2 (methylamine) might be either a nucleophile or an electrophile, depending on the circumstances. The lone pair of electrons on the nitrogen atom makes methylamine a potential nucleophile, while positively polarized N−H hydrogens make methylamine a potential acid (electrophile).
(d) (CH3)3S+ (trimethylsulfonium ion) is likely to be an electrophile because it is positively charged.
CH3Cl
(b) CH3S (c)
(d)
Problem 6-3
An electrostatic potential map of boron trifluoride is shown. Is BF3 likely to be a nucleophile or an electrophile? Draw a Lewis structure for BF3, and explain your answer. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.04%3A_Polar_Reactions.txt |
Let’s look at a typical polar process—the addition reaction of an alkene, such as ethylene, with hydrogen bromide. When ethylene is treated with HBr at room temperature, bromoethane is produced. Overall, the reaction can be formulated as
The reaction is an example of a polar reaction type known as an electrophilic addition reaction and can be understood using the general ideas discussed in the previous section. Let’s begin by looking at the two reactants.
What do we know about ethylene? We know from Section 1.8 that a carbon–carbon double bond results from the orbital overlap of two sp2-hybridized carbon atoms. The σ part of the double bond results from sp2sp2 overlap, and the π part results from pp overlap.
What kind of chemical reactivity might we expect from a $C═CFigure 6.3). As a result, the double bond is nucleophilic and the chemistry of alkenes is dominated by reactions with electrophiles.$
.
What about the second reactant, HBr? As a strong acid, HBr is a powerful proton (H+) donor and electrophile. Thus, the reaction between HBr and ethylene is a typical electrophile–nucleophile combination, characteristic of all polar reactions.
We’ll see more details about alkene electrophilic addition reactions shortly, but for the present we can imagine the reaction as taking place by the pathway shown in Figure 6.4. The reaction begins when the alkene nucleophile donates a pair of electrons from its $C═CC═C$ bond to HBr to form a new C−H bond plus Br, as indicated by the path of the curved arrows in the first step of Figure 6.4. One curved arrow begins at the middle of the double bond (the source of the electron pair) and points to the hydrogen atom in HBr (the atom to which a bond will form). This arrow indicates that a new C−H bond forms using electrons from the former $C═CC═C$ bond. Simultaneously, a second curved arrow begins in the middle of the H−Br bond and points to the Br, indicating that the H−Br bond breaks and the electrons remain with the Br atom, giving Br.
Figure 6.4 MECHANISM The electrophilic addition reaction of ethylene and HBr. The reaction takes place in two steps, both of which involve electrophile–nucleophile interactions.
When one of the alkene carbon atoms bonds to the incoming hydrogen, the other carbon atom, having lost its share of the double-bond electrons, now has only six valence electrons and is left with a positive charge. This positively charged species—a carbon-cation, or carbocation—is itself an electrophile that can accept an electron pair from nucleophilic Br anion in a second step, forming a C−Br bond and yielding the observed addition product. Once again, a curved arrow in Figure 6.4 shows the electron-pair movement from Br to the positively charged carbon.
The electrophilic addition of HBr to ethylene is only one example of a polar process; there are many others that we’ll study in depth in later chapters. But regardless of the details of individual reactions, all polar reactions take place between an electron-poor site and an electron-rich site and involve the donation of an electron pair from a nucleophile to an electrophile.
Problem 6-4
What product would you expect from reaction of cyclohexene with HBr? With HCl?
Problem 6-5
Reaction of HBr with 2-methylpropene yields 2-bromo-2-methylpropane. What is the structure of the carbocation formed during the reaction? Show the mechanism of the reaction. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.05%3A_An_Example_of_a_Polar_Reaction_-_Addition_of_HBr_to_Ethylene.txt |
It takes practice to use curved arrows properly in reaction mechanisms, but there are a few rules and a few common patterns you should look for that will help you become more proficient:
RULE 1
Electrons move from a nucleophilic source (Nu: or Nu:) to an electrophilic sink (E or E+). The nucleophilic source must have an electron pair available, usually either as a lone pair or in a multiple bond. For example:
The electrophilic sink must be able to accept an electron pair, usually because it has either a positively charged atom or a positively polarized atom in a functional group. For example:
RULE 2
The nucleophile can be either negatively charged or neutral. If the nucleophile is negatively charged, the atom that donates an electron pair becomes neutral. For example:
If the nucleophile is neutral, the atom that donates the electron pair acquires a positive charge. For example:
RULE 3
The electrophile can be either positively charged or neutral. If the electrophile is positively charged, the atom bearing that charge becomes neutral after accepting an electron pair. For example:
If the electrophile is neutral, the atom that ultimately accepts the electron pair acquires a negative charge. For this to happen, however, the negative charge must be stabilized by being on an electronegative atom such as oxygen, nitrogen, or a halogen. Carbon and hydrogen do not typically stabilize a negative charge. For example:
The result of Rules 2 and 3 together is that charge is conserved during the reaction. A negative charge in one of the reactants gives a negative charge in one of the products, and a positive charge in one of the reactants gives a positive charge in one of the products.
RULE 4
The octet rule must be followed. That is, no second-row atom can be left with ten electrons (or four for hydrogen). If an electron pair moves to an atom that already has an octet (or two electrons for hydrogen), another electron pair must simultaneously move from that atom to maintain the octet. When two electrons move from the $C═CC═C$ bond of ethylene to the hydrogen atom of H3O+, for instance, two electrons must leave that hydrogen. This means that the H−O bond must break and the electrons must stay with the oxygen, giving neutral water.
Worked Example 6.2 gives another example of drawing curved arrows.
Worked Example 6.2
Using Curved Arrows in Reaction Mechanisms
Add curved arrows to the following polar reaction to show the flow of electrons:
Strategy
Look at the reaction, and identify the bonding changes that have occurred. In this case, a C−Br bond has broken and a C−C bond has formed. The formation of the C−C bond involves donation of an electron pair from the nucleophilic carbon atom of the reactant on the left to the electrophilic carbon atom of CH3Br, so we draw a curved arrow originating from the lone pair on the negatively charged C atom and pointing to the C atom of CH3Br. At the same time that the C−C bond forms, the C−Br bond must break so that the octet rule is not violated. We therefore draw a second curved arrow from the C−Br bond to Br. The bromine is now a stable Br ion.
Solution
Problem 6-6 Add curved arrows to the following polar reactions to indicate the flow of electrons in each: (a)
(b)
(c)
Problem 6-7
Predict the products of the following polar reaction, a step in the citric acid cycle for food metabolism, by interpreting the flow of electrons indicated by the curved arrows:
6.07: Radical Reactions
Radical reactions are much less common than polar reactions but are nevertheless important in some industrial processes and biological pathways. We’ll look at them in more detail in Sections 10.2 and 10.3 but will briefly see how they occur at this point.
A radical is highly reactive because it contains an atom with an odd number of electrons (usually seven) in its valence shell, rather than a noble-gas octet. The radical can achieve a valence-shell octet in several ways however. For instance, it might abstract an atom and one bonding electron from another reactant, leaving behind a new radical. The net result is a radical substitution reaction.
Alternatively, a reactant radical might add to a double bond, taking one electron from the double bond and yielding a new radical. The net result is a radical addition reaction.
An example of an industrially useful radical reaction is the reaction of chlorine with methane to yield chloromethane, which is used to manufacture the solvents dichloromethane (CH2Cl2) and chloroform (CHCl3). The process begins with irradiation of Cl2 with ultraviolet light to break the relatively weak ClCl bond of Cl2 and produce chlorine radicals (·Cl).
Chlorine radicals then react with methane by abstracting a hydrogen atom to give HCl and a methyl radical (·CH3) that reacts further with Cl2 to give chloromethane plus a new chlorine radical that cycles back and repeats the first step. Thus, once the sequence has started, it becomes a self-sustaining cycle of repeating steps (a) and (b), making the overall process a chain reaction.
As a biological example of a radical reaction, look at the synthesis of prostaglandins, a large class of molecules found in virtually all body tissues and fluids. A number of pharmaceuticals are based on or derived from prostaglandins, including medicines that induce labor during childbirth, reduce intraocular pressure in glaucoma, control bronchial asthma, and help treat congenital heart defects.
Prostaglandin biosynthesis is initiated by the abstraction of a hydrogen atom from arachidonic acid by an iron–oxygen radical, thereby generating a carbon radical in a substitution reaction. Don’t be intimidated by the size of the molecules; focus on the changes that occur in each step. (To help you do that, the unchanged part of the molecule is “ghosted,” with only the reactive part clearly visible.)
Following the initial abstraction of a hydrogen atom, the carbon radical then reacts with O2 to give an oxygen radical, which reacts with a $C═CC═C$ bond within the same molecule in an addition reaction. Several further transformations ultimately yield prostaglandin H2.
Problem 6-8
Radical chlorination of alkanes is not generally useful because mixtures of products often result when more than one kind of C−H bond is present in the substrate. Draw and name all monochloro substitution products C6H13Cl you might obtain by reaction of 2-methylpentane with Cl2.
Problem 6-9
Using curved fishhook arrows, propose a mechanism for the formation of the cyclopentane ring of prostaglandin H2. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.06%3A_Using_Curved_Arrows_in_Polar_Reaction_Mechanisms.txt |
Every chemical reaction can go in either a forward or reverse direction. Reactants can go forward to products, and products can revert to reactants. As you may remember from your general chemistry course, the position of the resulting chemical equilibrium is expressed by an equation in which Keq, the equilibrium constant, is equal to the product concentrations multiplied together, divided by the reactant concentrations multiplied together, with each concentration raised to the power of its coefficient in the balanced equation. For the generalized reaction
$aA+bB⇌cC+dDaA+bB⇌cC+dD$
we have
$K eq = [C] c [D] d [A] a [B] b K eq = [C] c [D] d [A] a [B] b$
The value of the equilibrium constant tells which side of the reaction arrow is energetically favored. If Keq is much larger than 1, then the product concentration term [C]c[D]d is much larger than the reactant concentration term [A]a[B]b, and the reaction proceeds as written from left to right. If Keq is near 1, appreciable amounts of both reactant and product are present at equilibrium. And if Keq is much smaller than 1, the reaction does not take place as written but instead goes in the reverse direction, from right to left.
In the reaction of ethylene with HBr, for example, we can write the following equilibrium expression and determine experimentally that the equilibrium constant at room temperature is approximately 7.1 × 107:
Because Keq is relatively large, the reaction proceeds as written and more than 99.999 99% of the ethylene is converted into bromoethane. For practical purposes, an equilibrium constant greater than about 103 means that the amount of reactant left over will be barely detectable (less than 0.1%).
What determines the magnitude of the equilibrium constant? For a reaction to have a favorable equilibrium constant and proceed as written, the energy of the products must be lower than the energy of the reactants. In other words, energy must be released. This situation is analogous to that of a rock poised precariously in a high-energy position near the top of a hill. When it rolls downhill, the rock releases energy until it reaches a more stable, low-energy position at the bottom.
The energy change that occurs during a chemical reaction is called the Gibbs free-energy change (ΔG), which is equal to the free energy of the products minus the free energy of the reactants: ΔG = GproductsGreactants. For a favorable reaction, ΔG has a negative value, meaning that energy is lost by the chemical system and released to the surroundings, usually as heat. Such reactions are said to be exergonic. For an unfavorable reaction, ΔG has a positive value, meaning that energy is absorbed by the chemical system from the surroundings. Such reactions are said to be endergonic.
You might also recall from general chemistry that the standard free-energy change for a reaction is denoted as ΔG°, where the superscript ° means that the reaction is carried out under standard conditions, with pure substances in their most stable form at 1 atm pressure and a specified temperature, usually 298 K. For biological reactions, the standard free-energy change is denoted as ΔG°′ and refers to a reaction carried out at pH = 7.0 with solute concentrations of 1.0 M.
Because the equilibrium constant, Keq, and the standard free-energy change, ΔG°, both measure whether a reaction is favorable, they are mathematically related by the equation
$Δ G ∘ =−RTln K eq or K eq = e −Δ G ∘ /RT Δ G ∘ =−RTln K eq or K eq = e −Δ G ∘ /RT$
where
$R=8.314 J/(K⋅mol)=1.987 cal/(K⋅mol) T=Kelvin temperature e=2.718 ln K eq =natural logarithm of K eq R=8.314 J/(K⋅mol)=1.987 cal/(K⋅mol) T=Kelvin temperature e=2.718 ln K eq =natural logarithm of K eq$
For example, the reaction of ethylene with HBr has Keq = 7.1 × 107, so ΔG° = −44.8 kJ/mol (−10.7 kcal/mol) at 298 K:
$K eq =7.1× 10 7 and ln K eq =18.08 Δ G ∘ =−RTln K eq =−[8.314 J/(K⋅mol)](298 K)(18.08) =−44,800 J/mol=−44.8 kJ/mol K eq =7.1× 10 7 and ln K eq =18.08 Δ G ∘ =−RTln K eq =−[8.314 J/(K⋅mol)](298 K)(18.08) =−44,800 J/mol=−44.8 kJ/mol$
The free-energy change ΔG is made up of two terms, an enthalpy term, ΔH, and a temperature-dependent entropy term, TΔS. Of the two terms, the enthalpy term is often larger and more dominant.
$Δ G ∘ =Δ H ∘ −TΔ S ∘ Δ G ∘ =Δ H ∘ −TΔ S ∘$
For the reaction of ethylene with HBr at room temperature (298 K), the approximate values are
The enthalpy change (ΔH), also called the heat of reaction, is a measure of the change in total bonding energy during a reaction. If ΔH is negative, as in the reaction of HBr with ethylene, the products have less energy than the reactants. Thus, the products are more stable and have stronger bonds than the reactants, heat is released, and the reaction is said to be exothermic. If ΔH is positive, the products are less stable and have weaker bonds than the reactants, heat is absorbed, and the reaction is said to be endothermic. For example, if a reaction breaks reactant bonds with a total strength of 380 kJ/mol and forms product bonds with a total strength of 400 kJ/mol, then ΔH for the reaction is 400 kJ/mol – 380 kJ/mol = −20 kJ/mol and the reaction is exothermic.
The entropy change (ΔS) is a measure of the change in the amount of molecular randomness, or freedom of motion, that accompanies a reaction. For example, in an elimination reaction of the type
$A⟶B+CA⟶B+C$
there is more freedom of movement and molecular randomness in the products than in the reactant because one molecule has split into two. Thus, there is a net increase in entropy during the reaction and ΔS has a positive value.
On the other hand, for an addition reaction of the type
$A+B⟶CA+B⟶C$
the opposite is true. Because such reactions restrict the freedom of movement of two molecules by joining them together, the product has less randomness than the reactants and ΔS has a negative value. The reaction of ethylene and HBr to yield bromoethane, which has ΔS° = −0.132 kJ/(K · mol), is an example. Table 6.2 describes the thermodynamic terms more fully.
Table 6.2 Explanation of Thermodynamic Quantities: ΔG° = ΔH°TΔS°
Term Name Explanation
ΔG° Gibbs free-energy change The energy difference between reactants and products. When ΔG° is negative, the reaction is exergonic, has a favorable equilibrium constant, and can occur spontaneously. When ΔG° is positive, the reaction is endergonic, has an unfavorable equilibrium constant, and cannot occur spontaneously.
ΔH° Enthalpy change The heat of reaction, or difference in strength between the bonds broken in a reaction and the bonds formed. When ΔH° is negative, the reaction releases heat and is exothermic. When ΔH° is positive, the reaction absorbs heat and is endothermic.
ΔS° Entropy change The change in molecular randomness during a reaction. When ΔS° is negative, randomness decreases. When ΔS° is positive, randomness increases.
Knowing the value of Keq for a reaction is useful, but it’s important to realize its limitations. An equilibrium constant tells only the position of the equilibrium, or how much product is theoretically possible. It doesn’t tell the rate of reaction, or how fast the equilibrium is established. Some reactions are extremely slow even though they have favorable equilibrium constants. Gasoline is stable at room temperature, for instance, because the rate of its reaction with oxygen is slow at 298 K. Only at higher temperatures, such as contact with a lighted match, does gasoline react rapidly with oxygen and undergo complete conversion to the equilibrium products water and carbon dioxide. Rates (how fast a reaction occurs) and equilibria (how much a reaction occurs) are entirely different.
$Rate→Is the reaction fast or slow? Rate→Is the reaction fast or slow?$
$Equilibrium→In what direction does the reaction proceed? Equilibrium→In what direction does the reaction proceed?$
Problem 6-10
Which reaction is more energetically favored, one with ΔG° = −44 kJ/mol or one with ΔG° = +44 kJ/mol?
Problem 6-11
Which reaction is more exergonic, one with Keq = 1000 or one with Keq = 0.001? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.08%3A_Describing_a_Reaction_-_Equilibria__Rates_and_Energy_Changes.txt |
We’ve just seen that heat is released (negative ΔH) when a bond is formed because the products are more stable and have stronger bonds than the reactants. Conversely, heat is absorbed (positive ΔH) when a bond is broken because the products are less stable and have weaker bonds than the reactants. The amount of energy needed to break a given bond to produce two radical fragments when the molecule is in the gas phase at 25 °C is a quantity called the bond strength, or bond dissociation energy (D).
Each specific bond has its own characteristic strength, and extensive tables of such data are available. For example, a C−H bond in methane has a bond dissociation energy D = 439.3 kJ/mol (105.0 kcal/mol), meaning that 439.3 kJ/mol must be added to break a C−H bond of methane to give the two radical fragments ·CH3 and ·H. Conversely, 439.3 kJ/mol of energy is released when a methyl radical and a hydrogen atom combine to form methane. Table 6.3 lists some other bond strengths.
Table 6.3 Some Bond Dissociation Energies, D
Bond D (kJ/mol) Bond D (kJ/mol) Bond D (kJ/mol)
HH 436 (CH3)2CHH 410 C2H5CH3 370
HF 570 (CH3)2CHCI 354 (CH3)2CHCH3 369
HCI 431 (CH3)2CHBr 299 (CH3)3CCH3 363
HBr 366 (CH3)3CH 400 H2C═CHCH3 426
HI 298 (CH3)3CCI 352 H2C═CHCH2CH3 318
CICI 242 (CH3)3CBr 293 H2C═CH2 728
BrBr 194 (CH3)3CI 227 427
II 152 H2C═CHH 464 325
CH3H 439 H2C═CHCI 396 374
CH3CI 350 H2C═CHCH2H 369 HOH 497
CH3Br 294 H2C═CHCH2CI 298 HOOH 211
CH3I 239 472 CH3OH 440
CH3OH 385 400 CH3SH 366
CH3NH2 386 375 C2H5OH 441
C2H5H 421 300 352
C2H5CI 352 336 CH3CH2OCH3 355
C2H5Br 293 464 NH2H 450
C2H5I 233 $HC≡CHC≡C$HH$ 558 HCN 528
C2H5OH 391 CH3CH3 377
Think again about the connection between bond strengths and chemical reactivity. In an exothermic reaction, more heat is released than is absorbed. But because making bonds in the products releases heat and breaking bonds in the reactants absorbs heat, the bonds in the products must be stronger than the bonds in the reactants. In other words, exothermic reactions are favored by products with strong bonds and by reactants with weak, easily broken bonds.
Sometimes, particularly in biochemistry, reactive substances that undergo highly exothermic reactions, such as ATP (adenosine triphosphate), are referred to as “energy-rich” or “high-energy” compounds. Such a label doesn’t mean that ATP is special or different from other compounds, it only means that ATP has relatively weak bonds that require a relatively small amount of heat to break, thus leading to a larger release of heat when a strong new bond forms in a reaction. When a typical organic phosphate such as glycerol 3-phosphate reacts with water, for instance, only 9 kJ/mol of heat is released (ΔH°′ = −9 kJ/mol), but when ATP reacts with water, 30 kJ/mol of heat is released (ΔH°′ = −30 kJ/mol). The difference between the two reactions is due to the fact that the bond broken in ATP is substantially weaker than the bond broken in glycerol 3-phosphate. We’ll see the metabolic importance of this reaction in later chapters. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.09%3A_Describing_a_Reaction_-_Bond__Dissociation_Energies.txt |
For a reaction to take place, reactant molecules must collide and reorganization of atoms and bonds must occur. Let’s again look at the addition reaction of HBr and ethylene.
As the reaction proceeds, ethylene and HBr approach each other, the ethylene π bond and the H−Br bond break, a new C−H bond forms in step 1 and a new C−Br bond forms in step 2.
To depict graphically the energy changes that occur during a reaction, chemists use energy diagrams, such as that in Figure 6.5. The vertical axis of the diagram represents the total energy of all reactants, and the horizontal axis, called the reaction coordinate, represents the progress of the reaction from beginning to end. Let’s see how the addition of HBr to ethylene can be described in an energy diagram.
At the beginning of the reaction, ethylene and HBr have the total amount of energy indicated by the reactant level on the left side of the diagram in Figure 6.5. As the two reactants collide and reaction commences, their electron clouds repel each other, causing the energy level to rise. If the collision has occurred with enough force and proper orientation, however, the reactants continue to approach each other despite the rising repulsion until the new C−H bond starts to form. At some point, a structure of maximum energy is reached, a structure called the transition state.
The transition state represents the highest-energy structure involved in this step of the reaction. It is unstable and can’t be isolated, but we can imagine it to be an activated complex of the two reactants in which both the $C═CFigure 6.6).$
The energy difference between reactants and the transition state is called the activation energy, ΔG, and determines how rapidly the reaction occurs at a given temperature. (The double-dagger superscript, , always refers to the transition state.) A high activation energy results in a slow reaction because few collisions occur with enough energy for the reactants to reach the transition state. A low activation energy results in a rapid reaction because almost all collisions occur with enough energy for the reactants to reach the transition state.
As an analogy, you might think of reactants that need enough energy to climb the activation barrier to the transition state as hikers who need enough energy to climb to the top of a mountain pass. If the pass is a high one, the hikers need a lot of energy and surmount the barrier with difficulty. If the pass is low, the hikers need less energy and reach the top easily.
As a rough generalization, many organic reactions have activation energies in the range 40 to 150 kJ/mol (10–35 kcal/mol). The reaction of ethylene with HBr, for example, has an activation energy of approximately 140 kJ/mol (34 kcal/mol). Reactions with activation energies less than 80 kJ/mol take place at or below room temperature, while reactions with higher activation energies normally require a higher temperature to give the reactants enough energy to climb the activation barrier.
Once the transition state is reached, the reaction can either continue on to give the carbocation product or revert back to reactants. When reversion to reactants occurs, the transition-state structure comes apart and an amount of free energy corresponding to −ΔG is released. When the reaction continues on to give the carbocation, the new C−H bond forms fully and an amount of energy is released corresponding to the difference between the transition state and carbocation product. The net energy change for the step, ∆G°, is represented in the diagram as the difference in level between reactant and product. Since the carbocation is higher in energy than the starting alkene, the step is endergonic, has a positive value of ∆G°, and absorbs energy.
Not all energy diagrams are like that shown for the reaction of ethylene and HBr. Each reaction has its own energy profile. Some reactions are fast (small ΔG) and some are slow (large ΔG); some have a negative ΔG°, and some have a positive ΔG°. Figure 6.7 illustrates some different possibilities.
Problem 6-12
Which reaction is faster, one with ΔG = +45 kJ/mol or one with ΔG = +70 kJ/mol? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.10%3A_Describing_a_Reaction_-_Energy_Diagrams_and_Transition_States.txt |
How can we describe the carbocation formed in the first step of the reaction of ethylene with HBr? The carbocation is clearly different from the reactants, yet it isn’t a transition state and it isn’t a final product.
We call the carbocation, which exists only transiently during the course of the multistep reaction, a reaction intermediate. As soon as the intermediate is formed in the first step by reaction of ethylene with H+, it reacts further with Br in a second step to give the final product, bromoethane. This second step has its own activation energy (ΔG), its own transition state, and its own energy change (ΔG°). We can picture the second transition state as an activated complex between the electrophilic carbocation intermediate and the nucleophilic bromide anion, in which Br donates a pair of electrons to the positively charged carbon atom as the new C−Br bond just starts to form.
A complete energy diagram for the overall reaction of ethylene with HBr is shown in Figure 6.8. In essence, we draw a diagram for each of the individual steps and then join them so that the carbocation product of step 1 is the reactant for step 2. As indicated in Figure 6.8, the reaction intermediate lies at an energy minimum between steps. Because the energy level of the intermediate is higher than the level of either the reactant that formed it or the product it yields, the intermediate can’t normally be isolated. It is, however, more stable than its two neighboring transition states.
Each step in a multistep process can always be considered separately. Each step has its own ΔG and its own ΔG°. The overall activation energy that controls the rate of the reaction, however, is the energy difference between initial reactants and the highest transition state, regardless of which step it occurs in. The overall ΔG° of the reaction is the energy difference between reactants and final products.
The biological reactions that take place in living organisms have the same energy requirements as reactions that take place in the laboratory and can be described in similar ways. They are, however, constrained by the fact that they must have low enough activation energies to occur at moderate temperatures, and they must release energy in relatively small amounts to avoid overheating the organism. These constraints can be met through the use of large, structurally complex, enzyme catalysts that alter the mechanism of a reaction to a pathway that can proceed through a series of small steps rather than one or two large steps. Thus, an energy diagram for a biological reaction might look like that in Figure 6.9.
and an uncatalyzed laboratory reaction. The biological reaction involves many steps, each of which has a relatively small activation energy and small energy change. The end result is the same, however.
Worked Example 6.3
Drawing a Reaction Energy Diagram
Sketch an energy diagram for a one-step reaction that is fast and highly exergonic.
Strategy
A fast reaction has a small ΔG, and a highly exergonic reaction has a large negative ΔG°.
Worked Example 6.4
Drawing a Reaction Energy Diagram
Sketch an energy diagram for a two-step exergonic reaction whose second step has a higher-energy transition state than its first step. Show ΔG and ΔG° for the overall reaction.
Strategy
A two-step reaction has two transition states and an intermediate between them. The ΔG for the overall reaction is the energy change between reactants and the highest-energy transition state—the second one in this case. An exergonic reaction has a negative overall ΔG°.
Solution
Problem 6-13
Sketch an energy diagram for a two-step reaction in which both steps are exergonic and in which the second step has a higher-energy transition state than the first. Label the parts of the diagram corresponding to reactant, product, intermediate, overall ΔG, and overall ΔG°. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.11%3A_Describing_a_Reaction-_Intermediates.txt |
Beginning in the next chapter, we’ll be seeing a lot of reactions, some that are important in laboratory chemistry yet don’t occur in nature and others that have counterparts in biological pathways. In comparing laboratory reactions with biological reactions, several differences are apparent. For one, laboratory reactions are usually carried out in an organic solvent such as diethyl ether or dichloromethane to dissolve the reactants and bring them into contact, whereas biological reactions occur in the aqueous medium within cells. For another, laboratory reactions often take place over a wide range of temperatures without catalysts, while biological reactions take place at the temperature of the organism and are catalyzed by enzymes.
We’ll look at enzymes in more detail in Section 26.10, but you may already be aware that an enzyme is a large, globular, protein molecule that contains in its structure a protected pocket called its active site. The active site is lined by acidic or basic groups as needed for catalysis and has precisely the right shape to bind and hold a substrate molecule in the orientation necessary for reaction. Figure 6.10 shows a molecular model of hexokinase, along with an X-ray crystal structure of the glucose substrate and adenosine diphosphate (ADP) bound in the active site. Hexokinase is an enzyme that catalyzes the initial step of glucose metabolism—the transfer of a phosphate group from ATP to glucose, giving glucose 6-phosphate and ADP. The structures of ATP and ADP were shown at the end of Section 6.8.
Note how the hexokinase-catalyzed phosphorylation reaction of glucose is written. It’s common when writing biological equations to show only the structures of the primary reactant and product, while abbreviating the structures of various biological “reagents” and by-products such as ATP and ADP. A curved arrow intersecting the straight reaction arrow indicates that ATP is also a reactant and ADP also a product.
Yet a third difference between laboratory and biological reactions is that laboratory reactions are often done using relatively small, simple reagents such as Br2, HCl, NaBH4, CrO3, and so forth, while biological reactions usually involve relatively complex “reagents” called coenzymes. In the hexokinase-catalyzed phosphorylation of glucose just shown, ATP is the coenzyme. As another example, compare the H2 molecule, a laboratory reagent that adds to a carbon–carbon double bond to yield an alkane, with the reduced nicotinamide adenine dinucleotide (NADH) molecule, a coenzyme that effects an analogous addition of hydrogen to a double bond in many biological pathways. Of all the atoms in the coenzyme, only the one hydrogen atom shown in red is transferred to the double-bond substrate.
Don’t be intimidated by the size of the ATP or NADH molecule; most of the structure is there to provide an overall shape for binding to the enzyme and to provide appropriate solubility behavior. When looking at biological molecules, focus on the small part of the molecule where the chemical change takes place.
One final difference between laboratory and biological reactions is in their specificity. A catalyst might be used in the laboratory to catalyze the reaction of thousands of different substances, but an enzyme, because it can only bind a specific substrate molecule having a specific shape, will usually catalyze only a specific reaction. It’s this exquisite specificity that makes biological chemistry so remarkable and that makes life possible. Table 6.4 summarizes some of the differences between laboratory and biological reactions.
Table 6.4 A Comparison of Typical Laboratory and Biological Reactions A Comparison of Typical Laboratory and Biological Reactions
Laboratory reaction Biological reaction
Solvent Organic liquid, such as ether Aqueous environment in cells
Temperature Wide range; −80 to 150 °C Temperature of organism
Catalyst Either none, or very simple Large, complex enzymes needed
Reagent size Usually small and simple Relatively complex coenzymes
Specificity Little specificity for substrate Very high specificity for substrate | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.12%3A_A_Comparison_between_Biological_Reactions_and_Laboratory_Reactions.txt |
6 • Chemistry Matters 6 • Chemistry Matters
It has been estimated that major pharmaceutical companies in the United States spent some \$200 billion on drug research and development in 2020, while government agencies and private foundations spent another \$28 billion. What does this money buy? From 1983 to 2022, the money resulted in a total of 1237 new molecular entities (NMEs)—new biologically active chemical substances approved for sale as drugs by the U.S. Food and Drug Administration (FDA).
Where do the new drugs come from? According to a study carried out several years ago at the U.S. National Cancer Institute, only about 33% of new drugs are entirely synthetic and completely unrelated to any naturally occurring substance. The remaining 67% take their lead, to a greater or lesser extent, from nature. Vaccines and genetically engineered proteins of biological origin account for 15% of NMEs, but most new drugs come from natural products, a catchall term generally taken to mean small molecules found in bacteria, plants, algae, and other living organisms. Unmodified natural products isolated directly from the producing organism account for 24% of NMEs, while natural products that have been chemically modified in the laboratory account for the remaining 28%.
Many years of work go into screening many thousands of substances to identify a single compound that might ultimately gain approval as an NME. But after that single compound has been identified, the work has just begun because it takes an average of 9 to 10 years for a drug to make it through the approval process. First, the safety of the drug in animals must be demonstrated and an economical method of manufacture must be devised. With these preliminaries out of the way, an Investigational New Drug (IND) application is submitted to the FDA for permission to begin testing in humans.
Human testing takes, or should take, 5 to 7 years and is divided into three phases. Phase I clinical trials are carried out on a small group of healthy volunteers to establish safety and look for side effects. Several months to a year are needed, and only about 70% of drugs pass at this point. Phase II clinical trials next test the drug for 1 to 2 years in several hundred patients with the target disease or condition, looking both for safety and efficacy, and only about 33% of the original group pass. Finally, phase III trials are undertaken on a large sample of patients to document definitively the drug’s safety, dosage, and efficacy. If the drug is one of the 25% of the original group that make it to the end of phase III, all the data are then gathered into a New Drug Application (NDA) and sent to the FDA for review and approval, which can take another 2 years. Ten years have elapsed and at least \$500 million has been spent, with only a 20% success rate for the drugs that began testing. Finally, though, the drug will begin to appear in medicine cabinets. The following timeline shows the process.
6.14: Key Terms
6 • Key Terms 6 • Key Terms
• activation energy (ΔG)
• active site
• addition reaction
• bond dissociation energy (D)
• carbocation
• electrophile
• elimination reaction
• endergonic
• endothermic
• enthalpy change (ΔH)
• entropy change (ΔS)
• enzyme
• exergonic
• exothermic
• Gibbs free-energy change (ΔG)
• heat of reaction
• nucleophile
• polar reaction
• polarizability
• radical
• radical reaction
• reaction coordinate
• reaction intermediate
• reaction mechanism
• rearrangement reaction
• substitution reaction
• transition state
6.15: Summary
6 • Summary 6 • Summary
All chemical reactions, whether in the laboratory or in living organisms, follow the same chemical rules. To understand both organic and biological chemistry, it’s necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we’ve taken a brief look at the fundamental kinds of organic reactions, we’ve seen why reactions occur, and we’ve seen how reactions can be described.
There are four common kinds of reactions: addition reactions take place when two reactants add together to give a single product; elimination reactions take place when one reactant splits apart to give two products; substitution reactions take place when two reactants exchange parts to give two new products; and rearrangement reactions take place when one reactant undergoes a reorganization of bonds and atoms to give an isomeric product.
A full description of how a reaction occurs is called its mechanism. There are two general kinds of mechanisms by which most reactions take place: radical mechanisms and polar mechanisms. Polar reactions, the more common type, occur because of an attractive interaction between a nucleophilic (electron-rich) site in one molecule and an electrophilic (electron-poor) site in another molecule. A bond is formed in a polar reaction when the nucleophile donates an electron pair to the electrophile. This transfer of electrons is indicated by a curved arrow showing the direction of electron travel from the nucleophile to the electrophile. Radical reactions involve species that have an odd number of electrons. A bond is formed when each reactant donates one electron.
The energy changes that take place during reactions can be described by considering both rates (how fast the reactions occur) and equilibria (how much the reactions occur). The position of a chemical equilibrium is determined by the value of the free-energy change (ΔG) for the reaction, where ΔG = ΔHTΔS. The enthalpy term (ΔH) corresponds to the net change in strength of chemical bonds broken and formed during the reaction; the entropy term (ΔS) corresponds to the change in the amount of molecular randomness during the reaction. Reactions that have negative values of ΔG release energy, are said to be exergonic, and have favorable equilibria. Reactions that have positive values of ΔG absorb energy, are said to be endergonic, and have unfavorable equilibria.
A reaction can be described pictorially using an energy diagram that follows the reaction course from reactants through transition state to product. The transition state is an activated complex occurring at the highest-energy point of a reaction. The amount of energy needed by reactants to reach this high point is the activation energy, ΔG. The higher the activation energy, the slower the reaction.
Many reactions take place in more than one step and involve the formation of a reaction intermediate. An intermediate is a species that lies at an energy minimum between steps on the reaction curve and is formed briefly during the course of a reaction. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.13%3A_Chemistry_MattersWhere_Do_Drugs_Come_From.txt |
6 • Additional Problems 6 • Additional Problems
Visualizing Chemistry
Problem 6-14
The following alkyl halide can be prepared by addition of HBr to two different alkenes. Draw the structures of both (reddish-brown = Br).
Problem 6-15
The following structure represents the carbocation intermediate formed in the addition reaction of HBr to two different alkenes. Draw the structures of both.
Problem 6-16 Electrostatic potential maps of (a) formaldehyde (CH2O) and (b) methanethiol (CH3SH) are shown. Is the formaldehyde carbon atom likely to be electrophilic or nucleophilic? What about the methanethiol sulfur atom? Explain. (a) (b) Problem 6-17 Look at the following energy diagram: (a) Is ΔG° for the reaction positive or negative? Label it on the diagram. (b)
How many steps are involved in the reaction?
(c) How many transition states are there? Label them on the diagram.
Problem 6-18 Look at the following energy diagram for an enzyme-catalyzed reaction: (a)
How many steps are involved?
(b) Which step is most exergonic? (c) Which step is slowest?
Energy Diagrams and Reaction Mechanisms
Problem 6-19
What is the difference between a transition state and an intermediate?
Problem 6-20
Draw an energy diagram for a one-step reaction with Keq < 1. Label the parts of the diagram corresponding to reactants, products, transition state, ΔG°, and ΔG. Is ΔG° positive or negative?
Problem 6-21
Draw an energy diagram for a two-step reaction with Keq > 1. Label the overall ΔG°, transition states, and intermediate. Is ΔG° positive or negative?
Problem 6-22
Draw an energy diagram for a two-step exergonic reaction whose second step is faster than its first step.
Problem 6-23
Draw an energy diagram for a reaction with Keq = 1. What is the value of ΔG° in this reaction?
Problem 6-24 The addition of water to ethylene to yield ethanol has the following thermodynamic parameters: (a)
Is the reaction exothermic or endothermic?
(b) Is the reaction favorable (spontaneous) or unfavorable (nonspontaneous) at room temperature (298 K)?
Problem 6-25 When isopropylidenecyclohexane is treated with strong acid at room temperature, isomerization occurs by the mechanism shown below to yield 1-isopropylcyclohexene: At equilibrium, the product mixture contains about 30% isopropylidenecyclohexane and about 70% 1-isopropylcyclohexene. (a)
What is an approximate value of Keq for the reaction?
(b) Since the reaction occurs slowly at room temperature, what is its approximate ΔG? (c) Draw an energy diagram for the reaction.
Problem 6-26
Add curved arrows to the mechanism shown in Problem 6-25 to indicate the electron movement in each step.
Problem 6-27 Draw the complete mechanism for each of the following polar reactions. (a) (b) (c) Problem 6-28 Use curved arrows to show the flow of electrons, and draw the carbon radical that is formed when the halogen radicals below add to the corresponding alkenes. (a) (b) (c) Polar Reactions Problem 6-29 Identify the functional groups in the following molecules, and show the polarity of each: (a) (b) (c) (d) (e) (f) Problem 6-30 Identify the following reactions as additions, eliminations, substitutions, or rearrangements: (a) (b) (c) (d) Problem 6-31 Identify the likely electrophilic and nucleophilic sites in each of the following molecules: (a) (b) Problem 6-32 Identify the electrophile and the nucleophile. (a) (b) (c) Problem 6-33 Add curved arrows to the following polar reactions to indicate the flow of electrons in each: (a) (b) Problem 6-34 Follow the flow of electrons indicated by the curved arrows in each of the following polar reactions, and predict the products that result: (a) (b) (c) (d) Radical Reactions Problem 6-35 Radical chlorination of pentane is a poor way to prepare 1-chloropentane, but radical chlorination of neopentane, (CH3)4C, is a good way to prepare neopentyl chloride, (CH3)3CCH2Cl. Explain. Problem 6-36 Despite the limitations of radical chlorination of alkanes, the reaction is still useful for synthesizing certain halogenated compounds. For which of the following compounds does radical chlorination give a single monochloro product? (a) (b) (c) (d) (e) (f) Problem 6-37 Draw the different monochlorinated constitutional isomers you would obtain by the radical chlorination of the following compounds. (a) (b) (c) Problem 6-38 Answer question 6-37 taking all stereoisomers into account. Problem 6-39 Show the structure of the carbocation that would result when each of the following alkenes reacts with an acid, H+. (a) (b) (c) General Problems Problem 6-40 2-Chloro-2-methylpropane reacts with water in three steps to yield 2-methyl-2-propanol. The first step is slower than the second, which in turn is much slower than the third. The reaction takes place slowly at room temperature, and the equilibrium constant is approximately 1. (a) Give approximate values for ΔG‡ and ΔG° that are consistent with the above information. (b)
Draw an energy diagram for the reaction, labeling all points of interest and placing relative energy levels on the diagram consistent with the information given.
Problem 6-41
Add curved arrows to the mechanism shown in Problem 6-40 to indicate the electron movement in each step.
Problem 6-42
The reaction of hydroxide ion with chloromethane to yield methanol and chloride ion is an example of a general reaction type called a nucleophilic substitution reaction:
HO + CH3Cl ⇄ CH3OH + Cl
The value of ΔH° for the reaction is −75 kJ/mol, and the value of ΔS° is +54 J/(K·mol). What is the value of ΔG° (in kJ/mol) at 298 K? Is the reaction exothermic or endothermic? Is it exergonic or endergonic?
Problem 6-43
Methoxide ion (CH3O) reacts with bromoethane in a single step according to the following equation:
Identify the bonds broken and formed, and draw curved arrows to represent the flow of electrons during the reaction.
Problem 6-44
Ammonia reacts with acetyl chloride (CH3COCl) to give acetamide (CH3CONH2). Identify the bonds broken and formed in each step of the reaction, and draw curved arrows to represent the flow of electrons in each step.
Problem 6-45 The naturally occurring molecule α-terpineol is biosynthesized by a route that includes the following step: (a)
Propose a likely structure for the isomeric carbocation intermediate.
(b) Show the mechanism of each step in the biosynthetic pathway, using curved arrows to indicate electron flow.
Problem 6-46 Predict the product(s) of each of the following biological reactions by interpreting the flow of electrons as indicated by the curved arrows: (a) (b) (c) Problem 6-47
Reaction of 2-methylpropene with HBr might, in principle, lead to a mixture of two alkyl bromide addition products. Name them, and draw their structures.
Problem 6-48
Draw the structures of the two carbocation intermediates that might form during the reaction of 2-methylpropene with HBr (Problem 6-47). We’ll see in the next chapter that the stability of carbocations depends on the number of alkyl substituents attached to the positively charged carbon—the more alkyl substituents there are, the more stable the cation. Which of the two carbocation intermediates you drew is more stable? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/06%3A_An_Overview_of_Organic_Reactions/6.16%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 7, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. describe the importance of alkenes to the chemical industry.
3. use the concept of “degree of unsaturation” in determining chemical structures.
4. describe the electronic structure and geometry of alkenes.
5. describe the factors that influence alkene stability, and determine the relative stability of a number of given alkenes.
6. write the IUPAC name of a given alkene, and draw the structure of any alkene, given its IUPAC name.
7. determine whether a given alkene has an E configuration or a Z configuration.
8. explain why alkenes are more reactive than alkanes.
9. describe the reaction between an alkene and a hydrogen halide, and explain why one product is formed rather than another. Base your explanation on the concepts of carbocation stability and the Hammond postulate.
10. define, and use in context, the key terms introduced in this chapter.
This, the first of two chapters devoted to the chemistry of alkenes, describes how certain alkenes occur naturally, then shows the industrial importance of ethylene and propylene (the simplest members of the alkene family). The electronic structure of alkenes is reviewed, and their nomenclature discussed in detail. After dealing with the question of cis-trans isomerism in alkenes, Chapter 7 introduces the reactivity of the carbon-carbon double bond. The chapter then focuses on one specific reaction—the addition of hydrogen halides to alkenes—to raise a number of important concepts, including carbocation stability and the Hammond postulate.
• 7.1: Why This Chapter?
• 7.2: Industrial Preparation and Use of Alkenes
Among the most important and most abundant organic chemicals produced worldwide are the two simple alkenes, ethylene and propylene. They are used as the starting materials to synthesize numerous valuable compounds.
• 7.3: Calculating Degree of Unsaturation
Calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of π bonds and/or cyclic rings.
• 7.4: Naming Alkenes
Alkenes contain carbon-carbon double bonds and are unsaturated hydrocarbons with the molecular formula is CnH2n; this is also the same molecular formula as cycloalkanes. For straight chain alkenes, it is the same basic rules as nomenclature of alkanes except change the suffix to "-ene."
• 7.5: Cis-Trans Isomerism in Alkenes
Geometric isomerism (also known as cis-trans isomerism or E-Z isomerism) is a form of stereoisomerism. Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements (structural isomerism) which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Structural isomerism is not a form of stereoisomerism, and is dealt with in a separate Module.
• 7.6: Alkene Stereochemistry and the E,Z Designation
The traditional system for naming the geometric isomers of an alkene, in which the same groups are arranged differently, is to name them as cis or trans. However, it is easy to find examples where the cis-trans system is not easily applied. IUPAC has a more complete system for naming alkene isomers. The R-S system is based on a set of "priority rules", which allow you to rank any groups. The IUPAC system for naming alkene isomers, called the E-Z system, is based on the same priority rules.
• 7.7: Stability of Alkenes
Alkene hydrogenation is the syn-addition of hydrogen to an alkene, saturating the bond. The alkene reacts with hydrogen gas in the presence of a metal catalyst which allows the reaction to occur quickly. The energy released in this process, called the heat of hydrogenation, indicates the relative stability of the double bond in the molecule.
• 7.8: Electrophilic Addition Reactions of Alkenes
This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with hydrogen halides such as hydrogen chloride and hydrogen bromide. Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.
• 7.9: Orientation of Electrophilic Additions - Markovnikov's Rule
This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with hydrogen halides such as hydrogen chloride and hydrogen bromide. Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.
• 7.10: Carbocation Structure and Stability
It is a general principle in chemistry that the more a charge is dispersed, the more stable is the species carrying the charge. Put simply, a species in which a positive charge is shared between two atoms would be more stable than a similar species in which the charge is borne wholly by a single atom.
• 7.11: The Hammond Postulate
The Hammond postulate states that a transition state resembles the structure of the nearest stable species. For an exergonic reaction, therefore, the transition state resembles the reactants more than it does the products.
• 7.12: Evidence for the Mechanism of Electrophilic Additions - Carbocation Rearrangements
Whenever possible, carbocations will rearrange from a less stable isomer to a more stable isomer. This rearrangement can be achieved by either a hydride shift, where a hydrogen atom migrates from one carbon atom to the next, taking a pair of electrons with it; or an alkyl shift, in which an alkyl group undergoes a similar migration, again taking a bonding pair of electrons with it. These migrations usually occur between neighboring carbon atoms.
• 7.13: Chemistry Matters—Bioprospecting- Hunting for Natural Products
• 7.14: Key Terms
Alkenes are a class of hydrocarbons (i.e., containing only carbon and hydrogen). They are unsaturated compounds with at least one carbon-to-carbon double bond. The double bond makes Alkenes more reactive than alkanes. Olefin is another term used to describe alkenes.
• 7.15: Summary
• 7.16: Additional Problems
Thumbnail: Ball-and-stick model of the ethylene (ethene) molecule, \(\ce{C2H4}\). (Public Domain; Ben Mills via Wikipedia)
07: Alkenes- Structure and Reactivity
Chapter Contents
7.1 Industrial Preparation and Use of Alkenes
7.2 Calculating the Degree of Unsaturation 7.3 Naming Alkenes 7.4 Cis–Trans Isomerism in Alkenes 7.5 Alkene Stereochemistry and the E,Z Designation 7.6 Stability of Alkenes 7.7 Electrophilic Addition Reactions of Alkenes 7.8 Orientation of Electrophilic Additions: Markovnikov’s Rule 7.9 Carbocation Structure and Stability 7.10 The Hammond Postulate 7.11 Evidence for the Mechanism of Electrophilic Additions: Carbocation Rearrangements
7 • Why This Chapter?
Carbon–carbon double bonds are present in most organic and biological molecules, so a good understanding of their behavior is needed. In this chapter, we’ll look at some consequences of alkene stereoisomerism and then focus on the broadest and most general class of alkene reactions, the electrophilic addition reaction. Carbon-carbon triple bonds, by contrast, occur much less commonly, so we’ll not spend much time on their chemistry.
An alkene, sometimes called an olefin from the German term for oil forming, is a hydrocarbon that contains a carbon–carbon double bond, while an alkyne is a hydrocarbon that contains a carbon-carbon triple bond. Alkenes occur abundantly in nature. Ethylene, for instance, is a plant hormone that induces ripening in fruit, and α-pinene is the major component of turpentine. Lycopene, found in fruits such as watermelon and papaya as well as tomatoes, is an antioxidant with numerous health benefits such as sun protection and cardiovascular protection. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.01%3A_Why_This_Chapter.txt |
Ethylene and propylene, the simplest alkenes, are the two most important organic chemicals produced industrially. Approximately 220 million tons of ethylene and 138 million tons of propylene are produced worldwide each year for use in the synthesis of polyethylene, polypropylene, ethylene glycol, acetic acid, acetaldehyde, and a host of other substances (Figure 7.2).
Ethylene, propylene, and butene are synthesized from (C2–C8) alkanes by a process called steam cracking at temperatures up to 900 °C.
The cracking process is complex, although it undoubtedly involves radical reactions. The high-temperature reaction conditions cause spontaneous breaking of C−C and C−H bonds, with the resultant formation of smaller fragments. We might imagine, for instance, that a molecule of butane splits into two ethyl radicals, each of which then loses a hydrogen atom to generate two molecules of ethylene.
Steam cracking is an example of a reaction whose energetics are dominated by entropy (ΔS°) rather than by enthalpy (ΔH°) in the free-energy equation ΔG° = ΔH° − TΔS° discussed in Section 6.7. Although the bond dissociation energy D for a carbon–carbon single bond is relatively high (about 370 kJ/mol) and cracking is endothermic, the large positive entropy change resulting from the fragmentation of one large molecule into several smaller pieces, together with the high temperature, makes the TΔS° term larger than the ΔH° term, thereby favoring the cracking reaction.
7.03: Calculating Degree of Unsaturation
Because of its double bond, an alkene has fewer hydrogens than an alkane with the same number of carbons—CnH2n for an alkene versus CnH2n+2 for an alkane—and is therefore referred to as unsaturated. Ethylene, for example, has the formula C2H4, whereas ethane has the formula C2H6.
In general, each ring or double bond in a molecule corresponds to a loss of two hydrogens from the alkane formula CnH2n+2. Knowing this relationship, it’s possible to work backward from a molecular formula to calculate a molecule’s degree of unsaturation—the number of rings and/or multiple bonds present in the molecule.
Let’s assume that we want to find the structure of an unknown hydrocarbon. A molecular weight determination yields a value of 82 amu, which corresponds to a molecular formula of C6H10. Since the saturated C6 alkane (hexane) has the formula C6H14, the unknown compound has two fewer pairs of hydrogens (H14 − H10 = H4 = 2 H2) so its degree of unsaturation is 2. The unknown therefore contains either two double bonds, one ring and one double bond, two rings, or one triple bond. There’s still a long way to go to establish its structure, but the simple calculation has told us a lot about the molecule.
Similar calculations can be carried out for compounds containing elements other than just carbon and hydrogen.
• Organohalogen compounds (C, H, X, where X = F, Cl, Br, or I) A halogen substituent acts as a replacement for hydrogen in an organic molecule, so we can add the number of halogens and hydrogens to arrive at an equivalent hydrocarbon formula from which the degree of unsaturation can be found. For example, the formula C4H6Br2 is equivalent to the hydrocarbon formula C4H8 and thus corresponds to one degree of unsaturation.
• Organooxygen compounds (C, H, O) Oxygen forms two bonds, so it doesn’t affect the formula of an equivalent hydrocarbon and can be ignored when calculating the degree of unsaturation. You can convince yourself of this by seeing what happens when an oxygen atom is inserted into an alkane bond: C−C becomes C−O−C or C−H becomes C−O−H, and there is no change in the number of hydrogen atoms. For example, the formula C5H8O is equivalent to the hydrocarbon formula C5H8 and thus corresponds to two degrees of unsaturation.
• Organonitrogen compounds (C, H, N) Nitrogen forms three bonds, so an organonitrogen compound has one more hydrogen than a related hydrocarbon. We therefore subtract the number of nitrogens from the number of hydrogens to arrive at the equivalent hydrocarbon formula. Again, you can convince yourself of this by seeing what happens when a nitrogen atom is inserted into an alkane bond: C−C becomes C−NH−C or C−H becomes C−NH2, meaning that one additional hydrogen atom has been added. We must therefore subtract this extra hydrogen atom to arrive at the equivalent hydrocarbon formula. For example, the formula C5H9N is equivalent to C5H8 and thus has two degrees of unsaturation.
To summarize:
• Add the number of halogens to the number of hydrogens.
• Ignore the number of oxygens.
• Subtract the number of nitrogens from the number of hydrogens.
Problem 7-1 Calculate the degree of unsaturation in each of the following formulas, and then draw as many structures as you can for each: (a)
C4H8
(b) C4H6 (c) C3H4
Problem 7-2 Calculate the degree of unsaturation in each of the following formulas: (a)
C6H5N
(b) C6H5NO2 (c) C8H9Cl3 (d) C9H16Br2 (e) C10H12N2O3 (f) C20H32ClN
Problem 7-3
Diazepam, marketed as an antianxiety medication under the name Valium, has three rings, eight double bonds, and the formula C16H?ClN2O. How many hydrogens does diazepam have? (Calculate the answer; don’t count hydrogens in the structure.) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.02%3A_Industrial_Preparation_and_Use_of_Alkenes.txt |
Alkenes are named using a series of rules similar to those for alkanes (Section 3.4), with the suffix -ene used instead of -ane to identify the functional group. There are three steps to this process.
STEP 1
Name the parent hydrocarbon. Find the longest carbon chain containing the double bond, and name the compound accordingly, using the suffix -ene:
STEP 2
Number the carbon atoms in the chain. Begin at the end nearer the double bond or, if the double bond is equidistant from the two ends, begin at the end nearer the first branch point. This rule ensures that the double-bond carbons receive the lowest possible numbers.
STEP 3
Write the full name. Number the substituents according to their positions in the chain, and list them alphabetically. Indicate the position of the double bond by giving the number of the first alkene carbon and placing that number directly before the parent name. If more than one double bond is present, indicate the position of each and use one of the suffixes -diene, -triene, and so on.
We should also note that IUPAC changed their naming recommendations in 1993 to place the locant indicating the position of the double bond immediately before the -ene suffix rather than before the parent name: but-2-ene rather than 2-butene, for instance. This change has not been widely accepted by the chemical community in the United States, however, so we’ll stay with the older but more commonly used names. Be aware, though, that you may occasionally encounter the newer system.
Cycloalkenes are named similarly, but because there is no chain end to begin from, we number the cycloalkene so that the double bond is between C1 and C2 and the first substituent has as low a number as possible. It’s not necessary to indicate the position of the double bond in the name because it’s always between C1 and C2. As with open-chain alkenes, the newer but not yet widely accepted naming rules place the locant immediately before the suffix in a cyclic alkene.
For historical reasons, there are a few alkenes whose names are firmly entrenched in common usage but don’t conform to the rules. For example, the alkene derived from ethane should be called ethene, but the name ethylene has been used for so long that it is accepted by IUPAC. Table 7.1 lists several other common names that are often used and are recognized by IUPAC. Note also that a =CH2 substituent is called a methylene group, a $H2C═CH–H2C═CH–$ substituent is called a vinyl group, and a $H2C═CHCH2–H2C═CHCH2–$ substituent is called an allyl group.
Table 7.1 Common Names of Some Alkenes
Compound Systematic name Common name
$H2C═CH2H2C═CH2$ Ethene Ethylene
$CH3CH═CH2CH3CH═CH2$ Propene Propylene
2-Methylpropene Isobutylene
2-Methyl-1,3-butadiene Isoprene
(b)
(c)
(d)
Problem 7-5 Draw structures corresponding to the following IUPAC names: (a) 2-Methyl-1,5-hexadiene (b)
3-Ethyl-2,2-dimethyl-3-heptene
(c) 2,3,3-Trimethyl-1,4,6-octatriene (d) 3,4-Diisopropyl-2,5-dimethyl-3-hexene
Problem 7-6 Name the following cycloalkenes: (a) (b) (c) Problem 7-7 Change the following old names to new, post-1993 names, and draw the structure of each compound: (a) 2,5,5-Trimethyl-2-hexene (b)
2,3-Dimethyl-1,3-cyclohexadiene
7.05: Cis-Trans Isomerism in Alkenes
We saw in the chapter on Structure and Bonding that the carbon–carbon double bond can be described in two ways. In valence bond language (Section 1.8), the carbons are sp2-hybridized and have three equivalent hybrid orbitals that lie in a plane at angles of 120° to one another. The carbons form a σ bond by a head-on overlap of sp2 orbitals and form a π bond by sideways overlap of unhybridized p orbitals oriented perpendicular to the sp2 plane, as shown in Figure 1.15.
In molecular orbital language (Section 1.11), interaction between the p orbitals leads to one bonding and one antibonding π molecular orbital. The π bonding MO has no node between nuclei and results from a combination of p orbital lobes with the same algebraic sign. The π antibonding MO has a node between nuclei and results from a combination of lobes with different algebraic signs, as shown in Figure 1.19.
Although essentially free rotation around single bonds is possible (Section 3.6), the same is not true of double bonds. For rotation to occur around a double bond, the π bond must break and re-form (Figure 7.3). Thus, the barrier to double-bond rotation must be at least as great as the strength of the π bond itself, an estimated 350 kJ/mol (84 kcal/mol). Recall that the barrier to bond rotation in ethane is only 12 kJ/mol.
The lack of rotation around carbon–carbon double bonds is of more than just theoretical interest; it also has chemical consequences. Imagine the situation for a disubstituted alkene such as 2-butene. (Disubstituted means that two substituents other than hydrogen are bonded to the double-bond carbons.) The two methyl groups in 2-butene can either be on the same side of the double bond or on opposite sides, a situation similar to that in disubstituted cycloalkanes (Section 4.2).
Since bond rotation can’t occur, the two 2-butenes can’t spontaneously interconvert; they are different, isolable compounds. As with disubstituted cycloalkanes, we call such compounds cis–trans stereoisomers. The compound with substituents on the same side of the double bond is called cis-2-butene, and the isomer with substituents on opposite sides is trans-2-butene (Figure 7.4).
Cis–trans isomerism is not limited to disubstituted alkenes. It can occur whenever both double-bond carbons are attached to two different groups. If one of the double-bond carbons is attached to two identical groups, however, cis–trans isomerism is not possible (Figure 7.5).
Problem 7-8
The sex attractant of the common housefly is an alkene named cis-9-tricosene. Draw its structure. (Tricosane is the straight-chain alkane C23H48.)
Problem 7-9
Which of the following compounds can exist as pairs of cis–trans isomers? Draw each pair, and indicate the geometry of each isomer.
(a) CH3CH$\text{=}$CH2(b) (CH3)2C$\text{=}$CHCH3(c) CH3CH2CH$\text{=}$CHCH3(d) (CH3)2C$\text{=}$C(CH3)CH2CH3
(e) ClCH$\text{=}$CHCl (f) BrCH$\text{=}$CHCl
Problem 7-10 Name the following alkenes, including a cis or trans designation: (a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.04%3A_Naming_Alkenes.txt |
The cis–trans naming system used in the previous section works only with disubstituted alkenes—compounds that have two substituents other than hydrogen on the double bond. With trisubstituted and tetrasubstituted double bonds, a more general method is needed for describing double-bond geometry. (Trisubstituted means three substituents other than hydrogen on the double bond; tetrasubstituted means four substituents other than hydrogen.)
The method used for describing alkene stereochemistry is called the E,Z system and employs the same Cahn–Ingold–Prelog sequence rules given in Section 5.5 for specifying the configuration of a chirality center. Let’s briefly review the sequence rules and then see how they’re used to specify double-bond geometry. For a more thorough review, reread Section 5.5.
RULE 1
Considering each of the double-bond carbons separately, look at the two substituents attached and rank them according to the atomic number of the first atom in each (8 for oxygen, 6 for carbon, 1 for hydrogen, and so forth). An atom with higher atomic number ranks higher than an atom with lower atomic number.
RULE 2
If a decision can’t be reached by ranking the first atoms in the two substituents, look at the second, third, or fourth atoms away from the double-bond until the first difference is found.
RULE 3
Multiple-bonded atoms are equivalent to the same number of single-bonded atoms.
Once the two groups attached to each double-bonded carbon have been ranked as either higher or lower, look at the entire molecule. If the higher-ranked groups on each carbon are on the same side of the double bond, the alkene is said to have a Z configuration, for the German zusammen, meaning “together.” If the higher-ranked groups are on opposite sides, the alkene has an E configuration, for the German entgegen, meaning “opposite.” (For a simple way to remember which is which, note that the groups are on “ze zame zide” in the Z isomer.)
As an example, look at the following two isomers of 2-chloro-2-butene. Because chlorine has a higher atomic number than carbon, a −Cl substituent is ranked higher than a −CH3 group. Methyl is ranked higher than hydrogen, however, so isomer (a) is designated E because the higher-ranked groups are on opposite sides of the double bond. Isomer (b) has a Z configuration because its higher-ranked groups are on ze zame zide of the double bond.
For further practice, work through each of the following examples to convince yourself that the assignments are correct:
Worked Example 7.1
Assigning E and Z Configurations to Alkenes
Assign E or Z configuration to the double bond in the following compound:
Strategy
Look at the two substituents connected to each double-bonded carbon, and determine their ranking using the Cahn–Ingold–Prelog rules. Then, check whether the two higher-ranked groups are on the same or opposite sides of the double bond.
Solution
The left-hand carbon has −H and −CH3 substituents, of which −CH3 ranks higher by sequence rule 1. The right-hand carbon has −CH(CH3)2 and −CH2OH substituents, which are equivalent by rule 1. By rule 2, however, −CH2OH ranks higher than −CH(CH3)2 because the substituent −CH2OH has an oxygen as its highest second atom, but −CH(CH3)2 has a carbon as its highest second atom. The two higher-ranked groups are on the same side of the double bond, so we assign a Z configuration.
Problem 7-11 Which member in each of the following sets ranks higher? (a)
−H or −CH3
(b) −Cl or −CH2Cl (c) –CH2CH2Br or –CH$\text{=}$CH2 (d) −NHCH3 or −OCH3 (e) −CH2OH or −CH$\text{=}$O (f) −CH2OCH3 or −CH$\text{=}$O
Problem 7-12 Rank the substituents in each of the following sets according to the sequence rules: (a)
−CH3, −OH, −H, −Cl
(b) −CH3, −CH2CH3, −CH$\text{=}$CH2, −CH2OH (c) −CO2H, −CH2OH, −C$\text{≡}$N, −CH2NH2 (d) −CH2CH3, −C$\text{≡}$CH, −C$\text{≡}$N, −CH2OCH3
Problem 7-13 Assign E or Z configuration to the following alkenes: (a) (b) (c) (d) Problem 7-14
Assign stereochemistry (E or Z) to the double bond in the following compound, and convert the drawing into a skeletal structure (red = O): | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.06%3A_Alkene_Stereochemistry_and_the_EZ_Designation.txt |
Although the cis–trans interconversion of alkene isomers does not occur spontaneously, it can often be made to happen by treating the alkene with a strong acid catalyst. If we interconvert cis-2-butene with trans-2-butene and allow them to reach equilibrium, we find that they aren’t of equal stability. The trans isomer is more stable than the cis isomer by 2.8 kJ/mol (0.66 kcal/mol) at room temperature, corresponding to a 76 : 24 ratio.
Cis alkenes are less stable than their trans isomers because of steric strain between the two larger substituents on the same side of the double bond. This is the same kind of steric interference that we saw previously in the axial conformation of methylcyclohexane (Section 4.7).
Although it’s sometimes possible to find relative stabilities of alkene isomers by establishing a cis–trans equilibrium through treatment with strong acid, a more general method is to take advantage of the fact that alkenes undergo a hydrogenation reaction to give the corresponding alkane when treated with H2 gas in the presence of a catalyst such as palladium or platinum.
Energy diagrams for the hydrogenation reactions of cis- and trans-2-butene are shown in Figure 7.6. Because cis-2-butene is less stable than trans-2-butene by 2.8 kJ/mol, the energy diagram shows the cis alkene at a higher energy level. After reaction, however, both curves are at the same energy level (butane). It therefore follows that ΔG° for reaction of the cis isomer must be larger than ΔG° for reaction of the trans isomer by 2.8 kJ/mol. In other words, more energy is released in the hydrogenation of the cis isomer than the trans isomer because the cis isomer has more energy to begin with.
If we were to measure the so-called heats of hydrogenation (ΔH°hydrog) for two double-bond isomers and find their difference, we could determine the relative stabilities of cis and trans isomers without having to measure an equilibrium position. cis-2-Butene, for instance, has ΔH°hydrog = −119 kJ/mol(−28.3 kcal/mol), while trans-2-butene has ΔH°hydrog = −115 kJ/mol(−27.4 kcal/mol)—a difference of 4 kJ/mol.
The 4 kJ/mol energy difference between the 2-butene isomers calculated from heats of hydrogenation agrees reasonably well with the 2.8 kJ/mol energy difference calculated from equilibrium data, but the values aren’t exactly the same for two reasons. First, there is probably some experimental error, because heats of hydrogenation are difficult to measure accurately. Second, heats of reaction and equilibrium constants don’t measure exactly the same thing. Heats of reaction measure enthalpy changes, ΔH°, whereas equilibrium constants measure free-energy changes, ΔG°, so we might expect a slight difference between the two.
Table 7.2 Heats of Hydrogenation of Some Alkenes
Substitution Alkene ΔH°hydrog
(kJ/mol) (kcal/mol)
Ethylene $H2C═CH2H2C═CH2$ −136 −32.6
Monosubstituted $CH3CH═CH2CH3CH═CH2$ −125 −29.9
Disubstituted $CH3CH═CHCH3CH3CH═CHCH3$ (cis)
$CH3CH═CHCH3CH3CH═CHCH3$ (trans)
$(CH3)2C═CH2(CH3)2C═CH2$
−119
−115
−118
−28.3
−27.4
−28.2
Trisubstituted $(CH3)2C═CHCH3(CH3)2C═CHCH3$ −112 −26.7
Tetrasubstituted $(CH3)2C═C(CH3)2(CH3)2C═C(CH3)2$ −110 −26.4
Table 7.2 lists some representative data for the hydrogenation of different alkenes and shows that alkenes become more stable with increasing substitution. That is, alkenes follow the stability order:
The stability order of substituted alkenes is due to a combination of two factors. One is a stabilizing interaction between the $C═CFigure 7.7. The more substituents present on the double bond, the more hyperconjugation occurs and the more stable the alkene.$
A second factor that contributes to alkene stability involves bond strengths. A bond between a sp2 carbon and a sp3 carbon is somewhat stronger than a bond between two sp3 carbons. Thus, in comparing 1-butene and 2-butene, the monosubstituted isomer has one sp3sp3 bond and one sp3sp2 bond, while the disubstituted isomer has two sp3sp2 bonds. More highly substituted alkenes always have a higher ratio of sp3sp2 bonds to sp3sp3 bonds than less highly substituted alkenes and are therefore more stable.
Problem 7-15 Name the following alkenes, and tell which compound in each pair is more stable: (a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.07%3A_Stability_of_Alkenes.txt |
Before beginning a detailed discussion of alkene reactions, let’s review briefly some conclusions from the previous chapter. We said in Section 6.5 that alkenes behave as nucleophiles (Lewis bases) in polar reactions, donating a pair of electrons from their electron-rich $C═CC═C$ bond to an electrophile (Lewis acid). For example, reaction of 2-methylpropene with HBr yields 2-bromo-2-methylpropane. A careful study of this and similar reactions by the British chemist Christopher Ingold and others in the 1930s led to the generally accepted mechanism shown in Figure 7.8 for an electrophilic addition reaction.
Figure 7.8 MECHANISM Mechanism of the electrophilic addition of HBr to 2-methylpropene. The reaction occurs in two steps, protonation and bromide addition, and involves a carbocation intermediate.
The reaction begins with an attack on the hydrogen of the electrophile HBr by the electrons of the nucleophilic π bond. Two electrons from the π bond form a new σ bond between the entering hydrogen and an alkene carbon, as shown by the curved arrow at the top of Figure 7.8. The resulting carbocation intermediate is itself an electrophile, which can accept an electron pair from nucleophilic Br ion to form a C−Br bond and yield a neutral addition product.
An energy diagram for the overall electrophilic addition reaction (Figure 7.9) has two peaks (transition states) separated by a valley (carbocation intermediate). The energy level of the intermediate is higher than that of the starting alkene, but the reaction as a whole is exergonic (negative ΔG°). The first step, protonation of the alkene to yield the intermediate cation, is relatively slow. But once the cation intermediate is formed, it rapidly reacts to yield the final alkyl bromide product. The relative rates of the two steps are indicated in Figure 7.9 by the fact that ΔG1 is larger than ΔG2.
Electrophilic addition to alkenes is successful not only with HBr but with HCl, HI, and H2O as well. Note that HI is usually generated in the reaction mixture by treating potassium iodide with phosphoric acid and that a strong acid catalyst is needed for the addition of water.
Writing Organic Reactions
This is a good place to mention that the equations for organic reactions are sometimes written in different ways to emphasize different points. In describing a laboratory process, for instance, the reaction of 2-methylpropene with HCl might be written in the format A + B ⟶ C to emphasize that both reactants are equally important for the purposes of the discussion. The solvent and notes about other reaction conditions such as temperature are written either above or below the reaction arrow.
Alternatively, we might write the same reaction in a format to emphasize that 2-methylpropene is the reactant whose chemistry is of greater interest. The second reactant, HCl, is placed above the reaction arrow together with notes about solvent and reaction conditions.
In describing a biological process, the reaction is almost always written to show only the structures of the primary reactant and product, while abbreviating the structures of various biological “reagents” and by-products with a curved arrow that intersects the straight reaction arrow. As discussed in Section 6.11, the reaction of glucose with ATP to give glucose 6-phosphate plus ADP would then be written as | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.08%3A_Electrophilic_Addition_Reactions_of_Alkenes.txt |
Look carefully at the electrophilic addition reactions shown in the previous section. In each case, an unsymmetrically substituted alkene has given a single addition product rather than the mixture that might be expected. For example, 2-methylpropene might react with HCl to give both 2-chloro-2-methylpropane and 1-chloro-2-methylpropane, but it doesn’t. It gives only 2-chloro-2-methylpropane as the sole product. Similarly, it’s invariably the case in biological alkene addition reactions that only a single product is formed. We say that such reactions are regiospecific (ree-jee-oh-specific) when only one of two possible orientations of an addition occurs.
After looking at the results of many such reactions, the Russian chemist Vladimir Markovnikov proposed in 1869 what has become known as:
Markovnikov’s rule
In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents.
When both double-bonded carbon atoms have the same degree of substitution, a mixture of addition products results.
Because carbocations are involved as intermediates in these electrophilic addition reactions, Markovnikov’s rule can be restated in the following way:
Markovnikov’s rule restated
In the addition of HX to an alkene, the more highly substituted carbocation is formed as the intermediate rather than the less highly substituted one.
For example, addition of H+ to 2-methylpropene yields the intermediate tertiary carbocation rather than the alternative primary carbocation, and addition to 1-methylcyclohexene yields a tertiary cation rather than a secondary one. Why should this be?
Worked Example 7.2
Predicting the Product of an Electrophilic Addition Reaction
What product would you expect from reaction of HCl with 1-ethylcyclopentene?
Strategy
When solving a problem that asks you to predict a reaction product, begin by looking at the functional group(s) in the reactants and deciding what kind of reaction is likely to occur. In the present instance, the reactant is an alkene that will probably undergo an electrophilic addition reaction with HCl. Next, recall what you know about electrophilic addition reactions to predict the product. You know that electrophilic addition reactions follow Markovnikov’s rule, so H+ will add to the double-bond carbon that has one alkyl group (C2 on the ring) and the Cl will add to the double-bond carbon that has two alkyl groups (C1 on the ring).
Solution
The expected product is 1-chloro-1-ethylcyclopentane.
Worked Example 7.3
Synthesizing a Specific Compound
What alkene would you start with to prepare the following alkyl halide? There may be more than one possibility.
Strategy
When solving a problem that asks how to prepare a given product, always work backward. Look at the product, identify the functional group(s) it contains, and ask yourself, “How can I prepare that functional group?” In the present instance, the product is a tertiary alkyl chloride, which can be prepared by reaction of an alkene with HCl. The carbon atom bearing the −Cl atom in the product must be one of the double-bond carbons in the reactant. Draw and evaluate all possibilities.
Solution
There are three possibilities, all of which could give the desired product according to Markovnikov’s rule.
Problem 7-16 Predict the products of the following reactions:
(a)
(b)
(c)
(d)
Problem 7-17 What alkenes would you start with to prepare the following products? (a)
(b)
(c)
(d) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.09%3A_Orientation_of_Electrophilic_Additions_-_Markovnikov%27s_Rule.txt |
To understand why Markovnikov’s rule works, we need to learn more about the structure and stability of carbocations and about the general nature of reactions and transition states. The first point to explore involves structure.
A great deal of experimental evidence has shown that carbocations are planar. The trivalent carbon is sp2-hybridized, and the three substituents are oriented toward the corners of an equilateral triangle, as indicated in Figure 7.10. Because there are only six valence electrons on carbon and all six are used in the three σ bonds, the p orbital extending above and below the plane is unoccupied.
The second point to explore involves carbocation stability. 2-Methylpropene might react with H+ to form a carbocation having three alkyl substituents (a tertiary ion, 3°), or it might react to form a carbocation having one alkyl substituent (a primary ion, 1°). Since the tertiary alkyl chloride, 2-chloro-2-methylpropane, is the only product observed, formation of the tertiary cation is evidently favored over formation of the primary cation. Thermodynamic measurements show that, indeed, the stability of carbocations increases with increasing substitution so that the stability order is tertiary > secondary > primary > methyl.
One way of determining carbocation stabilities is to measure the amount of energy required to form a carbocation by dissociation of the corresponding alkyl halide, R − X → R+ + :X. As shown in Figure 7.11, tertiary alkyl halides dissociate to give carbocations more easily than secondary or primary ones. Thus, trisubstituted carbocations are more stable than disubstituted ones, which are more stable than monosubstituted ones. The data in Figure 7.11 are taken from measurements made in the gas phase, but a similar stability order is found for carbocations in solution. The dissociation enthalpies are much lower in solution because polar solvents can stabilize the ions, but the order of carbocation stability remains the same.
Why are more highly substituted carbocations more stable than less highly substituted ones? There are at least two reasons. Part of the answer has to do with inductive effects, and part has to do with hyperconjugation. Inductive effects, discussed in Section 2.1 in connection with polar covalent bonds, result from the shifting of electrons in a σ bond in response to the electronegativity of nearby atoms. In the present instance, electrons from a relatively larger and more polarizable alkyl group can shift toward a neighboring positive charge more easily than the electron from a hydrogen. Thus, the more alkyl groups attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs (Figure 7.12).
Hyperconjugation, discussed in Section 7.6 in connection with the stabilities of substituted alkenes, is the stabilizing interaction between a p orbital and properly oriented C−H σ bonds on neighboring carbons that are roughly parallel to the p orbital. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconjugation and the more stable the carbocation. Figure 7.13 shows the molecular orbital for the ethyl carbocation, CH3CH2+, and indicates the difference between the C−H bond perpendicular to the cation p orbital and the two C−H bonds more parallel to the cation p orbital. Only these roughly parallel C−H bonds are oriented properly to take part in hyperconjugation.
Problem 7-18 Show the structures of the carbocation intermediates you would expect in the following reactions: (a)
(b)
Problem 7-19
Draw a skeletal structure of the following carbocation. Identify it as primary, secondary, or tertiary, and identify the hydrogen atoms that have the proper orientation for hyperconjugation in the conformation shown. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.10%3A_Carbocation_Structure_and_Stability.txt |
Let’s summarize what we’ve learned of electrophilic addition reactions to this point:
• Electrophilic addition to an unsymmetrically substituted alkene gives the more substituted carbocation intermediate. A more substituted carbocation forms faster than a less substituted one and, once formed, rapidly goes on to give the final product.
• A more substituted carbocation is more stable than a less substituted one. That is, the stability order of carbocations is tertiary > secondary > primary > methyl.
What we have not yet seen is how these two points are related. Why does the stability of the carbocation intermediate affect the rate at which it’s formed and thereby determine the structure of the final product? After all, carbocation stability is determined by the free-energy change ΔG°, but reaction rate is determined by the activation energy ΔG. The two quantities aren’t directly related.
Although there is no simple quantitative relationship between the stability of a carbocation intermediate and the rate of its formation, there is an intuitive relationship. It’s generally true when comparing two similar reactions that the more stable intermediate forms faster than the less stable one. The situation is shown graphically in Figure 7.14, where the energy profile in part (a) represents the typical situation, as opposed to the profile in part (b). That is, the curves for two similar reactions don’t cross one another.
Called the Hammond postulate, the explanation of the relationship between reaction rate and intermediate stability goes like this: Transition states represent energy maxima. They are high-energy activated complexes that occur transiently during the course of a reaction and immediately go on to a more stable species. Although we can’t actually observe transition states because they have no finite lifetime, the Hammond postulate says that we can get an idea of a particular transition state’s structure by looking at the structure of the nearest stable species. Imagine the two cases shown in Figure 7.15, for example. The reaction profile in part (a) shows the energy curve for an endergonic reaction step, and the profile in part (b) shows the curve for an exergonic step.
In an endergonic reaction (Figure 7.15a), the energy level of the transition state is closer to that of the product than that of the reactant. Since the transition state is closer energetically to the product, we make the natural assumption that it’s also closer structurally. In other words, the transition state for an endergonic reaction step structurally resembles the product of that step. Conversely, the transition state for an exergonic reaction (Figure 7.15b), is closer energetically, and thus structurally, to the reactant than to the product. We therefore say that the transition state for an exergonic reaction step structurally resembles the reactant for that step.
Hammond postulate:
The structure of a transition state resembles the structure of the nearest stable species. Transition states for endergonic steps structurally resemble products, and transition states for exergonic steps structurally resemble reactants.
How does the Hammond postulate apply to electrophilic addition reactions? The formation of a carbocation by protonation of an alkene is an endergonic step. Thus, the transition state for alkene protonation structurally resembles the carbocation intermediate, and any factor that stabilizes the carbocation will also stabilize the nearby transition state. Since increasing alkyl substitution stabilizes carbocations, it also stabilizes the transition states leading to those ions, thus resulting in a faster reaction. In other words, more stable carbocations form faster because their greater stability is reflected in the lower-energy transition state leading to them (Figure 7.16).
because its increased stability lowers the energy of the transition state leading to it.
We can imagine the transition state for alkene protonation to be a structure in which one of the alkene carbon atoms has almost completely rehybridized from sp2 to sp3 and the remaining alkene carbon bears much of the positive charge (Figure 7.17). This transition state is stabilized by hyperconjugation and inductive effects in the same way as the product carbocation. The more alkyl groups that are present, the greater the extent of stabilization and the faster the transition state forms.
Problem 7-20
What about the second step in the electrophilic addition of HCl to an alkene—the reaction of chloride ion with the carbocation intermediate? Is this step exergonic or endergonic? Does the transition state for this second step resemble the reactant (carbocation) or product (alkyl chloride)? Make a rough drawing of what the transition-state structure might look like. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.11%3A_The_Hammond_Postulate.txt |
How do we know that the carbocation mechanism for electrophilic addition reactions of alkenes is correct? The answer is that we don’t know it’s correct; at least we don’t know with complete certainty. Although an incorrect reaction mechanism can be disproved by demonstrating that it doesn’t account for observed data, a correct reaction mechanism can never be entirely proven. The best we can do is to show that a proposed mechanism is consistent with all known facts. If enough facts are accounted for, the mechanism is probably correct.
One of the best pieces of evidence supporting the carbocation mechanism for the electrophilic addition reaction was discovered during the 1930s by F. C. Whitmore of Pennsylvania State University, who found that structural rearrangements often occur during the reaction of HX with an alkene. For example, reaction of HCl with 3-methyl-1-butene yields a substantial amount of 2-chloro-2-methylbutane in addition to the “expected” product, 2-chloro-3-methylbutane.
If the reaction takes place in a single step, it would be difficult to account for rearrangement, but if the reaction takes place in several steps, rearrangement is more easily explained. Whitmore suggested that it is a carbocation intermediate that undergoes rearrangement. The secondary carbocation intermediate formed by protonation of 3-methyl-1-butene rearranges to a more stable tertiary carbocation by a hydride shift—the shift of a hydrogen atom and its electron pair (a hydride ion, :H) between neighboring carbons.
Carbocation rearrangements can also occur by the shift of an alkyl group with its electron pair. For example, reaction of 3,3-dimethyl-1-butene with HCl leads to an equal mixture of unrearranged 3-chloro-2,2-dimethylbutane and rearranged 2-chloro-2,3-dimethylbutane. In this instance, a secondary carbocation rearranges to a more stable tertiary carbocation by the shift of a methyl group.
Note the similarities between the two carbocation rearrangements: in both cases, a group (:H or :CH3) moves to an adjacent positively charged carbon, taking its bonding electron pair with it. Also in both cases, a less stable carbocation rearranges to a more stable ion. Rearrangements of this kind are a common feature of carbocation chemistry and are particularly important in the biological pathways by which steroids and related substances are synthesized. An example is the following hydride shift that occurs during the biosynthesis of cholesterol.
As always, when looking at any complex chemical transformation, whether biochemical or not, focus on the part of the molecule where the change is occurring and don’t worry about the rest. The tertiary carbocation just pictured looks complicated, but all the chemistry is taking place in the small part of the molecule inside the red circle.
Problem 7-21
On treatment with HBr, vinylcyclohexane undergoes addition and rearrangement to yield 1-bromo-1-ethylcyclohexane. Using curved arrows, propose a mechanism to account for this result.
7.13: Chemistry MattersBioprospecting- Hunting for Natural Products
7 • Chemistry Matters 7 • Chemistry Matters
Most people know the names of the common classes of biomolecules—proteins, carbohydrates, lipids, and nucleic acids—but there are many more kinds of compounds in living organisms than just those four. All living organisms also contain a vast diversity of substances usually grouped under the heading natural products. The term natural product really refers to any naturally occurring substance but is generally taken to mean a so-called secondary metabolite—a small molecule that is not essential to the growth and development of the producing organism and is not classified by structure.
It has been estimated that well over 300,000 secondary metabolites exist, and it’s thought that their primary function is to increase the likelihood of an organism’s survival by repelling or attracting other organisms. Alkaloids, such as morphine; antibiotics, such as erythromycin and the penicillins; and immunosuppressive agents, such as rapamycin (sirolimus) prescribed for liver transplant recipients, are examples.
Where do these natural products come from, and how are they found? Although most chemists and biologists spend their working time in the laboratory, a few spend their days scuba diving on South Pacific islands or trekking through the rainforests of South America and Southeast Asia at work as bioprospectors. Their job is to hunt for new and unusual natural products that might be useful as drugs.
As noted in the Chapter 6 Chemistry Matters, more than half of all new drug candidates come either directly or indirectly from natural products. Morphine from the opium poppy, prostaglandin E1 from sheep prostate glands, erythromycin A from a Streptomyces erythreus bacterium cultured from a Philippine soil sample, and benzylpenicillin from the mold Penicillium notatum are examples. The immunosuppressive agent rapamycin, whose structure is shown previously, was first isolated from a Streptomyces hygroscopicus bacterium found in a soil sample from Rapa Nui (Easter Island), located 2200 miles off the coast of Chile.
With less than 1% of living organisms yet investigated, bioprospectors have a lot of work to do. But there is a race going on. Rainforests throughout the world are being destroyed at an alarming rate, causing many species of both plants and animals to become extinct before they can even be examined. The governments in many countries seem aware of the problem, but there is as yet no international treaty on biodiversity that could help preserve vanishing species.
7.14: Key Terms
7 • Key Terms 7 • Key Terms
• alkene $(R2C═CR2)(R2C═CR2)$
• allyl group
• degree of unsaturation
• E configuration
• E,Z system
• electrophilic addition reaction
• Hammond postulate
• hydride shift
• hyperconjugation
• Markovnikov’s rule
• methylene group
• regiospecific reaction
• unsaturated
• vinyl group
• Z configuration
7.15: Summary
7 • Summary 7 • Summary
Carbon–carbon double bonds are present in most organic and biological molecules, so a good understanding of their behavior is needed. In this chapter, we’ve looked at some consequences of alkene stereoisomerism and at the details of the broadest class of alkene reactions—the electrophilic addition reaction.
An alkene is a hydrocarbon that contains a carbon–carbon double bond. Because they contain fewer hydrogens than alkanes with the same number of carbons, alkenes are said to be unsaturated.
Because rotation around the double bond can’t occur, substituted alkenes can exist as cis–trans stereoisomers. The configuration of a double bond can be specified by applying the Cahn–Ingold–Prelog sequence rules, which rank the substituents on each double-bond carbon. If the higher-ranking groups on each carbon are on the same side of the double bond, the configuration is Z (zusammen, “together”); if the higher-ranking groups on each carbon are on opposite sides of the double bond, the configuration is E (entgegen, “apart”).
Alkene chemistry is dominated by electrophilic addition reactions. When HX reacts with an unsymmetrically substituted alkene, Markovnikov’s rule predicts that the H will add to the carbon having fewer alkyl substituents and the X group will add to the carbon having more alkyl substituents. Electrophilic additions to alkenes take place through carbocation intermediates formed by reaction of the nucleophilic alkene π bond with electrophilic H+. Carbocation stability follows the order
Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl
R3C+ > R2CH+ > RCH2+ > CH3+
Markovnikov’s rule can be restated by saying that, in the addition of HX to an alkene, a more stable carbocation intermediate is formed. This result is explained by the Hammond postulate, which says that the transition state of an exergonic reaction step structurally resembles the reactant, whereas the transition state of an endergonic reaction step structurally resembles the product. Since an alkene protonation step is endergonic, the stability of the more highly substituted carbocation is reflected in the stability of the transition state leading to its formation.
Evidence in support of a carbocation mechanism for electrophilic additions comes from the observation that structural rearrangements often take place during reaction. Rearrangements occur by shift of either a hydride ion, :H (a hydride shift), or an alkyl anion, :R, from a carbon atom to the neighboring positively charged carbon. This results in isomerization of a less stable carbocation to a more stable one. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.12%3A_Evidence_for_the_Mechanism_of_Electrophilic_Additions_-_Carbocation_Rearrangements.txt |
7 • Additional Problems 7 • Additional Problems
Visualizing Chemistry
Problem 7-22
Name the following alkenes, and convert each drawing into a skeletal structure: (a)
(b)
Problem 7-23 Assign E or Z stereochemistry to the double bonds in each of the following alkenes, and convert each drawing into a skeletal structure (red = O, green = Cl): (a) (b) Problem 7-24 The following carbocation is an intermediate in the electrophilic addition reaction of HCl with two different alkenes. Identify both, and tell which C−H bonds in the carbocation are aligned for hyperconjugation with the vacant p orbital on the positively charged carbon. Problem 7-25 The following alkyl bromide can be made by HBr addition to three different alkenes. Show their structures. Mechanism Problems Problem 7-26 Predict the major product and show the complete mechanism for each of the following electrophilic addition reactions. (a) (b) (c) Problem 7-27 Each of the following electrophilic addition reactions involves a carbocation rearrangement. Predict the product and draw the complete mechanism of each using curved arrows. (a) (b) (c) Problem 7-28 When 1,3-butadiene reacts with 1 mol of HBr, two isolable products result. Propose mechanisms for both. Problem 7-29 When methyl vinyl ether reacts with a strong acid, H+ adds to C2 instead of C1 or the oxygen atom. Explain. Problem 7-30 Addition of HCl to 1-isopropylcyclohexene yields a rearranged product. Propose a mechanism, showing the structures of the intermediates and using curved arrows to indicate electron flow in each step. Problem 7-31 Addition of HCl to 1-isopropenyl-1-methylcyclopentane yields 1-chloro-1,2,2-trimethylcyclohexane. Propose a mechanism, showing the structures of the intermediates and using curved arrows to indicate electron flow in each step. Problem 7-32 Limonene, a fragrant hydrocarbon found in lemons and oranges, is biosynthesized from geranyl diphosphate by the following pathway. Add curved arrows to show the mechanism of each step. Which step involves an alkene electrophilic addition? (The ion OP2O64− is the diphosphate ion, and “Base” is an unspecified base in the enzyme that catalyzes the reaction.) Problem 7-33 epi-Aristolochene, a hydrocarbon found in both pepper and tobacco, is biosynthesized by the following pathway. Add curved arrows to show the mechanism of each step. Which steps involve alkene electrophilic addition(s), and which involve carbocation rearrangement(s)? (The abbreviation H—A stands for an unspecified acid, and “Base” is an unspecified base in the enzyme.) Calculating a Degree of Unsaturation Problem 7-34 Calculate the degree of unsaturation in the following formulas, and draw five possible structures for each: (a) C10H16 (b)
C8H8O
(c) C7H10Cl2 (d) C10H16O2 (e) C5H9NO2 (f) C8H10ClNO
Problem 7-35 How many hydrogens does each of the following compounds have? (a)
C8H?O2, has two rings and one double bond
(b) C7H?N, has two double bonds (c) C9H?NO, has one ring and three double bonds
Problem 7-36
Loratadine, marketed as an antiallergy medication under the brand name Claritin, has four rings, eight double bonds, and the formula C22H?ClN2O2. How many hydrogens does loratadine have? (Calculate your answer; don’t count hydrogens in the structure.)
Naming Alkenes
Problem 7-37 Name the following alkenes: (a) (b) (c) (d) (e) (f) Problem 7-38 Draw structures corresponding to the following systematic names: (a) (4E)-2,4-Dimethyl-1,4-hexadiene (b)
cis-3,3-Dimethyl-4-propyl-1,5-octadiene
(c) 4-Methyl-1,2-pentadiene (d) (3E,5Z)-2,6-Dimethyl-1,3,5,7-octatetraene (e) 3-Butyl-2-heptene (f) trans-2,2,5,5-Tetramethyl-3-hexene
Problem 7-39 Name the following cycloalkenes: (a) (b) (c) (d) (e) (f) Problem 7-40 Ocimene is a triene found in the essential oils of many plants. What is its IUPAC name, including stereochemistry? Problem 7-41 α-Farnesene is a constituent of the natural wax found on apples. What is its IUPAC name, including stereochemistry? Problem 7-42 Menthene, a hydrocarbon found in mint plants, has the systematic name 1-isopropyl-4-methylcyclohexene. Draw its structure. Problem 7-43 Draw and name the six alkene isomers, C5H10, including E,Z isomers. Problem 7-44 Draw and name the 17 alkene isomers, C6H12, including E,Z isomers. Alkene Isomers and Their Stability Problem 7-45 Rank the following sets of substituents according to the Cahn–Ingold–Prelog sequence rules: (a) (b) (c) (d) (e) (f) Problem 7-46 Assign E or Z configuration to each of the following compounds: (a) (b) (c) (d) Problem 7-47 Which of the following E,Z designations are correct, and which are incorrect? (a) (b) (c) (d) (e) (f) Problem 7-48 Rank the double bonds according to their increasing stability. (a) (b) (c) Problem 7-49 trans-2-Butene is more stable than cis-2-butene by only 4 kJ/mol, but trans-2,2,5,5-tetramethyl-3-hexene is more stable than its cis isomer by 39 kJ/mol. Explain. Problem 7-50 Cyclodecene can exist in both cis and trans forms, but cyclohexene cannot. Explain. Problem 7-51 Normally, a trans alkene is more stable than its cis isomer. trans-Cyclooctene, however, is less stable than cis-cyclooctene by 38.5 kJ/mol. Explain. Problem 7-52 trans-Cyclooctene is less stable than cis-cyclooctene by 38.5 kJ/mol, but trans-cyclononene is less stable than cis-cyclononene by only 12.2 kJ/mol. Explain. Problem 7-53 Tamoxifen, a drug used in the treatment of breast cancer, and clomiphene, a drug used in fertility treatment, have similar structures but very different effects. Assign E or Z configuration to the double bonds in both compounds. Carbocations and Electrophilic Addition Reactions Problem 7-54 Rank the following carbocations according to their increasing stability. (a) (b) (c) Problem 7-55 Use the Hammond Postulate to determine which alkene in each pair would be expected to form a carbocation faster in an electrophilic addition reaction. (a) (b) (c) Problem 7-56 The following carbocations can be stabilized by resonance. Draw all the resonance forms that would stabilize each carbocation. (a) (b) (c) Problem 7-57 Predict the major product in each of the following reactions: (a) (b) (c) (d) Problem 7-58 Predict the major product from addition of HBr to each of the following alkenes: (a) (b) (c) Problem 7-59 Alkenes can be converted into alcohols by acid-catalyzed addition of water. Assuming that Markovnikov’s rule is valid, predict the major alcohol product from each of the following alkenes. (a) (b) (c) Problem 7-60 Each of the following carbocations can rearrange to a more stable ion. Propose structures for the likely rearrangement products. (a) (b) (c) General Problems Problem 7-61 Allene (1,2-propadiene), H2C$\text{=}$C$\text{=}$CH2, has two adjacent double bonds. What kind of hybridization must the central carbon have? Sketch the bonding π orbitals in allene. What shape do you predict for allene? Problem 7-62 The heat of hydrogenation for allene (Problem 7-61) to yield propane is −295 kJ/mol, and the heat of hydrogenation for a typical monosubstituted alkene, such as propene, is −125 kJ/mol. Is allene more stable or less stable than you might expect for a diene? Explain. Problem 7-63 Retin A, or retinoic acid, is a medication commonly used to reduce wrinkles and treat severe acne. How many different isomers arising from E,Z double-bond isomerizations are possible? Problem 7-64 Fucoserratene and ectocarpene are sex pheromones produced by marine brown algae. What are their systematic names? (Ectocarpene is difficult; make your best guess, and then check your answer in the Student Solutions Manual.) Problem 7-65 tert-Butyl esters [RCO2C(CH3)3] are converted into carboxylic acids (RCO2H) by reaction with trifluoroacetic acid, a reaction useful in protein synthesis (Section 26.7). Assign E,Z designation to the double bonds of both reactant and product in the following scheme, and explain why there is an apparent change in double-bond stereochemistry: Problem 7-66 Vinylcyclopropane reacts with HBr to yield a rearranged alkyl bromide. Follow the flow of electrons as represented by the curved arrows, show the structure of the carbocation intermediate in brackets, and show the structure of the final product. Problem 7-67 Calculate the degree of unsaturation in each of the following formulas: (a) Cholesterol, C27H46O (b)
DDT, C14H9Cl5
(c) Prostaglandin E1, C20H34O5 (d) Caffeine, C8H10N4O2 (e) Cortisone, C21H28O5 (f) Atropine, C17H23NO3
Problem 7-68
The isobutyl cation spontaneously rearranges to the tert-butyl cation by a hydride shift. Is the rearrangement exergonic or endergonic? Draw what you think the transition state for the hydride shift might look like according to the Hammond postulate.
Problem 7-69
Draw an energy diagram for the addition of HBr to 1-pentene. Let one curve on your diagram show the formation of 1-bromopentane product and another curve on the same diagram show the formation of 2-bromopentane product. Label the positions for all reactants, intermediates, and products. Which curve has the higher-energy carbocation intermediate? Which curve has the higher-energy first transition state?
Problem 7-70
Sketch the transition-state structures involved in the reaction of HBr with 1-pentene (Problem 7-69). Tell whether each structure resembles reactant or product.
Problem 7-71
Aromatic compounds such as benzene react with alkyl chlorides in the presence of AlCl3 catalyst to yield alkylbenzenes. This reaction occurs through a carbocation intermediate, formed by reaction of the alkyl chloride with AlCl3 (R−Cl + AlCl3 ⟶ R+ + AlCl4). How can you explain the observation that reaction of benzene with 1-chloropropane yields isopropylbenzene as the major product?
Problem 7-72
Reaction of 2,3-dimethyl-1-butene with HBr leads to an alkyl bromide, C6H13Br. On treatment of this alkyl bromide with KOH in methanol, elimination of HBr occurs and a hydrocarbon that is isomeric with the starting alkene is formed. What is the structure of this hydrocarbon, and how do you think it is formed from the alkyl bromide? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/07%3A_Alkenes-_Structure_and_Reactivity/7.16%3A_Additional_Problems.txt |
As you have seen, addition reactions dominate the chemistry of alkenes. This chapter shows how a variety of reagents can add to alkenes; how hydrogen bromide can be made to add to alkenes in a non-Markovnikov manner; and how alkene molecules can be cleaved into easily identifiable parts. First, you will examine the preparation of alkenes by elimination reactions.
08: Alkenes- Reactions and Synthesis
Figure 8.1 The Spectra fiber used to make the bulletproof vests used by police and military is made of ultra-high-molecular-weight polyethylene, a simple alkene polymer. (credit: modification of work “US Navy 081028-N-3857R-007 Seabees participate in a chemical, biological and radiological warfare drill” by U.S. Navy photo by Mass Communication Specialist 1st Class Chad Runge/Wikimedia Commons, Public Domain)
Chapter Contents
8.1 Preparing Alkenes: A Preview of Elimination Reactions
8.2 Halogenation of Alkenes: Addition of X2
8.3 Halohydrins from Alkenes: Addition of HO-X
8.4 Hydration of Alkenes: Addition of H2O by Oxymercuration
8.5 Hydration of Alkenes: Addition of H2O by Hydroboration
8.6 Reduction of Alkenes: Hydrogenation
8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation
8.8 Oxidation of Alkenes: Cleavage to Carbonyl Compounds
8.9 Addition of Carbenes to Alkenes: Cyclopropane Synthesis
8.10 Radical Additions to Alkenes: Chain-Growth Polymers
8.11 Biological Additions of Radicals to Alkenes
8.12 Reaction Stereochemistry: Addition of H2O to an Achiral Alkene
8.13 Reaction Stereochemistry: Addition of H2O to a Chiral Alkene
Much of the background needed to understand organic reactions has now been covered, and it’s time to begin a systematic description of the major functional groups. In this chapter on alkenes, and in future chapters on other functional groups, we’ll discuss a variety of reactions, but try to focus on the general principles and patterns of reactivity that tie organic chemistry together. There are no shortcuts; you have to know the reactions to understand organic and biological chemistry.
Alkene addition reactions occur widely, both in the laboratory and in living organisms. Although we’ve studied only the addition of HX thus far, many closely related reactions also take place. In this chapter, we’ll see briefly how alkenes are prepared and we’ll discuss further examples of alkene addition reactions. Particularly important are the addition of a halogen (X2) to give a 1,2-dihalide, addition of a hypohalous acid (HOX) to give a halohydrin, addition of water to give an alcohol, addition of hydrogen to give an alkane, addition of a single oxygen to give a three-membered cyclic ether called an epoxide, and addition of two hydroxyl groups to give a 1,2-diol.
Figure 8.2 Some useful alkene reactions.
8.02: Preparation of Alkenes - A Preview of Elimination Reactions
Before getting to the main subject of this chapter—the reactions of alkenes—let’s take a brief look at how alkenes are prepared. The subject is a bit complex, though, so we’ll return to it in Chapter 11 for a more detailed study. For the present, it’s enough to realize that alkenes are readily available from simple precursors—usually alcohols in biological systems and either alcohols or alkyl halides in the laboratory.
Just as the chemistry of alkenes is dominated by addition reactions, the preparation of alkenes is dominated by elimination reactions. Additions and eliminations are, in many respects, two sides of the same coin. That is, an addition reaction might involve the addition of HBr or H2O to an alkene to form an alkyl halide or alcohol, whereas an elimination reaction might involve the loss of HBr or H2O from an alkyl halide or alcohol to form an alkene.
The two most common elimination reactions are dehydrohalogenation—the loss of HX from an alkyl halide—and dehydration—the loss of water from an alcohol. Dehydrohalogenation usually occurs by reaction of an alkyl halide with strong base such as potassium hydroxide. For example, bromocyclohexane yields cyclohexene when treated with KOH in ethanol solution.
Dehydration is often carried out in the laboratory by treatment of an alcohol with a strong acid. For example, when 1-methylcyclohexanol is warmed with aqueous sulfuric acid in tetrahydrofuran (THF) solvent, loss of water occurs and 1-methylcyclohexene is formed.
In biological pathways, dehydrations rarely occur with isolated alcohols. Instead, they normally take place on substrates in which the −OH is positioned two carbons away from a $C═OC═O$ group. In the biosynthesis of fats, for instance, β-hydroxybutyryl ACP is converted by dehydration to trans-crotonyl ACP, where ACP is an abbreviation for acyl carrier protein. We’ll see the reason for this requirement in Section 11.10.
Problem 8-1
One problem with elimination reactions is that mixtures of products are often formed. For example, treatment of 2-bromo-2-methylbutane with KOH in ethanol yields a mixture of two alkene products. What are their likely structures?
Problem 8-2
How many alkene products, including E,Z isomers, might be obtained by dehydration of 3-methyl-3-hexanol with aqueous sulfuric acid? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.01%3A_Why_This_Chapter.txt |
Bromine and chlorine add rapidly to alkenes to yield 1,2-dihalides, a process called halogenation. For example, nearly 50 million tons of 1,2-dichloroethane (ethylene dichloride) are synthesized worldwide each year, much of it by addition of Cl2 to ethylene. The product is used both as a solvent and as starting material for the manufacture of poly(vinyl chloride), PVC, the third most widely synthesized polymer in the world afterpolyethelyne and polypropolyne. Fluorine is too reactive and difficult to control for most laboratory applications, and iodine does not react with most alkenes.
Based on what we’ve seen thus far, a possible mechanism for the reaction of bromine with alkenes might involve electrophilic addition of Br+ to the alkene, giving a carbocation intermediate that could undergo further reaction with Br to yield the dibromo addition product.
Although this mechanism seems plausible, it’s not fully consistent with known facts. In particular, it doesn’t explain the stereochemistry of the addition reaction. That is, the mechanism doesn’t account for which product stereoisomer is formed.
When the halogenation reaction is carried out on a cycloalkene, such as cyclopentene, only the trans stereoisomer of the dihalide addition product is formed, rather than the mixture of cis and trans isomers that might have been expected if a planar carbocation intermediate were involved. We say that the reaction occurs with anti stereochemistry, meaning that the two bromine atoms come from opposite faces of the double bond—one from the top face and one from the bottom face.
An explanation for the observed stereochemistry of addition was suggested in 1937 by George Kimball and Irving Roberts, who proposed that the reaction intermediate is not a carbocation but is instead a bromonium ion, $R2Br+R2Br+$, formed by electrophilic addition of Br+ to the alkene. (Similarly, a chloronium ion contains a positively charged, divalent chlorine, R2Cl+.) The bromonium ion is formed in a single step by interaction of the alkene with Br2 and the simultaneous loss of Br.
How does the formation of a bromonium ion account for the observed anti stereochemistry of addition to cyclopentene? If a bromonium ion is formed as an intermediate, we can imagine that the large bromine atom might “shield” one side of the molecule. Reaction with Br ion in the second step could then occur only from the opposite, unshielded side to give the trans product.
The bromonium ion postulate, made more than 85 years ago to explain the stereochemistry of halogen addition to alkenes, is a remarkable example of deductive logic in chemistry. Arguing from experimental results, chemists were able to make a hypothesis about the intimate mechanistic details of alkene electrophilic reactions. Subsequently, strong evidence supporting the mechanism came from the work of George Olah at the University of Southern California, who prepared and studied stable solutions of cyclic bromonium ions in liquid SO2. There’s no question that bromonium ions exist.
Alkene halogenation reactions occur in nature just as they do in the laboratory but are limited primarily to marine organisms living in halide-rich environments. These biological halogenation reactions are carried out by enzymes called haloperoxidases, which use H2O2 to oxidize Br or Cl ions to a biological equivalent of Br+ or Cl+. Electrophilic addition to the double bond of a substrate molecule then yields a bromonium or chloronium ion intermediate just as in the laboratory, and reaction with another halide ion completes the process. Halomon, for example, an antitumor pentahalide isolated from red alga, is thought to arise by a route that involves twofold addition of BrCl through the corresponding bromonium ions.
Problem 8-3
What product would you expect to obtain from addition of Cl2 to 1,2-dimethylcyclohexene? Show the stereochemistry of the product.
Problem 8-4
Addition of HCl to 1,2-dimethylcyclohexene yields a mixture of two products. Show the stereochemistry of each, and explain why a mixture is formed.
8.04: Halohydrins from Alkenes - Addition of HO-X
Another example of an electrophilic addition is the reaction an alkene with either Br2 or Cl2 in the presence of water to yield a 1,2-halo alcohol, called a halohydrin.
We saw in the previous section that when Br2 reacts with an alkene, the cyclic bromonium ion intermediate reacts with the only nucleophile present, Br ion. If the reaction is carried out in the presence of an additional nucleophile, however, the intermediate bromonium ion can be intercepted by the added nucleophile and diverted to a different product. In the presence of a high concentration of water, for instance, water competes with Br ion as a nucleophile and reacts with the bromonium ion intermediate to yield a bromohydrin. The net effect is addition of HO−Br to the alkene by the pathway shown in Figure 8.3.
Figure 8.3 MECHANISM Bromohydrin formation by reaction of an alkene with Br2 in the presence of water. Water acts as a nucleophile in step 2 to react with the intermediate bromonium ion.
In practice, few alkenes are soluble in water, and bromohydrin formation is often carried out in a solvent such as aqueous dimethyl sulfoxide, CH3SOCH3 (DMSO), using a reagent called N-bromosuccinimide (NBS) as a source of Br2. NBS is a stable, easily handled compound that slowly decomposes in water to yield Br2 at a controlled rate. Bromine itself can also be used in the addition reaction, but it is more dangerous and more difficult to handle than NBS.
Notice that the aromatic ring in the above example does not react with Br2, even though it appears to have three carbon–carbon double bonds. As we’ll see in Section 15.2, aromatic rings are a good deal more stable and less reactive than might be expected.
There are a number of biological examples of halohydrin formation, particularly in marine organisms. As with halogenation (Section 8.2), halohydrin formation is carried out by haloperoxidases. For example:
Problem 8-5
What product would you expect from the reaction of cyclopentene with NBS and water? Show the stereochemistry.
Problem 8-6
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation? (Section 7.8) Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.03%3A_Halogenation_of_Alkenes_-_Addition_of_X.txt |
Water adds to alkenes to yield alcohols, a process called hydration. The reaction takes place on treatment of the alkene with water and a strong acid catalyst, such as H2SO4, by a mechanism similar to that of HX addition. Thus, as shown in Figure 8.4, protonation of an alkene double bond yields a carbocation intermediate, which reacts with water to yield a protonated alcohol product, ROH2+. Loss of H+ from this protonated alcohol gives the neutral alcohol and regenerates the acid catalyst.
Figure 8.4 MECHANISM Mechanism of the acid-catalyzed hydration of an alkene to yield an alcohol. Protonation of the alkene gives a carbocation intermediate, which reacts with water. The initial product is then deprotonated.
Most ethanol throughout the world is now made by fermentation of biological precursors, such as corn and sugar, but acid-catalyzed alkene hydration is particularly suited to large-scale industrial procedures, and approximately 90,000 tons of ethanol is manufactured each year in the United States by hydration of ethylene. The reaction is of little value in the laboratory, however, because it requires high temperatures—250 °C in the case of ethylene—and strongly acidic conditions.
Acid-catalyzed hydration of double bonds is also uncommon in biological pathways. Instead, biological hydrations usually require that the double bond be adjacent to a carbonyl group for reaction to proceed. Fumarate, for instance, is hydrated to give malate as one step in the citric acid cycle of food metabolism. Note that the requirement for an adjacent carbonyl group in the addition of water is the same as in Section 8.1 for the elimination of water. We’ll see the reason for this requirement in Section 19.13, but will note for now that the reaction is not an electrophilic addition but instead occurs through a mechanism that involves formation of an anion intermediate followed by protonation by an acid HA.
When it comes to circumventing problems like those with acid-catalyzed alkene hydrations, laboratory chemists have a great advantage over the cellular “chemists” in living organisms. Laboratory chemists are not constrained to carry out their reactions in water solution; they can choose from any of a large number of solvents. Laboratory reactions don’t need to be carried out at a fixed temperature; they can take place over a wide range of temperatures. And laboratory reagents aren’t limited to containing carbon, oxygen, nitrogen, and a few other elements; they can contain any element in the periodic table.
In the laboratory, alkenes are often hydrated by the oxymercuration–demercuration procedure, which involves electrophilic addition of Hg2+ to the alkene on reaction with mercury(II) acetate [(CH3CO2)2Hg. The intermediate organomercury compound is then treated with sodium borohydride, NaBH4, and demercuration occurs to produce an alcohol. For example:
Alkene oxymercuration is closely analogous to halohydrin formation. The reaction is initiated by electrophilic addition of Hg2+ (mercuric) ion to the alkene to give an intermediate mercurinium ion, whose structure resembles that of a bromonium ion (Figure 8.5). Nucleophilic addition of water as in halohydrin formation, followed by the loss of a proton, then yields a stable organomercury product. The final step, demercuration of the organomercury compound by reaction with sodium borohydride, is complex and involves radicals. Note that the regiochemistry of the reaction corresponds to Markovnikov addition of water; that is, the −OH group attaches to the more highly substituted carbon atom, and the −H attaches to the less highly substituted carbon. The hydrogen that replaces mercury in the demercuration step can attach from either side of the molecule depending on the exact circumstances.
) Electrophilic addition of Hg2+ gives a mercurinium ion, which (2) reacts with water as in halohydrin formation. Loss of a proton gives an organomercury product, and (3) reaction with NaBH4 removes the mercury. The product of the reaction is a more highly substituted alcohol, corresponding to Markovnikov regiochemistry.
Problem 8-7 What products would you expect from oxymercuration–demercuration of the following alkenes?
(a)
(b)
Problem 8-8 From what alkenes might the following alcohols have been prepared?
(a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.05%3A_Hydration_of_Alkenes_-_Addition_of_HO_by_Oxymercuration.txt |
In addition to the oxymercuration–demercuration method, which yields the Markovnikov product, a complementary method that yields the non-Markovnikov product is also useful. Discovered in 1959 by H.C. Brown at Purdue University and called hydroboration, the reaction involves addition of a B−H bond of borane, BH3, to an alkene to yield an organoborane intermediate, RBH2. Oxidation of the organoborane by reaction with basic hydrogen peroxide, H2O2, then gives an alcohol. For example:
Borane is very reactive as a Lewis acid because the boron atom has only six electrons in its valence shell rather than an octet. In tetrahydrofuran solution, BH3 accepts an electron pair from a solvent molecule in a Lewis acid–base reaction to complete its octet and form a stable BH3–THF complex.
When an alkene reacts with BH3 in THF solution, rapid addition to the double bond occurs three times and a trialkylborane, R3B, is formed. For example, 1 molar equivalent of BH3 adds to 3 molar equivalents of cyclohexene to yield tricyclohexylborane. When tricyclohexylborane is then treated with aqueous hydrogen H2O2 in basic solution, an oxidation takes place. The three C−B bonds are broken, −OH groups bond to the three carbons, and 3 equivalents of cyclohexanol are produced. The net effect of the two-step hydroboration–oxidation sequence is hydration of the alkene double bond.
One of the features that makes the hydroboration reaction so useful is the regiochemistry that results when an unsymmetrical alkene is hydroborated. For example, hydroboration–oxidation of 1-methylcyclopentene yields trans-2-methylcyclopentanol. In this process, boron and hydrogen add to the alkene from the same face of the double bond—that is, with syn stereochemistry, the opposite of anti—with boron attaching to the less highly substituted carbon. During the oxidation step, the boron is replaced by an −OH with the same stereochemistry, resulting in an overall syn non-Markovnikov addition of water. This stereochemical result is particularly useful because it is complementary to the Markovnikov regiochemistry observed for oxymercuration–demercuration.
Why does alkene hydroboration take place with syn, non-Markovnikov regiochemistry to yield the less highly substituted alcohol? Hydroboration differs from many other alkene addition reactions in that it occurs in a single step without a carbocation intermediate (Figure 8.6). Because the C−H and C−B bonds form at the same time and from the same face of the alkene, syn stereochemistry results. Non-Markovnikov regiochemistry occurs because attachment of boron is favored at the less sterically crowded carbon atom of the alkene.
Worked Example 8.1
Predicting the Products of a Hydration Reaction
What products would you obtain from reaction of 2-methyl-2-pentene with:
(a) BH3, followed by H2O2, OH(b) Hg(OAc)2, followed by NaBH4
Strategy
When predicting the product of a reaction, you have to recall what you know about the kind of reaction being carried out and apply that knowledge to the specific case you’re dealing with. In the present instance, recall that the two methods of hydration—hydroboration–oxidation and oxymercuration–demercuration—give complementary products. Hydroboration–oxidation occurs with syn stereochemistry and gives the non-Markovnikov addition product; oxymercuration–demercuration gives the Markovnikov product.
Worked Example 8.2
Synthesizing an Alcohol
How might you prepare the following alcohol?
Strategy
Problems that require the synthesis of a specific target molecule should always be worked backward. Look at the target, identify its functional group(s), and ask yourself, “What are the methods for preparing that functional group?” In the present instance, the target molecule is a secondary alcohol (R2CHOH), and we’ve seen that alcohols can be prepared from alkenes by either hydroboration–oxidation or oxymercuration–demercuration. The −OH-bearing carbon in the product must have been a double-bond carbon in the alkene reactant, so there are two possibilities here: 4-methyl-2-hexene and 3-methyl-3-hexene.
4-Methyl-2-hexene has a disubstituted double bond, $RCH═CHR′RCH═CHR′$, and will probably give a mixture of two alcohols with either hydration method since Markovnikov’s rule does not apply to symmetrically substituted alkenes. 3-Methyl-3-hexene, however, has a trisubstituted double bond, and should give only the desired product on non-Markovnikov hydration using the hydroboration–oxidation method.
Solution
Problem 8-9 Show the structures of the products you would obtain by hydroboration–oxidation of the following alkenes: (a)
(b)
Problem 8-10 What alkenes might be used to prepare the following alcohols by hydroboration–oxidation?
(a)
(b)
(c)
Problem 8-11
The following cycloalkene gives a mixture of two alcohols on hydroboration followed by oxidation. Draw the structures of both, and explain the result. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.06%3A_Hydration_of_Alkenes_-_Addition_of_HO_by_Hydroboration.txt |
Alkenes react with H2 in the presence of a metal catalyst such as palladium or platinum to yield the corresponding saturated alkanes. We describe the result by saying that the double bond has been hydrogenated, or reduced. Note that the word reduction is used somewhat differently in organic chemistry from what you might have learned previously. In general chemistry, a reduction is defined as the gain of one or more electrons by an atom. In organic chemistry, however, a reduction is a reaction that results in a gain of electron density for carbon, caused either by bond formation between carbon and a less electronegative atom—usually hydrogen—or by bond-breaking between carbon and a more electronegative atom—usually oxygen, nitrogen, or a halogen. We’ll explore this topic in more detail in Section 10.8.
$ReductionIncreases electron density on carbon by: –forming this:C−H–or breaking one of these:C−OC−NC−X ReductionIncreases electron density on carbon by: –forming this:C−H–or breaking one of these:C−OC−NC−X$
Platinum and palladium are the most common laboratory catalysts for alkene hydrogenations. Palladium is normally used as a very fine powder “supported” on an inert material such as charcoal (Pd/C) to maximize surface area. Platinum is normally used as PtO2, a reagent known as Adams’ catalyst after its discoverer, Roger Adams at the University of Illinois.
Catalytic hydrogenation, unlike most other organic reactions, is a heterogeneous process rather than a homogeneous one. That is, the hydrogenation reaction does not occur in a homogeneous solution but instead takes place on the surface of solid catalyst particles. Hydrogenation usually occurs with syn stereochemistry: both hydrogens add to the double bond from the same face.
As shown in Figure 8.7, hydrogenation begins with adsorption of H2 onto the catalyst surface. Complexation between catalyst and alkene then occurs as a vacant orbital on the metal interacts with the filled alkene π orbital on the alkene. In the final steps, hydrogen is inserted into the double bond and the saturated product diffuses away from the catalyst. The stereochemistry of hydrogenation is syn because both hydrogens add to the double bond from the same catalyst surface.
Figure 8.7 MECHANISM Mechanism of alkene hydrogenation. The reaction takes place with syn stereochemistry on the surface of insoluble catalyst particles.
An interesting feature of catalytic hydrogenation is that the reaction is extremely sensitive to the steric environment around the double bond. As a result, the catalyst usually approaches the more accessible face of an alkene, giving rise to a single product. In α-pinene, for example, one of the methyl groups attached to the four-membered ring hangs over the top face of the double bond and blocks approach of the hydrogenation catalyst from that side. Reduction therefore occurs exclusively from the bottom face to yield the product shown.
Alkenes are much more reactive toward catalytic hydrogenation than most other unsaturated functional groups, and the reaction is therefore quite selective. Other functional groups, such as aldehydes, ketones, esters, and nitriles, often survive alkene hydrogenation conditions unchanged, although reaction with these groups does occur under more vigorous conditions. Note that, particularly in the hydrogenation of methyl 3-phenylpropenoate shown below, the aromatic ring is not reduced by hydrogen and palladium even though it contains apparent double bonds.
In addition to its usefulness in the laboratory, catalytic hydrogenation is also important in the food industry, where unsaturated vegetable oils are reduced on a large scale to produce the saturated fats used in margarine and cooking products (Figure 8.8). As we’ll see in Section 27.1, vegetable oils are triesters of glycerol, HOCH2CH(OH)CH2OH, with three long-chain carboxylic acids called fatty acids. The fatty acids are generally polyunsaturated, and their double bonds have cis stereochemistry. Complete hydrogenation yields the corresponding saturated fatty acids, but incomplete hydrogenation often results in partial cis–trans isomerization of a remaining double bond. When eaten and digested, the free trans fatty acids are released, raising blood cholesterol levels and potentially contributing to coronary problems.
Double-bond reductions are very common in biological pathways, although the mechanism is completely different from that of laboratory catalytic hydrogenation over palladium. As with biological hydrations (Section 8.4), biological reductions usually occur in two steps and require that the double bond be adjacent to a carbonyl group. In the first step, the biological reducing agent NADPH (reduced nicotinamide adenine dinucleotide phosphate), adds a hydride ion (H:) to the double bond to give an anion. In the second, the anion is protonated by acid HA, leading to overall addition of H2. An example is the reduction of trans-crotonyl ACP to yield butyryl ACP, a step involved in the biosynthesis of fatty acids (Figure 8.9).
is delivered from NADPH as a hydride ion, H:; the other hydrogen is delivered by protonation of the anion intermediate with an acid, HA.
Problem 8-12 What product would you obtain from catalytic hydrogenation of the following alkenes?
(a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.07%3A_Reduction_of_Alkenes_-_Hydrogenation.txt |
Like the word reduction used in the previous section for the addition of hydrogen to a double bond, the word oxidation has a slightly different meaning in organic chemistry than what you might have previously learned. In general chemistry, an oxidation is defined as the loss of one or more electrons by an atom. In organic chemistry, however, an oxidation is a reaction that results in a loss of electron density for carbon, caused either by bond formation between carbon and a more electronegative atom—usually oxygen, nitrogen, or a halogen—or by bond-breaking between carbon and a less electronegative atom—usually hydrogen. Note that an oxidation often adds oxygen, while a reduction often adds hydrogen.
$Oxidation Decreases electron density on carbon by: –forming one of these:C−OC−NC−X –or breaking this:C−H Oxidation Decreases electron density on carbon by: –forming one of these:C−OC−NC−X –or breaking this:C−H$
In the laboratory, alkenes are oxidized to give epoxides on treatment with a peroxyacid, RCO3H, such as meta-chloroperoxybenzoic acid. An epoxide, also called an oxirane, is a cyclic ether with an oxygen atom in a three-membered ring. For example:
Peroxyacids transfer an oxygen atom to the alkene with syn stereochemistry—both C−O bonds form on the same face of the double bond—through a one-step mechanism without intermediates. The oxygen atom farthest from the carbonyl group is the one transferred.
Another method for the synthesis of epoxides involves the use of halohydrins, prepared by electrophilic addition of HO−X to alkenes (Section 8.3). When a halohydrin is treated with base, HX is eliminated and an epoxide is produced.
Epoxides undergo an acid-catalyzed ring-opening reaction with water (a hydrolysis) to give the corresponding 1,2-dialcohol, or diol, also called a glycol. Thus, the net result of the two-step alkene epoxidation/hydrolysis is hydroxylation—the addition of an −OH group to each of the two double-bond carbons. In fact, approximately 204 million tons of ethylene glycol, HOCH2CH2OH, most of it used for automobile antifreeze, are produced worldwide each year by the epoxidation of ethylene and subsequent hydrolysis.
Acid-catalyzed epoxide opening begins with protonation of the epoxide to increase its reactivity, followed by nucleophilic addition of water. This nucleophilic addition is analogous to the final step of alkene bromination, in which a cyclic bromonium ion is opened by a nucleophile (Section 8.2). That is, a trans-1,2-diol results when an epoxycycloalkane is opened by aqueous acid, just as a trans-1,2-dibromide results when a cycloalkene is brominated. We’ll look at epoxide chemistry in more detail in Section 18.6.
Hydroxylation can also be carried out directly (without going through an intermediate epoxide) by treating an alkene with osmium tetroxide, OsO4. The reaction occurs with syn stereochemistry and does not involve a carbocation intermediate. Instead, it takes place through an intermediate cyclic osmate, which is formed in a single step by addition of OsO4 to the alkene. This cyclic osmate is then cleaved using aqueous sodium bisulfite, NaHSO3.
Because OsO4 is both very expensive and very toxic, the reaction is usually carried out using only a small, catalytic amount of OsO4 in the presence of a stoichiometric amount of a safe and inexpensive co-oxidant such as N-methylmorpholine N-oxide, abbreviated NMO. The initially formed osmate intermediate reacts rapidly with NMO to yield the product diol plus N-methylmorpholine and reoxidized OsO4, which reacts with more alkene in a catalytic cycle.
Problem 8-13
What product would you expect from reaction of cis-2-butene with meta-chloroperoxybenzoic acid? Show the stereochemistry.
Problem 8-14 Starting with an alkene, how would you prepare each of the following compounds?
(a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.08%3A_Oxidation_of_Alkenes_-_Epoxidation_and_Hydroxylation.txt |
In all the alkene addition reactions we’ve seen thus far, the carbon–carbon double bond has been converted into a single bond but the carbon skeleton has been unchanged. There are, however, powerful oxidizing reagents that will cleave $C═CC═C$ bonds and produce two carbonyl-containing fragments.
Ozone (O3) is perhaps the most useful double-bond cleavage reagent. Prepared by passing a stream of oxygen through a high-voltage electrical discharge, ozone adds rapidly to a $C═CC═C$ bond at low temperature to give a cyclic intermediate called a molozonide. Once formed, the molozonide spontaneously rearranges to form an ozonide. Although we won’t study the mechanism of this rearrangement in detail, it involves the molozonide coming apart into two fragments that then recombine in a different way.
Low-molecular-weight ozonides are explosive and therefore not isolated. Instead, the ozonide is immediately treated with a reducing agent, such as zinc metal in acetic acid, to produce carbonyl compounds. The net result of the ozonolysis/reduction sequence is that the $C═CC═C$ bond is cleaved and an oxygen atom becomes doubly bonded to each of the original alkene carbons. If an alkene with a tetrasubstituted double bond is ozonized, two ketone fragments result; if an alkene with a trisubstituted double bond is ozonized, one ketone and one aldehyde result; and so on.
Several oxidizing reagents other than ozone also cause double-bond cleavage, although such reactions are not often used. For example, potassium permanganate (KMnO4) in neutral or acidic solution cleaves alkenes to give carbonyl-containing products. If hydrogens are present on the double bond, carboxylic acids are produced; if two hydrogens are present on one carbon, CO2 is formed.
In addition to direct cleavage with ozone or KMnO4, an alkene can also be cleaved in a two-step process by initial hydroxylation to a 1,2-diol, as discussed in the previous section, followed by treatment of the diol with periodic acid, HIO4. If the two −OH groups are in an open chain, two carbonyl compounds result. If the two −OH groups are on a ring, a single, open-chain dicarbonyl compound is formed. As indicated in the following examples, the cleavage reaction takes place through a cyclic periodate intermediate.
Worked Example 8.3
Predicting the Reactant in an Ozonolysis Reaction
What alkene would yield a mixture of cyclopentanone and propanal on treatment with ozone followed by reduction with zinc?
Strategy
Reaction of an alkene with ozone, followed by reduction with zinc, cleaves the $C=CC=C$ bond and becomes two $C=OC=O$ bonds. Working backward from the carbonyl-containing products, the alkene precursor can be found by removing the oxygen from each product and joining the two carbon atoms.
Solution
Problem 8-15 What products would you expect from reaction of 1-methylcyclohexene with the following reagents? (a)
Aqueous acidic KMnO4
(b) O3, followed by Zn, CH3CO2H
Problem 8-16 Propose structures for alkenes that yield the following products on reaction with ozone followed by treatment with Zn: (a)
(CH3)2C$\text{=}$ O + H2C$\text{=}$ O
(b) 2 equiv CH3CH2CH$\text{=}$O
8.10: Addition of Carbenes to Alkenes - Cyclopropane Synthesis
Yet another kind of alkene addition is the reaction with a carbene to yield a cyclopropane. A carbene, R2C:, is a neutral molecule containing a divalent carbon with only six electrons in its valence shell. It is therefore highly reactive and generated only as a reaction intermediate, rather than as an isolable molecule. Because they’re electron-deficient, carbenes behave as electrophiles and react with nucleophilic $C═CC═C$ bonds. The reaction occurs in a single step without intermediates.
One of the simplest methods for generating a substituted carbene is by treatment of chloroform, CHCl3, with a strong base such as KOH. As shown in Figure 8.10, the loss of a proton from CHCl3 gives trichloromethanide anion, :CCl2, which spontaneously expels a Cl ion to yield dichlorocarbene, :CCl2.
Figure 8.10 MECHANISM Mechanism of the formation of dichlorocarbene by reaction of chloroform with strong base. Deprotonation of CHCl3 gives the trichloromethanide anion, $−:CCl3−:CCl3$, which spontaneously expels a Cl ion.
The carbon atom in dichlorocarbene is sp2-hybridized, with a vacant p orbital extending above and below the plane of the three atoms and with an unshared pair of electrons occupying the third sp2 lobe. Note that this electronic description of dichlorocarbene is similar to that of a carbocation (Section 7.9) with respect to both the sp2 hybridization of carbon and the vacant p orbital. Electrostatic potential maps further illustrate the similarity (Figure 8.11).
coincides with the empty p orbital in both dichlorocarbene and a carbocation (CH3+). The negative region in the dichlorocarbene map coincides with the lone-pair electrons.
If dichlorocarbene is generated in the presence of an alkene, addition to the double bond occurs and a dichlorocyclopropane is formed. As the reaction of dichlorocarbene with cis-2-pentene demonstrates, the addition is stereospecific, meaning that only a single stereoisomer is formed as product. Starting from a cis alkene, for instance, only cis-disubstituted cyclopropane is produced; starting from a trans alkene, only trans-disubstituted cyclopropane is produced.
The best method for preparing nonhalogenated cyclopropanes is by a process called the Simmons–Smith reaction. First investigated at the DuPont company, this reaction does not involve a free carbene. Rather, it utilizes a carbenoid—a metal-complexed reagent with carbene-like reactivity. When diiodomethane is treated with a specially prepared zinc–copper mix, (iodomethyl)zinc iodide, ICH2ZnI, is formed. In the presence of an alkene, ICH2ZnI transfers a CH2 group to the double bond to yield cyclopropane. For example, cyclohexene reacts cleanly and with good yield to give the corresponding cyclopropane. Although we won’t discuss the mechanistic details, carbene addition to an alkene is one of a general class of reactions called cycloadditions, which we’ll study more carefully in Chapter 30.
Problem 8-17 What products would you expect from the following reactions?
(a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.09%3A_Oxidation_of_Alkenes_-_Cleavage_to_Carbonyl_Compounds.txt |
In our brief introduction to radical reactions in Section 6.6, we said that radicals can add to $C═CC═C$ bonds, taking one electron from the double bond and leaving one behind to yield a new radical. Let’s now look at the process in more detail, focusing on the industrial synthesis of alkene polymers. A polymer is a large—sometimes very large—molecule, built up by repetitive joining together of many smaller molecules, called monomers.
Nature makes wide use of biological polymers. Cellulose, for instance, is a polymer built of repeating glucose monomer units; proteins are polymers built of repeating amino acid monomers; and nucleic acids are polymers built of repeating nucleotide monomers.
Synthetic polymers, such as polyethylene, are much simpler chemically than biopolymers, but there is still a great diversity to their structures and properties, depending on the identity of the monomers and on the reaction conditions used for polymerization. The simplest synthetic polymers are those that result when an alkene is treated with a small amount of a suitable catalyst. Ethylene, for example, yields polyethylene, an enormous alkane that may have a molecular weight up to 6 million u and may contain as many as 200,000 monomer units. Worldwide production of polyethylene is approximately 88 million tons per year.
Polyethylene and other simple alkene polymers are called chain-growth polymers because they are formed in a chain-reaction process in which an initiator adds to a carbon–carbon double bond to yield a reactive intermediate. The intermediate then reacts with a second molecule of monomer to yield a new intermediate, which reacts with a third monomer unit, and so on.
Historically, ethylene polymerization was carried out at high pressure (1000–3000 atm) and high temperature (100–250 °C) in the presence of a radical initiator such as benzoyl peroxide. Like many radical reactions, the mechanism of ethylene polymerization occurs in three steps: initiation, propagation, and termination:
• Initiation The polymerization reaction is initiated when a few radicals are generated on heating a small amount of benzoyl peroxide catalyst to break the weak O−O bond. The initially formed benzoyloxy radical loses CO2 and gives a phenyl radical (Ph·), which adds to the $C═CC═C$ bond of ethylene to start the polymerization process. One electron from the ethylene double bond pairs up with the odd electron on the phenyl radical to form a new C−C bond, and the other electron remains on carbon.
• Propagation Polymerization occurs when the carbon radical formed in the initiation step adds to another ethylene molecule to yield another radical. Repetition of the process for hundreds or thousands of times builds the polymer chain.
• Termination The chain process is eventually ended by a reaction that consumes the radical. The combination of two growing chains is one possible chain-terminating reaction.
$2R–CH2CH2·⟶R–CH2CH2CH2CH2–R2R–CH2CH2·⟶R–CH2CH2CH2CH2–R$
Ethylene is not unique in its ability to form a polymer. Many substituted ethylenes, called vinyl monomers, also undergo polymerization to yield polymers with substituent groups regularly spaced on alternating carbon atoms along the chain. Propylene, for example, yields polypropylene, and styrene yields polystyrene.
When an unsymmetrically substituted vinyl monomer such as propylene or styrene is polymerized, the radical addition steps can take place at either end of the double bond to yield either a primary radical intermediate (RCH2·) or a secondary radical (R2CH·). Just as in electrophilic addition reactions, however, we find that only the more highly substituted, secondary radical is formed.
Table 8.1 shows some commercially important alkene polymers, their uses, and the monomers from which they are made.
Table 8.1 Some Alkene Polymers and Their Uses
Monomer Formula Trade or common name of polymer Uses
Ethylene $H2C═CH2H2C═CH2$ Polyethylene Packaging, bottles
Propene (propylene) $H2C═CHCH3H2C═CHCH3$ Polypropylene Moldings, rope, carpets
Chloroethylene (vinyl chloride) $H2C═CHClH2C═CHCl$ Poly(vinyl chloride) Insulation, films, pipes
Styrene $H2C═CHC6H5H2C═CHC6H5$ Polystyrene Foam, moldings
Tetrafluoroethylene $F2C═CF2F2C═CF2$ Teflon Gaskets, nonstick coatings
Acrylonitrile $H2C═CHCNH2C═CHCN$ Orlon, Acrilan Fibers
Methyl methacrylate Plexiglas, Lucite Paint, sheets, moldings
Vinyl acetate $H2C═CHOCOCH3H2C═CHOCOCH3$ Poly(vinyl acetate) Paint, adhesives, foams
Worked Example 8.4
Predicting the Structure of a Polymer
Show the structure of poly(vinyl chloride), a polymer made from $H2C═CHClH2C═CHCl$, by drawing several repeating units.
Strategy
Mentally break the carbon–carbon double bond in the monomer unit, and form single bonds by connecting numerous units together.
Solution
The general structure of poly(vinyl chloride) is
Problem 8-18 Show the monomer units you would use to prepare the following polymers: (a)
(b)
Problem 8-19
One of the chain-termination steps that sometimes occurs to interrupt polymerization is the following reaction between two radicals. Propose a mechanism for the reaction, using fishhook arrows to indicate electron flow. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.11%3A_Radical_Additions_to_Alkenes_-__Chain-Growth_Polymers.txt |
The same high reactivity of radicals that enables the alkene polymerization we saw in the previous section also makes it difficult to carry out controlled radical reactions on complex molecules. As a result, there are severe limitations on the usefulness of radical addition reactions in the laboratory. In contrast to an electrophilic addition, where reaction occurs once and the reactive cation intermediate is rapidly quenched by a nucleophile, the reactive intermediate in a radical reaction is not usually quenched. Instead, it reacts again and again in a largely uncontrollable way.
In biological reactions, the situation is different from that in the laboratory. Only one substrate molecule at a time is present in the active site of an enzyme, and that molecule is held in a precise position, with other necessary reacting groups nearby. As a result, biological radical reactions are more controlled and more common than laboratory or industrial radical reactions. A particularly impressive example occurs in the biosynthesis of prostaglandins from arachidonic acid, where a sequence of four radical additions take place. Its reaction mechanism was discussed briefly in Section 6.6.
As shown in Figure 8.12, prostaglandin biosynthesis begins with abstraction of a hydrogen atom from C13 of arachidonic acid by an iron–oxy radical to give a carbon radical that reacts with O2 at C11 through a resonance form. The oxygen radical that results adds to the C8–C9 double bond to give a carbon radical at C8, which adds to the C12–C13 double bond and gives a carbon radical at C13. A resonance form of this carbon radical adds at C15 to a second O2 molecule, completing the prostaglandin skeleton. Reduction of the O−O bond then gives prostaglandin H2, called PGH2. The pathway looks complicated, but the entire process is catalyzed with exquisite control by a single enzyme.
and 5 are radical addition reactions to O2; steps 3 and 4 are radical additions to carbon–carbon double bonds.
8.13: Stereochemistry of Reactions - Addition of HO to an Achiral Alkene
Most of the biochemical reactions that take place in the body, as well as many organic reactions in the laboratory, yield products with chirality centers. For example, acid-catalyzed addition of H2O to 1-butene in the laboratory yields 2-butanol, a chiral alcohol. What is the stereochemistry of this chiral product? If a single enantiomer is formed, is it R or S? If a mixture of enantiomers is formed, how much of each? In fact, the 2-butanol produced is a racemic mixture of R and S enantiomers. Let’s see why.
To understand why a racemic product results from the reaction of H2O with 1-butene, think about the reaction mechanism. 1-Butene is first protonated to yield an intermediate secondary carbocation. Because the trivalent carbon is sp2-hybridized and planar, the cation has a plane of symmetry and is achiral. As a result, it can react with H2O equally well from either the top or the bottom. Reaction from the top leads to (S)-2-butanol through transition state 1 (TS 1) in Figure 8.13, and reaction from the bottom leads to (R)-2-butanol through TS 2. But the two transition states are mirror images, so they have identical energies, form at identical rates, and are equally likely to occur.
As a general rule, the formation of a new chirality center by achiral reactants always leads to a racemic mixture of enantiomeric products. Put another way, optical activity can’t appear from nowhere; an optically active product can only result by starting with an optically active reactant or chiral environment (Section 5.12).
In contrast to laboratory reactions, enzyme-catalyzed biological reactions often give a single enantiomer of a chiral product, even when the substrate is achiral. One step in the citric acid cycle of food metabolism, for instance, is the aconitase-catalyzed addition of water to (Z)-aconitate (usually called cis-aconitate) to give isocitrate.
Even though cis-aconitate is achiral, only the (2R,3S) enantiomer of the product is formed. As discussed in Section 5.11 and Section 5.12, cis-aconitate is a prochiral molecule, which is held in a chiral environment by the aconitase enzyme during the reaction. In this environment, the two faces of the double bond are chemically distinct, and addition occurs on only the Re face at C2. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.12%3A_Biological_Additions_of_Radicals_to__Alkenes.txt |
The reaction discussed in the previous section involves an addition to an achiral reactant and forms an optically inactive, racemic mixture of two enantiomeric products. What would happen, though, if we were to carry out the reaction on a single enantiomer of a chiral reactant? For example, what stereochemical result would be obtained from addition of H2O to a chiral alkene, such as (R)-4-methyl-1-hexene? The product of the reaction, 4-methyl-2-hexanol, has two chirality centers and so has four possible stereoisomers.
Let’s think about the two chirality centers separately. What about the configuration at C4, the methyl-bearing carbon atom? Since C4 has the R configuration in the starting material and this chirality center is unaffected by the reaction, its configuration is unchanged. Thus, the configuration at C4 in the product remains R (assuming that the relative rankings of the four attached groups are not changed by the reaction).
What about the configuration at C2, the newly formed chirality center? As shown in Figure 8.14, the stereochemistry at C2 is established by reaction of H2O with a carbocation intermediate in the usual manner. But this carbocation doesn’t have a plane of symmetry; it is chiral because of the chirality center at C4. Because the carbocation is chiral and has no plane of symmetry, it doesn’t react equally well from the top and bottom faces. One of the two faces is likely, for steric reasons, to be a bit more accessible than the other, leading to a mixture of R and S products in some ratio other than 50 : 50. Thus, two diastereomeric products, (2R,4R)-4-methyl-2-hexanol and (2S,4R)-4-methyl-2-hexanol, are formed in unequal amounts, and the mixture is optically active.
As a general rule, the formation of a new chirality center by a chiral reactant leads to unequal amounts of diastereomeric products. If the chiral reactant is optically active because only one enantiomer is used rather than a racemic mixture, then the products are also optically active.
Problem 8-20
What products are formed from acid-catalyzed hydration of racemic (±)-4-methyl-1-hexene? What can you say about the relative amounts of the products? Is the product mixture optically active?
Problem 8-21
What products are formed from hydration of 4-methylcyclopentene? What can you say about the relative amounts of the products?
8.15: Chemistry MattersTerpenes- Naturally Occurring Alkenes
8 • Chemistry Matters 8 • Chemistry Matters
Ever since its discovery in Persia around 1000 A.D., it has been known that steam distillation, the distillation of plant materials together with water, produces a fragrant mixture of liquids called essential oils. The resulting oils have long been used as medicines, spices, and perfumes, and their investigation played a major role in the emergence of organic chemistry as a science during the 19th century.
Chemically, plant essential oils consist largely of mixtures of compounds called terpenoids—small organic molecules with an immense diversity of structure. More than 60,000 different terpenoids are known. Some are open-chain molecules, and others contain rings; some are hydrocarbons, and others contain oxygen. Hydrocarbon terpenoids, in particular, are known as terpenes, and all contain double bonds. For example:
Regardless of their apparent structural differences, all terpenoids are related. According to a formalism called the isoprene rule, they can be thought of as arising from head-to-tail joining of 5-carbon isoprene units (2-methyl-1,3-butadiene). Carbon 1 is the head of the isoprene unit, and carbon 4 is the tail. For example, myrcene contains two isoprene units joined head to tail, forming an 8-carbon chain with two 1-carbon branches. α-Pinene similarly contains two isoprene units assembled into a more complex cyclic structure, and humulene contains three isoprene units. See if you can identify the isoprene units in α-pinene, humulene, and β-santalene.
Terpenes (and terpenoids) are further classified according to the number of 5-carbon units they contain. Thus, monoterpenes are 10-carbon substances derived from two isoprene units, sesquiterpenes are 15-carbon molecules derived from three isoprene units, diterpenes are 20-carbon substances derived from four isoprene units, and so on. Monoterpenes and sesquiterpenes are found primarily in plants, but the higher terpenoids occur in both plants and animals, and many have important biological roles. The triterpenoid lanosterol, for instance, is the biological precursor from which all steroid hormones are made.
Isoprene itself is not the true biological precursor of terpenoids. Nature instead uses two “isoprene equivalents”—isopentenyl diphosphate and dimethylallyl diphosphate—which are themselves made by two different routes depending on the organism. Lanosterol, in particular, is biosynthesized from acetic acid by a complex pathway that has been worked out in great detail. We’ll look at the subject more closely in Sections 27.5 and 27.7.
8.16: Key Terms
8 • Key Terms 8 • Key Terms
• anti stereochemistry
• bromonium ion
• carbene, R2C
• chain-growth polymer
• epoxide
• glycol
• halohydrin
• hydroboration
• hydrogenated
• hydroxylation
• monomer
• oxidation
• oxirane
• oxymercuration–demercuration
• ozonide
• polymer
• reduction
• Simmons–Smith reaction
• stereospecific
• syn stereochemistry | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.14%3A_Stereochemistry_of_Reactions_-_Addition_of_HO_to_a_Chiral_Alkene.txt |
8 • Summary 8 • Summary
With the background needed to understand organic reactions now covered, this chapter has begun the systematic description of major functional groups.
Alkenes are generally prepared by an elimination reaction, such as dehydrohalogenation, the elimination of HX from an alkyl halide, or dehydration, the elimination of water from an alcohol. The converse of this elimination reaction is the addition of various substances to the alkene double bond to give saturated products.
HCl, HBr, and HI add to alkenes by a two-step electrophilic addition mechanism. Initial reaction of the nucleophilic double bond with H+ gives a carbocation intermediate, which then reacts with halide ion. Bromine and chlorine add to alkenes via three-membered-ring bromonium ion or chloronium ion intermediates to give addition products having anti stereochemistry. If water is present during the halogen addition reaction, a halohydrin is formed.
Hydration of an alkene—the addition of water—is carried out by either of two procedures, depending on the product desired. Oxymercuration–demercuration involves electrophilic addition of Hg2+ to an alkene, followed by trapping of the cation intermediate with water and subsequent treatment with NaBH4. Hydroboration involves addition of borane (BH3) followed by oxidation of the intermediate organoborane with alkaline H2O2. The two hydration methods are complementary: oxymercuration–demercuration gives the product of Markovnikov addition, whereas hydroboration–oxidation gives the product with non-Markovnikov syn stereochemistry.
Alkenes are reduced by addition of H2 in the presence of a catalyst such as platinum or palladium to yield alkanes, a process called catalytic hydrogenation. Alkenes are also oxidized by reaction with a peroxyacid to give epoxides, which can be converted into trans-1,2-diols by acid-catalyzed hydrolysis. The corresponding cis-1,2-diols can be made directly from alkenes by hydroxylation with OsO4. Alkenes can also be cleaved to produce carbonyl compounds by reaction with ozone, followed by reduction with zinc metal. In addition, alkenes react with divalent substances called carbenes, R2C:, to give cyclopropanes. Nonhalogenated cyclopropanes are best prepared by treatment of the alkene with CH2I2 and zinc–copper, a process called the Simmons–Smith reaction.
Alkene polymers—large molecules resulting from repetitive bonding of many hundreds or thousands of small monomer units—are formed by chain-reaction polymerization of simple alkenes. Polyethylene, polypropylene, and polystyrene are examples. As a general rule, radical addition reactions are not common in the laboratory but occur frequently in biological pathways.
Many reactions give chiral products. If the reactants are optically inactive, the products are also optically inactive. If one or both of the reactants is optically active, the products can also be optically active.
Learning Reactions
What’s seven times nine? Sixty-three, of course. You didn’t have to stop and figure it out; you knew the answer immediately because you long ago learned the multiplication tables. Learning the reactions of organic chemistry requires the same approach: reactions have to be learned for quick recall if they are to be useful.
Different people take different approaches to learning reactions. Some people make flashcards; others find studying with friends to be helpful. To help guide your study, most chapters in this book end with a summary of the reactions just presented. In addition, the accompanying Study Guide and Student Solutions Manual has several appendixes that organize organic reactions from other perspectives. Fundamentally, though, there are no shortcuts. Learning organic chemistry does take effort.
8.18: Summary of Reactions
8 • Summary of Reactions 8 • Summary of Reactions
No stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.
1. Addition reactions of alkenes
1. Addition of HCl, HBr, and HI (Section 7.7 and Section 7.8)
Markovnikov regiochemistry occurs, with H adding to the less highly substituted alkene carbon and halogen adding to the more highly substituted carbon.
• Addition of halogens Cl2 and Br2 (Section 8.2)
Anti addition is observed through a halonium ion intermediate.
• Halohydrin formation (Section 8.3)
Markovnikov regiochemistry and anti stereochemistry occur.
• Addition of water by oxymercuration–demercuration (Section 8.4)
Markovnikov regiochemistry occurs.
• Addition of water by hydroboration–oxidation (Section 8.5)
Non-Markovnikov syn addition occurs.
• Catalytic hydrogenation (Section 8.6)
Syn addition occurs.
• Epoxidation with a peroxyacid (Section 8.7)
Syn addition occurs.
• Hydroxylation with OsO4 (Section 8.7)
Syn addition occurs.
• Addition of carbenes to yield cyclopropanes (Section 8.9)
(1) Dichlorocarbene addition
(2) Simmons–Smith reaction
• Hydroxylation by acid-catalyzed epoxide hydrolysis (Section 8.7)
Anti stereochemistry occurs.
• Oxidative cleavage of alkenes (Section 8.8)
1. Reaction with ozone followed by zinc in acetic acid
2. Reaction with KMnO4 in acidic solution
• Cleavage of 1,2-diols (Section 8.8) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.17%3A_Summary.txt |
8 • Additional Problems 8 • Additional Problems
Visualizing Chemistry
Problem 8-22
Name the following alkenes, and predict the products of their reaction with (1) meta-chloroperoxybenzoic acid, (2) KMnO4 in aqueous acid, (3) O3, followed by Zn in acetic acid: (a)
(b)
Problem 8-23 Draw the structures of alkenes that would yield the following alcohols on hydration (red = O). Tell in each case whether you would use hydroboration–oxidation or oxymercuration–demercuration. (a) (b) Problem 8-24 The following alkene undergoes hydroboration–oxidation to yield a single product rather than a mixture. Explain the result, and draw the product showing its stereochemistry. Problem 8-25 From what alkene was the following 1,2-diol made, and what method was used, epoxide hydrolysis or OsO4? Mechanism Problems Problem 8-26 Predict the products for the following reactions, showing the complete mechanism and appropriate stereochemistry: (a) (b) (c) Problem 8-27 Draw the structures of the organoboranes formed when borane reacts with the following alkenes, including the regiochemistry and stereochemistry as appropriate. Propose a mechanism for each reaction. (a) (b) (c) Problem 8-28 meta-Chlorobenzoic acid is not the only peroxyacid capable of epoxide formation. For each reaction below, predict the products and show the mechanism. (a) (b) Problem 8-29 Give the mechanism and products for the following acid-catalyzed epoxide-opening reactions, including appropriate stereochemistry. (a) (b) (c) Problem 8-30 Which of the reactions below would result in a product mixture that would rotate plane-polarized light? (a) (b) (c) Problem 8-31 Reaction of 2-methylpropene with CH3OH in the presence of H2SO4 catalyst yields methyl tert-butyl ether, CH3OC(CH3)3, by a mechanism analogous to that of acid-catalyzed alkene hydration. Write the mechanism, using curved arrows for each step. Problem 8-32 Iodine azide, IN3, adds to alkenes by an electrophilic mechanism similar to that of bromine. If a monosubstituted alkene such as 1-butene is used, only one product results: (a) Add lone-pair electrons to the structure shown for IN3, and draw a second resonance form for the molecule. (b)
Calculate formal charges for the atoms in both resonance structures you drew for IN3 in part (a).
(c) In light of the result observed when IN3 adds to 1-butene, what is the polarity of the I − N3 bond? Propose a mechanism for the reaction using curved arrows to show the electron flow in each step.
Problem 8-33
10-Bromo-α-chamigrene, a compound isolated from marine algae, is thought to be biosynthesized from γ-bisabolene by the following route:
Draw the structures of the intermediate bromonium and cyclic carbocation, and propose mechanisms for all three steps.
Problem 8-34
Isolated from marine algae, prelaureatin is thought to be biosynthesized from laurediol by the following route. Propose a mechanism.
Problem 8-35
Dichlorocarbene can be generated by heating sodium trichloroacetate. Propose a mechanism for the reaction, and use curved arrows to indicate the movement of electrons in each step. What relationship does your mechanism bear to the base-induced elimination of HCl from chloroform?
Problem 8-36
Reaction of cyclohexene with mercury(II) acetate in CH3OH rather than H2O, followed by treatment with NaBH4, yields cyclohexyl methyl ether rather than cyclohexanol. Suggest a mechanism.
Problem 8-37
Use your general knowledge of alkene chemistry to suggest a mechanism for the following reaction.
Problem 8-38
Treatment of 4-penten-1-ol with aqueous Br2 yields a cyclic bromo ether rather than the expected bromohydrin. Suggest a mechanism, using curved arrows to show electron movement.
Problem 8-39
Hydroboration of 2-methyl-2-pentene at 25 °C, followed by oxidation with alkaline H2O2, yields 2-methyl-3-pentanol, but hydroboration at 160 °C followed by oxidation yields 4-methyl-1-pentanol. Suggest a mechanism.
Reactions of Alkenes
Problem 8-40
Predict the products of the following reactions (the aromatic ring is unreactive in all cases). Indicate regiochemistry when relevant.
Problem 8-41 Suggest structures for alkenes that give the following reaction products. There may be more than one answer for some cases. (a) (b) (c) (d) (e) (f) Problem 8-42 Predict the products of the following reactions, showing both regiochemistry and stereochemistry where appropriate: (a) (b) (c) (d) Problem 8-43 Which reaction would you expect to be faster, addition of HBr to cyclohexene or to 1-methylcyclohexene? Explain. Problem 8-44 What product will result from hydroboration–oxidation of 1-methylcyclopentene with deuterated borane, BD3? Show both the stereochemistry (spatial arrangement) and the regiochemistry (orientation) of the product. Problem 8-45 The cis and trans isomers of 2-butene give different cyclopropane products in the Simmons–Smith reaction. Show the structures of both, and explain the difference. Problem 8-46 Predict the products of the following reactions. Don’t worry about the size of the molecule; concentrate on the functional groups. Problem 8-47 Addition of HCl to 1-methoxycyclohexene yields 1-chloro-1-methoxycyclohexane as a sole product. Use resonance structures of the carbocation intermediate to explain why none of the alternate regioisomer is formed. Synthesis Using Alkenes Problem 8-48 How would you carry out the following transformations? What reagents would you use in each case? (a) (b) (c) (d) (e) (f) Problem 8-49 Draw the structure of an alkene that yields only acetone, (CH3)2C$\text{=}$O, on ozonolysis followed by treatment with Zn. Problem 8-50 Show the structures of alkenes that give the following products on oxidative cleavage with KMnO4 in acidic solution: (a) (b) (c) (d) Problem 8-51 In planning the synthesis of one compound from another, it’s just as important to know what not to do as to know what to do. The following reactions all have serious drawbacks to them. Explain the potential problems of each. (a) (b) (c) (d) Problem 8-52 Which of the following alcohols could not be made selectively by hydroboration–oxidation of an alkene? Explain. (a) (b) (c) (d) Polymers Problem 8-53 Plexiglas, a clear plastic used to make many molded articles, is made by polymerization of methyl methacrylate. Draw a representative segment of Plexiglas. Problem 8-54 Poly(vinyl pyrrolidone), prepared from N-vinylpyrrolidone, is used both in cosmetics and as a component of a synthetic substitute for blood. Draw a representative segment of the polymer. Problem 8-55 When a single alkene monomer, such as ethylene, is polymerized, the product is a homopolymer. If a mixture of two alkene monomers is polymerized, however, a copolymer often results. The following structure represents a segment of a copolymer called Saran. What two monomers were copolymerized to make Saran? General Problems Problem 8-56 Compound A has the formula C10H16. On catalytic hydrogenation over palladium, it reacts with only 1 molar equivalent of H2. Compound A also undergoes reaction with ozone, followed by zinc treatment, to yield a symmetrical diketone, B (C10H16O2). (a) How many rings does A have? (b)
What are the structures of A and B?
(c) Write the reactions.
Problem 8-57
An unknown hydrocarbon A with the formula C6H12 reacts with 1 molar equivalent of H2 over a palladium catalyst. Hydrocarbon A also reacts with OsO4 to give diol B. When oxidized with KMnO4 in acidic solution, A gives two fragments. One fragment is propanoic acid, CH3CH2CO2H, and the other fragment is ketone C. What are the structures of A, B, and C? Write all reactions.
Problem 8-58
Using an oxidative cleavage reaction, explain how you would distinguish between the following two isomeric dienes:
Problem 8-59
Compound A, C10H18O, undergoes reaction with dilute H2SO4 at 50 °C to yield a mixture of two alkenes, C10H16. The major alkene product, B, gives only cyclopentanone after ozone treatment followed by reduction with zinc in acetic acid. Identify A and B, and write the reactions.
Problem 8-60
Draw the structure of a hydrocarbon that absorbs 2 molar equivalents of H2 on catalytic hydrogenation and gives only butanedial on ozonolysis.
Problem 8-61
Simmons–Smith reaction of cyclohexene with diiodomethane gives a single cyclopropane product, but the analogous reaction of cyclohexene with 1,1-diiodoethane gives (in low yield) a mixture of two isomeric methylcyclopropane products. What are the two products, and how do they differ?
Problem 8-62
The sex attractant of the common housefly is a hydrocarbon with the formula C23H46. On treatment with aqueous acidic KMnO4, two products are obtained, CH3(CH2)12CO2H and CH3(CH2)7CO2H. Propose a structure.
Problem 8-63
Compound A has the formula C8H8. It reacts rapidly with KMnO4 to give CO2 and a carboxylic acid, B (C7H6O2), but reacts with only 1 molar equivalent of H2 on catalytic hydrogenation over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4 equivalents of H2 are taken up and hydrocarbon C (C8H16) is produced. What are the structures of A, B, and C? Write the reactions.
Problem 8-64 How would you distinguish between the following pairs of compounds using simple chemical tests? Tell what you would do and what you would see. (a)
Cyclopentene and cyclopentane
(b) 2-Hexene and benzene
Problem 8-65 α-Terpinene, C10H16, is a pleasant-smelling hydrocarbon that has been isolated from oil of marjoram. On hydrogenation over a palladium catalyst, α-terpinene reacts with 2 molar equivalents of H2 to yield a hydrocarbon, C10H20. On ozonolysis, followed by reduction with zinc and acetic acid, α-terpinene yields two products, glyoxal and 6-methyl-2,5-heptanedione. (a)
How many degrees of unsaturation does α-terpinene have?
(b) How many double bonds and how many rings does it have? (c) Propose a structure for α-terpinene.
Problem 8-66
Evidence that cleavage of 1,2-diols by HIO4 occurs through a five-membered cyclic periodate intermediate is based on the measurement of reaction rates. When diols A and B were prepared and the rates of their reaction with HIO4 were measured, it was found that diol A cleaved approximately 1 million times faster than diol B. Make molecular models of A and B and of potential cyclic periodate intermediates, and then explain the results.
Problem 8-67
Reaction of HBr with 3-methylcyclohexene yields a mixture of four products: cis- and trans-1-bromo-3-methylcyclohexane and cis- and trans-1-bromo-2-methylcyclohexane. The analogous reaction of HBr with 3-bromocyclohexene yields trans-1,2-dibromocyclohexane as the sole product. Draw structures of the possible intermediates, and then explain why only a single product is formed in this reaction.
Problem 8-68
We’ll see in the next chapter that alkynes undergo many of the same reactions that alkenes do. What product might you expect from each of the following reactions?
Problem 8-69
Hydroxylation of cis-2-butene with OsO4 yields a different product than hydroxylation of trans-2-butene. Draw the structure, show the stereochemistry of each product, and explain the difference between them.
Problem 8-70
Compound A, C11H16O, was found to be an optically active alcohol. Despite its apparent unsaturation, no hydrogen was absorbed on catalytic reduction over a palladium catalyst. On treatment of A with dilute sulfuric acid, dehydration occurred and an optically inactive alkene B, C11H14, was the major product. Alkene B, on ozonolysis, gave two products. One product was identified as propanal, CH3CH2CHO. Compound C, the other product, was shown to be a ketone, C8H8O. How many degrees of unsaturation does A have? Write the reactions, and identify A, B, and C. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes-_Reactions_and_Synthesis/8.19%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 9, you should be able to
1. fulfillall of the detailed objectives listed under each individual section.
2. solve road-map problems involving any of the reactions introduced to this point.
3. design multistep syntheses using any of the reactions introduced to this point, and determine the viability of a given synthesis.
4. define, and use in context, the key terms introduced.
Addition reactions not only dominate the chemistry of alkenes, they are also the major class of reaction you will encounter. This chapter discusses an important difference between (terminal) alkynes and alkenes, that is, the acidity of the former; it also addresses the problem of devising organic syntheses. Once you have completed this chapter you will have increased the number of organic reactions in your repertoire, and should be able to design much more elaborate multistep syntheses. As you work through Chapter 9, you should notice the many similarities among the reactions described here and those in Chapters 7 and 8.
• 9.1: Why This Chapter?
• 9.2: Naming Alkynes
Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of CnH2n−2 . They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule.
• 9.3: Preparation of Alkynes - Elimination Reactions of Dihalides
Alkynes can be a useful functional group to synthesize due to some of their antibacterial, antiparasitic, and antifungal properties. One simple method for alkyne synthesis is by double elimination from a dihaloalkane.
• 9.4: Reactions of Alkynes - Addition of HX and X₂
• 9.5: Hydration of Alkynes
As with alkenes, hydration (addition of water) to alkynes requires a strong acid, usually sulfuric acid, and is facilitated by mercuric sulfate. However, unlike the additions to double bonds which give alcohol products, addition of water to alkynes gives ketone products ( except for acetylene which yields acetaldehyde ). The explanation for this deviation lies in enol-keto tautomerization.
• 9.6: Reduction of Alkynes
Reactions between alkynes and catalysts are a common source of alkene formation. Because alkynes differ from alkenes on account of their two procurable π bonds, alkynes are more susceptible to additions. Aside from turning them into alkenes, these catalysts affect the arrangement of substituents on the newly formed alkene molecule. Depending on which catalyst is used, the catalysts cause anti- or syn-addition of hydrogens. Alkynes can readily undergo additions because of their availability of tw
• 9.7: Oxidative Cleavage of Alkynes
Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenens, the reactions conditions can be gentler.
• 9.8: Alkyne Acidity - Formation of Acetylide Anions
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion, RC=C:-. The origin of the enhanced acidity can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character.
• 9.9: Alkylation of Acetylide Anions
The alkylation of acetylide ions is important in organic synthesis because it is a reaction in which a new carbon-carbon bond is formed; hence, it can be used when an organic chemist is trying to build a complicated molecule from much simpler starting materials.
• 9.10: An Introduction to Organic Synthesis
• 9.11: Chemistry Matters—The Art of Organic Synthesis
• 9.12: Key Terms
• 9.13: Summary
• 9.14: Summary of Reactions
• 9.15: Additional Problems
09: Alkynes - An Introduction to Organic Synthesis
Chapter Contents
9.1 Naming Alkynes
9.2 Preparation of Alkynes: Elimination Reactions of Dihalides 9.3 Reactions of Alkynes: Addition of HX and X2 9.4 Hydration of Alkynes 9.5 Reduction of Alkynes 9.6 Oxidative Cleavage of Alkynes 9.7 Alkyne Acidity: Formation of Acetylide Anions 9.8 Alkylation of Acetylide Anions 9.9 An Introduction to Organic Synthesis
Alkynes are less common than alkenes, both in the laboratory and in living organisms, so we won’t cover them in great detail. The real importance of this chapter is that we’ll use alkyne chemistry as a vehicle to begin looking at some of the general strategies used in organic synthesis—the construction of complex molecules in the laboratory. Without the ability to design and synthesize new molecules in the laboratory, many of the medicines we take for granted would not exist and few new ones would be made.
An alkyne is a hydrocarbon that contains a carbon–carbon triple bond. Acetylene, $H–C≡C–HH–C≡C–H$, the simplest alkyne, was once widely used in industry as a starting material for the preparation of acetaldehyde, acetic acid, vinyl chloride, and other high-volume chemicals, but more efficient routes to these substances using ethylene as starting material are now available. Acetylene is still used in the preparation of acrylic polymers, such as Plexiglas and Lucite, but is probably best known as the gas burned in high-temperature oxy–acetylene welding torches.
In addition to simple alkynes with one triple bond, research is also being carried out on polyynes—linear carbon chains of alternating single and triple bonds. Polyynes with up to eight triple bonds are thought to be present in interstellar space, and evidence has been presented for the existence of carbyne, an allotrope of carbon consisting of alternating single and triple bonds in long chains of indefinite length. The electronic properties of polyynes are being explored for potential use in nanotechnology applications. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.01%3A_Why_This_Chapter.txt |
Alkyne nomenclature follows the general rules for hydrocarbons discussed in Section 3.4 and Section 7.3. The suffix -yne is used, and the position of the triple bond is indicated by giving the number of the first alkyne carbon in the chain. Numbering the main chain begins at the end nearer the triple bond so that the triple bond receives as low a number as possible.
Compounds with more than one triple bond are called diynes, triynes, and so forth; compounds containing both double and triple bonds are called enynes (not ynenes). Numbering of an enyne chain starts from the end nearer the first multiple bond, whether double or triple. When there is a choice in numbering, double bonds receive lower numbers than triple bonds. For example:
As with alkyl and alkenyl substituents derived from alkanes and alkenes, respectively, alkynyl groups are also possible.
Problem 9-1 Name the following alkynes: (a)
(b)
(c)
(d)
(e)
Problem 9-2 There are seven isomeric alkynes with the formula C6H10. Draw and name them.
9.03: Preparation of Alkynes - Elimination Reactions of Dihalides
Alkynes can be prepared by the elimination of HX from alkyl halides in a similar manner as alkenes (Section 8.1). Treatment of a 1,2-dihaloalkane (called a vicinal dihalide) with an excess amount of a strong base such as KOH or NaNH2 results in a twofold elimination of HX and formation of an alkyne. As with the elimination of HX to form an alkene, we’ll defer a full discussion of this topic and the relevant reaction mechanisms to Chapter 11.
The starting vicinal dihalides are themselves readily available by addition of Br2 or Cl2 to alkenes. Thus, the overall halogenation/dehydrohalogenation sequence makes it possible to go from an alkene to an alkyne. For example, diphenylethylene is converted into diphenylacetylene by reaction with Br2 and subsequent base treatment.
The twofold dehydrohalogenation takes place through a vinylic halide intermediate, which suggests that vinylic halides themselves should give alkynes when treated with strong base. (A vinylic substituent is one that is attached to a double-bond.) This is indeed the case. For example:
9.04: Reactions of Alkynes - Addition of HX and X
You might recall from Section 1.9 that a carbon–carbon triple bond results from the interaction of two sp-hybridized carbon atoms. The two sp hybrid orbitals of carbon lie at an angle of 180° to each other along an axis perpendicular to the axes of the two unhybridized 2py and 2pz orbitals. When two sp-hybridized carbons approach each other, one spsp σ bond and two p–p π bonds are formed. The two remaining sp orbitals form bonds to other atoms at an angle of 180° from the carbon–carbon bond. Thus, acetylene is a linear molecule with $H–C≡CFigure 9.2). The length of the C≡CC≡C bond is 120 pm, and its strength is approximately 965 kJ/mol (231 kcal/mol), making it the shortest and strongest known carbon–carbon bond.$
around the molecule.
As a general rule, electrophiles undergo addition reactions with alkynes much as they do with alkenes. Take the reaction of alkynes with HX, for instance. The reaction often can be stopped with the addition of 1 equivalent of HX, but reaction with an excess of HX leads to a dihalide product. For example, reaction of 1-hexyne with 2 equivalents of HBr yields 2,2-dibromohexane. As the following examples indicate, the regiochemistry of addition follows Markovnikov’s rule, with halogen adding to the more highly substituted side of the alkyne bond and hydrogen adding to the less highly substituted side. Trans stereochemistry of H and X normally, although not always, occurs in the product.
Bromine and chlorine also add to alkynes to give addition products, and trans stereochemistry again results.
The mechanism of alkyne addition is similar but not identical to that of alkene addition. When an electrophile such as HBr adds to an alkene, the reaction takes place in two steps and involves an alkyl carbocation intermediate (Section 7.7 and Section 7.8). If HBr were to add by the same mechanism to an alkyne, an analogous vinylic carbocation would be formed as the intermediate.
A vinylic carbocation has an sp-hybridized carbon and generally forms less readily than an alkyl carbocation (Figure 9.3). As a rule, a secondary vinylic carbocation forms about as readily as a primary alkyl carbocation, but a primary vinylic carbocation is so difficult to form that there is no clear evidence it even exists. Thus, many alkyne additions occur through more complex mechanistic pathways.
coincide with lobes of the vacant p orbital and are perpendicular to the most negative regions associated with the π bond.
Problem 9-3 What products would you expect from the following reactions?
(a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.02%3A_Naming_Alkynes.txt |
Like alkenes (Section 8.4 and Section 8.5), alkynes can be hydrated by either of two methods. Direct addition of water catalyzed by mercury(II) ion yields the Markovnikov product, and indirect addition of water by a hydroboration–oxidation sequence yields the non-Markovnikov product.
Mercury(II)-Catalyzed Hydration of Alkynes
Alkynes don’t react directly with aqueous acid but will undergo hydration readily in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the −OH group adds to the more highly substituted carbon and the −H attaches to the less highly substituted one.
Interestingly, the actual product isolated from alkyne hydration is not a vinylic alcohol, or enol (ene + ol), but is instead a ketone. Although the enol is an intermediate in the reaction, it immediately rearranges into a ketone by a process called keto–enol tautomerism. The individual keto and enol forms are said to be tautomers, a word used to describe two isomers that undergo spontaneous interconversion accompanied by the change in position of a hydrogen. With few exceptions, the keto–enol tautomeric equilibrium lies on the side of the ketone; enols are almost never isolated. We’ll look more closely at this equilibrium in Section 22.1.
As shown in Figure 9.4, the mechanism of the mercury(II)-catalyzed alkyne hydration reaction is analogous to the oxymercuration reaction of alkenes (Section 8.4). Electrophilic addition of mercury(II) ion to the alkyne gives a vinylic cation, which reacts with water and loses a proton to yield a mercury-containing enol intermediate. In contrast with alkene oxymercuration, however, no treatment with NaBH4 is necessary to remove the mercury. The acidic reaction conditions alone are sufficient to effect replacement of mercury by hydrogen. Tautomerization then gives the ketone.
Figure 9.4 MECHANISM Mechanism of the mercury(II)-catalyzed hydration of an alkyne to yield a ketone. The reaction occurs through initial formation of an intermediate enol, which tautomerizes to the ketone.
A mixture of both possible ketones results when an unsymmetrically substituted internal alkyne ($RC≡CR′RC≡CR′$) is hydrated. The reaction is therefore most useful when applied to a terminal alkyne ($RC≡CHRC≡CH$) because only a methyl ketone is formed.
Problem 9-4 What products would you obtain by mercury-catalyzed hydration of the following alkynes?
(a)
(b)
Problem 9-5 What alkynes would you start with to prepare the following ketones?
(a)
(b)
Hydroboration–Oxidation of Alkynes
Borane adds rapidly to an alkyne just as it does to an alkene, and the resulting vinylic borane can be oxidized by H2O2 to give an enol, which tautomerizes to either a ketone or an aldehyde, depending on the alkyne. Hydroboration–oxidation of an internal alkyne such as 3-hexyne is straightforward and gives a ketone, but hydroboration–oxidation of a terminal alkyne is more complex because two molecules of borane often add to the triple bond, complicating the situation. To prevent this double addition, a bulky, sterically encumbered borane such as bis(1,2-dimethylpropyl)borane, known commonly as disiamylborane is used in place of BH3. When a terminal alkyne such as 1-butene reacts with disiamylborane, addition to the triple bond occurs normally, but a second addition is hindered by the bulk of the dialkylborane. Oxidation with H2O2 then gives an enol, which tautomerizes to the aldehyde.
The hydroboration–oxidation sequence is complementary to the direct, mercury(II)-catalyzed hydration reaction of a terminal alkyne because different products result. Direct hydration with aqueous acid and mercury(II) sulfate leads to a methyl ketone, whereas hydroboration–oxidation of the same terminal alkyne leads to an aldehyde.
Problem 9-6 What alkyne would you start with to prepare each of the following compounds by a hydroboration–oxidation reaction?
(a)
(b)
Problem 9-7 How would you prepare the following carbonyl compounds starting from an alkyne (reddish brown = Br)?
(a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.05%3A_Hydration_of_Alkynes.txt |
Alkynes are reduced to alkanes by addition of H2 over a metal catalyst. The reaction occurs in two steps through an alkene intermediate, and measurements show that the first step in the reaction is more exothermic than the second.
Complete reduction to the alkane occurs when palladium on carbon (Pd/C) is used as catalyst, but hydrogenation can be stopped at the alkene stage if the less active Lindlar catalyst is used. The Lindlar catalyst is a finely divided palladium metal that has been precipitated onto a calcium carbonate support and then deactivated by treatment with lead acetate and quinoline, an aromatic amine. The hydrogenation occurs with syn stereochemistry (Section 8.5), giving a cis alkene product.
The alkyne hydrogenation reaction has been explored extensively by the Hoffmann–LaRoche pharmaceutical company, where it is used in the commercial synthesis of vitamin A. The cis isomer of vitamin A produced initially on hydrogenation is converted to the trans isomer by heating.
An alternative method for the conversion of an alkyne to an alkene uses sodium or lithium metal as the reducing agent in liquid ammonia as solvent. This method is complementary to the Lindlar reduction because it produces trans rather than cis alkenes. For example, 5-decyne gives trans-5-decene on treatment with lithium in liquid ammonia. The mechanism is explained below.
Alkali metals dissolve in liquid ammonia at −33 °C to produce a deep blue solution containing the metal cation and ammonia-solvated electrons. When an alkyne is then added to the solution, reduction occurs by the mechanism shown in Figure 9.5. An electron first adds to the triple bond to yield an intermediate anion radical—a species that is both an anion (has a negative charge) and a radical (has an odd number of electrons). This anion radical is a strong base, able to remove H+ from ammonia to give a vinylic radical. Addition of a second electron to the vinylic radical gives a vinylic anion, which abstracts a second H+ from ammonia to give trans alkene product.
Figure 9.5 MECHANISM Mechanism of the lithium/ammonia reduction of an alkyne to produce a trans alkene.
Trans stereochemistry of the alkene product is established during the second reduction step (3) when the less-hindered trans vinylic anion is formed from the vinylic radical. Vinylic radicals undergo rapid cis–trans equilibration, but vinylic anions equilibrate much less rapidly. Thus, the more stable trans vinylic anion is formed rather than the less stable cis anion and is then protonated without equilibration.
Problem 9-8 Using any alkyne needed, how would you prepare the following alkenes? (a)
trans-2-Octene
(b) cis-3-Heptene (c) 3-Methyl-1-pentene
9.07: Oxidative Cleavage of Alkynes
Alkynes, like alkenes, can be cleaved by reaction with powerful oxidizing agents such as ozone or KMnO4, although the reaction is of little value and it is mentioned only for completeness. A triple bond is generally less reactive than a double bond, and yields of cleavage products can be low. The products obtained from cleavage of an internal alkyne are carboxylic acids; from a terminal alkyne, CO2 is formed as one product.
9.08: Alkyne Acidity - Formation of Acetylide Anions
The most striking difference between alkenes and alkynes is that terminal alkynes are relatively acidic. When a terminal alkyne is treated with a strong base, such as sodium amide, Na+ NH2, the terminal hydrogen is removed and the corresponding acetylide anion is formed.
According to the Brønsted–Lowry definition (Section 2.7), an acid is a substance that donates H+. Although we usually think of oxyacids (H2SO4, HNO3) or halogen acids (HCl, HBr) in this context, any compound containing a hydrogen atom can be an acid under the right circumstances. By measuring dissociation constants of different acids and expressing the results as pKa values, an acidity order can be established. Recall from Section 2.8 that a lower pKa corresponds to a stronger acid and a higher pKa corresponds to a weaker one.
Where do hydrocarbons lie on the acidity scale? As the data in Table 9.1 show, both methane (pKa ≈ 60) and ethylene (pKa = 44) are very weak acids and thus do not react with any of the common bases. Acetylene, however, has pKa = 25 and can be deprotonated by the conjugate base of any acid whose pKa is greater than 25. Amide ion (NH2), for example, the conjugate base of ammonia (pKa = 35), is often used to deprotonate terminal alkynes.
Table 9.1 Acidity of Simple Hydrocarbons
Family Example Ka pKa
Alkyne $HC≡CHHC≡CH$ 10−25 25
Alkene $H2C═CH2H2C═CH2$ 10−44 44
Alkane CH4 10−60 60
Why are terminal alkynes more acidic than alkenes or alkanes? In other words, why are acetylide anions more stable than vinylic or alkyl anions? The simplest explanation involves the hybridization of the negatively charged carbon atom. An acetylide anion has an sp-hybridized carbon, so the negative charge resides in an orbital that has 50% “s character.” A vinylic anion has a sp2-hybridized carbon with 33% s character, and an alkyl anion (sp3) has only 25% s character. Because s orbitals are nearer the positive nucleus and lower in energy than p orbitals, the negative charge is stabilized to a greater extent in an orbital with higher s character (Figure 9.6).
Problem 9-9
The pKa of acetone, CH3COCH3, is 19.3. Which of the following bases is strong enough to deprotonate acetone?
1. KOH (pKa of H2O = 15.7)
2. Na+ C $\text{≡}$ CH (pKa of C2H2 = 25)
3. NaHCO3 (pKa of H2CO3 = 6.4)
4. NaOCH3 (pKa of CH3OH = 15.6) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.06%3A_Reduction_of_Alkynes.txt |
The negative charge and unshared electron pair on carbon make an acetylide anion strongly nucleophilic. As a result, an acetylide anion can react with electrophiles, such as alkyl halides, in a process that replaces the halide and yields a new alkyne product.
We won’t study the details of this substitution reaction until Chapter 11, but for now you can picture it as happening by the pathway shown in Figure 9.7. The nucleophilic acetylide ion uses an electron pair to form a bond to the positively polarized, electrophilic carbon atom of bromomethane. As the new C−C bond forms, Br departs, taking with it the electron pair from the former C−Br bond and yielding propyne as product. We call such a reaction an alkylation because a new alkyl group has become attached to the starting alkyne.
Figure 9.7 MECHANISM A mechanism for the alkylation reaction of acetylide anion with bromomethane to give propyne.
Alkyne alkylation is not limited to acetylene itself. Any terminal alkyne can be converted into its corresponding anion and then allowed to react with an alkyl halide to give an internal alkyne product. Hex-1-yne, for instance, gives dec-5-yne when treated first with NaNH2 and then with 1-bromobutane.
Because of its generality, acetylide alkylation is a good method for preparing substituted alkynes from simpler precursors. A terminal alkyne can be prepared by alkylation of acetylene itself, and an internal alkyne can be prepared by further alkylation of a terminal alkyne.
The only limit to the alkylation reaction is that it can only use primary alkyl bromides and alkyl iodides because acetylide ions are sufficiently strong bases to cause elimination instead of substitution when they react with secondary and tertiary alkyl halides. For example, reaction of bromocyclohexane with propyne anion yields the elimination product cyclohexene rather than the substitution product 1-propynylcyclohexane.
Problem 9-10
Show the terminal alkyne and alkyl halide from which the following products can be obtained. If two routes look feasible, list both. (a)
(b)
(c)
Problem 9-11
How would you prepare cis-2-butene starting from propyne, an alkyl halide, and any other reagents needed? This problem can’t be worked in a single step. You’ll have to carry out more than one reaction.
9.10: An Introduction to Organic Synthesis
As mentioned in the introduction, one of the purposes of this chapter is to use alkyne chemistry as a vehicle to begin looking at some of the general strategies used in organic synthesis—the construction of complex molecules in the laboratory. There are many reasons for carrying out the laboratory synthesis of an organic compound. In the pharmaceutical industry, new molecules are designed and synthesized in the hope that some might be useful new drugs. In the chemical industry, syntheses are done to devise more economical routes to known compounds. In academic laboratories, the synthesis of extremely complex molecules is sometimes done just for the intellectual challenge involved in mastering so difficult a subject. The successful synthesis route is a highly creative work that is sometimes described by such subjective terms as elegant or beautiful.
In this book, too, we will often devise syntheses of molecules from simpler precursors, but the purpose here is to learn. The ability to plan a successful multistep synthetic sequence requires a working knowledge of the uses and limitations of many different organic reactions. Furthermore, it requires the practical ability to piece together the steps in a sequence such that each reaction does only what is desired without causing changes elsewhere in the molecule. Planning a synthesis makes you approach a chemical problem in a logical way, draw on your knowledge of chemical reactions, and organize that knowledge into a workable plan—it helps you learn organic chemistry.
There’s no secret to planning an organic synthesis: all it takes is a knowledge of the different reactions and some practice. The only real trick is to work backward in what is often called a retrosynthetic direction. Don’t look at a potential starting material and ask yourself what reactions it might undergo. Instead, look at the final product and ask, “What was the immediate precursor of that product?” For example, if the final product is an alkyl halide, the immediate precursor might be an alkene, to which you could add HX. If the final product is a cis alkene, the immediate precursor might be an alkyne, which you could hydrogenate using the Lindlar catalyst. Having found an immediate precursor, work backward again, one step at a time, until you get back to the starting material. You have to keep the starting material in mind, of course, so that you can work back to it, but you don’t want that starting material to be your main focus.
Let’s work several examples of increasing complexity.
Worked Example 9.1: Devising a Synthesis Route
How would you synthesize cis-2-hexene from 1-pentyne and an alkyl halide? More than one step is needed.
Strategy
When undertaking any synthesis problem, you should look at the product, identify the functional groups it contains, and then ask yourself how those functional groups can be prepared. Always work retrosynthetically, one step at a time.
The product in this case is a cis-disubstituted alkene, so the first question is, “What is an immediate precursor of a cis-disubstituted alkene?” We know that an alkene can be prepared from an alkyne by reduction and that the right choice of experimental conditions will allow us to prepare either a trans-disubstituted alkene (using lithium in liquid ammonia) or a cis-disubstituted alkene (using catalytic hydrogenation over the Lindlar catalyst). Thus, reduction of 2-hexyne by catalytic hydrogenation using the Lindlar catalyst should yield cis-2-hexene.
Next ask, “What is an immediate precursor of 2-hexyne?” We’ve seen that an internal alkyne can be prepared by alkylation of a terminal alkyne anion. In the present instance, we’re told to start with 1-pentyne and an alkyl halide. Thus, alkylation of the anion of 1-pentyne with iodomethane should yield 2-hexyne.
Solution
cis-2-Hexene can be synthesized from the given starting materials in three steps.
Worked Example 9.2: Devising a Synthesis Route
How would you synthesize 2-bromopentane from acetylene and an alkyl halide? More than one step is needed.
Strategy
Identify the functional group in the product (an alkyl bromide) and work the problem retrosynthetically. What is an immediate precursor of an alkyl bromide? Perhaps an alkene plus HBr. Of the two possibilities, Markovnikov addition of HBr to 1-pentene looks like a better choice than addition to 2-pentene because the latter reaction would give a mixture of isomers.
What is an immediate precursor of an alkene? Perhaps an alkyne, which could be reduced.
What is an immediate precursor of a terminal alkyne? Perhaps sodium acetylide and an alkyl halide.
Solution
The desired product can be synthesized in four steps from acetylene and 1-bromopropane.
Worked Example 9.3: Devising a Synthesis Route
How would you synthesize 5-methyl-1-hexanol (5-methyl-1-hydroxyhexane) from acetylene and an alkyl halide?
Strategy
What is an immediate precursor of a primary alcohol? Perhaps a terminal alkene, which could be hydrated with non-Markovnikov regiochemistry by reaction with borane followed by oxidation with H2O2.
What is an immediate precursor of a terminal alkene? Perhaps a terminal alkyne, which could be reduced.
What is an immediate precursor of 5-methyl-1-hexyne? Perhaps acetylene and 1-bromo-3-methylbutane.
Solution
The synthesis can be completed in four steps from acetylene and 1-bromo-3-methylbutane:
Problem 9-12
Beginning with 4-octyne as your only source of carbon, and using any inorganic reagents necessary, how would you synthesize the following compounds?
1. cis-4-Octene
2. Butanal
3. 4-Bromooctane
4. 4-Octanol
5. 4,5-Dichlorooctane
6. Butanoic acid
Problem 9-13
Beginning with acetylene and any alkyl halide needed, how would you synthesize the following compounds?
1. Decane
2. 2,2-Dimethylhexane
3. Hexanal
4. 2-Heptanone | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.09%3A_Alkylation_of_Acetylide_Anions.txt |
If you think some of the synthesis problems at the end of this chapter are difficult, try devising a synthesis of vitamin B12 starting only from simple substances you can buy in a chemical catalog. This extraordinary achievement was reported in 1973 as the culmination of a collaborative effort headed by Robert B. Woodward of Harvard University and Albert Eschenmoser of the Swiss Federal Institute of Technology in Zürich. More than 100 graduate students and postdoctoral associates contributed to the work, which took more than a decade to complete.
Why put such extraordinary effort into the laboratory synthesis of a molecule so easily obtained from natural sources? There are many reasons. On a basic human level, a chemist might be motivated primarily by the challenge, much as a climber might be challenged by the ascent of a difficult peak. Beyond the pure challenge, the completion of a difficult synthesis is also valuable in that it establishes new standards and raises the field to a new level. If vitamin B12 can be made, then why can’t any molecule found in nature be made? Indeed, the decades that have passed since the work of Woodward and Eschenmoser have seen the laboratory synthesis of many enormously complex and valuable substances. Sometimes these substances—for instance, the anticancer compound paclitaxel, trade named Taxol—are not easily available in nature, so laboratory synthesis is the only method for obtaining larger quantities.
But perhaps the most important reason for undertaking a complex synthesis is that, in so doing, new reactions and new chemistry are discovered. It invariably happens in a complex synthesis that a point is reached at which the planned route fails. At such a time, the only alternatives are either to quit or to devise a way around the difficulty. New reactions and new principles come from such situations, and it is in this way that the science of organic chemistry grows richer. In the synthesis of vitamin B12, for example, unexpected findings emerged that led to the understanding of an entire new class of reactions—the pericyclic reactions that are the subject of Chapter 30 in this book. From synthesizing vitamin B12 to understanding pericyclic reactions—no one could have possibly predicted such a link at the beginning of the synthesis, but that is the way of science.
9.12: Key Terms
• acetylide anion
• alkylation
• alkyne
• enol
• Lindlar Catalyst
• retrosynthetic
• tautomer
9.13: Summary
Alkynes are less common than alkenes, both in the laboratory and in living organisms, so we haven’t covered them in great detail. The real importance of this chapter is that alkyne chemistry is a useful vehicle for looking at the general strategies used in organic synthesis—the construction of complex molecules in the laboratory.
An alkyne is a hydrocarbon that contains a carbon–carbon triple bond. Alkyne carbon atoms are sp-hybridized, and the triple bond consists of one spsp σ bond and two p–p π bonds. There are relatively few general methods of alkyne synthesis. Two favorable ones are the alkylation of an acetylide anion with a primary alkyl halide and the twofold elimination of HX from a vicinal dihalide.
The chemistry of alkynes is dominated by electrophilic addition reactions, similar to those of alkenes. Alkynes react with HBr and HCl to yield vinylic halides and with Br2 and Cl2 to yield 1,2-dihalides (vicinal dihalides). Alkynes can be hydrated by reaction with aqueous sulfuric acid in the presence of mercury(II) catalyst. The reaction leads to an intermediate enol that immediately tautomerizes to yield a ketone. Because the addition reaction occurs with Markovnikov regiochemistry, a methyl ketone is produced from a terminal alkyne. Alternatively, hydroboration–oxidation of a terminal alkyne yields an aldehyde.
Alkynes can be reduced to yield alkenes and alkanes. Complete reduction of the triple bond over a palladium hydrogenation catalyst yields an alkane; partial reduction by catalytic hydrogenation over a Lindlar catalyst yields a cis alkene. Reduction of the alkyne with lithium in ammonia yields a trans alkene.
Terminal alkynes are weakly acidic. The alkyne hydrogen can be removed by a strong base such as Na+ NH2 to yield an acetylide anion. An acetylide anion acts as a nucleophile and can displace a halide ion from a primary alkyl halide in an alkylation reaction. Acetylide anions are more stable than either alkyl anions or vinylic anions because their negative charge is in a hybrid orbital with 50% s character, allowing the charge to be closer to the nucleus.
9.14: Summary of Reactions
No stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.
1. Preparation of alkynes
1. Dehydrohalogenation of vicinal dihalides (Section 9.2)
• Alkylation of acetylide anions (Section 9.8)
• Reactions of alkynes
1. Addition of HCl and HBr (Section 9.3)
2. Addition of Cl2 and Br2 (Section 9.3)
3. Hydration (Section 9.4)
(1) Mercuric sulfate catalyzed
(2) Hydroboration–oxidation
4. Reduction (Section 9.5)
(1) Catalytic hydrogenation
(2) Lithium in liquid ammonia
5. Conversion into acetylide anions (Section 9.7) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.11%3A_Chemistry_MattersThe_Art_of_Organic_Synthesis.txt |
Visualizing Chemistry
Problem 9-14
Name the following alkynes, and predict the products of their reaction with (1) H2 in the presence of a Lindlar catalyst and (2) H3O+ in the presence of HgSO4:
(a)
(b)
Problem 9-15 From what alkyne might each of the following substances have been made? (Green = Cl.)
(a)
(b)
Problem 9-16
How would you prepare the following substances, starting from any compounds having four carbons or fewer? (a)
(b)
Problem 9-17
The following cycloalkyne is too unstable to exist. Explain.
Mechanism Problems
Problem 9-18
Assuming that halogens add to alkynes in the same manner as they add to alkenes, propose a mechanism for and predict the product(s) of the reaction of phenylpropyne with Br2.
Problem 9-19 Assuming that strong acids add to alkynes in the same manner as they add to alkenes, propose a mechanism for each of the following reactions.
(a)
(b)
(c)
Problem 9-20
The mercury-catalyzed hydration of alkynes involves the formation of an organomercury enol intermediate. Draw the electron-pushing mechanism to show how each of the following intermediates is formed. (a)
(b)
(c)
Problem 9-21
The final step in the hydration of an alkyne under acidic conditions is the tautomerization of an enol intermediate to give the corresponding ketone. The mechanism involves a protonation followed by a deprotonation. Show the mechanism for each of the following tautomerizations. (a)
(b)
(c)
Problem 9-22
Predict the product(s) and show the complete electron-pushing mechanism for each of the following dissolving metal reductions. (a)
(b)
(c)
Problem 9-23
Identify the mechanisms for the following reactions as polar, radical, or both. (a)
(b)
(c)
Problem 9-24
Predict the product and provide the complete electron-pushing mechanism for the following two-step synthetic processes. (a)
(b)
(c)
Problem 9-25
Reaction of acetone with D3O+ yields hexadeuterioacetone. That is, all the hydrogens in acetone are exchanged for deuterium. Review the mechanism of mercuric-ion-catalyzed alkyne hydration, and then propose a mechanism for this deuterium incorporation.
Naming Alkynes
Problem 9-26 Give IUPAC names for the following compounds:
(a)
(b)
(c)
(d)
(e)
(f)
Problem 9-27
Draw structures corresponding to the following names: (a) 3,3-Dimethyl-4-octyne
(b) 3-Ethyl-5-methyl-1,6,8-decatriyne
(c) 2,2,5,5-Tetramethyl-3-hexyne (d) 3,4-Dimethylcyclodecyne (e) 3,5-Heptadien-1-yne (f) 3-Chloro-4,4-dimethyl-1-nonen-6-yne (g) 3-sec-Butyl-1-heptyne (h) 5-tert-Butyl-2-methyl-3-octyne
Problem 9-28 The following two hydrocarbons have been isolated from various plants in the sunflower family. Name them according to IUPAC rules.
(a) CH3CH$\text{=}$CHC$\text{≡}$CC$\text{≡}$CCH$\text{=}$CHCH$\text{=}$CHCH$\text{=}$CH2 (all trans)
(b) CH3C$\text{≡}$CC$\text{≡}$CC$\text{≡}$CC$\text{≡}$CC$\text{≡}$CCH$\text{=}$CH2
Reactions of Alkynes
Problem 9-29
Terminal alkynes react with Br2 and water to yield bromo ketones. For example:
Propose a mechanism for the reaction. To what reaction of alkenes is the process analogous?
Problem 9-30
Predict the products of the following reactions:
Problem 9-31 Predict the products from reaction of 1-hexyne with the following reagents:
(a) 1 equiv HBr
(b) 1 equiv Cl2 (c) H2, Lindlar catalyst (d) NaNH2 in NH3, then CH3Br (e) H2O, H2SO4, HgSO4 (f) 2 equiv HCl
Problem 9-32 Predict the products from reaction of 5-decyne with the following reagents:
(a) H2, Lindlar catalyst
(b) Li in NH3 (c) 1 equiv Br2 (d) BH3 in THF, then H2O2, OH (e) H2O, H2SO4, HgSO4 (f) Excess H2, Pd/C catalyst
Problem 9-33 Predict the products from reaction of 2-hexyne with the following reagents:
(a) 2 equiv Br2
(b) 1 equiv HBr (c) Excess HBr (d) Li in NH3 (e) H2O, H2SO4, HgSO4
(a)
(b)
(c)
(d)
(e)
Problem 9-35
Identify the reagents a–c in the following scheme:
Organic Synthesis
Problem 9-36
How would you carry out the following multistep conversions? More than one step may be needed in some instances.
Problem 9-37 How would you carry out the following reactions?
(a)
(b)
(c)
(d)
(e)
(f)
Problem 9-38
Each of the following syntheses requires more than one step. How would you carry them out? (a)
(b)
Problem 9-39
How would you carry out the following multistep transformation?
Problem 9-40
How would you carry out the following multistep conversions?
Problem 9-41 Synthesize the following compounds using 1-butyne as the only source of carbon, along with any inorganic reagents you need. More than one step may be needed.
(a) 1,1,2,2-Tetrachlorobutane
(b) 1,1-Dichloro-2-ethylcyclopropane
(a)
(b)
(c)
(d)
(e)
Problem 9-43 How would you carry out the following reactions to introduce deuterium into organic molecules?
(a)
(b)
(c)
(d)
Problem 9-44
How would you prepare cyclodecyne starting from acetylene and any required alkyl halide?
Problem 9-45
The sex attractant given off by the common housefly is an alkene named muscalure. Propose a synthesis of muscalure starting from acetylene and any alkyl halides needed. What is the IUPAC name for muscalure?
General Problems
Problem 9-46 A hydrocarbon of unknown structure has the formula C8H10. On catalytic hydrogenation over the Lindlar catalyst, 1 equivalent of H2 is absorbed. On hydrogenation over a palladium catalyst, 3 equivalents of H2 are absorbed.
(a) How many degrees of unsaturation are present in the unknown structure?
(b) How many triple bonds are present?
(c) How many double bonds are present? (d) How many rings are present? (e) Draw a structure that fits the data.
Problem 9-47
Compound A (C9H12) absorbed 3 equivalents of H2 on catalytic reduction over a palladium catalyst to give B (C9H18). On ozonolysis, compound A gave, among other things, a ketone that was identified as cyclohexanone. On treatment with NaNH2 in NH3, followed by addition of iodomethane, compound A gave a new hydrocarbon, C (C10H14). What are the structures of A, B, and C?
Problem 9-48
Hydrocarbon A has the formula C12H8. It absorbs 8 equivalents of H2 on catalytic reduction over a palladium catalyst. On ozonolysis, only two products are formed: oxalic acid (HO2CCO2H) and succinic acid (HO2CCH2CH2CO2H). Write the reactions, and propose a structure for A.
Problem 9-49 Occasionally, a chemist might need to invert the stereochemistry of an alkene—that is, to convert a cis alkene to a trans alkene, or vice versa. There is no one-step method for doing an alkene inversion, but the transformation can be carried out by combining several reactions in the proper sequence. How would you carry out the following reactions?
(a)
(b)
Problem 9-50
Organometallic reagents such as sodium acetylide undergo an addition reaction with ketones, giving alcohols:
How might you use this reaction to prepare 2-methyl-1,3-butadiene, the starting material used in the manufacture of synthetic rubber?
Problem 9-51
The oral contraceptive agent Mestranol is synthesized using a carbonyl addition reaction like that shown in Problem 9-50. Draw the structure of the ketone needed.
Problem 9-52
1-Octen-3-ol, a potent mosquito attractant commonly used in mosquito traps, can be prepared in two steps from hexanal, CH3CH2CH2CH2CH2CHO. The first step is an acetylide-addition reaction like that described in Problem 9-50. What is the structure of the product from the first step, and how can it be converted into 1-octen-3-ol?
Problem 9-53
Erythrogenic acid, C18H26O2, is an acetylenic fatty acid that turns a vivid red on exposure to light. On catalytic hydrogenation over a palladium catalyst, 5 equivalents of H2 are absorbed, and stearic acid, CH3(CH2)16CO2H, is produced. Ozonolysis of erythrogenic acid gives four products: formaldehyde, CH2O; oxalic acid, HO2CCO2H; azelaic acid, HO2C(CH2)7CO2H; and the aldehyde acid OHC(CH2)4CO2H. Draw two possible structures for erythrogenic acid, and suggest a way to tell them apart by carrying out some simple reactions.
Problem 9-54
Hydrocarbon A has the formula C9H12 and absorbs 3 equivalents of H2 to yield B, C9H18, when hydrogenated over a Pd/C catalyst. On treatment of A with aqueous H2SO4 in the presence of mercury(II), two isomeric ketones, C and D, are produced. Oxidation of A with KMnO4 gives a mixture of acetic acid (CH3CO2H) and the tricarboxylic acid E. Propose structures for compounds AD, and write the reactions.
Problem 9-55
A cumulene is a compound with three adjacent double bonds. Draw an orbital picture of a cumulene. What kind of hybridization do the two central carbon atoms have? What is the geometric relationship of the substituents on one end to the substituents on the other end? What kind of isomerism is possible? Make a model to help see the answer.
Problem 9-56
Which of the following bases could be used to deprotonate 1-butyne?
(a)(b)(c)(d)
Problem 9-57
Arrange the following carbocations in order of increasing stability.
(a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.15%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 10, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• design a multistep synthesis to prepare a given compound from a given starting material using any of the reactions studied up to this point in the course, including those which involve alkyl halides.
• solve road-map problems requiring a knowledge of any of the reactions or concepts studied up to this point, including those introduced in this chapter.
• define, and use in context, the key terms introduced.
10: Organohalides
Chapter Contents
10.1 Names and Structures of Alkyl Halides
10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation 10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination 10.4 Stability of the Allyl Radical: Resonance Revisited 10.5 Preparing Alkyl Halides from Alcohols 10.6 Reactions of Alkyl Halides: Grignard Reagents 10.7 Organometallic Coupling Reactions 10.8 Oxidation and Reduction in Organic Chemistry
Alkyl halides are encountered less frequently than their oxygen-containing relatives and are not often involved in the biochemical pathways of terrestrial organisms, but some of the kinds of reactions they undergo—nucleophilic substitutions and eliminations—are encountered frequently. Thus, alkyl halide chemistry is a relatively simple model for many mechanistically similar but structurally more complex reactions found in biomolecules. We’ll begin this chapter with a look at how to name and prepare alkyl halides, and we’ll see several of their reactions. Then, in the next chapter, we’ll make a detailed study of the substitution and elimination reactions of alkyl halides—two of the most important and well-studied reaction types in organic chemistry.
Now that we’ve covered the chemistry of hydrocarbons, it’s time to start looking at more complex substances that contain elements in addition to C and H. We’ll begin by discussing the chemistry of organohalides, compounds that contain one or more halogen atoms.
Halogen-substituted organic compounds are widespread in nature, and more than 5000 organohalides have been found in algae and various other marine organisms. Chloromethane, for instance, is released in large amounts by ocean kelp, as well as by forest fires and volcanoes. Halogen-containing compounds also have an array of industrial applications, including their use as solvents, inhaled anesthetics in medicine, refrigerants, and pesticides.
Still other halo-substituted compounds are used as medicines and food additives. The nonnutritive sweetener sucralose, marketed as Splenda, contains three chlorine atoms, for instance. Sucralose is about 600 times as sweet as sucrose, so only 1 mg is equivalent to an entire teaspoon of table sugar.
A large variety of organohalides are known. The halogen might be bonded to an alkynyl group ($C≡C−XC≡C−X$), a vinylic group ($C═C–XC═C–X$), an aromatic ring (Ar−X), or an alkyl group. In this chapter, however, we’ll be primarily concerned with alkyl halides, compounds with a halogen atom bonded to a saturated, sp3-hybridized carbon atom.
10.02: Names and Properties of Alkyl Halides
Although commonly called alkyl halides, halogen-substituted alkanes are named systematically as haloalkanes (Section 3.4), treating the halogen as a substituent on a parent alkane chain. There are three steps:
STEP 1
Find the longest chain, and name it as the parent. If a double or triple bond is present, the parent chain must contain it.
STEP 2
Number the carbons of the parent chain beginning at the end nearer the first substituent, whether alkyl or halo. Assign each substituent a number according to its position on the chain.
If different halogens are present, number each one and list them in alphabetical order when writing the name.
STEP 3
If the parent chain can be properly numbered from either end by step 2, begin at the end nearer the substituent that has alphabetical precedence.
In addition to their systematic names, many simple alkyl halides are also named by identifying first the alkyl group and then the halogen. For example, CH3I can be called either iodomethane or methyl iodide. Such names are well entrenched in the chemical literature and in daily usage, but they won’t be used in this book.
Halogens increase in size going down the periodic table, so the lengths of the corresponding carbon–halogen bonds increase accordingly (Table 10.1). In addition, C−X bond strengths decrease going down the periodic table. As we’ve been doing thus far, we’ll continue using an X to represent any of the halogens F, Cl, Br, or I.
Table 10.1 A Comparison of the Halomethanes
Halomethane Bond length (pm) Bond strength Dipole moment (D)
(kJ/mol) (kcal/mol)
CH3F 139 460 110 1.85
CH3Cl 178 350 84 1.87
CH3Br 193 294 70 1.81
CH3I 214 239 57 1.62
In our discussion of bond polarity in functional groups in Section 6.3, we noted that halogens are more electronegative than carbon. The C−X bond is therefore polar, with the carbon atom bearing a slight positive charge (δ+) and the halogen a slight negative charge (δ−). This polarity results in a dipole moment for all the halomethanes (Table 10.1) and implies that the alkyl halide C−X carbon atom should behave as an electrophile in polar reactions. We’ll soon see that this is indeed the case.
(b)
(c)
(d)
(e)
(f)
Problem 10-2 Draw structures corresponding to the following IUPAC names: (a) 2-Chloro-3,3-dimethylhexane (b)
3,3-Dichloro-2-methylhexane
(c) 3-Bromo-3-ethylpentane (d) 1,1-Dibromo-4-isopropylcyclohexane (e) 4-sec-Butyl-2-chlorononane (f) 1,1-Dibromo-4-tert-butylcyclohexane | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.01%3A_Why_This_Chapter.txt |
As we saw briefly in Section 6.6, simple alkyl halides can sometimes be prepared by the radical reaction of an alkane with Cl2 or Br2 in the presence of ultraviolet light. The detailed mechanism is shown in Figure 10.2 for chlorination.
Radical substitution reactions require three kinds of steps: initiation, propagation, and termination. Once an initiation step has started the process by producing radicals, the reaction continues in a self-sustaining cycle. The cycle requires two repeating propagation steps in which a radical, the halogen, and the alkane yield alkyl halide product plus more radical to carry on the chain. The chain is occasionally terminated by the combination of two radicals.
Unfortunately, alkane halogenation is a poor synthetic method for preparing alkyl halides because mixtures of products invariably result. For example, chlorination of methane does not stop cleanly at the monochlorinated stage but continues to give a mixture of dichloro, trichloro, and even tetrachloro products.
The situation is even worse for chlorination of alkanes that have more than one kind of hydrogen. Chlorination of butane, for instance, gives two monochlorinated products in a 30 : 70 ratio in addition to multiply chlorinated products such as dichlorobutane, trichlorobutane, and so on.
As another example, 2-methylpropane yields 2-chloro-2-methylpropane and 1-chloro-2-methylpropane in a 35 : 65 ratio, along with more highly chlorinated products.
From these and similar reactions, it’s possible to calculate a reactivity order toward chlorination for different kinds of hydrogen atoms in a molecule. Take the butane chlorination, for instance. Butane has six equivalent primary hydrogens (−CH3) and four equivalent secondary hydrogens (−CH2−). The fact that butane yields 30% of 1-chlorobutane product means that each one of the six primary hydrogens is responsible for 30% ÷ 6 = 5% of the product. Similarly, the fact that 70% of 2-chlorobutane is formed means that each of the four secondary hydrogens is responsible for 70% ÷ 4 = 17.5% of the product. Thus, a secondary hydrogen reacts 17.5% ÷ 5% = 3.5 times as often as a primary hydrogen.
A similar calculation for the chlorination of 2-methylpropane indicates that each of the nine primary hydrogens accounts for 65% ÷ 9 = 7.2% of the product, while the single tertiary hydrogen (R3CH) accounts for 35% of the product. Thus, a tertiary hydrogen is 35% ÷ 7.2% = 5 times as reactive as a primary hydrogen toward chlorination.
The observed reactivity order of alkane hydrogens toward radical chlorination can be explained by looking at the bond dissociation energies given previously in Table 6.3. The data show that a tertiary C−H bond (400 kJ/mol; 96 kcal/mol) is weaker than a secondary C−H bond (410 kJ/mol; 98 kcal/mol), which is in turn weaker than a primary C−H bond (421 kJ/mol; 101 kcal/mol). Since less energy is needed to break a tertiary C−H bond than to break a primary or secondary C−H bond, the resultant tertiary radical is more stable than a primary or secondary radical.
Problem 10-3
Draw and name all monochloro products you would expect to obtain from radical chlorination of 2-methylpentane. Which, if any, are chiral?
Problem 10-4
Taking the relative reactivities of 1°, 2°, and 3° hydrogen atoms into account, what product(s) would you expect to obtain from monochlorination of 2-methylbutane? What would the approximate percentage of each product be? (Don’t forget to take into account the number of each kind of hydrogen.)
10.04: Preparing Alkyl Halides from Alkenes - Allylic Bromination
We’ve already seen several methods for preparing alkyl halides from alkenes, including the reactions of HX and X2 with alkenes in electrophilic addition reactions (Section 7.7 and Section 8.2). The hydrogen halides HCl, HBr, and HI react with alkenes by a polar mechanism to give the product of Markovnikov addition. Bromine and chlorine undergo anti addition through halonium ion intermediates to give 1,2-dihalogenated products.
Another laboratory method for preparing alkyl halides from alkenes is by reaction with N-bromosuccinimide (abbreviated NBS), in the presence of ultraviolet light, to give products resulting from substitution of hydrogen by bromine at the position next to the double bond—the allylic position. Cyclohexene, for example, gives 3-bromocyclohexene.
This allylic bromination with NBS is analogous to the alkane chlorination reaction discussed in the previous section and occurs by a radical chain-reaction pathway (Figure 10.3). As in alkane halogenation, a Br· radical abstracts an allylic hydrogen atom, forming an allylic radical plus HBr. The HBr then reacts with NBS to form Br2, which in turn reacts with the allylic radical to yield the brominated product and a Br· radical that cycles back into the first step and carries on the chain.
) a Br· radical abstracts an allylic hydrogen atom of the alkene and gives an allylic radical plus HBr. (2) The HBr then reacts with NBS to form Br2, which (3) reacts with the allylic radical to yield the bromoalkene product and a Br· radical that continues the chain.
Why does bromination with NBS occur exclusively at an allylic position rather than elsewhere in the molecule? The answer, once again, is found by looking at bond dissociation energies to see the relative stabilities of various kinds of radicals. Although a typical secondary alkyl C−H bond has a strength of about 410 kJ/mol (98 kcal/mol) and a typical vinylic C−H bond has a strength of 465 kJ/mol (111 kcal/mol), an allylic C−H bond has a strength of only about 370 kJ/mol (88 kcal/mol). An allylic radical is therefore more stable than a typical alkyl radical with the same substitution by about 40 kJ/mol (9 kcal/mol).
We can thus expand the stability ordering to include vinylic and allylic radicals. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.03%3A_Preparing_Alkyl_Halides_from__Alkanes_-_Radical_Halogenation.txt |
To see why an allylic radical is so stable, look at the orbital picture in Figure 10.4. The radical carbon atom with an unpaired electron can adopt sp2 hybridization, placing the unpaired electron in a p orbital and giving a structure that is electronically symmetrical. The p orbital on the central carbon can therefore overlap equally well with a p orbital on either of the two neighboring carbons.
Because the allyl radical is electronically symmetrical, it has two resonance forms—one with the unpaired electron on the left and the double bond on the right and another with the unpaired electron on the right and the double bond on the left. Neither structure is correct by itself; the true structure of the allyl radical is a resonance hybrid of the two. (You might want to review Section 2.4 to Section 2.6 to brush up on resonance.) As noted in Section 2.5, the greater the number of resonance forms, the greater the stability of a compound, because bonding electrons are attracted to more nuclei. An allyl radical, with two resonance forms, is therefore more stable than a typical alkyl radical, which has only a single structure.
In molecular orbital terms, the stability of the allyl radical is due to the fact that the unpaired electron is delocalized, or spread out, over an extended π-orbital network rather than localized at only one site, as shown by the computer-generated MO in Figure 10.4. This delocalization is particularly apparent in the so-called spin-density surface in Figure 10.5, which shows the calculated location of the unpaired electron. The two terminal carbons share the unpaired electron equally.
and shows that it is equally shared between the two terminal carbons.
In addition to its effect on stability, delocalization of the unpaired electron in the allyl radical has other chemical consequences. Because the unpaired electron is delocalized over both ends of the π orbital system, reaction with Br2 can occur at either end. As a result, allylic bromination of an unsymmetrical alkene often leads to a mixture of products. For example, bromination of 1-octene gives a mixture of 3-bromo-1-octene and 1-bromo-2-octene. The two products are not formed in equal amounts, however, because the intermediate allylic radical is not symmetrical and reaction at the two ends is not equally likely. Reaction at the less hindered, primary end is favored.
The products of allylic bromination reactions are useful for conversion into dienes by dehydrohalogenation with base. Cyclohexene can be converted into 1,3-cyclohexadiene, for example.
Worked Example 10.1
Predicting the Product of an Allylic Bromination Reaction
What products would you expect from the reaction of 4,4-dimethylcyclohexene with NBS?
Strategy
Draw the alkene reactant, and identify the allylic positions. In this case, there are two different allylic positions; we’ll label them A and B. Now abstract an allylic hydrogen from each position to generate the two corresponding allylic radicals. Each of the two allylic radicals can add a Br atom at either end (A or A′; B or B′), to give a mixture of up to four products. Draw and name the products. In the present instance, the “two” products from reaction at position B are identical, so only three products are formed in this reaction.
Solution
Problem 10-5
Draw three resonance forms for the cyclohexadienyl radical.
Problem 10-6
The major product of the reaction of methylenecyclohexane with N-bromosuccinimide is 1-(bromomethyl)cyclohexene. Explain.
Problem 10-7 What products would you expect from reaction of the following alkenes with NBS? If more than one product is formed, show the structures of all.
(a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.05%3A_Stability_of_the_Allyl_Radical_-_Resonance_Revisited.txt |
The most generally useful method for preparing alkyl halides is to make them from alcohols, which themselves can be obtained from carbonyl compounds as we’ll see in Sections 17.4 and 17.5. Because of the importance of this process, many different methods have been developed to transform alcohols into alkyl halides. The simplest method is to treat the alcohol with HCl, HBr, or HI. For reasons that will be discussed in Section 11.5, this reaction works best with tertiary alcohols, R3COH. Primary and secondary alcohols react much more slowly and at higher temperatures.
The reaction of HX with a tertiary alcohol is so rapid that it’s often carried out simply by bubbling pure HCl or HBr gas into a cold ether solution of the alcohol. 1-Methylcyclohexanol, for example, is converted into 1-chloro-1-methylcyclohexane by treatment with HCl.
Primary and secondary alcohols are best converted into alkyl halides by treatment with either thionyl chloride (SOCl2) or phosphorus tribromide (PBr3). These reactions, which normally take place readily under mild conditions, are less acidic and less likely to cause acid-catalyzed rearrangements than the HX method.
As the preceding examples indicate, the yields of these SOCl2 and PBr3 reactions are generally high and other functional groups such as ethers, carbonyls, and aromatic rings don’t usually interfere. We’ll look at the mechanisms of these and other related substitution reactions in Section 11.3.
Alkyl fluorides can also be prepared from alcohols. Numerous alternative reagents are used for such reactions, including diethylaminosulfur trifluoride [(CH3CH2)2NSF3] and HF in pyridine solvent.
Problem 10-8 How would you prepare the following alkyl halides from the corresponding alcohols?
(a)
(b)
(c)
(d)
10.07: Reactions of Alkyl Halides - Grignard Reagents
Alkyl halides, RX, react with magnesium metal in ether or tetrahydrofuran (THF) solvent to yield alkylmagnesium halides, RMgX. The products, called Grignard reagents (RMgX) after their discoverer, Francois Auguste Victor Grignard, who received the 1912 Nobel Prize in Chemistry, are examples of organometallic compounds because they contain a carbon–metal bond. In addition to alkyl halides, Grignard reagents can also be made from alkenyl (vinylic) and aryl (aromatic) halides. The halogen can be Cl, Br, or I, although chlorides are less reactive than bromides and iodides. Organofluorides rarely react with magnesium.
As you might expect from the discussion of electronegativity and bond polarity in Section 6.3, the carbon–magnesium bond is polarized, making the carbon atom of Grignard reagents both nucleophilic and basic. An electrostatic potential map of methylmagnesium iodide, for instance, indicates the electron-rich (red) character of the carbon bonded to magnesium.
A Grignard reagent is formally the magnesium salt, R3C–+MgX, of a carbon acid, R3C–H, and is thus a carbon anion, or carbanion. But because hydrocarbons are such weak acids, with pKa’s in the range 44 to 60 (Section 9.7), carbon anions are very strong bases. Grignard reagents must therefore be protected from atmospheric moisture to prevent their being protonated and destroyed in acid–base reactions: R–Mg–X + H2O → R–H + HO–Mg–X.
Grignard reagents themselves don’t occur in living organisms, but they serve as useful carbon-based nucleophiles in several important laboratory reactions, which we’ll look at in detail in Section 17.5. In addition, they act as a simple model for other, more complex carbon-based nucleophiles that are important in biological chemistry. We’ll see many examples of these in Chapter 29.
Problem 10-9
How strong a base would you expect a Grignard reagent to be? Look at Table 9.1 and predict whether the following reactions will occur as written. (The pKa of NH3 is 35.)
(a) CH3MgBr + H–C$\text{≡}$C–H $\text{⟶}$ CH4 + H–C$\text{≡}$C–MgBr (b) CH3MgBr + NH3 $\text{⟶}$ CH4 + H2N–MgBr
Problem 10-10
How might you replace a halogen substituent by a deuterium atom if you wanted to prepare a deuterated compound? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.06%3A_Preparing_Alkyl_Halides_from_Alcohols.txt |
Many other kinds of organometallic compounds can be prepared in a manner similar to that of Grignard reagents. For instance, alkyllithium reagents, RLi, can be prepared by the reaction of an alkyl halide with lithium metal. Alkyllithiums are both nucleophiles and strong bases, and their chemistry is similar in many respects to that of alkylmagnesium halides.
One particularly valuable reaction of alkyllithiums occurs when making lithium diorganocopper compounds, R2CuLi, by reaction with copper(I) iodide in diethyl ether as solvent. Often called Gilman reagents (LiR2Cu), lithium diorganocopper compounds are useful because they undergo a coupling reaction with organochlorides, bromides, and iodides (but not fluorides). One of the alkyl groups from the lithium diorganocopper reagent replaces the halogen of the organohalide, forming a new carbon–carbon bond and yielding a hydrocarbon product. Lithium dimethylcopper, for instance, reacts with 1-iododecane to give undecane in a 90% yield.
This organometallic coupling reaction is useful in organic synthesis because it forms carbon–carbon bonds, thereby allowing the preparation of larger molecules from smaller ones. As the following examples indicate, the coupling reaction can be carried out on aryl and vinylic halides as well as on alkyl halides.
An organocopper coupling reaction is carried out commercially to synthesize muscalure, (9Z)-tricosene, the sex attractant secreted by the common housefly. Minute amounts of muscalure greatly increase the lure of insecticide-treated fly bait and provide an effective and species-specific means of insect control.
The mechanism of the coupling reaction involves initial formation of a triorganocopper intermediate, followed by coupling and loss of a mono-organocopper, RCu. The coupling is not a typical polar nucleophilic substitution reaction of the sort considered in the next chapter.
In addition to the coupling reaction of diorganocopper reagents with organohalides, related processes also occur with other organometallic reagents, particularly organopalladium compounds. One of the most commonly used procedures is the coupling reaction of an aromatic or vinyl substituted boronic acid [R—B(OH)2] with an aromatic or vinyl substituted organohalide in the presence of a base and a palladium catalyst. This reaction is less general than the diorganocopper reaction because it doesn’t work with alkyl substrates, but it is preferred when possible because it uses only a catalytic amount of metal rather than a full equivalent and because palladium compounds are less toxic than copper compounds. For example:
Called the Suzuki–Miyaura reaction, this process is particularly useful for preparing so-called biaryl compounds, which have two linked aromatic rings. A large number of commonly used drugs fit this description, so the Suzuki–Miyaura reaction is much-used in the pharmaceutical industry. As an example, valsartan, marketed as Diovan, is widely prescribed to treat high blood pressure, heart failure, and diabetic kidney disease. Its synthesis begins with a Suzuki–Miyaura coupling of ortho-chlorobenzonitrile with para-methylbenzeneboronic acid.
Shown in a simplified form in Figure 10.6, the mechanism of the Suzuki–Miyaura reaction involves initial reaction of the aromatic halide with the palladium catalyst to form an organopalladium intermediate, followed by reaction of that intermediate with the aromatic boronic acid. The resultant diorganopalladium complex then decomposes to the coupled biaryl product plus regenerated catalyst.
) reaction of the aromatic halide, ArX, with the catalyst to form an organopalladium intermediate, followed by (2) reaction with the aromatic boronic acid. (3) Subsequent decomposition of the diarylpalladium intermediate gives the biaryl product.
Problem 10-11 How would you carry out the following transformations using an organocopper coupling reaction? More than one step is required in each case.
(a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.08%3A_Organometallic_Coupling_Reactions.txt |
We’ve pointed out on several occasions that some of the reactions discussed in this and earlier chapters are either oxidations or reductions. As noted in Section 8.7, an organic oxidation results in a loss of electron density by carbon, caused either by bond formation between carbon and a more electronegative atom (usually O, N, or a halogen) or by bond-breaking between carbon and a less electronegative atom (usually H). Conversely, an organic reduction results in a gain of electron density by carbon, caused either by bond formation between carbon and a less electronegative atom or by bond-breaking between carbon and a more electronegative atom (Section 8.6).
$OxidationDecreases electron density on carbon by:–forming one of these:C−OC−NC−X–or breaking this:C–HReductionIncreases electron density on carbon by:–forming this:C–H–or breaking one of these:C–OC–NC–XOxidationDecreases electron density on carbon by:–forming one of these:C−OC−NC−X–or breaking this:C–HReductionIncreases electron density on carbon by:–forming this:C–H–or breaking one of these:C–OC–NC–X$
Based on these definitions, the chlorination reaction of methane to yield chloromethane is an oxidation because a C−H bond is broken and a C−Cl bond is formed. The conversion of an alkyl chloride to an alkane via a Grignard reagent followed by protonation is a reduction, however, because a C−Cl bond is broken and a C−H bond is formed.
As other examples, the reaction of an alkene with Br2 to yield a 1,2-dibromide is an oxidation because two C−Br bonds are formed, but the reaction of an alkene with HBr to yield an alkyl bromide is neither an oxidation nor a reduction because both a C−H and a C−Br bond are formed.
A list of compounds of increasing oxidation level is shown in Figure 10.7. Alkanes are at the lowest oxidation level because they have the maximum possible number of C−H bonds per carbon, and CO2 is at the highest level because it has the maximum possible number of C−O bonds per carbon. Any reaction that converts a compound from a lower level to a higher level is an oxidation, any reaction that converts a compound from a higher level to a lower level is a reduction, and any reaction that doesn’t change the level is neither an oxidation nor a reduction.
Worked Example 10.2 shows how to compare the oxidation levels of different compounds with the same number of carbon atoms.
Worked Example 10.2
Comparing Oxidation Levels
Rank the following compounds in order of increasing oxidation level:
Strategy
Compounds that have the same number of carbon atoms can be compared by adding the number of C−O, C−N, and C−X bonds in each and then subtracting the number of C−H bonds. The larger the resultant value, the higher the oxidation level.
Solution
The first compound (propene) has six C−H bonds, giving an oxidation level of −6; the second (2-propanol) has one C−O bond and seven C−H bonds, giving an oxidation level of −6; the third (acetone) has two C−O bonds and six C−H bonds, giving an oxidation level of −4; and the fourth (propane) has eight C−H bonds, giving an oxidation level of −8. Thus, the order of increasing oxidation level is
Problem 10-12 Rank both sets of compounds in order of increasing oxidation level:
(a)
(b)
Problem 10-13 Tell whether each of the following reactions is an oxidation, a reduction, or neither.
(a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.09%3A_Oxidation_and_Reduction_in_Organic_Chemistry.txt |
10 • Chemistry Matters 10 • Chemistry Matters
Just forty years ago in 1980, only about 30 naturally occurring organohalides were known. It was simply assumed that chloroform, halogenated phenols, chlorinated aromatic compounds called PCBs, and other such substances found in the environment were industrial pollutants. Now, less than half a century later, the situation is quite different. More than 5000 organohalides have been found to occur naturally, and tens of thousands more surely exist. From a simple compound like chloromethane to an extremely complex one like the antibiotic vancomycin, a remarkably diverse range of organohalides exists in plants, bacteria, and animals. Many even have valuable physiological activity. The pentahalogenated alkene halomon, for instance, has been isolated from the red alga Portieria hornemannii and found to have anticancer activity against several human tumor cell lines.
Some naturally occurring organohalides are produced in massive quantities. Forest fires, volcanic eruptions, and marine kelp release up to 5 million tons of CH3Cl per year, for example, while annual industrial emissions total about 26,000 tons. Termites are thought to release as much as 108 kg of chloroform per year. A detailed examination of the Okinawan acorn worm Ptychodera flava found that the 64 million worms living in a 1 km2 study area excreted nearly 8000 pounds per year of bromophenols and bromoindoles, compounds previously thought to be non-natural pollutants.
Why do organisms produce organohalides, many of which are undoubtedly toxic? The answer seems to be that many organisms use organohalogen compounds for self-defense, either as feeding deterrents, irritants to predators, or natural pesticides. Marine sponges, coral, and sea hares, for example, release foul-tasting organohalides that deter fish, starfish, and other predators. Even humans appear to produce halogenated compounds as part of their defense against infection. The human immune system contains a peroxidase enzyme capable of carrying out halogenation reactions on fungi and bacteria, thereby killing the pathogen. And most remarkable of all, even free chlorine—Cl2—has been found to be present in humans.
Much remains to be learned—only a few hundred of the more than 500,000 known species of marine organisms have been examined—but it’s clear that organohalides are an integral part of the world around us.
10.11: Key Terms
10 • Key Terms 10 • Key Terms
• alkyl halide
• allylic
• carbanion
• delocalized
• Gilman reagent (LiR2Cu)
• Grignard reagent (RMgX)
• organohalide
• organometallic
10.12: Summary
10 • Summary 10 • Summary
Alkyl halides are not often found in terrestrial organisms, but the kinds of reactions they undergo are among the most important and well-studied reaction types in organic chemistry. In this chapter, we saw how to name and prepare alkyl halides, and we’ll soon make a detailed study of their substitution and elimination reactions.
Simple alkyl halides can be prepared by radical halogenation of alkanes, but mixtures of products usually result. The reactivity order of alkanes toward halogenation is identical to the stability order of radicals: R3C· > R2CH· > RCH2·. Alkyl halides can also be prepared from alkenes by reaction with N-bromosuccinimide (NBS) to give the product of allylic bromination. The NBS bromination of alkenes takes place through an intermediate allylic radical, which is stabilized by resonance.
Alcohols react with HX to form alkyl halides, but the reaction works well only for tertiary alcohols, R3COH. Primary and secondary alkyl halides are normally prepared from alcohols using either SOCl2, PBr3, or HF in pyridine. Alkyl halides react with magnesium in ether solution to form organomagnesium halides, called Grignard reagents (RMgX), which are both nucleophilic and strongly basic.
Alkyl halides also react with lithium metal to form organolithium reagents, RLi. In the presence of CuI, these form diorganocoppers, or Gilman reagents (LiR2Cu). Gilman reagents react with organohalides to yield coupled hydrocarbon products.
10.13: Summary of Reactions
10 • Summary of Reactions 10 • Summary of Reactions
No stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.
1. Preparation of alkyl halides
1. From alkenes by allylic bromination (Section 10.3)
• From alcohols (Section 10.5)
(1) Reaction with HX
(2) Reaction of 1° and 2° alcohols with SOCl2
(3) Reaction of 1° and 2° alcohols with PBr3
(4) Reaction of 1° and 2° alcohols with HF–pyridine
• Reactions of alkyl halides
1. Formation of Grignard (organomagnesium) reagents (Section 10.6)
2. Formation of Gilman (diorganocopper) reagents (Section 10.7)
3. Organometallic coupling (Section 10.7)
(1) Diorganocopper reaction
(2) Palladium-catalyzed Suzuki–Miyaura reaction | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.10%3A_Chemistry_MattersNaturally_Occurring_Organohalides.txt |
10 • Additional Problems 10 • Additional Problems
Visualizing Chemistry
Problem 10-14
Give IUPAC names for the following alkyl halides (green = Cl): (a)
(b)
Problem 10-15 Show the product(s) of reaction of the following alkenes with NBS: (a) (b) Problem 10-16 The following alkyl bromide can be prepared by reaction of the alcohol (S)-2-pentanol with PBr3. Name the compound, assign (R) or (S) stereochemistry, and tell whether the reaction of the alcohol results in the same stereochemistry or a change in stereochemistry (reddish brown = Br). Mechanism Problems Problem 10-17 In light of the fact that tertiary alkyl halides undergo spontaneous dissociation to yield a carbocation plus halide ion (see Problem 10-41), propose a mechanism for the following reaction. Naming Alkyl Halides Problem 10-18 Name the following alkyl halides: (a) (b) (c) (d) (e) Problem 10-19 Draw structures corresponding to the following IUPAC names: (a) 2,3-Dichloro-4-methylhexane (b)
4-Bromo-4-ethyl-2-methylhexane
(c) 3-Iodo-2,2,4,4-tetramethylpentane (d) cis-1-Bromo-2-ethylcyclopentane
Problem 10-20 Draw and name all the monochlorination products you might obtain from radical chlorination of the following compounds. Which of the products are chiral? Are any of the products optically active? (a)
2-methylbutane
(b) methylcyclopropane (c) 2,2-dimethylpentane
Synthesizing Alkyl Halides
Problem 10-21 How would you prepare the following compounds, starting with cyclopentene and any other reagents needed? (a)
Chlorocyclopentane
(b) Methylcyclopentane (c) 3-Bromocyclopentene (d) Cyclopentanol (e) Cyclopentylcyclopentane (f) 1,3-Cyclopentadiene
Problem 10-22 Predict the product(s) of the following reactions: (a) (b) (c) (d) (e) (f) (g) Problem 10-23 A chemist requires a large amount of 1-bromo-2-pentene as starting material for a synthesis and decides to carry out an NBS allylic bromination reaction. What is wrong with the following synthesis plan? What side products would form in addition to the desired product? Problem 10-24 What product(s) would you expect from the reaction of 1-methylcyclohexene with NBS? Would you use this reaction as part of a synthesis? Problem 10-25 What product(s) would you expect from the reaction of 1,4-hexadiene with NBS? What is the structure of the most stable radical intermediate? Problem 10-26 What product would you expect from the reaction of 1-phenyl-2-butene with NBS? Explain. Oxidation and Reduction Problem 10-27 Rank the compounds in each of the following series in order of increasing oxidation level: (a) (b) Problem 10-28 Which of the following compounds have the same oxidation level, and which have different levels? Problem 10-29 Tell whether each of the following reactions is an oxidation, a reduction, or neither: (a) (b) (c) General Problems Problem 10-30 Arrange the following radicals from most stable to least stable. (a) (b) (c) Problem 10-31 Alkylbenzenes such as toluene (methylbenzene) react with NBS to give products in which bromine substitution has occurred at the position next to the aromatic ring (the benzylic position). Explain, based on the bond dissociation energies in Table 6.3. Problem 10-32 Draw resonance structures for the benzyl radical, C6H5CH2·, the intermediate produced in the NBS bromination reaction of toluene (Problem 10-31). Problem 10-33 Draw resonance structures for the following species: (a) (b) (c) Problem 10-34 (S)-3-Methylhexane undergoes radical bromination to yield optically inactive 3-bromo-3-methylhexane as the major product. Is the product chiral? What conclusions can you draw about the radical intermediate? Problem 10-35 Assume that you have carried out a radical chlorination reaction on (R)-2-chloropentane and have isolated (in low yield) 2,4-dichloropentane. How many stereoisomers of the product are formed, and in what ratio? Are any of the isomers optically active? (See Problem 10-34.) Problem 10-36 How would you carry out the following syntheses? Problem 10-37 The syntheses shown here are unlikely to occur as written. What is wrong with each? (a) (b) (c) Problem 10-38 Why do you suppose it’s not possible to prepare a Grignard reagent from a bromo alcohol such as 4-bromo-1-pentanol? Give another example of a molecule that is unlikely to form a Grignard reagent. Problem 10-39 Addition of HBr to a double bond with an ether (−OR) substituent occurs regiospecifically to give a product in which the −Br and −OR are bonded to the same carbon. Draw the two possible carbocation intermediates in this electrophilic addition reaction, and explain using resonance why the observed product is formed. Problem 10-40 Identify the reagents a–c in the following scheme: Problem 10-41 Tertiary alkyl halides, R3CX, undergo spontaneous dissociation to yield a carbocation, R3C+, plus halide ion. Which do you think reacts faster, (CH3)3CBr or H2C$\text{=}$CHC(CH3)2Br? Explain. Problem 10-42 Carboxylic acids (RCO2H; pKa ≈ 5) are approximately 1011 times more acidic than alcohols (ROH; pKa ≈ 16). In other words, a carboxylate ion (RCO2−) is more stable than an alkoxide ion (RO−). Explain, using resonance. Problem 10-43 How might you use a Suzuki–Miyaura reaction to prepare the biaryl compounds below? In each case, show the two potential reaction partners. (a) (b) (c) Problem 10-44 The relative rate of radical bromination is 1 : 82 : 1640 for 1° : 2° : 3° hydrogens, respectively. Draw all of the monobrominated products that you might obtain from the radical bromination of the compounds below. Calculate the relative percentage of each. (a) methylcyclobutane (b)
3,3-dimethylpentane
(c) 3-methylpentane
Problem 10-45
Choose the alcohol from each pair below that would react faster with HX to form the corresponding alkyl halide. (a)
(b)
(c)
Problem 10-46
Predict the product and provide the entire catalytic cycle for the following Suzuki–Miyaura reactions. (a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/10%3A_Organohalides/10.14%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 11, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• use the reactions studied in this chapter with those from earlier ones when designing multistep syntheses.
• use the reactions and concepts discussed in this chapter to solve road map problems.
• define, and use in context, the key terms introduced.
In this course, you have already seen several examples of nucleophilic substitution reactions; now you will see that these reactions can occur by two different mechanisms. You will study the factors that determine which mechanism will be in operation in a given situation, and examine possible ways for increasing or decreasing the rates at which such reactions occur. The stereochemical consequences of both mechanisms will also be discussed. Elimination reactions often accompany nucleophilic substitution; so these reactions are also examined in this chapter. Again you will see that two different mechanisms are possible, and, as in the case of nucleophilic substitution reactions, chemists have learned a great deal about the factors that determine which mechanism will be observed when a given alkyl halide undergoes such a reaction.
11: Reactions of Alkyl Halides- Nucleophilic Substitutions and Eliminations
Chapter Contents
11.1 The Discovery of Nucleophilic Substitution Reactions
11.2 The SN2 Reaction 11.3 Characteristics of the SN2 Reaction 11.4 The SN1 Reaction 11.5 Characteristics of the SN1 Reaction 11.6 Biological Substitution Reactions 11.7 Elimination Reactions: Zaitsev’s Rule 11.8 The E2 Reaction and the Deuterium Isotope Effect 11.9 The E2 Reaction and Cyclohexane Conformation 11.10 The E1 and E1cB Reactions 11.11 Biological Elimination Reactions 11.12 A Summary of Reactivity: SN1, SN2, E1, E1cB, and E2
Nucleophilic substitution and base-induced elimination are two of the most widely occurring and versatile reactions in organic chemistry, both in the laboratory and in biological pathways. We’ll look at them closely in this chapter to see how they occur, what their characteristics are, and how they can be used. We’ll begin with substitution reactions.
We saw in the preceding chapter that the carbon–halogen bond in an alkyl halide is polar and that the carbon atom is electron-poor. Thus, alkyl halides are electrophiles, and much of their chemistry involves polar reactions with nucleophiles and bases. Alkyl halides do one of two things when they react with a nucleophile/base such as hydroxide ion: they either undergo substitution of the X group by the nucleophile, or they undergo elimination of HX to yield an alkene.
11.02: The Discovery of Nucleophilic Substitution Reactions
Discovery of the nucleophilic substitution reaction of alkyl halides dates back to work carried out by the German chemist Paul Walden in 1896. Walden found that the pure enantiomeric (+)- and (–)-malic acids could be interconverted through a series of simple substitution reactions. When Walden treated (–)-malic acid with PCl5, he isolated (+)-chlorosuccinic acid. This, on treatment with wet Ag2O, gave (+)-malic acid. Similarly, reaction of (+)-malic acid with PCl5 gave (–)-chlorosuccinic acid, which was converted into (–)-malic acid when treated with wet Ag2O. The full cycle of reactions is shown in Figure 11.2.
At the time, the results were astonishing. The eminent chemist Emil Fischer called Walden’s discovery “the most remarkable observation made in the field of optical activity since the fundamental observations of Pasteur.” Because (–)-malic acid was converted into (+)-malic acid, some reactions in the cycle must have occurred with a change, or inversion, of configuration at the chirality center. But which ones, and how? (Remember from Section 5.5 that the direction of light rotation and the configuration of a chirality center aren’t directly related. You can’t tell by looking at the sign of rotation whether a change in configuration has occurred during a reaction.)
Today, we refer to the transformations taking place in Walden’s cycle as nucleophilic substitution reactions because each step involves the substitution of one nucleophile (chloride ion, Cl, or hydroxide ion, HO) by another. Nucleophilic substitution reactions are one of the most common and versatile reaction types in organic chemistry.
Following the work of Walden, further investigations were undertaken during the 1920s and 1930s to clarify the mechanism of nucleophilic substitution reactions and to find out how inversions of configuration occur. Among the first series studied was one that interconverted the two enantiomers of 1-phenyl-2-propanol (Figure 11.3). Although this particular series of reactions involves nucleophilic substitution of an alkyl para-toluenesulfonate (called a tosylate) rather than an alkyl halide, exactly the same type of reaction is involved as that studied by Walden. For all practical purposes, the entire tosylate group acts as if it were simply a halogen substituent. (In fact, when you see a tosylate substituent in a molecule, do a mental substitution and tell yourself that you’re dealing with an alkyl halide.)
, where acetate ion substitutes for tosylate ion.
In the three-step reaction sequence shown in Figure 11.3, (+)-1-phenyl-2-propanol is interconverted with its (–) enantiomer, so at least one of the three steps must involve an inversion of configuration at the chirality center. Step 1, formation of a tosylate, occurs by breaking the O–H bond of the alcohol rather than the C–O bond to the chiral carbon, so the configuration around the carbon is unchanged. Similarly, step 3, hydroxide-ion cleavage of the acetate, takes place without breaking the C–O bond at the chirality center. Thus, the inversion of stereochemical configuration must take place in step 2, the nucleophilic substitution of tosylate ion by acetate ion.
From this and nearly a dozen other series of similar reactions, researchers concluded that the nucleophilic substitution reaction of a primary or secondary alkyl halide or tosylate always proceeds with inversion of configuration. (Tertiary alkyl halides and tosylates, as we’ll see shortly, give different stereochemical results and react by a different mechanism than the primary and secondary ones.)
Worked Example 11.1
Predicting the Stereochemistry of a Nucleophilic Substitution Reaction
What product would you expect from a nucleophilic substitution reaction of (R)-1-bromo-1-phenylethane with cyanide ion, $−C≡N−C≡N$, as nucleophile? Show the stereochemistry of both reactant and product, assuming that inversion of configuration occurs.
Strategy
Draw the R enantiomer of the reactant, and then change the configuration of the chirality center while replacing the Br with a CN.
Solution
Problem 11-1
What product would you expect from a nucleophilic substitution reaction of (S)-2-bromohexane with acetate ion, CH3CO2? Assume that inversion of configuration occurs, and show the stereochemistry of both the reactant and product. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.01%3A_Why_This_Chapter.txt |
In almost all chemical reactions, there is a direct relationship between the rate at which the reaction occurs and the concentrations of the reactants. When we measure this relationship, we measure the kinetics of the reaction. For example, let’s look at the kinetics of a simple nucleophilic substitution—the reaction of CH3Br with OH to yield CH3OH plus Br.
With a given temperature, solvent, and concentration of reactants, the substitution occurs at a certain rate. If we double the concentration of OH, the frequency of encounters between reaction partners doubles and we find that the reaction rate also doubles. Similarly, if we double the concentration of CH3Br, the reaction rate again doubles. We call such a reaction, in which the rate is linearly dependent on the concentrations of two species, a second-order reaction. Mathematically, we can express this second-order dependence of the nucleophilic substitution reaction by setting up a rate equation. As either [RX] or [OH] changes, the rate of the reaction changes proportionately.
$Reaction rate =Rate of disappearance of reactant =k×[RX]× [ − OH] Reaction rate =Rate of disappearance of reactant =k×[RX]× [ − OH]$
where
$[RX]= CH 3 Br concentration in molarity [ − OH] = − OH concentration in molarity k=a constant value (the rate constant) [RX]= CH 3 Br concentration in molarity [ − OH] = − OH concentration in molarity k=a constant value (the rate constant)$
A mechanism that accounts for both the inversion of configuration and the second-order kinetics that are observed with nucleophilic substitution reactions was suggested in 1937 by the British chemists E. D. Hughes and Christopher Ingold, who formulated what they called the SN2 reaction—short for substitution, nucleophilic, bimolecular. (Bimolecular means that two molecules, nucleophile and alkyl halide, take part in the step whose kinetics are measured.)
The essential feature of the SN2 mechanism is that it takes place in a single step, without intermediates, when the incoming nucleophile reacts with the alkyl halide or tosylate (the substrate) from a direction opposite the group that is displaced (the leaving group). As the nucleophile comes in on one side of the substrate and bonds to the carbon, the halide or tosylate departs from the other side, thereby inverting the stereochemical configuration. The process is shown in Figure 11.4 for the reaction of (S)-2-bromobutane with HO to give (R)-2-butanol.
Figure 11.4 MECHANISM The mechanism of the SN2 reaction. The reaction takes place in a single step when the incoming nucleophile approaches from a direction 180° away from the leaving halide ion, thereby inverting the stereochemistry at carbon.
As shown in Figure 11.4, the SN2 reaction occurs when an electron pair on the nucleophile Nu: forces out the group X:, which takes with it the electron pair from the former C–X bond. This occurs through a transition state in which the new Nu–C bond is partially formed at the same time that the old C–X bond is partially broken and in which the negative charge is shared by both the incoming nucleophile and the outgoing halide ion. The transition state for this inversion has the remaining three bonds to carbon in a planar arrangement (Figure 11.5).
is delocalized in the transition state.
The mechanism proposed by Hughes and Ingold is fully consistent with experimental results, explaining both stereochemical and kinetic data. Thus, the requirement for a backside approach of the entering nucleophile (180° away from the departing X group) causes the stereochemistry of the substrate to invert, much like an umbrella turning inside-out in the wind. The Hughes–Ingold mechanism also explains why second-order kinetics are observed: the SN2 reaction occurs in a single step that involves both alkyl halide and nucleophile. Two molecules are involved in the step whose rate is measured.
Problem 11-2
What product would you expect to obtain from SN2 reaction of OH with (R)-2-bromobutane? Show the stereochemistry of both the reactant and product.
Problem 11-3
Assign configuration to the following substance, and draw the structure of the product that would result from nucleophilic substitution reaction with HS (reddish brown = Br): | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.03%3A_The_SN2_Reaction.txt |
Now that we know how SN2 reactions occur, we need to see how they can be used and what variables affect them. Some SN2 reactions are fast, and some are slow; some take place in high yield and others in low yield. Understanding the factors involved can be of tremendous value. Let’s begin by recalling a few things about reaction rates in general.
The rate of a chemical reaction is determined by the activation energy ∆G, the energy difference between reactant ground state and transition state. A change in reaction conditions can affect ∆G either by changing the reactant energy level or by changing the transition-state energy level. Lowering the reactant energy or raising the transition-state energy increases ∆G and decreases the reaction rate; raising the reactant energy or decreasing the transition-state energy decreases ∆G and increases the reaction rate (Figure 11.6). We’ll see examples of all these effects as we look at SN2 reaction variables.
Steric Effects in the SN2 Reaction
The first SN2 reaction variable to look at is the structure of the substrate. Because the SN2 transition state involves partial bond formation between the incoming nucleophile and the alkyl halide carbon atom, it seems reasonable that a hindered, bulky substrate should prevent easy approach of the nucleophile, making bond formation difficult. In other words, the transition state for reaction of a sterically hindered substrate, whose carbon atom is “shielded” from the approach of the incoming nucleophile, is higher in energy and forms more slowly than the corresponding transition state for a less hindered substrate (Figure 11.7).
As Figure 11.7 shows, the difficulty of nucleophile approach increases as the three substituents bonded to the halo-substituted carbon atom increase in size. Methyl halides are by far the most reactive substrates in SN2 reactions, followed by primary alkyl halides such as ethyl and propyl. Alkyl branching at the reacting center, as in isopropyl halides (2°), slows the reaction greatly, and further branching, as in tert-butyl halides (3°), effectively halts the reaction. Even branching one carbon away from the reacting center, as in 2,2-dimethylpropyl (neopentyl) halides, greatly hinders nucleophilic displacement. As a result, SN2 reactions occur only at relatively unhindered sites and are normally useful only with methyl halides, primary halides, and a few simple secondary halides. Relative reactivities for some different substrates are as follows:
Vinylic halides ($R2C═CRXR2C═CRX$) and aryl halides are not shown on this reactivity list because they are unreactive toward SN2 displacement. This lack of reactivity is due to steric factors: the incoming nucleophile would have to approach in the plane of the carbon–carbon double bond and burrow through part of the molecule to carry out a backside displacement.
The Nucleophile
Another variable that has a major effect on the SN2 reaction is the nature of the nucleophile. Any species, either neutral or negatively charged, can act as a nucleophile as long as it has an unshared pair of electrons; that is, as long as it is a Lewis base. If the nucleophile is negatively charged, the product is neutral; if the nucleophile is neutral, the product is positively charged.
A wide array of substances can be prepared using nucleophilic substitution reactions. In fact, we’ve already seen examples in previous chapters. For instance, the reaction of an acetylide anion with an alkyl halide, discussed in Section 9.8, is an SN2 reaction in which the acetylide nucleophile displaces a halide leaving group.
Table 11.1 lists some nucleophiles in the order of their reactivity, shows the products of their reactions with bromomethane, and gives the relative rates of their reactions. There are large differences in the rates at which various nucleophiles react.
Table 11.1 Some SN2 Reactions with Bromomethane
Nu: + CH3Br → CH3Nu + Br
Nucleophile Product Relative rate of reaction
Formula Name Formula Name
H2O Water CH3OH2+ Methylhydronium ion 1
CH3CO2 Acetate CH3CO2CH3 Methyl acetate 500
NH3 Ammonia CH3NH3+ Methylammonium ion 700
Cl Chloride CH3Cl Chloromethane 1,000
HO Hydroxide CH3OH Methanol 10,000
CH3O Methoxide CH3OCH3 Dimethyl ether 25,000
I Iodide CH3I Iodomethane 100,000
CN Cyanide CH3CN Acetonitrile 125,000
HS Hydrosulfide CH3SH Methanethiol 125,000
What are the reasons for the reactivity differences observed in Table 11.1? Why do some reactants appear to be much more “nucleophilic” than others? The answers to these questions aren’t straightforward. Part of the problem is that the term nucleophilicity is imprecise. The term is usually taken to be a measure of the affinity of a nucleophile for a carbon atom in the SN2 reaction, but the reactivity of a given nucleophile can change from one reaction to the next. The exact nucleophilicity of a species in a given reaction depends on the substrate, the solvent, and even the reactant concentrations. Detailed explanations for the observed nucleophilicities aren’t always simple, but some trends can be detected from the data of Table 11.1.
• Nucleophilicity roughly parallels basicity when comparing nucleophiles that have the same reacting atom. Thus, OH is both more basic and more nucleophilic than acetate ion, CH3CO2, which in turn is more basic and more nucleophilic than H2O. Since “nucleophilicity” is usually taken as the affinity of a Lewis base for a carbon atom in the SN2 reaction and “basicity” is the affinity of a base for a proton, it’s easy to see why there might be a correlation between the two kinds of behavior.
• Nucleophilicity usually increases going down a column of the periodic table. Thus, HS is more nucleophilic than HO, and the halide reactivity order is I > Br > Cl. Going down the periodic table, elements have their valence electrons in successively larger shells where they are successively farther from the nucleus, less tightly held, and consequently more reactive. This matter is complex, though, and the nucleophilicity order can change depending on the solvent.
• Negatively charged nucleophiles are usually more reactive than neutral ones. As a result, SN2 reactions are often carried out under basic conditions rather than neutral or acidic conditions.
Problem 11-4 What product would you expect from SN2 reaction of 1-bromobutane with each of the following? (a)
NaI
(b) KOH (c) H−C$\text{≡}$C−Li (d) NH3
Problem 11-5 Which substance in each of the following pairs is more reactive as a nucleophile? Explain. (a)
(CH3)2N or (CH3)2NH
(b) (CH3)3B or (CH3)3N (c) H2O or H2S
The Leaving Group
Still another variable that can affect the SN2 reaction is the nature of the group displaced by the incoming nucleophile, the leaving group. Because the leaving group is expelled with a negative charge in most SN2 reactions, the best leaving groups are those that best stabilize the negative charge in the transition state. The greater the extent of charge stabilization by the leaving group, the lower the energy of the transition state and the more rapid the reaction. But as we saw in Section 2.8, the groups that best stabilize a negative charge are also the weakest bases. Thus, weak bases such as Cl, Br, and tosylate ion make good leaving groups, while strong bases such as OH and NH2 make poor leaving groups.
It’s just as important to know which are poor leaving groups as to know which are good, and the preceding data clearly indicate that F, HO, RO, and H2N are not displaced by nucleophiles. In other words, alkyl fluorides, alcohols, ethers, and amines do not typically undergo SN2 reactions. To carry out an SN2 reaction with an alcohol, it’s necessary to convert the OH into a better leaving group. This, in fact, is just what happens when a primary or secondary alcohol is converted into either an alkyl chloride by reaction with SOCl2 or an alkyl bromide by reaction with PBr3 (Section 10.5).
Alternatively, an alcohol can be made more reactive toward nucleophilic substitution by treating it with para-toluenesulfonyl chloride to form a tosylate. As noted previously, tosylates are even more reactive than halides in nucleophilic substitutions. Note that tosylate formation does not change the configuration of the oxygen-bearing carbon because the C–O bond is not broken.
The one general exception to the rule that ethers don’t typically undergo SN2 reactions pertains to epoxides, the three-membered cyclic ethers that we saw in Section 8.7. Because of the angle strain in their three-membered ring, epoxides are much more reactive than other ethers. They react with aqueous acid to give 1,2-diols, as we saw in Section 8.7, and they react readily with many other nucleophiles as well. Propene oxide, for instance, reacts with HCl to give 1-chloro-2-propanol by an SN2 backside attack on the less hindered primary carbon atom. We’ll look at the process in more detail in Section 18.5.
Problem 11-6
Rank the following compounds in order of their expected reactivity toward SN2 reaction:
CH3Br, CH3OTos, (CH3)3CCl, (CH3)2CHCl
The Solvent
The rates of SN2 reactions are strongly affected by the solvent. Protic solvents—those that contain an –OH or –NH group—are generally the worst for SN2 reactions, while polar aprotic solvents, which are polar but don’t have an –OH or –NH group, are the best.
Protic solvents, such as methanol and ethanol, slow down SN2 reactions by solvation of the reactant nucleophile. The solvent molecules hydrogen-bond to the nucleophile and form a cage around it, thereby lowering its energy and reactivity.
In contrast with protic solvents—which decrease the rates of SN2 reactions by lowering the ground-state energy of the nucleophile—polar aprotic solvents increase the rates of SN2 reactions by raising the ground-state energy of the nucleophile. Acetonitrile (CH3CN), dimethylformamide [(CH3)2NCHO, abbreviated DMF], and dimethyl sulfoxide [(CH3)2SO, abbreviated DMSO] are particularly useful. A solvent known as hexamethylphosphoramide {[(CH3)2N]3PO, abbreviated HMPA} can also be useful but it should only be handled with great care and not be allowed to touch the eyes or skin. These solvents can dissolve many salts because of their high polarity, but they tend to solvate metal cations rather than nucleophilic anions. As a result, the bare, unsolvated anions have a greater nucleophilicity and SN2 reactions take place at correspondingly increased rates. For instance, a rate increase of 200,000 has been observed on changing from methanol to HMPA for the reaction of azide ion with 1-bromobutane.
Problem 11-7
Organic solvents like benzene, ether, and chloroform are neither protic nor strongly polar. What effect would you expect these solvents to have on the reactivity of a nucleophile in SN2 reactions?
A Summary of SN2 Reaction Characteristics
The effects on SN2 reactions of the four variables—substrate structure, nucleophile, leaving group, and solvent—are summarized in the following statements and in the energy diagrams of Figure 11.8:
Substrate Steric hindrance raises the energy of the SN2 transition state, increasing ∆G and decreasing the reaction rate (Figure 11.8a). As a result, SN2 reactions are best for methyl and primary substrates. Secondary substrates react slowly, and tertiary substrates do not react by an SN2 mechanism.
Nucleophile Basic, negatively charged nucleophiles are less stable and have a higher ground-state energy than neutral ones, decreasing ∆G and increasing the SN2 reaction rate (Figure 11.8b).
Leaving group Good leaving groups (more stable anions) lower the energy of the transition state, decreasing ∆G and increasing the SN2 reaction rate (Figure 11.8c).
Solvent Protic solvents solvate the nucleophile, thereby lowering its ground-state energy, increasing ∆G, and decreasing the SN2 reaction rate. Polar aprotic solvents surround the accompanying cation but not the nucleophilic anion, thereby raising the ground-state energy of the nucleophile, decreasing ∆G, and increasing the reaction rate (Figure 11.8d). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.04%3A_Characteristics_of_the_SN2_Reaction.txt |
Most nucleophilic substitutions take place by the SN2 pathway just discussed. The reaction is favored when carried out with an unhindered substrate and a negatively charged nucleophile in a polar aprotic solvent, but is disfavored when carried out with a hindered substrate and a neutral nucleophile in a protic solvent. You might therefore expect the reaction of a tertiary substrate (hindered) with water (neutral, protic) to be among the slowest of substitution reactions. Remarkably, however, the opposite is true. The reaction of the tertiary halide 2-bromo-2-methylpropane (CH3)3CBr with H2O to give the alcohol 2-methyl-2-propanol is more than 1 million times faster than the corresponding reaction of CH3Br to give methanol.
What’s going on here? A nucleophilic substitution reaction is occurring—a hydroxyl group is replacing a halogen—yet the reactivity order seems backward. These reactions can’t be taking place by the SN2 mechanism we’ve been discussing, so we must therefore conclude that they are occurring by an alternative substitution mechanism. This alternative mechanism is called the SN1 reaction, for substitution, nucleophilic, unimolecular.
In contrast to the SN2 reaction of CH3Br with OH, the SN1 reaction of (CH3)3CBr with H2O has a rate that depends only on the alkyl halide concentration and is independent of the H2O concentration. In other words, the process is a first-order reaction; the concentration of the nucleophile does not appear in the rate equation.
$Reaction rate =Rate of disappearance of alkyl halide =k×[RX] Reaction rate =Rate of disappearance of alkyl halide =k×[RX]$
To explain this result, we need to know more about kinetics measurements. Many organic reactions occur in several steps, one of which usually has a higher-energy transition state than the others and is therefore slower. We call this step with the highest transition-state energy the rate-limiting step, or rate-determining step. No reaction can proceed faster than its rate-limiting step, which acts as a kind of traffic jam, or bottleneck. In the SN1 reaction of (CH3)3CBr with H2O, the fact that the nucleophile concentration does not appear in the first-order rate equation means that it is not involved in the rate-limiting step and must therefore be involved in some other, non-rate-limiting step. The mechanism shown in Figure 11.9 accounts for these observations.
Figure 11.9 MECHANISM The mechanism of the SN1 reaction of 2-bromo-2-methylpropane with H2O involves three steps. Step 1 —the spontaneous, unimolecular dissociation of the alkyl bromide to yield a carbocation—is rate-limiting.
Unlike what occurs in an SN2 reaction, where the leaving group is displaced while the incoming nucleophile approaches, an SN1 reaction takes place by loss of the leaving group before the nucleophile approaches. 2-Bromo-2-methylpropane spontaneously dissociates to the tert-butyl carbocation (CH3)3C+, plus Br in a slow, rate-limiting step, and the intermediate carbocation is then immediately trapped by the nucleophile water in a faster second step. Thus, water is not a reactant in the step whose rate is measured. The energy diagram is shown in Figure 11.10.
Because an SN1 reaction occurs through a carbocation intermediate, its stereochemical outcome is different from that of an SN2 reaction. Carbocations, as we’ve seen, are planar, sp2-hybridized, and achiral. Thus, if we carry out an SN1 reaction on one enantiomer of a chiral reactant and go through an achiral carbocation intermediate, the product loses its optical activity (Section 8.12). That is, the symmetrical intermediate carbocation can react with a nucleophile equally well from either side, leading to a racemic, 50 : 50 mixture of enantiomers (Figure 11.11).
The conclusion that SN1 reactions on enantiomerically pure substrates should give racemic products is nearly, but not exactly, what is found. In fact, few SN1 displacements occur with complete racemization. Most give a minor (0–20%) excess of inversion. The reaction of (R)-6-chloro-2,6-dimethyloctane with H2O, for example, leads to an alcohol product that is approximately 80% racemized and 20% inverted (80% R,S + 20% S is equivalent to 40% R + 60% S).
This lack of complete racemization in SN1 reactions is due to the fact that ion pairs are involved. According to this explanation, first proposed by Saul Winstein at UCLA, dissociation of the substrate occurs to give a structure in which the two ions are still loosely associated and in which the carbocation is effectively shielded from reaction on one side by the departing anion. If a certain amount of substitution occurs before the two ions fully diffuse apart, then a net inversion of configuration will be observed (Figure 11.12).
Problem 11-8
What product(s) would you expect from reaction of (S)-3-chloro-3-methyloctane with acetic acid? Show the stereochemistry of both reactant and product.
Problem 11-9
Among the many examples of SN1 reactions that occur with incomplete racemization, the optically pure tosylate of 2,2-dimethyl-1-phenyl-1-propanol ([α]D = –30.3) gives the corresponding acetate ([α]D = +5.3) when heated in acetic acid. If complete inversion had occurred, the optically pure acetate would have had [α]D = +53.6. What percentage racemization and what percentage inversion occurred in this reaction?
Problem 11-10
Assign configuration to the following substrate, and show the stereochemistry and identity of the product you would obtain by SN1 reaction with water (reddish brown = Br): | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.05%3A_The_SN1_Reaction.txt |
Just as the SN2 reaction is strongly influenced by the structure of the substrate, the leaving group, the nucleophile, and the solvent, the SN1 reaction is similarly influenced. Factors that lower ∆G, either by lowering the energy level of the transition state or by raising the energy level of the ground state, favor faster SN1 reactions. Conversely, factors that raise ∆G, either by raising the energy level of the transition state or by lowering the energy level of the reactant, slow down the SN1 reaction.
The Substrate
According to the Hammond postulate (Section 7.10), any factor that stabilizes a high-energy intermediate also stabilizes the transition state leading to that intermediate. Because the rate-limiting step in an SN1 reaction is the spontaneous, unimolecular dissociation of the substrate to yield a carbocation, the reaction is favored whenever a stabilized carbocation intermediate is formed. The more stable the carbocation intermediate, the faster the SN1 reaction.
We saw in Section 7.9 that the stability order of alkyl carbocations is 3° > 2° > 1° > methyl. To this list we should also add the resonance-stabilized allyl and benzyl cations. Just as allylic radicals are unusually stable because the unpaired electron can be delocalized over an extended π orbital system (Section 10.4), so allylic and benzylic carbocations are unusually stable. (The word benzylic means “next to an aromatic ring.”) As Figure 11.13 indicates, an allylic cation has two resonance forms. In one form, the double bond is on the “left”; in the other form it’s on the “right.” A benzylic cation has five resonance forms, all of which contribute to the overall resonance hybrid.
are indicated by blue arrows.
Because of resonance stabilization, a primary allylic or benzylic carbocation is about as stable as a secondary alkyl carbocation, and a secondary allylic or benzylic carbocation is about as stable as a tertiary alkyl carbocation. This stability order of carbocations is the same as the order of SN1 reactivity for alkyl halides and tosylates.
We should also note parenthetically that primary allylic and benzylic substrates are particularly reactive in SN2 reactions as well as in SN1 reactions. Allylic and benzylic C–X bonds are about 50 kJ/mol (12 kcal/mol) weaker than the corresponding saturated bonds and are therefore more easily broken.
Problem 11-11
Rank the following substances in order of their expected SN1 reactivity:
Problem 11-12
3-Bromo-1-butene and 1-bromo-2-butene undergo SN1 reaction at nearly the same rate, even though one is a secondary halide and the other is primary. Explain.
The Leaving Group
We said during the discussion of SN2 reactivity that the best leaving groups are those that are most stable; that is, those that are the conjugate bases of strong acids. An identical reactivity order is found for the SN1 reaction because the leaving group is directly involved in the rate-limiting step. Thus, the SN1 reactivity order is
Note that in the SN1 reaction, which is often carried out under acidic conditions, neutral water is sometimes the leaving group. This occurs, for example, when an alkyl halide is prepared from a tertiary alcohol by reaction with HBr or HCl (Section 10.5). As shown in Figure 11.14, the alcohol is first protonated and then spontaneously loses H2O to generate a carbocation, which reacts with halide ion to give the alkyl halide. Knowing that an SN1 reaction is involved in the conversion of alcohols to alkyl halides explains why the reaction works well only for tertiary alcohols. Tertiary alcohols react fastest because they give the most stable carbocation intermediates.
Figure 11.14 MECHANISM The mechanism of the SN1 reaction of a tertiary alcohol with HBr to yield an alkyl halide. Neutral water is the leaving group (step 2).
The Nucleophile
The nature of the nucleophile plays a major role in the SN2 reaction but does not affect an SN1 reaction. Because the SN1 reaction occurs through a rate-limiting step in which the added nucleophile has no part, the nucleophile can’t affect the reaction rate. The reaction of 2-methyl-2-propanol with HX, for instance, occurs at the same rate regardless of whether X is Cl, Br, or I. Furthermore, neutral nucleophiles are just as effective as negatively charged ones, so SN1 reactions frequently occur under neutral or acidic conditions.
The Solvent
What about the solvent? Do solvents have the same effect in SN1 reactions that they have in SN2 reactions? The answer is both yes and no. Yes, solvents have a large effect on SN1 reactions, but no, the reasons for the effects on SN1 and SN2 reactions are not the same. Solvent effects in the SN2 reaction are due largely to stabilization or destabilization of the nucleophile reactant, while solvent effects in the SN1 reaction are due largely to stabilization or destabilization of the transition state.
The Hammond postulate says that any factor stabilizing the intermediate carbocation should increase the rate of an SN1 reaction. Solvation of the carbocation—the interaction of the ion with solvent molecules—has such an effect. Solvent molecules orient around the carbocation so that the electron-rich ends of the solvent dipoles face the positive charge (Figure 11.15), thereby lowering the energy of the ion and favoring its formation.
The properties of a solvent that contribute to its ability to stabilize ions by solvation are related to the solvent’s polarity. SN1 reactions take place much more rapidly in strongly polar solvents, such as water and methanol, than in less polar solvents, such as ether and chloroform. In the reaction of 2-chloro-2-methylpropane, for example, a rate increase of 100,000 is observed upon going from ethanol (less polar) to water (more polar). The rate increases when going from a hydrocarbon solvent to water are so large they can’t be measured accurately.
It should be emphasized again that both the SN1 and the SN2 reaction show solvent effects, but that they do so for different reasons. SN2 reactions are disfavored in protic solvents because the ground-state energy of the nucleophile is lowered by solvation. SN1 reactions are favored in protic solvents because the transition-state energy leading to carbocation intermediate is lowered by solvation.
A Summary of SN1 Reaction Characteristics
The effects on SN1 reactions of the four variables—substrate, leaving group, nucleophile, and solvent—are summarized in the following statements:
Substrate The best substrates yield the most stable carbocations. As a result, SN1 reactions are best for tertiary, allylic, and benzylic halides.
Leaving group Good leaving groups increase the reaction rate by lowering the energy level of the transition state for carbocation formation.
Nucleophile The nucleophile must be nonbasic to prevent a competitive elimination of HX (Section 11.7), but otherwise does not affect the reaction rate. Neutral nucleophiles work well.
Solvent Polar solvents stabilize the carbocation intermediate by solvation, thereby increasing the reaction rate.
Worked Example 11.2
Predicting the Mechanism of a Nucleophilic Substitution Reaction
Predict whether each of the following substitution reactions is likely to be SN1 or SN2:
Strategy
Look at the substrate, leaving group, nucleophile, and solvent. Then decide from the summaries at the ends of Section 11.3 and Section 11.5 whether an SN1 or an SN2 reaction is favored. SN1 reactions are favored by tertiary, allylic, or benzylic substrates, by good leaving groups, by nonbasic nucleophiles, and by protic solvents. SN2 reactions are favored by primary substrates, by good leaving groups, by good nucleophiles, and by polar aprotic solvents.
Solution
(a) This is likely to be an SN1 reaction because the substrate is secondary and benzylic, the nucleophile is weakly basic, and the solvent is protic.
(b) This is likely to be an SN2 reaction because the substrate is primary, the nucleophile is a good one, and the solvent is polar aprotic.
Problem 11-13 Predict whether each of the following substitution reactions is likely to be SN1 or SN2: (a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.06%3A_Characteristics_of_the_SN1_Reaction.txt |
Both SN1 and SN2 reactions are common in biological chemistry, particularly in the pathways for biosynthesis of the many thousands of plant-derived substances called terpenoids, which we saw briefly in the Chapter 8 Chemistry Matters and will discuss in Section 27.5. Unlike what typically happens in the laboratory, however, the substrate in a biological substitution reaction is usually an organodiphosphate rather than an alkyl halide. Thus, the leaving group is the diphosphate ion, abbreviated PPi, rather than a halide ion. In fact, it’s useful to think of the diphosphate group as the “biological equivalent” of a halide. The dissociation of an organodiphosphate in a biological reaction is typically assisted by complexation to a divalent metal cation such as Mg2+ to help neutralize charge and make the diphosphate a better leaving group.
As an example, two SN1 reactions occur during the biosynthesis of geraniol, a fragrant alcohol found in roses and used in perfumery. Geraniol biosynthesis begins with dissociation of dimethylallyl diphosphate to give an allylic carbocation, which reacts with isopentenyl diphosphate (Figure 11.16). From the viewpoint of isopentenyl diphosphate, the reaction is an electrophilic alkene addition, but from the viewpoint of dimethylallyl diphosphate, the process is an SN1 reaction in which the carbocation intermediate reacts with a double bond as the nucleophile.
Following this initial SN1 reaction, loss of the pro-R hydrogen gives geranyl diphosphate, itself an allylic diphosphate that dissociates a second time. Reaction of the geranyl carbocation with water in a second SN1 reaction, followed by loss of a proton, then yields geraniol.
As another example, SN2 reactions are involved in almost all biological methylations, which transfer a –CH3 group from an electrophilic donor to a nucleophile. The donor is S-adenosylmethionine (abbreviated SAM), which contains a positively charged sulfur (a sulfonium ion, Section 5.12), and the leaving group is the neutral S-adenosylhomocysteine molecule. In the biosynthesis of epinephrine (adrenaline) from norepinephrine, for instance, the nucleophilic nitrogen atom of norepinephrine attacks the electrophilic methyl carbon atom of S-adenosylmethionine in an SN2 reaction, displacing S-adenosylhomocysteine (Figure 11.17). In effect, S-adenosylmethionine is simply a biological equivalent of CH3Cl.
Problem 11-14
Review the mechanism of geraniol biosynthesis shown in Figure 11.16, and propose a mechanism for the biosynthesis of limonene from linalyl diphosphate.
11.08: Elimination Reactions- Zaitsev's Rule
We said at the beginning of this chapter that two kinds of reactions can take place when a nucleophile/Lewis base reacts with an alkyl halide. The nucleophile can either substitute for the halide by reaction at carbon or can cause elimination of HX by reaction at a neighboring hydrogen:
Elimination reactions are more complex than substitution reactions for several reasons. One is the problem of regiochemistry. What product results by loss of HX from an unsymmetrical halide? In fact, elimination reactions almost always give mixtures of alkene products, and the best we can usually do is to predict which will be the major product.
According to Zaitsev’s rule, formulated in 1875 by the Russian chemist Alexander Zaitsev, base-induced elimination reactions generally (although not always) give the more stable alkene product—that is, the alkene with more alkyl substituents on the double-bond carbons. In the following two cases, for example, the more highly substituted alkene product predominates.
ZAITSEV’S RULE
In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates.
Another factor that complicates a study of elimination reactions is that they can take place by different mechanisms, just as substitutions can. We’ll consider three of the most common mechanisms—the E1, E2, and E1cB reactions—which differ in the timing of C–H and C–X bond-breaking.
In the E1 reaction, the C–X bond breaks first to give a carbocation intermediate, which undergoes subsequent base abstraction of H+ to yield the alkene. In the E2 reaction, base-induced C–H bond cleavage is simultaneous with C–X bond cleavage, giving the alkene in a single step. In the E1cB reaction (cB for “conjugate base”), base abstraction of the proton occurs first, giving a carbanion (R:) intermediate. This anion, the conjugate base of the reactant “acid,” then undergoes loss of X in a subsequent step to give the alkene. All three mechanisms occur frequently in the laboratory, but the E1cB mechanism predominates in biological pathways.
Worked Example 11.3: Predicting the Product of an Elimination Reaction
What product would you expect from reaction of 1-chloro-1-methylcyclohexane with KOH in ethanol?
Strategy
Treatment of an alkyl halide with a strong base such as KOH yields an alkene. To find the products in a specific case, locate the hydrogen atoms on each carbon next to the leaving group, and then generate the potential alkene products by removing HX in as many ways as possible. The major product will be the one that has the most highly substituted double bond—in this case, 1-methylcyclohexene.
Solution
Problem 11-15
Ignoring double-bond stereochemistry, what products would you expect from elimination reactions of the following alkyl halides? Which product will be the major product in each case? (a)
(b)
(c)
Problem 11-16
What alkyl halides might the following alkenes have been made from?
(a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.07%3A_Biological_Substitution_Reactions.txt |
The E2 reaction (for elimination, bimolecular) occurs when an alkyl halide is treated with a strong base, such as hydroxide ion or alkoxide ion (RO). It is the most commonly occurring pathway for elimination and can be formulated as shown in Figure 11.18.
Figure 11.18 MECHANISM Mechanism of the E2 reaction of an alkyl halide. The reaction takes place in a single step through a transition state in which the double bond begins to form at the same time the H and X groups are leaving.
Like the SN2 reaction, the E2 reaction takes place in one step without intermediates. As the base begins to abstract H+ from a carbon next to the leaving group, the C–H bond begins to break, a $C═CC═C$ bond begins to form, and the leaving group begins to depart, taking with it the electron pair from the C–X bond. Among the pieces of evidence supporting this mechanism is the fact that E2 reactions show second-order kinetics and follow the rate law: rate = k × [RX] × [Base]. That is, both the base and alkyl halide take part in the rate-limiting step.
A second piece of evidence in support of the E2 mechanism is provided by a phenomenon known as the deuterium isotope effect. For reasons that we won’t go into, a carbon–hydrogen bond is weaker by about 5 kJ/mol (1.2 kcal/mol) than the corresponding carbon–deuterium bond. Thus, a C–H bond is more easily broken than an equivalent C–D bond, and the rate of C–H bond cleavage is faster. For instance, the base-induced elimination of HBr from 1-bromo-2-phenylethane proceeds 7.11 times faster than the corresponding elimination of DBr from 1-bromo-2, 2-dideuterio-2-phenylethane. This result tells us that the C–H (or C–D) bond is broken in the rate-limiting step, consistent with our picture of the E2 reaction as a one-step process. If it were otherwise, we wouldn’t observe a rate difference.
Yet a third piece of mechanistic evidence involves the stereochemistry of E2 eliminations. As shown by a large number of experiments, E2 reactions occur with periplanar geometry, meaning that all four reacting atoms—the hydrogen, the two carbons, and the leaving group—lie in the same plane. Two such geometries are possible: syn periplanar geometry, in which the H and the X are on the same side of the molecule, and anti periplanar geometry, in which the H and the X are on opposite sides of the molecule. Of the two, anti periplanar geometry is energetically preferred because it allows the substituents on the two carbons to adopt a staggered relationship, whereas syn geometry requires that the substituents be eclipsed.
What’s so special about periplanar geometry? Because the sp3 σ orbitals in the reactant C–H and C–X bonds must overlap and become p π orbitals in the alkene product, there must also be some overlap in the transition state. This can occur most easily if all the orbitals are in the same plane to begin with—that is, if they’re periplanar (Figure 11.19).
You can think of E2 elimination reactions with periplanar geometry as being similar to SN2 reactions with 180° geometry. In an SN2 reaction, an electron pair from the incoming nucleophile pushes out the leaving group on the opposite side of the molecule. In an E2 reaction, an electron pair from a neighboring C–H bond also pushes out the leaving group on the opposite side of the molecule.
Anti periplanar geometry for E2 eliminations has specific stereochemical consequences that provide strong evidence for the proposed mechanism. To take just one example, meso-1,2-dibromo-1,2-diphenylethane undergoes E2 elimination on treatment with base to give only the E alkene. None of the isomeric Z alkene is formed because the transition state leading to the Z alkene would have to have syn periplanar geometry and would thus be higher in energy.
Worked Example 11.4: Predicting the Double-Bond Stereochemistry of the Product in an E2 Reaction
What stereochemistry do you expect for the alkene obtained by E2 elimination of (1S,2S)-1,2-dibromo-1,2-diphenylethane?
Strategy
Draw (1S,2S)-1,2-dibromo-1,2-diphenylethane so that you can see its stereochemistry and so that the –H and –Br groups to be eliminated are anti periplanar. Then carry out the elimination while keeping all substituents in approximately the same positions, and see what alkene results.
Solution
Anti periplanar elimination of HBr gives (Z)-1-bromo-1,2-diphenylethylene.
Problem 11-17
What stereochemistry do you expect for the alkene obtained by E2 elimination of (1R,2R)-1,2-dibromo-1,2-diphenylethane? Draw a Newman projection of the reacting conformation.
Problem 11-18
What stereochemistry do you expect for the trisubstituted alkene obtained by E2 elimination of the following alkyl halide on treatment with KOH? (reddish brown = Br.) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.09%3A_The_E2_Reaction_and_the_Deuterium_Isotope_Effect.txt |
Anti periplanar geometry for E2 reactions is particularly important in cyclohexane rings, where chair geometry forces a rigid relationship between the substituents on neighboring carbon atoms (Section 4.8). The anti periplanar requirement for E2 reactions overrides Zaitsev’s rule and can be met in cyclohexanes only if the hydrogen and the leaving group are trans diaxial (Figure 11.20). If either the leaving group or the hydrogen is equatorial, E2 elimination can’t occur.
The elimination of HCl from the isomeric menthyl and neomenthyl chlorides shown in Figure 11.21 gives a good illustration of this trans-diaxial requirement. Neomenthyl chloride undergoes elimination of HCl on reaction with ethoxide ion 200 times faster than menthyl chloride. Furthermore, neomenthyl chloride yields 3-menthene as the major alkene product, whereas menthyl chloride yields 2-menthene.
The difference in reactivity between the isomeric menthyl chlorides is due to the difference in their conformations. Neomenthyl chloride has the conformation shown in Figure 11.21a, with the methyl and isopropyl groups equatorial and the chlorine axial—a perfect geometry for E2 elimination. Loss of the hydrogen atom at C4 occurs easily to yield the more substituted alkene product, 3-menthene, as predicted by Zaitsev’s rule.
Menthyl chloride, by contrast, has a conformation in which all three substituents are equatorial (Figure 11.21b). To achieve the necessary geometry for elimination, menthyl chloride must first ring-flip to a higher-energy chair conformation, in which all three substituents are axial. E2 elimination then occurs with loss of the only trans-diaxial hydrogen available, leading to the non-Zaitsev product 2-menthene. The net effect of the simple change in chlorine stereochemistry is a 200-fold change in reaction rate and a complete change of product. The chemistry of the molecule is controlled by its conformation.
Problem 11-19
Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer.
11.11: The E1 and E1cB Reactions
The E1 Reaction
Just as the E2 reaction is analogous to the SN2 reaction, the SN1 reaction has a close analog called the E1 reaction (for elimination, unimolecular). The E1 reaction can be formulated as shown in Figure 11.22, with the elimination of HCl from 2-chloro-2-methylpropane.
Figure 11.22 MECHANISM Mechanism of the E1 reaction. Two steps are involved, the first of which is rate-limiting, and a carbocation intermediate is present.
E1 eliminations begin with the same unimolecular dissociation to give a carbocation that we saw in the SN1 reaction, but the dissociation is followed by loss of H+ from the adjacent carbon rather than by substitution. In fact, the E1 and SN1 reactions normally occur together whenever an alkyl halide is treated in a protic solvent with a nonbasic nucleophile. Thus, the best E1 substrates are also the best SN1 substrates, and mixtures of substitution and elimination products are usually obtained. For example, when 2-chloro-2-methylpropane is warmed to 65 °C in 80% aqueous ethanol, a 64 : 36 mixture of 2-methyl-2-propanol (SN1) and 2-methylpropene (E1) results.
Much evidence has been obtained in support of the E1 mechanism. For example, E1 reactions show first-order kinetics, consistent with a rate-limiting, unimolecular dissociation process. Furthermore, E1 reactions show no deuterium isotope effect because rupture of the C–H (or C–D) bond occurs after the rate-limiting step rather than during it. Thus, we can’t measure a rate difference between a deuterated and nondeuterated substrate.
A final piece of evidence involves the stereochemistry of elimination. Unlike the E2 reaction, where anti periplanar geometry is required, there is no geometric requirement on the E1 reaction because the halide and the hydrogen are lost in separate steps. We might therefore expect to obtain the more stable (Zaitsev’s rule) product from E1 reaction, which is just what we find. To return to a familiar example, menthyl chloride loses HCl under E1 conditions in a polar solvent to give a mixture of alkenes in which the Zaitsev product, 3-menthene, predominates (Figure 11.23).
The E1cB Reaction
In contrast to the E1 reaction, which involves a carbocation intermediate, the E1cB reaction takes place through a carbanion intermediate. Base-induced abstraction of a proton in a slow, rate-limiting step gives an anion, which expels a leaving group on the adjacent carbon. The reaction is particularly common in substrates that have a poor leaving group, such as –OH, two carbons removed from a carbonyl group, as in $HOC–CH–C═OHOC–CH–C═O$. The poor leaving group disfavors the alternative E1 and E2 possibilities, and the carbonyl group makes the adjacent hydrogen unusually acidic by resonance stabilization of the anion intermediate. We’ll look at this acidifying effect of a carbonyl group in Section 22.5. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.10%3A_The_E2_Reaction_and_Cyclohexane__Conformation.txt |
All three elimination reactions—E2, E1, and E1cB—occur in biological pathways, but the E1cB mechanism is particularly common. The substrate is usually an alcohol rather than an alkyl halide, and the H atom removed is usually adjacent to a carbonyl group, just as in laboratory reactions. Thus, 3-hydroxy carbonyl compounds are frequently converted to unsaturated carbonyl compounds by elimination reactions. A typical example occurs during the biosynthesis of fats and oils when a 3-hydroxybutyryl thioester is dehydrated to the corresponding unsaturated (crotonyl) thioester. The base in this reaction is a histidine amino acid in the enzyme, and the loss of the –OH group is assisted by simultaneous protonation.
11.13: A Summary of Reactivity - SN1 SN2 E1 E1cB and E2
SN1, SN2, E1, E1cB, E2—how can you keep it all straight and predict what will happen in any given case? Will substitution or elimination occur? Will the reaction be bimolecular or unimolecular? There are no rigid answers to these questions, but it’s possible to recognize some trends and make some generalizations.
• Primary alkyl halides SN2 substitution occurs if a good nucleophile is used, E2 elimination occurs if a strong, sterically hindered base is used, and E1cB elimination occurs if the leaving group is two carbons away from a carbonyl group.
• Secondary alkyl halides SN2 substitution occurs if a weakly basic nucleophile is used in a polar aprotic solvent, E2 elimination predominates if a strong base is used, and E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group. Secondary allylic and benzylic alkyl halides can also undergo SN1 and E1 reactions if a weakly basic nucleophile is used in a protic solvent.
• Tertiary alkyl halides E2 elimination occurs when a base is used, but SN1 substitution and E1 elimination occur together under neutral conditions, such as in pure ethanol or water. E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group.
Worked Example 11.5
Predicting the Product and Mechanism of Reactions
Tell whether each of the following reactions is likely to be SN1, SN2, E1, E1cB, or E2, and predict the product of each:
Strategy
Look carefully in each reaction at the structure of the substrate, the leaving group, the nucleophile, and the solvent. Then decide from the preceding summary which kind of reaction is likely to be favored.
Solution
(a) A secondary, nonallylic substrate can undergo an SN2 reaction with a good nucleophile in a polar aprotic solvent but will undergo an E2 reaction on treatment with a strong base in a protic solvent. In this case, E2 reaction is likely to predominate.
(b) A secondary benzylic substrate can undergo an SN2 reaction on treatment with a nonbasic nucleophile in a polar aprotic solvent and will undergo an E2 reaction on treatment with a base. Under protic conditions, such as aqueous formic acid (HCO2H), an SN1 reaction is likely, along with some E1 reaction.
Problem 11-20 Tell whether each of the following reactions is likely to be SN1, SN2, E1, E1cB, or E2: (a)
(b)
(c)
(d) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.12%3A_Biological_Elimination_Reactions.txt |
11 • Chemistry Matters 11 • Chemistry Matters
Organic chemistry in the 20th century changed the world, giving us new medicines, food preservatives, insecticides, adhesives, textiles, dyes, building materials, composites, and all manner of polymers. But these advances did not come without a cost: Almost every chemical process produces waste that must be dealt with, including reaction solvents and toxic by-products that might evaporate into the air or be leached into groundwater if not disposed of properly. Even apparently harmless by-products must be safely buried or otherwise sequestered. As always, there’s no such thing as a free lunch. With the good also comes the bad.
It may never be possible to make organic chemistry completely benign, but awareness of the environmental problems caused by many chemical processes has grown dramatically in recent years, giving rise to a movement called green chemistry. Green chemistry is the design and implementation of chemical products and processes that reduce waste and attempt to eliminate the generation of hazardous substances. There are 12 principles of green chemistry:
• Prevent waste – Waste should be prevented rather than treated or cleaned up after it has been created.
• Maximize atom economy – Synthetic methods should maximize the incorporation of all materials used in a process into the final product so that waste is minimized.
• Use less hazardous processes – Synthetic methods should use reactants and generate wastes with minimal toxicity to health and the environment.
• Design safer chemicals – Chemical products should be designed to have minimal toxicity.
• Use safer solvents – Minimal use should be made of solvents, separation agents, and other auxiliary substances in a reaction.
• Design for energy efficiency – Energy requirements for chemical processes should be minimized, with reactions carried out at room temperature if possible.
• Use renewable feedstocks – Raw materials should come from renewable sources when feasible.
• Minimize derivatives – Syntheses should be designed with minimal use of protecting groups to avoid extra steps and reduce waste.
• Use catalysis – Reactions should be catalytic rather than stoichiometric.
• Design for degradation – Products should be designed to be biodegradable at the end of their useful lifetimes.
• Monitor pollution in real time – Processes should be monitored in real time for the formation of hazardous substances.
• Prevent accidents – Chemical substances and processes should minimize the potential for fires, explosions, or other accidents.
The foregoing 12 principles may not all be met in most real-world applications, but they provide a worthy goal and they can make chemists think more carefully about the environmental implications of their work. Real success stories have occurred, and more are in progress. Approximately 7 million pounds per year of ibuprofen (6 billion tablets!) are now made by a “green” process that produces approximately 99% less waste than the process it replaces. Only three steps are needed, the anhydrous HF solvent used in the first step is recovered and reused, and the second and third steps are catalytic.
11.15: Key Terms
11 • Key Terms 11 • Key Terms
• anti periplanar
• benzylic
• bimolecular
• deuterium isotope effect
• E1 reaction
• E1cB reaction
• E2 reaction
• first-order reaction
• ion pair
• kinetics
• leaving group
• nucleophilic substitution reaction
• periplanar
• rate-determining step
• rate-limiting step
• second-order reaction
• SN1 reaction
• SN2 reaction
• solvation
• syn periplanar
• tosylate
• unimolecular
• Zaitsev’s rule
11.16: Summary
11 • Summary 11 • Summary
The reaction of an alkyl halide or tosylate with a nucleophile/base results either in substitution or in elimination. The resultant nucleophilic substitution and base-induced elimination reactions are two of the most widely occurring and versatile reaction types in organic chemistry, both in the laboratory and in biological pathways.
Nucleophilic substitutions are of two types: SN2 reactions and SN1 reactions. In the SN2 reaction, the entering nucleophile approaches the halide from a direction 180° away from the leaving group, resulting in an umbrella-like inversion of configuration at the carbon atom. The reaction is kinetically second-order and is strongly inhibited by increasing steric bulk of the reactants. Thus, SN2 reactions are favored for primary and secondary substrates.
In the SN1 reaction, the substrate spontaneously dissociates to a carbocation in a slow rate-limiting step, followed by a rapid reaction with the nucleophile. As a result, SN1 reactions are kinetically first-order and take place with substantial racemization of configuration at the carbon atom. They are most favored for tertiary substrates. Both SN1 and SN2 reactions occur in biological pathways, although the leaving group is typically a diphosphate ion rather than a halide.
Eliminations of alkyl halides to yield alkenes occur by three mechanisms: E2 reactions, E1 reactions, and E1cB reactions, which differ in the timing of C–H and C–X bond-breaking. In the E2 reaction, C–H and C–X bond-breaking occur simultaneously when a base abstracts H+ from one carbon while the leaving group departs from the neighboring carbon. The reaction takes place preferentially through an anti periplanar transition state in which the four reacting atoms—hydrogen, two carbons, and leaving group—are in the same plane. The reaction shows second-order kinetics and a deuterium isotope effect, and occurs when a secondary or tertiary substrate is treated with a strong base. These elimination reactions usually give a mixture of alkene products in which the more highly substituted alkene predominates (Zaitsev’s rule).
In the E1 reaction, C–X bond-breaking occurs first. The substrate dissociates to yield a carbocation in the slow rate-limiting step before losing H+ from an adjacent carbon in a second step. The reaction shows first-order kinetics and no deuterium isotope effect and occurs when a tertiary substrate reacts in polar, nonbasic solution.
In the E1cB reaction, C–H bond-breaking occurs first. A base abstracts a proton to give a carbanion, followed by loss of the leaving group from the adjacent carbon in a second step. The reaction is favored when the leaving group is two carbons removed from a carbonyl, which stabilizes the intermediate anion by resonance. Biological elimination reactions typically occur by this E1cB mechanism.
In general, substrates react in the following way:
11.17: Summary of Reactions
11 • Summary of Reactions 11 • Summary of Reactions
1. Nucleophilic substitutions
1. SN1 reaction of 3°, allylic, and benzylic halides (Section 11.4 and Section 11.5)
• SN2 reaction of 1° and simple 2° halides (Section 11.2 and Section 11.3)
• Eliminations
1. E1 reaction (Section 11.10)
• E1cB reaction (Section 11.10)
• E2 reaction (Section 11.8) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.14%3A_Chemistry_MattersGreen_Chemistry.txt |
11 • Additional Problems 11 • Additional Problems
Visualizing Chemistry
Problem 11-21
Write the product you would expect from reaction of each of the following alkyl halides with (1) Na+ SCH3 and (2) Na+ –OH (green = Cl): (a)
(b)
(c)
Problem 11-22 From what alkyl bromide was the following alkyl acetate made by SN2 reaction? Write the reaction, showing all stereochemistry. Problem 11-23 Assign R or S configuration to the following molecule, write the product you would expect from SN2 reaction with NaCN, and assign R or S configuration to the product (green = Cl): Problem 11-24 Draw the structure and assign Z or E stereochemistry to the product you expect from E2 reaction of the following molecule with NaOH (green = Cl): Mechanism Problems Problem 11-25 Predict the product(s) and show the mechanism for each of the following reactions. What do the mechanisms have in common? Why? (a) (b) (c) (d) Problem 11-26 Show the mechanism for each of the following reactions. What do the mechanisms have in common? Why? (a) (b) (c) Problem 11-27 Predict the product(s) for each of the following elimination reactions. In each case show the mechanism. What do the mechanisms have in common? Why? (a) (b) (c) Problem 11-28 Predict the product(s) for each of the following elimination reactions. In each case show the mechanism. What do the mechanisms have in common? Why? (a) (b) (c) Problem 11-29 Predict the product(s) for each of the following elimination reactions. In each case show the mechanism. What do the mechanisms have in common? Why? (a) (b) (c) Problem 11-30 Predict the product of each of the following reactions, and indicate if the mechanism is likely to be SN1, SN2, E1, E2, or E1cB. (a) (b) (c) (d) Problem 11-31 We saw in Section 8.7 that bromohydrins are converted into epoxides when treated with base. Propose a mechanism, using curved arrows to show the electron flow. Problem 11-32 The following tertiary alkyl bromide does not undergo a nucleophilic substitution reaction by either SN1 or SN2 mechanisms. Explain. Problem 11-33 Metabolism of S-adenosylhomocysteine (Section 11.6) involves the following sequence. Propose a mechanism for the second step. Problem 11-34 Reaction of iodoethane with CN– yields a small amount of isonitrile, CH3CH2N$\text{≡}$C, along with the nitrile CH3CH2C$\text{≡}$N as the major product. Write electron-dot structures for both products, assign formal charges as necessary, and propose mechanisms to account for their formation. Problem 11-35 One step in the urea cycle for ridding the body of ammonia is the conversion of argininosuccinate to the amino acid arginine plus fumarate. Propose a mechanism for the reaction, and show the structure of arginine. Problem 11-36 Methyl esters (RCO2CH3) undergo a cleavage reaction to yield carboxylate ions plus iodomethane on heating with LiI in dimethylformamide: The following evidence has been obtained: (1) The reaction occurs much faster in DMF than in ethanol. (2) The corresponding ethyl ester (RCO2CH2CH3) cleaves approximately 10 times more slowly than the methyl ester. Propose a mechanism for the reaction. What other kinds of experimental evidence could you gather to support your hypothesis? Problem 11-37 SN2 reactions take place with inversion of configuration, and SN1 reactions take place with racemization. The following substitution reaction, however, occurs with complete retention of configuration. Propose a mechanism. (Hint: two inversions = retention.) Problem 11-38 Propose a mechanism for the following reaction, an important step in the laboratory synthesis of proteins: Nucleophilic Substitution Reactions Problem 11-39 Draw all isomers of C4H9Br, name them, and arrange them in order of decreasing reactivity in the SN2 reaction. Problem 11-40 The following Walden cycle has been carried out: Explain the results, and indicate where inversion occurs. Problem 11-41 Which compound in each of the following pairs will react faster in an SN2 reaction with OH–? (a) CH3Br or CH3I (b)
CH3CH2I in ethanol or in dimethyl sulfoxide
(c) (CH3)3CCl or CH3Cl (d) H2C$\text{=}$CHBr or H2C$\text{=}$CHCH2Br
Problem 11-42 Which reactant in each of the following pairs is more nucleophilic? Explain. (a)
NH2 or NH3
(b) H2O or CH3CO2 (c) BF3 or F (d) (CH3)3P or (CH3)3N (e) I or Cl (f) C$\text{≡}$N or OCH3
Problem 11-43 What effect would you expect the following changes to have on the rate of the SN2 reaction of 1-iodo-2-methylbutane with cyanide ion? (a)
The CN concentration is halved, and the 1-iodo-2-methylbutane concentration is doubled.
(b) Both the CN and the 1-iodo-2-methylbutane concentrations are tripled.
Problem 11-44 What effect would you expect the following changes to have on the rate of the reaction of ethanol with 2-iodo-2-methylbutane? (a)
The concentration of the halide is tripled.
(b) The concentration of the ethanol is halved by adding diethyl ether as an inert solvent.
Problem 11-45 How might you prepare each of the following using a nucleophilic substitution reaction at some step? (a) (b) (c) (d) Problem 11-46 Which reaction in each of the following pairs would you expect to be faster? (a) The SN2 displacement by I– on CH3Cl or on CH3OTos (b)
The SN2 displacement by CH3CO2 on bromoethane or on bromocyclohexane
(c) The SN2 displacement on 2-bromopropane by CH3CH2O or by CN (d) The SN2 displacement by HC$\text{≡}$C on bromomethane in benzene or in acetonitrile
Problem 11-47 Predict the product and give the stereochemistry resulting from reaction of each of the following nucleophiles with (R)-2-bromooctane: (a)
CN
(b) CH3CO2 (c) CH3S
Problem 11-48
(R)-2-Bromooctane undergoes racemization to give (±)-2-bromooctane when treated with NaBr in dimethyl sulfoxide. Explain.
Elimination Reactions
Problem 11-49 Propose structures for compounds that fit the following descriptions: (a)
An alkyl halide that gives a mixture of three alkenes on E2 reaction
(b) An organohalide that will not undergo nucleophilic substitution (c) An alkyl halide that gives the non-Zaitsev product on E2 reaction (d) An alcohol that reacts rapidly with HCl at 0 °C
Problem 11-50 What products would you expect from the reaction of 1-bromopropane with each of the following? (a)
NaNH2
(b) KOC(CH3)3 (c) NaI (d) NaCN (e) NaC$\text{≡}$CH (f) Mg, then H2O
Problem 11-51
1-Chloro-1,2-diphenylethane can undergo E2 elimination to give either cis- or trans-1,2-diphenylethylene (stilbene). Draw Newman projections of the reactive conformations leading to both possible products, and suggest a reason why the trans alkene is the major product.
Problem 11-52
Predict the major alkene product of the following E1 reaction:
Problem 11-53
There are eight diastereomers of 1,2,3,4,5,6-hexachlorocyclohexane. Draw each in its more stable chair conformation. One isomer loses HCl in an E2 reaction nearly 1000 times more slowly than the others. Which isomer reacts so slowly, and why?
General Problems
Problem 11-54 The following reactions are unlikely to occur as written. Tell what is wrong with each, and predict the actual product. (a) (b) (c) Problem 11-55 Arrange the following carbocations in order of increasing stability. (a) (b) (c) Problem 11-56 Order each of the following sets of compounds with respect to SN1 reactivity: (a) (b) (c) Problem 11-57 Order each of the following sets of compounds with respect to SN2 reactivity: (a) (b) (c) Problem 11-58 Predict the major product(s) of each of the following reactions. Identify those reactions where you would expect the product mixture to rotate plane-polarized light. (a) (b) (c) Problem 11-59 Reaction of the following S tosylate with cyanide ion yields a nitrile product that also has S stereochemistry. Explain. Problem 11-60 Ethers can often be prepared by SN2 reaction of alkoxide ions, RO–, with alkyl halides. Suppose you wanted to prepare cyclohexyl methyl ether. Which of the following two possible routes would you choose? Explain. Problem 11-61 How can you explain the fact that trans-1-bromo-2-methylcyclohexane yields the non-Zaitsev elimination product 3-methylcyclohexene on treatment with base? Problem 11-62 Predict the product(s) of the following reaction, indicating stereochemistry where necessary: Problem 11-63 Alkynes can be made by dehydrohalogenation of vinylic halides in a reaction that is essentially an E2 process. In studying the stereochemistry of this elimination, it was found that (Z)-2-chloro-2-butenedioic acid reacts 50 times as fast as the corresponding E isomer. What conclusion can you draw about the stereochemistry of eliminations in vinylic halides? How does this result compare with eliminations of alkyl halides? Problem 11-64 Based on your answer to Problem 11-63, predict the product(s) and show the mechanism for each of the following reactions. (a) (b) (c) Problem 11-65 (S)-2-Butanol slowly racemizes on standing in dilute sulfuric acid. Explain. Problem 11-66 Reaction of HBr with (R)-3-methyl-3-hexanol leads to racemic 3-bromo-3-methylhexane. Explain. Problem 11-67 Treatment of 1-bromo-2-deuterio-2-phenylethane with strong base leads to a mixture of deuterated and nondeuterated phenylethylenes in an approximately 7 : 1 ratio. Explain. Problem 11-68 Propose a structure for an alkyl halide that gives only (E)-3-methyl-2-phenyl-2-pentene on E2 elimination. Make sure you indicate the stereochemistry. Problem 11-69 Although anti periplanar geometry is preferred for E2 reactions, it isn’t absolutely necessary. The following deuterated bromo compound reacts with strong base to yield an undeuterated alkene. A syn elimination has occurred. Make a molecular model of the reactant, and explain the result. Problem 11-70 The reaction of 1-chlorooctane with CH3CO2– to give octyl acetate is greatly accelerated by adding a small quantity of iodide ion. Explain. Problem 11-71 Compound X is optically inactive and has the formula C16H16Br2. On treatment with strong base, X gives hydrocarbon Y, C16H14. Compound Y absorbs 2 equivalents of hydrogen when reduced over a palladium catalyst and reacts with ozone to give two fragments. One fragment, Z, is an aldehyde with formula C7H6O. The other fragment is glyoxal, (CHO)2. Write the reactions involved, and suggest structures for X, Y, and Z. What is the stereochemistry of X? Problem 11-72 When a primary alcohol is treated with p-toluenesulfonyl chloride at room temperature in the presence of an organic base such as pyridine, a tosylate is formed. When the same reaction is carried out at higher temperature, an alkyl chloride is often formed. Explain. Problem 11-73 The amino acid methionine is formed by a methylation reaction of homocysteine with N-methyltetrahydrofolate. The stereochemistry of the reaction has been probed by carrying out the transformation using a donor with a “chiral methyl group,” which contains protium (H), deuterium (D), and tritium (T) isotopes of hydrogen. Does the methylation reaction occur with inversion or retention of configuration? Problem 11-74 Amines are converted into alkenes by a two-step process called the Hofmann elimination. SN2 reaction of the amine with an excess of CH3I in the first step yields an intermediate that undergoes E2 reaction when treated with silver oxide as base. Pentylamine, for example, yields 1-pentene. Propose a structure for the intermediate, and explain why it readily undergoes elimination. Problem 11-75 The antipsychotic drug flupentixol is prepared by the following scheme: (a) What alkyl chloride B reacts with amine A to form C? (b)
Compound C is treated with SOCl2, and the product is allowed to react with magnesium metal to give a Grignard reagent D. What is the structure of D?
(c) We’ll see in Section 19.7 that Grignard reagents add to ketones, such as E, to give tertiary alcohols, such as F. Because of the newly formed chirality center, compound F exists as a pair of enantiomers. Draw both, and assign R,S configurations. (d) Two stereoisomers of flupentixol are subsequently formed from F, but only one is shown. Draw the other isomer, and identify the type of stereoisomerism. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.18%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 12, you should be able to
1. fulfillall of the detailed objectives listed under each individual section.
2. solve road-map problems that include mass spectral data, infrared data, or both.
3. define, and use in context, the key terms introduced.
The processes of identifying and characterizing organic compounds are of great importance to the working organic chemist. With the use of modern instrumental techniques, these tasks can now be accomplished much more readily than in the past. In this chapter, you will learn about two spectroscopic techniques (mass spectroscopy and infrared spectroscopy) that are used to identify organic compounds.
12: Structure Determination - Mass Spectrometry and Infrared Spectroscopy
Finding the structures of new molecules, whether small ones synthesized in the laboratory or large proteins and nucleic acids found in living organisms, is central to progress in chemistry and biochemistry. We can only scratch the surface of structure determination in this book, but after reading this and the following two chapters, you should have a good idea of the range of structural techniques available and of how and when each is used.
Every time a reaction is run, the products must be identified, and every time a new compound is found in nature, its structure must be determined. Determining the structure of an organic compound was a difficult and time-consuming process until the mid-20th century, but powerful techniques and specialized instruments are now routinely used to simplify the problem. In this and the next two chapters, we’ll look at four such techniques—mass spectrometry (MS), infrared (IR) spectroscopy, ultraviolet spectroscopy (UV), and nuclear magnetic resonance spectroscopy (NMR)—and we’ll see the kind of information that can be obtained from each.
Mass spectrometry What is the size and formula?
Infrared spectroscopy What functional groups are present?
Ultraviolet spectroscopy Is a conjugated π electron system present?
Nuclear magnetic resonance spectroscopy What is the carbon–hydrogen framework?
12.02: Mass Spectrometry of Small Molecules - Magnetic-Sector Instruments
At its simplest, mass spectrometry (MS) is a technique for measuring the mass, and therefore the molecular weight (MW), of a molecule. In addition, it’s often possible to gain structural information about a molecule by measuring the masses of the fragments produced when molecules are broken apart.
More than 20 different kinds of commercial mass spectrometers are available depending on the intended application, but all have three basic parts: an ionization source in which sample molecules are given an electrical charge, a mass analyzer in which ions are separated by their mass-to-charge ratio, and a detector in which the separated ions are observed and counted.
Among the most common mass spectrometers used for routine purposes in the laboratory is the electron-impact, magnetic-sector instrument shown schematically in Figure 12.2. A small amount of sample is vaporized into the ionization source, where it is bombarded by a stream of high-energy electrons. The energy of the electron beam can be varied but is commonly around 70 electron volts (eV), or 6700 kJ/mol. When a high-energy electron strikes an organic molecule, it dislodges a valence electron from the molecule, producing a cation radicalcation because the molecule has lost an electron and now has a positive charge; radical because the molecule now has an odd number of electrons.
Electron bombardment transfers so much energy that most of the cation radicals fragment after formation. They break apart into smaller pieces, some of which retain the positive charge and some of which are neutral. The fragments then flow through a curved pipe in a strong magnetic field, which deflects them into different paths according to their mass-to-charge ratio (m/z). Neutral fragments are not deflected by the magnetic field and are lost on the walls of the pipe, but positively charged fragments are sorted by the mass spectrometer onto a detector, which records them as peaks at the various m/z ratios. Since the number of charges z on each ion is usually 1, the value of m/z for each ion is simply its mass m. Masses up to approximately 2500 atomic mass units (amu) can be analyzed by this type of instrument.
Another common type of mass spectrometer uses what is called a quadrupole mass analyzer, which has a set of four solid rods is arranged parallel to the direction of the ion beam, with an oscillating electrostatic field is generated in the space between the rods. For a given field, only one m/z value will make it through the quadrupole region. The others will crash into the rods or the walls of the instrument and never reach the detector Figure 12.3.
The mass spectrum of a compound is typically presented as a bar graph, with masses (m/z values) on the x axis and intensity, or relative abundance of ions of a given m/z striking the detector, on the y axis. The tallest peak, assigned an intensity of 100%, is called the base peak, and the peak that corresponds to the unfragmented cation radical is called the parent peak, or the molecular ion (M+, or simply M). Figure 12.4 shows the mass spectrum of propane.
Mass spectral fragmentation patterns are usually complex, and the molecular ion is often not the base peak. The mass spectrum of propane in Figure 12.4, for instance, shows a molecular ion at m/z = 44 that is only about 30% as high as the base peak at m/z = 29. In addition, many other fragment ions are present. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.01%3A_Why_This_Chapter.txt |
What kinds of information can we get from a mass spectrum? The most obvious information is the molecular weight of the sample, which in itself can be invaluable. If we were given samples of hexane (MW = 86), 1-hexene (MW = 84), and 1-hexyne (MW = 82), for example, mass spectrometry would easily distinguish them.
Some instruments, called double-focusing mass spectrometers, have two magnetic sectors in their mass analyzers, giving these spectrometers have such high resolution that they provide mass measurements accurate to 5 ppm, or about 0.0005 amu, making it possible to distinguish between two formulas with the same nominal mass. For example, both C5H12 and C4H8O have MW = 72, but they differ slightly beyond the decimal point: C5H12 has an exact mass of 72.0939 amu, whereas C4H8O has an exact mass of 72.0575 amu. A high-resolution instrument can easily distinguish between them. Note, however, that exact mass measurements refer to molecules with specific isotopic compositions. Thus, the sum of the exact atomic masses of the specific isotopes in a molecule is measured—1.007 83 amu for 1H, 12.000 00 amu for 12C, 14.003 07 amu for 14N, 15.994 91 amu for 16O, and so on—rather than the sum of the average atomic masses of elements, as found on a periodic table.
Unfortunately, not every compound shows a molecular ion in its electron-impact mass spectrum. Although M+ is usually easy to identify if it’s abundant, some compounds, such as 2,2-dimethylpropane, fragment so easily that no molecular ion is observed (Figure 12.5). In such cases, alternative “soft” ionization methods that don’t use electron bombardment can prevent or minimize fragmentation (see Section 12.4).
Knowing the molecular weight makes it possible to narrow considerably the choices of molecular formula. For example, if the mass spectrum of an unknown compound shows a molecular ion at m/z = 110, the molecular formula is likely to be C8H14, C7H10O, C6H6O2, or C6H10N2. There are always a number of molecular formulas possible for all but the lowest molecular weights, and a computer can easily generate a list of the choices.
A further point about mass spectrometry, noticeable in the spectra of both propane (Figure 12.4) and 2,2-dimethylpropane (Figure 12.5), is that the peak for the molecular ion is not at the highest m/z value. There is also a small peak at M + 1 due to the presence of different isotopes in the molecules. Although 12C is the most abundant carbon isotope, a small amount (1.10% natural abundance) of 13C is also present. Thus, a certain percentage of the molecules analyzed in the mass spectrometer are likely to contain a 13C atom, giving rise to the observed M + 1 peak. In addition, a small amount of 2H (deuterium; 0.015% natural abundance) is present, making a further contribution to the M + 1 peak.
Mass spectrometry would be useful even if molecular weight and formula were the only information that could be obtained, but in fact it provides much more. For one thing, the mass spectrum of a compound serves as a kind of “molecular fingerprint.” Every organic compound fragments in a unique way depending on its structure, and the likelihood of two compounds having identical mass spectra is small. Thus, it’s sometimes possible to identify an unknown by computer-based matching of its mass spectrum to one of the more than 785,061 searchable spectra recorded in a database called the Registry of Mass Spectral Data.
It’s also possible to derive structural information about a molecule by interpreting its fragmentation pattern. Fragmentation occurs when the high-energy cation radical flies apart by spontaneous cleavage of a chemical bond. One of the two fragments retains the positive charge and is a carbocation, while the other fragment is a neutral radical.
Not surprisingly, the positive charge often remains with the fragment that is best able to stabilize it. In other words, a relatively stable carbocation is often formed during fragmentation. For example, 2,2-dimethylpropane tends to fragment in such a way that the positive charge remains with the tert-butyl group. 2,2-Dimethylpropane therefore has a base peak at m/z = 57, corresponding to C4H9+ (Figure 12.5).
Because mass-spectral fragmentation patterns are usually complex, it’s often difficult to assign structures to fragment ions. Most hydrocarbons fragment in many ways, as demonstrated by the mass spectrum of hexane in Figure 12.6. The hexane spectrum shows a moderately abundant molecular ion at m/z = 86 and fragment ions at m/z = 71, 57, 43, and 29. Since all the carbon–carbon bonds of hexane are electronically similar, all break to a similar extent, giving rise to the observed mixture of ions.
Figure 12.7 shows how the hexane fragments might arise. The loss of a methyl radical (CH3, M = 15) from the hexane cation radical (M+ = 86) gives rise to a fragment of mass 86 – 15 = 71; the loss of an ethyl radical (C2H5, M = 29) accounts for a fragment of mass 86 – 29 = 57; the loss of a propyl radical (C3H7, M = 43) accounts for a fragment of mass 86 – 43 = 43; and the loss of a butyl radical accounts for a fragment of mass 29. With practice, it’s sometimes possible to analyze the fragmentation pattern of an unknown compound and work backward to a structure that is compatible with the data.
We’ll see in the next section and in later chapters that specific functional groups, such as alcohols, ketones, aldehydes, and amines, show specific kinds of mass spectral fragmentations that can be interpreted to provide structural information.
Worked Example 12.1: Using Mass Spectra to Identify Compounds
Assume that you have two unlabeled samples, one of methylcyclohexane and the other of ethylcyclopentane. How could you use mass spectrometry to tell them apart? The mass spectra of both are shown in Figure 12.8.
Strategy
Look at the possible structures and decide on how they differ. Then think about how any of these differences in structure might give rise to differences in mass spectra. Methyl cyclohexane, for instance, has a –CH3 group, and ethylcyclopentane has a –CH2CH3 group, which should affect the fragmentation patterns.
Solution
Both mass spectra show molecular ions at M+ = 98, corresponding to C7H14, but they differ in their fragmentation patterns. Sample A has its base peak at m/z = 69, corresponding to the loss of a CH2CH3 group (29 mass units), but B has a rather small peak at m/z = 69. Sample B shows a base peak at m/z = 83, corresponding to the loss of a CH3 group (15 mass units), but sample A has only a small peak at m/z = 83. We can therefore be reasonably certain that A is ethylcyclopentane and B is methylcyclohexane.
Problem 12-1
The sex hormone testosterone contains only C, H, and O and has a mass of 288.2089 amu, as determined by high-resolution mass spectrometry. What is the likely molecular formula of testosterone?
Problem 12-2
Two mass spectra are shown in Figure 12.9. One spectrum is that of 2-methyl-2-pentene; the other is of 2-hexene. Which is which? Explain.
(a)
(b)
Figure 12.9 Mass spectra for Problem 12-2. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.03%3A_Interpreting_Mass_Spectra.txt |
As each functional group is discussed in future chapters, mass-spectral fragmentations characteristic of that group will be described. As a preview, though, we’ll point out some distinguishing features of several common functional groups.
Alcohols
Alcohols undergo fragmentation in a mass spectrometer by two pathways: alpha cleavage and dehydration. In the α-cleavage pathway, a C–C bond nearest the hydroxyl group is broken, yielding a neutral radical plus a resonance-stabilized, oxygen-containing cation. This type of fragmentation is seen in the spectrum of 2-pentanol in Figure 12.10.
In the dehydration pathway, water is eliminated, yielding an alkene radical cation with a mass 18 amu less than M+. For simplicity, we have drawn the dehydration below as an E2-type process. Often the hydrogen that is lost is not beta to the hydroxyl. Only a small peak from dehydration is observed in the spectrum of 2-pentanol (Figure 12.10).
Amines
The nitrogen rule of mass spectrometry says that a compound with an odd number of nitrogen atoms has an odd-numbered molecular weight. The logic behind the rule comes from the fact that nitrogen is trivalent, thus requiring an odd number of hydrogen atoms. The presence of nitrogen in a molecule is often detected simply by observing its mass spectrum. An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms.
Aliphatic amines undergo a characteristic α cleavage in a mass spectrometer, similar to that observed for alcohols. A C–C bond nearest the nitrogen atom is broken, yielding an alkyl radical and a resonance-stabilized, nitrogen-containing cation.
The mass spectrum of triethylamine has a base peak at m/z = 86, which arises from an alpha cleavage resulting in the loss of a methyl group (Figure 12.11).
Halides
The fact that some elements have two common isotopes gives their mass spectra a distinctive appearance. Chlorine, for example, exists as two isotopes, 35Cl and 37Cl, in roughly a 3 : 1 ratio. In a sample of chloroethane, three out of four molecules contain a 35Cl atom and one out of four has a 37Cl atom. In the mass spectrum of chloroethane (Figure 12.12 we see the molecular ion (M) at m/z = 64 for ions that contain a 35Cl and another peak at m/z = 66, called the M + 2 peak, for ions containing a 37Cl. The ratio of the relative abundance of M : M + 2 is about 3 : 1, a reflection of the isotopic abundances of chlorine.
In the case of bromine, the isotopic distribution is 50.7% 79Br and 49.3% 81Br. In the mass spectrum of 1-bromohexane (Figure 12.13) the molecular ion appears at m/z = 164 for 79Br-containing ions and the M + 2 peak is at m/z = 166 for 81Br-containing ions. The ions at m/z = 135 and 137 are informative as well. The two nearly equally large peaks tell us that the ions at those m/z values still contain the bromine atom. The peak at m/z = 85, on the other hand, does not contain bromine because there is not a large peak at m/z = 87.
Carbonyl Compounds
Ketones and aldehydes that have a hydrogen on a carbon three atoms away from the carbonyl group undergo a characteristic mass-spectral cleavage called the McLafferty rearrangement. The hydrogen atom is transferred to the carbonyl oxygen, a C–C 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, ketones and aldehydes frequently undergo α cleavage of the bond between the carbonyl carbon and the neighboring carbon to yield a neutral radical and a resonance-stabilized acyl cation. Because the carbon neighboring the carbonyl carbon is called the alpha carbon, the reaction is called an alpha cleavage.
(To be more general about neighboring positions in carbonyl compounds, Greek letters are used in alphabetical order: alpha, beta, gamma, delta, and so on.)
The mass spectrum of butyrophenone illustrates both alpha cleavage and the McLafferty rearrangement (Figure 12.14). Alpha cleavage of the propyl substituent results in the loss of C3H7 = 43 mass units from the parent ion at m/z = 148 to give the fragment ion at m/z = 105. A McLafferty rearrangement of butyrophenone results in the loss of ethylene, C2H4 = 28 mass units, from the parent leaving the ion at m/z = 120.
Worked Example 12.2: Identifying Fragmentation Patterns in a Mass Spectrum
The mass spectrum of 2-methyl-3-pentanol is shown in Figure 12.15. What fragments can you identify?
Strategy
Calculate the mass of the molecular ion, and identify the functional groups in the molecule. Then write the fragmentation processes you might expect, and compare the masses of the resultant fragments with the peaks present in the spectrum.
Solution
2-Methyl-3-pentanol, an open-chain alcohol, has M+ = 102 and might be expected to fragment by α cleavage and by dehydration. These processes would lead to fragment ions of m/z = 84, 73, and 59. Of the three expected fragments, dehydration is not observed (no m/z = 84 peak), but both α cleavages take place (m/z = 73, 59).
Problem 12-3
What are the masses of the charged fragments produced in the following cleavage pathways? (a)
1. Alpha cleavage of 2-pentanone (CH3COCH2CH2CH3)
2. Dehydration of cyclohexanol (hydroxycyclohexane)
3. McLafferty rearrangement of 4-methyl-2-pentanone [CH3COCH2CH(CH3)2]
4. Alpha cleavage of triethylamine [(CH3CH2)3N]
Problem 12-4
List the masses of the parent ion and of several fragments you might expect to find in the mass spectrum of the following molecule: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.04%3A_Mass_Spectrometry_of_Some_Common_Functional_Groups.txt |
MS analyses of sensitive biological samples rarely use magnetic sector ionization. Instead, they typically use either electrospray ionization (ESI) or matrix-assisted laser desorption ionization (MALDI), typically linked to a time-of-flight (TOF) mass analyzer. Both ESI and MALDI are soft ionization methods that produce charged molecules with little fragmentation, even with sensitive biological samples of very high molecular weight.
In an ESI source, as a sample solution exits the tube, it is subjected to a high voltage that causes the droplets to become charged. The sample molecules gain one or more protons from charged solvent molecules in the droplet. The volatile solvent quickly evaporates, giving variably protonated sample molecules (M + Hnn+). In a MALDI source, the sample is adsorbed onto a suitable matrix compound, such as 2,5-dihydroxybenzoic acid, which is ionized by a short burst of laser light. The matrix compound then transfers the energy to the sample and protonates it, forming M + Hnn+ ions.
Following ion formation, the variably protonated sample molecules are electrically focused into a small packet with a narrow spatial distribution, and the packet is given a sudden kick of energy by an accelerator electrode. As each molecule in the packet is given the same energy, E = mv2/2, it begins moving with a velocity that depends on the square root of its mass, $v=2E/mv=2E/m$. Lighter molecules move faster, and heavier molecules move slower. The analyzer itself—the drift tube—is simply an electrically grounded metal tube inside which the different charged molecules become separated as they move at different velocities and take different amounts of time to complete their flight.
The Time of Flight technique is considerably more sensitive than the magnetic sector alternative, and protein samples of up to 100 kilodaltons (100,000 amu) can be separated with a mass accuracy of 3 ppm. Figure 12.16 shows a MALDI–TOF spectrum of chicken egg-white lysozyme, MW = 14,306.7578 daltons. Biochemists generally use the unit dalton, abbreviated Da, instead of amu, although the two are equivalent (1 dalton = 1 amu).
12.06: Spectroscopy and the Electromagnetic Spectrum
Infrared, ultraviolet, and nuclear magnetic resonance spectroscopies differ from mass spectrometry in that they are nondestructive and involve the interaction of molecules with electromagnetic energy rather than with an ionizing source. Before beginning a study of these techniques, however, let’s briefly review the nature of radiant energy and the electromagnetic spectrum.
Visible light, X rays, microwaves, radio waves, and so forth are all different kinds of electromagnetic radiation. Collectively, they make up the electromagnetic spectrum, shown in Figure 12.17. The electromagnetic spectrum is arbitrarily divided into regions, with the familiar visible region accounting for only a small portion, from 3.8 × 10–7 m to 7.8 × 10–7 m in wavelength. The visible region is flanked by the infrared and ultraviolet regions.
Electromagnetic radiation is often said to have dual behavior. In some respects, it has the properties of a particle, called a photon, yet in other respects it behaves as an energy wave. Like all waves, electromagnetic radiation is characterized by a wavelength, a frequency, and an amplitude (Figure 12.18). The wavelength, λ (Greek lambda), is the distance from one wave maximum to the next. The frequency, ν (Greek nu), is the number of waves that pass by a fixed point per unit time, usually given in reciprocal seconds (s–1), or hertz, Hz (1 Hz = 1 s–1). The amplitude is the height of a wave, measured from midpoint to peak. The intensity of radiant energy, whether a feeble glow or a blinding glare, is proportional to the square of the wave’s amplitude.
Multiplying the wavelength of a wave in meters (m) by its frequency in reciprocal seconds (s–1) gives the speed of the wave in meters per second (m/s). The rate of travel of all electromagnetic radiation in a vacuum is a constant value, commonly called the “speed of light” and abbreviated c. Its numerical value is defined as exactly 2.997 924 58 × 108 m/s, usually rounded off to 3.00 × 108 m/s.
$Wavelength×Frequency=SpeedWavelength×Frequency=Speed$
$λ(m)×ν(s−1)=c(m/s)λ(m)×ν(s−1)=c(m/s)$
$λ=cvorv=cλλ=cvorv=cλ$
Just as matter comes only in discrete units called atoms, electromagnetic energy is transmitted only in discrete amounts called quanta. The amount of energy ε corresponding to 1 quantum of energy (1 photon) of a given frequency ν is expressed by the Planck equation
$ε=hv=hcλε=hv=hcλ$
where h = Planck’s constant (6.62 × 10–34 J ‧ s = 1.58 × 10–34 cal ‧ s).
The Planck equation says that the energy of a given photon varies directly with its frequency ν but inversely with its wavelength λ. High frequencies and short wavelengths correspond to high-energy radiation such as gamma rays; low frequencies and long wavelengths correspond to low-energy radiation such as radio waves. Multiplying ε by Avogadro’s number NA gives the same equation in more familiar units, where E represents the energy of Avogadro’s number (one “mole”) of photons of wavelength λ:
$E=NAhcλ=1.20×10−4kJ/molλ(m)or2.86×10−5kcal/molλ(m)E=NAhcλ=1.20×10−4kJ/molλ(m)or2.86×10−5kcal/molλ(m)$
When an organic compound is exposed to a beam of electromagnetic radiation, it absorbs energy of some wavelengths but passes, or transmits, energy of other wavelengths. If we irradiate the sample with energy of many different wavelengths and determine which are absorbed and which are transmitted, we can measure the absorption spectrum of the compound.
An example of an absorption spectrum—that of ethanol exposed to infrared radiation—is shown in Figure 12.19. The horizontal axis records the wavelength, and the vertical axis records the intensity of the various energy absorptions in percent transmittance. The baseline corresponding to 0% absorption (or 100% transmittance) runs along the top of the chart, so a downward spike means that energy absorption has occurred at that wavelength.
The energy a molecule gains when it absorbs radiation must be distributed over the molecule in some way. With infrared radiation, the absorbed energy causes bonds to stretch and bend more vigorously. With ultraviolet radiation, the energy causes an electron to jump from a lower-energy orbital to a higher-energy one. Different radiation frequencies affect molecules in different ways, but each provides structural information when the results are interpreted.
There are many kinds of spectroscopies, which differ according to the region of the electromagnetic spectrum used. We’ll look at three: infrared spectroscopy, ultraviolet spectroscopy, and nuclear magnetic resonance spectroscopy. Let’s begin by seeing what happens when an organic sample absorbs infrared energy.
Worked Example 12.3
Correlating Energy and Frequency of Radiation
Which is higher in energy, FM radio waves with a frequency of 1.015 × 108 Hz (101.5 MHz) or visible green light with a frequency of 5 × 1014 Hz?
Strategy
Remember the equations ε = and ε = hc/λ, which say that energy increases as frequency increases and as wavelength decreases.
Solution
Since visible light has a higher frequency than radio waves, it is higher in energy.
Problem 12-5
Which has higher energy, infrared radiation with λ = 1.0 × 10–6 m or an X ray with λ = 3.0 × 10–9 m? Radiation with ν = 4.0 × 109 Hz or with λ = 9.0 × 10–6 m?
Problem 12-6 It’s useful to develop a feeling for the amounts of energy that correspond to different parts of the electromagnetic spectrum. Calculate the energies in kJ/mol of each of the following kinds of radiation: (a)
A gamma ray with λ = 5.0 × 10–11 m
(b) An X ray with λ = 3.0 × 10–9 m (c) Ultraviolet light with ν = 6.0 × 1015 Hz (d) Visible light with ν = 7.0 × 1014 Hz (e) Infrared radiation with λ = 2.0 × 10–5 m (f) Microwave radiation with ν = 1.0 × 1011 Hz | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.05%3A_Mass_Spectrometry_in_Biological_-_Time-of-flight_%28TOF%29_Instruments.txt |
In infrared (IR) spectroscopy, the IR region of the electromagnetic spectrum covers the range from just above the visible (7.8 × 10–7 m) to approximately 10–4 m, but only the midportion from 2.5 × 10–6 m to 2.5 × 10–5 m is used by organic chemists (Figure 12.20. Wavelengths within the IR region are usually given in micrometers (1 μm = 10–6 m), and frequencies are given in wavenumbers rather than in hertz. The wavenumber $v˜v˜$ is the reciprocal of wavelength in centimeters and is therefore expressed in units of cm–1.
$Wavenumber:v˜(cm−1)=1λ(cm)Wavenumber:v˜(cm−1)=1λ(cm)$
Thus, the useful IR region is from 4000 to 400 cm–1, corresponding to energies of 48.0 kJ/mol to 4.80 kJ/mol (11.5–1.15 kcal/mol).
Why does an organic molecule absorb some wavelengths of IR radiation but not others? All molecules have a certain amount of energy and are in constant motion. Their bonds stretch and contract, atoms wag back and forth, and other molecular vibrations occur. Some of the kinds of allowed vibrations are shown below:
The amount of energy a molecule contains is not continuously variable but is quantized. That is, a molecule can stretch or bend only at specific frequencies corresponding to specific energy levels. Take bond stretching, for example. Although we usually speak of bond lengths as if they were fixed, the numbers given are really averages. In fact, a typical C–H bond with an average bond length of 110 pm is actually vibrating at a specific frequency, alternately stretching and contracting as if there were a spring connecting the two atoms.
When a molecule is irradiated with electromagnetic radiation, energy is absorbed if the frequency of the radiation matches the frequency of the vibration. The result of this energy absorption is an increased amplitude for the vibration; in other words, the “spring” connecting the two atoms stretches and compresses a bit further. Since each frequency absorbed by a molecule corresponds to a specific molecular motion, we can find what kinds of motions a molecule has by measuring its IR spectrum. By interpreting these motions, we can find out what kinds of bonds (functional groups) are present in the molecule.
$IR spectrum→What molecular motions?→What functional groups?IR spectrum→What molecular motions?→What functional groups?$
12.08: Interpreting Infrared Spectra
The complete interpretation of an IR spectrum is difficult because most organic molecules have dozens of different bond stretching and bending motions, and thus have dozens of absorptions. On the one hand, this complexity is a problem because it generally limits the laboratory use of IR spectroscopy to pure samples of fairly small molecules—little can be learned from IR spectroscopy about large, complex biomolecules. On the other hand, this complexity is useful because an IR spectrum acts as a unique fingerprint of a compound. In fact, the complex region of the IR spectrum, from 1500 cm–1 to around 400 cm–1, is called the fingerprint region. If two samples have identical IR spectra, they are almost certainly identical compounds.
Fortunately, we don’t need to interpret an IR spectrum fully to get useful structural information. Most functional groups have characteristic IR absorption bands that don’t change much from one compound to another. The C=O absorption of a ketone is almost always in the range 1680 to 1750 cm–1; the O–H absorption of an alcohol is almost always in the range 3400 to 3650 cm–1; the C=C absorption of an alkene is almost always in the range 1640 to 1680 cm–1; and so forth. By learning where characteristic functional-group absorptions occur, it’s possible to get structural information from IR spectra. Table 12.1 lists the characteristic IR bands of some common functional groups.
Table 12.1 Characteristic IR Absorptions of Some Functional Groups
Functional Group Absorption (cm–1) Intensity
Alkane C–H 2850–2960 Medium
Alkene =C–H 3020–3100 Medium
C=C 1640–1680 Medium
Alkyne $≡C–H≡C–H$ 3300 Strong
$C≡CC≡C$ 2100–2260 Medium
Alkyl halide C–Cl 600–800 Strong
C–Br 500–600 Strong
Alcohol O–H 3400–3650 Strong, broad
C–O 1050–1150 Strong
Arene C–H 3030 Weak
Aromatic ring 1660–2000 Weak
1450–1600 Medium
Amine N–H 3300–3500 Medium
C–N 1030–1230 Medium
Carbonyl compound $C═OC═O$ 1670–1780 Strong
Aldehyde 1730 Strong
Ketone 1715 Strong
Ester 1735 Strong
Amide 1690 Strong
Carboxylic acid 1710 Strong
Carboxylic acid O–H 2500–3100 Strong, broad
Nitrile $C≡NC≡N$ 2210–2260 Medium
Nitro NO2 1540 Strong
Look at the IR spectra of hexane, 1-hexene, and 1-hexyne in Figure 12.21 to see an example of how IR spectroscopy can be used. Although all three IR spectra contain many peaks, there are characteristic absorptions of $C═CC═C$ and $C≡CC≡C$ functional groups that allow the three compounds to be distinguished. Thus, 1-hexene shows a characteristic $C═CC═C$ absorption at 1660 cm–1 and a vinylic =C–H absorption at 3100 cm–1, whereas 1-hexyne has a $C≡CC≡C$ absorption at 2100 cm–1 and a terminal alkyne $≡C–H≡C–H$ absorption at 3300 cm–1.
It helps in remembering the position of specific IR absorptions to divide the IR region from 4000 cm–1 to 400 cm–1 into four parts, as shown in Figure 12.22.
• The region from 4000 to 2500 cm–1 corresponds to absorptions caused by N–H, C–H, and O–H single-bond stretching motions. N–H and O–H bonds absorb in the 3300 to 3600 cm–1 range; C–H bond stretching occurs near 3000 cm–1.
• The region from 2500 to 2000 cm–1 is where triple-bond stretching occurs. Both $C≡NC≡N$ and $C≡CC≡C$ bonds absorb here.
• The region from 2000 to 1500 cm–1 is where double bonds ($C═OC═O$, $C═NC═N$, and $C═CC═C$) absorb. Carbonyl groups generally absorb in the range 1680 to 1750 cm–1, and alkene stretching normally occurs in the narrow range of 1640 to 1680 cm–1.
• The region below 1500 cm–1 is the fingerprint portion of the IR spectrum. A large number of absorptions due to a variety of C–C, C–O, C–N, and C–X single-bond vibrations occur here.
Why do different functional groups absorb where they do? As noted previously, a good analogy is that of two weights (atoms) connected by a spring (a bond). Short, strong bonds vibrate at a higher energy and higher frequency than do long, weak bonds, just as a short, strong spring vibrates faster than a long, weak spring. Thus, triple bonds absorb at a higher frequency than double bonds, which in turn absorb at a higher frequency than single bonds. In addition, C–H, O–H, and N–H bonds vibrate at a higher frequency than bonds between heavier C, O, and N atoms.
Worked Example 12.4
Distinguishing Isomeric Compounds by IR Spectroscopy
Acetone (CH3COCH3) and 2-propen-1-ol ($H2C═CHCH2OHH2C═CHCH2OH$) are isomers. How could you distinguish them by IR spectroscopy?
Strategy
Identify the functional groups in each molecule, and refer to Table 12.1.
Solution
Acetone has a strong C=O absorption at 1715 cm–1, while 2-propen-1-ol has an –OH absorption at 3500 cm–1 and a C=C absorption at 1660 cm–1.
Problem 12-7 What functional groups might the following molecules contain? (a)
A compound with a strong absorption at 1710 cm–1
(b) A compound with a strong absorption at 1540 cm–1 (c) A compound with strong absorptions at 1720 cm–1 and 2500 to 3100 cm–1
Problem 12-8 How might you use IR spectroscopy to distinguish between the following pairs of isomers? (a)
CH3CH2OH and CH3OCH3
(b) Cyclohexane and 1-hexene (c) CH3CH2CO2H and HOCH2CH2CHO | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.07%3A_Infrared_Spectroscopy.txt |
As each functional group is discussed in future chapters, the spectroscopic properties of that group will be described. For the present, we’ll point out some distinguishing features of the hydrocarbon functional groups already studied and briefly preview some other common functional groups. We should also point out, however, that in addition to interpreting absorptions that are present in an IR spectrum, it’s also possible to get structural information by noticing which absorptions are not present. If the spectrum of a compound has no absorptions at 3300 and 2150 cm–1, the compound is not a terminal alkyne; if the spectrum has no absorption near 3400 cm–1, the compound is not an alcohol; and so on.
Alkanes
The IR spectrum of an alkane is fairly uninformative because no functional groups are present and all absorptions are due to C–H and C–C bonds. Alkane C–H bonds show a strong absorption from 2850 to 2960 cm–1, and saturated C–C bonds show a number of bands in the 800 to 1300 cm–1 range. Since most organic compounds contain saturated alkane-like portions, most organic compounds have these characteristic IR absorptions. The C–H and C–C bands are clearly visible in the three spectra shown previously in Figure 12.21.
Alkenes
Alkenes show several characteristic stretching absorptions. Vinylic =C–H bonds absorb from 3020 to 3100 cm–1, and alkene $C═CFigure 12.21b.$
Alkenes have characteristic =C–H out-of-plane bending absorptions in the 700 to 1000 cm–1 range, thereby allowing the substitution pattern on a double bond to be determined (Figure 12.23). For example, monosubstituted alkenes such as 1-hexene show strong characteristic bands at 910 and 990 cm–1, and 1,1-disubstituted alkenes ($R2C═CH2R2C═CH2$) have an intense band at 890 cm–1.
Alkynes
Alkynes show a $C≡CFigure 12.21c). This band is diagnostic for terminal alkynes because it is fairly intense and quite sharp.$
Aromatic Compounds
Aromatic compounds, such as benzene, have a weak C–H stretching absorption at 3030 cm–1, just to the left of a typical saturated C–H band. In addition, they have a series of weak absorptions in the 1660 to 2000 cm–1 range and a series of medium-intensity absorptions in the 1450 to 1600 cm–1 region. These latter absorptions are due to complex molecular motions of the entire ring. The C–H out-of-plane bending region for benzene derivatives, between 650 to 1000 cm–1, gives valuable information about the ring’s substitution pattern, as it does for the substitution pattern of alkenes (Figure 12.24).
The IR spectrum of phenylacetylene, shown in Figure 12.29 at the end of this section, gives an example, clearly showing the following absorbances: $≡C–H≡C–H$ stretch at 3300 cm–1, C–H stretches from the benzene ring at 3000 to 3100 cm–1, $C═CC═C$ stretches of the benzene ring between 1450 and 1600 cm–1, and out-of-plane bending of the ring’s C–H groups, indicating monosubstitution at 750 cm–1.
Alcohols
The O–H functional group of alcohols is easy to spot. Alcohols have a characteristic band in the range 3400 to 3650 cm–1 that is usually broad and intense. Hydrogen bonding between O–H groups is responsible for making the absorbance so broad. If an O–H stretch is present, it’s hard to miss this band or to confuse it with anything else.
Cyclohexanol (Figure 12.25) is a good example.
Amines
The N–H functional group of amines is also easy to spot in the IR, with a characteristic absorption in the 3300 to 3500 cm–1 range. Although alcohols absorb in the same range, an N–H absorption band is much sharper and less intense than an O–H band.
Primary amines (R–NH2) have two absorbances—one for the symmetric stretching mode and one for the asymmetric mode (Figure 12.26). Secondary amines (R2N–H) only have one N–H stretching absorbance in this region.
Carbonyl Compounds
Carbonyl functional groups are the easiest to identify of all IR absorptions because of their sharp, intense peak in the range 1670 to 1780 cm–1. Most important, the exact position of absorption within this range can often be used to identify the exact kind of carbonyl functional group—aldehyde, ketone, ester, and so forth.
ALDEHYDES
Saturated aldehydes absorb at 1730 cm–1; aldehydes next to either a double bond or an aromatic ring absorb at 1705 cm–1.
The C–H group attached to the carbonyl is responsible for the characteristic IR absorbance for aldehydes at 2750 and 2850 cm–1 (Figure 12.27). Although these are not very intense, the absorbance at 2750 cm–1 is helpful when trying to distinguish between an aldehyde and a ketone.
KETONES
Saturated open-chain ketones and six-membered cyclic ketones absorb at 1715 cm–1. Ring strain stiffens the $C═OC═O$ bond, making five-membered cyclic ketones absorb at 1750 cm–1 and four-membered cyclic ketones absorb at 1780 cm–1, about 20 to 30 cm–1 lower than the corresponding saturated ketone.
ESTERS
Saturated esters have a $C═OC═O$ absorbance at 1735 cm–1 and two strong absorbances in the 1300 to 1000 cm–1 range from the C–O portion of the functional group. Like other carbonyl functional groups, esters next to either an aromatic ring or a double bond absorb at 1715 cm–1, about 20 to 30 cm–1 lower than a saturated ester.
Worked Example 12.5
Predicting IR Absorptions of Compounds
Where might the following compounds have IR absorptions?
Strategy
Identify the functional groups in each molecule, and then check Table 12.1 to see where those groups absorb.
Solution
(a) Absorptions: 3400 to 3650 cm–1 (O–H), 3020 to 3100 cm–1 (=C–H), 1640 to 1680 cm–1 ($C═CC═C$). This molecule has an alcohol O–H group and an alkene double bond.
(b) Absorptions: 3300 cm–1 ($≡C–H≡C–H$), 2100 to 2260 cm–1 ($C≡CC≡C$), 1735 cm–1 ($C═OC═O$). This molecule has a terminal alkyne triple bond and a saturated ester carbonyl group.
Worked Example 12.6
Identifying Functional Groups from an IR Spectrum
The IR spectrum of an unknown compound is shown in Figure 12.28. What functional groups does the compound contain?
Strategy
All IR spectra have many absorptions, but those useful for identifying specific functional groups are usually found in the region from 1500 cm–1 to 3300 cm–1. Pay particular attention to the carbonyl region (1670 to 1780 cm–1), the aromatic region (1660 to 2000 cm–1), the triple-bond region (2000 to 2500 cm–1), and the C–H region (2500 to 3500 cm–1).
Solution
The spectrum shows an intense absorption at 1725 cm–1 due to a carbonyl group (perhaps an aldehyde, –CHO), a series of weak absorptions from 1800 to 2000 cm–1 characteristic of aromatic compounds, and a C–H absorption near 3030 cm–1, also characteristic of aromatic compounds. In fact, the compound is phenylacetaldehyde.
Problem 12-9
The IR spectrum of phenylacetylene is shown in Figure 12.29. What absorption bands can you identify?
Figure 12.29 The IR spectrum of phenylacetylene, Problem 12-9.
Problem 12-10 Where might the following compounds have IR absorptions? (a) (b) (c) Problem 12-11
Where might the following compound have IR absorptions? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.09%3A_Infrared_Spectra_of_Some_Common_Functional_Groups.txt |
12 • Chemistry Matters 12 • Chemistry Matters
The various spectroscopic techniques described in this and the next two chapters are enormously important in chemistry and have been fine-tuned to such a degree that the structure of almost any molecule can be found. Nevertheless, wouldn’t it be nice if you could simply look at a molecule and “see” its structure with your eyes?
Determining the three-dimensional shape of an object around you is easy—you just look at it, let your eyes focus the light rays reflected from the object, and let your brain assemble the data into a recognizable image. If the object is small, you use a microscope and let the microscope lens focus the visible light. Unfortunately, there is a limit to what you can see, even with the best optical microscope. Called the diffraction limit, you can’t see anything smaller than the wavelength of light you are using for the observation. Visible light has wavelengths of several hundred nanometers, but atoms in molecules have dimensions on the order of 0.1 nm. Thus, to see a molecule—whether a small one in the laboratory or a large, complex enzyme with a molecular weight in the tens of thousands—you need wavelengths in the 0.1 nm range, which corresponds to X rays.
Let’s say that we want to determine the structure and shape of an enzyme or other biological molecule. The technique used is called X-ray crystallography. First, the molecule is crystallized (which often turns out to be the most difficult and time-consuming part of the entire process) and a small crystal of 0.4 to 0.5 mm on its longest axis is glued to the end of a glass fiber. The fiber and attached crystal are then mounted in an instrument called an X-ray diffractometer, which consists of a radiation source, a sample positioning and orienting device that can rotate the crystal in any direction, a detector, and a controlling computer.
Once mounted in the diffractometer, the crystal is irradiated with X rays, usually so-called CuKα radiation with a wavelength of 0.154 nm. When the X rays strike the enzyme crystal, they interact with electrons in the molecule and are scattered into a diffraction pattern which, when detected and visualized, appears as a series of intense spots against a null background.
Manipulation of the diffraction pattern to extract three-dimensional molecular data is a complex process, but the final result is an electron-density map of the molecule. Because electrons are largely localized around atoms, any two centers of electron density located within bonding distance of each other are assumed to represent bonded atoms, leading to a recognizable chemical structure. So important is this structural information for biochemistry that an online database of approximately 145,000 biological substances has been created. Operated by Rutgers University and funded by the U.S. National Science Foundation, the Protein Data Bank (PDB) is a worldwide repository for processing and distributing three-dimensional structural data for biological macromolecules. We’ll see how to access the PDB in the Chapter 26 Chemistry Matters.
12.11: Key Terms
12 • Key Terms 12 • Key Terms
• absorption spectrum
• amplitude
• base peak
• cation radical
• electromagnetic spectrum
• frequency, ν
• hertz, Hz,
• infrared (IR) spectroscopy
• MALDI
• mass spectrometry (MS)
• McLafferty Rearrangement
• parent peak
• photon
• quadrupole mass analyzer
• Time of Flight (TOF)
• wavelength, λ
• wavenumber, $v˜v˜$
12.12: Summary
12 • Summary 12 • Summary
Finding the structure of a new molecule, whether a small one synthesized in the laboratory or a large protein found in living organisms, is central to the progression of chemistry and biochemistry. The structure of an organic molecule is usually determined using spectroscopic methods, including mass spectrometry and infrared spectroscopy. Mass spectrometry (MS) tells the molecular weight and formula of a molecule; infrared (IR) spectroscopy identifies the functional groups present in the molecule.
In small-molecule mass spectrometry, molecules are first ionized by collision with a high-energy electron beam. The ions then fragment into smaller pieces, which are magnetically sorted according to their mass-to-charge ratio (m/z). The ionized sample molecule is called the molecular ion, M+, and measurement of its mass gives the molecular weight of the sample. Structural clues about unknown samples can be obtained by interpreting the fragmentation pattern of the molecular ion. Mass-spectral fragmentations are usually complex, however, and interpretation is often difficult. In biological mass spectrometry, molecules are protonated using either electrospray ionization (ESI) or matrix-assisted laser desorption ionization (MALDI), and the protonated molecules are separated by time-of-flight (TOF) mass analysis.
Infrared spectroscopy involves the interaction of a molecule with electromagnetic radiation. When an organic molecule is irradiated with infrared energy, certain frequencies are absorbed by the molecule. The frequencies absorbed correspond to the amounts of energy needed to increase the amplitude of specific molecular vibrations such as bond stretching and bending. Since every functional group has a characteristic combination of bonds, every functional group has a characteristic set of infrared absorptions. For example, the terminal alkyne $≡C–H≡C–H$ bond absorbs IR radiation of 3300 cm–1, and the alkene $C═CC═C$ bond absorbs in the range 1640 to 1680 cm–1. By observing which frequencies of infrared radiation are absorbed by a molecule and which are not, it’s possible to determine the functional groups a molecule contains. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.10%3A_Chemistry_MattersX-Ray_Crystallography.txt |
12 • Additional Problems 12 • Additional Problems
Visualizing Chemistry
Problem 12-12
Where in the IR spectrum would you expect each of the following molecules to absorb? (a)
(b)
(c)
Problem 12-13 Show the structures of the fragments you would expect in the mass spectra of the following molecules: (a) (b) Mass Spectrometry Problem 12-14 Propose structures for compounds that fit the following mass-spectral data: (a) A hydrocarbon with M+ = 132 (b)
A hydrocarbon with M+ = 166
(c) A hydrocarbon with M+ = 84
Problem 12-15 Write molecular formulas for compounds that show the following molecular ions in their high-resolution mass spectra, assuming that C, H, N, and O might be present. The exact atomic masses are: 1.007 83 (1H), 12.000 00 (12C), 14.003 07 (14N), 15.994 91 (16O). (a)
M+ = 98.0844
(b) M+ = 123.0320
Problem 12-16
Camphor, a saturated monoketone from the Asian camphor tree, is used among other things as a moth repellent and as a constituent of embalming fluid. If camphor has M+ = 152.1201 by high-resolution mass spectrometry, what is its molecular formula? How many rings does camphor have?
Problem 12-17
The nitrogen rule of mass spectrometry says that a compound containing an odd number of nitrogens has an odd-numbered molecular ion. Conversely, a compound containing an even number of nitrogens has an even-numbered M+ peak. Explain.
Problem 12-18
In light of the nitrogen rule mentioned in Problem 12-17, what is the molecular formula of pyridine, M+ = 79?
Problem 12-19
Nicotine is a diamino compound isolated from dried tobacco leaves. Nicotine has two rings and M+ = 162.1157 by high-resolution mass spectrometry. Give a molecular formula for nicotine, and calculate the number of double bonds.
Problem 12-20
The hormone cortisone contains C, H, and O, and shows a molecular ion at M+ = 360.1937 by high-resolution mass spectrometry. What is the molecular formula of cortisone? (The degree of unsaturation for cortisone is 8.)
Problem 12-21 Halogenated compounds are particularly easy to identify by their mass spectra because both chlorine and bromine occur naturally as mixtures of two abundant isotopes. Recall that chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%); and bromine occurs as 79Br (50.7%) and 81Br (49.3%). At what masses do the molecular ions occur for the following formulas? What are the relative percentages of each molecular ion? (a)
Bromomethane, CH3Br
(b) 1-Chlorohexane, C6H13Cl
Problem 12-22
By knowing the natural abundances of minor isotopes, it’s possible to calculate the relative heights of M+ and M + 1 peaks. If 13C has a natural abundance of 1.10%, what are the relative heights of the M+ and M + 1 peaks in the mass spectrum of benzene, C6H6?
Problem 12-23 Propose structures for compounds that fit the following data: (a)
A ketone with M+ = 86 and fragments at m/z = 71 and m/z = 43
(b) An alcohol with M+ = 88 and fragments at m/z = 73, m/z = 70, and m/z = 59
Problem 12-24
2-Methylpentane (C6H14) has the mass spectrum shown. Which peak represents M+? Which is the base peak? Propose structures for fragment ions of m/z = 71, 57, 43, and 29. Why does the base peak have the mass it does?
Problem 12-25
Assume that you are in a laboratory carrying out the catalytic hydrogenation of cyclohexene to cyclohexane. How could you use a mass spectrometer to determine when the reaction is finished?
Problem 12-26 What fragments might you expect in the mass spectra of the following compounds? (a) (b) (c) Infrared Spectroscopy Problem 12-27 How might you use IR spectroscopy to distinguish among the three isomers 1-butyne, 1,3-butadiene, and 2-butyne? Problem 12-28 Would you expect two enantiomers such as (R)-2-bromobutane and (S)-2-bromobutane to have identical or different IR spectra? Explain. Problem 12-29 Would you expect two diastereomers such as meso-2,3-dibromobutane and (2R,3R)-dibromobutane to have identical or different IR spectra? Explain. Problem 12-30 Propose structures for compounds that meet the following descriptions: (a) C5H8, with IR absorptions at 3300 and 2150 cm–1 (b)
C4H8O, with a strong IR absorption at 3400 cm–1
(c) C4H8O, with a strong IR absorption at 1715 cm–1 (d) C8H10, with IR absorptions at 1600 and 1500 cm–1
Problem 12-31 How could you use infrared spectroscopy to distinguish between the following pairs of isomers? (a)
HC$\text{≡}$CCH2NH2 and CH3CH2C$\text{≡}$N
(b) CH3COCH3 and CH3CH2CHO
Problem 12-32 Two infrared spectra are shown. One is the spectrum of cyclohexane, and the other is the spectrum of cyclohexene. Identify them, and explain your answer. (a) (b) Problem 12-33 At what approximate positions might the following compounds show IR absorptions? (a) (b) (c) (d) (e) Problem 12-34 How would you use infrared spectroscopy to distinguish between the following pairs of constitutional isomers? (a) (b) (c) Problem 12-35 At what approximate positions might the following compounds show IR absorptions? (a) (b) (c) (d) (e) (f) Problem 12-36 Assume that you are carrying out the dehydration of 1-methylcyclohexanol to yield 1-methylcyclohexene. How could you use infrared spectroscopy to determine when the reaction is complete? Problem 12-37 Assume that you are carrying out the base-induced dehydrobromination of 3-bromo-3-methylpentane (Section 11.7) to yield an alkene. How could you use IR spectroscopy to tell which of three possible elimination products is formed, if one includes E/Z isomers? General Problems Problem 12-38 Which is stronger, the C$\text{=}$O bond in an ester (1735 cm–1) or the C$\text{=}$O bond in a saturated ketone (1715 cm–1)? Explain. Problem 12-39 Carvone is an unsaturated ketone responsible for the odor of spearmint. If carvone has M+ = 150 in its mass spectrum and contains three double bonds and one ring, what is its molecular formula? Problem 12-40 Carvone (Problem 12-39) has an intense infrared absorption at 1690 cm–1. What kind of ketone does carvone contain? Problem 12-41 The mass spectrum (a) and the infrared spectrum (b) of an unknown hydrocarbon are shown. Propose as many structures as you can. (a) (b) Problem 12-42 The mass spectrum (a) and the infrared spectrum (b) of another unknown hydrocarbon are shown. Propose as many structures as you can. (a) (b) Problem 12-43 Propose structures for compounds that meet the following descriptions: (a) An optically active compound C5H10O with an IR absorption at 1730 cm–1 (b)
A non-optically active compound C5H9N with an IR absorption at 2215 cm–1
Problem 12-44
4-Methyl-2-pentanone and 3-methylpentanal are isomers. Explain how you could tell them apart, both by mass spectrometry and by infrared spectroscopy.
Problem 12-45
Grignard reagents (alkylmagnesium halides) undergo a general and very useful reaction with ketones. Methylmagnesium bromide, for example, reacts with cyclohexanone to yield a product with the formula C7H14O. What is the structure of this product if it has an IR absorption at 3400 cm–1?
Problem 12-46
Ketones undergo a reduction when treated with sodium borohydride, NaBH4. What is the structure of the compound produced by reaction of 2-butanone with NaBH4 if it has an IR absorption at 3400 cm–1 and M+ = 74 in the mass spectrum?
Problem 12-47
Nitriles, R–C$\text{≡}$N, undergo a hydrolysis reaction when heated with aqueous acid. What is the structure of the compound produced by hydrolysis of propanenitrile, CH3CH2C$\text{≡}$N, if it has IR absorptions from 2500–3100 cm–1 and at 1710 cm–1, and has M+ = 74?
Problem 12-48
The infrared spectrum of the compound with the following mass spectrum lacks any significant absorption above 3000 cm–1. There is a prominent peak near 1740 cm–1 and another strong peak near 1200 cm–1. Propose a structure.
Problem 12-49
The infrared spectrum of the compound with the following mass spectrum has a medium-intensity peak at about 1650 cm–1. There is also a C–H out-of-plane bending peak near 880 cm–1. Propose a structure.
Problem 12-50
The infrared spectrum of the compound with the following mass spectrum has strong absorbances at 1584, 1478, and 1446 cm–1. Propose a structure. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.13%3A_Additional_Problems.txt |
Learning Objectives
• fulfillall of the detailed objectives listed under each individual section.
• solve road-map problems which may require the interpretation of 1H NMR spectra in addition to other spectral data.
• define, and use in context, the key terms introduced.
In Chapter 12, you learned how an organic chemist could use two spectroscopic techniques, mass spectroscopy and infrared spectroscopy, to assist in determining the structure of an unknown compound. This chapter introduces a third technique, nuclear magnetic resonance (NMR). The two most common forms of NMR spectroscopy, 1H NMR and 13C NMR, will be discussed, the former in much more detail than the latter. Nuclear magnetic resonance spectroscopy is a very powerful tool, particularly when used in combination with other spectroscopic techniques.
13: Structure Determination - Nuclear Magnetic Resonance Spectroscopy
Figure 13.1 NMR spectroscopy is an invaluable aid in carrying out the design and synthesis of new drugs. (credit: modification of work by Unknown/Pxhere, CC0 1.0)
Nuclear magnetic resonance (NMR) spectroscopy has far-reaching applications in many scientific fields, particularly in chemical structure determination. Although we’ll just give an overview of the subject in this chapter, focusing on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding.
Nuclear magnetic resonance (NMR) spectroscopy is the most valuable spectroscopic technique available to organic chemists. It’s the method of structure determination that organic chemists usually turn to first.
We saw in the chapter on Structure Determination: Mass Spectrometry and Infrared Spectroscopy that mass spectrometry gives a molecule’s formula and infrared spectroscopy identifies a molecule’s functional groups. Nuclear magnetic resonance spectroscopy complements these other techniques by mapping a molecule’s carbon–hydrogen framework. Taken together, MS, IR, and NMR make it possible to determine the structures of even very complex molecules.
Mass spectrometry Molecular size and formula
Infrared spectroscopy Functional groups present
NMR spectroscopy Map of carbon–hydrogen framework
13.02: Nuclear Magnetic Resonance Spectroscopy
Many kinds of nuclei behave as if they were spinning about an axis, somewhat as the earth spins daily. Because they’re positively charged, these spinning nuclei act like tiny magnets and can interact with an external magnetic field, denoted B0. Not all nuclei act this way, but fortunately for organic chemists, both the proton (1H) and the 13C nucleus do have spins. The more common 12C isotope, however, does not have nuclear spin. (In speaking about NMR, the words proton and hydrogen are often used interchangeably, since a hydrogen nucleus is just a proton.) Let’s see what the consequences of nuclear spin are and how we can use the results.
In the absence of an external magnetic field, the spins of magnetic nuclei are oriented randomly. When a sample containing these nuclei is placed between the poles of a strong magnet, however, the nuclei adopt specific orientations, much as a compass needle orients in the earth’s magnetic field. A spinning 1H or 13C nucleus can orient so that its own tiny magnetic field is aligned either with (parallel to) or against (antiparallel to) the external field. The two orientations don’t have the same energy, however, and aren’t equally likely. The parallel orientation is slightly lower in energy by an amount that depends on the strength of the external field, making this spin state very slightly favored over the antiparallel orientation (Figure 13.2).
If the oriented nuclei are irradiated with electromagnetic radiation of the proper frequency, energy absorption occurs and the lower-energy spin state “flips” to the higher-energy state. When this spin-flip occurs, the magnetic nuclei are said to be in resonance with the applied radiation—hence the name nuclear magnetic resonance.
The exact frequency necessary for resonance depends both on the strength of the external magnetic field, the identity of the nucleus, and the electronic environment of the nucleus. If a very strong magnetic field is applied, the energy difference between the two spin states is larger and higher-frequency (higher-energy) radiation is required for a spin-flip. If a weaker magnetic field is applied, less energy is required to effect the transition between nuclear spin states (Figure 13.3).
In practice, superconducting magnets that produce enormously powerful fields up to 23.5 tesla (T) are sometimes used, but field strengths in the range of 4.7 to 7.0 T are more common. At a magnetic field strength of 4.7 T, so-called radiofrequency (rf) energy in the 200 MHz range (1 MHz = 106 Hz) brings a 1H nucleus into resonance, and rf energy of 50 MHz brings a 13C nucleus into resonance. At the highest field strength currently available in commercial instruments (23.5 T), 1000 MHz energy is required for 1H spectroscopy. These energies needed for NMR are much smaller than those required for IR spectroscopy; 200 MHz rf energy corresponds to only 8.0 × 10–5 kJ/mol versus the 4.8 to 48 kJ/mol needed for IR spectroscopy.
1H and 13C nuclei are not unique in their ability to exhibit the NMR phenomenon. All nuclei with an odd number of protons (1H, 2H, 14N, 19F, 31P, for example) and all nuclei with an odd number of neutrons (13C, for example) show magnetic properties. Only nuclei with even numbers of both protons and neutrons (12C, 16O, 32S) do not give rise to magnetic phenomena (Table 13.1).
Table 13.1 The NMR Behavior of Some Common Nuclei
Magnetic nuclei Nonmagnetic nuclei
1H 12C
13C 16O
2H 32S
14N
19F
31P
Problem 13-1
The amount of energy required to spin-flip a nucleus depends both on the strength of the external magnetic field and on the nucleus. At a field strength of 4.7 T, rf energy of 200 MHz is required to bring a 1H nucleus into resonance, but energy of only 187 MHz will bring a 19F nucleus into resonance. Calculate the amount of energy required to spin-flip a 19F nucleus. Is this amount greater or less than that required to spin-flip a 1H nucleus?
Problem 13-2
Calculate the amount of energy required to spin-flip a proton in a spectrometer operating at 300 MHz. Does increasing the spectrometer frequency from 200 to 300 MHz increase or decrease the amount of energy necessary for resonance? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.01%3A_Why_This_Chapter.txt |
From the description thus far, you might expect all 1H nuclei in a molecule to absorb energy at the same frequency and all 13C nuclei to absorb at the same frequency. If so, we would observe only a single NMR absorption signal in the 1H or 13C spectrum of a molecule, a situation that would be of little use. In fact, the absorption frequency is not the same for all 1H or all 13C nuclei.
All nuclei in molecules are surrounded by electrons. When an external magnetic field is applied to a molecule, the electrons moving around nuclei set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field so that the effective field actually felt by the nucleus is a bit weaker than the applied field.
$Beffective=Bapplied=BlocalBeffective=Bapplied=Blocal$
In describing this effect of local fields, we say that nuclei experience shielding from the full effect of the applied field by the surrounding electrons. Because each chemically distinct nucleus in a molecule is in a slightly different electronic environment, each nucleus is shielded to a slightly different extent and the effective magnetic field felt by each is slightly different. These tiny differences in the effective magnetic fields experienced by different nuclei can be detected, and we thus see a distinct NMR signal for each chemically distinct 13C or 1H nucleus in a molecule. As a result, an NMR spectrum effectively maps the carbon–hydrogen framework of an organic molecule. With practice, it’s possible to read this map and derive structural information.
Figure 13.4 shows both the 1H and the 13C NMR spectra of methyl acetate, CH3CO2CH3. The horizontal axis shows the effective field strength felt by the nuclei, and the vertical axis indicates the intensity of absorption of rf energy. Each peak in the NMR spectrum corresponds to a chemically distinct 1H or 13C nucleus in the molecule. Note that NMR spectra are formatted with the zero absorption line at the bottom, whereas IR spectra are formatted with the zero absorption line at the top; Section 12.5. Note also that 1H and 13C spectra can’t be observed simultaneously on the same spectrometer because different amounts of energy are required to spin-flip the different kinds of nuclei. The two spectra must be recorded separately.
The 13C NMR spectrum of methyl acetate in Figure 13.4b shows three peaks, one for each of the three chemically distinct carbon atoms in the molecule. The 1H NMR spectrum in Figure 13.4a shows only two peaks, however, even though methyl acetate has six hydrogens. One peak is due to the CH3C═O hydrogens, and the other to the −OCH3 hydrogens. Because the three hydrogens in each methyl group have the same electronic environment, they are shielded to the same extent and are said to be equivalent. Chemically equivalent nuclei always show the same absorption. The two methyl groups themselves, however, are not equivalent, so the two sets of hydrogens absorb at different positions.
The operation of a basic NMR spectrometer is illustrated in Figure 13.5. An organic sample is dissolved in a suitable solvent (usually deuteriochloroform, CDCl3, which has no hydrogens) and placed in a thin glass tube between the poles of a magnet. The strong magnetic field causes the 1H and 13C nuclei in the molecule to align in one of the two possible orientations, and the sample is irradiated with rf energy. If the frequency of the rf irradiation is held constant and the strength of the applied magnetic field is varied, each nucleus comes into resonance at a slightly different field strength. A sensitive detector monitors the absorption of rf energy, and its electronic signal is then amplified and displayed as a peak.
NMR spectroscopy differs from IR spectroscopy (Section 12.6–Section 12.8) in that the timescales of the two techniques are quite different. The absorption of infrared energy by a molecule giving rise to a change in vibrational amplitude is an essentially instantaneous process (about 10–13 s), but the NMR process is much slower (about 10–3 s). This difference in timescales between IR and NMR spectroscopy is analogous to the difference between cameras operating at very fast and very slow shutter speeds. The fast camera (IR) takes an instantaneous picture and freezes the action. If two rapidly interconverting species are present, IR spectroscopy records the spectrum of both. The slow camera (NMR), however, takes a blurred, time-averaged picture. If two species interconverting faster than 103 times per second are present in a sample, NMR records only a single, averaged spectrum, rather than separate spectra of the two discrete species.
Because of this blurring effect, NMR spectroscopy can be used to measure the rates and activation energies of very fast chemical processes. In cyclohexane, for example, a ring-flip (Section 4.6) occurs so rapidly at room temperature that axial and equatorial hydrogens can’t be distinguished by NMR; only a single, averaged 1H NMR absorption is seen for cyclohexane at 25 °C. At –90 °C, however, the ring-flip is slowed down enough that two absorption peaks are visible, one for the six axial hydrogens and one for the six equatorial hydrogens. Knowing the temperature and the rate at which signal blurring begins to occur, it’s possible to calculate that the activation energy for the cyclohexane ring-flip is 45 kJ/mol (10.8 kcal/mol).
Problem 13-3
2-Chloropropene shows signals for three kinds of protons in its 1H NMR spectrum. Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.03%3A_The_Nature_of_NMR_Absorptions.txt |
NMR spectra are displayed on charts that show the applied field strength increasing from left to right (Figure 13.6). Thus, the left part of the chart is the low-field, or downfield, side, and the right part is the high-field, or upfield, side. Nuclei that absorb on the downfield side of the chart require a lower field strength for resonance, implying that they have less shielding. Nuclei that absorb on the upfield side require a higher field strength for resonance, implying that they have more shielding.
To define the position of an absorption, the NMR chart is calibrated and a reference point is used. In practice, a small amount of tetramethylsilane [TMS; (CH3)4Si] is added to the sample so that a reference absorption peak is produced when the spectrum is run. TMS is used as reference for both 1H and 13C measurements because in both cases it produces a single peak that occurs upfield of other absorptions normally found in organic compounds. The 1H and 13C spectra of methyl acetate in Figure 13.4 have the TMS reference peak indicated.
The position on the chart at which a nucleus absorbs is called its chemical shift. The chemical shift of TMS is set as the zero point, and other absorptions normally occur downfield, to the left on the chart. NMR charts are calibrated using an arbitrary scale called the delta (δ) scale, where 1 δ equals 1 part-per-million (1 ppm) of the spectrometer operating frequency. For example, if we were measuring the 1H NMR spectrum of a sample using an instrument operating at 200 MHz, 1 δ would be 1 part per million of 200,000,000 Hz, or 200 Hz. If we were measuring the spectrum using a 500 MHz instrument, 1 δ = 500 Hz. The following equation can be used for any absorption:
$δ=Observed chemical shift (number of Hz away from TMS)Spectrometer frequency in MHzδ=Observed chemical shift (number of Hz away from TMS)Spectrometer frequency in MHz$
Although this method of calibrating NMR charts may seem complex, there’s a good reason for it. As we saw earlier, the rf frequency required to bring a given nucleus into resonance depends on the spectrometer’s magnetic field strength. But because there are many different kinds of spectrometers with many different magnetic field strengths available, chemical shifts given in frequency units (Hz) vary from one instrument to another. Thus, a resonance that occurs at 120 Hz downfield from TMS on one spectrometer might occur at 600 Hz downfield from TMS on another spectrometer with a more powerful magnet.
By using a system of measurement in which NMR absorptions are expressed in relative terms (parts per million relative to spectrometer frequency) rather than absolute terms (Hz), it’s possible to compare spectra obtained on different instruments. The chemical shift of an NMR absorption in δ units is constant, regardless of the operating frequency of the spectrometer. A 1H nucleus that absorbs at 2.0 δ on a 200 MHz instrument also absorbs at 2.0 δ on a 500 MHz instrument.
The range in which most NMR absorptions occur is quite narrow. Almost all 1H NMR absorptions occur from 0 to 10 δ downfield from the proton absorption of TMS, and almost all 13C absorptions occur from 1 to 220 δ downfield from the carbon absorption of TMS. Thus, there is a likelihood that accidental overlap of nonequivalent signals will occur. The advantage of using an instrument with higher field strength (say, 500 MHz) rather than lower field strength (200 MHz) is that different NMR absorptions are more widely separated at the higher field strength. The chances that two signals will accidentally overlap are therefore lessened, and interpretation of spectra becomes easier. For example, two signals that are only 20 Hz apart at 200 MHz (0.1 ppm) are 50 Hz apart at 500 MHz (still 0.1 ppm).
Problem 13-4 The following 1H NMR peaks were recorded on a spectrometer operating at 200 MHz. Convert each into δ units. (a)
CHCl3; 1454 Hz
(b) CH3Cl; 610 Hz (c) CH3OH; 693 Hz (d) CH2Cl2; 1060 Hz
Problem 13-5 When the 1H NMR spectrum of acetone, CH3COCH3, is recorded on an instrument operating at 200 MHz, a single sharp resonance at 2.1 δ is seen. (a)
How many hertz downfield from TMS does the acetone resonance correspond to?
(b) If the 1H NMR spectrum of acetone is recorded at 500 MHz, what would the position of the absorption be in δ units? (c) How many hertz downfield from TMS does this 500 MHz resonance correspond to? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.04%3A_Chemical_Shifts.txt |
As mentioned previously, differences in chemical shifts are caused by the small local magnetic field of electrons surrounding different nuclei. Nuclei that are more strongly shielded by electrons require a higher applied field to bring them into resonance so they absorb on the right side of the NMR chart. Nuclei that are less strongly shielded need a lower applied field for resonance so they absorb on the left of the NMR chart.
Most 1H chemical shifts fall within the range 0 to 10 δ, which can be divided into the five regions shown in Table 13.2. By remembering the positions of these regions, it’s often possible to tell at a glance what kinds of protons a molecule contains.
Table 13.2 Regions of the 1H NMR Spectrum
Table 13.3 shows the correlation of 1H chemical shift with electronic environment in more detail. In general, protons bonded to saturated, sp3-hybridized carbons absorb at higher fields, whereas protons bonded to sp2-hybridized carbons absorb at lower fields. Protons on carbons that are bonded to electronegative atoms, such as N, O, or halogen, also absorb at lower fields.
Table 13.3 Correlation of 1H Chemical Shifts with Environment
Type of hydrogen Chemical shift (δ)
Reference Si(CH3)4 0
Alkyl (primary) —CH3 0.7–1.3
Alkyl (secondary) —CH2 1.2–1.6
Alkyl (tertiary) 1.4–1.8
Allylic 1.6–2.2
Methyl ketone 2.0–2.4
Aromatic methyl Ar—CH3 2.4–2.7
Alkynyl $—C≡C—H—C≡C—H$ 2.5–3.0
Alkyl halide 2.5–4.0
Alcohol 2.5–5.0
Alcohol, ether 3.3–4.5
Vinylic 4.5–6.5
Aryl Ar—H 6.5–8.0
Aldehyde 9.7–10.0
Carboxylic acid 11.0–12.0
Worked Example 13.1
Predicting Chemical Shifts in 1H NMR Spectra
Methyl 2,2-dimethylpropanoate (CH3)2CCO2CH3 has two peaks in its 1H NMR spectrum. What are their approximate chemical shifts?
Strategy
Identify the types of hydrogens in the molecule, and note whether each is alkyl, vinylic, or next to an electronegative atom. Then predict where each absorbs, using Table 13.3 if necessary.
Solution
The –OCH3 protons absorb around 3.5 to 4.0 δ because they are on carbon bonded to oxygen. The (CH3)3C– protons absorb near 1.0 δ because they are typical alkane-like protons.
Problem 13-6 Each of the following compounds has a single 1H NMR peak. Approximately where would you expect each compound to absorb? (a)
(b)
(c)
(d)
(e)
(f)
Problem 13-7
Identify the different types of protons in the following molecule, and tell where you would expect each to absorb:
13.06: Integration of H NMR Absorptions- Proton Counting
Look at the 1H NMR spectrum of methyl 2,2-dimethylpropanoate in Figure 13.7. There are two peaks, corresponding to the two kinds of protons, but the peaks aren’t the same size. The peak at 1.2 δ, due to the (CH3)3C– protons, is larger than the peak at 3.7 δ, due to the –OCH3 protons.
The area under each peak is proportional to the number of protons causing that peak. By electronically measuring, or integrating, the area under each peak, it’s possible to measure the relative numbers of the different kinds of protons in a molecule.
Modern NMR instruments provide a digital readout of relative peak areas, but an older, more visual method displays the integrated peak areas as a stair-step line, with the height of each step proportional to the area under the peak, and therefore proportional to the relative number of protons causing the peak. For example, the two steps for the peaks in methyl 2,2-dimethylpropanoate have a 1 : 3 (or 3 : 9) height ratio when integrated—exactly what we expect because the three −OCH3 protons are equivalent and the nine (CH3)3C– protons are equivalent.
Problem 13-8
How many peaks would you expect in the 1H NMR spectrum of 1,4-dimethylbenzene (para-xylene, or p-xylene)? What ratio of peak areas would you expect on integration of the spectrum? Refer to Table 13.3 for approximate chemical shifts, and sketch what the spectrum would look like. (Remember from Section 2.4 that aromatic rings have two resonance forms.) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.05%3A_Chemical_Shifts_in_H_NMR__Spectroscopy.txt |
In the 1H NMR spectra we’ve seen thus far, each different kind of proton in a molecule has given rise to a single peak. It often happens, though, that the absorption of a proton splits into multiple peaks, called a multiplet. For example, in the 1H NMR spectrum of bromoethane shown in Figure 13.8, the –CH2Br protons appear as four peaks (a quartet) centered at 3.42 δ and the –CH3 protons appear as three peaks (a triplet) centered at 1.68 δ.
Called spin–spin splitting, multiple absorptions of a nucleus are caused by the interaction, or coupling, of the spins of nearby nuclei. In other words, the tiny magnetic field produced by one nucleus affects the magnetic field felt by a neighboring nucleus. Look at the –CH3 protons in bromoethane, for example. The three equivalent –CH3 protons are neighbored by two other magnetic nuclei—the two protons on the adjacent –CH2Br group. Each of the neighboring –CH2Br protons has its own nuclear spin, which can align either with or against the applied field, producing a tiny effect that is felt by the –CH3 protons.
There are three ways in which the spins of the two –CH2Br protons can align, as shown in Figure 13.9. If both proton spins align with the applied field, the total effective field felt by the neighboring –CH3 protons is slightly larger than it would be otherwise. Consequently, the applied field necessary to cause resonance is slightly reduced. Alternatively, if one of the –CH2Br proton spins aligns with the field and one aligns against the field, there is no effect on the neighboring –CH3 protons. (This arrangement can occur in two ways, depending on which of the two proton spins aligns which way.) Finally, if both –CH2Br proton spins align against the applied field, the effective field felt by the –CH3 protons is slightly smaller than it would be otherwise, and the applied field needed for resonance is slightly increased.
Any given molecule has only one of the three possible alignments of –CH2Br spins, but in a large collection of molecules, all three spin states are represented in a 1 : 2 : 1 statistical ratio. We therefore find that the neighboring –CH3 protons come into resonance at three slightly different values of the applied field, and we see a 1 : 2 : 1 triplet in the NMR spectrum. One resonance is a little above where it would be without coupling, one is at the same place it would be without coupling, and the third resonance is a little below where it would be without coupling.
In the same way that the –CH3 absorption of bromoethane is split into a triplet, the –CH2Br absorption is split into a quartet. The three spins of the neighboring –CH3 protons can align in four possible combinations: all three with the applied field, two with and one against (three ways), one with and two against (three ways), or all three against. Thus, four peaks are produced for the –CH2Br protons in a 1 : 3 : 3 : 1 ratio.
As a general rule, called the n + 1 rule, protons that have n equivalent neighboring protons show n + 1 peaks in their NMR spectrum. For example, the spectrum of 2-bromopropane in Figure 13.10 shows a doublet at 1.71 δ and a seven-line multiplet, or septet, at 4.28 δ. The septet is caused by splitting of the –CHBr– proton signal by six equivalent neighboring protons on the two methyl groups (n = 6 leads to 6 + 1 = 7 peaks). The doublet is due to signal splitting of the six equivalent methyl protons by the single –CHBr– proton (n = 1 leads to 2 peaks). Integration confirms the expected 6 : 1 ratio.
The distance between peaks in a multiplet is called the coupling constant and is denoted J. Coupling constants are measured in hertz and generally fall in the range 0 to 18 Hz. The exact value of the coupling constant between two neighboring protons depends on the geometry of the molecule, but a typical value for an open-chain alkane is J = 6 to 8 Hz. The same coupling constant is shared by both groups of hydrogens whose spins are coupled and is independent of spectrometer field strength. In bromoethane, for instance, the –CH2Br protons are coupled to the –CH3 protons and appear as a quartet with J = 7 Hz. The –CH3 protons appear as a triplet with the same J = 7 Hz coupling constant.
Because coupling is a reciprocal interaction between two adjacent groups of protons, it’s sometimes possible to tell which multiplets in a complex NMR spectrum are related to each other. If two multiplets have the same coupling constant, they are probably related, and the protons causing those multiplets are therefore adjacent in the molecule.
The most commonly observed coupling patterns and the relative intensities of lines in their multiplets are listed in Table 13.4. Note that it’s not possible for a given proton to have five equivalent neighboring protons. (Why not?) A six-line multiplet, or sextet, is therefore found only when a proton has five nonequivalent neighboring protons that coincidentally happen to be coupled with an identical coupling constant J.
Table 13.4 Some Common Spin Multiplicities
Number of equivalent adjacent protons Multiplet Ratio of intensities
0 Singlet 1
1 Doublet 1 : 1
2 Triplet 1 : 2 : 1
3 Quartet 1 : 3 : 3 : 1
4 Quintet 1 : 4 : 6 : 4 : 1
6 Septet 1 : 6 : 15 : 20 : 15 : 6 : 1
Spin–spin splitting in 1H NMR can be summarized by three rules.
RULE 1
Chemically equivalent protons don’t show spin–spin splitting. Equivalent protons may be on the same carbon or on different carbons, but their signals don’t split.
RULE 2
The signal of a proton with n equivalent neighboring protons is split into a multiplet of n + 1 peaks with coupling constant J. Protons that are farther than two carbon atoms apart don’t usually couple, although they sometimes show weak coupling when they are separated by a π bond.
RULE 3
Two groups of protons coupled to each other have the same coupling constant, J.
The spectrum of para-methoxypropiophenone in Figure 13.11 further illustrates these three rules. The downfield absorptions at 6.91 and 7.93 δ are due to the four aromatic-ring protons. There are two kinds of aromatic protons, each of which gives a signal that is split into a doublet by its neighbor. The –OCH3 signal is unsplit and appears as a sharp singlet at 3.84 δ. The –CH2– protons next to the carbonyl group appear at 2.93 δ in the region expected for protons on carbon next to an unsaturated center, and their signal is split into a quartet by coupling with the protons of the neighboring methyl group. The methyl protons appear as a triplet at 1.20 δ in the usual upfield region.
Worked Example 13.2
Assigning a Chemical Structure from a 1H NMR Spectrum
Propose a structure for a compound, C5H12O, that fits the following 1H NMR data: 0.92 δ (3 H, triplet, J = 7 Hz), 1.20 δ (6 H, singlet), 1.50 δ (2 H, quartet, J = 7 Hz), 1.64 δ (1 H, broad singlet).
Strategy
It’s best to begin solving structural problems by calculating a molecule’s degree of unsaturation (we’ll see this again in Worked Example 13.4). In the present instance, a formula of C5H12O corresponds to a saturated, open-chain molecule, either an alcohol or an ether.
To interpret the NMR information, let’s look at each absorption individually. The three-proton absorption at 0.92 δ is due to a methyl group in an alkane-like environment, and the triplet-splitting pattern implies that the CH3 is next to a CH2. Thus, our molecule contains an ethyl group, CH3CH2–. The six-proton singlet at 1.20 δ is due to two equivalent alkane-like methyl groups attached to a carbon with no hydrogens, (CH3)2C, and the two-proton quartet at 1.50 δ is due to the CH2 of the ethyl group. All 5 carbons and 11 of the 12 hydrogens in the molecule are now accounted for. The remaining hydrogen, which appears as a broad one-proton singlet at 1.64 δ, is probably due to an OH group, since there is no other way to account for it. Putting the pieces together gives the structure: 2-methyl-2-butanol.
Solution
(a)
(b)
(c)
(d)
(e)
(f)
Problem 13-10 Draw structures for compounds that meet the following descriptions: (a) C2H6O; one singlet (b)
C3H7Cl; one doublet and one septet
(c) C4H8Cl2O; two triplets (d) C4H8O2; one singlet, one triplet, and one quartet
Problem 13-11
The integrated 1H NMR spectrum of a compound of formula C4H10O is shown in Figure 13.12. Propose a structure.
Figure 13.12 An integrated 1H NMR spectrum for Problem 11. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.07%3A_Spin-Spin_Splitting_in_H_NMR__Spectra.txt |
Because each electronically distinct hydrogen in a molecule has its own unique absorption, one use of 1H NMR is to find out how many kinds of electronically nonequivalent hydrogens are present. In the 1H NMR spectrum of methyl acetate shown previously in Figure 13.4a, for instance, there are two signals, corresponding to the two kinds of nonequivalent protons present, CH3C═O protons and –OCH3 protons.
For relatively small molecules, a quick look at the structure is often enough to decide how many kinds of protons are present and thus how many NMR absorptions might appear. If in doubt, though, the equivalence or nonequivalence of two protons can be determined by comparing the structures that would be formed if each hydrogen were replaced by an X group. There are four possibilities.
• One possibility is that the protons are chemically unrelated and thus nonequivalent. If so, the products formed on substitution of H by X would be different constitutional isomers. In butane, for instance, the –CH3 protons are different from the –CH2– protons. They therefore give different products on substitution by X than the –CH2 protons and would likely show different NMR absorptions.
• A second possibility is that the protons are chemically identical and thus electronically equivalent. If so, the same product would be formed regardless of which H is substituted by X. In butane, for instance, the six –CH3 hydrogens on C1 and C4 are identical, would give the identical structure on substitution by X, and would show an identical NMR absorption. Such protons are said to be homotopic.
• The third possibility is a bit more subtle. Although they might at first seem homotopic, the two –CH2– hydrogens on C2 in butane (and the two –CH2– hydrogens on C3) are in fact not identical. Substitution by X of a hydrogen at C2 (or C3) would form a new chirality center, so different enantiomers (Section 5.1) would result, depending on whether the pro-R or pro-S hydrogen had been substituted (Section 5.11). Such hydrogens, whose substitution by X would lead to different enantiomers, are said to be enantiotopic. Enantiotopic hydrogens, even though not identical, are nevertheless electronically equivalent and thus have the same NMR absorption.
• The fourth possibility arises in chiral molecules, such as (R)-2-butanol. The two –CH2– hydrogens at C3 are neither homotopic nor enantiotopic. Because substitution of a hydrogen at C3 would form a second chirality center, different diastereomers (Section 5.6) would result, depending on whether the pro-R or pro-S hydrogen had been substituted. Such hydrogens, whose substitution by X leads to different diastereomers, are said to be diastereotopic. Diastereotopic hydrogens are neither chemically nor electronically equivalent. They are completely different and would likely show different NMR absorptions.
(b)
(c)
(d)
(e)
(f)
Problem 13-13 How many kinds of electronically nonequivalent protons are present in each of the following compounds, and thus how many NMR absorptions might you expect in each? (a) CH3CH2Br (b)
CH3OCH2CH(CH3)2
(c) CH3CH2CH2NO2 (d) Methylbenzene (e) 2-Methyl-1-butene (f) cis-3-Hexene
Problem 13-14
How many absorptions would you expect (S)-malate, an intermediate in carbohydrate metabolism, to have in its 1H NMR spectrum? Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.08%3A_H_NMR_Spectroscopy_and_Proton_Equivalence.txt |
In the 1H NMR spectra we’ve seen so far, the chemical shifts of different protons have been distinct and the spin–spin splitting patterns have been straightforward. It often happens, however, that different kinds of hydrogens in a molecule have accidentally overlapping signals. The spectrum of toluene (methylbenzene) in Figure 13.13, for example, shows that the five aromatic ring protons give a complex, overlapping pattern, even though they aren’t all equivalent.
Yet another complication in 1H NMR spectroscopy arises when a signal is split by two or more nonequivalent kinds of protons, as is the case with trans-cinnamaldehyde, isolated from oil of cinnamon (Figure 13.14). Although the n + 1 rule predicts splitting caused by equivalent protons, splittings caused by nonequivalent protons are more complex.
To understand the 1H NMR spectrum of trans-cinnamaldehyde, we have to isolate the different parts and look at the signal of each proton individually.
• The five aromatic proton signals (black in Figure 13.14) overlap into a complex pattern with a large peak at 7.42 δ and a broad absorption at 7.57 δ.
• The aldehyde proton signal at C1 (red) appears in the normal downfield position at 9.69 δ and is split into a doublet with J = 6 Hz by the adjacent proton at C2.
• The vinylic proton at C3 (green) is next to the aromatic ring and is therefore shifted downfield from the normal vinylic region. This C3 proton signal appears as a doublet centered at 7.49 δ. Because it has one neighbor proton at C2, its signal is split into a doublet, with J = 12 Hz.
• The C2 vinylic proton signal (blue) appears at 6.73 δ and shows an interesting, four-line absorption pattern. It is coupled to the two nonequivalent protons at C1 and C3 with two different coupling constants: J1-2 = 6 Hz and J2-3 = 12 Hz.
A good way to understand the effect of multiple coupling, such as that occurring for the C2 proton of trans-cinnamaldehyde, is to draw a tree diagram, like that in Figure 13.15. The diagram shows the individual effect of each coupling constant on the overall pattern. Coupling with the C3 proton splits the signal of the C2 proton in trans-cinnamaldehyde into a doublet with J =12 Hz. Further coupling with the aldehyde proton then splits each peak of the doublet into new doublets with J = 6 Hz, and we therefore observe a four-line spectrum for the C2 proton.
One further trait evident in the cinnamaldehyde spectrum is that the four peaks of the C2 proton signal are not all the same size. The two left-hand peaks are somewhat larger than the two right-hand peaks. Such a size difference occurs whenever coupled nuclei have similar chemical shifts—in this case, 7.49 δ for the C3 proton and 6.73 δ for the C2 proton. The peaks nearer the signal of the coupled partner are always larger, and the peaks farther from the signal of the coupled partner are always smaller. Thus, the left-hand peaks of the C2 proton multiplet at 6.73 δ are closer to the C3 proton absorption at 7.49 δ and are larger than the right-hand peaks. At the same time, the right-hand peak of the C3 proton doublet at 7.49 δ is larger than the left-hand peak because it is closer to the C2 proton multiplet at 6.73 δ. This skewing effect on multiplets can often be useful because it tells where to look in the spectrum to find the coupled partner: look in the direction of the larger peaks.
Problem 13-15
3-Bromo-1-phenyl-1-propene shows a complex NMR spectrum in which the vinylic proton at C2 is coupled with both the C1 vinylic proton (J = 16 Hz) and the C3 methylene protons (J = 8 Hz). Draw a tree diagram for the C2 proton signal, and account for the fact that a five-line multiplet is observed. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.09%3A_More_Complex_Spin-Spin_Splitting_Patterns.txt |
NMR is used to help identify the product of nearly every reaction run in the laboratory. For example, we said in Section 8.5 that hydroboration–oxidation of alkenes occurs with non-Markovnikov regiochemistry to yield the less highly substituted alcohol. With the help of NMR, we can now prove this statement.
Does hydroboration–oxidation of methylenecyclohexane yield cyclohexylmethanol or 1-methylcyclohexanol?
The 1H NMR spectrum of the reaction product is shown in Figure 13.16a. The spectrum shows a two-proton peak at 3.40 δ, indicating that the product has a –CH2– group bonded to an electronegative oxygen atom (–CH2OH). Furthermore, the spectrum shows no large three-proton singlet absorption near 1 δ, where we would expect the signal of a quaternary –CH3 group to appear. (Figure 13.16b) gives the spectrum of 1-methylcyclohexanol, the alternative product.) Thus, it’s clear that cyclohexylmethanol is the reaction product.
Problem 13-16
How could you use 1H NMR to determine the regiochemistry of electrophilic addition to alkenes? For example, does addition of HCl to 1-methylcyclohexene yield 1-chloro-1-methylcyclohexane or 1-chloro-2-methylcyclohexane?
13.11: C NMR Spectroscopy - Signal Averaging and FT-NMR
In some ways, it’s surprising that carbon NMR is even possible. After all, 12C, the most abundant carbon isotope, has no nuclear spin and can’t be seen by NMR. Carbon-13 is the only naturally occurring carbon isotope with a nuclear spin, but its natural abundance is only 1.1%. Thus, only about 1 of every 100 carbon atoms in an organic sample can be observed by NMR. The problem of low abundance has been overcome, however, by the use of signal averaging and Fourier-transform NMR (FT–NMR). Signal averaging increases instrument sensitivity, and FT–NMR increases instrument speed.
The low natural abundance of 13C means that any individual NMR spectrum is extremely “noisy.” That is, the signals are so weak that they are cluttered with random background electronic noise, as shown in Figure 13.17a. If, however, hundreds or thousands of individual runs are added together by a computer and then averaged, a greatly improved spectrum results (Figure 13.17b). Background noise, because of its random nature, increases very slowly as the runs are added, while the nonzero signals stand out clearly. Unfortunately, the value of signal averaging is limited when using the method of NMR spectrometer operation described in Section 13.2, because it takes about 5 to 10 minutes to obtain a single spectrum. Thus, a faster way to obtain spectra is needed if signal averaging is to be used.
In the method of NMR spectrometer operation described in Section 13.2, the rf frequency is held constant while the strength of the magnetic field is varied so that all signals in the spectrum are recorded sequentially. In the FT–NMR technique used by modern spectrometers, however, all the signals are recorded simultaneously. A sample is placed in a magnetic field of constant strength and is irradiated with a short pulse of rf energy that covers the entire range of useful frequencies. All 1H or 13C nuclei in the sample resonate at once, giving a complex, composite signal that is mathematically manipulated using so-called Fourier transforms and then displayed in the usual way. Because all resonance signals are collected at once, it takes only a few seconds rather than a few minutes to record an entire spectrum.
Combining the speed of FT–NMR with the sensitivity enhancement of signal averaging is what gives modern NMR spectrometers their power. Literally thousands of spectra can be taken and averaged in a few hours, resulting in sensitivity so high that a 13C NMR spectrum can be obtained from less than 0.1 mg of sample and a 1H spectrum can be recorded from only a few micrograms.
One further question needs to be answered before moving forward with our discussion of 13C NMR. Why is spin–spin splitting seen only for 1H NMR? Why is there no splitting of carbon signals into multiplets in 13C NMR? After all, you might expect that the spin of a given 13C nucleus would couple with the spin of an adjacent magnetic nucleus, either 13C or 1H.
No coupling of a 13C nucleus with nearby carbons is seen because their low natural abundance makes it unlikely that two 13C nuclei will be adjacent. No coupling of a 13C nucleus with nearby hydrogens is seen because 13C spectra are normally recorded using broadband decoupling. At the same time that the sample is irradiated with a pulse of rf energy to cover the carbon resonance frequencies, it is also irradiated by a second band of rf energy covering all the hydrogen resonance frequencies. This second irradiation makes the hydrogens spin-flip so rapidly that their local magnetic fields average to zero and no coupling with carbon spins occurs. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.10%3A_Uses_of_H_NMR_Spectroscopy.txt |
At its simplest, 13C NMR makes it possible to count the number of different carbon atoms in a molecule. Look at the 13C NMR spectra of methyl acetate and 1-pentanol shown previously in Figure 13.4b and Figure 13.17b. In each case, a single sharp resonance line is observed for each different carbon atom.
Most 13C resonances are between 0 and 220 ppm downfield from the TMS reference line, with the exact chemical shift of each 13C resonance dependent on that carbon’s electronic environment within the molecule. Figure 13.18 shows the correlation of chemical shift with environment.
The factors that determine chemical shifts are complex, but it’s possible to make some generalizations from the data in Figure 13.18. One trend is that a carbon atom’s chemical shift is affected by the electronegativity of nearby atoms. Carbons bonded to oxygen, nitrogen, or halogen absorb downfield (to the left) of typical alkane carbons. Because electronegative atoms attract electrons, they pull electrons away from neighboring carbon atoms, causing those carbons to be deshielded and to come into resonance at a lower field.
Another trend is that sp3-hybridized carbons generally absorb from 0 to 90 δ, while sp2 carbons absorb from 110 to 220 δ. Carbonyl carbons (C=O) are particularly distinct in 13C NMR and are always found at the low-field end of the spectrum, from 160 to 220 δ. Figure 13.19 shows the 13C NMR spectra of 2-butanone and para-bromoacetophenone and indicates the peak assignments. Note that the C=O carbons are at the left edge of the spectrum in each case.
The 13C NMR spectrum of para-bromoacetophenone is interesting in several ways. Note particularly that only six carbon absorptions are observed, even though the molecule contains eight carbons. para-bromoacetophenone has a symmetry plane that makes ring carbons 4 and 4′, and ring carbons 5 and 5′ equivalent. (Remember from Section 2.4 that aromatic rings have two resonance forms.) Thus, the six ring carbons show only four absorptions in the range 128 to 137 δ.
A second interesting point about both spectra in Figure 13.19 is that the peaks aren’t uniform in size. Some peaks are larger than others even though they are one-carbon resonances (except for the two 2-carbon peaks of para-bromoacetophenone). This difference in peak size is a general feature of broadband-decoupled 13C NMR spectra, and explains why we can’t integrate 13C NMR spectra in the same way we integrate the resonances in a 1H NMR spectrum. The local environment of each carbon atom determines not only its chemical shift but also the time it takes for the nuclei to return to their equilibrium state after receiving a pulse of rf radiation and flipping their spins. Quaternary carbons, regardless of their hybridization state or substituents, typically give smaller resonances than primary, secondary, or tertiary carbons.
Worked Example 13.3
Predicting Chemical Shifts in 13C NMR Spectra
At what approximate positions would you expect ethyl acrylate, H2C═CHCO2CH2CH3, to show 13C NMR absorptions?
Strategy
Identify the distinct carbons in the molecule, and note whether each is alkyl, vinylic, aromatic, or in a carbonyl group. Then predict where each absorbs, using Figure 13.18 as necessary.
Solution
Ethyl acrylate has five chemically distinct carbons: two different C=C, one C=O, one O–C, and one alkyl C. From Figure 13.18, the likely absorptions are
The actual absorptions are at 14.1, 60.5, 128.5, 130.3, and 166.0 δ.
Problem 13-17 Predict the number of carbon resonance lines you would expect in the 13C NMR spectra of the following compounds: (a)
Methylcyclopentane
(b) 1-Methylcyclohexene (c) 1,2-Dimethylbenzene (d) 2-Methyl-2-butene (e)
(f)
Problem 13-18 Propose structures for compounds that fit the following descriptions: (a)
A hydrocarbon with seven lines in its 13C NMR spectrum
(b) A six-carbon compound with only five lines in its 13C NMR spectrum (c) A four-carbon compound with three lines in its 13C NMR spectrum
Problem 13-19
Classify the resonances in the 13C NMR spectrum of methyl propanoate, CH3CH2CO2CH3 (Figure 13.20).
Figure 13.20 13C NMR spectrum of methyl propanoate, Problem 19. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.12%3A_Characteristics_of_C_NMR_Spectroscopy.txt |
Numerous techniques developed in recent years have made it possible to obtain enormous amounts of information from 13C NMR spectra. Among these techniques is one called DEPT–NMR, for distortionless enhancement by polarization transfer, which makes it possible to distinguish between signals due to CH3, CH2, CH, and quaternary carbons. That is, the number of hydrogens attached to each carbon in a molecule can be determined.
A DEPT experiment is usually done in three stages, as shown in Figure 13.21 for 6-methyl-5-hepten-2-ol. The first stage is to run an ordinary spectrum (called a broadband-decoupled spectrum) to locate the chemical shifts of all carbons. Next, a second spectrum called a DEPT-90 is run, using special conditions under which only signals due to CH carbons appear. Signals due to CH3, CH2, and quaternary carbons are absent. Finally, a third spectrum called a DEPT-135 is run, using conditions under which CH3 and CH resonances appear as positive signals, CH2 resonances appear as negative signals—that is, as peaks below the baseline—and quaternary carbons are again absent.
Putting together the information from all three spectra makes it possible to tell the number of hydrogens attached to each carbon. The CH carbons are identified in the DEPT-90 spectrum, the CH2 carbons are identified as negative peaks in the DEPT-135 spectrum, the CH3 carbons are identified by subtracting the CH peaks from the positive peaks in the DEPT-135 spectrum, and quaternary carbons are identified by subtracting all peaks in the DEPT-135 spectrum from the peaks in the broadband-decoupled spectrum.
Worked Example 13.4
Assigning a Chemical Structure from a 13C NMR Spectrum
Propose a structure for an alcohol, C4H10O, that has the following 13C NMR spectral data:
Broadband decoupled 13C NMR: 19.0, 31.7, 69.5 δ;
DEPT-90: 31.7 δ;
DEPT-135: positive peak at 19.0 δ, negative peak at 69.5 δ.
Strategy
As noted in Section 7.2, it usually helps with compounds of known formula but unknown structure to calculate the compound’s degree of unsaturation. In the present instance, a formula of C4H10O corresponds to a saturated, open-chain molecule.
To gain information from the 13C data, let’s begin by noting that the unknown alcohol has four carbon atoms, yet has only three NMR absorptions, which implies that two of the carbons must be equivalent. Looking at chemical shifts, two of the absorptions are in the typical alkane region (19.0 and 31.7 δ), while one is in the region of a carbon bonded to an electronegative atom (69.5 δ)—oxygen in this instance. The DEPT-90 spectrum tells us that the alkyl carbon at 31.7 δ is tertiary (CH); the DEPT-135 spectrum tells us that the alkyl carbon at 19.0 δ is a methyl (CH3) and that the carbon bonded to oxygen (69.5 δ) is secondary (CH2). The two equivalent carbons are probably both methyls bonded to the same tertiary carbon, (CH3)2CH–. We can now put the pieces together to propose a structure: 2-methyl-1-propanol.
Solution
Problem 13-20
Assign a chemical shift to each carbon in 6-methyl-5-hepten-2-ol (Figure 13.21).
Problem 13-21
Estimate the chemical shift of each carbon in the following molecule. Predict which carbons will appear in the DEPT-90 spectrum, which will give positive peaks in the DEPT-135 spectrum, and which will give negative peaks in the DEPT-135 spectrum.
Problem 13-22
Propose a structure for an aromatic hydrocarbon, C11H16, that has the following 13C NMR spectral data:
• Broadband decoupled: 29.5, 31.8, 50.2, 125.5, 127.5, 130.3, 139.8 δ
• DEPT-90: 125.5, 127.5, 130.3 δ
• DEPT-135: positive peaks at 29.5, 125.5, 127.5, 130.3 δ; negative peak at 50.2 δ | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.13%3A_DEPT_C_NMR_Spectroscopy.txt |
The information derived from 13C NMR spectroscopy is extraordinarily useful for structure determination. Not only can we count the number of nonequivalent carbon atoms in a molecule, we can also get information about the electronic environment of each carbon and find how many protons are attached to each. As a result, we can address many structural questions that go unanswered by IR spectroscopy or mass spectrometry.
Here’s an example: how do we know that the E2 reaction of an alkyl halide follows Zaitsev’s rule (Section 11.7)? Does treatment of 1-chloro-1-methylcyclohexane with a strong base give predominantly the trisubstituted alkene 1-methylcyclohexene or the disubstituted alkene methylenecyclohexane?
1-Methylcyclohexene will have five sp3-carbon resonances in the 20 to 50 δ range and two sp2-carbon resonances in the 100 to 150 δ range. Methylenecyclohexane, however, because of its symmetry, will have only three sp3-carbon resonance peaks and two sp2-carbon peaks. The spectrum of the actual reaction product, shown in Figure 13.22, clearly identifies 1-methylcyclohexene as the product of this E2 reaction.
Problem 13-23
We saw in Section 9.3 that addition of HBr to a terminal alkyne leads to the Markovnikov addition product, with the Br bonding to the more highly substituted carbon. How could you use 13C NMR to identify the product of the addition of 1 equivalent of HBr to 1-hexyne?
13.15: Chemistry MattersMagnetic Resonance Imaging (MRI)
13 • Chemistry Matters 13 • Chemistry Matters
As practiced by organic chemists, NMR spectroscopy is a powerful method of structure determination. A small amount of sample, typically a few milligrams or less, is dissolved in a small amount of solvent, the solution is placed in a thin glass tube, and the tube is placed into the narrow (1–2 cm) gap between the poles of a strong magnet. Imagine, though, that a much larger NMR instrument were available. Instead of a few milligrams, the sample size could be tens of kilograms; instead of a narrow gap between magnet poles, the gap could be large enough for a whole person to climb into so that an NMR spectrum of body parts could be obtained. That large instrument is exactly what’s used for magnetic resonance imaging (MRI), a diagnostic technique of enormous value to the medical community.
Like NMR spectroscopy, MRI takes advantage of the magnetic properties of certain nuclei, typically hydrogen, and of the signals emitted when those nuclei are stimulated by radiofrequency energy. Unlike what happens in NMR spectroscopy, though, MRI instruments use data manipulation techniques to look at the three-dimensional location of magnetic nuclei in the body rather than at the chemical nature of the nuclei. As noted, most MRI instruments currently look at hydrogen, present in abundance wherever there is water or fat in the body.
The signals detected by MRI vary with the density of hydrogen atoms and with the nature of their surroundings, allowing identification of different types of tissue and even allowing the visualization of motion. For example, the volume of blood leaving the heart in a single stroke can be measured, and heart motion can be observed. Soft tissues that don’t show up well on X-ray images can be seen clearly, allowing diagnosis of brain tumors, strokes, and other conditions. This technique is also valuable in diagnosing damage to knees or other joints and is a noninvasive alternative to surgical explorations.
Several types of atoms in addition to hydrogen can be detected by MRI, and the applications of images based on 31P atoms are being explored. This approach holds great promise for studies of metabolism.
13.16: Key Terms
13 • Key Terms 13 • Key Terms
• chemical shift
• coupling
• coupling constant (J)
• delta (δ) scale
• DEPT-NMR
• diastereotopic
• downfield
• enantiotopic
• FT–NMR
• homotopic
• integration
• multiplet
• n + 1 rule
• nuclear magnetic resonance (NMR) spectroscopy
• shielding
• spin–spin splitting
• upfield
13.17: Summary
13 • Summary 13 • Summary
Nuclear magnetic resonance spectroscopy, or NMR, is the most valuable of the numerous spectroscopic techniques used for structure determination. Although we focused in this chapter on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding.
When magnetic nuclei, such as 1H and 13C, are placed in a strong magnetic field, their spins orient either with or against the field. On irradiation with radiofrequency (rf) waves, energy is absorbed and the nuclei “spin-flip” from the lower energy state to the higher energy state. This absorption of rf energy is detected, amplified, and displayed as an NMR spectrum.
Each electronically distinct 1H or 13C nucleus in a molecule comes into resonance at a slightly different value of the applied field, thereby producing a unique absorption signal. The exact position of each peak is called the chemical shift. Chemical shifts are caused by electrons setting up tiny local magnetic fields that shield a nearby nucleus from the applied field.
The NMR chart is calibrated in delta units (δ), where 1 δ = 1 ppm of spectrometer frequency. Tetramethylsilane (TMS) is used as a reference point because it shows both 1H and 13C absorptions at unusually high values of applied magnetic field. The TMS absorption occurs on the right-hand (upfield) side of the chart and is arbitrarily assigned a value of 0 δ.
13C spectra are run on Fourier-transform NMR (FT–NMR) spectrometers using broadband decoupling of proton spins so that each chemically distinct carbon shows a single unsplit resonance line. As with 1H NMR, the chemical shift of each 13C signal provides information about a carbon’s chemical environment in the sample. In addition, the number of protons attached to each carbon can be determined using the DEPT–NMR technique.
In 1H NMR spectra, the area under each absorption peak can be electronically integrated to determine the relative number of hydrogens responsible for each peak. In addition, neighboring nuclear spins can couple, causing the spin–spin splitting of NMR peaks into multiplets. The NMR signal of a hydrogen neighbored by n equivalent adjacent hydrogens splits into n + 1 peaks (the n + 1 rule) with coupling constant J. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.14%3A_Uses_of_C_NMR_Spectroscopy.txt |
13 • Additional Problems 13 • Additional Problems
Visualizing Chemistry
Problem 13-24
Into how many peaks would you expect the 1H NMR signals of the indicated protons to be split? (Green = Cl.) (a)
(b)
956 Hz
(c) 1504 Hz
Problem 13-30 The following 1H NMR absorptions were obtained on a spectrometer operating at 300 MHz. Convert the chemical shifts from δ units to hertz downfield from TMS. (a)
2.1 δ
(b) 3.45 δ (c) 6.30 δ (d) 7.70 δ
Problem 13-31 When measured on a spectrometer operating at 200 MHz, chloroform (CHCl3) shows a single sharp absorption at 7.3 δ. (a)
How many parts per million downfield from TMS does chloroform absorb?
(b) How many hertz downfield from TMS would chloroform absorb if the measurement were carried out on a spectrometer operating at 360 MHz? (c) What would be the position of the chloroform absorption in δ units when measured on a 360 MHz spectrometer?
Problem 13-32
Why do you suppose accidental overlap of signals is much more common in 1H NMR than in 13C NMR?
Problem 13-33
Is a nucleus that absorbs at 6.50 δ more shielded or less shielded than a nucleus that absorbs at 3.20 δ? Does the nucleus that absorbs at 6.50 δ require a stronger applied field or a weaker applied field to come into resonance than the one that absorbs at 3.20 δ?
1H NMR Spectroscopy
C5H10
(c) C4H8O2
Problem 13-38 Predict the splitting pattern for each kind of hydrogen in the following molecules: (a)
(CH3)3CH
(b) CH3CH2CO2CH3 (c) trans-2-Butene
Problem 13-39
Predict the splitting pattern for each kind of hydrogen in isopropyl propanoate, CH3CH2CO2CH(CH3)2.
Problem 13-40 Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic: (a) (b) (c) Problem 13-41 Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic: (a) (b) (c) Problem 13-42 The acid-catalyzed dehydration of 1-methylcyclohexanol yields a mixture of two alkenes. How could you use 1H NMR to help you decide which was which? Problem 13-43 How could you use 1H NMR to distinguish between the following pairs of isomers? (a) (b) (c) (d) Problem 13-44 Propose structures for compounds that fit the following 1H NMR data: (a) C5H10O 0.95 δ (6 H, doublet, J = 7 Hz) 2.10 δ (3 H, singlet) 2.43 δ (1 H, multiplet) (b)
• C3H5Br
• 2.32 δ (3 H, singlet)
• 2.32 δ (3 H, singlet)
• 2.32 δ (3 H, singlet)
Problem 13-45 Propose structures for the two compounds whose 1H NMR spectra are shown. (a)
C4H9Br
(b)
C4H8Cl2
13C NMR Spectroscopy
Problem 13-46
How many 13C NMR absorptions would you expect for cis-1,3-dimethylcyclohexane? For trans-1,3-dimethylcyclohexane? Explain.
Problem 13-47 How many absorptions would you expect to observe in the 13C NMR spectra of the following compounds? (a)
1,1-Dimethylcyclohexane
(b) CH3CH2OCH3 (c) tert-Butylcyclohexane (d) 3-Methyl-1-pentyne (e) cis-1,2-Dimethylcyclohexane (f) Cyclohexanone
Problem 13-48
Suppose you ran a DEPT-135 spectrum for each substance in Problem 13.47. Which carbon atoms in each molecule would show positive peaks, and which would show negative peaks?
Problem 13-49
How could you use 1H and 13C NMR to help distinguish the following isomeric compounds of the formula C4H8?
Problem 13-50
How could you use 1H NMR, 13C NMR, and IR spectroscopy to help you distinguish between the following structures?
Problem 13-51
Assign as many resonances as you can to specific carbon atoms in the 13C NMR spectrum of ethyl benzoate.
General Problems
Problem 13-52 Assume that you have a compound with the formula C3H6O. (a)
How many double bonds and/or rings does your compound contain?
(b) Propose as many structures as you can that fit the molecular formula. (c) If your compound shows an infrared absorption peak at 1715 cm–1, what functional group does it have? (d) If your compound shows a single 1H NMR absorption peak at 2.1 δ, what is its structure?
Problem 13-53
The compound whose 1H NMR spectrum is shown has the molecular formula C3H6Br2. Propose a structure.
Problem 13-54
The compound whose 1H NMR spectrum is shown has the molecular formula C4H7O2Cl and has an infrared absorption peak at 1740 cm–1. Propose a structure.
Problem 13-55 Propose structures for compounds that fit the following 1H NMR data: (a)
• C4H6Cl2
• 2.18 δ (3 H, singlet)
• 4.16 δ (2 H, doublet, J = 7 Hz)
• 5.71 δ (1 H, triplet, J = 7 Hz)
(b)
• C10H14
• 1.30 δ (9 H, singlet)
• 7.30 δ (5 H, singlet)
(c)
• C4H7BrO
• 2.11 δ (3 H, singlet)
• 3.52 δ (2 H, triplet, J = 6 Hz)
• 4.40 δ (2 H, triplet, J = 6 Hz)
(d)
• C9H11Br
• 2.15 δ (2 H, quintet, J = 7 Hz)
• 2.75 δ (2 H, triplet, J = 7 Hz)
• 3.38 δ (2 H, triplet, J = 7 Hz)
• 7.22 δ (5 H, singlet)
Problem 13-56
Long-range coupling between protons more than two carbon atoms apart is sometimes observed when π bonds intervene. An example is found in 1-methoxy-1-buten-3-yne. Not only does the acetylenic proton, Ha, couple with the vinylic proton Hb, it also couples with the vinylic proton Hc, four carbon atoms away. The data are:
Construct tree diagrams that account for the observed splitting patterns of Ha, Hb, and Hc.
Problem 13-57
The 1H and 13C NMR spectra of compound A, C8H9Br, are shown. Propose a structure for A, and assign peaks in the spectra to your structure.
C5H10O
(b)
C7H7Br
(c)
C8H9Br
Problem 13-59
The mass spectrum and 13C NMR spectrum of a hydrocarbon are shown. Propose a structure for this hydrocarbon, and explain the spectral data.
Problem 13-60
Compound A, a hydrocarbon with M+ = 96 in its mass spectrum, has the following 13C spectral data. On reaction with BH3, followed by treatment with basic H2O2, A is converted into B, whose 13C spectral data are also given. Propose structures for A and B.
Compound A
• Broadband-decoupled 13C NMR: 26.8, 28.7, 35.7, 106.9, 149.7 δ
• DEPT-90: no peaks
• DEPT-135: no positive peaks; negative peaks at 26.8, 28.7, 35.7, 106.9 δ
Compound B
• Broadband-decoupled 13C NMR: 26.1, 26.9, 29.9, 40.5, 68.2 δ
• DEPT-90: 40.5 δ
• DEPT-135: positive peak at 40.5 δ; negative peaks at 26.1, 26.9, 29.9, 68.2 δ
Problem 13-61
Propose a structure for compound C, which has M+ = 86 in its mass spectrum, an IR absorption at 3400 cm–1, and the following 13C NMR spectral data:
Compound C
• Broadband-decoupled 13C NMR: 30.2, 31.9, 61.8, 114.7, 138.4 δ
• DEPT-90: 138.4 δ
• DEPT-135: positive peak at 138.4 δ; negative peaks at 30.2, 31.9, 61.8, 114.7 δ
Problem 13-62
Compound D is isomeric with compound C (Problem 13.61) and has the following 13C NMR spectral data. Propose a structure.
Compound D
• Broadband-decoupled 13C NMR: 9.7, 29.9, 74.4, 114.4, 141.4 δ
• DEPT-90: 74.4, 141.4 δ
• DEPT-135: positive peaks at 9.7, 74.4, 141.4 δ; negative peaks at 29.9, 114.4 δ
Problem 13-63
Propose a structure for compound E, C7H12O2, which has the following 13C NMR spectral data:
Compound E
• Broadband-decoupled 13C NMR: 19.1, 28.0, 70.5, 129.0, 129.8, 165.8 δ
• DEPT-90: 28.0, 129.8 δ
• DEPT-135: positive peaks at 19.1, 28.0, 129.8 δ; negative peaks at 70.5, 129.0 δ
Problem 13-64
Compound F, a hydrocarbon with M+ = 96 in its mass spectrum, undergoes reaction with HBr to yield compound G. Propose structures for F and G, whose 13C NMR spectral data are given below.
Compound F
• Broadband-decoupled 13C NMR: 27.6, 29.3, 32.2, 132.4 δ
• DEPT-90: 132.4 δ
• DEPT-135: positive peak at 132.4 δ; negative peaks at 27.6, 29.3, 32.2 δ
Compound G
• Broadband-decoupled 13C NMR: 25.1, 27.7, 39.9, 56.0 δ
• DEPT-90: 56.0 δ
• DEPT-135: positive peak at 56.0 δ; negative peaks at 25.1, 27.7, 39.9 δ
Problem 13-65
3-Methyl-2-butanol has five signals in its 13C NMR spectrum at 17.90, 18.15, 20.00, 35.05, and 72.75 δ. Why are the two methyl groups attached to C3 nonequivalent? Making a molecular model should be helpful.
Problem 13-66
A 13C NMR spectrum of commercially available 2,4-pentanediol, shows five peaks at 23.3, 23.9, 46.5, 64.8, and 68.1 δ. Explain.
Problem 13-67
Carboxylic acids (RCO2H) react with alcohols (R′OH) in the presence of an acid catalyst. The reaction product of propanoic acid with methanol has the following spectroscopic properties. Propose a structure.
MS: M+ = 88
IR: 1735 cm–1
1H NMR: 1.11 δ (3 H, triplet, J = 7 Hz); 2.32 δ (2 H, quartet, J = 7 Hz);
3.65 δ (3 H, singlet)
13C NMR: 9.3, 27.6, 51.4, 174.6 δ
Problem 13-68
Nitriles (RC$\text{≡}$N) react with Grignard reagents (R′MgBr). The reaction product from 2-methylpropanenitrile with methylmagnesium bromide has the following spectroscopic properties. Propose a structure.
MS: M+ = 86
IR: 1715 cm–1
1H NMR: 1.05 δ (6 H, doublet, J = 7 Hz); 2.12 δ (3 H, singlet);
2.67 δ (1 H, septet, J = 7 Hz)
13C NMR: 18.2, 27.2, 41.6, 211.2 δ
Problem 13-69
The proton NMR spectrum is shown for a compound with the formula C5H9NO4. The infrared spectrum displays strong bands at 1750 and 1562 cm–1 and a medium-intensity band at 1320 cm–1. The normal carbon-13 and the DEPT experimental results are tabulated. Draw the structure of this compound.
Normal Carbon DEPT-135 DEPT-90
14 ppm Positive No peak
16 Positive No peak
63 Negative No peak
83 Positive Positive
165 No peak No peak
Problem 13-70
The proton NMR spectrum of a compound with the formula C5H10O is shown. The normal carbon-13 and the DEPT experimental results are tabulated. The infrared spectrum shows a broad peak at about 3340 cm–1 and a medium-sized peak at about 1651 cm–1. Draw the structure of this compound.
Normal Carbon DEPT-135 DEPT-90
22.2 ppm Positive No peak
40.9 Negative No peak
60.2 Negative No peak
112.5 Negative No peak
142.3 No peak No peak
Problem 13-71
The proton NMR spectrum of a compound with the formula C7H12O2 is shown. The infrared spectrum displays a strong band at 1738 cm–1 and a weak band at 1689 cm–1. The normal carbon-13 and the DEPT experimental results are tabulated. Draw the structure of this compound.
Normal Carbon DEPT-135 DEPT-90
18 ppm Positive No peak
21 Positive No peak
26 Positive No peak
61 Negative No peak
119 Positive Positive
139 No peak No peak
171 No peak No peak | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.18%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 14, you should be able to
1. fulfillall of the detailed objectives listed under each individual section.
2. use the reactions discussed, along with those from previous chapters, when designing multi-step syntheses.
3. use the reactions and concepts discussed to solve road-map problems.
4. use ultraviolet-spectral data, in conjunction with other spectral data, to elucidate the structure of an unknown compound.
5. define, and use in context, the key terms introduced.
You have already studied the chemistry of compounds that contain one carbon-carbon double bond. In this chapter, you will focus your attention on compounds that contain two or more such bonds. In particular you will study the properties of those compounds that contain two carbon-carbon double bonds which are separated by one carbon-carbon single bond. These compounds are called “conjugated dienes.”
To understand the properties exhibited by conjugated dienes, you must first examine their bonding in terms of the molecular orbital theory introduced in Section 1.11. Then, you must learn how the products of a reaction are dependent on both thermodynamic and kinetic considerations. Which of these two factors is the most important can sometimes determine which of two possible products will predominate when a reaction is carried out under specific conditions. Although we shall not make extensive use of ultraviolet spectroscopy, this technique can often provide important information when conjugated compounds are being investigated. In general, ultraviolet spectroscopy is less useful than the other spectroscopic techniques introduced earlier.
14: Conjugated Compounds and Ultraviolet Spectroscopy
Chapter Contents
14.1 Stability of Conjugated Dienes: Molecular Orbital Theory
14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations 14.3 Kinetic versus Thermodynamic Control of Reactions 14.4 The Diels–Alder Cycloaddition Reaction 14.5 Characteristics of the Diels–Alder Reaction 14.6 Diene Polymers: Natural and Synthetic Rubbers 14.7 Ultraviolet Spectroscopy 14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation 14.9 Conjugation, Color, and the Chemistry of Vision
Conjugated compounds of many different sorts are common in nature. Many of the pigments responsible for the brilliant colors of fruits, flowers, and even animals have numerous alternating single and double bonds. β-Carotene, for instance, the orange pigment responsible for the color of carrots and an important source of vitamin A, is a conjugated polyene with 11 double bonds. Conjugated enones (alkene + ketone) are common structural features of many biologically important molecules such as progesterone, the hormone that prepares the uterus for implantation of a fertilized ovum. Cyclic conjugated molecules such as benzene are a major field of study in themselves. In this chapter, we’ll look at some of the distinctive properties of conjugated molecules and at the reasons for those properties.
Most of the unsaturated compounds we looked at in the chapters on Alkenes: Structure and Reactivity and Alkenes: Reactions and Synthesis had only one double bond, but many compounds have numerous sites of unsaturation. If the different unsaturations are well separated in a molecule, they react independently, but if they’re close together, they may interact. In particular, compounds that have alternating single and double bonds—so-called conjugated compounds—have some distinctive characteristics. The conjugated diene 1,3-butadiene, for instance, has some properties quite different from those of the nonconjugated 1,4-pentadiene. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.01%3A_Why_This_Chapter.txt |
Conjugated dienes can be prepared by some of the methods previously discussed for preparing alkenes (Section 11.7–Section 11.11). The base-induced elimination of HX from an allylic halide is one such reaction.
Simple conjugated dienes used in polymer synthesis include 1,3-butadiene, chloroprene (2-chloro-1,3-butadiene), and isoprene (2-methyl-1,3-butadiene). Isoprene has been prepared industrially by several methods, including the acid-catalyzed double dehydration of 3-methyl-1,3-butanediol.
One of the properties that distinguishes conjugated from nonconjugated dienes is that the central single bond is shorter than might be expected. The C2–C3 single bond in 1,3-butadiene, for instance, has a length of 147 pm, some 6 pm shorter than the C2–C3 bond in butane (153 pm).
Another distinctive property of conjugated dienes is their unusual stability, as evidenced by their heats of hydrogenation (Table 14.1). We saw in Section 7.6 that monosubstituted alkenes, such as 1-butene, have $ΔH∘hydrogΔH∘hydrog$ near –126 kJ/mol (–30.1 kcal/mol), whereas disubstituted alkenes, such as 2-methylpropene, have $ΔH∘hydrogΔH∘hydrog$ near –119 kJ/mol (–28.4 kcal/mol), which is approximately 7 kJ/mol less negative. We concluded from these data that more highly substituted alkenes are more stable than less substituted ones. That is, more highly substituted alkenes release less heat on hydrogenation because they contain less energy to start with. A similar conclusion can be drawn for conjugated dienes.
Table 14.1 Heats of Hydrogenation for Some Alkenes and Dienes
Alkene or diene Product H°hydrog
(kJ/mol) (kcal/mol)
–126 –30.1
–119 –28.4
–253 –60.5
–236 –56.4
–229 –54.7
Because a monosubstituted alkene has a $ΔH∘hydrogΔH∘hydrog$ of approximately –126 kJ/mol, we might expect that a compound with two monosubstituted double bonds would have a $ΔH∘hydrogΔH∘hydrog$ approximately twice that value, or –252 kJ/mol. Nonconjugated dienes, such as 1,4-pentadiene ($ΔH∘hydrog=−253 kJ/molΔH∘hydrog=−253 kJ/mol$), meet this expectation, but the conjugated diene 1,3-butadiene ($ΔH∘hydrog=−236 kJ/molΔH∘hydrog=−236 kJ/mol$) does not. 1,3-Butadiene is approximately 16 kJ/mol (3.8 kcal/mol) more stable than expected.
What accounts for the stability of conjugated dienes? According to valence bond theory (Section 1.5 and Section 1.8), their stability is due to orbital hybridization. Typical C–C single bonds, like those in alkanes, result from σ overlap of sp3 orbitals on both carbons, but in a conjugated diene, the central C–C single bond results from σ overlap of sp2 orbitals on both carbons. Because sp2 orbitals have more s character (33% s) than sp3 orbitals (25% s), the electrons in sp2 orbitals are closer to the nucleus and the bonds they form are somewhat shorter and stronger. Thus, the “extra” stability of a conjugated diene results in part from the greater amount of s character in the orbitals forming the C–C single bond.
According to molecular orbital theory (Section 1.11), the stability of conjugated dienes arises because of an interaction between the π orbitals of the two double bonds. To review briefly, when two p atomic orbitals combine to form a π bond, two π molecular orbitals (MOs) result. One is lower in energy than the starting p orbitals and is therefore bonding; the other is higher in energy, has a node between nuclei, and is antibonding. The two π electrons occupy the low-energy, bonding orbital, resulting in formation of a stable bond between atoms (Figure 14.2).
Now let’s combine four adjacent p atomic orbitals, as occurs in a conjugated diene. In so doing, we generate a set of four π molecular orbitals, two of which are bonding and two of which are antibonding (Figure 14.3). The four π electrons occupy the two bonding orbitals, leaving the antibonding orbitals vacant.
The lowest-energy π molecular orbital (denoted ψ1, Greek psi) has no nodes between the nuclei and is therefore bonding. The π MO of next-lowest energy, ψ2, has one node between nuclei and is also bonding. Above ψ1 and ψ2 in energy are the two antibonding π MOs, $ψ3*ψ3*$ and $ψ4*ψ4*$. (The asterisks indicate antibonding orbitals.) Note that the number of nodes between nuclei increases as the energy level of the orbital increases. The $ψ3*ψ3*$ orbital has two nodes between nuclei, and $ψ4*ψ4*$, the highest-energy MO, has three nodes between nuclei.
Comparing the π molecular orbitals of 1,3-butadiene (two conjugated double bonds) with those of 1,4-pentadiene (two isolated double bonds) shows why the conjugated diene is more stable. In a conjugated diene, the lowest-energy π MO (ψ1) has a favorable bonding interaction between C2 and C3 that is absent in a nonconjugated diene. As a result, there is a certain amount of double-bond character to the C2–C3 single bond, making that bond both stronger and shorter than a typical single bond. Electrostatic potential maps show clearly the additional electron density in the central single bond (Figure 14.4).
is present in the central C−C bond of 1,3-butadiene, corresponding to partial double-bond character.
In describing 1,3-butadiene, we say that the π electrons are spread out, or delocalized, over the entire π framework, rather than localized between two specific nuclei. Delocalization allows the bonding electrons to be closer to more nuclei, thus leading to lower energy and greater stability.
Problem 14-1
Allene, H2C$\text{=}$C$\text{=}$CH2, has a heat of hydrogenation of –298 kJ/mol (–71.3 kcal/mol). Rank a conjugated diene, a nonconjugated diene, and an allene in order of stability. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.02%3A_Stability_of_Conjugated_Dienes-_Molecular_Orbital_Theory.txt |
One of the most striking differences between conjugated dienes and typical alkenes is their behavior in electrophilic addition reactions. To review briefly, the addition of an electrophile to a carbon–carbon double bond is a general reaction of alkenes (Section 7.7). Markovnikov regiochemistry is observed because the more stable carbocation is formed as an intermediate. Thus, addition of HCl to 2-methylpropene yields 2-chloro-2-methylpropane rather than 1-chloro-2-methylpropane, and addition of 2 equivalents of HCl to the nonconjugated diene 1,4-pentadiene yields 2,4-dichloropentane.
Conjugated dienes also undergo electrophilic addition reactions readily, but mixtures of products are invariably obtained. Addition of HBr to 1,3-butadiene, for instance, yields a mixture of two products (not counting cis–trans isomers). 3-Bromo-1-butene is the typical Markovnikov product of 1,2-addition to a double bond, but 1-bromo-2-butene seems unusual. The double bond in this product has moved to a position between carbons 2 and 3, and HBr has added to carbons 1 and 4, a result described as 1,4-addition.
Many other electrophiles besides HBr add to conjugated dienes, and mixtures of products are usually formed. For example, Br2 adds to 1,3-butadiene to give a mixture of 3,4-dibromo-1-butene and 1,4-dibromo-2-butene.
How can we account for the formation of 1,4-addition products? The answer is that allylic carbocations are involved as intermediates (recall that the word allylic means “next to a double bond”). When 1,3-butadiene reacts with an electrophile such as H+, two carbocation intermediates are possible—a primary nonallylic carbocation and a secondary allylic cation. Because an allylic cation is stabilized by resonance between two forms (Section 11.5), it is more stable and forms faster than a nonallylic carbocation.
When the allylic cation reacts with Br to complete the electrophilic addition, the reaction can occur either at C1 or at C3 because both carbons share the positive charge (Figure 14.5). Thus, a mixture of 1,2- and 1,4-addition products results. You might recall that a similar product mixture was seen for NBS bromination of alkenes in Section 10.3, a reaction that proceeds through an allylic radical.
is shared by carbons 1 and 3. Reaction of Br with the more positive carbon (C3) predominantly yields the 1,2-addition product.
Worked Example 14.1
Predicting the Product of an Electrophilic Addition Reaction of a Conjugated Diene
Give the structures of the likely products from reaction of 1 equivalent of HCl with 2-methyl-1,3-cyclohexadiene. Show both 1,2 and 1,4 adducts.
Strategy
Electrophilic addition of HCl to a conjugated diene involves the formation of allylic carbocation intermediate. Thus, the first step is to protonate the two ends of the diene and draw the resonance forms of the two allylic carbocations that result. Then, allow each resonance form to react with Cl, generating a maximum of four possible products.
In the present instance, protonation of the C1–C2 double bond gives a carbocation that can react further to give the 1,2 adduct 3-chloro-3-methylcyclohexene and the 1,4 adduct 3-chloro-1-methylcyclohexene. Protonation of the C3–C4 double bond gives a symmetrical carbocation, whose two resonance forms are equivalent. Thus, the 1,2 adduct and the 1,4 adduct have the same structure: 6-chloro-1-methylcyclohexene. Of the two possible modes of protonation, the first is more likely because it yields a more stable, tertiary allylic cation rather than a less-stable, secondary allylic cation.
Solution
Problem 14-2
Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HCl with 1,3-pentadiene.
Problem 14-3
Look at the possible carbocation intermediates produced during addition of HCl to 1,3-pentadiene (Problem 14-2), and predict which 1,2 adduct predominates. Which 1,4 adduct predominates?
Problem 14-4
Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HBr with the following compound: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.03%3A_Electrophilic_Additions_to_Conjugated_Dienes-_Allylic_Carbocations.txt |
Electrophilic addition to a conjugated diene at or below room temperature normally leads to a mixture of products in which the 1,2 adduct predominates over the 1,4 adduct. When the same reaction is carried out at higher temperatures, however, the product ratio often changes and the 1,4 adduct predominates. For example, addition of HBr to 1,3-butadiene at 0 °C yields a 71 : 29 mixture of 1,2 and 1,4 adducts, but the same reaction carried out at 40 °C yields a 15 : 85 mixture. Furthermore, when the product mixture formed at 0 °C is heated to 40 °C in the presence of HBr, the ratio of adducts slowly changes from 71 : 29 to 15 : 85. Why?
To understand the effect of temperature on product distribution, let’s briefly review what we said in Section 6.7 about rates and equilibria. Imagine a reaction that can give either or both of two products, B and C.
Let’s assume that B forms faster than C (in other words, $ΔG‡B<ΔG‡CFigure 14.6.$
forms faster than the more stable product C.
Let’s first carry out the reaction at a lower temperature so that both processes are irreversible and no equilibrium is reached. Because B forms faster than C, B is the major product. It doesn’t matter that C is more stable than B, because the two are not in equilibrium. The product of an irreversible reaction depends only on relative rates, not on stability. Such reactions are said to be under kinetic control.
Now let’s carry out the same reaction at some higher temperature so that both processes are readily reversible and an equilibrium is reached. Because C is more stable than B, C is the major product obtained. It doesn’t matter that C forms more slowly than B, because the two are in equilibrium. The product of a readily reversible reaction depends only on stability, not on relative rates. Such reactions are said to be under equilibrium control, or thermodynamic control.
We can now explain the effect of temperature on the electrophilic addition reactions of conjugated dienes. At low temperature (0 °C), HBr adds to 1,3-butadiene under kinetic control to give a 71 : 29 mixture of products, with the more rapidly formed 1,2 adduct predominating. Because these low-temperature conditions don’t allow the reaction to reach equilibrium, the product that forms faster predominates. At higher temperature (40 °C), however, the reaction occurs under thermodynamic control to give a 15 : 85 mixture of products, with the more stable 1,4 adduct predominating. The higher temperature allows the addition process to become reversible, so an equilibrium mixture of products results. Figure 14.7 shows this situation in an energy diagram.
The electrophilic addition of HBr to 1,3-butadiene is a good example of how a change in experimental conditions can change the product of a reaction. The concept of thermodynamic control versus kinetic control is a useful one that we can sometimes take advantage of in the laboratory.
Problem 14-5
The 1,2 adduct and the 1,4 adduct formed by reaction of HBr with 1,3-butadiene are in equilibrium at 40 °C. Propose a mechanism by which the interconversion of products takes place.
Problem 14-6
Why do you suppose 1,4 adducts of 1,3-butadiene are generally more stable than 1,2 adducts?
14.05: The Diels-Alder Cycloaddition Reaction
Perhaps the most striking difference between conjugated and nonconjugated dienes is that conjugated dienes undergo an addition reaction with alkenes to yield substituted cyclohexene products. For example, 1,3-butadiene and 3-buten-2-one give 3-cyclohexenyl methyl ketone.
This process, named the Diels–Alder cycloaddition reaction after its discoverers, is extremely useful in the laboratory because it forms two carbon–carbon bonds in a single step and is one of the few general methods available for making cyclic molecules. (As the name implies, a cycloaddition reaction is one in which two reactants add together to give a cyclic product.) The 1950 Nobel Prize in Chemistry was awarded to Otto Diels and Kurt Alder in recognition of the importance of their discovery.
The mechanism of Diels–Alder cycloaddition is different from that of other reactions we’ve studied because it is neither polar nor radical. Rather, the Diels–Alder reaction is a pericyclic process. Pericyclic reactions, which we’ll discuss in more detail in Chapter 30, take place in a single step by a cyclic redistribution of bonding electrons. The two reactants simply join together through a cyclic transition state in which the two new C–C bonds form at the same time.
We can picture a Diels–Alder addition as occurring by head-on (σ) overlap of the two alkene p orbitals with the two p orbitals on carbons 1 and 4 of the diene (Figure 14.8). This is, of course, a cyclic orientation of the reactants.
In the Diels–Alder transition state, the two alkene carbons and carbons 1 and 4 of the diene rehybridize from sp2 to sp3 to form two new single bonds, while carbons 2 and 3 of the diene remain sp2-hybridized to form the new double bond in the cyclohexene product. We’ll study this mechanism in more detail in Section 30.5 but will concentrate for the present on learning about the characteristics and uses of the Diels–Alder reaction. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.04%3A_Kinetic_vs._Thermodynamic_Control_of_Reactions.txt |
The Dienophile
The Diels–Alder cycloaddition reaction occurs most rapidly if the alkene component, called the dienophile (“diene lover”), has an electron-withdrawing substituent group. Thus, ethylene itself reacts sluggishly, but propenal, ethyl propenoate, maleic anhydride, benzoquinone, propenenitrile, and similar compounds are highly reactive. Note also that alkynes, such as methyl propynoate, can act as Diels–Alder dienophiles.
In all cases, the double or triple bond of the dienophile is adjacent to the positively polarized carbon of an electron-withdrawing substituent. As a result, the double-bond carbons in these substances are substantially less electron-rich than the carbons in ethylene, as indicated by the electrostatic potential maps in Figure 14.9.
.
One of the most useful features of the Diels–Alder reaction is that it is stereospecific, meaning that a single product stereoisomer is formed. Furthermore, the stereochemistry of the dienophile is retained. If we carry out cycloaddition with methyl cis-2-butenoate, only the cis-substituted cyclohexene product is formed. With methyl trans-2-butenoate, only the trans-substituted cyclohexene product is formed.
Another stereochemical feature of the Diels–Alder reaction is that the diene and dienophile partners orient so that the endo product, rather than the alternative exo product, is formed. The words endo and exo are used to indicate relative stereochemistry when referring to bicyclic structures like substituted norbornanes (Section 4.9). A substituent on one bridge is said to be endo if it is syn (cis) to the larger of the other two bridges and is said to be exo if it is anti (trans) to the larger of the other two.
Endo products result from Diels–Alder reactions because the amount of orbital overlap between diene and dienophile is greater when the reactants lie directly on top of one another, so that the electron-withdrawing substituent on the dienophile is underneath the diene double bonds. In the reaction of 1,3-cyclopentadiene with maleic anhydride, for instance, the following result is obtained:
Worked Example 14.2
Predicting the Product of a Diels–Alder Reaction
Predict the product of the following Diels–Alder reaction:
Strategy
Draw the diene so that the ends of its two double bonds are near the dienophile double bond. Then form two single bonds between the partners, convert the three double bonds into single bonds, and convert the former single bond of the diene into a double bond. Because the dienophile double bond is cis to begin with, the two attached hydrogens must remain cis in the product.
Solution
Problem 14-7
Predict the product of the following Diels–Alder reaction:
The Diene
Just as the dienophile component has certain constraints that affect its reactivity, so too does the conjugated diene component. The diene must adopt what is called an s-cis conformation, meaning “cis-like” about the single bond, to undergo a Diels–Alder reaction. Only in the s-cis conformation are carbons 1 and 4 of the diene close enough to react through a cyclic transition state.
In the alternative s-trans conformation, the ends of the diene partner are too far apart to overlap with the dienophile p orbitals.
Two examples of dienes that can’t adopt an s-cis conformation, and thus don’t undergo Diels–Alder reactions, are shown in Figure 14.10. In the bicyclic diene, the double bonds are rigidly fixed in an s-trans arrangement by geometric constraints of the rings. In (2Z,4Z)-2,4-hexadiene, steric strain between the two methyl groups prevents the molecule from adopting s-cis geometry.
In contrast to these unreactive dienes that can’t achieve an s-cis conformation, other dienes are fixed only in the correct s-cis geometry and are therefore highly reactive in Diels–Alder cycloaddition. 1,3-Cyclopentadiene, for example, is so reactive that it reacts with itself. At room temperature, 1,3-cyclopentadiene dimerizes. One molecule acts as diene and a second molecule acts as dienophile in a self-Diels–Alder reaction.
Biological Diels–Alder reactions are also known but are uncommon. One example occurs in the biosynthesis of the cholesterol-lowering drug lovastatin (trade name Mevacor) isolated from the bacterium Aspergillus terreus. The key step is the intramolecular Diels–Alder reaction of a triene, in which the diene and dienophile components are within the same molecule.
Problem 14-8
Which of the following alkenes would you expect to be good Diels–Alder dienophiles?
(a)(b)(c)(d)(e)
Problem 14-9
Which of the following dienes have an s-cis conformation, and which have an s-trans conformation? Of the s-trans dienes, which can readily rotate to s-cis?
(a)(b)(c)
Problem 14-10
Predict the product of the following Diels–Alder reaction:
14.07: Diene Polymers- Natural and Synthetic Rubbers
Conjugated dienes can be polymerized just as simple alkenes can (Section 8.10). Diene polymers are structurally more complex than simple alkene polymers, however, because double bonds occur every four carbon atoms along the chain, leading to the possibility of cis–trans isomers. The initiator (In) for the reaction can be either a radical, as occurs in ethylene polymerization, or an acid. Note that the polymerization is a 1,4 addition of the growing chain to a conjugated diene monomer.
Rubber is a naturally occurring diene polymer of isoprene (2-methyl-1,3-butadiene) and is produced by more than 400 different plants. The major source is the so-called rubber tree, Hevea brasiliensis, from which the crude material, called latex, is harvested as it drips from a slice made through the bark. The double bonds of rubber have Z stereochemistry, but gutta-percha, the E isomer of rubber, also occurs naturally. Harder and more brittle than rubber, gutta-percha has a variety of applications, including use in dental endodontics and as the covering on some golf balls.
A number of different synthetic rubbers are produced commercially by diene polymerization. Both cis- and trans-polyisoprene can be made, and the synthetic rubber thus produced is similar to the natural material. Chloroprene (2-chloro-1,3-butadiene) is polymerized to yield neoprene, an excellent, although expensive, synthetic rubber with good weather resistance. Neoprene is used in the production of industrial hoses and gloves, among other things.
Both natural and synthetic rubbers are too soft and tacky to be useful until they are hardened by heating with elemental sulfur, a process called vulcanization. Vulcanization cross-links the rubber chains by forming carbon–sulfur bonds between them, thereby hardening and stiffening the polymer. The exact degree of hardening can be varied, yielding material soft enough for automobile tires or hard enough for bowling balls (ebonite).
The unusual ability of rubber to stretch and then contract to its original shape is due to the irregular structure of the polymer chains caused by the double bonds. These double bonds introduce bends and kinks into the polymer chains, thereby preventing neighboring chains from nestling together. When stretched, the randomly coiled chains straighten out and orient along the direction of the pull but are kept from sliding over one another by the cross-links. When the tension is released, the polymer reverts to its original random state.
Problem 14-11
Draw a segment of the polymer that might be prepared from 2-phenyl-1,3-butadiene.
Problem 14-12
Show the mechanism of the acid-catalyzed polymerization of 1,3-butadiene. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.06%3A_Characteristics_of_the_Diels-Alder_Reaction.txt |
As we’ve seen, mass spectrometry, infrared spectroscopy, and nuclear magnetic resonance spectroscopy are techniques of structure determination applicable to all organic molecules. In addition to these three generally useful methods, there is a fourth—ultraviolet (UV) spectroscopy—that is applicable only to conjugated compounds. UV is less commonly used than the other three spectroscopic techniques because of the specialized information it gives, so we’ll only discuss it briefly.
Mass spectrometry Molecular size and formula
IR spectroscopy Functional groups present
NMR spectroscopy Carbon–hydrogen framework
UV spectroscopy Conjugated π electron systems
The ultraviolet region of the electromagnetic spectrum extends from the short-wavelength end of the visible region (4 × 10–7 m) to the long-wavelength end of the X-ray region (10–8 m), but the narrow range from 2 × 10–7 m to 4 × 10–7 m is the part of greatest interest to organic chemists. Absorptions in this region are usually measured in nanometers (nm), where 1 nm = 10–9 m. Thus, the ultraviolet range of interest is from 200 to 400 nm (Figure 14.11).
We saw in Section 12.5 that when an organic molecule is irradiated with electromagnetic energy, the radiation either passes through the sample or is absorbed, depending on its energy. With IR irradiation, the energy absorbed corresponds to the amount needed to increase molecular vibrations. With UV radiation, the energy absorbed corresponds to the amount needed to promote an electron from a lower-energy orbital to a higher-energy one in a conjugated molecule. The conjugated diene 1,3-butadiene, for instance, has four π molecular orbitals, as shown previously in Figure 14.3. The two lower-energy, bonding MOs are occupied in the ground state, and the two higher-energy, antibonding MOs are unoccupied.
On irradiation with ultraviolet light (), 1,3-butadiene absorbs energy and a π electron is promoted from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). Because the electron is promoted from a bonding π molecular orbital to an antibonding π* molecular orbital, we call this a ππ* excitation (read as “pi to pi star”). The energy gap between the HOMO and the LUMO of 1,3-butadiene is such that UV light of 217 nm wavelength is required to effect the ππ* electronic transition (Figure 14.12).
An ultraviolet spectrum is recorded by irradiating a sample with UV light of continuously changing wavelength. When the wavelength corresponds to the energy level required to excite an electron to a higher level, energy is absorbed. This absorption is detected and displayed on a chart that plots wavelength versus absorbance (A), defined as
$A=logI0IA=logI0I$
where I0 is the intensity of the incident light and I is the intensity of the light transmitted through the sample.
Note that UV spectra differ from IR spectra in how they are presented. For historical reasons, IR spectra are usually displayed so that the baseline corresponding to zero absorption runs across the top of the chart and a valley indicates an absorption, whereas UV spectra are displayed with the baseline at the bottom of the chart so that a peak indicates an absorption (Figure 14.13).
The amount of UV light absorbed is expressed as the sample’s molar absorptivity (ϵ), defined by the equation
$ε=Ac×lε=Ac×l$
where
$A=Absorbance c=Concentration in mol/L l=Sample pathlength in cm A=Absorbance c=Concentration in mol/L l=Sample pathlength in cm$
Molar absorptivity is a physical constant, characteristic of the particular substance being observed and thus characteristic of the particular π electron system in the molecule. Typical values for conjugated dienes are in the range ε = 10,000 to 25,000. The units for molar absorptivity, L/(mol · cm), are usually dropped.
A particularly important use of this equation comes from rearranging it to the form c = A/(ε · l), which lets us measure the concentration of a sample in solution when A, ε, and l are known. As an example, β-carotene, the pigment responsible for the orange color of carrots, has ε = 138,000 L/(mol · cm). If a sample of β-carotene is placed in a cell with a pathlength of 1.0 cm and the UV absorbance reads 0.37, then the concentration of β-carotene in the sample is
$c= A εl = 0.37 1.38× 10 5 L mol⋅cm (1.00 cm) =2.7× 10 −6 mol/L c= A εl = 0.37 1.38× 10 5 L mol⋅cm (1.00 cm) =2.7× 10 −6 mol/L$
Unlike IR and NMR spectra, which show many absorptions for a given molecule, UV spectra are usually quite simple—often only a single peak. The peak is usually broad, and we identify its position by noting the wavelength at the top of the peak—λmax, read as “lambda max.”
Problem 14-13
Calculate the energy range of electromagnetic radiation in the UV region of the spectrum from 200 to 400 nm (see Section 12.5). How does this value compare with the values calculated previously for IR and NMR spectroscopy?
Problem 14-14
If pure vitamin A has λmax = 325 (ε = 50,100), what is the vitamin A concentration in a sample whose absorbance at 325 nm is A = 0.735 in a cell with a pathlength of 1.00 cm? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.08%3A_Ultraviolet_Spectroscopy.txt |
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