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Objectives
After completing this section, you should be able to
1. Draw the chair conformation of cyclohexane, with axial and equatorial hydrogen atoms clearly shown and identified.
2. identify the axial and equatorial hydrogens in a given sketch of the cyclohexane molecule.
3. explain how chair conformations of cyclohexane and its derivatives can interconvert through the process of ring flip.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• axial position
• equatorial position
• ring flip
Axial and Equatorial Positions in Cyclohexane
Careful examination of the chair conformation of cyclohexane, shows that the twelve hydrogens are not structurally equivalent. Six of them are located about the periphery of the carbon ring, and are termed equatorial. The other six are oriented above and below the approximate plane of the ring (three in each location), and are termed axial because they are aligned parallel to the symmetry axis of the ring.
In the figure above, the equatorial hydrogens are colored blue, and the axial hydrogens are black. Since there are two equivalent chair conformations of cyclohexane in rapid equilibrium, all twelve hydrogens have 50% equatorial and 50% axial character.
How To Draw Axial and Equatorial Bonds
How not to draw the chair:
Aside from drawing the basic chair, the key points are:
• Axial bonds alternate up and down, and are shown "vertical".
• Equatorial groups are approximately horizontal, but actually somewhat distorted from that (slightly up or slightly down), so that the angle from the axial group is a bit more than a right angle -- reflecting the common 109.5o bond angle.
• Each carbon has an axial and an equatorial bond.
• Each face of the cyclohexane ring has three axial and three equatorial bonds.
• Each face alternates between axial and equatorial bonds. Then looking at the "up" bond on each carbon in the cyclohexane ring they will alternate axial-equatorial-axial ect.
• When looking down at a cyclohexane ring:
• the equatorial bonds will form an "equator" around the ring.
• The axial bonds will either face towards you or away. These will alternate with each axial bond. The first axial bond will be coming towards with the next going away. There will be three of each type.
• Note! The terms cis and trans in regards to the stereochemistry of a ring are not directly linked to the terms axial and equatorial. It is very common to confuse the two. It typically best not to try and directly inter convert the two naming systems.
Axial vs. Equatorial Substituents
When a substituent is added to cyclohexane, the ring flip allows for two distinctly different conformations. One will have the substituent in the axial position while the other will have the substituent in the equatorial position. In the next section will discuss the energy differences between these two possible conformations. Below are the two possible chair conformations of methylcyclohexane created by a ring-flip. Although the conformation which places the methyl group in the equatorial position is more stable by 7 kJ/mol, the energy provided by ambient temperature allows the two conformations to rapidly interconvert.
The figure below illustrates how to convert a molecular model of cyclohexane between two different chair conformations - this is something that you should practice with models. Notice that a 'ring flip' causes equatorial groups to become axial, and vice-versa.
Example \(1\)
For the following please indicate if the substituents are in the axial or equatorial positions.
Solution
Due to the large number of bonds in cyclohexane it is common to only draw in the relevant ones (leaving off the hydrogens unless they are involved in a reaction or are important for analysis). It is still possible to determine axial and equatorial positioning with some thought. With problems such as this it is important to remember that each carbon in a cyclohexane ring has one axial and one equatorial bond. Also, remember that axial bonds are perpendicular with the ring and appear to be going either straight up or straight down. Equatorial bonds will be roughly in the plane of the cyclohexane ring (only slightly up or down). Sometimes it is valuable to draw in the additional bonds on the carbons of interest.
With this it can be concluded that the bromine and chlorine substituents are attached in equatorial positions and the CH3 substituent is attached in an axial position.
Exercises
1) Draw two conformations of cyclohexyl amine (C6H11NH2). Indicate axial and equatorial positions.
2) Draw the two isomers of 1,4-dihydroxylcyclohexane, identify which are equatorial and axial.
3) In the following molecule, label which are equatorial and which are axial, then draw the chair flip (showing labels 1,2,3).
Solutions
1)
2)
3) Original conformation: 1 = axial, 2 = equatorial, 3 = axial
Flipped chair now looks like this. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.06%3A_Axial_and_Equatorial_Bonds_in_Cyclohexane.txt |
Objectives
After completing this section, you should be able to
1. account for the greater stability of the equatorial conformers of monosubstituted cyclohexanes compared to their axial counterparts, using the concept of 1,3‑diaxial interaction.
2. compare the gauche interactions in butane with the 1,3‑diaxial interactions in the axial conformer of methylcyclohexane.
3. arrange a given list of substituents in increasing or decreasing order of 1,3‑diaxial interactions.
Key Terms
Make certain that you can define, and use in context, the key term below.
• 1,3‑diaxial interaction
Study Notes
1,3-Diaxial interactions are steric interactions between an axial substituent located on carbon atom 1 of a cyclohexane ring and the hydrogen atoms (or other substituents) located on carbon atoms 3 and 5.
Be prepared to draw Newman-type projections for cyclohexane derivatives as the one shown for methylcyclohexane. Note that this is similar to the Newman projections from chapter 3 such as n-butane.
Newman projections of methylcyclohexane and n‑butane
When a substituent is added to a cyclohexane ring, the two possible chair conformations created during a ring flip are not equally stable. In the example of methylcyclohexane the conformation where the methyl group is in the equatorial position is more stable than the axial conformation by 7.6 kJ/mol at 25o C. The percentages of the two different conformations at equilibrium can be determined by solving the following equation for K (the equilibrium constant): ΔE = -RTlnK. In this equation ΔE is the energy difference between the two conformations, R is the gas constant (8.314 J/mol•K), T is the temperature in Kelvin, and K is the equilibrium constant for the ring flip conversion. Using this equation, we can calculate a K value of 21 which means about 95% methylcyclohexane molecules have the methyl group in the equatorial position at 25o C.
The energy difference between the two conformations comes from strain, called 1,3-diaxial interactions, created when the axial methyl group experiences steric crowding with the two axial hydrogens located on the same side of the cyclohexane ring. Because axial bonds are parallel to each other, substituents larger than hydrogen experience greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents are in the equatorial orientation. When the methyl group is in the equatorial position this strain is not present which makes the equatorial conformer more stable and favored in the ring flip equilibrium.
Actually, 1,3-diaxial steric strain is directly related to the steric strain created in the gauche conformer of butane discussed in Section: 3-7. When butane is in the gauche conformation 3.8 kJ/mol of strain was created due the steric crowding of two methyl group with a 60o dihedral angle. When looking at the a Newman projection of axial methylcyclohexane the methyl group is at a 60o dihedral angle with the ring carbon in the rear. This creates roughly the same amount of steric strain as the gauche conformer of butante. Given that there is actually two such interactions in axial methylcyclohexane, it makes sense that there is 2(3.8 kJ/mol) = 7.6 kJ/mol of steric strain in this conformation. The Newman projection of equatorial methylcyclohexane shows no such interactions and is therefore more stable.
Newman projections of methyl cyclohexane and butane showing similarity of 1,3-diaxial and gauche interactions.
Strain values for other cyclohexane substituents can also be considered. The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane determined the amount of strain generated. The strain generated can be used to evaluate the relative tendency of substituents to exist in an equatorial or axial location. Looking at the energy values in this table, it is clear that as the size of the substituent increases, the 1,3-diaxial energy tends to increase, also. Note that it is the size and not the molecular weight of the group that is important. Table 4.7.1 summarizes some of these strain values values.
Table 4.7.1: A Selection of ΔG° Values for the Change from Axial to Equatorial Orientation of Substituents for Monosubstituted Cyclohexanes
Substituent -ΔG° (kcal/mol) Substituent -ΔG° (kcal/mol)
\(\ce{CH_3\bond{-}}\) 1.7 \(\ce{O_2N\bond{-}}\) 1.1
\(\ce{CH_2H_5\bond{-}}\) 1.8 \(\ce{N#C\bond{-}}\) 0.2
\(\ce{(CH_3)_2CH\bond{-}}\) 2.2 \(\ce{CH_3O\bond{-}}\) 0.5
\(\ce{(CH_3)_3C\bond{-}}\) \(\geq 5.0\) \(\ce{HO_2C\bond{-}}\) 0.7
\(\ce{F\bond{-}}\) 0.3 \(\ce{H_2C=CH\bond{-}}\) 1.3
\(\ce{Cl\bond{-}}\) 0.5 \(\ce{C_6H_5\bond{-}}\) 3.0
\(\ce{Br\bond{-}}\) 0.5
\(\ce{I\bond{-}}\) 0.5
Exercises
1) In the molecule, cyclohexyl ethyne there is little steric strain, why?
2) Calculate the energy difference between the axial and equatorial conformations of bromocyclohexane?
3) Using your answer from Question 2) estimate the percentages of axial and equatorial conformations of bromocyclohexane at 25o C.
4) There very little in 1,3-diaxial strain when going from a methyl substituent (3.8 kJ/mol) to an ethyl substituent (4.0 kJ/mol), why? It may help to use molecular model to answer this question.
Solutions
1) The ethyne group is linear and therefore does not affect the hydrogens in the 1,3 positions to say to the extent as a bulkier or a bent group (e.g. ethene group) would. This leads to less of a strain on the molecule.
2) The equatorial conformation of bromocyclohexane will have two 1,3 diaxial interactions. The table above states that each interaction accounts for 1.2 kJ/mol of strain. The total strain in equatorial bromocyclohexane will be 2(1.2 kJ/mol) = 2.4 kJ/mol.
3) Remembering that the axial conformation is higher in energy, the energy difference between the two conformations is ΔE = (E equatorial - E axial) = (0 - 2.4 kJ/mol) = -2.4 kJ/mol. After converting oC to Kelvin and kJ/mol to J/mol we can use the equation ΔE = -RT lnK to find that -ΔE/RT = lnK or (2.4 x 103 J/mol) / (8.313 kJ/mol K • 298 K) = lnK. From this we calculate that K = 2.6. Because the ring flip reaction is an equilibrium we can say that K = [Equatorial] / [Axial]. If assumption is made that [Equatorial] = X then [Axial] must be 1-X. Plugging these values into the equilibrium expression produces K = [X] / [1-X]. After plugging in the calculated value for K, X can be solved algebraically. 2.6 = [X] / [1-X] → 2.6 - 2.6X = X → 2.6 = 3.6X → 2.6/3.6 = X = 0.72. This means that bromocyclohexane is in the equatorial position 72% of the time and in the axial position 28% of the time.
4) The fact that C-C sigma bonds can freely rotate allows the ethyl subsistent to obtain a conformation which places the bulky CH3 group away from the cyclohexane ring. This forces the ethyl substituent to have only have 1,3- diaxial interactions between hydrogens, which only provides a slight difference to a methyl group. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.07%3A_Conformations_of_Monosubstituted_Cyclohexanes.txt |
Objective
After completing this section, you should be able to use conformational analysis to determine the most stable conformation of a given disubstituted cyclohexane.
Key Terms
Make certain that you can define, and use in context, the key term below.
• conformational analysis
Study Notes
When faced with the problem of trying to decide which of two conformers of a given disubstituted cyclohexane is the more stable, you may find the following generalizations helpful.
1. A conformation in which both substituents are equatorial will always be more stable than a conformation with both groups axial.
2. When one substituent is axial and the other is equatorial, the most stable conformation will be the one with the bulkiest substituent in the equatorial position. Steric bulk decreases in the order
tert-butyl > isopropyl > ethyl > methyl > hydroxyl > halogens
Monosubstituted Cyclohexanes
In the previous section, it was stated that the chair conformation in which the methyl group is equatorial is more stable because it minimizes steric repulsion, and thus the equilibrium favors the more stable conformer. This is true for all monosubstituted cyclohexanes. The chair conformation which places the substituent in the equatorial position will be the most stable and be favored in the ring flip equilibrium.
Disubstituted Cyclohexanes
Determining the more stable chair conformation becomes more complex when there are two or more substituents attached to the cyclohexane ring. To determine the stable chair conformation, the steric effects of each substituent, along with any additional steric interactions, must be taken into account for both chair conformations.
In this section, the effect of conformations on the relative stability of disubstituted cyclohexanes is examined using the two principles:
1. Substituents prefer equatorial rather than axial positions in order to minimize the steric strain created of 1,3-diaxial interactions.
2. The more stable conformation will place the larger substituent in the equatorial position.
1,1-Disubstituted Cyclohexanes
The more stable chair conformation can often be determined empirically or by using the energy values of steric interactions previously discussed in this chapter. Note, in some cases there is no discernable energy difference between the two chair conformations which means they are equally stable.
1,1-dimethylcyclohexane does not have cis or trans isomers, because both methyl groups are on the same ring carbon. Both chair conformers have one methyl group in an axial position and one methyl group in an equatorial position giving both the same relative stability. The steric strain created by the 1,3-diaxial interactions of a methyl group in an axial position (versus equatorial) is 7.6 kJ/mol (from Table 4.7.1), so both conformers will have equal amounts of steric strain. Thus, the equilibrium between the two conformers does not favor one or the other. Note, that both methyl groups cannot be equatorial at the same time without breaking bonds and creating a different molecule.
However, if the two groups are different, as in 1-tert-butyl-1-methylcyclohexane, then the equilibrium favors the conformer in which the larger group (tert-butyl in this case) is in the more stable equatorial position. The energy cost of having one tert-butyl group axial (versus equatorial) can be calculated from the values in table 4.7.1 and is approximately 22.8 kJ/mol. The conformer with the tert-butyl group axial is approximately 15.2 kJ/mol (22.8 kJ/mol - 7.6 kJ/mol) less stable then the conformer with the tert-butyl group equatorial. Solving for the equilibrium constant K shows that the equatorial is preferred about 460:1 over axial. This means that 1-tert-butyl-1-methylcyclohexane will spend the majority of its time in the more stable conformation, with the tert-butyl group in the equatorial position.
Cis and trans stereoisomers of 1,2-dimethylcyclohexane
In cis-1,2-dimethylcyclohexane, both chair conformations have one methyl group equatorial and one methyl group axial. As previously discussed, the axial methyl group creates 7.6 kJ/mol of steric strain due to 1,3-diaxial interactions. It is important to note, that both chair conformations also have an additional 3.8 kJ/mol of steric strain created by a gauche interaction between the two methyl groups. Overall, both chair conformations have 11.4 kJ/mol of steric strain and are of equal stability.
In trans-1,2-dimethylcyclohexane, one chair conformer has both methyl groups axial and the other conformer has both methyl groups equatorial. The conformer with both methyl groups equatorial has no 1,3-diaxial interactions however there is till 3.8 kJ/mol of strain created by a gauche interaction. The conformer with both methyl groups axial has four 1,3-Diaxial interactions which creates 2 x 7.6 kJ/mol (15.2 kJ/mol) of steric strain. This conformer is (15.2 kJ/mol -3.8 kJ/mol) 11.4 kJ/mol less stable than the other conformer. The equilibrium will therefore favor the conformer with both methyl groups in the equatorial position.
Cis and trans stereoisomers of 1,3-dimethylcyclohexane
A similar conformational analysis can be made for the cis and trans stereoisomers of 1,3-dimethylcyclohexane. For cis-1,3-dimethylcyclohexane one chair conformation has both methyl groups in axial positions creating 1,3-diaxial interactions. The other conformer has both methyl groups in equatorial positions thus creating no 1,3-diaxial interaction. Because the methyl groups are not on adjacent carbons in the cyclohexane rings gauche interactions are not possible. Even without energy calculations it is simple to determine that the conformer with both methyl groups in the equatorial position will be the more stable conformer.
For trans-1,3-dimethylcyclohexane both conformations have one methyl axial and one methyl group equatorial. Each conformer has one methyl group creating a 1,3-diaxial interaction so both are of equal stability.
Summary of Disubstitued Cyclohexane Chair Conformations
When considering the conformational analyses discussed above a pattern begins to form. There are only two possible relationships which can occur between ring-flip chair conformations:
1) AA/EE: One chair conformation places both substituents in axial positions creating 1,3-diaxial interactions. The other conformer places both substituents in equatorial positions creating no 1,3-diaxial interactions. This diequatorial conformer is the more stable regardless of the substituents.
2) AE/EA: Each chair conformation places one substituent in the axial position and one substituent in the equatorial position. If the substituents are the same, there will be equal 1,3-diaxial interactions in both conformers making them equal in stability. However, if the substituents are different then different 1,3-diaxial interactions will occur. The chair conformation which places the larger substituent in the equatorial position will be favored.
Substitution type Chair Conformation Relationship
cs-1,2-disubstituted cyclohexanes AE/EA
trans-1,2-disubstituted cyclohexanes AA/EE
cis-1,3-disubstituted cyclohexanes AA/EE
trans-1,3-disubstituted cyclohexanes AE/EA
cis-1,4-disubstituted cyclohexanes AE/EA
trans-1,4-disubstituted cyclohexanes AA/EE
Example \(1\)
For cis-1-chloro-4-methylcyclohexane, draw the most stable chair conformation and determine the energy difference between the two chair conformers.
Solution
Based on the table above, cis-1,4-disubstitued cyclohexanes should have two chair conformations each with one substituent axial and one equatorial. Based on this, we can surmise that the energy difference of the two chair conformations will be based on the difference in the 1,3-diaxial interactions created by the methyl and chloro substituents.
As predicted, each chair conformer places one of the substituents in the axial position. Because the methyl group is larger and has a greater 1,3-diaxial interaction than the chloro, the most stable conformer will place it the equatorial position, as shown in the structure on the right. Using the 1,3-diaxial energy values given in the previous sections we can calculate that the conformer on the right is (7.6 kJ/mol - 2.0 kJ/mol) 5.6 kJ/mol more stable than the other.
Example \(2\)
For trans-1-chloro-2-methylcyclohexane, draw the most stable chair conformation and determine the energy difference between the two chair conformers.
Solution
Based on the table above, trans-1,2-disubstitued cyclohexanes should have one chair conformation with both substituents axial and one conformation with both substituents equatorial. Based on this, we can predict that the conformer which places both substituents equatorial will be the more stable conformer. The energy difference of the two chair conformations will be based on the 1,3-diaxial interactions created by both the methyl and chloro substituents.
As predicted, one chair conformer places both substituents in the axial position and other places both substituents equatorial. The more stable conformer will place both substituents in the equatorial position, as shown in the structure on the right. Using the 1,3-diaxial energy values given in the previous sections we can calculate that the conformer on the right is (7.6 kJ/mol + 2.0 kJ/mol) 9.6 kJ/mol more stable than the other.
Conformational Analysis of Complex Six Membered Ring Structures
Cyclohexane can have more than two substituents. Also, there are multiple six membered rings which contain atoms other than carbon. All of these systems usually form chair conformations and follow the same steric constraints discussed in this section. Because the most commonly found rings in nature are six membered, conformational analysis can often help in understanding the usual shapes of some biologically important molecules. In complex six membered ring structures a direct calculation of 1,3-diaxial energy values may be difficult. In these cases a determination of the more stable chair conformer can be made by empirically applying the principles of steric interactions.
A later chapter will discuss how many sugars can exist in cyclic forms which are often six remembered rings. When in an aqueous solution the six carbon sugar, glucose, is usually a six membered ring adopting a chair conformation. When looking at the two possible ring-clip chair conformations, one has all of the substituents axial and the other has all the substutents equatorial. Even without a calculation, it is clear that the conformation with all equatorial substituents is the most stable and glucose will most commonly be found in this conformation.
Example \(3\)
The six carbon sugar, fructose, in aqueous solution is also a six-membered ring in a chair conformation. Which of the two possible chair conformations would be expected to be the most stable?
Solution
The lower energy chair conformation is the one with three of the five substituents (including the bulky –CH2OH group) in the equatorial position (pictured on the right). The left structure has 3 equatorial substituents while the structure on the right only has two equatorial substituents.
Exercises
1. Draw the two chair conformations for cis-1-ethyl-2-methylcyclohexane using bond-line structures and indicate the more energetically favored conformation.
2. Draw the most stable conformation for trans-1-ethyl-3-methylcyclohexane using bond-line structures.
3. Draw the most stable conformation for trans-1-t-butyl-4-methylcyclohexane using bond-line structures.
4. Draw the most stable conformation fo trans-1-isopropyl-3-methylcyclohexane.
5. Can a ‘ring flip’ change a cis-disubstituted cyclohexane to trans? Explain.
6. Draw the two chair conformations of the six-carbon sugar mannose, being sure to clearly show each non-hydrogen substituent as axial or equatorial. Predict which conformation is likely to be more stable, and explain why.
Solutions
4.
The bulkier isopropyl groups is in the equatorial position.
5. No. In order to change the relationship of two substituents on a ring from cis to trans, you would need to break and reform two covalent bonds. Ring flips involve only rotation of single bonds.
6. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.08%3A_Conformations_of_Disubstituted_Cyclohexanes.txt |
Objective
After completing this section, you should be able to draw the structures and construct molecular models of simple polycyclic molecules.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bridgehead carbon atom
• polycyclic molecule
Study Notes
A bridgehead carbon atom is a carbon atom which is shared by at least two rings. The hydrogen atom which is attached to a bridgehead carbon may be referred to as a bridgehead hydrogen.
Note that bicyclo[2.2.1]heptane is the systematic name of norborane. You need not be concerned over the IUPAC name of norbornane. The nomenclature of compounds of this type is beyond the scope of this course.
Nomenclature of Bicyclic Ring Systems
There are many hydrocarbons and hydrocarbon derivatives with two rings having common carbon atoms. There are three main ways that the two rings can be connected. The first is called a fused bicyclic ring structure where the two rings share a covalent bond and a have two bridgehead carbons (marked in red on the structures below). A bridgehead is defined as a carbon that is part of two or more rings. Hydrogens attached to bridge head carbons are often referred to as bridge head hydrogens. The two rings can also be connected by a bridge containing one or more carbons to form a bridged bicyclic molecule. Lastly, the two rings can be joined with a singe bridge head carbon to form spiro bicyclic molecules.
Bicyclic Isomers of C10H18
Naming Fused and Bridged Compounds
Fused and bridged bicyclic compounds are follow similar naming conventions:
1. Count the total number of carbons in both rings. This is the parent name. (eg. ten carbons in the system would be decane)
2. Count the number of carbons between the bridgeheads, then place the numbers in square brackets in descending order separated by periods. Fused and bridged bycyclic compounds should have three numbers such as [2.2.0]. For fused compounds one of the numbers should be zero.
3. Place the word bicyclo at the beginning of the name.
Examples with carbons and hydrogens explicitly shown:
Naming Spiro Compounds
Spiro bicyclics are named using the same basic rules. Because there is only one bridgehead carbon only two numbers will be required in the brackets. Also, the word spiro is placed at the beginning.
Conformations in Bicyclic Ring Systems
As expected, the connection of two rings has defined effects on the possible conformations. However, the ideas previously discussed in this chapter can be used for conformational analysis. Fused rings have the possibility of two isomers where the bridgehead hydrogens are either cis or trans along the shared bond. These two isomers have significant differences in flexibility and stability as seen in bicyclo[4,4,0]decane more commonly known as decalin. If the positioning of the bridgehead hydrogens are shown in a fused ring the prefix cis or trans should be included in the name.
The trans-isomer is the easiest to describe because the fusion of the two rings creates a rigid, roughly planar, structure made up of two chair conformations. Unlike cyclohexane, the two rings cannot flip from one chair form to another. Accordingly, the orientation of the any substituents is fixed in either an axial or equatorial position in trans-decalin. This means that the C-C bonds coming away from the fused edge are held in equatorial positions relative to each ring thus preventing the possibility of any 1,3-diaxial interactions occurring between ring atoms.
Interactive Element
The 3D Structure of Trans-Decalin
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The two rings in cis-decalin are also both held in a chair conformations. In comparison, the chair-chair forms of cis-decalin are relatively flexible, and inversion of both rings at once occurs fairly easily.
Interactive Element
The 3D Structure of Cis-Decalin
The flexibility of cis-decalin allows for a substituent to interconvert between axial and equatorial conformations. In much the same fashion as cyclohexane, equatorial substituents tend to create less steric strain and create a more stable conformer.
A major difference in cis-decalin is the fact that one of C-C bonds coming away from the fused edge is held an an axial position. This is true in both ring-flip conformations. This axial C-C bond causes 1,3-diaxial interactions to occur in cis-decalin making it roughly 8.4 kJ/mol less stable than trans-decalin. This amount of 1,3-diaxial steric strain is roughly equivalent to that of an ethyl substituent attached to a cyclohexane ring (8.0 kJ/mol)
Bicyclic compounds with a bridge typically have very little flexibility and are often held in a ridged conformation. The molecule norbornane represent a cyclohexane ring connected by a single carbon bridge.
Interactive Element
The 3D Structure of Norbornane
Norbornane is estimated to have 72 kJ/mol of ring strain which can be understood when viewing the contained rings. The carbon bridge in norbornane holds the cyclohexane ring at the bottom in a boat conformation creating torsional strain from eclipsing bonds along the edge.
Also, the carbon bridge forms a cyclopentane ring (shown in red below making up the right side of the structure) with increased angle strain throughout the whole molecule.
Polycyclic Systems in Nature
Fused ring systems like decalin are very common in natural products. In fact, similar ring systems are found in steroids, which are an important class of lipids. Steroids generally have structures that include three six-membered rings and a five-membered ring connected by three fused bonds. Most natural steroids have a trans configuration at all three fusion points. This tends to give steroids a rigid and semi-flat structure.
Sex hormones are an example of steroids. The primary male hormone, testosterone, is responsible for the development of secondary sex characteristics. Two female sex hormones, progesterone and estrogen (or estradiol) control the ovulation cycle. Notice that the male and female hormones have only slight differences in structures, but yet have very different physiological effects. Testosterone promotes the normal development of male genital organs and is synthesized from cholesterol in the testes. It also promotes secondary male sexual characteristics such as deep voice, facial and body hair.
Interactive Element
The 3D Structure of Estradiol
The best known and most abundant steroid in the body is cholesterol. Cholesterol is formed in brain tissue, nerve tissue, and the blood stream. It is the major compound found in gallstones and bile salts. Cholesterol also contributes to the formation of deposits on the inner walls of blood vessels. These deposits harden and obstruct the flow of blood. This condition, known as atherosclerosis, results in various heart diseases, strokes, and high blood pressure.
Exercises
1)
i)
j)
3) The following molecule is cholic acid. Determine if the three fused bonds have a cis or trans configuration.
Solutions
1)
a) Bicyclo[2.1.1]hexane
b) Bicyclo[3.2.1]octane
c) Bicyclo[2.1.0]pentane (more commonly called "housane")
d) Bicyclo[2.2.2]octane
e) cis-Bicyclo[3.3.0]octane
f) cis-Bicyclo[1.1.0]butane
g) Bicyclo[1.1.1]pentane
h) Bicyclo[4.3.3]dodecane
i) Spiro[5.2]octane
j) Spiro[3.3]heptane
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.09%3A_Conformations_of_Polycyclic_Molecules.txt |
Concepts & Vocabulary
4.1: Naming Cycloalkanes
• Cycloalkanes are saturated hydrocarbons that have the generic formula CnH2n , where n is the number of carbons in the ring.
• The IUPAC rules for naming cycloalkanes is very similar to the rules used for naming alkanes.
4.2: Cis-Trans Isomerism in Cycloalkanes
• Stereoisomers are molecules that have the same molecular formula, the same atom connectivity, but they differ in the relative spatial orientation of the atoms.
• Di-substituted cycloalkanes exhibit cis- / trans- stereoisomerism. The cis- isomer has both substituents on the same face of the ring, while the trans- isomer has groups on opposite faces of the ring.
4.3: Stability of Cycloalkanes - Ring Strain
• Ring strain is the total strain in a ring due to torsional strain, steric strain and angle strain.
• Angle strain is when the C-C-C bond angles in rings are different than 109.5o, the optimal bond angle for sp3 hybridized carbons.
• Ring strain causes small cycloalkanes, like cyclopropane and cyclobutane, to be much less stable than other cycloalkanes.
4.4: Conformations of Cycloalkanes
• Cyclopentane has less ring strain than cyclopropane and cyclobutane, because its ring carbons have more flexibility to rotate away from planarity, resulting in lower angle and torsional strains.
4.5: Conformations of Cyclohexane
• Cyclohexane has significantly lower ring strain than smaller cycloalkanes, because cyclohexane can adopt non-planar structures, which minimize angle strain and torsional strain.
• The common non-planar structures of cyclohexane are the boat, twist-boat, and chair conformations. The most stable, and hence, the most common, is the chair conformation.
4.6: Axial and Equatorial Bonds in Cyclohexane
• The two chair conformations of cyclohexane interconvert rapidly at room temperature in a process called chair flip or ring flip.
• In the chair conformation of cyclohexane, of the two groups attached to each ring carbon, one of the groups occupies the axial position, while the other group occupies the equatorial position.
• A group that was axial will switch to the equatorial position during a ring flip, and vice versa.
4.7: Conformations of Monosubstituted Cyclohexanes
• To minimize the steric effects of 1,3-diaxial interactions, the single group on a monosubstituted cyclohexane ring will prefer to be in the equatorial position over the axial position. The larger the group, the greater is the preference shifts.
4.8: Conformations of Disubstituted Cyclohexanes
• The preference for large groups to be in the equatorial position effects the relative stability of the cis and trans isomers of disubstituted cyclohexanes. Conformational analysis is the process used to determine which isomer, cis or trans, is most stable.
4.9: Conformations of Polycyclic Molecules
Summary Problems
Exercise \(1\)
The following molecule, quinic acid, is a natural product that can be obtained from a variety of sources including the coffee bean. Draw both chair conformations for this molecule, identify each substituent in both structures as axial or equatorial, and clearly indicate which chair conformation is the most stable.
Answer
The circled conformation is more stable because it has more equatorial substituents (3 versus 2) and the largest group (the carboxylic acid) is equatorial.
Exercise \(2\)
Convert the following name to a skeletal structure: cis-1-t-butyl-2-ethylcyclohexane. Then, draw this molecule in a chair conformation and perform a ring flip. Circle the most stable of the two conformations.
Answer
Remember that, due to its large size, the t-butyl substituent locks the cyclohexane ring into one conformation with the t-butyl in the equatorial position. Thus, this isn't an equilibrium. It exists only as the circled conformation.
Exercise \(3\)
Convert the following name to a skeletal structure: trans-3-isobutylcyclohexanol. Then, draw the two chair conformations, label substituents as axial or equatorial, and circle the more stable conformation.
Answer
The circled molecule is most stable because the larger substituent is equatorial.
Skills to Master
• Skill 4.1 Be able to name and draw cycloalkanes
• Skill 4.2 Identify and draw the cis- and trans- stereoisomers of disubstituted cycloalkanes.
• Skill 4.3 Determine the effects of torsional strain, steric strain, and angle strain on the overall ring strain of a cycloalkane.
• Skill 4.4 Draw the chair conformers of cyclohexane.
• Skill 4.5 Draw and identify the axial and equatorial positions in a chair conformer of cyclohexane and its ring-flip conformer.
• Skill 4.6 Use conformational analysis to determine the most stable stereoisomer in disubstituted and polysubstituted cyclohexanes.
Contributors
• Dr. Kelly Matthews, Harrisburg Area Community College
• Kevin M. Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.S%3A__Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry_%28Summary%29.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.
• 5.0: Chapter Objectives and Introduction
• 5.1: 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.
• 5.2: The Reason for Handedness in Molecules - Chirality
Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands. Your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image.
• 5.3: 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.
• 5.4: 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
• 5.5: 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.
• 5.6: Diastereomers
Diastereomers are two molecules which are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers. Unlike enantiomers which are mirror images of each other and non-superimposable, diastereomers are not mirror images of each other and non-superimposable. Diastereomers can have different physical properties and reactivity. They have different melting points and boiling points and different densities.
• 5.7: 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.
• 5.8: 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.
• 5.9: A Review of Isomerism
• 5.10: Chirality at Nitrogen, Phosphorus, and Sulfur
• 5.11: 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.
• 5.12: Chirality in Nature and Chiral Environments
• 5.S: Stereochemistry at Tetrahedral Centers (Summary)
• 5.xx: Enantiomers and Diastereomers
Thumbnail: Two enantiomers of a generic amino acid that are chiral. (Public Domain; unknonw author via Wikipedia)
05: Stereochemistry at Tetrahedral Centers
The opposite of chiral is achiral. Achiral objects are superimposable with their mirror images. If the molecules are superimposable, they are identical to each other. For example, two pieces of paper are achiral. In contrast, chiral objects, like our hands, are non-superimposable mirror images of each other. Try to line up your left hand perfectly with your right hand, so that the palms are both facing in the same directions. Spend about a minute doing this. Do you see that they cannot line up exactly?
The same thing applies to some molecules. A chiral molecule has a mirror image that cannot line up with it perfectly - the mirror images are non-superimposable. This pair of non-superimposable mirror image molecules are called enantiomers. But why are chiral molecules so interesting? Just like your left hand will not fit properly in your right glove, one of the enantiomers of a molecule may not work the same way in your body, as the other. It turns out that many of the biological molecules such as our DNA, amino acids and sugars, are chiral molecules. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.00%3A_Chapter_Objectives.txt |
Objectives
After completing this section, you should be able to
1. use molecular models to show that only a tetrahedral carbon atom satisfactorily accounts for the lack of isomerism in molecules of the type CH2XY, and for the existence of optical isomerism in molecules of the type CHXYZ.
2. determine whether two differently oriented wedge-and-broken-line structures are identical or represent a pair of enantiomers.
Key Terms
Make certain that you can define, and use in context, the key term below.
• enantiomer
Study Notes
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. The word “chiral” was derived from the Greek word for hand, because our hands are a good example of chirality since they are non-superimposable mirror images of each other.
Chiral Molecules
The term chiral, from the Greek work for 'hand', refers to anything which cannot be superimposed on its own mirror image. Certain organic molecules are chiral meaning that they are not superimposable on their mirror image. Chiral molecules contain one or more chiral centers, which are almost always tetrahedral (sp3-hybridized) carbons with four different substituents. Consider the molecule A below: a tetrahedral carbon, with four different substituents denoted by balls of four different colors.
The mirror image of A, which we will call B, is drawn on the right side of the figure, and an imaginary mirror is in the middle. Notice that every point on A lines up through the mirror with the same point on B: in other words, if A looked in the mirror, it would see B looking back.
Now, if we flip compound A over and try to superimpose it point for point on compound B, we find that we cannot do it: if we superimpose any two colored balls, then the other two are misaligned.
A is not superimposable on its mirror image (B), thus by definition A is a chiral molecule. It follows that B also is not superimposable on its mirror image (A), and thus it is also a chiral molecule.
A and B are called stereoisomers or optical isomers: molecules with the same molecular formula and the same bonding arrangement, but a different arrangement of atoms in space. Enantiomers are pairs of stereoisomers which are mirror images of each other: thus, A and B are enantiomers. It should be self-evident that a chiral molecule will always have one (and only one) enantiomer: enantiomers come in pairs. Enantiomers have identical physical properties (melting point, boiling point, density, and so on). However, enantiomers do differ in how they interact with polarized light (we will learn more about this soon) and they may also interact in very different ways with other chiral molecules - proteins, for example. We will begin to explore this last idea in later in this chapter, and see many examples throughout the remainder of our study of biological organic chemistry.
The Many Synonyms of the Chiral Carbon
Be aware - all of the following terms can be used to describe a chiral carbon.
chiral carbon = asymmetric carbon = optically active carbon = stereo carbon = stereo center = chiral center
Let's apply our chirality discussion to real molecules.
Consider 2-butanol, drawn in two dimensions below.
Carbon #2 is a chiral center: it is sp3-hybridized and tetrahedral (even though it is not drawn that way above), and the four substituents attached to is are different: a hydrogen (H) , a methyl (-CH3) group, an ethyl (-CH2CH3) group, and a hydroxyl (OH) group. If the bonding at C2 of 2-butanol is drawn in three dimensions and this structure called A. Then the mirror image of A can be drawn to form structure B.
When we try to superimpose A onto B, we find that we cannot do it. Because structure A and B are not superimposable on their mirror image they are both chiral molecules. Because A and B are different due only to the arrangement of atoms in space they are stereoisomers. Because A and B are mirror images of each other they are also enantiomers. When looking at simplified line structures is clear that there are two distinct ways of drawing 2-butanol which only differ in their spatial arrangement around a chiral carbon.
The 3D Structures of the Two Enantiomers of 2-Butanol
For comparison, 2-propanol, is an achiral molecule because is lacks a chiral carbon. Carbon #2 is bonded to two identical substituents (methyl groups), and so it is not a chiral carbon. Being achiral means that 2-propanol should be superimposable on its mirror image which is shown in the figure below. A more detailed explaination on why 2-propanol is achiral will be given in the next section.
2-propanol is achiral:
Stereoisomers
Stereoisomers have been defined as molecules with the same connectivity but different arrangements of the atoms in space. It is important to note that there are two types of stereoisomers: geometric and optical.
Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral. 25.7.1a" id="MathJax-Element-1-Frame" role="presentation" style="position:relative;" tabindex="0">
Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement. Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds.mers have the same connectivity, but different arrangements of atoms in space. Geometric isomers will be discussed in more detain in Sections 7.4 and 7.5.
Exercise $1$
Identify the following molecules as chiral or achiral.
Answer
a) chiral (4 different groups off C)
b) achiral (2 identical -CH3 substituents off central C)
c) achiral (2 identical -CH2CH3 substituents off central C)
d) achiral (2 identical CH3 substituents off carbon 2)
e) chiral (4 different groups off carbon 2)
f) achiral (2 identical CH3 substituents off central C)
Exercise $2$
Determine if the following sets of compounds in each group are enantiomers or the same compound.
Answer
a) enantiomers – non superimposable mirror images
b) same compound – when you rotate the molecule on the right it is identical to the one on the left
c) enantiomers – non superimposable mirror images
d) enantiomers – non superimposable mirror images | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.01%3A_Enantiomers_and_the_Tetrahedral_Carbon.txt |
Objectives
After completing this section, you should be able to
1. determine whether or not a compound is chiral, given its Kelulé, condensed or shorthand structure, with or without the aid of molecular models.
2. label the chiral centres (carbon atoms) in a given Kelulé, condensed or shorthand structure.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• achiral
• chiral
• chiral (stereogenic) centre
• plane of symmetry
Symmetry and Chirality
Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands. Your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks 25.7.1b in the figure below. 25.7.1b.
An an important questions is why is one chiral and the other not? The answer is that the flask has a plane of symmetry and your hand does not. A plane of symmetry is a plane or a line through an object which divides the object into two halves that are mirror images of each other. When looking at the flask, a line can be drawn down the middle which separates it into two mirror image halves. However, a similar line down the middle of a hand separates it into two non-mirror image halves. This idea can be used to predict chirality. If an object or molecule has a plane of symmetry it is achiral. If if lacks a plane of symmetry it is chiral.
Symmetry can be used to explain why a carbon bonded to four different substituents is chiral. When a carbon is bonded to fewer than four different substituents it will have a plane of symmetry making it achiral. A carbon atom that is bonded to four different substituents loses all symmetry, and is often referred to as an asymmetric carbon. The lack of a plane of symmetry makes the carbon chiral. The presence of a single chiral carbon atom sufficient to render the molecule chiral, and modern terminology refers to such groupings as chiral centers or stereo centers.
An example is shown in the bromochlorofluoromethane molecule shown in part (a) of the figure below. This carbon, is attached to four different substituents making it chiral. which is often designated by an asterisk in structural drawings. If the bromine atom is replaced by another chlorine to make dichlorofluoromethane, as shown in part (b) below, the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Upon comparison, bromochlorofluoromethane lacks a plane of symmetry while dichlorofluoromethane has a plane of symmetry.
Identifying Chiral carbons
Identifying chiral carbons in a molecule is an important skill for organic chemists. The presence of a chiral carbon presents the possibility of a molecule having multiple stereoisomers. Most of the chiral centers we shall discuss in this chapter are asymmetric carbon atoms, but it should be recognized that other tetrahedral or pyramidal atoms may become chiral centers if appropriately substituted. Also, when more than one chiral center is present in a molecular structure, care must be taken to analyze their relationship before concluding that a specific molecular configuration is chiral or achiral. This aspect of stereoisomerism will be treated later. Because an carbon requires four different substituents to become asymmertric, it can be said, with few exceptions, that sp2 and sp hybridized carbons involved in multiple bonds are achiral. Also, any carbon with more than one hydrogen, such as a -CH3 or -CH2- group, are also achiral.
Looking for planes of symmetry in a molecule is useful, but often difficult in practice. It is difficult to illustrate on the two dimensional page, but you will see if you build models of these achiral molecules that, in each case, there is at least one plane of symmetry, where one side of the plane is the mirror image of the other. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions, the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral.
Determining if a carbon is bonded to four distinctly different substituents can often be difficult to ascertain. Remember even the slightest difference makes a substituent unique. Often these difference can be distant from the chiral carbon itself. Careful consideration and often the building of molecular models may be required. A good example is shown below. It may appear that the molecule is achiral, however, when looking at the groups directly attached to the possible chiral carbon, it is clear that they all different. The two alkyl groups are differ by a single -CH2- group which is enough to consider them different.
Example $1$
Predict if the following molecule would be chiral or achiral:
Answer
Achiral. When determining the chirality of a molecule, it best to start by locating any chiral carbons. An obvious candidate is the ring carbon attached to the methyl substituent. The question then becomes: does the ring as two different substituents making the substituted ring carbon chiral? With an uncertainty such as this, it is then helpful try to identify any planes of symmetry in the molecule. This molecule does have a plane of symmetry making the molecule achiral. The plane of symmetry would be easier see if the molecule were view from above. Typically, monosubstituted cycloalkanes have a similar plane of symmetry making them all achiral.
Exercise 5.2.1
Determine if each of the following molecules are chiral or achiral. For chiral molecules indicate any chiral carbons.
Answer
Explanation
Structures F and G are achiral. The former has a plane of symmetry passing through the chlorine atom and bisecting the opposite carbon-carbon bond. The similar structure of compound E does not have such a symmetry plane, and the carbon bonded to the chlorine is a chiral center (the two ring segments connecting this carbon are not identical). Structure G is essentially flat. All the carbons except that of the methyl group are sp2 hybridized, and therefore trigonal-planar in configuration. Compounds C, D & H have more than one chiral center, and are also chiral.
Note
In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women.
Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects.
Researchers later realized the problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms.
One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp3-hybridized carbon. These two forms of thalidomide are stereoisomers. If you make models of the two stereoisomers of thalidomide, you will see that they too are mirror images, and cannot be superimposed.
As a historical note, thalidomide was never approved for use in the United States. This was thanks in large part to the efforts of Dr. Frances Kelsey, a Food and Drug officer who, at peril to her career, blocked its approval due to her concerns about the lack of adequate safety studies, particularly with regard to the drug's ability to enter the bloodstream of a developing fetus. Unfortunately, though, at that time clinical trials for new drugs involved widespread and unregulated distribution to doctors and their patients across the country, so families in the U.S. were not spared from the damage caused.
Very recently a close derivative of thalidomide has become legal to prescribe again in the United States, with strict safety measures enforced, for the treatment of a form of blood cancer called multiple myeloma. In Brazil, thalidomide is used in the treatment of leprosy - but despite safety measures, children are still being born with thalidomide-related defects.
Example 5.2.2
Label the molecules below as chiral or achiral, and locate all stereocenters.
Answer
Exercise 5.2.2
1) For the following compounds, star (*) each chiral center, if any.
2) Explain why the following compound is chiral.
3) Determine which of the following objects is chiral.
a) A Glove.
b) A nail.
c) A pair of sunglasses.
d) The written word "Chiral".
4) Place an "*" by all of the chrial carbons in the following molecules.
a)
Erythrose, a four carbon sugar.
b) Isoflurane, an anestetic. Bright green = Chlorine, Pale green = Fluorine.
Answer
1)
2) Though the molecule does not contain a chiral carbon, it is chiral as it is non-superimposable on its mirror image due to its twisted nature (the twist comes from the structure of the double bonds needing to be at 90° angles to each other, preventing the molecule from being planar).
3)
a) Just as hands are chiral a glove must also be chiral.
b) A nail has a plane of symmetry which goes down the middle making it a achiral.
c) A pair of sunglasses has a plane of symmetry which goes through the nose making it achiral.
d) Most written words are chiral. Look one in a mirror to confirm this.
4
a)
b)
Exercise 5.2.3
Circle all of the carbon stereocenters in the molecules below.
Answer
Exercise 5.2.4
Circle all of the carbon stereocenters in the molecules below.
Answer
Here are some more examples of chiral molecules that exist as pairs of enantiomers. In each of these examples, there is a single stereocenter, indicated with an arrow. (Many molecules have more than one stereocenter, but we will get to that that a little later!)
Here are some examples of molecules that are achiral (not chiral). Notice that none of these molecules has a stereocenter.
It is difficult to illustrate on the two dimensional page, but you will see if you build models of these achiral molecules that, in each case, there is at least one plane of symmetry, where one side of the plane is the mirror image of the other. Chirality is tied conceptually to the idea of asymmetry, and any molecule that has a plane of symmetry cannot be chiral. When looking for a plane of symmetry, however, we must consider all possible conformations that a molecule could adopt. Even a very simple molecule like ethane, for example, is asymmetric in many of its countless potential conformations – but it has obvious symmetry in both the eclipsed and staggered conformations, and for this reason it is achiral.
Looking for planes of symmetry in a molecule is useful, but often difficult in practice. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions (see section 3.7B), the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Carbon stereocenters are also referred to quite frequently as chiral carbons.
When evaluating a molecule for chirality, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not stereocenters – look, for example, at the drawings of glycine and citrate in the figure above. Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a stereocenter.
Other elements in addition to carbon can be stereocenters. The phosphorus center of phosphate ion and organic phosphate esters, for example, is tetrahedral, and thus is potentially a stereocenter.
We will see in chapter 10 how researchers, in order to investigate the stereochemistry of reactions at the phosphate center, incorporated sulfur and/or 17O and 18O isotopes of oxygen (the ‘normal’ isotope is 16O) to create chiral phosphate groups. Phosphate triesters are chiral if the three substituent groups are different.
Asymmetric quaternary ammonium groups are also chiral. Amines, however, are not chiral, because they rapidly invert, or turn ‘inside out’, at room temperature.
Exercise 5.2.5
Label the molecules below as chiral or achiral, and circle all stereocenters.
a) fumarate (a citric acid cycle intermediate)
b) malate (a citric acid cycle intermediate)
b) malate (a citric acid cycle intermediate)
Answer
a) achiral (no stereocenters)
b) chiral
c) chiral
Exercise 5.2.6
Label the molecules below as chiral or achiral, and circle all stereocenters.
a) acetylsalicylic acid (aspirin)
b) acetaminophen (active ingredient in Tylenol)
c) thalidomide (drug that caused birth defects in pregnant mothers in the 1960’s)
Answer
a) achiral (no stereocenters)
b) achiral (no stereocenters)
c) chiral
Exercise 5.2.7
Draw both enantiomers of the following chiral amino acids.
a) Cysteine
b) Proline
Answer
Exercise 5.2.8
Draw both enantiomers of the following compounds from the given names.
a) 2-bromobutane
b) 2,3-dimethyl-3-pentanol
Answer
Exercise 5.2.9
Which of the following body parts are chiral?
a) Hands b) Eyes c) Feet d) Ears
Answer
a) Hands- chiral since the mirror images cannot be superimposed (think of the example in the beginning of the section)
b) Eyes- achiral since mirror images that are superimposable
c) Feet- chiral since the mirror images cannot be superimposed (Does your right foot fit in your left shoe?)
d) Ears- chiral since the mirror images cannot be superimposed
Exercise 5.2.10
Circle the chiral centers in the following compounds.
Answer
Exercise 5.2.11
Identify the chiral centers in the following compounds.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.02%3A_The_Reason_for_Handedness_in_Molecules_-_Chirality.txt |
Objectives
After completing this section, you should be able to
1. describe the nature of plane-polarized light.
2. describe the features and operation of a simple polarimeter.
3. calculate the specific rotation of a compound, given the relevant experimental data.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• analyzer
• dextrorotatory
• levorotatory
• optically active
• plane-polarized light
• polarimeter
• polarizer
• specific rotation, $[\alpha]^{20}_{D}$
Study Notes
A polarizer is a device through which only light waves oscillating in a single plane may pass. A polarimeter is an instrument used to determine the angle through which plane-polarized light has been rotated by a given sample. You will have the opportunity to use a polarimeter in the laboratory component of the course. An analyzer is the component of a polarimeter that allows the angle of rotation of plane-polarized light to be determined.
Specific rotations are normally measured at 20°C, and this property may be indicated by the symbol $[\alpha]^{20}_{D}$. Sometimes the solvent is specified in parentheses behind the specific rotation value, for example,
$[\alpha]^{20}_{D} = +12^o\,(\text{chloroform}) \nonumber$
For liquids, the specific rotation may be obtained using the neat liquid rather than a solution; in such cases the formula is
$[ α ]_D^{\text{temp}} \,(\text{neat})= α \times l \times d \nonumber$
where $α$ is the observed rotation, $l$ is the path length of the cell (measured in decimetres, dm), and $d$ is the density of the liquid.
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.
Polarimetry
Plane-polarized light is created by passing ordinary light through a polarizing device, which may be as simple as a lens taken from polarizing sun-glasses. Such devices transmit selectively only that component of a light beam having electrical and magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by an instrument called a polarimeter (Figure $1$).
Monochromatic (single wavelength) light, is polarized by a fixed polarizer next to the light source. A sample cell holder is located in line with the light beam, followed by a movable polarizer (the analyzer) and an eyepiece through which the light intensity can be observed. In modern instruments an electronic light detector takes the place of the human eye. In the absence of a sample, the light intensity at the detector is at a maximum when the second (movable) polarizer is set parallel to the first polarizer (α = 0º). If the analyzer is turned 90º to the plane of initial polarization, all the light will be blocked from reaching the detector.
Chemists use polarimeters to investigate the influence of compounds (in the sample cell) on plane polarized light. Samples composed only of achiral molecules (e.g. water or hexane), have no effect on the polarized light beam. However, if a single enantiomer is examined (all sample molecules being right-handed, or all being left-handed), the plane of polarization is rotated in either a clockwise (positive) or counter-clockwise (negative) direction, and the analyzer must be turned an appropriate matching angle, α, if full light intensity is to reach the detector. In the above illustration, the sample has rotated the polarization plane clockwise by +90º, and the analyzer has been turned this amount to permit maximum light transmission.
The observed rotations ($\alpha$) of enantiomers are opposite in direction. One enantiomer will rotate polarized light in a clockwise direction, termed dextrorotatory or (+), and its mirror-image partner in a counter-clockwise manner, termed levorotatory or (–). The prefixes dextro and levo come from the Latin dexter, meaning right, and laevus, for left, and are abbreviated d and l respectively. If equal quantities of each enantiomer are examined , using the same sample cell, then the magnitude of the rotations will be the same, with one being positive and the other negative. To be absolutely certain whether an observed rotation is positive or negative it is often necessary to make a second measurement using a different amount or concentration of the sample. In the above illustration, for example, α might be –90º or +270º rather than +90º. If the sample concentration is reduced by 10%, then the positive rotation would change to +81º (or +243º) while the negative rotation would change to –81º, and the correct α would be identified unambiguously.
Since it is not always possible to obtain or use samples of exactly the same size, the observed rotation is usually corrected to compensate for variations in sample quantity and cell length. Thus it is common practice to convert the observed rotation, $α$, to a specific rotation, by the following formula:
$[\alpha]_D = \dfrac{\alpha}{l c} \tag{5.3.1}$
where
• $[\alpha]_D$ is the specific rotation
• $l$ is the cell length in dm
• $c$ is the concentration in g/ml
• $D$ designates that the light used is the 589 line from a sodium lamp
Compounds that rotate the plane of polarized light are termed optically active. Each enantiomer of a stereoisomeric pair is optically active and has an equal but opposite-in-sign specific rotation. Specific rotations are useful in that they are experimentally determined constants that characterize and identify pure enantiomers. For example, the lactic acid enantiomers have the following specific rotations:
• Carvone from caraway: $[\alpha]^{20}_{D} = +62.5^o$ (this isomer may be referred to as (+)-carvone or d-carvone)
• Carvone from spearmint: $[\alpha]^{20}_{D} = -62.5^o$ (this isomer may be referred to as (–)-carvone or l-carvone)
and carvone enantiomers have the following specific rotations:
• Lactic acid from muscle tissue: $[\alpha]^{20}_{D} = +2.5^o$ (this isomer may be referred to as (+)-lactic acid or d-lactic acid)
• Lactic acid from sour milk: $[\alpha]^{20}_{D} = -2.5^o$ (this isomer may be referred to as (–)-lactic acid or l-lactic acid)
A 50:50 mixture of enantiomers has no observable optical activity. Such mixtures are called racemates or racemic modifications, and are designated (±). When chiral compounds are created from achiral compounds, the products are racemic unless a single enantiomer of a chiral co-reactant or catalyst is involved in the reaction. The addition of $\ce{HBr}$ to either cis- or trans-2-butene is an example of racemic product formation (the chiral center is colored red).
$\ce{CH3CH=CHCH3 + HBr -> (±) CH3CH2}\textcolor{red}{\ce{C}} \ce{HBrCH3} \nonumber$
Chiral organic compounds isolated from living organisms are usually optically active, indicating that one of the enantiomers predominates (often it is the only isomer present). This is a result of the action of chiral catalysts we call enzymes, and reflects the inherently chiral nature of life itself. Chiral synthetic compounds, on the other hand, are commonly racemates, unless they have been prepared from enantiomerically pure starting materials.
There are two ways in which the condition of a chiral substance may be changed:
1. A racemate may be separated into its component enantiomers. This process is called resolution.
2. A pure enantiomer may be transformed into its racemate. This process is called racemization.
Enantiomeric Excess
The "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%.
$\% \text { optical purity }=\frac{\text { specific rotation of mixture }}{\text { specific rotation of pure enantiomer }} \times 100\% \nonumber$
Because R and S enantiomers have equal but opposite optical activity, it naturally follows that a 50:50 racemic mixture of two enantiomers will have no observable optical activity. If we know the specific rotation for a chiral molecule, however, we can easily calculate the ratio of enantiomers present in a mixture of two enantiomers, based on its measured optical activity. When a mixture contains more of one enantiomer than the other, chemists often use the concept of enantiomeric excess (ee) to quantify the difference. Enantiomeric excess can be expressed as:
$e e=\frac{(\% \text { more abundant enantiomer }-50) \times 100\%}{50} \nonumber$
For example, a mixture containing 60% R enantiomer (and 40% S enantiomer) has a 20% enantiomeric excess of R: ((60-50) x 100) / 50 = 20 %.
Exercise $1$
The specific rotation of (S)-carvone is (+)61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-)23°. Which enantiomer is in excess, and what is its ee? What are the percentages of (R)- and (S)-carvone in the sample?
Answer
The observed rotation of the mixture is levorotary (negative, counter-clockwise), and the specific rotation of the pure S enantiomer is given as dextrorotary (positive, clockwise), meaning that the pure R enantiomer must be levorotary, and the mixture must contain more of the R enantiomer than of the S enantiomer.
Rotation (R/S Mix) = [Fraction(S) × Rotation (S)] + [Fraction(R) × Rotation (R)]
Let Fraction (S) = x, therefore Fraction (R) = 1 – x.
Rotation (R/S Mix) = x[Rotation (S)] + (1 – x)[Rotation (R)].
–23 = x(+61) + (1 – x)(–61)
Solve for x: x = 0.3114 and (1 – x) = 0.6885
Therefore the percentages of (R)- and (S)-carvone in the sample are 68.9% and 31.1%, respectively.
ee = [(% more abundant enantiomer – 50) × 100]/50. = [68.9 – 50) × 100]/50 = 37.8%.
Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration
Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration.
Separation of Chiral Compounds
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used of this technique is known as chiral resolution.
Exercise $2$
A 3.20 g sample of morphine ([α]D = -132) was dissolved in 10.0 mL of acetic acid ([α]D = 0). If it is put into a sample tube with a path length of 2.00 cm, what would be its observed rotation (α)?
Answer
The specific rotation, [α]D = (observed rotation, α (degrees))/ [(pathlength, l (dm)) x (concentration, c (g/cm3))] = α/(l x c)
Solving for α, α = [α]D x l x c
([α]D = -132) x (l = 2.00 cm = 0.200 dm) x (c = 3.20 g / 10.0 cm3 = 0.320 g/cm3)
α = -132 x 0.200 dm x 0.320 g/cm3 = -8.45 o
Exercise $3$
Is the morphine in the previous excercise dextrorotatory or levorotatory?
Answer
Since morphine has a (-) rotation, it indicates that it rotates light to the left (counterclockwise) and morphine is levorotatory.
Exercise $4$
Label the following compounds as dextrorotatory or levorotatory.
1. sucrose ([α]D = + 66.7)
2. cholesterol ([α]D = - 31.5)
3. cocaine ([α]D = - 16)
4. chloroform ([α]D = 0)
Answer
1. sucrose ([α]D = + 66.7) dextrorotatory
2. cholesterol ([α]D = - 31.5) levorotatory
3. cocaine ([α]D = - 16) levorotatory
4. chloroform ([α]D = 0) neither, not optically active
Exercise $\PageIndex{5a}$
The specific rotation of (S)-carvone is (+) 61o when measured neat (pure liquid sample with no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-) 23 o.
a) Which enantiomer is in excess?
Answer
Since the pure S enantiomer ((+) 61o) is dextrorotatory (positive, clockwise), the R enantiomer must be levorotatory. The observed rotation of the mixture is levorotatory since its negative (counterclockwise). This means the mixture must contain more of the R enantiomer than the S enantiomer.
Exercise $\PageIndex{5b}$
b) What are the percentages of (S)- and (R)- carvone in the sample mixture?
Answer
Optical rotation (α) of the (R/S mixture) = [fraction (S) x [α]D (S)] + [fraction (R) x [α]D (R)]
To determine the fraction of S and R, we make y = fraction (S) and 1 – y = fraction (R)
-23o = y x (61o) + (1 – y) x (-61o) solving for y: y = 0.3114 and (1-y) = 0.6885
Therefore the percentage of (S)-carvone is 31.1 % and (R)-carvone is 68.9 %
Exercise $\PageIndex{5c}$
c) What is the ee (enantiomeric excess)?
Answer
ee = [(% more abundant isomer – 50) x 100]/50 = [(68.9 – 50) x100]/50 = 37.8 % ee
Exercise $\PageIndex{6a}$
Determine the ee’s of the following from the percentages
• 95 % (R)- tartaric acid and 5.0 % (S)- tartaric acid
Answer
[(95 – 50) x 100] / 50 = 90 % ee (R)-tartaric acid
Exercise $\PageIndex{6b}$
Determine the ee’s of the following from the percentages
• 75 % (S)- limonene and 25 % (R)- limonene
Answer
[(75 – 50) x 100] / 50 = 50 % ee (S)- limonene
Exercise $\PageIndex{6c}$
Determine the ee’s of the following from the percentages
• 85 % (R) cysteine
Answer
(85 – 50) x 100] / 50 = 70 % ee (R)-cysteine
Exercise $\PageIndex{6d}$
Determine the ee’s of the following from the percentages
• 50 % (S) alanine
Answer
(50 – 50) x 100] / 50 = 0 % ee, racemic mixture
5.04: Pasteur's Discovery of Enantiomers
Objective
After completing this section, you should be able to discuss how the results of work carried out by Biot and Pasteur contributed to the development of the concept of the tetrahedral carbon atom.
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. As background for the following example, it is recommended that the section on crystal properties be reviewed. Tartaric acid, its potassium salt known in antiquity as "tartar", has served as the locus of several landmark events in the history of stereochemistry. In 1832 the French chemist Jean Baptiste Biot observed that tartaric acid obtained from tartar was optically active, rotating the plane of polarized light clockwise (dextrorotatory). An optically inactive, higher melting, form of tartaric acid, called racemic acid was also known.
A little more than a decade later, young Louis Pasteur conducted a careful study of the crystalline forms assumed by various salts of these acids. He noticed that under certain conditions, the sodium ammonium mixed salt of the racemic acid formed a mixture of enantiomorphic hemihedral crystals; a drawing of such a pair is shown below. Pasteur reasoned that the dissymmetry of the crystals might reflect the optical activity and dissymmetry of its component molecules. After picking the different crystals apart with a tweezer, he found that one group yielded the known dextrorotatory tartaric acid measured by Biot; the second led to a previously unknown levorotatory tartaric acid, having the same melting point as the dextrorotatory acid. Today we recognize that Pasteur had achieved the first resolution of a racemic mixture, and laid the foundation of what we now call stereochemistry.
Optical activity was first observed by the French physicist Jean-Baptiste Biot. He concluded that the change in direction of plane-polarized light when it passed through certain substances was actually a rotation of light, and that it had a molecular basis. His work was supported by the experimentation of Louis Pasteur. Pasteur observed the existence of two crystals that were mirror images in tartaric acid, an acid found in wine. Through meticulous experimentation, he found that one set of molecules rotated polarized light clockwise while the other rotated light counterclockwise to the same extent. He also observed that a mixture of both, a racemic mixture (or racemic modification), did not rotate light because the optical activity of one molecule canceled the effects of the other molecule. Pasteur was the first to show the existence of chiral molecules. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.03%3A_Optical_Activity.txt |
Objectives
After completing this section, you should be able to
1. assign Cahn-Ingold-Prelog priorities to a given set of substituents.
2. determine whether a given wedge-and-broken-line structure corresponds to an R or an S configuration, with or without the aid of molecular models.
3. draw the wedge-and-broken-line structure for a compound, given its IUPAC name, complete with R or S designation.
4. construct a stereochemically accurate model of a given enantiomer from either a wedge-and-broken-line structure or the IUPAC name of the compound, complete with R or S designation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• absolute configuration
• R configuration
• S configuration
Study Notes
When designating a structure as R or S, you must ensure that the atom or group with the lowest priority is pointing away from you, the observer. The easiest way to show this is to use the wedge-and-broken-line representation. You can then immediately determine whether you are observing an R configuration or an S configuration.
To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature.
Introduction
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. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer:
1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.
However, for non-laboratory purposes, it is beneficial to focus on the (R)/(S) system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule,at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.
Stereocenters are labeled (R) or (S)
The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as (R) or (S).
The Cahn-Ingold-Prelog rules of assign priorities the groups directly bonded to the chiral carbon. 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) is given a dash bond to indicate it points directly away from us. We then look at the three remaining substituents, which now appear to radiate toward us which is shown by using wedge bonds. If a curved arrow drawn from the highest to second-highest to third-highest ranked substituent is clockwise, we say that the chirality center has the (R) configuration (Latin rectus, meaning “right”). If an arrow from 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. Note, the (R) or (S) configurations represent the two enantiomers of a chiral molecule. The (R) or (S) configuration is often added as a prefix, in parenthesis, to a chiral molecule's name to indicate which enantiomer is being discussed (e.g., (R)-2-bromobutane). If more than chiral carbon is present in a chiral molecule, each carbon's number is included before the (R) or (S) configuration. Ex: (2R,4S,6R)-2-bromo-6-chloro-4-methylheptane.
Sequence Rules to Assign Priorities to Substituents
Before applying the (R) and (S) nomenclature to a stereocenter, the substituents must be prioritized according to the following rules:
Rule 1
First, examine at the atoms directly attached to the stereocenter of the compound. A atom with a higher atomic number takes precedence over a atom with a lower atomic number. Hydrogen is the lowest possible priority atom, because it has the lowest atomic number.
1. The atom with higher atomic number has higher priority (I > Br > Cl > S > P > F > O > N > C > H).
2. When comparing isotopes, the atom with the higher mass number has higher priority [18O > 16O or 15N > 14N or 13C > 12C or T (3H) > D (2H) > H].
Rule 2
If there are two or more substituents which have the same element directly attached to chiral carbon, proceed along the substituent chains until a point of difference is found. Determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority.
For example: an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has two hydrogen atoms and a carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl.
The "-H" (left) ranks lower than the "C-" (right) based on the relative molecular weights at the first point of difference.
Worked Exercise \(1\)
For the following pairs of substituents, determine which would have the higher and lower priority based on the Cahn-Ingold-Prelog rules. Explain your answer.
Answer
A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:
The "C-" (right) ranks higher than the "H-" (left) based on
the first point of difference and their relative atomic numbers.
However:
In this case, even though the bold carbon on the right structure has two
connections to a non-hydrogen atom (C), it is the lower priority.
This is because one of the atoms attached to the bold carbon on the left molecule
ranks higher than any of the atoms attached to the bold carbon on the right structure,
since Br has a higher atomic number than C.
Caution!!
Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.
When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.
Rule 3
For assigning priority, multiple bonds are treated as if each bond of the multiple bond is bonded to a unique atom. For example, an alkene substituent (CH2=CH-) has higher priority than an ethyl substituent (CH3CH2-). The alkene carbon priority is "two" bonds to carbon atoms and one bond to a hydrogen atom compared with the ethyl carbon that has only one bond to a carbon atom and two bonds to two hydrogen atoms. Similarly, alkyne substituent (HCC-) would have an even higher priority because the alkyne carbon is treated as if it is bonded to three carbons. This method remains the same with compounds containing a carbonyl (C=O) group. The carbon of an aldehyde substituent (O=CH-) is treated as if it is bonded to a hydrogen and two oxygen atoms.
Determining (R) or (S) Configuration Using a Molecular Model
In order to demonstrate how to determine the (R)/(S) configuration of the chiral carbon in the following molecule using molecular models, first construct a model of the bromoethanol structure:
First make a molecular model of a tetrahedral carbon with four different substituents. In many cases, this will appear as a carbon with four bonds with a different colored ball attached to each bond.
For the molecule in question, determine the location of the chiral carbon and assign CIP priorities to the substituents. In this case, Br gets the highest priority because it has the highest atomic number. The O in the OH substituent gets priority 2 and the C in CH3 gets priority 3. Lastly, H gets the lowest priority, 4, because it has the smallest atomic number.
Now take your molecular model and orientate it to match the molecule in question. Remember in the dash/wedge representation, two regular bonds are in the plane of the page. The wedge bond is coming toward you and the dashed bond is going away from you. If you were to hold a piece of paper directly in front of you, the substituents with the regular bond should both be touching the piece of paper. The dashed bond should be pointing behind the piece of paper and the wedge bond should be pointing in front.
In this structure, the bromine is going away from you, the hydrogen is coming toward you and the hydroxide and methyl groups are in the plane of the page.
Then based on the position, assign each substituent on the chiral carbon a colored ball on your molecular model. In this case, bromine is going away so it is assigned the green ball. The hydrogen is coming toward you so it is assigned the blue ball. The last two substituents are in the plane of the page, however, the CH3 is positioned higher so it is assigned the red ball which leaves OH being assigned the black ball.
Lastly, grab onto the ball for the lowest priority substitutent, in this case the blue one, and point the other three substituents towards you. The three bonds should be angled towards you as if they all have wedge bonds. Assign the original substituents and their corresponding CIP priorities to the three colored balls. The green ball was assigned to bromine which was given priority one. The OH was assigned to the black ball and given priority two. The CH3 was assigned to the red ball and given priority three. In this case the priorities are going counter clockwise so the chiral carbon has an (S) configuration.
Determining (R) or (S) Configuration Without a Molecular Model
If a molecular model cannot be used there are a couple of simple methods which can be applied if the dash/wedge bond system is being used.
After assigning CIP priorities, if the lowest priority substituent (4) is on the dash bond the configuration of substituents 1-3 can be assigned directly. As shown in the figure below, the configuration of substituents 1-3 does not change when moving to sight down the bond of substituent 4. In both cases, substituents 1-3 are ordered in a counterclockwise fashion which gives the chiral carbon an (S) configuration.
The opposite is true if the lowest priority substituent (4) is on the wedge bond. As shown in the figure below, the configuration of substituents 1-3 is inverted when moving to sight down the bond of substituent 4. When the lowest priority substituent is on the wedge bond, the configuration of substituents 1-3 can be assigned directly only if the direction is inverted. i.e. clockwise = (S) and counterclockwise = (R).
With the lowest priority group in front, drawing an arc from 1 to 2 to 3 gives the reverse of the configuration.
However, if the lowest priority substituent is on one of the regular bonds when the dash/wedge system is being used then configurations are best assigned by changing perspectives. This method can also be used if the three-dimensional configuration of the chiral carbon is represented. First, locate the chiral carbon and assign CIP priorities to its substituents. Then while perceiving the drawn molecule as a three-dimensional image, mentally change your perspective such that you are looking down the bond between the chiral carbon and the lowest CIP ranked substituent (#4). If done correctly, the bonds for substituents 1-3 should be coming towards you as wedge bonds. You can then follow the direction of the CIP priority numbers to determine the (R)/(S) configuration of the chiral carbon.
Locate the chiral carbon and assign CIP priorities to its substituents.
Mentally sight down the bond between the chiral carbon and the lowest CIP ranked substituent.
This bond is shown in magenta.
clockwise
(R)-configuration
Follow the direction of the CIP priority numbers to determine the (R)/(S) configuration.
Drawing the Structure of a Chiral Molecule from its Name
Draw the structure of (S)-2-Bromobutane:
1) Draw the basic structure of the molecule and determine the location of the chiral carbon.
2) Determine the chiral carbon's substituents and assign them a CIP priority.
-H (Priority 4)
-CH3 (Priority 3)
-CH2CH3 (Priority 2)
-Br (Priority 1)
3) Draw the chiral carbon in a dash/wedge form and add the lowest priority substituent to the dash bond. In this case, the lowest priority substituent is -H.
4) Add the remaining substituents in a clockwise fashion for (R) and a counterclockwise fashion for (S).
The molecule posed in this question has an (S) configuration so the remaining substituents are added in a counterclockwise fashion.
Exercise \(1\)
1) Orient the following so that the least priority (4) atom is paced behind, then assign stereochemistry ((R) or (S)).
2) Draw (R)-2-bromobutan-2-ol.
3) Assign (R)/(S) to the following molecule.
4) Which in the following pairs would have a higher CIP priority?
1. -H or -Cl
2. -Br or -I
3. -CH2OH or -OCH3
4. -CH2CH3 or -CH=CH2
5. -NH2 or -OH
5) Rank the following substituents in order of their CIP priority:
1. -H, -OCH3, -CH2OH, -OH
2. -OH, -CO2H, -CH=O, -CH2OH
3. -CN, -NH2, -CH=O, -NHCH3
4. -SH, -SCH3, -OH, -OOCH3
6) Determine if the chiral carbon in the following molecules have an (R) or (S) configuration. Red = Oxygen & Blue = Nitrogen.
a)
b)
Answer
1) A is (S) and B is (R).
2)
3) The stereocenter is (R).
4)
1. -Cl
2. -I
3. -OCH3
4. -CH=CH2
5. -OH
5) Rank the following substituents in order of their CIP priority:
1. -OCH3, -OH, -CH2OH, -H,
2. -OH, -CO2H, -CH=O, -CH2OH
3. -NHCH3, -NH2, -CH=O, -CN
4. -SCH3, -SH, -OOCH3, -OH
6)
a) The chiral carbon is (R). The four substituents of the chiral carbon are -OH (1), -NH2 (2), -CH3 (3), and -H (4). Then looking down the lowest priority bond, you should roughly see what appears in the picture below. The substituents with priorities 1-3 are ordered in a clockwise fashion so the chiral carbon is (R).
b) The chiral carbon is (S). The four substituents of the chiral carbon are -CO2H (1), -OH (2), -CH2CH2CH3 (3), and -H (4). Then looking down the lowest priority bond, you should roughly see what appears in the picture below. The substituents with priorities 1-3 are ordered in a counterclockwise fashion so the chiral carbon is (S).
Exercise \(2\)
Identify which substituent in the following sets has a higher ranking.
1. -H or -CH3
2. -CH2CH2CH3 or CH2CH3
3. -CH2Cl or CH2OH
Answer
1. -CH3
2. -CH2CH2CH3
3. -CH2Cl
Exercise \(3\)
Identify which substituent in the following sets has a higher ranking.
1. -NH2 or -N=NH
2. -CH2CH2OH or -CH2OH
3. -CH=CH2 or -CH2CH3
Answer
a) -N=NH
b) -CH2OH
c) -CH=CH2
Exercise \(4\)
Place the following sets of substituents in each group in order of lowest priority (1st) to highest priority (4th)
1. -NH2, -F, -Br, -CH3
2. -SH, -NH2, -F, -H
Answer
a) -CH3 < -NH2 < -F, < -Br
b) -H < -NH2 < -F, < -SH
Exercise \(5\)
Place the following sets of substituents in each group in order of lowest priority (1st) to highest priority (4th)
1. -CH2CH3, -CN, -CH2CH2OH, -CH2CH2CH2OH
2. -CH2NH2, -CH2SH, -C(CH3)3, -CN
Answer
a) -CH2CH3 < -CH2CH2OH < -CH2CH2CH2OH, < -CN
b) -C(CH3)3 < -CH2NH2 < -CN < -CH2SH
Exercise \(6\)
Assign the following chiral centers as (R) or (S).
Answer
a) (S): I > Br > F > H. The lowest priority substituent is going backwards so following the highest priority, it goes left (counterclockwise).
b) (R): Br > Cl > CH3 > H. Using a model kit, you need to rotate the H to the back position where the Br is. This causes the priority to go to the left (clockwise) when looking at it with the H in the back position. Alternatively, if you do not have a model kit, you can imagine the structure 3-dimensionally and since the lowest priority (H) is facing up (as drawn), if you look at it from below, starting with Br (1st priority) and moving towards Cl (2nd priority), you are moving right (clockwise) which represents (R) stereochemistry.
c) Neither (R) or (S): Since there are two identical substituents (H’s) the molecule is achiral and cannot be assigned (R) or (S).
Exercise \(7\)
Assign the following chiral centers as (R) or (S).
Answer
a) (R): OH > CN (C triple bonded to N) > CH2NH2 > H. The H needs to be moved to the back position which causes the priority to go to the right (clockwise) which indicates (R).
b) (S): COOH > CH2OH > CCH > H. Since the H is coming forward, you can assign the priority and it goes to the right (clockwise which would be (R)) but since the lowest priority is forward, you have to switch it to (S). Alternatively, you can rotate the molecule to put the lowest priority to the back and you’ll see that it rotates left (or counterclockwise) for (S).
c) (S): Br > OH > NH2 > CH3. Since the lowest priority is going back, you can follow the priority and see that it is going left (counterclockwise) and therefore (S).
Exercise \(8\)
Draw the structure of (R)-2-bromohexane.
Answer
Exercise \(9\)
Draw the structure of (S)-2-methyl-3-pentanol.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.05%3A_Sequence_Rules_for_Specifying_Configuration.txt |
Objectives
After completing this section, you should be able to
1. calculate the maximum number of stereoisomers possible for a compound containing a specified number of chiral carbon atoms.
2. draw wedge-and-broken-line structures for all possible stereoisomers of a compound containing two chiral carbon atoms, with or without the aid of molecular models.
3. assign (R)/(S) configurations to wedge-and-broken-line structures containing two chiral carbon atoms, with or without the aid of molecular models.
4. determine, with or without the aid of molecular models, whether two wedge-and-broken-line structures containing two chiral carbon atoms are identical, represent a pair of enantiomers, or represent a pair of diastereomers.
5. draw the wedge-and-broken-line structure of a specific stereoisomer of a compound containing two chiral carbon atoms, given its IUPAC name and (R)/(S) configuration.
Key Terms
Make certain that you can define, and use in context, the key term below.
• diastereomer
Diastereomers are two molecules which are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers. Unlike enantiomers which are mirror images of each other and non-superimposable, diastereomers are not mirror images of each other and non-superimposable. Diastereomers can have different physical properties and reactivity. They have different melting points and boiling points and different densities. In order for diastereomer stereoisomers to occur, a compound must have two or more stereocenters.
Introduction
So far, we have been analyzing compounds with a single chiral center. Next, we turn our attention to those which have multiple chiral centers. We'll start with some stereoisomeric four-carbon sugars with two chiral centers.
We will start with a common four-carbon sugar called D-erythrose.
A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators 'D' and 'L'. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system.
As you can see, D-erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the (R) configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).
Notice that both chiral centers in L-erythrose both have the (S) configuration. To avoid confusion, we will simply refer to the different stereoisomers by capital letters.
Now let's consider all the possible stereoisomers.
Look first at compound A below. Both chiral centers in have the (R) configuration (you should confirm this for yourself!). The mirror image of Compound A is compound B, which has the (S) configuration at both chiral centers. If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not superimposable (again, confirm this for yourself with your models!). A and B are nonsuperimposable mirror images: in other words, enantiomers.
Now, look at compound C, in which the configuration is (S) at chiral center 1 and (R) at chiral center 2. Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term 'stereoisomer). However, they are not mirror images of each other (confirm this with your models!), and so they are not enantiomers. By definition, they are diastereomers of each other.
Notice that compounds C and B also have a diastereomeric relationship, by the same definition.
So, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them. Does compound C have its own enantiomer? Compound D is the mirror image of compound C, and the two are not superimposable. Therefore, C and D are a pair of enantiomers. Compound D is also a diastereomer of compounds A and B.
This can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers:
Stereoisomer Shortcuts
If all of the chiral centers are of opposite (R)/(S) configuration between two stereoisomers, they are enantiomers.
If at least one, but not all of the chiral centers are opposite between two stereoisomers, they are diastereomers.
These shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups: we will come to that later
Here's another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue. Pairs of enantiomers are stacked together.
We know, using the shortcut above, that the enantiomer of (R,R) must be (S,S) - both chiral centers are different. We also know that (R,S) and (S,R) are diastereomers of (R,R), because in each case one - but not both - chiral centers are different.
Determining the Maximum Number of Stereoisomers for a Compound
In general, a structure with n stereocenters will have a maximum of 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).
In L-glucose, all of the stereocenters are inverted relative to D-glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D-galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.
Example \(1\)
• Draw the structure of L-galactose, the enantiomer of D-galactose.
• Draw the structure of two more diastereomers of D-glucose. One should be an epimer.
Answer
Erythronolide B, a precursor to the 'macrocyclic' antibiotic erythromycin, has 10 stereocenters. It’s enantiomer is that molecule in which all 10 stereocenters are inverted.
In total, there are 210 = 1024 stereoisomers in the erythronolide B family: 1022 of these are diastereomers of the structure above, one is the enantiomer of the structure above, and the last is the structure above.
We know that enantiomers have identical physical properties and equal but opposite degrees of specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to separate them. In addition, the specific rotations of diastereomers are unrelated – they could be the same sign or opposite signs, and similar in magnitude or very dissimilar.
Exercise \(1\)
Determine the number of stereoisomers a molecule can have with…
1. 3 chiral centers
2. 1 chiral center
3. 6 chiral centers
Answer
Since a molecule with n chiral centers can have 2n stereoisomers…
1. 23 = 8 possible stereoisomers
2. 21 = 2 possible stereoisomers
3. 26 = 64 possible stereoisomers
Exercise \(\PageIndex{2a}\)
What is the relationship between enantiomers?
Answer
They are mirror images of each other and when 2 or more chiral centers are present, every stereocenter is the opposite in its enantiomer.
Exercise \(\PageIndex{2b}\)
How does the stereochemistry in diastereomers differ from each other?
Answer
In diastereomers, one or more of the chiral centers is the opposite but they all can’t be the opposite or else they’d be enantiomers.
Exercise \(\PageIndex{2c}\)
What are epimers?
Answer
Epimers are when only one chiral center is the opposite (in molecules with 2 or more chiral centers) in its diastereomer.
Exercise \(\PageIndex{3a}\)
Draw the structure of (2R,3R) 2-fluoro-3-methylhexane.
Answer
Exercise \(\PageIndex{3b}\)
Draw both diastereomers of (2R,3R) 2-fluoro-3-methylhexane.
Answer
Exercise \(\PageIndex{3c}\)
Draw the enantiomer of (2R,3R) 2-fluoro-3-methylhexane.
Answer
Exercise \(\PageIndex{4a}\)
Draw the structure of L-galactose, the enantiomer of D-galactose.
Answer
Exercise \(\PageIndex{4b}\)
Draw a diastereomer of D-galactose that is an epimer.
Answer
You can draw an epimer by drawing D-galactose with 1 (and only 1) of its chiral centers reversed. Here’s an example when you switch only the first chiral center (in red). (There are 3 other epimers that could be drawn as long as you only swap a single chiral center in the diastereomer that you use.)
Exercise \(\PageIndex{4c}\)
Identify if the following diastereomer of galactose is an epimer of D- galactose or L- galactose.
Answer
Since the diastereomer above only varies from L-galactose by 1 chiral center, the above is an epimer in relationship to L-galactose. Since it varies from D-galactose by 3 chiral centers, it is not an epimer but a diastereomer. Since not all of the chiral centers are swapped, it is not an enantiomer!
Exercise \(\PageIndex{5a}\)
For the compound shown below, label each chiral center as R or S.
Answer
Exercise \(\PageIndex{5b}\)
How many stereoisomers are possible for the compound in part a)?
Answer
Since there are 3 chiral centers, 23 = 8 possible stereoisomers.
Exercise \(6\)
Consider the stereoisomers below.
1. Which is/are an enantiomer of i?
2. Which is/are a diastereomer of ii?
3. Which is/are an epimer of i?
Answer
1. iv is an enantiomer of i since both chiral centers are switched and they are non superimposable mirror images.
2. i & iv are diastereomers of ii since they are stereoisomers that are not mirror images.
3. ii and iii are epimers of i since they are diastereomers with only 1 chiral center switched and the other one the same.
Exercise \(7\)
Consider the 8 stereoisomers below.
1. Which is/are an enantiomer of i?
2. Which is/are a diastereomer of i?
3. Which is/are an epimer of i?
Answer
1. v is an enantiomer since all three chiral centers are switched and they are non superimposable mirror images.
2. ii, iii, iv, vi, vii & viii are diastereomers of i since they are stereoisomers that are not mirror images.
3. ii,iii & viii are epimers of i since they are diastereomers with only 1 chiral center switched and the other chiral centers the same. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.06%3A_Diastereomers.txt |
Objectives
After completing this section, you should be able to
1. determine whether or not a compound containing two chiral carbon atoms will have a meso form, given its Kekulé, condensed or shorthand structure, or its IUPAC name.
2. draw wedge-and-broken-line structures for the enantiomers and meso form of a compound such as tartaric acid, given its IUPAC name, or its Kekulé, condensed or shorthand structure.
3. make a general comparison of the physical properties of the enantiomers, meso form and racemic mixture of a compound such as tartaric acid.
Key Terms
Make certain that you can define, and use in context, the key term below.
• meso compound
Study Notes
You may be confused by the two sets of structures showing “rotations.” Of course in each case the two structures shown are identical, they represent the same molecule looked at from two different perspectives. In the first case, there is a 120° rotation around the single carbon-carbon bond. In the second, the whole molecule is rotated 180° top to bottom.
Introduction
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 in Section 5.2.
In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of reflected stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, resulting the molecule being optically inactive.
Identification
A meso compound must have:
1. Two or more stereocenters.
2. An internal plane of symmetry, or internal mirror, that lies in the compound.
3. Stereochemistry that cancels out. This means reflected stereocenter should have the same substituents and be inverted. For instance, in a meso compound with two stereocenters one should be R and the other S.
The compounds 2,3-dichlorobutane contains two chiral carbons and therefore would be expected to provide 22 = 4 different stereoisomers. These stereoisomers should be made up of two pairs of enantiomers.
After drawing out all the possible stereoisomers of 2,3-dichlorobutane, the pair on the right in the figure below are mirror images. Also, they are non-superimposable because they have distinctly different conformation (R,R & S,S). This makes the pair enantiomers of each other. However, the pair on the left represent a meso compound, they both are identical despite being mirror images.
Upon further investigation, the meso compound has an internal plane of symmetry which is not present in the pair of enantiomers. The plane of symmetry in the meso compound comes about because there are two chiral carbons present, both chiral carbons are identically substituted (Cl, H, CH3), and one chiral carbon is R and the other is S. Despite being represented as mirror images, both structures represent the same compound. This is best proven by making molecular models of both representations and then superimposing them. Overall, 2,3-dichlorobutane only has three possible stereosiomers, the pair of enantiomers and the meso compound.
When looking for an internal plane of symmetry, it is important to remember that sigma bonds (single bonds) can rotate. Just because the immediate representation of a molecule does not have a plane of symmetry does not mean that one cannot be obtained through rotation. Often the substituents attached to a stereocenter need to be rotated to recognize the internal plane of symmetry. As the stereocenter is rotated, its configuration does not change. Building a molecular model when considering a possible meso compound is an invaluable tool because it allows for easy rotation of chiral carbons. An example of how rotation of a chiral carbon can reveal an internal plane of symmetry is shown below.
Example \(1\)
Below are the two mirror images of (meso)-2,3-Butanediol. Because it is a meso compound, the two structures are identical. Show that both mirror images can be obtained by simply rotating the three-dimensional structure provided below.
Example \(2\)
1 has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; 1 has chiral centers. Thus, 1 is a meso compound.
Example \(3\)
This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound.
Other Examples of Meso Compounds
Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cycloalkanes. Although two chiral carbons must be present, meso compounds can have many more. Notice that in every case a plane of symmetry is present.
In general, a disubstituted cycloalkane is meso if the two substituents are the same and they are in a cis conformation. Trans disubstituted cycloalkanes are not meso regardless if the two substituents are the same.
Optical Activity Analysis of a Meso Compound
When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+) because a meso compound is achiral (optically inactive). Investigations of isomeric tartaric acid (2,3-dihydroxybutanedioic acid), carried out by Louis Pasteur in the mid 19th century, were instrumental in elucidating some of the subtleties of stereochemistry. Tartaric acid, has two chiral but only three stereoisomers. Two of these stereoisomers are enantiomers and the third is an achiral a meso compound. Some physical properties of these stereoisomers of tartaric acid are given in the table below. Notice that the enantiomers have the same amount of optical rotation but in different directons. Meso-tartaric acid produces no optical rotation because it is achiral and not optically active. Meso-tartaric acid is actually a diastereomer of both (-) and (+)-tartaric acid, which gives it a distinctly different melting point.
(+)-tartaric acid: [α]D = +13º m.p. 172 ºC
(–)-tartaric acid: [α]D = –13º m.p. 172 ºC
meso-tartaric acid: [α]D = 0º m.p. 140 ºC
Exercise \(1\)
1) Determine which of the following molecules are meso.
2) Explain why 2,3-dibromobutane has the possibility of being a meso compound while 2,3-dibromopentane does not.
3) Observe the following compound and determine if it is a meso compound. If so indicate the plane of symmetry. Red = oxygen. Remember sigma bonds are able to rotate.
Answer
1) A C, D, E are meso compounds.
2) One of the requirements of a meso compound is that the reflected chiral carbons have the same substituents. The compound 2,3-dibromobutane, fulfills this requirement (Br, H, CH3) and can possibly be a meso compound if the two chiral carbons have the appropriate configuration (R & S). The substituents of the two chiral carbons in 2,3-dibromopentane do not have the same substituents (Br, H, CH3 vs. Br, H, CH2CH3). This 2,3-dibromopentane cannot form a meso compound regardless of the configurations of its chiral carbons.
3) The compound is meso.
Exercise \(1\)
Which of the following are meso compounds?
Answer
a) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) between the C’s (and it has stereochemistry of S & R).
b) This is not a meso compound. No matter how you rotate the C-C bond, you do not see a plane of symmetry (and its stereochemistry is S & S)
c) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) that can be seen when you rotate the C-C bond (and it has stereochemistry of S & R).
Exercise \(2\)
Which of the following are meso compounds?
Answer
a) This is not a meso compound. There is no plane of symmetry and has stereochemistry of S & S.
b) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) and it has stereochemistry of R & S.
c) This is not a meso compound (even though it has planes of symmetry). The plane of symmetry shown in red makes it so that both chiral centers have symmetrical groups (the ring) and thus the compound is not chiral (so it can’t be a meso compound).
Exercise \(3\)
Which of the following are meso compounds?
Answer
a) This is a meso compound. If you rotate between the C-C bond, you can see that it has a mirror plane between the C’s (shown in red on the structure to the right). Notice how rotating a C-C bond doesn’t change the stereochemistry of the molecule (S & R).
b) This is a meso compound. You can see the plane of symmetry in red and the compound has stereochemistry of S & R.
c) This is not a meso compound. There is no plane of symmetry and it has stereochemistry of S & S.
Exercise \(4\)
Determine (and draw) if any of the forms of 3,4-dichlorohexane are a meso compound.
Answer
Looking at the 4 different possibilities below, i & ii are equivalent structures (with R & S stereochemistry) so it is a meso compound. iii & iv are not meso compounds but are enantiomers to each other. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.07%3A_Meso_Compounds.txt |
Objectives
After completing this section, you should be able to
1. describe a common process for separating a mixture of enantiomers.
2. explain why racemic mixtures do not rotate plane-polarized light.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• racemic mixture (or racemate)
• resolve
Study Notes
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.
The dramatic biochemical consequences of chirality are illustrated by the use, in the 1950s, of the drug Thalidomide, a sedative given to pregnant women to relieve morning sickness. It was later realized that while the (+)‑form of the molecule, was a safe and effective sedative, the (−)‑form was an active teratogen. The drug caused numerous birth abnormalities when taken in the early stages of pregnancy because it contained a mixture of the two forms.
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as 'Moscher's esters', after Harry Stone Moscher, a chemist who pioneered the method at Stanford University.
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent.
Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. Figure 5.8.1 illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.
Because the physical properties of enantiomers are identical, they seldom can be separated by simple physical methods, such as fractional crystallization or distillation. It is only under the influence of another chiral substance that enantiomers behave differently, and almost all methods of resolution of enantiomers are based upon this fact. We include here a discussion of the primary methods of resolution.
Chiral Amines as Resolving Agents and Resolution of Racemic Acids
The most commonly used procedure for separating enantiomers is to convert them to a mixture of diastereomers that will have different physical properties: melting point, boiling point, solubility, and so on (Section 5-5). For example, if you have a racemic or (R)/(S) mixture of enantiomers of a carboxylic acid and convert this to a salt with a chiral amine base having the (R) configuration, the salt will be a mixture of two diastereomers, ((R)-acid・(R)-base) and ((S)-acid・(R)-base). These diastereomeric salts are not identical and they are not mirror images. Therefore they will differ to some degree in their physical properties, and a separation by physical methods, such as crystallization, may be possible. If the diastereomeric salts can be completely separated, the carboxylic acid regenerated from each salt will be either exclusively the (R) or the (S) enantiomer.
Resolution of chiral acids through the formation of diastereomeric salts requires adequate supplies of suitable chiral bases. Brucine, strychnine, and quinine frequently are used for this purpose because they are readily available, naturally occurring chiral bases. Simpler amines of synthetic origin, such as 2-amino-1-butanol, amphetamine, and 1-phenylethanamine, also can be used, but first they must be resolved themselves.
Worked Example $1$
Show how (S)-1-phenylethylamine can be used to resolve a racemic mixture of lactic acid. Please draw all the structures involved.
Answer
Resolution of Racemic Bases
Chiral acids, such as (+)-tartaric acid, (-)-malic acid, (-)-mandelic acid, and (+)-camphor- 10-sulfonic acid, are used for the resolution of a racemic base.
The principle is the same as for the resolution of a racemic acid with a chiral base, and the choice of acid will depend both on the ease of separation of the diastereomeric salts and, of course, on the availability of the acid for the scale of the resolution involved. Resolution methods of this kind can be tedious, because numerous recrystallizations in different solvents may be necessary to progressively enrich the crystals in the less-soluble diastereomer. To determine when the resolution is complete, the mixture of diastereomers is recrystallized until there is no further change in the measured optical rotation of the crystals. At this stage it is hoped that the crystalline salt is a pure diastereomer from which one pure enantiomer can be recovered. The optical rotation of this enantiomer will be a maximum value if it is "optically" pure because any amount of the other enantiomer could only reduce the magnitude of the measured rotation $\alpha$.
Resolution of Racemic Alcohols
To resolve a racemic alcohol, a chiral acid can be used to convert the alcohol to a mixture of diastereomeric esters. This is not as generally useful as might be thought because esters tend to be liquids unless they are very high-molecularweight compounds. If the diastereomeric esters are not crystalline, they must be separated by some other method than fractional crystallization (for instance, by chromatography methods, Section 9-2). Two chiral acids that are useful resolving agents for alcohols are:
The most common method of resolving an alcohol is to convert it to a half-ester of a dicarboxylic acid, such as butanedioic (succinic) or 1,2-benzenedicarboxylic (phthalic) acid, with the corresponding anhydride. The resulting half-ester has a free carboxyl function and may then be resolvable with a chiral base, usually brucine:
Other Methods of Resolution
One of the major goals in the field of organic chemistry is the development of reagents with the property of "chiral recognition" such that they can affect a clean separation of enantiomers in one operation without destroying either of the enantiomers. We have not achieved that ideal yet, but it may not be far in the future. Chromatographic methods, whereby the stationary phase is a chiral reagent that adsorbs one enantiomer more strongly than the other, have been used to resolve racemic compounds, but such resolutions seldom have led to both pure enantiomers on a preparative scale. Other methods, called kinetic resolutions, are excellent when applicable. The procedure takes advantage of differences in reaction rates of enantiomers with chiral reagents. One enantiomer may react more rapidly, thereby leaving an excess of the other enantiomer behind. For example, racemic tartaric acid can be resolved with the aid of certain penicillin molds that consume the dextrorotatory enantiomer faster than the levorotatory enantiomer. As a result, almost pure (-)-tartaric acid can be recovered from the mixture:
(±)-tartaric acid + mold $\rightarrow$ (-)-tartaric acid + more mold
The crystallization procedure employed by Pasteur for his classical resolution of (±)-tartaric acid (Section 5.4) has been successful only in a very few cases. This procedure depends on the formation of individual crystals of each enantiomer. Thus if the crystallization of sodium ammonium tartrate is carried out below 27", the usual racemate salt does not form; a mixture of crystals of the (+) and (-) salts forms instead. The two different kinds of crystals, which are related as an object to its mirror image, can be separated manually with the aid of a microscope and subsequently may be converted to the tartaric acid enantiomers by strong acid. A variation on this method of resolution is the seeding of a saturated solution of a racemic mixture with crystals of one pure enantiomer in the hope of causing crystallization of just that one enantiomer, thereby leaving the other in solution. Unfortunately, very few practical resolutions have been achieved in this way.
Predicating the Chirality of the Product of a Reaction
It important to understand the changes in chirality which occur during the formation of product during a reaction. A chiral reaction product, has the possibility of forming multiple stereoisomers which all need to be considered. Changes in chirality, if possible, will be discussed with each individual reaction as this textbook moves forward. Some possible situations which can occur are:
• A new chiral carbon is formed during a reaction. This commonly occurs when an sp2 hybridized carbon in the reactant is converted to sp3 hybridized chiral carbon in the product. When this occurs, a racemic mixture of the new chiral carbon is formed.
• A chiral carbon is lost during a reaction. This commonly occurs when an sp3 hybridized chiral carbon in the reactant is converted to either a sp2 or sp hybridized carbon in the product.
• An enantiomerically pure starting material is converted to a racemic mixture in the product. This commonly occurs when a sp3 hybridized chiral carbon is temporarily converted to an sp2 hybridized carbon during a reaction's mechanism. The chiral carbon is reformed as a racemic mixture.
• Chiral carbons remain unchanged during a reaction. If a chiral carbon is not directly involved in a reaction, it will move from a reactant to a product unchanged.
Determining if a chiral carbon is involved in a given reaction is vital for determining which of these four situations is occurring.
Worked Example $2$
The following reaction involves the conversion of a carboxylic acid reacting with an alcohol to form an ester. If a pure sample of (R)-2-methylbutanoic acid is reacted with methanol to form an ester, what would be the stereochemistry of the product?
Answer
First it is important to identify the location of the chiral carbon and determine if it is directly involved in the reaction. In this case, the chiral carbon is not involved so the stereochemistry will be carried over into the product unchanged.
Exercise $1$
Indicate the reagents you could use to resolve the following compounds. Show the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 1 -phenyl-2-propanamine.
Answer
You could react the 1-phenyl-2-propanamine racemic mixture with a chiral acid such as (+)-tartaric acid (R, R). The reaction will produce a mixture of diastereomeric salts (i.e. R, R, R and S, R, R). You can separate the diastereomers through crystallization and treat the salt with a strong base (e.g. KOH) to recover the pure enantiomeric amine.
Exercise $2$
Indicate the reagents you would use to resolve the following and discuss the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 2,3-pentadienedioic acid.
Answer
You could react the 2,3-pentadienedioic acid mixture with a chiral base such as (R)‑1‑phenylethylamine. The reaction will produce a mixture of diastereomeric salts. Separate the diastereomers through crystallization and treat the resulting salt with strong acid (e.g. HCl) to recover the pure enantiomeric acid.
Exercise $3$
Indicate the reagents you would use to resolve the following and discuss the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 1 -phenylethanol.
Answer
You could react the 1-phenylethanol mixture with 1,2-benzenedicarboxylic anhydride. The reaction will produce a mixture of diastereomeric salts. You could then separate the diastereomers through crystallization and then alkaline hydrolysis treatment should recover the pure enantiomeric alcohol. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.08%3A_Racemic_Mixtures_and_the_Resolution_of_Enantiomers.txt |
Objective
After completing this section, you should be able to explain the differences among constitutional (structural) isomers and stereoisomers (geometric isomers).
Key Terms
Make certain that you can define, and use in context, the key terms below.
• constitutional
• (structural) isomers
• stereoisomers
The following flow chart can be used to identify the relationship of two compounds with respect to isomerization:
Conformational Isomers
The C–C single bonds in ethane, propane, and other alkanes are formed by the overlap of an sp3 hybrid orbital on one carbon atom with an sp3 hybrid orbital on another carbon atom, forming a σ bond. Each sp3 hybrid orbital is cylindrically symmetrical (all cross-sections are circles), resulting in a carbon–carbon single bond that is also cylindrically symmetrical about the C–C axis. Because rotation about the carbon–carbon single bond can occur without changing the overlap of the sp3 hybrid orbitals, there is no significant electronic energy barrier to rotation. Consequently, many different arrangements of the atoms are possible, each corresponding to different degrees of rotation. Differences in three-dimensional structure resulting from rotation about a σ bond are called differences in conformation, and each different arrangement is called a conformational isomer (or conformer).
Conformational Isomers of Pentane
Structural Isomers
Unlike conformational isomers, which do not differ in connectivity, structural isomers differ in connectivity, as illustrated here for 1-propanol and 2-propanol. Although these two alcohols have the same molecular formula (C3H8O), the position of the –OH group differs, which leads to differences in their physical and chemical properties.
In the conversion of one structural isomer to another, at least one bond must be broken and reformed at a different position in the molecule. Consider, for example, the following five structures represented by the formula C5H12:
Of these structures, (a) and (d) represent the same compound, as do (b) and (c). No bonds have been broken and reformed; the molecules are simply rotated about a 180° vertical axis. Only three—n-pentane (a) and (d), 2-methylbutane (b) and (c), and 2,2-dimethylpropane (e)—are structural isomers. Because no bonds are broken in going from (a) to (d) or from (b) to (c), these alternative representations are not structural isomers. The three structural isomers—either (a) or (d), either (b) or (c), and (e)—have distinct physical and chemical properties.
Stereoisomers
Stereoisomers have the same connectivity in their atoms but a different arrangement in three-dimensional space. There are different classifications of stereoisomers depending on how the arrangements differ from one another. Notice that in the structural isomers, there was some difference in the connection of atoms. For example, 1-butene has a double bond followed by two single bonds while 2-butene has a single bond, then a double bond, then a single bond. A stereoisomer will have the same connectivity among all atoms in the molecule.
Geometric Isomers
With a molecule such as 2-butene, a different type of isomerism called geometric isomerism can be observed. Geometric isomers are isomers in which the order of atom bonding is the same but the arrangement of atoms in space is different. The double bond in an alkene is not free to rotate because of the nature of the bond. Therefore, there are two different ways to construct the 2-butene molecule (see figure below). The image below shows the two geometric isomers, called cis-2-butene and trans-2-butene.
Geometric Isomers of 2-Butene
The cis isomer has the two single hydrogen atoms on the same side of the molecule, while the trans isomer has them on opposite sides of the molecule. In both molecules, the bonding order of the atoms is the same. In order for geometric isomers to exist, there must be a rigid structure in the molecule to prevent free rotation around a bond. This occurs with a double bond or a ring. In addition, the two carbon atoms must each have two different groups attached in order for there to be geometric isomers. Propene (see figure below) has no geometric isomers because one of the carbon atoms (the one on the far left) involved in the double bond has two single hydrogens bonded to it.
Physical and chemical properties of geometric isomers are generally different. As with alkenes, alkynes display structural isomerism beginning with 1-butyne and 2-butyne. However, there are no geometric isomers with alkynes because there is only one other group bonded to the carbon atoms that are involved in the triple bond.
Optical Isomers
Stereoisomers that are not geometric isomers are known as optical isomers. Optical isomers differ in the placement of substituted groups around one or more atoms of the molecule. They were given their name because of their interactions with plane-polarized light. Optical isomers are labeled enantiomers or diastereomers.
Enantiomers are non-superimposable mirror images. A common example of a pair of enantiomers is your hands. Your hands are mirror images of one another but no matter how you turn, twist, or rotate your hands, they are not superimposable.
Two models that are mirror images and superimposable.
Since they are superimposable, they are the same molecule and are not isomers.
Your hands and some molecules are mirror images but are not superimposable.
These pairs of molecules are called enantiomers.
Objects that have non-superimposable mirror images are called chiral. When examining a molecule, carbon atoms with four unique groups attached are considered chiral. Look at the figure below to see an example of a chiral molecule. Note that we have to look beyond the first atom attached to the central carbon atom. The four circles indicate the four unique groups attached to the central carbon atom, which is chiral.
A chiral carbon has four unique groups attached to it
Another type of optical isomer are diastereomers, which are non-mirror image optical isomers. Diastereomers have a different arrangement around one or more atoms while some of the atoms have the same arrangement. As shown in the figure below, note that the orientation of groups on the first and third carbons are different but the second one remains the same so they are not the same molecule. The solid wedge indicates a group coming out of the page/screen towards you and the dashed line indicates that a group is going away from you "behind" the page/screen.
Diastereomers differ at one or more atom. These molecules are not mirror images and they are not superimposable.
They are optical isomers because they have the same connectivity between atoms but a different arrangement of substituent groups.
Epimers are a sub-group of diastereomers that differ at only one location. All epimers are diastereomers but not all diastereomers are epimers.
Epimers have a different arrangement around one atom, while arrangements around the other atoms are the same
Diagram showing the division of stereoisomers (also known as optical isomers)
Exercise \(1\)
What kind of isomers are the following pairs?
Answer
1. Since both structures have the same formula (C6H12) and they have the same connectivity, but are in a different arrangement of the atoms in space, they are conformational isomers.
2. Since both structures have the same formula (C2H6O) but different connectivity, they are structural isomers.
3. Since both structures have the same formula (C3H9Br) and the same connectivity but the structure on the left has R stereochemistry and the structure on the right has S stereochemistry, they are stereoisomers called enantiomers.
Exercise \(2\)
What kind of isomers are the following pairs?
Answer
a) Since both structures have the same formula (C2H6O) but different connectivity of where the double bond is, they are structural isomers.
b) Both structures have the same formula and the same connectivity but the structure on the left has (R,S) stereochemistry and the structure on the right has (S,S) stereochemistry, they are stereoisomers that are diastereomers.
c) Both structures and have the same connectivity but their stereochemistry differs so they are stereoisomers. Since the one on the left is (R,S) and the one on the right is (S,R), they are non-super imposable mirror images of each other and enantiomers.
Exercise \(3\)
What is the relationship between the following pairs?
c) (R) – 2 chlorohexane & 1-chlorohexane
d) (2R, 3R) dichlorohexane & (2S, 3R) dichlorohexane
Answer
a) Both structures have the same connectivity and same chemical formula and when you rotate the structure on the right you can see that these are the same compound both with (R,S) stereochemistry.
b) ) Since both structures have the same formula (C4H8) and the same connectivity they are stereoisomers since they differ in what is known as cis/trans isomerism (covered more in Ch 7).Due to the pi bond between carbons 2 & 3, there is not free rotation between the C2-C3 bond so the structure on the left has the carbon chains on the same side of the double bond (cis) and the structure on the right has the carbon chains on opposite sides of the double bond (trans).
c) Looking at the structures below, you can see that they have the same formula (C6H13Cl) and different connectivity, so they are structural isomers.
d) Looking at the structures below, you can see that they have the same formula (C6H12Br2) and the same connectivity, but the structure on the left is (R,R) and the structure on the right is (S,R) so since they differ by only one stereocenter, they are stereoisomers that are diastereomers (& epimers in this case). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.09%3A_A_Review_of_Isomerism.txt |
Objectives
After completing this section, you should be able to
1. recognize that atoms other than carbon can be chiral centres.
2. explain why enantiomers of compounds such as ethylmethylamine cannot normally be isolated.
Study Notes
The first example of a resolvable compound containing a chiral nitrogen atom was resolved by William Pope and Stanley Peachey in 1899. It had the structure shown below.
Chiral sulfoxides find application in certain drugs such as esomeprazole and armodafinil and are a good example of a stereogenic sulfur center.
Chirality at Nitrogen
Due to their tetrahedral configuration, amines with three different substituents are chiral. The R and S enantiomeric forms of chiral amines cannot be resolved due to their rapid interconversion by a process called pyramidal or nitrogen inversion. During the inversion, the sp3 hybridized amine momentarily rehybridizes to a sp2 hybridized, trigonal planar, transition state where the lone pair electrons occupy a p orbital. The nitrogen then returns to tetrahedral sp3 hybridization causing the lone pair electrons to enter to a hybrid orbital on the opposite side of the nitrogen. During this process substituents invert to form the enantiomer, analogous to the Walden inversion seen in SN2 reactions. The thermodynamic barrier for this inversion (~25 kJ/mol) is low enough to allow rapid inversion at room temperature, leading to a mixture of interconverting R and S configurations. At room temperature a nitrogen atom exists as a racemic mixture of R and S configurations.
Quaternary amines lack lone pair electrons and therefore do not undergo pyramidal inversions. Quaternary amines with four different substituents are chiral and are readily resolved into separate enantiomers.
Example \(1\)
The Enantiomers of a chiral quaternary amine.
Chirality at Phosphorus
Trivalent phosphorus compounds called phosphines have a tetrahedral electron-group geometry which makes them structurally analogous to amines. The rate of inversion for phosphines are so much slower than amines that chiral phosphines can be isolated. In this case the set of lone pair electrons are considered a substituent and given the lowest Cahn-Ingold-Prelog priority.
The phosphorus center of phosphate ions and organic phosphate esters is also tetrahedral, and thus potentially a stereocenter. In order to investigate the stereochemistry of reactions at the phosphate center, 17O and 18O isotopes of oxygen (the ‘normal’ isotope is 16O) can be incorporated to create chiral phosphate groups. Phosphate triesters are chiral if the all four substituent groups are different (including the carbonyl oxygen).
Chirality at Sulfur
Trivalent sulfur compounds called sulfonium salts (R3S+) have a tetrahedral electron-group geometry similar to amines and can be chiral if the R groups are all different. In a similar fashion as phosphorus, the inversion rates are slow enough for chiral sulfonium salts to be isolated. Here again the set of lone pair electrons are considered a substituent and given the lowest CIP priority. Sulfonium salts will be discussed in greater detail in Section 18.8.
An excellent example of a chiral sulfonium salt in biological systems is the coenzyme (S)-adenosylmethionine (SAM). The presence of a sulfonium allows SAM to be a biological methyl group donor in many metabolic pathways. Note that SAM has an (S) configuration at the sulfur atom.
The sulfur in sulfoxides (R'SOR'') can be chiral if both R groups are different. Here again the inversion rate is slow enough to allow chiral sulfoxides to be isolated. An excellent example is methyl phenyl sulfoxide. Sulfoxides will also be discussed in greater detail in Section 18.8. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.10%3A_Chirality_at_Nitrogen_Phosphorus_and_Sulfur.txt |
Objectives
After completing this section, you should be able to
1. identify a compound as being prochiral.
2. identify the Re and Si faces of prochiral sp2 centre.
3. identify atoms (or groups of atoms) as pro-R or pro-S on a prochiral sp3 centre.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• prochiral
• pro-R
• pro-S
• Re
• Si
Prochiral Carbons
When a tetrahedral carbon can be converted to a chiral center by changing only one of the attached groups, it is referred to as a ‘prochiral' carbon. The two hydrogens on the prochiral carbon can be described as 'prochiral hydrogens'.
Note that if, in a 'thought experiment', we were to change either one of the prochiral hydrogens on a prochiral carbon center to a deuterium (the 2H isotope of hydrogen), the carbon would now have four different substituents and thus would be a chiral center.
Prochirality is an important concept in biological chemistry, because enzymes can distinguish between the two ‘identical’ groups bound to a prochiral carbon center due to the fact that they occupy different regions in three-dimensional space. Consider the isomerization reaction below, which is part of the biosynthesis of isoprenoid compounds. We do not need to understand the reaction itself (it will be covered in chapter 14); all we need to recognize at this point is that the isomerase enzyme is able to distinguish between the prochiral 'red' and the 'blue' hydrogens on the isopentenyl diphosphate (IPP) substrate. In the course of the left to right reaction, IPP specifically loses the 'red' hydrogen and keeps the 'blue' one.
Prochiral hydrogens can be unambiguously designated using a variation on the R/S system for labeling chiral centers. For the sake of clarity, we'll look at a very simple molecule, ethanol, to explain this system. To name the 'red' and 'blue' prochiral hydrogens on ethanol, we need to engage in a thought experiment. If we, in our imagination, were to arbitrarily change red H to a deuterium, the molecule would now be chiral and the chiral carbon would have the R configuration (D has a higher priority than H).
For this reason, we can refer to the red H as the pro-R hydrogen of ethanol, and label it HR. Conversely, if we change the blue H to D and leave red H as a hydrogen, the configuration of the molecule would be S, so we can refer to blue H as the pro-S hydrogen of ethanol, and label it HS.
Looking back at our isoprenoid biosynthesis example, we see that it is specifically the pro-R hydrogen that the isopentenyl diphosphate substrate loses in the reaction.
Prochiral hydrogens can be designated either enantiotopic or diastereotopic. If either HR or HS on ethanol were replaced by a deuterium, the two resulting isomers would be enantiomers (because there are no other stereocenters anywhere on the molecule).
Thus, these two hydrogens are referred to as enantiotopic.
In (R)-glyceraldehyde-3-phosphate ((R)-GAP), however, we see something different:
R)-GAP already has one chiral center. If either of the prochiral hydrogens HR or HS is replaced by a deuterium, a second chiral center is created, and the two resulting molecules will be diastereomers (one is S,R, one is R,R). Thus, in this molecule, HR and HS are referred to as diastereotopic hydrogens.
Finally, hydrogens that can be designated neither enantiotopic nor diastereotopic are called homotopic. If a homotopic hydrogen is replaced by deuterium, a chiral center is not created. The three hydrogen atoms on the methyl (CH3) group of ethanol (and on any methyl group) are homotopic. An enzyme cannot distinguish among homotopic hydrogens.
Example \(1\)
Identify in the molecules below all pairs/groups of hydrogens that are homotopic, enantiotopic, or diastereotopic. When appropriate, label prochiral hydrogens as HR or HS.
Answer
Groups other than hydrogens can be considered prochiral. The alcohol below has two prochiral methyl groups - the red one is pro-R, the blue is pro-S. How do we make these designations? Simple - just arbitrarily assign the red methyl a higher priority than the blue, and the compound now has the R configuration - therefore red methyl is pro-R.
Citrate is another example. The central carbon is a prochiral center with two 'arms' that are identical except that one can be designated pro-R and the other pro-S.
In an isomerization reaction of the citric acid (Krebs) cycle, a hydroxide is shifted specifically to the pro-R arm of citrate to form isocitrate: again, the enzyme catalyzing the reaction distinguishes between the two prochiral arms of the substrate (we will study this reaction in chapter 13).
Exercise \(1\)
Assign pro-R and pro-S designations to all prochiral groups in the amino acid leucine. (Hint: there are two pairs of prochiral groups!). Are these prochiral groups diastereotopic or enantiotopic?
Answer
Prochiral Carbonyl and Imine Groups
Trigonal planar, sp2-hybridized carbons are not, as we well know, chiral centers– but they can be prochiral centers if they are bonded to three different substitutuents. We (and the enzymes that catalyze reactions for which they are substrates) can distinguish between the two planar ‘faces’ of a prochiral sp2 - hybridized group. These faces are designated by the terms re and si. To determine which is the re and which is the si face of a planar organic group, we simply use the same priority rankings that we are familiar with from the R/S system, and trace a circle: re is clockwise and si is counterclockwise.
When the two groups adjacent to a carbonyl (C=O) are not the same, we can distinguish between the re and si 'faces' of the planar structure. The concept of a trigonal planar group having two distinct faces comes into play when we consider the stereochemical outcome of a nucleophilic addition reaction. Nucleophilic additions to carbonyls will be covered in greater detail in Chapter 19. Notice that in the course of a carbonyl addition reaction, the hybridization of the carbonyl carbon changes from sp2 to sp3, meaning that the bond geometry changes from trigonal planar to tetrahedral. If the two R groups are not equivalent, then a chiral center is created upon addition of the nucleophile. The configuration of the new chiral center depends upon which side of the carbonyl plane the nucleophile attacks from. Reactions of this type often result in a 50:50 racemic mixture of stereoisomers, but it is also possible that one stereoisomer may be more abundant, depending on the structure of the reactants and the conditions under which the reaction takes place.
Below, for example, we are looking down on the re face of the ketone group in pyruvate. If we flipped the molecule over, we would be looking at the si face of the ketone group. Note that the carboxylate group does not have re and si faces, because two of the three substituents on that carbon are identical (when the two resonance forms of carboxylate are taken into account).
As we will see in chapter 10, enzymes which catalyze reactions at carbonyl carbons act specifically from one side or the other.
We need not worry about understanding the details of the reaction pictured above at this point, other than to notice the stereochemistry involved. The pro-R hydrogen (along with the two electrons in the C-H bond) is transferred to the si face of the ketone (in green), forming, in this particular example, an alcohol with the R configuration. If the transfer had taken place at the re face of the ketone, the result would have been an alcohol with the S configuration.
Exercise \(2\)
For each of the carbonyl groups in uracil, state whether we are looking at the re or the si face in the structural drawing below.
Answer
Exercise \(3\)
a) State which of the following hydrogen atoms are pro-R or pro-S.
b) Identify which side is Re or Si.
Answer
a) Left compound: Ha = pro-S and Hb = pro-R; Right compound: Ha = pro-R and Hb = pro-S
b) A – Re; B – Si; C – Re; D – Si\)
Exercise \(4\)
State whether the H's indicated below are pro-R or pro-S for the following structures.
Answer
a) Ha is pro-R; Hb is pro-S
Ha is pro-R; Hb is pro-S
Exercise \(5\)
In the structures below, determine if the H's are homotopic, enantiotopic, or diastereotopic.
Answer
In a), the CH2 is diastereotopic since there is another chiral center on the molecule. Both CH3's are homotopic since replacing one of them doesn't create a chiral center.
IN b), the CH2's are enantiotopic since it would create the only chiral center on the molecule. Both CH3's are homotopic since replacing one of them doesn't create a chiral center.
Exercise \(6\)
State whether you are looking down at the molecule from the re face or si face.
Answer
1. You are looking at the si face. The re face would be if you were facing the molecule from the back.
2. You are looking at the re face. The si face would be if you were facing the molecule from the back. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.11%3A_Prochirality.txt |
Objective
After completing this section, you should be able to explain how chiral molecules in nature can have such dramatically different biological properties.
Some Chiral Organic Molecules
There are a number of important biomolecules that could occur as enantiomers, including amino acids and sugars. In most cases, only one enantiomer occurs naturally (although some fungi, for example, are able to produce mirror-image forms of these compounds). We will look later at some of these biomolecules, but first we will look at a compound that occurs naturally in both enantiomeric forms.
Carvone is a secondary metabolite. That means it is a naturally-occurring compound that is not directly connected to the very basic functions of a cell, such as self-replication or the production of energy. The role of secondary metabolites in nature is often difficult to determine. However, these compounds often play roles in self-defense, acting as deterrents against competitor species in a sort of small-scale chemical warfare scenario. They are also frequently used in communications; this role has been studied most extensively among insects, which use lots of compounds to send information to each other.
Carvone is produced in two enantiomeric forms. One of these forms, called (-)-carvone, is found in mint leaves, and it is a principal contributor to the distinctive odor of mint. The other form, (+)-carvone, is found in caraway seeds. This form has a very different smell, and is typically used to flavor rye bread and other Eastern European foods.
Note that (+)-carvone is another name for (S)-carvone. The (+) designation is based on its positive optical rotation value, which is experimentally measured. This means the (-)-carvone is (R)-carvone and would have a negative optical rotation value.
How different, exactly, are these two compounds, (+)- and (-)-carvone? Are they completely different isomers, with different physical properties? In most ways, the answer is no. These two compounds have the same appearance (colorless oil), the same boiling point (230 °C), the same refractive index (1.499) and specific gravity (0.965). However, they have optical rotations that are almost exactly opposite values.
• Two enantiomers have the same physical properties.
• Enantiomers have opposite optical rotations.
Clearly they have different biological properties; since they have slightly different odors, they must fit into slightly different nasal receptors, signaling to the brain whether the person next to you is chewing a stick of gum or a piece of rye bread. This is much like how a left hand only fits into a left handed baseball glove and not into a right handed one. Receptors which allow for a biological effect, in this case to perceive a certain smell, are often chiral and will only allow one enantiomer of a chiral substrate to fit. An example of this is shown in the figure below. The receptor as a complementary three-dimension shape to allow the R configuration of the chiral substrate to fit. When the S configuration of the chiral substrate attempts to fit the configuration does not match that of the receptor as shown in the second drawing where the bottom two groups do not fit the receptor site.
Chiral Environments
In the previous section, enzymes were shown to be capable of converting a prochiral substrate into a single enantiomer product. Enzymes can provides such an effect because they create a chiral environment. The figure below shows a prochiral substrate. The magenta and blue balls represent substituents while the two green balls represent two of the same substituent which is available for a given reaction. One of these substituents is pro-R and the other is pro-S. Without the presence of a chiral environment the two green substituents are chemically identical. However, as the prochiral molecule interacts with the chiral environment provided by an enzyme the two green substituents become chemically distinct. Despite being achiral, the prochiral molecule can only interact with the chiral environment in one specific position. The figure below shows that the pro-S green substituent of the prochiral molecule is protected by the enzyme, while the pro-R green substituent is exposed and can undergo a reaction. In this case the enzyme would provide a product that is predominantly the R enantiomer.
Exercise \(1\)
The following are three molecules found in nature. Please identify four chiral centers in each, mark them with asterisks, and identify each center as having a R or S configuration. Each molecule contains more than four chiral centers.
a) The following is the structure of dysinosin A, a potent thrombin inhibitor that consequently prevents blood clotting.
b) Ginkgolide B (below) is a secondary metabolite of the ginkgo tree, extracts of which are used in Chinese medicine.
c) Sanglifehrin A, shown below, is produced by a bacteria that may be found in the soil of coffee plantations in Malawi. It is also a promising candidate for the treatment of organ transplant patients owing to its potent immuno-suppressant activity.
Answer
a)
b)
c)
There are other reasons that we might concern ourselves with an understanding of enantiomers, apart from dietary and olfactory preferences. Perhaps the most dramatic example of the importance of enantiomers can be found in the case of thalidomide. Thalidomide was a drug commonly prescribed during the 1950's and 1960's in order to alleviate nausea and other symptoms of morning sickness. In fact, only one enantiomer of thalidomide had any therapeutic effect in this regard. The other enantiomer, apart from being therapeutically useless in this application, was subsequently found to be a teratogen, meaning it produces pronounced birth defects. This was obviously not a good thing to prescribe to pregnant women. Workers in the pharmaceutical industry are now much more aware of these kinds of consequences, although of course not all problems with drugs go undetected even through the extensive clinical trials required in the United States. Since the era of thalidomide, however, a tremendous amount of research in the field of synthetic organic chemistry has been devoted to methods of producing only one enantiomer of a useful compound and not the other. This effort probably represents the single biggest aim of synthetic organic chemistry through the last quarter century.
• Enantiomers may have very different biological properties.
• Obtaining enantiomerically pure compounds is very important in medicine and the pharmaceutical industry. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.12%3A_Chirality_in_Nature_and_Chiral_Environments.txt |
Concepts & Vocabulary
5.1: Enantiomers and the Tetrahedral Carbon
• Every molecule is either chiral (not superimposable on its mirror image) or achiral (superimposable on its mirror image).
• Chiral molecules do not have a plane of symmetry, while achiral molecules have one or more planes of symmetry.
• Stereoisomers vary by spatial arrangement of atoms but have the same atom connectivity.
• Stereoisomers that are mirror images of one another but are not superimposable are called enantiomers.
5.2: The Reason for Handedness in Molecules - Chirality
• A Tetrahedral carbon atom bonded to four different substituents is an asymetric carbon (also called a stereocenter or chiral carbon), which typically leads to a chiral molecule (meso compounds are the exception in section 5.7).
5.3: Optical Activity
• Enantiomers cause rotation of plane-polarized light in equal amounts in opposite directions. This is called optical activity. Clockwise rotation is called dextrorotatory (+) and counter-clockwise is called levorotatory (-).
• Specific rotation is the amount that a sample of a chemical rotates planne-polarized light. It can be used to calculate the purity of a mixture of enantiomers called the enantiomeric excess.
• Resolution is the separation of a mixture of enatiomers.
• Racemates are defined as a 50:50 mixture of enantiomers, resulting in a sample that is not optically active. The process of forming a racemic mixture is called racemization.
5.4: Pasteur's Discovery of Enantiomers
5.5: Sequence Rules for Specifying Configuration
• Use the CIP rules to determine the priority of each substituent attached to a chiral carbon to determine whether configuration is R or S. With the lowest priority group facing away from you, draw an arc connecting groups 1-2-3. If that arc is clockwise, the configuration is R. If counterclockwise, the configuration is S.
5.6: Diastereomers
• Stereoisomers that are not mirror images of one another are called diastereomers.
• Diastereomers have two or more stereocenters. The configurations of the stereocenters cannot be inverse of each other (example R,R and S,S) because that defines a pair of enantiomers.
5.7: Meso Compounds
• Meso compounds are achiral but have chiral centers. This is caused by having an internal plane of symmetry that allows the two molecules to be superimposable on one another and be optically inactive.
5.8: Racemic Mixtures and the Resolution of Enantiomers
• Each component of a racemic mixture rotates plane polarized light an equal amount in opposite directions, so there is no optical activity.
• Racemic mixtures can be separated into the component enantiomers by reaction with a chiral reagent, which will form diastereomer intermediates of the molecules which can then be separated. Following separation the chiral reagent is removed to yield the two pure enantiomers.
5.9: A Review of Isomerism
• There are several categories of isomers with the largest distinction between:
• constitutional (structural) isomers that contain the same number of each atom but differ in connectivity
• stereoisomers that have all the same atoms with the same connectivity, but only differ in how the atoms are arranged three dimensionally
• In addition to the diastereomers and enantiomers that have been discussed at length in this chapter, stereoisomers can also be:
• cis/trans or E/Z isomers which differ by spatial arrangement around a double bond
• conformational isomers (conformers) which occur due to free rotation of sigma bonds
5.10: Chirality at Nitrogen, Phosphorus, and Sulfur
• Nitrogen when bonded to three different atoms is chiral, however the lone pair of electrons moves freely between positions on the Nitrogen causing these molecules to become a racemic mixture.
• When bonded to four different atoms in quaternary ammonium salts, nitrogen atoms lead to chiral molecules.
• Organic phosphates with four different groups can also be chiral.
5.11: Prochirality
• When a carbon can be converted to a chiral center by changing only one of its attached groups, it is called prochiral.
• If a molecule has two hydrogens on the same atom and replacement of either one with deuterium would lead to enantiomers, the hydrogens are enantiotopic.
• Similarly if this replacement would lead to diastereomer molecules, the hydrogens are diastereotopic.
• If replacement of a hydrogen would not lead to a chiral center being created, they are termed homotopic.
5.12: Chirality in Nature and Chiral Environments
Skills to Master
• Skill 5.1 Identify stereocenters in molecular structures.
• Skill 5.2 Identify whether two structures are identical (not meso), constitutional isomers, enantiomers, diastereomers or meso and identical.
• Skill 5.3 Explain how plane polarized light is used to show optical activity.
• Skill 5.4 Calculate specific rotation from experimental data.
• Skill 5.5 Calculate optical purity and enantiomeric excess from experimental data.
• Skill 5.6 Determine configuration of stereocenters as R or S.
• Skill 5.7 Draw the enantiomer and diastereomers of a given compound with one or more stereocenters.
• Skill 5.8 Identify planes of symmetry in meso compounds.
• Skill 5.9 Describe a process for separating a mixture of enantiomers.
• Skill 5.10 Explain why racemic mixtures are optically inactive.
• Skill 5.11 Explain the difference between constitutional and stereoisomers.
• Skill 5.12 Give an example of a chiral center that is not carbon.
Memorization Tasks
MT 5.1 Memorize the rules for determining R and S configuration.
MT 5.2 Memorize the types of isomers and how to identify them. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.S%3A_Stereochemistry_at_Tetrahedral_Centers_%28Summary%29.txt |
Objective
After completing this section, you should be able to discuss how the results of work carried out by Biot and Pasteur contributed to the development of the concept of the tetrahedral carbon atom.
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. As background for the following example, it is recommended that the section on crystal properties be reviewed.
Tartaric acid, its potassium salt known in antiquity as "tartar", has served as the locus of several landmark events in the history of stereochemistry. In 1832 the French chemist Jean Baptiste Biot observed that tartaric acid obtained from tartar was optically active, rotating the plane of polarized light clockwise (dextrorotatory). An optically inactive, higher melting, form of tartaric acid, called racemic acid was also known. A little more than a decade later, young Louis Pasteur conducted a careful study of the crystalline forms assumed by various salts of these acids. He noticed that under certain conditions, the sodium ammonium mixed salt of the racemic acid formed a mixture of enantiomorphic hemihedral crystals; a drawing of such a pair is shown on the right. Pasteur reasoned that the dissymmetry of the crystals might reflect the optical activity and dissymmetry of its component molecules. After picking the different crystals apart with a tweezer, he found that one group yielded the known dextrorotatory tartaric acid measured by Biot; the second led to a previously unknown levorotatory tartaric acid, having the same melting point as the dextrorotatory acid. Today we recognize that Pasteur had achieved the first resolution of a racemic mixture, and laid the foundation of what we now call stereochemistry.
Optical activity was first observed by the French physicist Jean-Baptiste Biot. He concluded that the change in direction of plane-polarized light when it passed through certain substances was actually a rotation of light, and that it had a molecular basis. His work was supported by the experimentation of Louis Pasteur. Pasteur observed the existence of two crystals that were mirror images in tartaric acid, an acid found in wine. Through meticulous experimentation, he found that one set of molecules rotated polarized light clockwise while the other rotated light counterclockwise to the same extent. He also observed that a mixture of both, a racemic mixture (or racemic modification), did not rotate light because the optical activity of one molecule canceled the effects of the other molecule. Pasteur was the first to show the existence of chiral molecules.
Exercise \(1\)
When you have one enantiomer that rotates plane polarized light clockwise, why does a racemic mixture of that compound not rotate plane polarized light?
Answer
Since a racemic mixture consists of equal amounts of both enantiomers, one enantiomer will rotate plane polarized light clockwise, while the other enantiomer will rotate plane polarized light counterclockwise an equal amount, causing them to cancel out, and the racemic mixture being not optically active. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.xx%3A_Enantiomers_and_Diastereomers.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
Objective
After completing this section, you should be able to list and describe the four important “kinds” of reactions that occur in organic chemistry.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• addition reaction
• elimination reaction
• rearrangement reaction
• substitution reaction
Study Notes
It is sufficient that you know the general form of each kind of reaction. However, given a chemical equation, you should be able to recognize which kind of reaction it involves.
If you scan any organic textbook you will encounter what appears to be a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute one organizational scheme that helps us catalog and remember reactions.
Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course.
First, we identify four broad classes of reactions based solely on the structural change occurring in the reactant molecules. This classification does not require knowledge or speculation concerning reaction paths or mechanisms. The four main reaction classes are additions, eliminations, substitutions, and rearrangements.
Addition Reaction
Elimination Reaction
Substitution Reaction
Rearrangement Reaction
In an addition reaction the number of σ-bonds in the substrate molecule increases, usually at the expense of one or more π-bonds. The reverse is true of elimination reactions, i.e.the number of σ-bonds in the substrate decreases, and new π-bonds are often formed. Substitution reactions, as the name implies, are characterized by replacement of an atom or group (Y) by another atom or group (Z). Aside from these groups, the number of bonds does not change. A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.
The examples illustrated above involve simple alkyl and alkene systems, but these reaction types are general for most functional groups, including those incorporating carbon-oxygen double bonds and carbon-nitrogen double and triple bonds. Some common reactions may actually be a combination of reaction types.
Example \(1\): Reaction of an Ester with Ammonia
The reaction of an ester with ammonia to give an amide, as shown below, appears to be a substitution reaction ( Y = CH3O & Z = NH2 ); however, it is actually two reactions, an addition followed by an elimination.
Example \(2\): The Addition of water to a Nitrile
The addition of water to a nitrile does not seem to fit any of the above reaction types, but it is simply a slow addition reaction followed by a rapid rearrangement, as shown in the following equation. Rapid rearrangements of this kind are called tautomerizations. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.01%3A_Kinds_of_Organic_Reactions.txt |
Objectives
After completing this section, you should be able to
1. explain the difference between heterolytic and homolytic bond breakage, and between heterogenic and homogenic bond formation.
2. state the two reaction types involved in symmetrical and unsymmetrical processes.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• heterogenic
• heterolytic
• homogenic
• homolytic
• polar reaction
• radical reaction
• reaction mechanism
Study Notes
Upon first reading first four key terms, it is easy to be puzzled. The ending of the word tells you whether a bond is being formed (‑genic) or broken (‑lytic), while the root of the word describes the nature of that formation or decomposition. So hetero (meaning different) reactions involve asymmetrical bond making (or breaking) and homo (meaning same) involve symmetrical processes.
Because one pair of electrons constitutes a single bond, the unsymmetrical making or breaking of that bond in a hetero processes are described as polar reactions. Similarly, symmetrical homo processes of bond making and breaking are called radical reactions. Radicals (sometimes referred to as free radicals) are highly reactive neutral chemical species with one unpaired electron. In later sections we discuss radical and polar reactions in more detail.
The Arrow Notation in Mechanisms
Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding (and non-bonding) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism. In general, two kinds of curved arrows are used in drawing mechanisms:
A full head on the arrow indicates the movement or shift of an electron pair:
A partial head (fishhook) on the arrow indicates the shift of a single electron:
The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis. If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis.
Bond-Breaking Bond-Making
Other Arrow Symbols
Chemists also use arrow symbols for other purposes, and it is essential to use them correctly.
The Reaction Arrow The Equilibrium Arrow The Resonance Arrow
The following equations illustrate the proper use of these symbols:
Reactive Intermediates
The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates, and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below.
Charged Intermediates Uncharged Intermediates
a carbocation
a radical
a carbanion
a carbene
A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here.
• Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
• Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid).
Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals.
The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions, since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined below.
The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH4, has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies.
Ionic Reactions
The principles and terms introduced in the previous sections can now be summarized and illustrated by the following three examples. Reactions such as these are called ionic or polar reactions, because they often involve charged species and the bonding together of electrophiles and nucleophiles. Ionic reactions normally take place in liquid solutions, where solvent molecules assist the formation of charged intermediates.
The substitution reaction shown on the left can be viewed as taking place in three steps. The first is an acid-base equilibrium, in which HCl protonates the oxygen atom of the alcohol. The resulting conjugate acid then loses water in a second step to give a carbocation intermediate. Finally, this electrophile combines with the chloride anion nucleophile to give the final product.
The addition reaction shown on the left can be viewed as taking place in two steps. The first step can again be considered an acid-base equilibrium, with the pi-electrons of the carbon-carbon double bond functioning as a base. The resulting conjugate acid is a carbocation, and this electrophile combines with the nucleophilic bromide anion.
The elimination reaction shown on the left takes place in one step. The bond breaking and making operations that take place in this step are described by the curved arrows. The initial stage may also be viewed as an acid-base interaction, with hydroxide ion serving as the base and a hydrogen atom component of the alkyl chloride as an acid.
rearrangement (tautomerism)
There are many kinds of molecular rearrangements. The examples shown on the left are from an important class called tautomerization or, more specifically, keto-enol tautomerization. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.02%3A_How_Organic_Reactions_Occur_-_Mechanisms.txt |
Objectives
After completing this section, you should be able to
1. give an example of a radical substitution reaction.
2. identify the three steps (initiation, propagation and termination) that occur in a typical radical substitution reaction.
3. write out the steps involved in a simple radical substitution reaction, such as the chlorination of methane.
4. explain why the halogenation of an alkane is not a particularly useful method of preparing specific alkyl halides.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• chain reaction
• initiation step
• propagation step
• radical substitution
• termination step
Study Notes
A radical substitution reaction is a reaction which occurs by a free radical mechanism and results in the substitution of one or more of the atoms or groups present in the substrate by different atoms or groups.
The initiation step in a radical chain reaction is the step in which a free radical is first produced. A termination step of a radical chain reaction is one in which two radicals react together in some way so that the chain can no longer be propagated.
While radical halogenation of very simple alkanes can be an effective synthetic strategy, it cannot be employed for larger more complex alkanes to yield specific alkyl halides, since the reactive nature of radicals always leads to mixtures of single- and multiple-halogenated products.
The Three Phases of Radical Chain Reactions
Because of their high reactivity, free radicals have the potential to be both extremely powerful chemical tools and extremely harmful contaminants. Much of the power of free radical species stems from the natural tendency of radical processes to occur in a chain reaction fashion. Radical chain reactions have three distinct phases: initiation, propagation, and termination.
The initiation phase describes the step that initially creates a radical species. In most cases, this is a homolytic cleavage event, and takes place very rarely due to the high energy barriers involved. Often the influence of heat, UV radiation, or a metal-containing catalyst is necessary to overcome the energy barrier.
Molecular chlorine and bromine will both undergo homolytic cleavage to form radicals when subjected to heat or light. Other functional groups which also tend to form radicals when exposed to heat or light are chlorofluorocarbons, peroxides, and the halogenated amide N-bromosuccinimide (NBS).
The propagation phase describes the 'chain' part of chain reactions. Once a reactive free radical is generated, it can react with stable molecules to form new free radicals. These new free radicals go on to generate yet more free radicals, and so on. Propagation steps often involve hydrogen abstraction or addition of the radical to double bonds.
Chain termination occurs when two free radical species react with each other to form a stable, non-radical adduct. Although this is a very thermodynamically downhill event, it is also very rare due to the low concentration of radical species and the small likelihood of two radicals colliding with one another. In other words, the Gibbs free energy barrier is very high for this reaction, mostly due to entropic rather than enthalpic considerations. The active sites of enzymes, of course, can evolve to overcome this entropic barrier by positioning two radical intermediates adjacent to one another. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.03%3A_Radical_Reactions.txt |
Objectives
After completing this section, you should be able to
1. identify the positive and negative ends of the bonds present in the common functional groups.
2. explain how bond polarity can be enhanced by the interaction of a functional group with a solvent, metal cation or acid.
3. explain how the polarizability of an atom can be an important factor in determining the reactivity of a bond.
4. describe the heterolytic bond-breaking process.
5. use curved (curly) arrows to indicate the movement of electron pairs during bond breakage and bond formation.
6. predict whether a given species (compound or ion) is likely to behave as a nucleophile or as an electrophile.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electrophile
• nucleophile
• polar reaction
• polarizability
Study Notes
You may wish to review Section 2.1 before you begin this section. The relative electronegativities of the elements shown in the periodic table should already be familiar. Remember that it is the relative electronegativities that are important, not the actual numerical values.
Make sure that you understand the polarity patterns of the common functional groups. Do not try to memorize these polarities; rather, concentrate on why they arise. You will encounter these group polarities so frequently that they will soon become “second nature” to you.
Halogens and the Character of the Carbon-Halogen Bond
With respect to electronegativity, most halogens are more electronegative than carbon. This results in a carbon-halogen bond that is polarized. As shown in the image below, the carbon atom has a partial positive charge, while the halogen has a partial negative charge.
The following table shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
Electronegativity I < Br < Cl < F
The following table shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
Bond length C–F < C–Cl < C–Br < C–I
Bond strength C–I < C–Br < C–Cl < C–F
Molecular size F < Cl < Br < I
The influence of bond polarity
Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples:
The electronegativities of carbon and iodine are not very different, and so there will be little separation of charge on the bond. One of the important set of reactions of alkyl halides involves replacing the halogen by something else - substitution reactions. These reactions can involve the carbon-halogen bond breaking to give positive and negative ions. The ion with the positively charged carbon atom then reacts with something either fully or slightly negatively charged. Alternatively, something either fully or negatively charged is attracted to the slightly positive carbon atom and pushes off the halogen atom.
You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. It is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one.
The Carbonyl Group
C=O is prone to additions and nucleophillic attack because or carbon's positive charge and oxygen's negative charge. The resonance of the carbon partial positive charge allows the negative charge on the nucleophile to attack the Carbonyl group and become a part of the structure and a positive charge (usually a proton hydrogen) attacks the oxygen. Just a reminder, the nucleophile is a good acid therefore "likes protons" so it will attack the side with a positive charge.
Before we consider in detail the reactivity of aldehydes and ketones, we need to look back and remind ourselves of what the bonding picture looks like in a carbonyl. Carbonyl carbons are sp2 hybridized, with the three sp2 orbitals forming overlaps with orbitals on the oxygen and on the two carbon or hydrogen atoms. These three bonds adopt trigonal planar geometry. The remaining unhybridized 2p orbital on the central carbonyl carbon is perpendicular to this plane, and forms a ‘side-by-side’ pi bond with a 2p orbital on the oxygen.
The carbon-oxygen double bond is polar: oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Recall that bond polarity can be depicted with a dipole arrow, or by showing the oxygen as holding a partial negative charge and the carbonyl carbon a partial positive charge.
A third way to illustrate the carbon-oxygen dipole is to consider the two main resonance contributors of a carbonyl group: the major form, which is what you typically see drawn in Lewis structures, and a minor but very important contributor in which both electrons in the pbond are localized on the oxygen, giving it a full negative charge. The latter depiction shows the carbon with an empty 2p orbital and a full positive charge.
Some Carbonyl Compounds
Compound aldehyde ketone formaldehyde carboxylic acid ester amide enone acyl halide acid anhydride
Structure
General Formula RCHO RCOR' CH2O RCOOH RCOOR' RCONR'R'' RC(O)C(R')CR''R''' RCOX RCO2COR'
Nucleophilic Addition to Aldehydes and Ketones
The result of carbonyl bond polarization, however it is depicted, is straightforward to predict. The carbon, because it is electron-poor, is an electrophile: it is a great target for attack by an electron-rich nucleophilic group. Because the oxygen end of the carbonyl double bond bears a partial negative charge, anything that can help to stabilize this charge by accepting some of the electron density will increase the bond’s polarity and make the carbon more electrophilic. Very often a general acid group serves this purpose, donating a proton to the carbonyl oxygen.
The same effect can also be achieved if a Lewis acid, such as a magnesium ion, is located near the carbonyl oxygen. Unlike the situation in a nucleophilic substitution reaction, when a nucleophile attacks an aldehyde or ketone carbon there is no leaving group – the incoming nucleophile simply ‘pushes’ the electrons in the pi bond up to the oxygen.
Alternatively, if you start with the minor resonance contributor, you can picture this as an attack by a nucleophile on a carbocation.
After the carbonyl is attacked by the nucleophile, the negatively charged oxygen has the capacity to act as a nucleophile. However, most commonly the oxygen acts instead as a base, abstracting a proton from a nearby acid group in the solvent or enzyme active site.
This very common type of reaction is called a nucleophilic addition. In many biologically relevant examples of nucleophilic addition to carbonyls, the nucleophile is an alcohol oxygen or an amine nitrogen, or occasionally a thiol sulfur. In one very important reaction type known as an aldol reaction, the nucleophile attacking the carbonyl is a resonance-stabilized carbanion. In this chapter, we will concentrate on reactions where the nucleophile is an oxygen or nitrogen.
Nucleophile?
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates.
More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
Of course, carbons can also be nucleophiles - otherwise how could new carbon-carbon bonds be formed in the synthesis of large organic molecules like DNA or fatty acids? Enolate ions (section 7.5) are the most common carbon nucleophiles in biochemical reactions, while the cyanide ion (CN-) is just one example of a carbon nucleophile commonly used in the laboratory. Reactions with carbon nucleophiles will be dealt with in chapters 13 and 14, however - in this chapter and the next, we will concentrate on non-carbon nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Neutral Nucleophiles Charged Nucleophiles
H2O, NH3, RNH2, R2NH, R3N, ROH, RCOOH, RSH, and PR3 OH, OR, NH2, NHR, NR2, SH, SR, SeR, Cl, Br, I, F, CN, RCOO
Electrophiles
In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, i.e. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the SN1 vs. SN2 character of a nucleophilic substitution reaction.
Consider two hypothetical SN2 reactions: one in which the electrophile is a methyl carbon and another in which it is tertiary carbon.
Because the three substituents on the methyl carbon electrophile are tiny hydrogens, the nucleophile has a relatively clear path for backside attack. However, backside attack on the tertiary carbon is blocked by the bulkier methyl groups. Once again, steric hindrance - this time caused by bulky groups attached to the electrophile rather than to the nucleophile - hinders the progress of an associative nucleophilic (SN2) displacement.
The factors discussed in the above paragraph, however, do not prevent a sterically-hindered carbon from being a good electrophile - they only make it less likely to be attacked in a concerted SN2 reaction. Nucleophilic substitution reactions in which the electrophilic carbon is sterically hindered are more likely to occur by a two-step, dissociative (SN1) mechanism. This makes perfect sense from a geometric point of view: the limitations imposed by sterics are significant mainly in an SN2 displacement, when the electrophile being attacked is a sp3-hybridized tetrahedral carbon with its relatively ‘tight’ angles of 109.4o. Remember that in an SN1 mechanism, the nucleophile attacks an sp2-hybridized carbocation intermediate, which has trigonal planar geometry with ‘open’ 120 angles.
With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.04%3A_Polar_Reactions.txt |
Objectives
After completing this section, you should be able to
1. give an example of a simple polar reaction (e.g., a electrophilic addition).
2. identify the electrophile and nucleophile in a simple polar reaction.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electrophilic addition
• carbocation
Study Notes
The curved arrows introduced in this section are used throughout the course to indicate the movement of electron pairs. It takes practice for the beginning student to feel comfortable using these arrows. Remember that the head of the arrow indicates where the electron pair moves to; its tail shows where the electron pair comes from. (Chemists often refer to the use of curved arrows as “electron pushing.")
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.
Addition to symmetrical alkenes
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other. For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Mechanism
The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both an electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of pi electrons, attaching it to the molecule (1). This is basically the reverse of the last step in the E1 reaction (deprotonation step). The resulting molecule will have a single carbon-carbon bond with a positive charge on one of them (carbocation). The next step is when the nucleophile (halide) bonds to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant (2). The second step will only occur if a good nucleophile is used.
Mechanism of Electrophilic Addition of Hydrogen Halide to Ethene
Mechanism of Electrophilic Addition of Hydrogen Halide to Propene
All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.
Reaction rates
Variation of rates when you change the halogen
Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions. When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
Variation of rates when you change the alkene
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be. Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond.
For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this. Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.
The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride. The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Addition to unsymmetrical alkenes
In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition - in other words, which way around the hydrogen and the halogen add across the double bond.
Orientation of addition
If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product.
This is in line with Markovnikov's Rule which says:
When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.
In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant.
Exercises
1. Supply the missing curved arrows in the equations given below.
Answers: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.05%3A_An_Example_of_a_Polar_Reaction_-_Addition_of_HBr_to_Ethylene.txt |
Objective
After completing this section, you should be able to use curved (curly) arrows, in conjunction with a chemical equation, to show the movement of electron pairs in a simple polar reaction, such as electrophilic addition.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electrophlic
• nucleophlic
Pushing Electrons and Curved Arrows
Understanding the location of electrons and being able to draw the curved arrows that depict the mechanisms by which the reactions occur is one of the most critical tools for learning organic chemistry since they allow you to understand what controls reactions, and how reactions proceed.
Before you can do this you need to understand that a bond is due to a pair of electrons shared between atoms.
When asked to draw a mechanism, curved arrows should be used to show all the bonding changes that occur.
A few simple lessons that illustrate these concepts can be found below.
Lesson 1
If we remove the pair of electrons in a bond, then we BREAK that bond. This is true for single and multiple bonds as shown below:
Notice that since the starting materials were neutral, the products are also neutral. In general terms, the sum of the charges on the starting materials MUST equal the sum of the charges on the products since we have the same number of electrons.
The first example is a REACTION since we broke a sigma bond. In the second two examples, we moved pi electrons into long pairs. This is RESONANCE.
If we move electrons between two atoms, then we MAKE a new bond:
We always show electrons moving from electron rich to electron poor.
Lesson 2
This is a simple acid/base reaction, showing the formation of the hydronium ion produced when hydrochloric acid is dissolved in water. It is useful to analyze the bond changes that are occurring. Water is functioning as a base and hydrochloric acid as an acid. Consider the differences in bonding between the starting materials and the products:
One of the lone pairs on the oxygen atom of water was used to form a bond to a hydrogen atom, creating the hydronium ion (H3O+) seen in the products. The hydrogen-chlorine bond of HCl was broken, and the electrons in this bond became a lone pair on the chlorine atom, thus generating a chloride ion. We can illustrate these changes in bonding using the curved arrows shown below.
Note that in this diagram, the overall charge of the reactants is the same as the overall charge of the products. We can also show the curved arrows for the reverse reaction:
This shows the formation of the new H-Cl bond by using a lone pair of electrons from the electron-rich chloride ion to form a bond to an electron poor hydrogen atom of the hydronium ion. Because hydrogen can only form one bond, the oxygen-hydrogen bond is broken and its electrons become a lone pair on the electron-poor oxygen atom. Notice that the charges balance!
Lesson 3
In this section, we will look at the curved arrows for some nucleophilic substitution reactions. Overall, the processes involved are similar to those for the acid/base reactions described above. In a nucleophilic substitution reaction, an electron-rich nucleophile (Nu) becomes bonded to an electron-poor carbon atom, and a leaving group (LG) is displaced. In bonding terms, we must make a Nu-C bond and break a C-LG bond.
Let's consider the stepwise SN1 reaction between (1-chloroethyl)benzene and sodium cyanide. The first step of this process is breaking the C-Cl bond, where the electrons in that bond become a lone pair on the chlorine atom. The carbon atom has lost electrons and therefore becomes positive, generating a secondary carbocation. Because the chlorine atom gained an additional lone pair of electrons, it becomes a negatively charged chloride ion.
In the second step, the electron-rich nucleophile donates electrons to form a new C-C bond with the electron-poor secondary carbocation.
In an SN2 reaction, the bond forming and breaking processes occur simultaneously. The scheme below shows the Nu donating electrons to form a new C-C bond at the same time that the C-Cl bond is breaking. The electrons in the C-Cl bond become a long pair on the chlorine atom, generating a chloride ion. Forming and breaking the bonds simultaneously allows carbon to obey the octet rule throughout this process.
Notice that in all steps for the processes above, the overall charges of the starting materials match those of the products.
Lesson 4
This section will dissect another substitution reaction, although it is more involved. Let's consider the SN1 reaction of tert-butyl bromide with water.
It can be helpful to take inventory of which bonds have been formed, and which bonds have been broken.
The curved arrows we draw must account for ALL of these bonding changes. Since we are dealing with an SN1 reaction process, the first step will be cleavage of the C-Br bond to give a carbocation and and a bromide anion.
Water then acts as a nucleophile, using one of its lone pairs to form a bond to the electron-poor t-butyl cation. This generates an oxonium ion, where oxygen has three bonds and a positive formal charge.
The final step is an acid/base reaction between the bromide anion generated in step 1 and the oxonium product of step 2. The bromide anion acts as a base, using a lone pair to form a bond to one of the hydrogen atoms. The O-H bond then breaks, and its electrons become a lone pair on oxygen. This gives the final products of HBr and t-butyl alcohol.
Notice that in each of the mechanistic steps above, the overall charge of the reactant side balances with the overall charge of the product side.
While the above process was broken down into distinct steps, however it is important to note that mechanisms are almost always shown as a continuous process. The overall mechanism for this processes can be found below:
Now consider the reverse reaction, i.e. the reaction of t-butyl alcohol with hydrobromic acid to generate t-butyl bromide and water. The scheme is shown below, along with an analysis of the bonds formed and broken in this process:
The mechanism must occur via the same pathway as shown above (Law of Macroscopic Reversibility), however this mechanism can still be deduced without knowing that. First, it is known that HBr is a strong acid and can donate a proton to a base. The most basic sites in the whole system are the lone pairs on the oxygen atom of t-butanol. Since the lone pairs are the electron-rich area of the molecule, the arrow starts at a lone pair and ends at the proton of HBr. The H-Br bond breaks, pushing its electrons onto the bromine atom and generating a bromide ion.
The bromide ion generated in the first step can then react with the t-butyl cation to generate t-butyl bromide.
Once again, the above the overall process is broken down into individual steps, however it is more common to illustrate this as one overall process:
Curved Arrow Summary
• Curved arrows flow from electron rich to electron poor.
• Therefore they start from lone pairs or bonds.
• The charges in any particular step should always be balanced.
• Remember to obey the rules of valence (eg. octet rule for C,N,O,F etc.)
• If electrons are taken out of a bond, then that bond is broken.
• If electrons are placed between two atoms then it implies a bond is being made.
Exercises
Draw curved arrows to indicate mechanisms for the following reactions: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.06%3A_Using_Curved_Arrows_in_Polar_Reaction_Mechanisms.txt |
Objectives
After completing this section, you should be able to
1. write the equilibrium constant expression for a given reaction.
2. assess, qualitatively, how far a reaction will proceed in a given direction, given the value of Keq.
3. explain the difference between rate and equilibrium.
4. state the relationship between ΔG° and Keq, and use this relationship to determine the value of either of the two variables, given the other.
5. state the relationship between Gibbs free-energy, enthalpy and entropy, and use the relationship to calculate any one of ΔG°, ΔH° and ΔS°, given the other two.
6. make a qualitative assessment of whether ΔS° for a given process is expected to be positive or negative.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• exergonic
• endergonic
• exothermic
• endothermic
• enthalpy change (heat of reaction), ΔH°
• entropy change, ΔS°
• reaction mechanism
• standard Gibbs free-energy change, ΔG°
Study Notes
Throughout this course you will be paying a great deal of attention to the mechanisms of the reactions that you study. Some students see this as a laborious task of little practical use. However, you will find that a knowledge of reaction mechanisms can help reduce the number of reactions to memorize, provide a connecting link between apparently unrelated reactions, and enable someone with a basic knowledge of organic chemistry to deduce how a previously unseen reaction might proceed. The investigation of reaction mechanisms is a popular research area for organic chemists.
Equilibrium Constant
For the hypothetical chemical reaction:
$aA + bB \rightleftharpoons cC + dD \nonumber$
the equilibrium constant is defined as:
$K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber$
The notation [A] signifies the molar concentration of species A. An alternative expression for the equilibrium constant involves partial pressures:
$K_P = \dfrac{P_C^c P_D^d}{P_A^aP_B^b} \nonumber$
Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction
$\ce{CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g)} \nonumber$
is the following:
$K_C = \dfrac{[H_2]^2}{[H_2O]^2} \nonumber$
Observe that the gas-phase species $\ce{H2O}$ and $\ce{H2}$ appear in the expression but the solids $\ce{CaH2}$ and $\ce{Ca(OH)2}$ do not appear.
The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action.
Free Energy
The interaction between enthalpy and entropy changes in chemical reactions is best observed by studying their influence on the equilibrium constants of reversible reactions. To this end a new thermodynamic function called Free Energy (or Gibbs Free Energy), symbol $ΔG$, is defined as shown in the first equation below. Two things should be apparent from this equation. First, in cases where the entropy change is small, $ΔG ≅ ΔH$. Second, the importance of $ΔS$ in determining $ΔG$ increases with increasing temperature.
$ΔGº = ΔHº – TΔSº \nonumber$
where the temperature is measured in absolute temperature (K).
The free energy function provides improved insight into the thermodynamic driving forces that influence reactions. A negative $ΔGº$ is characteristic of an exergonic reaction, one which is thermodynamically favorable and often spontaneous, as is the melting of ice at 1 ºC. Likewise a positive $ΔGº$ is characteristic of an endergonic reaction, one which requires an input of energy from the surroundings.
For an example of the relationship of free energy to enthalpy consider the decomposition of cyclobutane to ethene, shown in the following equation. The standard state for all the compounds is gaseous.
This reaction is endothermic, but the increase in number of molecules from one (reactants) to two (products) results in a large positive ΔSº.
At 25 ºC (298 K):
ΔGº = 19 kcal/mol – 298(43.6) cal/mole = 19 – 13 kcal/mole = +6 kcal/mole.
Thus, the entropy change opposes the enthalpy change, but is not sufficient to change the sign of the resulting free energy change, which is endergonic. Indeed, cyclobutane is perfectly stable when kept at room temperature.
Because the entropy contribution increases with temperature, this energetically unfavorable transformation can be made favorable by raising the temperature. At 200 ºC (473 K),
\begin{align} ΔGº &= 19\, kcal/mol – 473(43.6)\, cal/mole \[4pt] &= 19 – 20.6\, kcal/mole \[4pt] &= –1.6 kcal/mole.\end{align} \nonumber
This is now an exergonic reaction, and the thermal cracking of cyclobutane to ethene is known to occur at higher temperatures.
$ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}$
where R = 1.987 cal/ K mole T = temperature in K and K = equilibrium constant
Note
Equation \ref{eq2} is important because it demonstrates the fundamental relationship of $ΔGº$ to the equilibrium constant, $K$. Because of the negative logarithmic relationship between these variables, a negative ΔGº generates a K>1, whereas a positive ΔGº generates a K<1. When ΔGº = 0, K = 1. Furthermore, small changes in ΔGº produce large changes in K. A change of 1.4 kcal/mole in ΔGº changes K by approximately a factor of 10. This interrelationship may be explored with the calculator on the right. Entering free energies outside the range -8 to 8 kcal/mole or equilibrium constants outside the range 10-6 to 900,000 will trigger an alert, indicating the large imbalance such numbers imply.
Exercises
1. At 155°C, the equilibrium constant, Keq, for the reaction
$\ce{\sf{CH3CO2H + CH3CH2OH <=> CH3CO2CH2CH3 + H2O}} \nonumber$
has a value of 4.0. Calculate ΔG° for this reaction at 155°C.
2. Acetylene (C2H2) can be converted into benzene (C6H6) according to the equation:
$\ce{\sf{3H-C#C-H (g) <=> C6H6 (l)}} \nonumber$
At 25°C, ΔG° for this reaction is −503 kJ and ΔH° is −631 kJ. Determine ΔS° and indicate whether the size of ΔS° agrees with what you would have predicted simply by looking at the chemical equation.
Answers:
1. The entropy change is negative, as one would expect from looking at the chemical equation, since three moles of reactants yield one mole of product; that is, the system becomes much more “ordered” as it goes from reactants to products. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.07%3A_Describing_a_Reaction_-_Equilibria__Rates_and_Energy_Changes.txt |
Objectives
After completing this section, you should be able to
1. predict the value of ΔH° for a gas-phase reaction, given the necessary bond dissociation energy data.
2. predict the dissociation energy of a particular bond, given ΔH° for a reaction involving the bond and any other necessary bond dissociation energy data.
3. outline the limitations of using bond dissociation energies to predict whether or not a given reaction will occur.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bond dissociation energy
• solvation
Study Notes
The idea of calculating the standard enthalpy of a reaction from the appropriate bond dissociation energy data should be familiar to you from your first-year chemistry course.
Solvation is the interaction between solvent molecules and the ions or molecules dissolved in that solvent.
The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The SI units used to describe bond energy are kiloJoules per mole of bonds (kJ/Mol). It indicates how strongly the atoms are bonded to each other.
Introduction
Breaking a covalent bond between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another
$A-B \rightarrow A^+ + B:^- \nonumber$
or
$A-B \rightarrow A:^- + B^+ \nonumber$
or homolytically, where one electron stays with each partner.
$A-B \rightarrow A^• + B^• \nonumber$
The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond.
Calculation of the BDE
The BDE for a molecule A-B is calculated as the difference in the enthalpies of formation of the products and reactants for homolysis
$BDE = \Delta_fH(A^•) + \Delta_fH(B^•) - \Delta_fH(A-B) \nonumber$
Officially, the IUPAC definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $D_o$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation enthalpy, which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes.
Bond Breakage/Formation
Bond dissociation energy (or enthalpy) is a state function and consequently does not depend on the path by which it occurs. Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using Hess's Law can be used to estimate reaction enthalpies.
Example 6.8.1: Chlorination of Methane
Consider the chlorination of methane
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl \nonumber$
the overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed
CH4 → CH3• + H• BDE(CH3-H)
Cl2 → 2Cl• BDE(Cl2) \nonumber \]
H• + Cl• → HCl -BDE(HCl)
CH3• + Cl• → CH3Cl -BDE(CH3-Cl)
---------------------------------------------------
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl \nonumber$
$\Delta H = BDE(R-H) + BDE(Cl_2) - BDE(HCl) - BDE(CH_3-Cl) \nonumber$
Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available.
Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in free radical chlorination of alkanes is the abstraction of hydrogen from the alkane to form a free radical.
RH + Cl• → R• + HCl
The energy change for this step is equal to the difference in the BDEs in RH and HCl
$\Delta H = BDE(R-H) - BDE(HCl) \nonumber$
This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller. The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds.
Table 6.8.1: Representative C-H BDEs in Organic Molecules
R-H Do, kJ/mol D298, kJ/mol R-H Do, kJ/mol D298, kJ/mol
CH3-H 432.7±0.1 439.3±0.4 H2C=CH-H 456.7±2.7 463.2±2.9
CH3CH2-H 423.0±1.7 C6H5-H 465.8±1.9 472.4±2.5
(CH3)2CH-H 412.5±1.7 HCCH 551.2±0.1 557.8±0.3
(CH3)3C-H 403.8±1.7
H2C=CHCH2-H 371.5±1.7
HC(O)-H 368.6±0.8 C6H5CH2-H 375.3±2.5
CH3C(O)-H 374.0±1.2
Trends in C-H BDEs
It is important to remember that C-H BDEs refer to the energy it takes to break the bond, and is the difference in energy between the reactants and the products. Therefore, it is not appropriate to interpret BDEs solely in terms of the "stability of the radical products" as is often done.
Analysis of the BDEs shown in the table above shows that there are some systematic trends:
1. BDEs vary with hybridization: Bonds with sp3 hybridized carbons are weakest and bonds with sp hybridized carbons are much stronger. The vinyl and phenyl C-H bonds are similar, reflecting their sp2 hybridization. The correlation with hybridization can be viewed as a reflection of the C-H bond lengths. Longer bonds formed with sp3 orbitals are consequently weaker. Shorter bonds formed with orbitals that have more s-character are similarly stronger.
2. C-H BDEs vary with substitution: Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons.
Interpretation of C-H BDEs for sp3 Hybridized Carbons
The interpretation of the BDEs in saturated molecules has been subject of recent controversy. As indicated above, the variation in BDEs with substitution has traditionally been interpreted as reflecting the stabilities of the alkyl radicals, with the assessment that more highly substituted radicals are more stable, as with carbocations. Although this is a popular explanation, it fails to account fo the fact the bonds to groups other than H do not show the same types of variation.
R BDE(R-CH3) BDE(R-Cl) BDE(R-Br) BDE(R-OH)
CH3- 377.0±0.4 350.2±0.4 301.7±1.3 385.3±0.4
CH3CH2- 372.4±1.7 354.8±2.1 302.9±2.5 393.3±1.7
(CH3)2CH- 370.7±1.7 356.5±2.1 309.2±2.9 399.6±1.7
(CH3)3C- 366.1±1.7 355.2±2.9 303.8±2.5 400.8±1.7
Therefore, although C-CH3 bonds get weaker with more substitution, the effect is not nearly as large as that observed with C-H bonds. The strengths of C-Cl and C-Br bonds are not affected by substitution, despite the fact that the same radicals are formed as when breaking C-H bonds, and the C-OH bonds in alcohols actually increase with more substitution.
Gronert has proposed that the variation in BDEs is alternately explained as resulting from destabilization of the reactants due to steric repulsion of the substituents, which is released in the nearly planar radicals.1 Considering that BDEs reflect the relative energies of reactants and products, either explanation can account for the trend in BDEs.
Another factor that needs to be considered is the electronegativity. The Pauling definition of electronegativity says that the bond dissociation energy between unequal partners is going to be dependent on the difference in electrongativities, according to the expression
$D_o(A-B) = \dfrac{D_o(A-A) + D_o(B-B)}{2} + (X_A - X_B)^2 \nonumber$
where $X_A$ and $X_B$ are the electronegativities and the bond energies are in eV. Therefore, the variation in BDEs can be interpreted as reflecting variation in the electronegativities of the different types of alkyl fragments.
There is likely some merit in all three interpretations. Since Gronert's original publication of his alternate explanation, there have been many desperate attempts to defend the radical stability explanation.
Further Reading
MasterOrganicChemistry
Bond Strengths And Radical Stability
Exercises
1. Given that ΔH° for the reaction
CH4 (g) + 4F2 (g) → CF4 (g) + 4HF (g)
is −1936 kJ, use the following data to calculate the average bond energy of the C-F bonds in CF4.
Bond Average Bond Energy
$\ce{\sf{C-H}}$ 413 kJ · mol−1
$\ce{\sf{F-F}}$ 155 kJ · mol−1
$\ce{\sf{H-F}}$ 567 kJ · mol−1
2. Calculate ΔH° for the reactions given below. [Access bond energy tables via left-hand blue column Resources > Reference Tables > Reference Tables > Thermodynamic Tables > T3: Bond Energies]
1. CH3CH2OCH3 + HI -> CH3CH2OH + CH3I
2. CH3Cl + NH3 -> CH3NH2 + HCl
Answers:
1. Bonds broken:
4 mol C-H bonds $×\text{}\frac{\left(413\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=1652\text{\hspace{0.17em}}\text{kJ}$
4 mol F-F bonds $×\text{}\frac{\left(155\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=620\text{\hspace{0.17em}}\text{kJ}$
Bonds formed:
4 mol CF bonds $×\text{}\frac{\left(x\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=4x\text{\hspace{0.17em}}\text{kJ}$
(where x = the average energy of one mole of C-F bonds in CF4, expressed in kJ)
4 mol H-F bonds $×\text{}\frac{\left(567\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=2268\text{\hspace{0.17em}}\text{kJ}$
$\begin{array}{rl}\Delta H°& =\Delta H°\left(\text{bonds broken}\right)-\Delta H°\left(\text{bonds formed}\right)\ & =\left(1652\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}\right)-\left(4x+2268\text{\hspace{0.17em}}\text{kJ}\right)\ & =1652\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}-4x-2268\text{\hspace{0.17em}}\text{kJ}\ & =-1936\text{\hspace{0.17em}}\text{kJ}\end{array}$
Thus,
$\begin{array}{rl}4x& =1936\text{\hspace{0.17em}}\text{kJ}-2268\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}+1652\text{\hspace{0.17em}}\text{kJ}\ & =1940\text{\hspace{0.17em}}\text{kJ}\end{array}$
and
$\begin{array}{rl}x& =\frac{1940\text{\hspace{0.17em}}\text{kJ}}{4\text{\hspace{0.17em}}\text{mol}}\ & =385\text{\hspace{0.17em}}\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$
The average energy of a C-F bond in CF4 is 385 kJ · mol-1
1. CH3CH2OCH3 + HI → CH3CH2OH + CH3I
$\dfrac { \begin{array}{cc} \textrm{Reactant bonds broken} & D \[6pt] \ce{CH3CH2O-CH3} & {339 \,\, \textrm{kJ/mol}} \ \ce{H-I} & {298 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Reactant bonds broken}}} & {637 \,\, \textrm{kJ/mol}} \end{array} } \quad \dfrac { \begin{array}{cc} \textrm{Product bonds formed} & D \[6pt] \ce{CH3CH2O-H} & {438 \,\, \textrm{kJ/mol}} \ \ce{CH3-I} & {238 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Product bonds formed}}} & {676 \,\, \textrm{kJ/mol}} \end{array} }$
\begin{align*} \Delta H^{\circ} & = D_{\textrm{bonds broken}} + D_{\textrm{bonds formed}}\ & = {637 \,\, \textrm{kJ/mol} - 676 \,\, \textrm{kJ/mol}} \ & = {-39 \,\, \textrm{kJ/mol}} \end{align*}
2. CH3Cl + NH3 → CH3NH2 + HCl
$\dfrac { \begin{array}{cc} \textrm{Reactant bonds broken} & D \[6pt] \ce{CH3-Cl} & {356 \,\, \textrm{kJ/mol}} \ \ce{NH2-H} & {450 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Reactant bonds broken}}} & {806 \,\, \textrm{kJ/mol}} \end{array} } \quad \dfrac { \begin{array}{cc} \textrm{Product bonds formed} & D \[6pt] \ce{CH3-NH2} & {364 \,\, \textrm{kJ/mol}} \ \ce{H-Cl} & {432 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Product bonds formed}}} & {796 \,\, \textrm{kJ/mol}} \end{array} }$
\begin{align*} \Delta H^{\circ} & = D_{\textrm{bonds broken}} + D_{\textrm{bonds formed}}\ & = {806 \,\, \textrm{kJ/mol} - 796 \,\, \textrm{kJ/mol}} \ & = {+10 \,\, \textrm{kJ/mol}} \end{align*} | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.08%3A_Describing_a_Reaction_-_Bond__Dissociation_Energies.txt |
Objectives
After completing this section, you should be able to
1. sketch the reaction energy diagram for a single-step reaction, given some indication of whether the reaction is fast or slow, exothermic or endothermic.
2. interpret the reaction energy diagram for a single-step process (e.g., use the diagram to decide whether the reaction is exothermic or endothermic).
3. suggest possible transition-state structures for simple one-step processes.
4. assess the likelihood of a reaction occurring at room temperature, given the value of the activation energy ΔG.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• activation energy, ΔG
• reaction energy diagram
• transition state
Study Notes
You may have been taught to use the term “activated complex” rather than “transition state,” as the two are often used interchangeably. Similarly, the activation energy of a reaction is often represented by the symbol Eact or Ea.
You may recall from general chemistry that it is often convenient to describe chemical reactions with energy diagrams. In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products. The energy diagram for a typical one-step reaction might look like this:
Despite its apparent simplicity, this energy diagram conveys some very important ideas about the thermodynamics and kinetics of the reaction. Recall that when we talk about the thermodynamics of a reaction, we are concerned with the difference in energy between reactants and products, and whether a reaction is ‘downhill’ (exergonic, energy releasing) or ‘uphill (endergonic, energy absorbing). When we talk about kinetics, on the other hand, we are concerned with the rate of the reaction, regardless of whether it is uphill or downhill thermodynamically.
First, let’s review what this energy diagram tells us about the thermodynamics of the reaction illustrated by the energy diagram above. The energy level of the products is lower than that of the reactants. This tells us that the change in standard Gibbs Free Energy for the reaction (Δrnx) is negative. In other words, the reaction is exergonic, or ‘downhill’. Recall that the Δrnx term encapsulates both Δrnx, the change in enthalpy (heat) and Δrnx , the change in entropy (disorder):
$ΔG˚ = ΔH˚- TΔS˚ \nonumber$
where T is the absolute temperature in Kelvin. For chemical processes where the entropy change is small (~0), the enthalpy change is essentially the same as the change in Gibbs Free Energy. Energy diagrams for these processes will often plot the enthalpy (H) instead of Free Energy for simplicity.
The standard Gibbs Free Energy change for a reaction can be related to the reaction's equilibrium constant ($K_{eq}\_) by a simple equation: $ΔG˚ = -RT \ln K_{eq} \nonumber$ where: • Keq = [product] / [reactant] at equilibrium • R = 8.314 J×K-1×mol-1 or 1.987 cal× K-1×mol-1 • T = temperature in Kelvin (K) If you do the math, you see that a negative value for Δrnx (an exergonic reaction) corresponds - as it should by intuition - to Keq being greater than 1, an equilibrium constant which favors product formation. In a hypothetical endergonic (energy-absorbing) reaction the products would have a higher energy than reactants and thus Δrnx would be positive and Keq would be less than 1, favoring reactants. Now, let's move to kinetics. Look again at the energy diagram for exergonic reaction: although it is ‘downhill’ overall, it isn’t a straight downhill run. First, an ‘energy barrier’ must be overcome to get to the product side. The height of this energy barrier, you may recall, is called the ‘activation energy’ (ΔG). The activation energy is what determines the kinetics of a reaction: the higher the energy hill, the slower the reaction. At the very top of the energy barrier, the reaction is at its transition state (TS), which is the point at which the bonds are in the process of breaking and forming. The transition state is an ‘activated complex’: a transient and dynamic state that, unlike more stable species, does not have any definable lifetime. It may help to imagine a transition state as being analogous to the exact moment that a baseball is struck by a bat. Transition states are drawn with dotted lines representing bonds that are in the process of breaking or forming, and the drawing is often enclosed by brackets. Here is a picture of a likely transition state for a substitution reaction between hydroxide and chloromethane: Note that this species is absent from the chemical equation (that is it is neither a reactant nor product) $\ce{CH_3Cl + HO^{-} \rightarrow CH_3OH + Cl^{-}} \nonumber$ This reaction involves a collision between two molecules: for this reason, we say that it has second order kinetics. The rate expression for this type of reaction is: $\text{rate} = k[\text{reactant 1}][\text{reactant 2}] \nonumber$ ... which tells us that the rate of the reaction depends on the rate constant \(k$ as well as on the concentration of both reactants. The rate constant can be determined experimentally by measuring the rate of the reaction with different starting reactant concentrations. The rate constant depends on the activation energy, of course, but also on temperature: a higher temperature means a higher k and a faster reaction, all else being equal. This should make intuitive sense: when there is more heat energy in the system, more of the reactant molecules are able to get over the energy barrier.
Here is one more interesting and useful expression. Consider a simple reaction where the reactants are $\ce{A}$ and $\ce{B}$, and the product is $\ce{AB}$ (this is referred to as a condensation reaction, because two molecules are coming together, or condensing). If we know the rate constant $k$ for the forward reaction and the rate constant $k_{reverse}$ for the reverse reaction (where AB splits apart into $\ce{A}$ and $\ce{B}$), we can simply take the quotient to find our equilibrium constant $K_{eq}$:
$A + B \rightleftharpoons AB \nonumber$
with
$K _{ eq }=\frac{[ AB ]}{[ A ][ B ]}=\frac{k_{\text {forward }}}{k_{\text {reverse }}} \nonumber$
This too should make some intuitive sense; if the forward rate constant is higher than the reverse rate constant, equilibrium should lie towards products. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.09%3A_Describing_a_Reaction_-_Energy_Diagrams_and_Transition_States.txt |
Objectives
After completing this section, you should be able to
1. explain the difference between a transition state and an intermediate.
2. draw a reaction energy diagram for a given multistep process.
3. interpret the reaction energy diagram of a multistep process (e.g., determine which of the steps is rate-determining).
Key Terms
Make certain that you can define, and use in context, the key term below.
• reaction intermediate
Study Notes
Each step in a multistep reaction has its own activation energy. The overall activation energy is the difference in energy between the reactants and the transition state of the slowest (rate-determining) step. The rate-determining step, that is, the one that controls the overall rate of reaction, is the step with the highest activation energy.
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise 6.10.1
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
Answer
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise 6.10.2
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Answer
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding
We come now to the subject of catalysis. Our hypothetical bowl of sugar (from section 6.2) is still stubbornly refusing to turn into carbon dioxide and water, even though by doing so it would reach a much more stable energy state. There are, in fact, two ways that we could speed up the process so as to avoid waiting several millennia for the reaction to reach completion. We could supply enough energy, in the form of heat from a flame, to push some of the sugar molecules over the high energy hill. Heat would be released from the resulting exothermic reaction, and this energy would push more molecules over their energy hills, and so on - the sugar would literally burn up.
A second way to make the reaction go faster is to employ a catalyst. You probably already know that a catalyst is an agent that causes a chemical reaction to go faster by lowering its activation energy.
How might you catalyze the conversion of sugar to carbon dioxide and water? It’s not too hard – just eat the sugar, and let your digestive enzymes go to work catalyzing the many biochemical reactions involved in breaking it down. Enzymes are proteins, and are very effective catalysts. ‘Very effective’ in this context means very specific, and very fast. Most enzymes are very selective with respect to reactant molecules: they have evolved over millions of years to catalyze their specific reactions. An enzyme that attaches a phosphate group to glucose, for example, will not do anything at all to fructose (the details of these reactions are discussed in section 10.2B).
Glucose kinase is able to find and recognize glucose out of all of the other molecules floating around in the 'chemical soup' of a cell. A different enzyme, fructokinase, specifically catalyzes the phosphorylation of fructose.
We have already learned (section 3.9) that enzymes are very specific in terms of the stereochemistry of the reactions that they catalyze . Enzymes are also highly regiospecific, acting at only one specific part of a molecule. Notice that in the glucose kinase reaction above only one of the alcohol groups is phosphorylated.
Finally, enzymes are capable of truly amazing rate acceleration. Typical enzymes will speed up a reaction by anywhere from a million to a billion times, and the most efficient enzyme currently known to scientists is believed to accelerate its reaction by a factor of about 1017 (see Chemical and Engineering News, March 13, 2000, p. 42 for an interesting discussion about this enzyme, orotidine monophosphate decarboxylase).
We will now begin an exploration of some of the basic ideas about how enzymes accomplish these amazing feats of catalysis, and these ideas will be revisited often throughout the rest of the text as we consider various examples of enzyme-catalyzed organic reactions. But in order to begin to understand how enzymes work, we will first need to learn (or review, as the case may be) a little bit about protein structure. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.10%3A_Describing_a_Reaction-_Intermediates.txt |
Objectives
• No objectives have been identified for this section
Key Terms
Make certain that you can define, and use in context, the key term below.
• enzyme
Study Notes
This section is a brief (but perhaps interesting) overview of some of the key differences between reactions performed in the lab and those in living systems. At this point, do not concern yourself with memorizing large biological molecules and reactions.
The active site
A critical element in the three-dimensional structure of any enzyme is the presence of an ‘active site’, which is a pocket, usually located in the interior of the protein, that serves as a docking point for the enzyme’s substrate(s) (‘substrate’ is the term that biochemists use for a reactant molecule in an enzyme-catalyzed reaction). It is inside the active site pocket that enzymatic catalysis occurs. Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase, with its substrate bound inside the active site pocket.
When the substrate binds to the active site, a large number of noncovalent interactions form with the amino acid residues that line the active site. The shape of the active site, and the enzyme-substrate interactions that form as a result of substrate binding, are specific to the substrate-enzyme pair: the active site has evolved to 'fit' one particular substrate and to catalyze one particular reaction. Other molecules do not fit in this active site nearly so well as fructose 1,6-bisphosphate.
Here are two close-up views of the same active site pocket, showing some of the specific hydrogen-bonding interactions between the substrate and active site amino acids. The first image below is a three-dimensional rendering directly from the crystal structure data. The substrate is shown in 'space-filling' style, while the active site amino acids are shown in the 'ball and stick' style. Hydrogens are not shown. The color scheme is grey for carbon, red for oxygen, blue for nitrogen, and orange for phosphorus.
Below is a two-dimensional picture of the substrate (colored blue) surrounded by hydrogen-bonding active site amino acids. Notice that both main chain and side chain groups contribute to hydrogen bonding: in this figure, main chain H-bonding groups are colored purple, and side chain H-bonding groups are colored green.
Looking at the last three images should give you some appreciation for the specific manner in which a substrate fits inside its active site.
Transition state stabilization
One of the most important ways that an enzyme catalyzes any given reaction is through entropy reduction: by bringing order to a disordered situation (remember that entropy is a component of Gibbs Free Energy, and thus a component of the activation energy). Let’s turn again to our previous example (from section 6.1C) of a biochemical nucleophilic substitution reaction, the methylation of adenosine in DNA. The reaction is shown below with non-reactive sections of the molecules depicted by variously shaped 'bubbles' for the sake of simplicity.
In order for this reaction to occur, the two substrates (reactants) must come into contact in precisely the right way. If they are both floating around free in solution, the likelihood of this occurring is very small – the entropy of the system is simply too high. In other words, this reaction takes place very slowly without the help of a catalyst.
Here’s where the enzyme’s active site pocket comes into play. It is lined with various functional groups from the amino acid main and side chains, and has a very specific three-dimensional architecture that has evolved to bind to both of the substrates. If the SAM molecule, for example, diffuses into the active site, it can replace its (favorable) interactions with the surrounding water molecules with (even more favorable) new interactions with the functional groups lining the active site.
Depiction of SAM bound in active site of enzyme
In a sense, SAM is moving from one solvent (water) to another 'solvent' (the active site), where many new energetically favorable interactions are possible. Remember: these new contacts between SAM and the active site groups are highly specific to SAM and SAM alone – no other molecule can ‘fit’ so well in this precise active site environment, and thus no other molecule will be likely to give up its contacts to water and bind to the active site.
The second substrate also has a specific spot reserved in the active site. (Because in this case the second substrate is a small segment of a long DNA molecule, the DNA-binding region of the active site is more of a 'groove' than a 'pocket').
So now we have both substrates bound in the active site. But they are not just bound in any random orientation – they are specifically positioned relative to one another so that the nucleophilic nitrogen is held very close to the electrophilic carbon, with a free path of attack. What used to be a very disordered situation – two reactants diffusing freely in solution – is now a very highly ordered situation, with everything set up for the reaction to proceed. This is what is meant by entropy reduction: the entropic component of the energy barrier has been lowered.
Looking a bit deeper, though, it is not really the noncovalent interaction between enzyme and substrate that are responsible for catalysis. Remember: all catalysts, enzymes included, accelerate reactions by lowering the energy of the transition state. With this in mind, it should make sense that the primary job of an enzyme is to maximize favorable interactions with the transition state, not with the starting substrates. This does not imply that enzyme-substrate interactions are not strong, rather that enzyme-TS interactions are far stronger, often by several orders of magnitude. Think about it this way: if an enzyme were to bind to (and stabilize) its substrate(s) more tightly than it bound to (and stabilized) the transition state, it would actually slow down the reaction, because it would be increasing the energy difference between starting state and transition state. The enzyme has evolved to maximize favorable noncovalent interactions to the transition state: in our example, this is the state in which the nucleophilic nitrogen is already beginning to attack the electrophilic carbon, and the carbon-sulfur bond has already begun to break.
In many enzymatic reactions, certain active site amino acid residues contribute to catalysis by increasing the reactivity of the substrates. Often, the catalytic role is that of acid and/or base. In our DNA methylation example, the nucleophilic nitrogen is deprotonated by a nearby aspartate side chain as it begins its nucleophilic attack on the methyl group of SAM. We will study nucleophilicity in greater detail in chapter 8, but it should make intuitive sense that deprotonating the amine increases the electron density of the nitrogen, making it more nucleophilic. Notice also in the figure below that the main chain carbonyl of an active site proline forms a hydrogen bond with the amine, which also has the effect of increasing the nitrogen's electron density and thus its nucleophilicity (Nucleic Acids Res. 2000, 28, 3950).
How does our picture of enzyme catalysis apply to multi-step reaction mechanisms? Although the two-step nucleophilic substitution reaction between 2-chloro-2-methylpropane and the cyanide anion is not a biologically relevant process, let’s pretend just for the sake of illustration that there is a hypothetical enzyme that catalyzes this reaction.
The same basic principles apply here: the enzyme binds best to the transition state. But therein lies the problem: there are two transition states! To which TS does the enzyme maximize its contacts?
Recall that the first step – the loss of the chloride leaving group to form the carbocation intermediate – is the slower, rate-limiting step. It is this step that our hypothetical enzyme needs to accelerate if it wants to accelerate the overall reaction, and it is thus the energy of TS1 that needs to be lowered.
By Hammond’s postulate, we also know that the intermediate I is a close approximation of TS1. So the enzyme, by stabilizing the intermediate, will also stabilize TS1 (as well as TS2) and thereby accelerate the reaction.
If you read scientific papers about enzyme mechanisms, you will often see researchers discussing how an enzyme stabilizes a reaction intermediate. By virtue of Hammond's postulate, they are, at the same time, talking about how the enzyme lowers the energy of the transition state.
An additional note: although we have in this section been referring to SAM as a 'substrate' of the DNA methylation reaction, it is also often referred to as a coenzyme, or cofactor. These terms are used to describe small (relative to protein and DNA) biological organic molecules that bind specifically in the active site of an enzyme and help the enzyme to do its job. In the case of SAM, the job is methyl group donation. In addition to SAM, we will see many other examples of coenzymes in the coming chapters, a number of which - like ATP (adenosine triphosphate), coenzyme A, thiamine, and flavin - you have probably heard of before. The full structures of some common coenzymes are shown in table 6 in the tables section. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.11%3A_A_Comparison_between_Biological_Reactions_and_Laboratory_Reactions.txt |
Concepts & Vocabulary
6.1: Kinds of Organic Reactions
• Addition reactions increase the number of sigma bonds in a molecule.
• Elimination reactions reduce the number of sigma bonds in a molecule.
• Substitution reactions incorporate replacement of an atom or group with another.
• Rearrangement reactions cause a molecule to be converted to a constitutional isomer without gaining or losing any atoms.
6.2: How Organic Reactions Occur: Mechanisms
• A reaction mechanism describes movement of electrons by using curved arrows to show bonds that are breaking and forming.
• Homolysis occurs when a bond breaks with each atom keeping one electron.
• Heterolysis occurs when a bond breaks and both electrons remain with one of the atoms.
• Some reactions occur in more than one step with a reactive intermediate formed briefly on the way to the new product.
• Reactive intermediates can be charged species such as carbocations and carbanions or uncharged species such as radicals.
• In organic chemistry Lewis acids are more often referred to as electrophiles, having an affinity for an electron pair.
• In organic chemistry Lewis bases are more often referred to as nucleophiles, having an electron pair that is available to bond to an electrophile.
• Ionic reactions involve charged species.
• Polar reactions involve bonds with unequally shared electrons.
6.3: Radical Reactions
• Radical chain reactions have three distinct phases: initiation, propagation and termination.
• Initiation causes radicals to be created from non-radical species.
• During the Propagation phase, radicals react with stable molecules to form new radicals.
• Termination occurs when two radicals react together to form a stable molecule.
6.4: Polar Reactions
• Carbon when bonded to a halogen, oxygen, nitrogen, sulfur, or metal has a partial positive charge. This allows these carbons to react with many nucleophiles.
• For carbonyl groups bond polarity is reinforced by resonance making the carbon even more positive than in other molecules. This makes carbonyl groups prone to addition and substitution reactions with nucleophiles.
• Nucleophiles have electron rich atoms that are able to donate a pair of electrons.
• In nucleophilic substitution reactions, the electrophile is typically carbon bonded to a more electronegative atom.
6.5: An Example of a Polar Reaction: Addition of HBr to Ethylene
• Alkene addition reaction with HBr occurs through the pi bond reacting as a nucleophile and abstracting a proton from the acid. This creates a carbocation intermediate which reacts with the bromide ion to form the final product.
• Reaction rates for this alkene addition reaction increase with larger halogens and more substituted alkenes.
• Markovnikov's Rule states that addition reactions of unsymmetrical alkenes yield the more substituted product.
6.6: Using Curved Arrows in Polar Reaction Mechanisms
• Curved arrows in mechanism drawings always represent electrons moving, starting at either a bond or lone pair of electrons.
• Electrons flow from electron rich to electron poor.
6.7: Describing a Reaction: Equilibria, Rates, and Energy Changes
• Exergonic reactions have a negative free energy meaning they are thermodynamically favorable and give off energy.
• Endergonic reactions have a positive free energy and require energy from the surroundings to occur.
6.8: Describing a Reaction: Bond Dissociation Energies
• Bond dissociation energy for a molecule is the difference in enthalpy of formation (homolytic) for the products and reactants.
• Bond dissociation energies are independent of path of reaction, so they do not give direct information on mechanisms. However, they can be used to evaluate the results of individual steps of a mechanism.
• Bond dissociation energies show that sigma bonds formed with sp hybridized carbon are stronger than sp2 which are stronger than bonds formed with sp3 carbons.
• Bond dissociation energies show that carbon-hydrogen bonds on primary carbons are stronger than secondary, which are stronger than tertiary.
6.9: Describing a Reaction: Energy Diagrams and Transition States
• Reaction coordinate diagrams are a special type of energy diagram that has the reaction coordinate (or reaction progress) on the x-axis.
• Thermodynamics of a reaction is conveyed on a reaction coordinate diagram by the difference in energy between the reactants and products.
• Activation energy is the energy barrier to a reaction occurring.
• A transition state is the highest energy point during the process of bonds forming and breaking in a reaction step.
• Kinetics of a reaction is conveyed on a reaction coordinate diagram by the difference in energy between the reactants and transition state.
• A rate expression relates rate to the rate constant and concentration of reactants.
6.10: Describing a Reaction: Intermediates
• A reaction intermediate is a short-lived species that goes on to react in a subsequent reaction step.
• Reaction intermediates appear as a local minimum (or valley) on a reaction coordinate diagram.
• Catalysts cause reaction rates to increase by lowing activation energy.
6.11: A Comparison between Biological Reactions and Laboratory Reactions
• An enzyme active site is the location where the enzyme interacts with its substrate and where catalysis occurs.
• Substrates are reactant molecules in enzymatic reactions.
Skills to Master
• Skill 6.1 Identify organic reactions by type (addition, elimination, substitution, rearrangement).
• Skill 6.2 Draw homolytic and heterolytic bond breaking as part of reaction mechanisms.
• Skill 6.3 Identify radical and ionic reactions.
• Skill 6.4 Identify and write out steps in a typical radical substitution reaction (initiation, propagation, termination).
• Skill 6.5 Identify polarity of bonds in organic molecules.
• Skill 6.6 Use curved arrows to indicate movement of electrons in resonance and reaction mechanisms.
• Skill 6.7 Predict whether a chemical species will act as an electrophile or nucleophile.
• Skill 6.8 Write an equilibrium expression for a reaction.
• Skill 6.9 Determine the direction of a reaction based on the equilibrium constant.
• Skill 6.10 Explain how rate and equilibrium are related to \(ΔG^o\) and \(K_{eq}\).
• Skill 6.11 Calculate bond dissociation energy given enthalpies of formation for reactants and products.
• Skill 6.12 Describe order of bond strength based on bond dissociation energy.
• Skill 6.13 Explain activation energy, kinetics, thermodynamics and transition states based on energy diagrams (reaction coordinate diagrams).
• Skill 6.14 Predict possible transition state structures for single reaction steps.
• Skill 6.15 Differentiate between transition states and intermediates.
• Skill 6.16 Draw a reaction coordinate diagram for a given multi-step process.
• Skill 6.17 Interpret a reaction coordinate diagram for a multi-step process.
• Skill 6.18 Briefly explain how enzymes catalyze reactions.
Memorization Tasks
MT 6.1 Memorize that arrows in reaction mechanisms always define movement of electrons.
MT 6.2 Memorize the relative electronegativities of common atoms (necessary for determining polarity of bonds).
MT 6.3 Memorize the equations that relate equilibrium, free energy, enthalpy and entropy.
ΔGº=–RTlnK
ΔGº=ΔHº–TΔSº | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/06%3A_An_Overview_of_Organic_Reactions/6.S%3A_An_Overview_of_Organic_Reactions_%28Summary%29.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: Introduction to Alkenes
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.
• 7.2: 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: Sequence Rules - 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.S: Alkenes- Structure and Reactivity (Summary)
Thumbnail: Ball-and-stick model of the ethylene (ethene) molecule, \(\ce{C2H4}\). (Public Domain; Ben Mills via Wikipedia)
07: Alkenes- Structure and Reactivity
Objective
After completing this section, you should be able to give an example of a naturally occurring compound that contains at least one double bond.
Key Terms
Make certain that you can define, and use in context, the key term below.
• olefin
Study Notes
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.
The graphic shows three alkenes. The more complex alkene is commonly known as 1-menthene, but its full proper IUPAC name is 1-methyl-2-(1-methylethyl)-cyclohexene.
Alkenes
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. The alkene group can also be called a vinyl group and the carbons sharing the double bond can be called vinyl carbons.
Structure of Ethene - the Simplest Alkene
Ethene is often written as CH2=CH2 which stands for:
The double bond is shared by the two carbon atoms and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
Ethene is not a very complicated molecule. It is made up of four 1s1 hydrogen atoms and two 2s2 2px1 2py1 carbon atoms. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight (octet) in the valence shell. Having eight valence electrons around carbon gives the atom the same electron configuration as neon, a noble gas. Carbon wants to have the same configuration as neon because when it has eight valence electrons carbon is at its most stable, lowest energy state.
This forms a total of three bonds to each carbon atom, giving them an $sp^2$ hybridization. Since the carbon atom is forming three sigma bonds instead of the four that it can, it only needs to hybridize three of its outer orbitals, instead of four. It does this by using the $2s$ electron and two of the $2p$ electrons, leaving the other unchanged. This new orbital is called an $sp^2$ hybrid because it is made from one s orbital and two p orbitals.
When atoms use $sp^2$ hybrids they have a trigonal planar structure. These structures are very similar to a 'peace' sign, there is a central atom with three atoms around it, all on one plane. Trigonal planar molecules have an ideal bond angle of 120° on each side.
The H-C-H bond angle is 117°, which is very close to the ideal 120° of a carbon with $sp^2$ hybridization. The other two angles (H-C=C) are both 121.5°.
Rigidity in Ethene
There is rigidity in the ethene molecule due to the double-bonded carbons. A double bond consists of one sigma bond formed by overlap of sp2 hybrid orbitals and one pi bond formed by overlap of parallel 2 p orbitals. In ethene there is no free rotation about the carbon-carbon sigma bond because these two carbons also share a $\pi$ bond. A $\pi$ bond is only formed when there is adequate overlap between both top and bottom p-orbitals. Rotation of the p-orbitals causes them to be 90° from each other breaking the $\pi$ bond because there would be no overlap. There is a much larger energy barrier to rotation than there is about a carbon-carbon sigma bond.
(a) The σ-bonded framework is formed by the overlap of two sets of singly occupied carbon sp2 hybrid orbitals and four singly occupied hydrogen 1s orbitals to form electron-pair bonds. This uses 10 of the 12 valence electrons to form a total of five σ bonds (four C–H bonds and one C–C bond).
(b) One singly occupied unhybridized 2pz orbital remains on each carbon atom to form a carbon–carbon π bond. (Note: by convention, in planar molecules the axis perpendicular to the molecular plane is the z-axis.)
Exercise
1. Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8. Draw all of the possible bond line structures for alkenes with the formula C4H8 including all possible structural and stereoisomers.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.01%3A_Introduction_to_Alkenes.txt |
Objectives
After completing this section, you should be able to
1. discuss the industrial importance of ethylene (ethene) and propylene (propene).
2. describe, briefly, the industrial process known as thermal cracking.
Study Notes
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.
Produced from ethylene (ethene)
Chemical Uses
ethanol solvent; constituent of cleaning preparations; in synthesis of esters
acetaldehyde slug killer, in the form of methaldehyde (CH3CHO)4
acetic acid manufacture of vinyl acetate polymers, ethyl acetate solvent and cellulose acetate polymers
ethylene oxide “cellosolves” (industrial solvents)
ethylene glycol anti-freeze; production of DacronOR
ethylene dichloride solvent; production of vinyl chloride
vinyl chloride manufacture of poly (vinyl chloride)—PVC
vinyl acetate manufacture of poly (vinyl acetate) used in paint emulsions, plywood adhesives and textiles
polyethylene “plastic” bags; toys; packaging
Produced from propylene (propene)
Chemical Uses
isopropyl alcohol rubbing alcohol; cosmetics; synthesis of acetone
propylene oxide manufacture of polyurethanes; polyesters
cumene industrial preparation of phenol and acetone
polypropylene molded articles (e.g., kitchenware); fibres for indoor-outdoor carpeting
Industrial Preparation of Ethylene and Propylene
Ethene (CH2CH2) and propene (CH3CHCH2), are most often called by their common names—ethylene and propylene. Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products. Both ethylene and propylene are the feedstock for the industrial synthesis of a wide variety of small organic molecules.
Ethylene, propylene, and butylene (CH3CH2CHCH2) are typically industrially synthesized through the steam cracking of light alkanes (C < 8) obtained from fractional distillation of crude oil. Cracking is the name given to a number of petroleum refining processes which break up large hydrocarbon molecules into smaller fragment. Steam cracking is achieved without a catalyst by using high temperatures (~900 oC) and produces a mixtures of products containing high proportions of hydrocarbons with double bonds. There is not any single unique reaction happening during steam cracking. The hydrocarbon molecules are broken up in a fairly random way but the process can be generically represented by the reaction below.
The mechanism of the steam cracking is complex and believed to involve the formation of free radicals. The high temperatures of steam cracking is enough to cause the homolytic cleavage of C-C and C-H bonds in the starting material. The cleavage of C-C bonds inherently creates smaller hydrocarbons as represented below.
Industrial Preparations of Alcohols from Ethylene and Propylene
Ethanol is manufactured by reacting ethylene with steam. The catalyst used is solid silicon dioxide coated with phosphoric acid. The reaction is reversible.
Only 5% of the ethylene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethylene, it is possible to achieve an overall 95% conversion.
Isopropyl alcohol is synthesized on an industrial scale using a similar reaction with propylene.
Ethylene is also used in the industrial synthesis of ethylene glycol. Ethylene glycol is a major industrial compound with a wide range of applications. The traditional method to synthesize ethylene glycol is a two step process in which ethylene is directly oxidized to ethylene oxide. Next the three-membered ring of ethylene oxide is opened using water at high temperatures and pressures to form ethylene glycol.
Industrial Preparations of Plastics from Ethylene and Propylene
One of the primary uses of industrial synthesized ethylene is the production of the polymer polyethylene (PE). Polyethylene is the most common plastic. As of 2017, over 100 million tons of polyethylene resins are produced annually, accounting for 34% of the total plastics market. Its primary use is in packaging (plastic bags, plastic films, geomembranes, containers including bottles, etc.). Many kinds of polyethylene are known, with most having the chemical formula (C2H4)n .
It is the simplest polymer, consisting of random-length (but generally very long) chains made up of two-carbon units. During the polymerization process many ethylene molecules are linked together to form the carbon backbone of polyethylene. In the polyethylene structures are represented below. The squiggly lines at the ends of the long structure indicate that the same pattern extends indefinitely. The more compact notation on the right shows the minimal repeating unit enclosed in brackets; this means the same thing and is the preferred way of depicting polymer structures. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.02%3A_Industrial_Preparation_and_Use_of_Alkenes.txt |
Objectives
After completing this section, you should be able to
1. determine the degree of unsaturation of an organic compound, given its molecular formula, and hence determine the number of double bonds, triple bonds and rings present in the compound.
2. draw all the possible isomers that correspond to a given molecular formula containing only carbon (up to a maximum of six atoms) and hydrogen.
3. draw a specified number of isomers that correspond to a given molecular formula containing carbon, hydrogen, and possibly other elements, such as oxygen, nitrogen and the halogens.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• degree of unsaturation
• saturated
• unsaturated
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information because it easily provides information about molecular structure.
Saturated and Unsaturated Molecules
Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons. The presence of a double bond causes alkenes to have less hydrogens than an alkane with the same number of carbons. Likewise, compounds containing a carbon-to-carbon triple bonds (R–C≡C–R) called alkynes (Discussed in Chapter 9), also have fewer hydrogens than the corresponding alkane. Collectively, compounds which have fewer hydrogen atoms than an alkane with the same number of carbon atoms are called unsaturated hydrocarbons. The relationship between the number of carbons (n) and hydrogens in the molecular formula for alkanes, alkenes, and alkynes are listed below.
For example, the three carbon alkane, propane has the molecular formula of C3H8. While the unsaturated compounds propene (C3H6) and propyne (C3H4) both have fewer hydrogens. Also, it is important to note that cycloalkanes with one ring have a general molecular formula of CnH2n just like alkenes. Because they also have fewer than maximum number of hydrogens possible, cyclic compounds are also considered unsaturated.
Identifying Degrees of Unsaturation
Every ring or pi bond in a compound is said to represent one degree of unsaturation. Being able to determine the degrees of unsaturation in a given compound is an important skill. Each of the following compounds are isomers of C5H7 and contain two degrees of unsaturation.
Exercise $1$
How many degrees of unsaturation do the following compounds have?
Answer
a) 0
b) 1
c) 1
d) 2
e) 2
f) 2
Calculating the Degree of Unsaturation (DoU)
As noted above, every degree of unsaturation causes the loss of two hydrogens from a compound's molecular formula when compared to an alkane with the same number of carbons. Understanding this relationship allows for the degrees of unsaturation of a compound to be calculated from its molecular formula. First, the maximum number of hydrogens possible for a given compound (2C + 2) is calculated and then the actual number of hydrogens present in the compound (H) is subtracted. If this difference is then divided by 2 the answer will be equal to the degrees of unsaturation for the compound.
For a compound which only contains carbon and hydrogen:
DoU = (2C + 2) - H / 2
As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. Because the compound only has 4 hydrogens in its molecular formula, it would have to gain 4 more hydrogens in order to be fully saturated (8-4 = 4). Degrees of unsaturation is equal to half the number of hydrogens the molecule needs to be fully saturated. This compound has 2 degrees of unsaturation (4/2 = 2).
The DoU of compounds containing elements other than carbon and hydrogen can also be calculated in a similar fashion. However, different elements can affect the formula used to calculate DoU.
For a compound which contains elements other than carbon and hydrogen:
$DoU= \dfrac{2C+2+N-X-H}{2} \tag{7.2.1}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
A halogen (X) replaces a hydrogen in a compound because both form one single bond. Therefore the DoU formula subtracts the number of halogens (X) present in a compound. For instance, 1,1-dichloroethene (C2H2Cl2) has two fewer hydrogens than ethene (C2H4) yet they both have one degree of unsaturation.
Oxygen and sulfur are not included in the DoU formula because saturation is unaffected by these elements. The inclusion of an alcohol or sulfur in a compound does not change the number of hydrogens to obtain saturation. As seen in alcohols, the number of hydrogens in cyclohexanol (C6H12O) matches the number of hydrogens in cyclohexane (C6H12) and they both have one degree of unsaturation.
When a nitrogen is present in a compound one more hydrogen is required to reach saturation. Therefore, we add the number of nitrogens (N). Propyl amine (C3H9N) has one more hydrogen compared to propane (C3H8) both of which are saturated compounds with 0 DoU.
With the degrees of unsaturation comes information about the possible number of rings and multiple bonds in a given compound. Remember, the degrees of unsaturation only gives the sum of pi bonds and/or rings.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
Example 7.3.1: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.
References
1. Vollhardt, K. P.C. & Shore, N. (2007). Organic Chemistry (5thEd.). New York: W. H. Freeman. (473-474)
2. Shore, N. (2007). Study Guide and Solutions Manual for Organic Chemistry (5th Ed.). New York: W.H. Freeman. (201)
Exercise $2$
Determine whether the following molecules are saturated or unsaturated.
Answer
a) unsaturated (Even though the rings only contain single bonds, rings are considered unsaturated.)
b) unsaturated
c) saturated
d) unsaturated
e) unsaturated
f) saturated
Exercise $3$
Determine the degrees of unsaturation for each of the following compounds.
Answer
If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
a) 2 (2 rings)
b) 2 (one double bond and the double bond from the carbonyl)
c) 0 (no double bonds or rings)
d) 10 (2(10) + 2 + 4 - 0 - 6)/2 = 10
e) 1 (2(5) + 2 + 0 - 0 - 10)/2 = 1
f) 0 (2(6) + 2 + 0 - 2 - 12)/2 = 0
Exercise $4$
Calculate the degrees of unsaturation for the following molecular formulas:
a) C9H20 b) C7H8 c) C5H7Cl d) C9H9NO4
Answer
Use the formula to solve (O not involved in the formula)
(a.) 0 (2(9) + 2 - 20)/2 = 0
(b.) 4 (2(7) + 2 - 8)/2 = 4
(c.) 2 (2(5) + 2 - 1 -7)/2 = 2
(d.) 6 (2(9) + 1 - 7)/2 = 6 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.03%3A_Calculating_Degree_of_Unsaturation.txt |
Objectives
After completing this section, you should be able to
1. provide the correct IUPAC name for an acyclic or cyclic alkene, given its Kekulé, condensed or shorthand structure.
2. draw the Kekulé, condensed or shorthand structure of an alkene (cyclic or acyclic), given its IUPAC name.
3. give the IUPAC equivalent of the following trivial names: ethylene, propylene, isobutylene and isoprene.
4. draw the structure of a vinyl (ethenyl) and allyl (2-propenyl) group, and use these names in alkene nomenclature.
Study Notes
This course uses IUPAC nomenclature; therefore, you need not usually memorize a large number of trivial names. However, you will encounter some trivial names so frequently in books and articles that they soon become familiar.
An alkene that can exhibit geometric isomerism has not been properly named unless its name specifies whether the double bond (or bonds) is (or are) cis or trans. The most effective way of giving this information is discussed, and more details of cis and trans follow in Section 7.4.
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. The parent chain of an alkene is the longest chain containing both carbon atoms of the double bond. Alkenes are named by dropping the -ane ending of the parent and adding -ene. Also, the position of double bond in the parent chain of the alkene is indicated with a number.
The Basic Rules for Naming Alkenes
For straight chain alkenes, it is the same basic rules as nomenclature of alkanes apply except the -ane suffix is changed to -ene.
1) Find the longest carbon chain that contains both carbons of the double bond.
2) Start numbering from the end of the parent chain which gives the lowest possible number to the double bond. If the double bond is equidistant from both ends of the parent chain, number from the end which gives the substituents the lowest possible number. The double bond in cycloalkenes do not need to number because it is understood that they are in the one position.
3) Place the location number of the double bond directly before the parent name. The location number indicates the position of the first carbon of the double bond. Add substituents and their position to the alkene as prefixes. Remember substituents are written in alphabetical order.
The presence of multiple double bonds is indicated by using the appropriate suffix such as -diene, -triene, ect. Each of the multiple bonds receives a location number. Also, only -ne is removed from the parent alkane chain name leaving an "a" in the name.
Overall, the name of an alkene should look like:
(Location number of substituent)-(Name of substituent)-(Location number of double bond)-(Name of parent chain) + ene
Newer IUPAC Nomenclature
In 1993 IUPAC updated their naming recommendation to place the location number of the double bond before the -ene suffix of alkene names. The provides names such as hex-2-ene rather than 2-hexene. The newer system is slowly being accepted so it may occasionally be encountered.
Naming Cycloalkenes
Because there are no chain ends in cycloalkenes, the double bond is assumed to numbered C1 and C2 and its location number is not required in the name. The direction of the numbering is determined by which will give the substituent closest to the double bond the lowest number. If multiple double bonds are present, it may be necessary to include their location numbers in the name. One of the double bonds will be number C1 and C2 and the numbering direction is determined by which gives the remaining double bonds the lowest possible number.
Endocyclic vs. Exocyclic Alkenes
Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring.
Cyclopentene is an example of an endocyclic double bond.
Methylenecylopentane is an example of an exocyclic double bond.
Name the following compounds...
1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene.
1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene.
Try to draw structures for the following compounds...
• 3-allylcyclohex-1-ene
• 2-vinyl-1,3-cyclohexadiene
Common Names of Alkene Fragments
Some alkene containing fragments have common names which should be recognized. These common names can be used to simplify naming much the alkyl fragments discussed in Section: 3.3. Some of these fragments are the methylene group (H2C=), the vinyl group (H2C=CH-), and the allyl group (H2C=CH-CH2-).
In addition, the common name some small alkene compounds are still accepted by IUPAC. It is important to be able to identify them.
Exercise \(1\)
Name the following compounds using common fragment names.
a) b) C)
Answer
a) 2-Vinyl-1,3-cyclohexadiene
b) Methylenecylopentane
c) 3-Allylcyclohexene
Examples
Both these compounds have double bonds, making them alkenes. In example (1) the longest chain consists of six carbons, so the root name of this compound will be hexene. Three methyl substituents (colored red) are present. Numbering the six-carbon chain begins at the end nearest the double bond (the left end), so the methyl groups are located on carbons 2 & 5. The IUPAC name is therefore: 2,5,5-trimethyl-2-hexene.
In example (2) the longest chain incorporating both carbon atoms of the double bond has a length of five. There is a seven-carbon chain, but it contains only one of the double bond carbon atoms. Consequently, the root name of this compound will be pentene. There is a propyl substituent on the inside double bond carbon atom (#2), so the IUPAC name is: 2-propyl-1-pentene.
The double bond in example (3) is located in the center of a six-carbon chain. The double bond would therefore have a locator number of 3 regardless of the end chosen to begin numbering. The right hand end is selected because it gives the lowest first-substituent number (2 for the methyl as compared with 3 for the ethyl if numbering were started from the left). The IUPAC name is assigned as shown.
Example (4) is a diene (two double bonds). Both double bonds must be contained in the longest chain, which is therefore five- rather than six-carbons in length. The second and fourth carbons of this 1,4-pentadiene are both substituted, so the numbering begins at the end nearest the alphabetically first-cited substituent (the ethyl group).
These examples include rings of carbon atoms as well as some carbon-carbon triple bonds. Example (6) is best named as an alkyne bearing a cyclobutyl substituent. Example (7) is simply a ten-membered ring containing both a double and a triple bond. The double bond is cited first in the IUPAC name, so numbering begins with those two carbons in the direction that gives the triple bond carbons the lowest locator numbers. Because of the linear geometry of a triple bond, a-ten membered ring is the smallest ring in which this functional group is easily accommodated. Example (8) is a cyclooctatriene (three double bonds in an eight-membered ring). The numbering must begin with one of the end carbons of the conjugated diene moiety (adjacent double bonds), because in this way the double bond carbon atoms are assigned the smallest possible locator numbers (1, 2, 3, 4, 6 & 7). Of the two ways in which this can be done, we choose the one that gives the vinyl substituent the lower number.
Problems
Exercise \(2\)
Name the following alkenes.
Answer
a) 2-ethylhept-1-ene or 2-ethyl-1-heptene
b) 1,2-dimethylcycloheptene
c) 2,5-dimethyloct-2-ene or 2,5-dimethyl-1-octene
Exercise \(3\)
Draw structures for the following compounds from the given names.
1. 3-butylhept-2-ene (3-butyl-2-heptene)
2. 1,4-pentadiene (penta-1,4-diene)
3. 3-vinyl-1,4-cyclohexadiene (cyclohexa-1,4-diene)
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.04%3A_Naming_Alkenes.txt |
Objectives
After completing this section, you should be able to
1. discuss the formation of carbon-carbon double bonds using the concept of sp2 hybridization.
2. describe the geometry of compounds containing carbon-carbon double bonds.
3. compare the molecular parameters (bond lengths, strengths and angles) of a typical alkene with those of a typical alkane.
4. explain why free rotation is not possible about a carbon-carbon double bond.
5. explain why the lack of free rotation about a carbon-carbon double bond results in the occurrence of cis-trans isomerism in certain alkenes.
6. decide whether or not cis-trans isomerism is possible for a given alkene, and where such isomerism is possible, draw the Kekulé structure of each isomer.
Key Terms
Make certain that you can define, and use in context, the key term below.
• cis-trans stereoisomers
Study Notes
Your previous studies in chemistry may have prepared you to discuss the nature of a carbon-carbon double bond. If not, you should review Section 1.8 of this course before beginning the present section. It is particularly important that you make molecular models of some simple alkenes to gain insight into the geometry of these compounds.
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 which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is not a form of stereoisomerism, and is dealt with in a separate Module.
Geometric (cis / trans) isomerism
These isomers occur where you have restricted rotation somewhere in a molecule. At an introductory level in organic chemistry, examples usually just involve the carbon-carbon double bond - and that's what this page will concentrate on. Think about what happens in molecules where there is unrestricted rotation about carbon bonds - in other words where the carbon-carbon bonds are all single. The next diagram shows two possible configurations of 1,2-dichloroethane.
These two models represent exactly the same molecule. You can get from one to the other just by twisting around the carbon-carbon single bond. These molecules are not isomers. If you draw a structural formula instead of using models, you have to bear in mind the possibility of this free rotation about single bonds. You must accept that these two structures represent the same molecule:
But what happens if you have a carbon-carbon double bond - as in 1,2-dichloroethene?
These two molecules are not the same. The carbon-carbon double bond won't rotate and so you would have to take the models to pieces in order to convert one structure into the other one. That is a simple test for isomers. If you have to take a model to pieces to convert it into another one, then you've got isomers. If you merely have to twist it a bit, then you haven't!
Drawing structural formulae for the last pair of models gives two possible isomers:
1. In one, the two chlorine atoms are locked on opposite sides of the double bond. This is known as the trans isomer. (trans : from latin meaning "across" - as in transatlantic).
2. In the other, the two chlorine atoms are locked on the same side of the double bond. This is know as the cis isomer. (cis : from latin meaning "on this side")
trans-1,2-dichloroethene
cis-1,2-dichloroethene
The most likely example of geometric isomerism you will meet at an introductory level is but-2-ene. In one case, the CH3 groups are on opposite sides of the double bond, and in the other case they are on the same side.
trans-but-2-ene
cis-but-2-ene
The importance of drawing geometric isomers properly
It's very easy to miss geometric isomers in exams if you take short-cuts in drawing the structural formulae. For example, it is very tempting to draw but-2-ene as
CH3CH=CHCH3
If you write it like this, you will almost certainly miss the fact that there are geometric isomers. If there is even the slightest hint in a question that isomers might be involved, always draw compounds containing carbon-carbon double bonds showing the correct bond angles (120°) around the carbon atoms at the ends of the bond. In other words, use the format shown in the last diagrams above.
How to recognize the possibility of geometric isomerism
You obviously need to have restricted rotation somewhere in the molecule. Compounds containing a carbon-carbon double bond have this restricted rotation. (Other sorts of compounds may have restricted rotation as well, but we are concentrating on the case you are most likely to meet when you first come across geometric isomers.) If you have a carbon-carbon double bond, you need to think carefully about the possibility of geometric isomers.
What needs to be attached to the carbon-carbon double bond?
Think about this case:
Although we've swapped the right-hand groups around, these are still the same molecule. To get from one to the other, all you would have to do is to turn the whole model over. You won't have geometric isomers if there are two groups the same on one end of the bond - in this case, the two pink groups on the left-hand end. So there must be two different groups on the left-hand carbon and two different groups on the right-hand one. The cases we've been exploring earlier are like this:
But you could make things even more different and still have geometric isomers:
Here, the blue and green groups are either on the same side of the bond or the opposite side. Or you could go the whole hog and make everything different. You still get geometric isomers, but by now the words cis and trans are meaningless. This is where the more sophisticated E-Z notation comes in.
Rigidity of C=C Bonds
As discussed in Section 1.8 the double bond in the molecule ethene (H2C=CH2) is created by the overlap of two different sets of orbitals. The C-C σ bond is formed when an sp2 orbital from each carbon atom overlaps end to end. Also, the C-C pi bond is created by the side-to-side overlap of a pz orbital from each carbon atom.
Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), pi bonds are not free to rotate. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. If free rotation were to occur the p-orbitals would have to go through a phase where they are 90° from each other, which would break the pi bond because there would be no overlap. Since the pi bond is essential to the structure of ethene it must not break, so there can be no free rotation about the carbon-carbon sigma bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane.
Restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism in double bonds. Consider the alkene with the condensed structural formula CH3CH=CHCH3. We could name it 2-butene, but there are actually two such compounds due to this isomerism. The isomer in which the two methyl (CH3) groups lie on the same side of the molecule is called the cis isomer (Latin cis, meaning “on this side”) and is named cis-2-butene. The isomer with the two (CH3) groups on opposite sides of the molecule is the trans isomer (Latin trans, meaning “across”) and is named trans-2-butene. These two compounds are cis-trans isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule. In general these isomers have different physical, chemical, and physiological properties.
It is important to note that the presence of a double bond does not necessarily lead to cis-trans isomerism. Being able to tell if a double bond has the possibility of isomerism is a very important skill. Cis-trans can occur whenever both double-bond carbons are directly attached to a carbon and a hydrogen. In this case, interchanging the substituents on one of the double-bond carbons creates a different isomer.
If one of the double-bond carbons of an alkene is attached to two identical groups, cis-trans isomerism is not possible. Here, interchanging the substituents on one of the double-bond carbons forms an identical molecule.
Worked Example \(1\)
Are the following molecules cis-trans isomers?
Answer
Although the two molecules are seemingly different propenes, these two structures are not really different from each other. Because the one of the double-bond carbons is attached to two identical groups (Hydrogens) it is incapable of forming cis-trans isomers. The interchange of two substituens seen here does not create a new isomer. If either molecule were flipped over top to bottom, the two would you would look identical.
Exercises
Exercise \(1\)
Classify each compound as a cis isomer, a trans isomer, or neither.
Answer
a) trans isomer
b) neither
c) cis isomer
d) cis isomer
Exercise \(2\)
Which of the following compounds could exist as cis/trans isomers? Draw (& label) both of the isomers for the ones that can.
a) CH3CH=CHCH3
b) (CH3)2C=CHCH3
c) H2C=CHCH3
d) CH3CH2CH=CHBr
Answer
Exercise \(3\)
Draw (& label) the cis and trans isomer for each of the following compound names. If no cis/trans isomerism is possible, write none.
1. 3-hexene
2. 1-hexene
3. 4-methylpent-2-ene (4-methyl-2-pentene)
4. 1,1-dibromobut-2-ene (1,1-dibromo-2-butene)
Answer
Exercise \(4\)
Name the following compounds using cis/trans nomenclature
Answer
a) trans-4-methylhex-2-ene (trans-4-methyl-2-hexene)
b) cis-2,5-dibromohex-3-ene (cis-2,5-dibromo-3-hexene) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.05%3A_Cis-Trans_Isomerism_in_Alkenes.txt |
Objectives
After completing this section, you should be able to
1. illustrate, by means of a suitable example, the limitations of the terms cis and trans in naming isomeric alkenes.
2. use the E/Z designation to describe the geometry of a given alkene structure.
3. incorporate the E/Z designation into the IUPAC name of a given alkene.
4. draw the correct Kekulé, condensed or shorthand structure of an alkene, given its E/Z designation plus other necessary information (e.g., molecular formula, IUPAC name).
Key Terms
Make certain that you can define, and use in context, the key term below.
• sequence rules (Cahn-Ingold-Prelog rules)
Study Notes
The limitations of the cis/trans system are illustrated in the examples given below.
1. From your study of the IUPAC system, you should be able to identify this compound as 4-ethyl-3-methyl-3-heptene, but is it cis or trans?
At first you might say cis, because it appears that two ethyl groups appear on the same side of the double bond. However, the correct answer is trans. The rule is that the designation cis or trans must correspond to the configuration of the longest carbon chain. Tracing out the seven-carbon chain in the compound shown above, you change sides as you pass through the double bond:
So, the full name for this compound is trans-4-ethyl-3methyl-3-heptene.
2. The cis/trans system breaks down completely in a compound such as that shown below. The E/Z system, which is the subject of this section, is designed to accommodate such situations.
In cases where two or more double bonds are present, you must be prepared to assign an E or Z designation to each of the double bonds. For example:
Another use for these sequence rules will be part of the discussion of optical isomerism in Section 9.5.
E/Z nomenclature
When each carbon in a double bond is attached to a hydrogen and and a non-hydrogen substituent, the geometric isomers can be identified by using the cis-trans nomenclature discussed in the previous section. However, when a double bond is attached to three or four non-hydrogen substituents there are some examples where cis-trans nomenclature is ineffective in describing the substituents orientation in geometric isomers. In these situations the rigorous IUPAC system for naming alkene isomers, called the E/Z system, is used. The E/Z system analyzes the two substituents attached to each carbon in the double bond and assigns each either a high or low priority. If the higher priority group on both carbons in the double bond the same side the alkene is said to have a Z isomer (from German zusammen = together). You could think of Z as Zame Zide to help memorize it. If the higher priority group on opposite sides the alkene has an E isomer (from German entgegen = opposite).
Note, if both substituents an a double bond carbon are exactly the same there is no E/Z isomerizem is possible. Also, if E/Z isomerism is possible, interchanging the substituents attached on double-bond carbon converts one isomer to the other.
Substituent priority for the E,Z system is assigned using the Cahn–Ingold–Prelog (CIP) sequence rules. These are the same rules used to assign R/S configurations to chiral centers in Section 5.5. A brief overview of using CIP rules to determine alkene configuration is given here but CIP rules are discussed in greater detail in Section 5.5.
Note
The priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system
Rule 1) The "first point of difference" rule
First, determine the two stubstitents on each double-bond carbon separately. Rank these substituents based on the atom which directly attached to the double-bond carbon. The substituent whose atom has a higher atomic number takes precedence over the substituent whose atom has a lower atomic number.
Which is higher priority, by the CIP rules: a C with an O and 2 H attached to it or a C with three C? The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done. The simple answer is that the first point of difference is what matters; the O wins.
To illustrate this, consider the molecule at the left. Is the double bond here E or Z? At the left end of the double bond, Br > C. But the right end of the double bond requires a careful analysis.
At the right hand end, the first atom attached to the double bond is a C at each position. A tie, so we look at what is attached to this first C. For the upper C, it is CCC (since the triple bond counts three times). For the lower C, it is OHH -- listed in order from high priority atom to low. OHH is higher priority than CCC, because of the first atom in the list. That is, the O of the lower group beats the C of the upper group. In other words, the O is the highest priority atom of any in this comparison; thus the O "wins".
Therefore, the high priority groups are "up" on the left end (the -Br) and "down" on the right end (the -CH2-O-CH3). This means that the isomer shown is opposite = entgegen = E. And what is the name? The "name" is (E)-2-Bromo-3-(methoxymethyl)hex-2-en-4-yne.
Rule 2)
If the first atom on both substituents are the identical, then proceed along both substituent chains until the first point of difference is determined.
Rule 3)
Remember that atoms involved in multiple bonds are considered with a specific set of rules. These atoms are treated as if they have the same number of single-bond atoms as they have attached to multiply bonded atoms.
An easy example which shows the necessity of the E/Z system is the alkene, 1-bromo-2-chloro-2-fluoro-1-iodoethene, which has four different substituents attached to the double bond. The figure below shows that there are two distinctly different geometric isomers for this molecule neither of which can be named using the cis-trans system.
Consider the left hand structure. On the double bond carbon on the left, the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at carbon on the right. The attached atoms are Cl and F, with Cl having the higher atomic number and the higher priority.
When considering the relative positions of the higher priority groups, the higher priority group is "down" on the left double bond carbon and "down" at right double bond carbon. Since the two higher priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E).
Example 7..61: Butene
cis-2-butene
(Z)-2-butene
trans-2-butene
(E)-2-butene
The Figure above shows the two isomers of 2-butene. You should recognize them as cis and trans. Let's analyze them to see whether they are E or Z. Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is (Z)-2-butene.
Now look at the right hand structure (the trans isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on opposite sides of the double bond, they are entgegen = opposite. Therefore, this is (E)-2-butene.
E/Z will work – even when cis/trans fails
In simple cases, such as 2-butene, Z corresponds to cis and E to trans. However, that is not a rule. This section and the following one illustrate some idiosyncrasies that happen when you try to compare the two systems. The real advantage of the E/Z system is that it will always work. In contrast, the cis/trans system breaks down with many ambiguous cases.
Example 7.6.2
The following figure shows two isomers of an alkene with four different groups on the double bond, 1-bromo-2-chloro-2-fluoro-1-iodoethene.
(Z)-1-bromo-2-chloro-2-fluoro-1-iodoethene
(E)-1-bromo-2-chloro-2-fluoro-1-iodoethene
It should be apparent that the two structures shown are distinct chemicals. However, it is impossible to name them as cis or trans. On the other hand, the E/Z system works fine... Consider the left hand structure. On C1 (the left end of the double bond), the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at C2. The atoms are Cl and F, with Cl being higher priority. We see that the higher priority group is "down" at C1 and "down" at C2. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E).
E/Z will work, but may not agree with cis/trans
Consider the molecule shown at the left.
This is 2-bromo-2-butene -- ignoring the geometric isomerism for now. Cis or trans? This molecule is clearly cis. The two methyl groups are on the same side. More rigorously, the "parent chain" is cis.
E or Z? There is a methyl at each end of the double bond. On the left, the methyl is the high priority group -- because the other group is -H. On the right, the methyl is the low priority group -- because the other group is -Br. That is, the high priority groups are -CH3 (left) and -Br (right). Thus the two priority groups are on opposite sides = entgegen = E.
Note
This example should convince you that cis and Z are not synonyms. Cis/trans and E/Z are determined by distinct criteria. There may seem to be a simple correspondence, but it is not a rule. Be sure to determine cis/trans or E/Z separately, as needed.
Multiple double bonds
If the compound contains more than one double bond, then each one is analyzed and declared to be E or Z.
Example 7.6.3
The configuration at the left hand double bond is E; at the right hand double bond it is Z. Thus this compound is (1E,4Z)-1,5-dichloro-1,4-hexadiene.
The double-bond rule in determining priorities
Example 7.6.4
Consider the compound below
This is 1-chloro-2-ethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism. There is no geometric isomerism at the second double bond, at 3-4, because it has 2 H at its far end.
What about the first double bond, at 1-2? On the left hand end, there is H and Cl; Cl is higher priority (by atomic number). On the right hand end, there is -CH2-CH3 (an ethyl group) and -CH=CH2 (a vinyl or ethenyl group). Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H. CCH is higher than CHH; therefore, the ethenyl group is higher priority. Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the Z-isomer; the compound is (Z)-1-chloro-2-ethyl-1,3-butadiene.
Example 7.6.5
The configuration about double bonds is undoubtedly best specified by the cis/trans notation when there is no ambiguity involved. Unfortunately, many compounds cannot be described adequately by the cis/trans system. Consider, for example, configurational isomers of 1-fluoro-1-chloro-2-bromo-2-iodo-ethene, 9 and 10. There is no obvious way in which the cis/trans system can be used:
A system that is easy to use and which is based on the sequence rules already described for the R,S system works as follows:
1. An order of precedence is established for the two atoms or groups attached to each end of the double bond according to the sequence rules of Section 19-6. When these rules are applied to 1-fluoro- 1-chloro-2-bromo-2- iodoethene, the priority sequence is:
• at carbon atom 1, C1 > F
• at carbon atom 2, I > Br
1. Examination of the two configurations shows that the two priority groups- one on each end- are either on the same side of the double bond or on opposite sides:
priority groups on opposite sides
(E) configuration
priority groups on same side
(Z) configuration
The Z isomer is designated as the isomer in which the top priority groups are on the same side (Z is taken from the German word zusammen- together). The E isomer has these groups on opposite sides (E, German for entgegen across). Two further examples show how the nomenclature is used:
(Z)-1-chloro-1-butene
(1Z,3E)-1,3-butadiene-1,4-d2
Exercises
Exercise \(1\)
Which of the following sets has a higher ranking?
a) -CH3 or -CH2Br
b) -Br or -Cl
c) -CH=CH2 or -CH=O
Answer
a) -CH2Br
b) -Br
c) -CH=O
Exercise \(2\)
Place the following sets of substituents in each group in order of lowest priority (1st) to highest priority (4th)
a) -CH(CH3)2, -CH2CH3, -C(CH3)3, -CH3
b) -NH2, -F, -Br, -CH3
c) -SH, -NH2, -F, -H
Answer
a) (lowest priority) -CH3 < -CH2CH3 < -CH(CH3)2 < -C(CH3)3 (highest priority)
b) (lowest priority) -CH3 < -NH2 < -F < -Br (highest priority)
c) (lowest priority) -H < -NH2 < -F < -SH (highest priority)
Exercise \(3\)
Label the following alkenes as E, Z, or neither.
Answer
The higher priority group is highlighted in red.
a) E
b) E
c) E
d) Z
e) neither (both isopropyls on the right have the same priority)
Exercise \(4\)
Name the following alkenes.
Answer
The higher priority group is highlighted in red.
a) (Z)-4-ethyl-5-methyloct-3-ene or (Z)-4-ethyl-5-methyl-3-octene
b) 3-ethyl-6-methyl-4-propylhept-3-ene or 3-ethyl-6-methyl-4-propyl-3-heptene
c) (E)-2-chloro-4-bromo-5-ethyl-7-methyldec-4-ene or (E)-2-chloro-4-bromo-5-ethyl-7-methyl-4-decene | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.06%3A_Sequence_Rules_-_The_EZ_Designation.txt |
Objectives
After completing this section, you should be able to
1. explain why cis alkenes are generally less stable than their trans isomers.
2. explain that catalytic reduction of a cis alkene produces the same alkane as the catalytic reduction of the trans isomer.
3. explain how heats of hydrogenation (ΔH°hydrog) can be used to show that cis alkenes are less stable than their trans isomers, and discuss, briefly, the limitations of this approach.
4. arrange a series of alkenes in order of increasing or decreasing stability.
5. describe, briefly, two of the hypotheses proposed to explain why alkene stability increases with increased substitution. [Note: This problem is a typical example of those instances in science where there is probably no single “correct” explanation for an observed phenomenon.]
Key Terms
Make certain that you can define, and use in context, the key terms below.
• catalytic hydrogenation
• heat of hydrogenation, (ΔH°hydrog)
• hyperconjugation
Study Notes
The two alkenes, cis-CH3CH=CHCH3 and (CH3)2C=CH2 have similar heats of hydrogenation (−120 kJ/mol and −119 kJ/mol, respectively), and are therefore of similar stability. However, they are both less stable than trans-CH3CH=CHCH3 (−116 kJ/mol).
You may wonder why an sp2 -sp3 bond is stronger than an sp3-sp3 bond. Bond strength depends on the efficiency with which orbitals can overlap. In general, s orbitals overlap more efficiently than do p orbitals; therefore, the s-s bond in the hydrogen molecule is stronger than the p-p bond in fluorine. In hybrid orbitals, the greater the s character of the orbital, the more efficiently it can overlap: an sp2 orbital, which has a 33% s character, can overlap more effectively than an sp3 orbital, with only 25% s character.
Hydrogenation
Alkene hydrogenation is the addition of hydrogen gas (H2) to an alkene which saturates the bond and forms an alkane. Alkene hydrogenation reactions require a transition metal catalyst, such as Pt or Pd, to speed up the reaction. The hydrogenation reaction is used in this section to investigate the stability of alkenes, however, it will be discussed in greater detail in Section 8.7. Hydrogenation reactions are exothermic and the enthalpy change in this reaction is called the heat of hydrogenation (ΔH°hydrog). Since the double bond is breaking in this reaction, the energy released during hydrogenation is proportional to the energy in the double bond of the molecule. By comparing the heat of hydrogenations from a series of alkenes that produce the same alkane, a quantitative measure of relative alkene stabilities can be produced. These experiments will lead to an general understanding of structural features which tend to stabilize or destabilize alkenes.
The Catalyst
A catalyst increases the reaction rate by lowering the activation energy of the reaction. Although the catalyst is not consumed in the reaction, it is required to accelerate the reaction sufficiently to be observed in a reasonable amount of time. Catalysts commonly used in alkene hydrogenation are: platinum, palladium, and nickel. The metal catalyst acts as a surface on which the reaction takes place. This increases the rate by putting the reactants in close proximity to each other, facilitating interactions between them. With this catalyst present, the sigma bond of H2 breaks, and the two hydrogen atoms instead bind to the metal (see #2 in the figure below). The $\pi$ bond of the alkene weakens as it also interacts with the metal (see #3 below).
Since both the reactants are bound to the metal catalyst, the hydrogen atoms can easily add, one at a time, to the previously double-bonded carbons (see #4 and #5 below). The position of both of the reactants bound to the catalyst makes it so the hydrogen atoms are only exposed to one side of the alkene. This explains why the hydrogen atoms add to same side of the molecule, called syn-addition.
Note
The catalyst remains intact and unchanged throughout the reaction.
Heats of Hydrogenation
The stability of alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule. This is a useful tool because heats of hydrogenation can be measured very accurately. The $\Delta H^o$ is usually around -30 kcal/mol for alkenes. Stability is simply a measure of energy. Lower energy molecules are more stable than higher energy molecules. More substituted alkenes are more stable than less substituted ones due to hyperconjugation. They have a lower heat of hydrogenation. The following illustrates stability of alkenes with various substituents:
Cis/Trans Isomers
Between cis and trans isomers of an alkene, the cis isomer tends to be less stable due to the molecular crowding created nonbonding interaction between two alky groups on the same side of the double bond. The crowding creates steric strain which distorts bond angles creating less effective bond orbital overlap and desabilizing the molecule. Steric strain has previously been seen in gauche interactions in Newman projections (Section 3.7) and 1,3-diaxial interactions in substituted cyclohexanes (Section 4.7). Steric strain is directly related to the size of the species being crowded. The difference in energy between cis and trans 2-butene is 5 kJ/mol, however, this difference would be greater if larger group were being held in the cis position. Two cis-tert-butyl group can create over 40 kJ/mol of steric strain.
See the following isomers of butene:
Alkene Stabilization by Alkyl Substituents
In general, the stability of an alkene increases with the number of alkyl substituents. This effect is due by the combination of two factors:
Hyperconjugation
In classical valence-bond theory, electron delocalization can only occur by the parallel overlap of adjacent p orbitals. According to hyperconjugation theory, electron delocalization could also occur by the parallel overlap of p orbitals with adjacent hybridized orbitals participating in sigma bonds. This electron delocalization serves to stabilize the alkene. As the number of alkyl substituents increases, the number of sigma bonds available for hyperconjugation increases, and the alkene tends to become more stabilized. In the example of propene shown below, a p orbital from a sp2 hybridized carbon involved in the double bond interacts with a sp3 hybridized orbital participating in an adjacent C-H sigma bond.
In a molecular orbital description of hyperconjugation, the electrons in sigma molecular orbitals (C-H or C-C) of alkyl substituens, interact with adjacent unpopulated non-bonding or antibonding molecular orbitals from the double bond. The interaction creates a bonding molecular orbital which extends over the four atom chain (C=C-C-H) involved in hyperconjugation. The expanded molecular orbital helps to stabilize the double bond.
Bond Stability
Bond strengths play an important part in determining the overall stability of a molecule. A C-C bond between a sp3 carbon and a sp2 carbon is slightly stronger than a C-C bond between two sp3 carbons. Increasing the number alkyl substituents of a double bond also increases the number of sp3-sp2 C-C bonds making the alkene more stable. This is idea can be clearly seen when comparing the isomers 1-butene and 2-butene. The molecule 1-butene is monosubstituted and contains a sp3-sp3 C-C and a sp3-sp2 C-C bond. The disubstituted, 2-butene, contains 2 sp3-sp2 C-C bonds which contributes to its greater stability.
Note
In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain.
Exercise $1$
1) Of the three following isomers which would be expected to be the most stable?
a)
b)
c)
Answer
1)
a)
b)
c)
Exercise $2$
3-Bromobut-1-ene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product?
Answer
2-Bromobutane (numbering changes when alkene is no longer present)
Exercise $3$
Cyclohexene reacts with hydrogen gas in the presence of a palladium catalyst. What is the name of the product?
Answer
Cyclohexane
Exercise $4$
What is the stereochemistry (syn or anti addition) of an alkene hydrogenation reaction?
Answer
Syn-addition
Exercise $5$
When looking at their heats of hydrogenation, is the cis or the trans isomer generally more stable?
Answer
Trans
Exercise $1$
Show the product for the following
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.07%3A_Stability_of_Alkenes.txt |
Objectives
After completing this section, you should be able to
1. explain the term “electrophilic addition reaction,” using the reaction of a protic acid, HX, with an alkene as an example.
2. write the mechanism for the reaction of a protic acid, HX, with an alkene.
3. sketch a reaction energy diagram for the electrophilic addition of an acid, HX, to an alkene.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• carbocation (carbonium ion)
• electrophilic addition reaction
Study Notes
An electrophilic addition reaction is a reaction in which a substrate is initially attacked by an electrophile, and the overall result is the addition of one or more relatively simple molecules across a multiple bond.
The mechanism for the addition of hydrogen halide to propene shown in the reading is quite detailed. Normally, an organic chemist would write the reaction scheme as follows:
However, the more detailed mechanism shown in the reading does allow you to see the exact fate of all the electrons involved in the reaction.
In your previous chemistry course, you were probably taught the importance of balancing chemical equations. It may come as a surprise to you that organic chemists usually do not balance their equations, and often represent reactions using a format which is quite different from the carefully written, balanced equations encountered in general chemistry courses. In fact, organic chemists are rarely interested in the inorganic products of their reactions; furthermore, most organic reactions are non-quantitative in nature.
In many of the reactions in this course, the percentage yield is indicated beneath the products: you are not expected to memorize these figures. The question of yield is very important in organic chemistry, where two, five, ten or even twenty reactions may be needed to synthesize a desired product. For example, if a chemist wishes to prepare compound D by the following reaction sequence:
$A → B → C → D \nonumber$
and each of the individual steps gives only a 50% yield, one mole of A would give only
1 mol $×\frac{50%}{100%}×\frac{50%}{100%}×\frac{50%}{100%}=$ 0.125 mol of D
You will gain first-hand experience of such situations in the laboratory component of this course.
Introduction
One of the most important reactions for alkenes is called electrophilic addition. In this chapter several variations of the electrophilic addition reaction will be discussed. Each case will have aspects common among all electrophilic addition. In this section, the electrophilic addition reaction will be discussed in general to provide a better understanding of subsequent alkene reactions.
As discussed in Section 6-5, the double bond in alkenes is electron rich due to the prescience of 4 electrons instead of the two in a single bond. Also, the pi electrons are positioned above and below the double bond making them more accessibly for reactions. Overall, double bonds can easily donate lone pair electrons to act like a nucleophile (nucleus-loving, electron rich, a Lewis acid). During an electrophilic addition reactions double bonds donate lone pair electrons to an electrophile (Electron-loving, electron poor, a Lewis base). There are many types of electrophilic addition, but this section will focus on the addition of hydrogen halides (HX). Many of the basic ideas discussed will aplicable to subsequent electrophilic addition reactions.
General Reaction
Overall during this reaction the pi bond of the alkene is broken to form two single, sigma bonds. As shown in the reaction mechanism, one of these sigma bonds is connected to the H and the other to the X of the hydrogen halide. This reaction works well with HBr and HCl. HI can also but used but is is usually generated during the reaction by reacting potassium iodidie (KI) with phosphoric acid (H3PO4).
Addition to symmetrical alkenes
What happens?
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.
For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Mechanism
Step 1) Electrophilic Attack
During the first step of the mechanism, the 2 pi electrons from the double bond attack the H in the HBr electrophile which is shown by a curved arrow. The two pi electrons form a C-H sigma bond between the hydrogen from HBr and a carbon from the double bond. Simultaneously the electrons from the H-X bond move onto the halogen to form a halide anion. The removal of pi electrons form the double bond makes one of the carbons become an electron deficient carbocation intermediate. This carbon is sp2 hybridized and the positive charge is contained in an unhybridized p orbital.
Step 2) Nucleophilic attack by halide anion
The formed carbocation now can act as an electrophile and accept an electron pair from the nucleophilic halide anion. The electron pair becomes a X-C sigma bond to create the neutral alkyl halide product of electrophilic addition.
All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.
Reaction Energy Diagram
An energy diagram for the two-step electrophilic addition mechanism is shown below. The energy diagram has two peaks which represent the transition state for each mechanistic step. The peaks are separated by a valley which represents the high energy carbocation reaction intermediate. Because the energy of activation for the first step of the mechanism (ΔE1) is much larger than the second (ΔE1), the first step of the mechanism is the rate-determining step. Both the alkene and the hydrogen halide are reactants in the first step of the mechanism, this electrophilic addition is a second order reaction and the rate law expression can be written rate = k[Alkene][HX]. Also, any structural feature which can stabilize the transition state between the reactants the carbocation intermediate will lower ΔE1 and thereby increase the reaction rate. Overall, the alkyl halide product of this reaction more stable than the reactants making the reaction exothermic.
Reaction rates
Variation of rates when you change the halogen
Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.
When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
Variation of rates when you change the alkene
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be.
Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.
Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.
The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.
The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Representing Organic Reactions
Organic reaction equations are often written in one of two ways. The reactant for the reaction is written to the left of the reaction arrow. The products are written to the right of the arrow. The reagent for the reaction is written above the arrow. Other reaction conditions such as the solvent or the temperature can be written above or below the reaction arrow.
Alternativley the reactant and reagent can both be written to the left of the reaction arrow. This is typically done to highlight the importance of the reactant. The solvent and reaction temperature are still written above or below the reaction arrow. The reaction products are still written to the right of the reaction arrow. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.08%3A_Electrophilic_Addition_Reactions_of_Alkenes.txt |
Objectives
After completing this section, you should be able to
1. use Markovnikov’s rule to predict the product formed when a protic acid, HX, reacts with an alkene.
2. identify the protic acid, HX, and the alkene that must be reacted together to produce a given alkyl halide. [Note: Special conditions are needed if an alkyl iodide is to be produced.]
3. distinguish among primary, secondary and tertiary carbocations.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Markovnikov’s rule
• regioselective (regiospecific)
Study Notes
Recall the definitions of primary, secondary and tertiary hydrogen atoms given in Section 3.3. It follows that a “primary carbocation” is a carbocation in which the carbon atom carrying the positive charge is bonded to only one other carbon atom, a “secondary carbocation” is one in which the carbon atom carrying the positive charge is bonded to two other carbon atoms, and so on.
Regioselectivity
If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if one product forms in greater amounts than the others, the overall reaction is said to be regioselective.
Say three reactions could occur between the hypothetical reactants A and B under the same conditions giving the constitutionally isomeric products C, D, and E.
There are two possibilities:
1. The three products form in equal amounts, i.e., of the total product 33% is C, another 33% D, the remaining 33% E. (These percentages are called relative yields of the products.)
If this is what is observed, the overall reaction between A and B is not regioselective.
2. One product forms in greater amounts than the others. Say, for example, the relative yields of C, D, and E are 25%, 50%, and 25%, respectively.
If this is what is observed, the overall reaction between A and B is regioselective.
eg:
Experimentally, 2 is the major product; 3 is the minor product. Thus, the overall reaction between 1 and HBr is regioselective toward 2.
If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if only one product is observed, the overall reaction is said to be 100% regioselective or regiospecific.
eg:
The only observed product is 5. (Relative yields of 5 and 6 are 100% and 0%, respectively.) Thus the overall reaction between 4 and HBr is regiospecific toward 5.
Regiospecificity is merely the limiting case of regioselectivity. All regiospecific reactions are regioselective, but not all regioselective reactions are regiospecific.
Addition to unsymmetrical alkenes
During the electrophilic addition of HX to an alkene, the halide (X) could attach to either carbon in the double bond producing two different isomers as products. But, when an unsymmertrically alkyl substituted alkene undergoes an electrophillic addition with HX a single isomer is typically produced. For example, if propene were reacted with HBr, two products could possibly form: 2-bromolpropane and 1-bromopropane. However, 2-bromopropane is produced as the reaction's only product. Reactions are called regiospecific when only one of multiple possible isomers is exclusively formed.
Example \(1\)
If HCl adds to an unsymmetrical alkene like propene what will the major product be?
Solution
The regiospecificity of electrophilic additions to alkenes is commonly known as Markovnikov's rule, after the Russian chemist Vladimir Markovnikov who proposed it in 1869. The electrophilic addition of HX to an alkene is said to follow Markovnikov's rule.
Markovnikov's rule: During the electrophilic addition of HX to an alkene, the H adds to the carbon of the double bond with the fewest number of alkyl substitutent. The halide (X) adds to the double bond carbon with the most alkyl substituents. Although Markovnikov's rule has been specifically stated for the electrophilic addition of HX, later in this chapter many more reaction will be shown to also follow Markovnikov's rule.
It is important to point out that Markovnikov's rule only truley applies when there is a difference between the number of alky groups attached to each carbon in the double bond. When both carbon of the double bond have the same degree of alkyl substitution, Markovnikovs's rule becomes void and a mixture of both possible isomers is produced.
To consider an explanation for why Markovnikov's rule holds true, the mechanism of the reaction needs to be considered. As seen in the previous section, the mechanism starts with the addition of H to a carbon in the double bond. This in turn caused the other double bond carbon to become a carbocation intermediate. In the second step of the mechanism the halide ion attacks the carbocation to form a C-X. Because Markovnikov's rule says that the halogen adds to the carbon in the double bond with the most alky substituents, it can be said the carbocation also prefers to form on the carbon with the most alkyl substituents. The reason for this holds the explaination for Markovnikov's rule. To simplify this idea, Markovnikov's rule can be restated in a different form.
Markovnikov's rule: During the electrophilic addition of HX to an alkene, the carbocation intermediate forms on the double bond carbon with the greatest number of alkyl substitutents.
Predicting the Product of an Electrophilic Addition with HX
Overall, during the electrophilic addition of HX to an alkene there are two major changes in the bonding. First, the pi bond of the alkene is broken. Second, a single bond is formed on each carbon that was originally in the double bond. The two single bonds will become attached to and H and an X. If the alkene is unsymmertrically alkyl substituted, Markovnikov's rule is followed and the X will be bonded to the more alky substituted carbon and the H to the less substiuted. If the alkene is symmertrically alkyl substituted a mixture of isomers will be produced in the product.
Worked Example\(1\)
Please draw the product of the following reaction:
Answer
In answering these types of questions it is always important to first determine which reaction is occurring. Because an alkene is the reactant and HBr is the product this reaction is an electrophilic addition. Overall, the double bond will be broken as H and Br are added. The next step is to determine if Markovnikov's rule needs to be applied. In the reactant's double bond the upper carbon has two alkyl substituents and the lower carbon has only two. Markovnikov's rule says that Br will attach to the upper carbon and H to the lower.
Planning the Synthesis of an Alkyl Halide using Electrophilic Addition
Understanding the starting material and reaction required for the synthesis a specific target molecule is an important concept in organic chemistry. The preferred method for answering these types of questions is to work backwards from the target molecule. Often there will be multiple variations of reactions and starting materials capable synthesizing the target molecule. Further analysis of these variations can draw out strengths and weaknesses and lead to the pathway with the best chance of success.
To create a possible starting material for a given alkyl halide simply reverse the bonding changes expected during an electrophilic addition. Remove the C-X single bond and a C-H bond from an adjacent carbon. Then connect these two carbons with a double bond.
Worked Example \(1\)
What alkene would be required to make the following 2-bromopentane using an electrophilic addition?
Answer
Analysis: Possible starting materials can be made by first removing the C-Br bond and an adjacent C-H bond. Then connecting the two carbons with a double bond. Because there are two different locations where an adjacent C-H bond can be removed there will be two possible starting material.
Solution: When comparing the the possible starting materials, #1 will follow Markovnikov's rule and produce 2-bromopentane as the sole product. Starting material #2 is symmertrically alkyl substituted, so Markovnikov's rule does not apply. A mixture of 2-bromopentane and 3-bromopentane would be produced. Between the two possible starting material #1 would be preferred because only one product is produced.
Exercise \(1\)
Predict the product(s) for the following reactions:
Answer
Exercise \(2\)
In each case, suggest an alkene that would give the product shown.
Answer
Exercise \(3\)
Give the IUPAC name for the product of the following reaction.
Answer
Exercise \(4\)
Draw the reaction mechanism of the previous problem
Answer
Exercise \(5\)
Identify the products of the following reactions.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.09%3A_Orientation_of_Electrophilic_Additions_-_Markovnikov%27s_Rule.txt |
Objectives
After completing this section, you should be able to
1. describe the geometry of a given carbocation.
2. arrange a given series of carbocations in order of increasing or decreasing stability.
3. explain the relative stability of methyl, primary, secondary and tertiary carbocations in terms of hyperconjugation and inductive effects.
Study Notes
Although hyperconjugation can be used to explain the relative stabilities of carbocations, this explanation is certainly not the only one, and is by no means universally accepted. A more common explanation, involving the concept of an inductive effect, is given below.
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. In a tertiary carbocation, the positively charged carbon atom attracts the bonding electrons in the three carbon-carbon sigma (σ) bonds, and thus creates slight positive charges on the carbon atoms of the three surrounding alkyl groups (and, indeed, on the hydrogen atoms attached to them). Chemists sometimes use an arrow to represent this inductive release:
Note: These diagrams do not reflect the geometry of the carbocation. The overall charge on the carbocation remains unchanged, but some of the charge is now carried by the alkyl groups attached to the central carbon atom; that is, the charge has been dispersed.
In the tertiary carbocation shown above, the three alkyl groups help to stabilize the positive charge. In a secondary carbocation, only two alkyl groups would be available for this purpose, while a primary carbocation has only one alkyl group available. Thus the observed order of stability for carbocations is as follows:
tertiary > secondary > primary > methyl.
Stability of carbocation intermediates
The next step in understanding why Markovnikov's rule is often followed in electrophilic additions, involves understanding the structure and stability of the carboncation intermediate formed during the mechanism.
Carbocation Structure
Carbocations typically have three substituents which makes the carbon sp2 hybridized and gives the overall molecule a trigonal planar geometry. The carbocation's substituents are all in the same plane and have a bond angle of 120o between them. The carbon atom in the carbocation is electron deficient; it only has six valence electrons which are used to form three sigma covalent bonds with the substituents. The carbocation carbon has an unoccupied p orbital which is perpendicular to the plane created by the substituents. The p orbital can easily accept electron pairs during reactions making carbocations excellent Lewis acids.
Stability of Carbocation Intermediates
By being a reactive intermediate of the electrophilic addition mechanism, the stability of a carbocation has a direct effect on the reaction.
The critical question now becomes, what stabilizes a carbocation?
A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
electron donating group stabilizes a carbocation
electron withdrawing group destabilizes a carbocation
Extensive experimental evidence has shown that a carbocation becomes more stable as the number of alkyl substituents increases. Carbocations can be given a designation based on the number of alkyl groups attached to the carbocation carbon. Three alkyl groups is called a tertiary (3o) carbocation, 2 alkyl groups is called secondary (2o), and 1 alkyl group is called primary (1o). No alkyl groups are attached (3 hydrogen substituents) is called a methyl carbocation.
The overall order of stability is as follows:
Alkyl groups stabilized carbocations for two reasons. The first is through inductive effects. As discussed in Section 2-1, inductive effects occur when the electrons in covalent bonds are shifted towards an nearby atom with a higher electronegativity. In this case, the positively charged carbocation draws in electron density from the surrounding substituents thereby gaining stabilization by slightly reducing its positive charge. Alkyl groups are more effective at inductively donating electron density than a hydrogen because they are larger, more polarizable, and contain more bonding electrons. As more alkyl groups are attached to the carbocation more inductive electron donation occurs and the carbocation becomes more stable.
The second reason alkyl groups stabilize carbocations is through hyperconjugation. As previously discussed in Section 7.6, hyperconjugation is an electron donation that occurs from the parallel overlap of p orbitals with adjacent hybridized orbitals participating in sigma bonds. This electron donation serves to stabilize the carbocation. As the number of alkyl substituents increases, the number of sigma bonds available for hyperconjugation increases, and the carbocation tends to become more stabilized.
In the example of ethyl carbocation shown below, the p orbital from a sp2 hybridized carbocation carbon involved interacts with a sp3 hybridized orbital participating in an adjacent C-H sigma bond. Electron density from the C-H sigma bond is donated into carbocation's p orbital providing stabilization.
The molecular orbital of the ethyl carbocation shows the interaction of electrons in methyl group's C-H sigma bonds with the adjacent empty p orbital from the carbocation. The interaction creates a bonding molecular orbital which extends over the three atom chain (C-C-H) involved in hyperconjugation. The expanded molecular orbital helps to stabilize the carbocation.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
In the next chapter we will see how the carbocation-destabilizing effect of electron-withdrawing fluorine substituents can be used in experiments designed to address the question of whether a biochemical nucleophilic substitution reaction is SN1 or SN2.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
a vinylic carbocation (very unstable)
Example \(1\)
In which of the structures below is the carbocation expected to be more stable? Explain.
Answer
In the carbocation on the left, the positive charge is located in a position relative to the nitrogen such that the lone pair of electrons on the nitrogen can be donated to fill the empty orbital. This is not possible for the carbocation species on the right.
Example \(2\)
Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
Answer
When considering the possibility that a nucleophilic substitution reaction proceeds via an SN1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate. If this intermediate is not sufficiently stable, an SN1 mechanism must be considered unlikely, and the reaction probably proceeds by an SN2 mechanism. In the next chapter we will see several examples of biologically important SN1 reactions in which the positively charged intermediate is stabilized by inductive and resonance effects inherent in its own molecular structure.
Example \(3\)
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Answer
a) 1 (tertiary vs. secondary carbocation)
b) 1 (tertiary vs. secondary carbocation)
c) 2 (positive charge is further from electron-withdrawing fluorine)
d) 1 (lone pair on nitrogen can donate electrons by resonance)
e) 1 (allylic carbocation – positive charge can be delocalized to a second carbon) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.10%3A_Carbocation_Structure_and_Stability.txt |
Objective
After completing this section, you should be able to use the Hammond postulate to explain the formation of the most stable carbocation during the addition of a protic acid, HX, to an alkene.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Hammond postulate
So far in this chapter the following points have been made about the electrophilic addition of HX to a double bond.
• The reaction takes place through a two step mechanism which forms a carbocation intermediate.
• During electrophilic addition the carbocation intermediate, and the subsequent H-X bond, forms on the double bond carbon with the most alkyl stubstituents (Markovnikov's rule)
• Carbocations become more stable as the number of alkyl substituents increases.
It appears that the stability of the carbocation reactive intermediate has a direct effect on the products of a reaction. However, it is the activation energy required to reach the transition state of the reaction's rate determine step which determines which determines which product is produced. This implies that there is a relationship between the transition state and the carbocation reactive intermediate in the mechanism of electrophilic addition.
The Hammond Postulate
Chemists are often very interested in the structures of the transition states in a reaction's mechanism. In particular, the transition state for a mechanism's rate determining step directly determines the energy of activation barrier and thereby the rate for the overall reaction. Understanding the structure of a transition state allows chemists to consider structural features which might stabilize or destabilize the transition state causing a corresponding change in the rate of reaction. However, transition state structures cannot be directly observed because they are highly unstable activated complexes which instantly convert to a more stable species. In order to gain some insight into the structure of particular transition state, chemists often invoke the Hammond postulate, which states that a transition state resembles the structure of the nearest stable species (reactant, intermediate or product).
For an exergonic reaction, the transition state is closer in energy to the reactants. Therefore, the structure of the transition state can assumed to resemble the reactants more than the products. Shown below is a hypothetical exergonic reaction between reactant compounds A and B to form the product AB. The Hammond postulate would theorize that the distance between A and B in the transition state would be relatively large thus resembling the reactants where A and B are two isolated species.
For an endergonic reaction, the transition state is closer in energy to the product. Therefore, the structure of the transition state can assumed to resemble the products more than the reactants. In the hypothetical endergonic reaction shown below, reactant compounds C and D react to form the product CD. The Hammond postulate would predict that the distance between C and D in the transition state would be relatively small thus resembling the products where C and D are bonded together as a single product CD.
The Hammond Postulate and Electrophilic Addition
By applying the Hammond postulate and other ideas cultivated in this chapter the reason why electrophilic additions tend to follow Markovnikov's rule. When the energy diagram of an electrophilic addition was discussed in Section 7.2, it was noted the first step of the mechanism was the rate determining step. The first step of the mechanism also is endergonic and results in the formation of a carbocation intermediate.
The Hammond postulate suggests that the transition state structure for the first step of the mechanism resembles that of the carbocation intermediate because they are the closest in energy. A transitions state, seen below, is typically drawn as a theoretical structure part way between the reactants and the product. For this transition state the pi bonds and the H-Br bond are in the process of being broken and are represented with a dashed line. The C-H bond is in the process of being formed so it also represented with a dashed line. The bromine is shown with a partial negative charge (sigma-) because it is becoming a bromide ion (Br-) which has a full negative charge. Most importantly, the carbon is in the process of becoming a carbocation so it is shown to have a partial positive change (sigma+).
Because the Hammond postulate predicts this transitions state closely resembles the carbocation intermediate, the partial positive charge can said to closely resemble the full positive charge of the carbocation. Consequently, any structural feature that stabilized the carbocation intermediate will also stabilize the transition state. The partial positive charge of the transition state is stabilized by adjacent alkyl groups thorough inductive effects and hyperconjugation much like the carbocation intermediate. Adding more alkyl substituents to the partially positive charged carbon stabilizes the transition state, causing it to become lower in energy. This in turn, decreases the energy of activation and increases the rate of the reaction. In short, during an electrophilic addition, the double bond carbon with the most alkyl substituents will for a carbocation intermediate and therefore its C-X bond faster than the double bond carbon with fewer alkyl substituents. These effects cause electrophillic additions to follow Markovnikov's rule and place the halogen (X) group on the more substituted carbon of asymmetrically alkyl substituted double bond.
Carbocation
Stability
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Exercises
Exercise \(1\)
1) Consider the second step in the electrophilic addition of HBr to an alkene. Is this step exergonic or endergonic and does the transition state represent the product or the reactant (cation)? Draw out an energy diagram of this step reaction.
Answer
1) Exergonic and the transition state (second step) represents the reactant (cation).
As shown to go from intermediate cation to final product the step is exergonic. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.11%3A_The_Hammond_Postulate.txt |
Objective
After completing this section, you should be able to explain the “unusual” products formed in certain reactions in terms of the rearrangement of an intermediate carbocation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alkyl shift
• hydride shift
Study Notes
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 neighbouring carbon atoms, and hence are termed 1,2-hydride shifts or 1,2-alkyl shifts.
[A hydride ion consists of a proton and two electrons, that is, [H:]. Hydride ions exist in compounds such as sodium hydride, NaH, and calcium hydride, CaH2.]
An electrophilic reaction such as HX with an alkene will often yield more than one product. This is strong evidence that the mechanism includes intermediate rearrangement steps of the cation.
Throughout this textbook many reaction mechanisms will be presented. It is impossible to know with absolute certainty that a mechanism is correct. At best a proposed mechanism can be shown to be consistent with existing experimental data. Virtually all of the mechanisms in this textbook have been carefully studied by experiments designed to test their validity although the details are not usually discussed. An excellent example of experimental evidence which supports the carbocation based mechanism for electrophilic addition, is that structural rearrangements often occur during the reaction.
1,2-Hydride Shift
A 1,2-hydride shift is a carbocation rearrangement in which a hydrogen atom in a carbocation migrates to the carbon atom bearing the formal charge of +1 (carbon 2 in the example below) from an adjacent carbon (carbon 1).
An example of this structural rearrangment occurs during the reaction of 3-methyl-1-butene with HBr. Markovnikov's rule predicts that the preferred product would be 2-bromo-3-methylbutane, however, very little of this product forms. The predominant product is actually 2-bromo-2-methylbutane.
Mechanism of Hydride Shift
This result comes from a Hydride Shift during the reaction mechanism. The mechanism begins with protonation of the alkene which places a positive charge on the more alkyl substituted double bond carbon resulting in a secondary carbocation. In step 2, The electrons in the C-H bond on carbon #3 are attracted by the positive charge on carbon #2, and they simply shift over to fill the carbocation's empty p orbital, pulling the proton over with them. The process called a carbocation rearrangement, and more specifically, a hydride shift. A hydride ion (H:-) is a proton plus two electrons which not to be confused with H+, which is just a proton without any electrons. Notice that the hydride, in shifting, is not acting as an actual leaving group - a hydride ion is a very strong base and a very poor leaving group.
As the hydride shift proceeds, a new $C-H$ $\sigma$ bond is formed at carbon #2, and carbon #3 is left with an empty $p$ orbital and a positive charge.
What is the thermodynamic driving force for this process? Notice that the hydride shift results in the conversion of a secondary carbocation (on carbon 2) to a (more stable) tertiary carbocation (on carbon 3) - a thermodynamically downhill step. As it turns out, the shift occurs so quickly that it is accomplished before the bromide nucleophile has time to attack at carbon #2. Rather, the bromide will attack after the hydride shift (step 3) at carbon #3 to complete the addition.
1,2-Alkyl Shift
A 1,2-alkyl shift is a carbocation rearrangement in which an alkyl group migrates to the carbon atom bearing the formal charge of +1 (carbon 2) from an adjacent carbon atom (carbon 1), e.g.
Consider another example. When HBr is added to 3,3-dimethyl-1-butene the preferred product is 2-bromo-2,3-dimethylbutane and not 3-bromo-2,2-dimethylbutane as predicted by Markovnikov's rule.
Notice that in the observed product, the carbon framework has been rearranged: a methyl carbon has been shifted. This is an example of another type of carbocation rearrangement, called an alkyl shift or more specifically a methyl shift.
Mechanism of Alkyl Shift
Below is the mechanism for the reaction. Once again a secondary carbocation intermediate is formed in step 1. In this case, there is no hydrogen on carbon #3 available to shift over create a more stable tertiary carbocation. Instead, it is a methyl group that does the shifting, as the electrons in the carbon-carbon $\sigma$ bond shift over to fill the empty orbital on carbon #2 (step 2 below). The methyl shift results in the conversion of a secondary carbocation to a more stable tertiary carbocation. It is this tertiary carbocation which is attacked by the bromide nucleophile to make the rearranged end product. The end result is a rearrangement of the carbon framework of the molecule.
Electrophilic addition with methyl shift:
Predicting the Product of a Carbocation Rearrangement
Carbocation shifts occur in many more reactions than just electrophilic additions as some of which will be discussed in subsequent chapters of this textbook. Whenever a carbocation is produced in a reaction's mechanism the possibility of rearrangements should be considered. As discussed in Section 7.9, there are multiple ways to stabilize a carbocation all of which could induce a rearrangement.
The most common situation for a rearrangement to occur during electrophilic addition is:
A 2o Carbocation with a 3o or 4o Alkyl Substituent
When considering the possibility of a carbocation rearrangement the most important factors are the designation of the carbocation formed and the designation of the alkyl groups attached to the carbocation. When a 2o carbocation has a 3o alkyl substituent a hydride shift will occur to create a more stable 3o carbocation. When a 2o carbocation has a 4o alkyl substituent an alkyl shift will occur to create a more stable 3o carbocation.
Drawing the Rearranged Product
First, draw the unrearranged product. Add HX to the double bond following Markovnikov's rule if necessary. Then determine if a hydride or alkyl shift is occurring by observing the designation of the alkyl substituent. The switch X ⇔H for a hydride shift and X ⇔ CH3 for an alkyl shift this will produce the rearranged product.
Exercise $1$
Draw the expectred produts of the following reaction.
Answer
Biological Carbocation Rearrangement
Carbocation rearrangements are involved in many known biochemical reactions. Rearrangements are particularly important in carbocation-intermediate reactions in which isoprenoid molecules cyclize to form complex multi-ring structures. For example, one of the key steps in the biosynthesis of cholesterol is the electrophilic cyclization of oxidosqualene to form a steroid called lanosterol.
This complex but fascinating reaction has two phases. The first phase is where the actual cyclization takes place, with the formation of four new carbon-carbon bonds and a carbocation intermediate. The second phase involves a series of hydride and methyl shifts culminating in a deprotonation. In the exercise below, you will have the opportunity to work through the second phase of the cyclase reaction mechanism.
Exercise $1$
The second phase of the cyclase reaction mechanism involves multiple rearrangement steps and a deprotonation. Please supply the missing mechanistic arrows.
Answer
Exercise $1$
1) The following reaction shows a rearrangement within the mechanism. Propose a mechanism that shows this.
2) Propose a mechanism for the following reaction. It involves an electrophilic addition and the shift of a C-C and a C-H bond.
Answer
1)
2) In most examples of carbocation rearrangements that you are likely to encounter, the shifting species is a hydride or methyl group. However, pretty much any alkyl group is capable of shifting. Sometimes, the entire side of a ring will shift over in a ring-expanding rearrangement.
The first 1,2-alkyl shift is driven by the expansion of a five-membered ring to a six-membered ring, which has slightly less ring strain. A hydride shift then converts a secondary carbocation to a tertiary carbocation, which is the electrophile ultimately attacked by the bromide nucleophile.
Once again, the driving force for this process is an increase in stability of the carbocation. Initially, there is a primary carbocation at C2, and this becomes a tertiary carbocation at C1 as a result of the (1,2)-methyl shift. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.12%3A_Evidence_for_the_Mechanism_of_Electrophilic_Additions_-_Carbocation_Rearrangements.txt |
Concepts & Vocabulary
7.1 Industrial Preparation and Use of Alkenes
• Breaking up of large hydrocarbon molecules into smaller, useful molecules is called cracking.
7.2 Calculating Degree of Unsaturation
• Saturated molecules contain only single bonds and no rings.
• Saturated hydrocarbons have the formula CnH2n+2, where n can be any integer.
• Degrees of unsaturation account for the total number of rings and pi bonds in a molecule.
• Each degree of unsaturation reduces the number of hydrogens in the molecule by 2.
7.3 Naming Alkenes
• When the two largest groups are on the same side of the double bond (top or bottom) they are called cis or Z.
• When the two largest groups are on opposite sides of the double bond (top or bottom) they are called trans or E.
• Endocyclic double bonds occur when there is a pi bond within a ring.
7.4 Cis-Trans Isomerism in Alkenes
7.5 Alkene Stereochemistry and the E, Z Designation
• E and Z are less limited than cis and trans in naming.
• E and Z configurations use the same priority rules as R and S (CIP rules).
7.6 Stability of Alkenes
• Relative stability of alkenes can be measured by using heats of hydrogenation upon reduction to the related alkane.
• More substituted alkenes are more stable than less substituted.
• Alkenes with the largest groups trans are more stable than cis.
7.7 Electrophilic Addition Reactions of Alkenes
• In electrophilic addition reactions, the pi bond of the alkene acts as the nucleophile.
• Electrophilic addition reactions occur faster with larger hydrogen halides as well as more substituted alkenes.
7.8 Orientation of Electrophilic Additions: Markovnikov's Rule
• The more substituted carbocation intermediate forms during electrophilic addition reactions, since more substituted carbocations are more stable. This is known as Markovnikov's rule.
7.9 Carbocation Structure and Stability
• Molecules or ions that can disperse (delocalize) charge are more stable than structures with charge localized on a single atom.
• Due to inductive stabilization, carbocation stability follows the order:
tertiary > secondary > primary > methyl
• Electron donating groups stabilize carbocations.
• Electron withdrawing groups destabilize carbocations.
• Resonance effects can stabilize a carbocation (some examples include benzylic and allylic carbocations).
• Vinylic carbocations are unstable and are unlikely to form.
7.10 The Hammond Postulate
• The Hammond Postulate states that transition state structure most resembles the nearest stable species.
• Based on the Hammond Postulate, transition states for exothermic reaction steps resemble reactants, while endergonic step transition states resemble products.
7.11 Evidence for the Mechanism of ELectrophilic Additions: Carbocation Rearrangements
• Carbocations will rearrange from less stable to more stable isomers through hydride shifts or alkyl shifts.
Skills to Master
• Skill 7.1 Calculate degree of unsaturation for organic molecular formulae.
• Skill 7.2 Draw isomers from a molecular formula.
• Skill 7.3 Name alkenes following IUPAC rules, including configuration (E, Z).
• Skill 7.4 Draw structures from IUPAC name.
• Skill 7.5 Describe bonding in alkenes including bond length, strength, angle and restricted rotation.
• Skill 7.6 Explain stability of alkenes.
• Skill 7.7 Rank alkenes in order of stability.
• Skill 7.8 Draw mechanism for electrophilic addition of HX to alkenes, including regiochemistry.
• Skill 7.9 Explain stability of carbocations.
• Skill 7.10 Explain transition states related to the Hammond Postulate.
• Skill 7.11 Explain products formed by carbocation rearrangements.
Memorization Tasks
MT 7.1 Memorize formula for saturated hydrocarbons CnH2n+2.
MT 7.2 Memorize basic IUPAC naming rules.
MT 7.3 Memorize relative stability of alkenes.
MT 7.4 Memorize relative stability of carbocations. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.S%3A_Alkenes-_Structure_and_Reactivity_%28Summary%29.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
After you have completed Chapter 8, you should be able to
1. fulfill all of the detailed objectives listed under each section.
2. design a relatively simple, multistep synthesis using the reactions introduced in this chapter, given the structure, name, or both, of the starting material and product. For example, show how you would convert 1-bromobutane to
3. deduce the structures of a number of compounds involved in a certain reaction sequence, given sufficient information. In other words, solve so-called road-map problems.
4. define, and use in context, the key terms introduced in this chapter.
8.01: Preparation of Alkenes - A Preview of Elimination Reactions
Objectives
After completing this section, you should be able to
1. explain the relationship between an addition reaction and an elimination reaction.
2. write an equation to describe the dehydrohalogenation of an alkyl halide.
3. identify the reagents required to bring about dehydrohalogenation of an alkyl halide.
4. write an equation to represent the dehydration of an alcohol.
5. identify the reagents required to dehydrate a given alcohol.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• dehydration
• dehydrohalogenation
• elimination reaction
Study Notes
An elimination reaction is a reaction in which two or more atoms, one of which is usually hydrogen, are removed from adjacent atoms in the reactant, resulting in the formation of a multiple bond.
The relationship between addition reactions and elimination reactions is shown in Figure 8.1, below.
Alkenes can be readily prepared from the alkylhalide (X = Cl, Br, I) or the alcohol.
Electrophilic Addition
Alkenes are found throughout nature. They form the basis of many natural products, such as terpenes, which play a variety of roles in the lives of plants and insects. The C=C bonds of alkenes are very different from the C=O bonds that are also common in nature. The C=C bonds of alkenes are electron-rich and nucleophilic, in contrast to the electron-poor C=O bonds of carbohydrates, fatty acids and proteins. That difference plays a role in how terpenes form in nature.
Alkenes, or olefins, are also a major product of the petroleum industry. Reactions of alkenes form the basis for a significant portion of our manufacturing economy. Commonly used plastics such as polyethylene, polypropylene and polystyrene are all formed through the reactions of alkenes. These materials continue to find use in our society because of their valuable properties, such as high strength, flexibility and low weight.
Alkenes undergo addition reactions as you will see carbonyls do as well. Most commonly they add a proton to one end of the double bond and another group to the other end. These reactions happen in slightly different ways, however.
Alkenes are reactive because they have a high-lying pair of π-bonding electrons. These electrons are loosely held, being high in energy compared to σ-bonds. The fact that they are not located between the carbon nuclei, but are found above and below the plane of the double bond, also makes these electrons more accessible.
Alkenes can donate their electrons to strong electrophiles other than protons, too. Sometimes their reactivity pattern is a little different than the simple addition across the double bond, but that straightforward pattern is what we will focus on in this chapter.
Elimination reactions
Elimination reactions are possible by abstraction of a proton at positions that are next to a potential leaving group. This type of elimination can be described by two model mechanisms: it can occur in a single concerted step (proton abstraction at Cα occurring at the same time as Cβ-X bond cleavage), or in two steps (Cβ-X bond cleavage occurring first to form a carbocation intermediate, which is then 'quenched' by proton abstraction at the alpha-carbon).
The most common elimination reactions are dehydrohalogenation and dehydration. In the mechanism above, X could be Cl, Br, or I for the dehydrohalogenation where there is a loss of HX from an alkyl halide. For dehydration, X would be an OH group in the above mechanism where the overall loss is water from an alcohol. These mechanisms, termed E2 and E1, respectively, are important in laboratory organic chemistry, but are less common in biological chemistry. As explained below, which mechanism actually occurs in a laboratory reaction will depend on the identity of the R groups (ie., whether the alkyl halide is primary, secondary, tertiary, etc.) as well as on the characteristics of the base. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.00%3A_Chapter_Objectives.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the reaction of chlorine or bromine with a given alkene.
2. identify the conditions under which an addition reaction occurs between an alkene and chlorine or bromine.
3. draw the structure of the product formed when a given alkene undergoes an addition reaction with chlorine or bromine.
4. write the mechanism for the addition reaction that occurs between an alkene and chlorine or bromine, and account for the stereochemistry of the product.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anti stereochemistry
• bromonium ion
Study Notes
In the laboratory you will test a number of compounds for the presence of a carbon-carbon double bond. A common test is the decolourization of a reddish-brown bromine solution by an alkene.
The two-step mechanism shown in the LibreText pages gives you an idea of how the reaction between an alkene and a halogen occurs. Note the formation of the bridged bromonium ion intermediate and the anti stereochemistry of the final product because the two bromine atoms come from opposite faces of the double bond.
Additional evidence in support of the bromonium ion mechanism comes from the results obtained when an alkene (such as cyclopentene) reacts with bromine in the presence of sodium chloride (see Figure 8.2: Reaction of an alkene with bromine in the presence of sodium chloride, below).
Once formed, the bromonium ion is susceptible to attack by two nucleophiles—chloride ion and bromide ion—and, in fact, a mixture of two products (both produced by anti attack) is formed.
Halogens can act as electrophiles to which can be attacked by a pi bond from an alkene. Pi bonds represents a region of electron density and therefore function as a nucleophiles. How is it possible for a halogen to obtain positive charge to be an electrophile?
Introduction
A halogen molecule, for example Br2, approaches a double bond of the alkene, electrons in the double bond repel electrons in the bromine molecule causing polarization of the halogen-halogen bond. This creates a dipole moment in the halogen-halogen bond. Heterolytic bond cleavage occurs and one of the halogens obtains a positive charge and reacts as an electrophile. The reaction of the addition is not regioselective but is stereoselective. Stereochemistry of this addition can be explained by the mechanism of the reaction. In the first step, the electrophilic halogen (with the positive charge) approaches the pi bond and 2p orbitals of the halogen bond with two carbon atoms creating a cyclic ion with a halogen as the intermediate. In the second step, the remaining halide ion (halogen with the negative charge) attacks either of the two carbons in the cyclic ion from the back side of the cycle as in the SN2 reaction. Therefore stereochemistry of the product is anti addition of vicinal dihalides.
$\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X} \tag{8.2.1}$
Step 1: In the first step of the addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a cyclic intermediate with the two carbons from the alkene.
Step 2: In the second step, bromide anion attacks either carbon of the bridged bromonium ion from the back side of the ring. The bromine atom in the bromonium ion acts as a shield in a way, forcing the bromonium anion to attack from the opposite side as it. The result of this is the ring opening up with the two halogens on opposite sides as each other. This is anti stereochemistry, which is defined as the two bromine atoms come from opposite faces of the double bond. The product is that the bromines add on trans to each other.
Halogens that are commonly used in this type of the reaction are: $Br$ and $Cl$. In thermodynamical terms $I$ is too slow for this reaction because of the size of its atom, and $F$ is too vigorous and explosive.
Because the halide ion can attack any carbon from the opposite side of the ring it creates a mixture of steric products. Optically inactive starting material produce optically inactive achiral products (meso) or a racemic mixture.
Electrophilic addition mechanism consists of two steps.
Before constructing the mechanism let us summarize conditions for this reaction. We will use Br2 in our example for halogenation of ethylene.
Nucleophile Double bond in alkene
Electrophile Br2, Cl2
Regiochemistry not relevant
Stereochemistry ANTI
Summary
Halogens can act as electrophiles due to polarizability of their covalent bond. Addition of halogens is stereospecific and produces vicinal dihalides with anti addition.
Problems
1.What is the mechanism of adding Cl2 to the cyclohexene?
2.A reaction of Br2 molecule in an inert solvent with alkene follows?
a) syn addition
b) anti addition
c) Morkovnikov rule
3)
4)
Key:
1.
2. b
3.enantiomer
4.
Contributors and Attributions
• Jim Clark (Chemguide.co.uk)
• Layne Morsch (University of Illinois Springfield)
• Dr. Krista Cunningham
• Lauren Reutenauer (Amherst College) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.02%3A_Halogenation_of_Alkenes_-_Addition_of_X.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the formation of a halohydrin from an alkene.
2. write the mechanism for the formation of a halohydrin from an alkene and a mixture of halogen and water.
3. predict the mechanism of the addition reaction that occurs between a given reagent and an alkene, basing your prediction on mechanisms you have studied in this chapter.
4. identify the alkene, the reagents, or both, that should be used to produce a given halohydrin by an addition reaction.
5. identify N-bromosuccinimide in aqueous dimethyl sulphoxide as an alternative source of bromine for producing bromohydrins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bromohydrin
• halohydrin
Study Notes
Bromohydrin and chlorohydrin are examples of halohydrins (where X = Br or Cl).
Chemists often abbreviate the names of frequently used chemicals: DMSO for dimethyl sulfoxide, NBS for N-bromosuccinimide, etc. You should already be familiar with some similar examples from everyday life: DDT for dichlorodiphenyltrichloroethane, PCB for polychlorinated biphenyl, and ASA for acetylsalicylic acid (aspirin). You can see how someone with a limited knowledge of chemistry could misinterpret the abbreviation NBS—it is not a compound containing nitrogen, boron and sulfur!
NBS can serve as a less dangerous and easier to handle replacement for Br2 in the formation of bromohydrins.
The proton is not the only electrophilic species that initiates addition reactions to the double bond of alkenes. Lewis acids like the halogens, boron hydrides and certain transition metal ions are able to bond to the alkene pi-electrons, and the resulting complexes rearrange or are attacked by nucleophiles to give addition products.The electrophilic character of the halogens is well known as discussed in the previous section. Chlorine (Cl2) and bromine(Br2) react selectively with the double bond of alkenes, and these reactions are what we will focus on. Fluorine adds uncontrollably with alkenes,and the addition of iodine is unfavorable, so these are not useful preparative methods.
The addition of chlorine and bromine to alkenes, as shown below, proceeds by an initial electrophilic attack on the pi-electrons of the double bond (Section 8.2). Dihalo-compounds in which the halogens are bound to adjacent carbons are called vicinal, from the Latin vicinalis, meaning neighboring.
R2C=CR2 + X2 ——> R2CX-CR2X
Another electrophilic addition to an alkene is the reaction of an alkene with the other halogen-containing reagents like hypohalous acids, HOX, to form halohydrins. However, halohydrins are not formed by directly adding a hypohalous acid, instead the alkene is reacted with Br2 or Cl2 in the presence of water. These reagents are unsymmetrical (unlike Br2 or Cl2), so their addition to unsymmetrical double bonds may in principle take place in two ways. In practice, these addition reactions are regioselective, with one of the two possible constitutionally isomeric products being favored. The electrophilic moiety in both of these reagents is the halogen.
(CH3)2C=CH2 + HOBr ——> (CH3)2COH-CH2Br
The regioselectivity of the above reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Thus, bonding of an electrophilic species to the double bond of an alkene should result in preferential formation of the more stable (more highly substituted) carbocation, and this intermediate should then combine rapidly with a nucleophilic species to produce the addition product.
The regioselectivity of the above reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Thus, bonding of an electrophilic species to the double bond of an alkene should result in preferential formation of the more stable (more highly substituted) carbocation, and this intermediate should then combine rapidly with a nucleophilic species to produce the addition product.
Step one is the same as halogenation of alkenes. The halogen, in this case bromine, reacts with the alkene to form a cyclic bromonium ion.
Step two does not have the bromine ion attack, but instead water acts as the nucleophile. Water is available at a higher concentration than the bromine ion, so is more likely to be available in the correct orientation for the nucleophilic attack. Water is also a better nucleophile than the bromine ion. The better a nucleophile, the more likely it is to attack. You may notice that as oxygen makes a bond it gets a positive formal charge at the completion of this step.
Step 3 is a final acid-base step, where there is a loss of a proton from the oxygen to the solvent (water) to form the neutral halohydrin. In this case, a bromohydrin is the product of the addition reaction.
To apply this mechanism we need to determine the electrophilic moiety in each of the reagents. By using electronegativity differences we can dissect common addition reagents into electrophilic and nucleophilic moieties, as shown on the right. In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pKa's ~ 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation. The nucleophilic species that bonds to the intermediate carbocation is then hydroxide ion, or more likely water (the usual solvent for these reagents), and the products are called halohydrins. Sulfenyl chlorides add in the opposite manner because the electrophile is a sulfur cation, RS(+), whereas the nucleophilic moiety is chloride anion (chlorine is more electronegative than sulfur).
Below are some examples illustrating the addition of various electrophilic halogen reagents to alkene groups. Notice the specific regiochemistry of the products, as explained above. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.03%3A_Halohydrins_from_Alkenes_-_Addition_of_HOX.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the hydration of an alkene with sulfuric acid.
2. write an equation for the formation of an alcohol from an alkene by the oxymercuration-demercuration process.
3. identify the alkene, the reagents, or both, that should be used to produce a given alcohol by the oxymercuration-demercuration process.
4. write the mechanism for the reaction of an alkene with mercury(II) acetate in aqueous tetrahydrofuran (THF).
Key Terms
Make certain that you can define, and use in context, the key terms below.
• hydration
• oxymercuration
Study Notes
Oxymercuration is the reaction of an alkene with mercury(II) acetate in aqueous THF, followed by reduction with sodium borohydride. The final product is an alcohol.
It is important that you recognize the similarity between the mechanisms of bromination and oxymercuration. Recognizing these similarities helps you to reduce the amount of factual material that you need to remember.
Mercuric acetate, or mercury(II) acetate, to give it the preferred IUPAC name, is written as Hg(OAc)2; by comparing this formula with the formula Hg(O2CCH3)2, you can equate Ac with -COCH3. In fact, Ac is an abbreviation used for the acetyl group with the structure shown below as are other similar abbreviations that you will encounter.
Ac (acetyl)
Me (methyl)
Et (ethyl)
Prn (n-propyl)
Pri (isopropyl)
But (tert-butyl)
Ph (phenyl)
What Is Electrophilic Hydration?
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane. Electrophilic hydration is the reverse dehydration of alcohols and has practical application in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:
Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts (see SN2 for more details).
How Does Electrophilic Hydration Work?
Mechanism for 3º Alcohol (1º and 2º mechanisms are similar):
Hydration is the process where water is added to an alkene to yield an alcohol. Acid-catalyzed hydration is when a strong acid is used as a catalyst to begin the reaction, but let's look at the mechanism below and break down the steps.
Step 1: A hydrogen atom from the acid is attacked by the nucleophilic Pi-electrons in the double bond. This is similar to the other alkene reactions we have seen so far. In this process, a new C-H bond is formed to create the more stable carbocation.
Step 2: A nucleophilic water attacks or donates a lone pair to the positively charged carbon in the carbocation intermediate created in the first step. There is new C-O bond with the O having a formal charge of +1. The product is a protonated alcohol.
Step 3: To obtain the neutral alcohol product, the final step is to deprotonate the oxygen atom with the +1 formal charge using the acid. This final step regenerates the acid catalyst and yields the neutral alcohol product.
Temperatures for Types of Alcohol Synthesis
Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the product being formed.
What is Regiochemistry and How Does It Apply?
Regiochemistry deals with where the substituent bonds on the product. Zaitsev's and Markovnikov's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation
In the mechanism for a 3º alcohol shown above, the H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugationelectrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction.
If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products:
• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.
• Alkyl shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift.
The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. After one H atom is removed from the water molecule, the alcohol is attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.
What is Stereochemistry and How Does It Apply?
Stereochemistry deals with how the substituent bonds on the product directionally. Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that the product is a racemic mixture.
Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product of the following reaction could look like either of the following products:
There is no stereochemical control in acid-catalyzed hydration reactions. This is due to the trigonal planar, sp2 nature of the carbocation intermediate. Water can act as a nucleophile to form a bond to either face of the carbocation, resulting in a mixture of stereochemical outcomes.
Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity.
Is this a Reversible Synthesis?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example:
• Less sulfuric or phosphoric acid and an excess of water help synthesize more alcohol product.
• Lower temperatures help synthesize more alcohol product.
Is There a Better Way to Add Water to Synthesize an Alcohol From an Alkene?
A more efficient pathway does exist: Oxymercuration - Demercuration, a special type of electrophilic addition. Oxymercuration does not allow for rearrangements, but it does require the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include:
• Allowing for carbocation rearrangements
• Poor yields due to the reactants and products being in equilibrium
• Allowing for product mixtures (such as an (R)-enantiomer and an (S)-enantiomer)
• Using sulfuric or phosphoric acid
Oxymercuration is a special electrophilic addition. It is anti-stereospecific and regioselective. Regioselectivity is a process in which the substituents choses one direction it prefers to be attached to over all the other possible directions. The good thing about this reaction is that there are no carbocation rearrangement due to stabilization of the reactive intermediate. Similar stabilization is also seen in bromination addition to alkenes.
Introduction to Oxymercuration
One of the major advantages to oxymercuration is that carbocation rearrangements cannot occur under these conditions (Hg(OAc)2, H2O). Carbocation rearrangement is a process in which the carbocation intermediate can undergo a methyl or alkyl shift to form a more stable ion. Due to a possible carbocation rearrangement, the reaction below would not generate the product shown in high yields. In contrast, the oxymercuration reaction would proceed to form the desired product.
This reaction involves a mercury acting as a reagent attacking the alkene double bond to form a Mercurinium Ion Bridge. A water molecule will then attack the most substituted carbon to open the mercurium ion bridge, followed by proton transfer to solvent water molecule.
The organomercury intermediate is then reduced by sodium borohydride - the mechanism for this final step is beyond the scope of our discussion here. Notice that overall, the oxymercuration - demercuration mechanism follows Markovnikov's regioselectivity with the OH group is attached to the most substituted carbon and the H is attach to the least substituted carbon. The reaction is useful, however, because strong acids are not required, and carbocation rearrangements are avoided because no discreet carbocation intermediate forms.
It is important to note that for the mechanism shown above, the enantiomer of the product shown is also formed. This is the result of formation of the mercurium ion below the alkene in the first step.
Some Practice Problems
Practice problems
What are the end products of these reactants?
Answer
The end product to these practice problems are pretty much very similar. First, you locate where the double bond is on the reactant side. Then, you look at what substituents are attached to each side of the double bond and add the OH group to the more substituent side and the hydrogen on the less substituent side.
More Problems
Predict the product of each reaction.
1)
2) How does the cyclopropane group affect the reaction?
3) (Hint: What is different about this problem?)
4) (Hint: Consider stereochemistry.)
5) Indicate any shifts as well as the major product:
Answer
1) This is a simple electrophilic hydration.
2) The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).
3) A hydride shift actually occurs from the 3 carbon of the 3-methylcyclopentene to where the carbocation had formed.
4) This reaction will have poor yields due to a very unstable intermediate. For a brief moment, carbocations can form on the two center carbons, which are more stable than the outer two carbons. The carbocations have an sp2 hybridization, and when the water is added on, the carbons change their hybridization to sp3. This makes the methyl and alcohol groups equally likely to be found going into or out of the plane of the paper- the product is racemic.
5) In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.04%3A_Hydration_of_Alkenes_-_Addition_of_HO_by_Oxymercuration.txt |
Objectives
After completing this section, you should be able to
1. identify hydroboration (followed by oxidation) as a method for bringing about the (apparently) non-Markovnikov addition of water to an alkene.
2. write an equation for the formation of a trialkylborane from an alkene and borane.
3. write an equation for the oxidation of a trialkylborane to an alcohol.
4. draw the structure of the alcohol produced by the hydroboration, and subsequent oxidation, of a given alkene.
5. determine whether a given alcohol should be prepared by oxymercuration-demercuration or by hydroboration-oxidation, and identify the alkene and reagents required to carry out such a synthesis.
6. write the detailed mechanism for the addition of borane to an alkene, and explain the stereochemistry and regiochemistry of the reaction.
Key Terms
Make certain that you can define, and use in context, the key term below.
• hydroboration
Study Notes
The two most important factors influencing organic reactions are polar (or electronic) effects and steric effects.
Hydroboration-oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an anti-Markovnikov manner, where the hydrogen (from BH3 or BHR2) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene double bond. Furthermore, the borane acts as a Lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The hydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
Introduction
Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An additional feature of this reaction is that it occurs without rearrangement.
The Borane Complex
First off it is very important to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B2H6 (diborane). Additionally, the dimer B2H6 ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a Lewis acid-base complex, which allows boron to have an electron octet.BH3B2H6
The Mechanism
Step 1
• Part 1: Hydroboration of the alkene. In this first step the addition of the borane to the alkene is initiated and proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time. This part consists of the vacant 2p orbital of the boron electrophile pairing with the electron pair of the pi bond of the nucleophile.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
• Part 2: The Anti Markovnikov addition of Boron. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition).
Oxidation of the Trialkylborane by Hydrogen Peroxide
Step 2
• Part 1: The first part of this mechanism deals with the donation of a pair of electrons from the hydrogen peroxide ion. the hydrogen peroxide is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from hydroboration.
• Part 2: In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the removal of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
• Part 3: This is the final part of the oxidation process. In this part the trialkylborate reacts with aqueous NaOH to give the alcohol and sodium borate.
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well.
Stereochemistry of Hydroboration
The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. As noted above, this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that the geometry of this addition must be syn. Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene.
Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with retention of configuration so the overall addition of water is also syn.
The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored purple in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl.
References
1. Vollhardt, Peter, and Neil Shore. Organic Chemistry: Structure and Function. 5th. New York: W.H. Freeman and Company, 2007.
2. Foote, S. Christopher, and William H. Brown. Organic Chemistry. 5th. Belmont, CA: Brooks/Cole Cengage Learning, 2005.
3. Bruice, Paula Yurkanis. Oragnic Chemistry. 5th. CA. Prentice Hall, 2006.
4. Bergbreiter E. David , and David P. Rainville. Stereochemistry of hydroboration-oxidation of terminal alkenes. J. Org. Chem., 1976, 41 (18), pp 3031–3033
5. Ilich, Predrag-Peter; Rickertsen, Lucas S., and Becker Erienne. Polar Addition to C=C Group: Why Is Anti-Markovnikov Hydroboration-Oxidation of Alkenes Not "Anti-"? Journal of Chemical Education., 2006, v83, n11, pg 1681-1685
Problems
What are the products of these following reactions?
1.
2.
3.
Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown.
4.
5. (E)-3-methyl-2-pentene
If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the alkenes.
Answer
1.
2.
3.
4.
5.
Exercises
Questions
Q8.5.1
Write out the reagents or products (A–D) shown in the following reaction schemes.
S8.5.1 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.05%3A_Hydration_of_Alkenes_-_Addition_of_HO_by_Hydroboration.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the catalytic hydrogenation of an alkene.
2. identify the product obtained from the hydrogenation of a given alkene.
3. identify the alkene, the reagents, or both, required to prepare a given alkane by catalytic hydrogenation.
4. describe the mechanism of the catalytic hydrogenation of alkenes.
5. explain the difference between a heterogeneous reaction and a homogeneous reaction.
6. recognize that other types of compounds containing multiple bonds, such as ketones, esters, nitriles and aromatic compounds, do not react with hydrogen under the conditions used to hydrogenate alkenes.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Adams’ catalyst
• hydrogenation
Study Notes
Chemical reactions that are heterogeneous have reactants that are in at least two different phases (e.g. gas with a solid), whereas homogeneous reactions occur in a single phase (e.g. gas with another gas).
Some confusion may arise from the description of the catalyst used in the reaction between alkenes and hydrogen. Three metals—nickel, platinum and palladium—are commonly used, but a chemist cannot simply place a piece of one of these metals in a mixture of the alkene and hydrogen and get a reaction. Each metal catalyst must be prepared in a special way:
• nickel is usually used in a finely divided form called “Raney nickel.” It is prepared by reacting a Ni-Al alloy with NaOH.
• palladium is obtained commercially “supported” on an inert substance, such as charcoal, (Pd/C). The alkene is usually dissolved in ethanol when Pd/C is used as the catalyst.
• platinum is used as PtO2, Adams’ catalyst, although it is actually platinum metal that is the catalyst. The hydrogen used to add to the carbon-carbon double bond also reduces the platinum(IV) oxide to finely divided platinum metal. Ethanol or acetic acid is used as the solvent for the alkene.
Other types of compounds containing multiple bonds, such as ketones, esters, and nitriles, do not react with hydrogen under the conditions used to hydrogenate alkenes. The examples below show reduction of an alkene, but the ketone and nitrile groups present remain intact and are not reduced.
Aromatic rings are also not reduced under the conditions used to reduce alkenes, although these rings appear to contain three carbon-carbon double bonds. As you will see later, aromatic rings do not really contain any double bonds, and many chemists prefer to represent the benzene ring as a hexagon with a circle inside it
rather than as a hexagon with three alternating double bonds.
The representation of the benzene ring will be discussed further in Section 15.2.
The reaction between carbon-carbon double bonds and hydrogen provides a method of determining the number of double bonds present in a compound. For example, one mole of cyclohexene reacts with one mole of hydrogen to produce one mole of cyclohexane:
but one mole of 1,4-cyclohexadiene reacts with two moles of hydrogen to form one mole of cyclohexane:
A chemist would say that cyclohexene reacts with one equivalent of hydrogen, and 1,4-cyclohexadiene reacts with two equivalents of hydrogen. If you take a known amount of an unknown, unsaturated hydrocarbon and determine how much hydrogen it will absorb, you can readily determine the number of double bonds present in the hydrocarbon (see question 2, below).
Addition of hydrogen to a carbon-carbon double bond is called hydrogenation. The overall effect of such an addition is the reductive removal of the double bond functional group. Regioselectivity is not an issue, since the same group (a hydrogen atom) is bonded to each of the double bond carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H2), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst, as shown in the reaction coordinate diagram below.
An example of an alkene addition reaction is a process called hydrogenation. In a hydrogenation reaction, two hydrogen atoms are added across the double bond of an alkene, resulting in a saturated alkane. Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability.
Catalysts are substances that changes the rate (velocity) of a chemical reaction without being consumed or appearing as part of the product. Catalysts act by lowering the activation energy of reactions, but they do not change the relative potential energy of the reactants and products. Finely divided metals, such as platinum, palladium and nickel, are among the most widely used hydrogenation catalysts. Catalytic hydrogenation takes place in at least two stages, as depicted in the diagram. First, the alkene must be adsorbed on the surface of the catalyst along with some of the hydrogen. Next, two hydrogens shift from the metal surface to the carbons of the double bond, and the resulting saturated hydrocarbon, which is more weakly adsorbed, leaves the catalyst surface. The exact nature and timing of the last events is not well understood.
As shown in the energy diagram, the hydrogenation of alkenes is exothermic, and heat is released corresponding to the ΔH in the diagram. This heat of reaction can be used to evaluate the thermodynamic stability of alkenes having different numbers of alkyl substituents on the double bond. For example, the following table lists the heats of hydrogenation for three C5H10 alkenes which give the same alkane product (2-methylbutane). Since a larger heat of reaction indicates a higher energy reactant, these heats are inversely proportional to the stabilities of the alkene isomers. To a rough approximation, we see that each alkyl substituent on a double bond stabilizes this functional group by a bit more than 1 kcal/mole.
Alkene Isomer (CH3)2CHCH=CH2
3-methyl-1-butene
CH2=C(CH3)CH2CH3
2-methyl-1-butene
(CH3)2C=CHCH3
2-methyl-2-butene
Heat of Reaction
( ΔHº )
–30.3 kcal/mole –28.5 kcal/mole –26.9 kcal/mole
Catalytic Hydrogenation Mechanism
From the mechanism shown here we would expect the addition of hydrogen to occur with syn-stereoselectivity. This is often true, but the hydrogenation catalysts may also cause isomerization of the double bond prior to hydrogen addition, in which case stereoselectivity may be uncertain.
Exercise 1
1. In the reaction
1. 0.500 mol of ethene reacts with _______ mol of hydrogen. Thus a chemist might say that ethene reacts with one _______ of hydrogen.
2. ethene is being _______; while _______ is being oxidized.
3. the oxidation number of carbon in ethene is _______; in ethane it is _______.
Answer
1. 0.500 mol of ethene reacts with 0.500 mol of hydrogen. Thus a chemist might say that ethene reacts with one equivalent of hydrogen.
2. ethene is being reduced; while hydrogen is being oxidized.
3. the oxidation number of carbon in ethene is −2; in ethane it is −3.
Exercise 2
When 1.000 g of a certain triglyceride (fat) is treated with hydrogen gas in the presence of Adams’ catalyst, it is found that the volume of hydrogen gas consumed at 99.8 kPa and 25.0°C is 162 mL. A separate experiment indicates that the molar mass of the fat is 914 g mol−1. How many carbon-carbon double bonds does the compound contain?
Answer
Amount of hydrogen consumed
$\begin{array}{l}=n\text{\hspace{0.17em}}\text{mol}\ \text{=}\frac{PV}{RT}\ =\frac{99.8\text{\hspace{0.17em}}\text{kPa}×0.162\text{\hspace{0.17em}}\text{L}}{8.31\text{\hspace{0.17em}}\text{kPa}\cdot {\text{mol}}^{}}\end{array}$1 K1 ×298K =6.53× 103 mol H2
Amount of fat used
$\begin{array}{l}=\frac{\left(1.000\text{\hspace{0.17em}}\text{g}\right)×\left(1\text{\hspace{0.17em}}\text{mol}\right)}{\left(914\text{\hspace{0.17em}}\text{g}\right)}\ =1.09×{10}^{-3}\text{\hspace{0.17em}}\text{mol}\text{\hspace{0.17em}}\text{fat}\end{array}$
Ratio of moles of hydrogen consumed to moles of fat
$\begin{array}{l}=6.53×{10}^{-3}:1.09×{10}^{-3}\ =6:1\end{array}$
Thus, the fat contains six carbon-carbon double bonds per molecule. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.06%3A_Reduction_of_Alkenes_-_Hydrogenation.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the epoxidation of an alkene using meta-chloroperoxybenzoic acid.
2. identify the alkene, reagents, or both, that must be used to prepare a given epoxide.
3. write the equation for the hydroxylation of an alkene using osmium tetroxide, and draw the structure of the cyclic intermediate.
4. draw the structure of the diol formed from the reaction of a given alkene with osmium tetroxide.
5. identify the alkene, the reagents, or both, that must be used to prepare a given 1,2-diol.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• diol
• glycol
• hydroxylation
The previous section discussed the reduction of a double bond, so adding hydrogen to the the double bond. This section will discuss oxidation. In organic chemistry, this is a reaction that where the carbon atom loses electron density, which happens when new bond to a more electronegative atom occurs or when a double bond is broken between a carbon and a less electronegative atom. A simplified to say this is in organic chemistry a reduction is more bonds to hydrogen and oxidation is more bonds to oxygen often.
Oxacyclopropane Synthesis by Peroxycarboxylic Acid
One way to oxidized a double bond is to produce an oxacyclopropane ring. Oxacyclopropane rings, also called epoxide rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with peroxycarboxylic acid. Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH.
Mechanism
Peroxycarboxylic acids are generally unstable. An exception is meta-chloroperoxybenzoic acid, shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions.
Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as chloroform, ether, acetone, or dioxane is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal diol, a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions.
Example 8.7.1
Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the epoxide ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal diol, a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is anti since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one (S) and one (R) stereocenter. But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of (S,S) and (R,R) enantiomers.
Epoxidation exercises
1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid)
a) in acetone solvent.
b) in an aqueous medium with acid or base catalyst present.
2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent.
3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent.
4. Predict the reaction of propene with MCPBA.
a) in acetone solvent
b) after aqueous work-up.
5. Predict the reaction of cis-2-butene in chloroform solvent.
Answers
1. a) Cis-2-methyl-3-propyloxacyclopropane
b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
2. Trans-3-ethyl-2-methyloxacyclopropane.
3. Trans-3,4-diethyloxacyclopropane.
4. a) 2-methyl-oxacyclopropane
b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol
5. Cis-2,3-dimethyloxacyclopropane
Anti Dihydroxylation
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups.
Syn Dihydroxylation
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
Introduction
The reaction with $OsO_4$ is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an anti dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures.
$OsO_4$ is formed slowly when osmium powder reacts with gasoues $O_2$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
$Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4 \nonumber$
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO4 was potassium permanganate, $KMnO_4$. Although syn diols will result from the reaction of KMnO4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation.
Mechanism
• Electrophilic attack on the alkene
• Pi bond of the alkene acts as the nucleophile and reacts with osmium (VIII) tetroxide (OsO4)
• 2 electrons from the double bond flows toward the osmium metal
• In the process, 3 electron pairs move simultaneously
• Cyclic ester with Os (VI) is produced
• Reduction
• H2S reduces the cyclic ester
• NaHSO4 with H2O may be used
• Forms the syn-1,2-diol (glycol)
Example: Dihydroxylation of 1-ethyl-1-cycloheptene
Hydroxylation of alkenes
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown above.
Chemical Highlight
Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate.
References
1. Dehestani, Ahmad et al. (2005). Ligand-assisted reduction of osmium tetroxide with molecular hydrogen via a [3+2] mechanism. Journal of the American Chemical Society, 2005, 127 (10), 3423-3432.
2. Sorrell, Thomas, N. Organic Chemistry. New York: University Science Books, 2006.
3. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
Dihydroxylation Exercises
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
Answers
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.07%3A_Oxidation_of_Alkenes_-_Epoxidation_and_Hydroxylation.txt |
Objective
After completing this section, you should be able to
1. write an equation to describe the cleavage of an alkene by ozone, followed by reduction of the ozonide so formed with either sodium borohydride or zinc and acetic acid.
2. predict the products formed from the ozonolysis-reduction of a given alkene.
3. write an equation to describe the cleavage of an alkene by potassium permanganate.
4. predict the products from the oxidative cleavage of a given alkene by potassium permanganate.
5. use the results of ozonolysis-reduction, or cleavage with permanganate, to deduce the structure of an unknown alkene.
6. identify the reagents that should be used in the oxidative cleavage of an alkene to obtain a given product or products.
7. write the equation for the cleavage of a 1,2-diol by periodic acid, and draw the structure of the probable intermediate.
8. predict the product or products that will be formed from the treatment of a given 1,2-diol with periodic acid.
9. use the results of hydroxylation/1,2-diol cleavage to deduce the structure of an unknown alkene.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• molozonide
• ozonide
• ozonolysis
Study Notes
Ozonolysis, or ozonolysis-reduction, refers to the treatment of an alkene with ozone followed by a suitable reducing agent to break down complex double-bond-containing compounds into smaller, more easily identified products. From the identity of the products formed, it may be possible to deduce the structure of the original double-bond-containing substance. Ozonolysis will feature prominently in many of the road-map problems that you will encounter in this course.
A molozonide is an unstable, cyclic intermediate that is initially formed when an alkene reacts with ozone.
Alkenes can also be cleaved by other oxidizing agents such as potassium permanganate. However, KMnO4 will carry the oxidation further than ozonolysis, so products can be slightly different. Note within the summary of the following reactions that ozonolysis produces aldehydes and ketones, while KMnO4 can oxidize all the way to to carbon dioxide and carboxylic acid.
Diol cleavage is another example of a redox reaction; periodic acid, HIO4, is reduced to iodic acid, HIO3.
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes. by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis.
Introduction
The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the Carbon-Carbon double bond and is replaced by a Carbon-Oxygen double bond instead.
Reaction Mechanism
Step 1:
The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the Carbon-Carbon double bond, which then form the molozonide intermediate. Due to the unstable molozonide molecule, it continues further with the reaction and breaks apart to form a carbonyl and a carbonyl oxide molecule.
Step 2:
The carbonyl and the carbonyl oxide rearranges itself and reforms to create the stable ozonide intermediate. Since some ozonides are explosive, it is immediately reacted with a reductive workup to then convert the ozonide molecule into the desired carbonyl products. A typical reductive workup is zinc metal in acetic acid. A variety of carbonyl products could result depending on the starting alkene. For example, a tetrasubstituted alkene would yield two ketone products, while a trisubstituted alkene would yield one ketone product and one aldehyde product.
While there are other options for oxidative cleavage of the double bond, this is the most commonly used reaction.
References
1. Vollhardt, K., Schore, N. Organic Chemistry: Structure and Function. 5th ed. New York, NY: W. H. Freeman and Company, 2007.
2. Shore, N. Study Guide and Solutions Manual for Organic Chemistry. 5th ed. New York, NY: W.H. Freeman and Company, 2007.
Answer
Exercises
1. Draw the structure of the product or products obtained in each of the following reactions:
2. What important point did you learn from questions 1(a) and 1(b), above?
Answers:
1. Exercises 1(a) and 1(b), above, indicate that it is not possible to distinguish between cis and trans isomers of alkenes using oxidative cleavage. Both isomers give the same product or products.
Contributors and Attributions
• Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
• Lauren Reutenauer (Amherst College)
• Dr. Krista Cunningham | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.08%3A_Oxidation_of_Alkenes_-_Cleavage_to_Carbonyl_Compounds.txt |
Objectives
After completing this section, you should be able to
1. describe, and write the detailed mechanism for, the formation of a carbene, such as dichlorocarbene.
2. describe the structure of a carbene in terms of the hybridization of the central carbon atom.
3. write an equation for the formation of a substituted cyclopropane from an alkene and a carbene.
4. identify the reagents, the alkene, or both, needed to prepare a given substituted cyclopropane by addition of a carbene to a double bond.
5. identify the substituted cyclopropane formed from the reaction of a given alkene with the reagents necessary to form a carbene.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• carbene (R2C:)
• carbenoid
• Simmons-Smith reaction
• stereospecific
Study Notes
A carbenoid is best considered to be a reagent which, while not actually a carbene, behaves as if it were an intermediate of this type.
Dichlorocarbenes can also form cyclopropane structures and are created in situ from reagents such as chloroform and KOH.
The detailed mechanism of the formation of dichlorocarbene is given below. Note that the deprotonation of chloroform generates the trichloromethanide anion, which spontaneously expels the chloride anion.
The highly strained nature of cyclopropane compounds makes them very reactive and interesting synthetic targets. Additionally cyclopropanes are present in numerous biological compounds. One common method of cyclopropane synthesis is the reaction of carbenes with the double bond in alkenes or cycloalkenes. Methylene, H2C, is simplest carbene, and in general carbenes have the formula R2C. Other species that will also react with alkenes to form cyclopropanes but do not follow the formula of carbenes are referred to as carbenoids.
Introduction
Carbenes were once only thought of as short lived intermediates. The reactions of this section only deal with these short lived carbenes which are mostly prepared in situ, in conjunction with the main reaction. However, there do exist so called persistent carbenes. These persistent carbenes are stabilized by a variety of methods often including aromatic rings or transition metals. In general a carbene is neutral and has 6 valence electrons, 2 of which are non bonding. These electrons can either occupy the same sp2 hybridized orbital to form a singlet carbene (with paired electrons), or two different sp2 orbitals to from a triplet carbene (with unpaired electrons). The chemistry of triplet and singlet carbenes is quite different but can be oversimplified to the statement: singlet carbenes usually retain stereochemistry while triplet carbenes do not. The carbenes discussed in this section are singlet and thus retain stereochemistry.
The reactivity of a singlet carbene is concerted and similar to that of electrophilic or nucleophilic addition wheras, triplet carbenes react like biradicals, explaining why sterochemistry is not retained. The highly reactive nature of carbenes leads to very fast reactions in which the rate determining step is generally carbene formation.
Preparation of methylene
The preparation of methylene starts with the yellow gas diazomethane, CH2N2. Diazomethane can be exposed to light, heat or copper to facilitate the loss of nitrogen gas and the formation of the simplest carbene methylene. The process is driven by the formation of the nitrogen gas which is a very stable molecule.
Carbene reaction with alkenes
A carbene such as methlyene will react with an alkene which will break the double bond and result with a cyclopropane. The reaction will usually leave stereochemistry of the double bond unchanged. As stated before, carbenes are generally formed along with the main reaction; hence the starting material is diazomethane not methylene.
In the above case cis-2-butene is converted to cis-1,2-dimethylcyclopropane. Likewise, below the trans configuration is maintained. This shows that the reactions are stereospecific, only a single stereoisomer is obtained as the product.
Additional Types of Carbenes and Carbenoids
In addition to the general carbene with formula R2C there exist a number of other compounds that behave in much the same way as carbenes in the synthesis of cyclopropane. Halogenated carbenes are formed from halomethanes. An example is dicholorcarbene, Cl2C. The mechanism for the formation of dichlorocarbene is above in the study notes. These halogenated carbenes will form cyclopropanes in the same manner as methylene but with the interesting presence of two halogen atoms in place of the hydrogen atoms.
Carbenoids are substances that form cyclopropanes like carbenes but are not technically carbenes. One common example is the stereospecific Simmon-Smith reaction which utilizes the carbenoid - ICH2ZnI. The (iodomethyl) zinc iodide is formed in situ via the mixing of Zn-Cu with CH2I2. If this ICH2ZnI is in the presence of an alkene, a CH2 group is transferred to the double bond to create cyclopropane. Since this reacts as a carbene, the same methods can be applied to determine the product.
Outside links
• en.Wikipedia.org/wiki/Simmons-Smith_reaction
• en.Wikipedia.org/wiki/Carbene
Problems
1. Knowing that cycloalkenes react much the same as regular alkenes what would be the expected structure of the product of cyclohexene and diazomethane facilitated by copper metal?
2. What would be the result of a Simmons-Smith reaction that used trans-2-pentene as a reagent?
3. What starting material could be used to form cis-1,2-diethylcyclopropane?
4. What would the following reaction yield?
5. Draw the product of this reaction. What type of reaction is this?
Answers
1. The product will be a bicyclic ring, Bicyclo[4.1.0]heptane.
2. The stereochemistry will be retained making a cyclopropane with trans methyl and ethyl groups. Trans-1-ethyl-2-methylcyclopropane
3. The cis configuration will be maintained from reagent to product so we would want to start with cis-3-hexene. A Simmons Smith reagent, or methylene could be used as the carbene or carbenoid.
4. The halogenated carbene will react the same as methylene yielding, cis-1,1-dichloro-2,3-dimethylcyclopropane.
5. This is a Simmons-Smith reaction which uses the carbenoid formed by the CH2I2 and Zu-Cu. The reaction results in the same product as if methylene was used and retains stereospecificity. Iodine metal and the Zn-Cu are not part of the product. The product is trans-1,2-ethyl-methylcyclopropane.
Contributors and Attributions
• Paul Tisher
• Lauren Reutenauer (Amherst College)
• Dr. Krista Cunningham | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.09%3A_Addition_of_Carbenes_to_Alkenes_-_Cyclopropane_Synthesis.txt |
Objectives
After completing this section, you should be able to
1. write the detailed mechanism for the radical polymerization of an alkene.
2. give examples of some common alkene monomers used in the manufacture of chain-growth polymers.
3. identify the alkene monomer used to prepare a specific chain-growth polymer, given the structure of the polymer.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• monomer
• polymer
• vinyl monomer
Study Notes
Vinyl monomers are monomers of the type \$\ce{\sf{CH2=CHX}}\$. Recall that the vinyl group is \$\ce{\sf{CH2=CH-}}\$.
Although benzoyl peroxide is commonly used as an initiator in free-radical polymerization reactions, an alternative reagent is azobisisobutyronitrile, shown below.
A polymer is a large molecule built of repeat units of many monomers (smaller molecules). An example of this in the biological world is cellulose, which is a polymer composed of repeating glucose monomer units. A synthetic example would be of polyethylene, which is a polymer with ethylene as the repeating unit for the monomer. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions. Many of these addition reactions are known to proceed in a stepwise fashion by way of reactive intermediates, and this is the mechanism followed by most polymerizations. A general diagram illustrating this assembly of linear macromolecules, which supports the name chain growth polymers, is presented here. Since a pi-bond in the monomer is converted to a sigma-bond in the polymer, the polymerization reaction is usually exothermic by 8 to 20 kcal/mol. Indeed, cases of explosively uncontrolled polymerizations have been reported.
It is useful to distinguish four polymerization procedures fitting this general description.
• Radical Polymerization The initiator is a radical, and the propagating site of reactivity (*) is a carbon radical.
• Cationic Polymerization The initiator is an acid, and the propagating site of reactivity (*) is a carbocation.
• Anionic Polymerization The initiator is a nucleophile, and the propagating site of reactivity (*) is a carbanion.
• Coordination Catalytic Polymerization The initiator is a transition metal complex, and the propagating site of reactivity (*) is a terminal catalytic complex.
Radical Chain-Growth Polymerization
Chain-growth polymers are formed in a chain-reaction process. The first step involves an initiator to add to a carbon-carbon double bond in the first monomer, which results in a reactive intermediate. This intermediate reactive intermediate then goes and reacts with a second alkene monomer to yield another reactive intermediate. This process continues to grow the polymer from one monomer to many monomers until the termination step when the radical is consumed.
Note: In radical mechanisms, the arrow showing the movement of the electron looks like a "fish hook" or half-arrow as opposed to a full arrow, which indicates the movement of an electron pair.
Initiation: Virtually all of the monomers described above are subject to radical polymerization. Since radical polymerization can be initiated by traces of oxygen or other minor impurities, pure samples of these compounds are often "stabilized" by small amounts of radical inhibitors to avoid unwanted reaction. When radical polymerization is desired, it must be started by using a radical initiator, such as a peroxide or certain azo compounds. The formulas of some common initiators, and equations showing the formation of radical species from these initiators are presented below.
Propagation: Once the initiator has reacted with the alkene to create the carbon radical intermediate, it adds to another alkene molecule to yield another radical. This process repeats building the polymer chain.
Termination: The chain growing process ends when a reaction that consumes the radical happens. The most common processes are radical combination (two growing chains combine together) and disproportionation.
One example of this radical chain-growth polymerization is the conversion of styrene to polystyrene, shown in the following diagram. The first two equations illustrate the initiation process, and the last two equations are examples of chain propagation. Each monomer unit adds to the growing chain in a manner that generates the most stable radical. Since carbon radicals are stabilized by substituents of many kinds, the preference for head-to-tail regioselectivity in most addition polymerizations is understandable. Because radicals are tolerant of many functional groups and solvents (including water), radical polymerizations are widely used in the chemical industry.
In principle, once started a radical polymerization might be expected to continue unchecked, producing a few extremely long chain polymers. In practice, larger numbers of moderately sized chains are formed, indicating that chain-terminating reactions must be taking place. The most common termination processes are Radical Combination and Disproportionation. These reactions are illustrated by the following equations. The growing polymer chains are colored blue and red, and the hydrogen atom transferred in disproportionation is colored green. Note that in both types of termination two reactive radical sites are removed by simultaneous conversion to stable product(s). Since the concentration of radical species in a polymerization reaction is small relative to other reactants (e.g. monomers, solvents and terminated chains), the rate at which these radical-radical termination reactions occurs is very small, and most growing chains achieve moderate length before termination.
Chain Termination Reactions
The relative importance of these terminations varies with the nature of the monomer undergoing polymerization. For acrylonitrile and styrene combination is the major process. However, methyl methacrylate and vinyl acetate are terminated chiefly by disproportionation.
Another reaction that diverts radical chain-growth polymerizations from producing linear macromolecules is called chain transfer. As the name implies, this reaction moves a carbon radical from one location to another by an intermolecular or intramolecular hydrogen atom transfer (colored green). These possibilities are demonstrated by the following equations
Chain Transfer Reactions
Chain transfer reactions are especially prevalent in the high pressure radical polymerization of ethylene, which is the method used to make LDPE (low density polyethylene). The 1º-radical at the end of a growing chain is converted to a more stable 2º-radical by hydrogen atom transfer. Further polymerization at the new radical site generates a side chain radical, and this may in turn lead to creation of other side chains by chain transfer reactions. As a result, the morphology of LDPE is an amorphous network of highly branched macromolecules. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.10%3A_Radical_Additions_to_Alkenes_-__Chain-Growth_Polymers.txt |
Objective
After completing this section, you should be able to discuss, briefly, some of the addition reactions that take place in nature, and the role of enzymes in such processes.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• enzyme
• coenzyme
Study Notes
You need not memorize the reaction described in this section of the textbook. However, you should note how the names of enzymes are derived from the reactions they catalyze; for example, ascorbic acid oxidase is the enzyme that catalyzes the oxidation of ascorbic acid (vitamin C).
In the previous section (8-10: Radical Additions to Alkenes: Chain-Growth Polymers), very reactive carbon radical intermediates were formed for the alkene polymerizations. The high reactivity of radicals makes it hard to control, though many chemists are currently working on this. Therefore, there is a limit to the usefulness of radical polymerization. Biological reactions do not face this same issue since the enzyme controls the reaction. In an enzyme active site, one monomer at a time enters to react in the proper orientation with all the nearby groups necessary for the reaction to proceed nearby. An example of a radical chain-growth polymerization is discussed next.
Radical mechanisms for flavin-dependent reactions
Flavin coenzymes, like their nicotinamide adenine dinucleotide (NAD) counterparts, can act as hydride acceptors and donors. In these redox reactions, two electrons are transferred together in the form of a hydride ion. Flavin, however, is also capable of mediating chemical steps in which a single unpaired electron is transferred - in other words, radical chemistry. This is due to the ability of the flavin system to form a stabilized radical intermediate called a semiquinone, formed when FADH2 (or FMNH2) donates a single electron, or when FAD (or FMN) accepts a single electron.
This single-electron transfer capability of flavins is critical to their metabolic role as the entry point of electrons into the electron transport phase of respiration. Electrons 'harvested' from the oxidation of fuel molecules are channeled, one by one, by FMNH2 into the electron transport chain, where they eventually reduce molecular oxygen. NADH is incapable of single electron transfer - all it can do is transfer two electrons, in the form of a hydride, to FMN; the regenerated FMNH2 is then able to continue sending single electrons into the transport chain.
You will learn more details about this process in a biochemistry class.
Because flavins are capable of single-electron as well as two-electron chemistry, the relevant mechanisms of flavoenzyme-catalyzed reactions are often more difficult to determine. Recall the dehydrogenation reaction catalyzed by acyl-CoA dehydrogenase (section 16.5C) - it involves the transfer of two electrons and two protons (ie. a hydrogen molecule) to FAD. Both electrons could be transferred together, with the FAD coenzyme simply acting as a hydride acceptor (this is the mechanism we considered previously). However, because the oxidizing coenzyme being used is FAD rather than NAD+, it is also possible that the reaction could proceed by a single-electron, radical intermediate process. In the alternate radical mechanism proposed below, for example, the enolate intermediate first donates a single electron to FAD, forming a radical semiquinone intermediate (step 2). The second electron is transferred when the semiquinone intermediate abstracts a hydrogen from Cb in a homolytic fashion (step 3).
Scientists are still not sure which mechanism - the hydride transfer mechanism that we saw in section 16.5B or the single electron transfer detailed above - more accurately depicts what is going on in this reaction.
The conjugated elimination catalyzed by chorismate synthase (section 14.3B) is another example of a reaction where the participation of flavin throws doubt on the question of what is the relevant mechanism. This could simply be a conjugated E1' reaction, with formation of an allylic carbocation intermediate. The question plaguing researchers studying this enzyme, however, is why FADH2 is required. This is not a redox reaction, and correspondingly, FADH2 is not used up in the course of the transformation - it just needs to be bound in the active site in order for the reaction to proceed. Given that flavins generally participate in single-electron chemistry, this is an indication that radical intermediates may be involved. Recently an alternative mechanism, involving a flavin semiquinone intermediate, has been proposed (J. Biol. Chem 2004, 279, 9451). Notice that a single electron is transferred from substrate to coenzyme in step 2, then transferred back in step 4. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.11%3A_Biological_Additions_of_Radicals_to__Alkenes.txt |
Objective
After completing this section, you should be able to account for the stereochemistry of the product of the addition of water to an alkene in terms of the formation of a planar carbocation.
Study Notes
Organic reactions in the laboratory or in living systems can produce chiral centres. Consider reaction of 1-butene with water (acid catalyzed). Markovnikov regiochemistry occurs and the OH adds to the second carbon. However, both (R) and (S) products occur giving a racemic (50/50) mixture of 2‑butanol. How does this occur? The proton addition to 1‑butene results in a planar carbocation intermediate. A molecule of water is then equally likely to attack from the top (path a) or the bottom (path b) of this cation to produce either (S)‑2‑butanol or (R)‑2‑butanol, respectively.
As a reminder, a chiral center is a carbon that is bonded to four different groups. Organic reactions whether taking place in the body or in the laboratory can result in the product having a chiral center. The example in the study notes uses 1-butene to yield a product that contains a carbon with four different groups attached. Does this mean we get just one stereoisomer? Do we get a mixture of enantiomers? What is the stereochemistry of the reaction?
The product formed from 1-butene acid-catalyzed hydration reaction is a racemic mixture of 2-butanol. So, both R and S enantiomers are present. For more clarity, we can look into the mechanism for this reaction. The first step protonates 1-butene to yield a carbocation. The carbocation has an sp2-hybridized carbon in a trigonal planar geometry. The planarity of the carbocation allows the nucleophlic water to attack from either side of the plane equally. Many refer to this as top and bottom attack.
One thing to consider is you cannot create chirality from something that is achiral. 1-butene is achiral as is the carbocation intermediate. Therefore, our product must also be achiral and to do this with a molecule that does contain a chiral center means that both enantiomers must be present. In other words, the product is formed as a racemic mixture. While this may be true in the laboratory, biological reactions can give a single enantiomeric product. This is because the enzyme catalyzing the reaction itself is chiral and can therefore yield a chiral product. An example of this is cis-aconitate, which is achiral, to (2R,3S)-isocitrate, which is chiral. In this case, aconitase is the enzyme that holds the cis-actonitate in a chiral environment, which does create a chemically distinct way for the addition to occur yielding the chiral product, (2R,3S)-isocitrate.
Contributors and Attributions
• Lauren Reutenauer (Amherst College)
• Dr. Krista Cunningham
8.13: Stereochemistry of Reactions - Addition of HO to a Chiral Alkene
Objective
After completing this section, you should be able to explain why the addition of H2O to a chiral alkene leads to unequal amounts of diastereomeric products.
Study Notes
In the previous section, the addition of water to the achiral alkene produced a racemic mixture of two enantiomeric alcohols. They are produced in equal amounts so the mixture is optically inactive. What would occur if we carried out a similar reaction on a chiral alkene? Consider (S)-3-methyl-1-pentene reacting with water (acid catalyzed). Proton addition produces a carbocation intermediate that is chiral (* denotes stereogenic centre). That intermediate does not have a plane of symmetry and therefore attack by water is not equal from the top and bottom. This ultimately produces R and S products in a non 50:50 ratio.
In the previous section (8.12), an achiral alkene yielded a racemic mixture product. In this section, the starting alkene is chiral. If we consider (S)-3-methyl-1-pentene, it contains a chiral center, thus is optically active. When (S)-3-methyl-1-pentene undergoes acid-catalyzed hydration, it creates a second chiral center. There is the possibility of four products. Do they all form? The do not. Let's see why.
In the reaction, only one site is being reacted at. The S-configuration at C3 just goes along for the ride never participating in the actual mechanism. If you do not react at that site, then the stereochemistry will remain the unchanged, so it will stay with the S-configuration. However, at C2 we are reacting and water can still approach the planar carbocation from either side. Therefore at the C2 site, there will be some R-configuration and some S-configuration. The final product with be a mixture of enantiomers of 2-pentanol. Since the carbocation does not have a plane of symmetry (as was the case in the previous section), there will not be equal attack on either face. One of the faces may be slightly hindered due to sterics, which would result in a little less nucleophilic attack. Instead of obtaining a 50:50 diasteriomeric mixture of products, it would have a slightly different ratio of R to S due to this unequal attack.
Contributors and Attributions
• Lauren Reutenauer (Amherst College)
• Dr. Krista Cunningham | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.12%3A_Stereochemistry_of_Reactions_-_Addition_of_HO_to_an_Achiral_Alkene.txt |
Concepts & Vocabulary
8.1 Preparing Alkenes: A Preview of Elimination Reactions
• Alkenes can be prepared by either E1 or E2 elimination reactions of alkyl halides.
8.2 Halogenation of Alkenes: Addition of X2
• Halogen molecules can react as electrophiles due to polarization of the halogen-halogen bond.
• During electrophilic addition of halogens to pi bonds, an intermediate halonium ion is formed.
• During electrophilic halogenation, ring opening of the halonium intermediate causes anti stereochemistry of the halogen atoms in the dihalide product.
8.3 Halohydrins from Alkenes: Addition of HOX
• Halohydrins have a halogen and a hydroxide on adjacent carbon atoms. Bromohydrin and chlorohydrin are the specific types of halohydrins where the halogen is bromine or chlorine respectively.
• In halohydrin formation a carbocation intermediate is formed on the more substituted carbon (when available). This causes the hydroxide to be added to the more substituted carbon of the original alkene and the halogen to add to the less substituted carbon.
8.4 Hydration of Alkenes: Addition of H2O by Oxymercuration
• Electrophilic hydration is the addition of water to an alkene with one carbon adding a hydrogen and the other carbon a hydroxide.
• The mechanism begins with addition of a proton, yielding the more substituted carbocation.
• Carbocations can undergo hydride shifts and alkyl shifts to form a more stable carbocation when possible.
• Markovnikov addition through acid and water or oxymercuration-demercuration yields the more substituted alcohol product (when the two sides of the alkene are not equally substituted).
• Oxymercuration-demercuration avoids carbocation rearrangements through mercurinium ion bridge.
8.5 Hydration of Alkenes: Addition of H2O by Hydroboration
• Hydroboration-oxidation proceeds through anti-Markovnikov addition of water to an alkene, yielding the less substituted alcohol.
8.6 Reduction of Alkenes: Hydrogenation
• Hydrogenation reactions increase the number of carbon-hydrogen bonds, therefore are reduction reactions.
• Addition of hydrogen to carbon-carbon pi bonds is called hydrogenation.
• Hydrogenation requires a catalyst to lower the activation energy allowing the reaction to proceed (commonly nickel, palladium or platinum).
• Hydrogenation reactions occur primarily with syn addition of the two hydrogen atoms, though potential for isomerization makes this uncertain.
8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation
• Epoxidation can be carried out by reacting an alkene with a peroxy acid such as MCPBA.
• Anti dihydroxylation is achieved by ring opening an epoxide with water under acidic or basic conditions.
• Vicinal diols have hydroxy groups on adjacent carbon atoms.
• Syn dihydroxylation occurs through reaction with osmium tetraoxide, followed by reduction of the intermediate with sulfur compounds.
8.8 Oxidation of Alkenes: Cleavage to Carbonyl Compounds
• Ozonolysis is the cleavage of an alkene resulting in carbonyls at each carbon of the alkene.
• Alkenes can be cleaved by potassium permanganate, which also results in carbonyls at each alkene carbon, though potassium permanganate will oxidize every carbon-hydrogen bonds on the alkene to a carbon-oxygen bond.
8.9 Addition of Carbenes to Alkenes: Cyclopropane Synthesis
• Organic molecules that have a carbon with only two bonds and a lone pair of electrons are called carbenes.
• Most carbenes are highly reactive and short-lived and are often created in situ.
• Carbenes can be formed from diazo compounds by reacting wtih a copper catalyst.
• Carbenes will react with alkenes to form cyclopropane rings.
8.10 Radical Addition to Alkenes: Chain-Growth Polymers
• Monomers are units that repeat to form a polymer.
• In radical polymerization, the polymer chain reaction is initiated by a radical.
• Polymer chain reactions occur through a series of steps beginning with initiation, continuing through propagation, and ending in termination.
8.11 Biological Additions of Radicals to Alkenes
8.12 Reaction Stereochemistry: Addition of H2O to an Achiral Alkene
• Since addition of water to an alkene proceeds through a planar carbocation intermediate, achiral alkenes lead to a racemic mixture of alcohol products.
8.13 Reaction Stereochemistry: Additon of H2O to a Chiral Alkene
• Addition of water to alkenes which also contain a stereocenter does not lead to a 50:50 mixture of R and S products as the chiral center can reduce reactivity from one side of the carbocation. The products of this type of reaction will be diastereomers, since the original stereocenter will not change and the product will have an additional stereocenter.
Skills to Master
• Skill 8.1 Draw accurate Electrophilic Addition Mechanisms incorporating halonium intermediates.
• Skill 8.2 Draw accurate Electrophilic Addition Mechanisms incorporating carbocation intermediates.
• Skill 8.3 Draw Markovnikov products of alkene additions based on the most substituted carbocation intermediate.
• Skill 8.4 Draw hydrogenation products of alkenes.
• Skill 8.5 Draw appropriate epoxidation products including sterechemistry.
• Skill 8.6 Describe how to prepare syn and anti diols from alkenes.
• Skill 8.7 Draw products of oxidative cleavage reactions.
• Skill 8.8 Describe radical chain reactions to form polymers.
Memorization Tasks
MT 8.1 Memorize reagents for alkene reactions.
MT 8.2 Memorize stability order of carbocations.
Summary of Reactions
Preparation of Alkenes
Addition Reactions
Anti-hydroxylation
Oxidative Cleavage
Diol Cleavage
Contributors
• Layne Morsch (University of Illinois Springfield)
• Dr. Krista Cunningham | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/08%3A_Alkenes-_Reactions_and_Synthesis/8.S%3A_Alkenes_-_Reactions_and_Synthesis_%28Summary%29.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: 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.
• 9.2: 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.
• 9.3: Reactions of Alkynes - Addition of HX and X₂
• 9.4: 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.
• 9.5: 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
• 9.6: 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.
• 9.7: 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.
• 9.8: 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.
• 9.9: An Introduction to Organic Synthesis
• 9.S: Alkynes - An Introduction to Organic Synthesis (Summary)
09: Alkynes - An Introduction to Organic Synthesis
Objectives
After completing this section, you should be able to
1. provide the correct IUPAC name of an alkyne, given its Kekulé, condensed or shorthand structure.
2. provide the correct IUPAC name of a compound containing both double and triple bonds, given its Kekulé, condensed or shorthand structure.
3. draw the structure of a compound containing one or more triple bonds, and possibly one or more double bonds, given its IUPAC name.
4. name and draw the structure of simple alkynyl groups, and where appropriate, use these names as part of the IUPAC system of nomenclature.
Study Notes
Simple alkynes are named by the same rules that are used for alkenes (see Section 7.3), except that the ending is -yne instead of -ene. Alkynes cannot exhibit E,Z (cis‑trans) isomerism; hence, in this sense, their nomenclature is simpler than that of alkenes.
Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). 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.
Introduction
Here are the molecular formulas and names of the first ten carbon straight chain alkynes.
Name Molecular Formula
Ethyne C2H2
Propyne C3H4
1-Butyne C4H6
1-Pentyne C5H8
1-Hexyne C6H10
1-Heptyne C7H12
1-Octyne C8H14
1-Nonyne C9H16
1-Decyne C10H18
The more commonly used name for ethyne is acetylene, which used industrially.
Naming Alkynes
Like previously mentioned, the IUPAC rules are used for the naming of alkynes.
Rule 1
Find the longest carbon chain that includes both carbons of the triple bond.
Rule 2
Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes.
For example:
Rule 3
After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order.
For example:
If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond.
When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne.
For example:
Rule 4
Substituents containing a triple bond are called alkynyl.
For example:
Here is a table with a few of the alkynyl substituents:
Name Molecule
Ethynyl -C≡CH
2- Propynyl -CH2C≡CH
2-Butynyl -CH3C≡CH2CH3
Rule 5
A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example:
If both functional groups are the exact same distance from the ending of the parent chain, the alkene takes precedence in the numbering.
Exercise \(1\)
1) Name the following compounds:
2) How many isomers are possible for C5H8? Draw them.
3) Draw the following compounds:
a) 4,4-dimethyl-2-pentyne
b) 3-octyne
c) 3-methyl-1-hexyne
d) trans 3-hepten-1-yne
4) Do alkynes show cis-trans isomerism? Explain.
Answer
1)
A – 3,6-diethyl-4-octyne
B – 3-methylbutyne
C – 4-ethyl-2-heptyne
D – cyclodecyne
2) 2 possible isomers
3)
4) No. A triply bonded carbon atom can form only one other bond and has linear electron geometry so there are no "sides". Allkenes have two groups attached to each inyl carbon with a trigonal planar electron geometry that creates the possibility of cis-trans isomerism.
Reference
1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.01%3A_Naming_Alkynes.txt |
Objectives
After completing this section, you should be able to
1. write an equation to describe the preparation of an alkyne by the dehydrohalogenation of a vicinal dihalide or vinylic halide.
2. identify the alkyne produced from the dehydrohalogenation of a given vicinal dihalide or vinylic halide.
3. write a reaction sequence to show how the double bond of an alkene can be transformed into a triple bond.
4. identify the vicinal dihalide (or vinylic halide) needed to synthesize a given alkyne by dehydrohalogenation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• vicinal dihalide
• vinylic halide
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.
E2 Mechanism
Section 8.2 discussed that alkenes can be formed through an elimination reaction. In particular, the synthesis of alkynes will utilize the E2 elimination reaction. During the mechanism of an E2 reaction, a strong base removes a hydrogen adjacent to a halogen. The electrons from the broken C-H bond move to form the C=C double bond. Doing this causes the halogen to be ejected from the compound. Overall, a hydrogen and a halogen are eliminated from the compound to form an alkene. During this mechanism there is a stereoelectronic requirement that the adjacent hydrogen and the halogen be adjacent to each other.
E2 reaction will be discussed in greater detail in Section 11.10.
Alkyne Formation Through Dihaloalkane Elimination
Alkynes are frequently prepared through a double E2 reaction using 2 halides that are vicinal (meaning on adjacent carbons) or geminal (meaning on the same carbon). Because the E2 reaction takes place twice 2 $\pi$ bonds are formed thus creating an Alkyne. Although hydroxide and alkoxide bases could be used for the strong base required for an E2 reaction, their used opens the possibility of position rearrangement in the alkyne product. Because of this, the stronger base sodium amide in ammonia (NaNH2/NH3) is commonly used.
General Reaction
Vicinal dihalide converted to an alkyne
or
Geminal dihalide converted to an alkyne
Note! If a terminal alkyne is formed during the reaction, 3 equivalents of base are required instead of 2 as discussed below.
Mechanism
The following mechanism represents the reaction between 2,3-Dibromopentane with sodium amide in liquid ammonia to form pent-2-yne. During this mechanism an intermediate alkene is formed. Notice that in the alkene intermediate, the remaining hydrogen and halogen are anti to each other due to the stereoelectronic requirements of the E2 mechanism. The intermediate alkene is converted to an alkyne by a second E2 elimination of a hydrogen and halogen.
Terminal Alkynes
The acidity of terminal alkynes also plays a role in product determination when vicinal (or geminal) dihalides undergo base induced dielimination reactions. The following example illustrates eliminations of this kind starting from 1,2-dibromopentane, prepared from 1-pentene by addition of bromine. The initial elimination presumably forms 1-bromo-1-pentene, since base attack at the more acidic and less hindered 1º-carbon should be favored. The second elimination then produces 1-pentyne. If the very strong base such as sodium amide is used, the terminal alkyne is trapped as its sodium salt, from which it may be released by mild acid treatment. However, if the weaker base KOH with heat is used for the elimination, the terminal alkyne salt is not formed, or is formed reversibly, and the initially generated 1-pentyne rearranges to the more stable 2-pentyne via an allene intermediate. Even though terminal alkynes can be generated using sodium amide as a base, most chemists will prefer to use SN2 nucleophilic substitution instead of elimination when trying to form a terminal alkyne.
Preparation of Alkynes from Alkenes
An simple method for the preparation of alkynes utilizes alkenes as starting material. The process begins with the electrophilic addition of a halogen to the alkene bond to form the dihaloalkane. Then the double E2 elimination process is used to form the 2 $\pi$ bonds of an alkyne.
This first process is gone over in much greater detail in the page on halogenation of an alkene. In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is the reactant needed to produce the alkyne using double elimination, as covered previously on this page.
Exercise $1$
1) Why would we need 3 bases for every terminal dihaloalkane instead of 2 in order to form an alkyne?
2) What are the major products of the following reactions:
a.) 1,2-Dibromopentane with sodium amide in liquid ammonia
b.) 1-Pentene first with Br2 and chloromethane, followed by sodium ethoxide (Na+ -O-CH2CH3)
3) What would be good starting molecules for the synthesis of the following molecules:
4) Use a 6 carbon diene to synthesize a 6 carbon molecule with 2 terminal alkynes.
5) Identify the vinyl halide or halides and the vicinal dihalide or dihalides that could be used in the synthesis of:
a) 2,2,5,5-Tetramethyl-3-hexyne.
b) 4-Methyl-2-hexyne.
Answer
1) Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want.
2)
a) 1-Pentyne
b) 1-Pentyne
3)
4) Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up.
5)
a)
b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.02%3A_Preparation_of_Alkynes_-_Elimination_Reactions_of_Dihalides.txt |
Objectives
After completing this section, you should be able to
1. describe the bonding and geometry of the carbon-carbon triple bond in terms of the sp-hybridization of the carbon atoms involved.
2. explain the reactivity of alkynes based on the known strengths of carbon-carbon single, double and triple bonds.
3. write equations for the reaction of an alkyne with one or two equivalents of halogen (chlorine or bromine) or halogen acid (HCl, HBr or HI).
4. draw the structure of the product formed when an alkyne reacts with one equivalent of the halogens and halogen acids listed in Objective 3.
5. identify the alkyne which must have been used in an addition reaction with a halogen or halogen acid, given the product of such a reaction.
Study Notes
You might find it useful to review Section 1.9 before you begin work on this chapter. If necessary, construct a molecular model of a simple alkyne. Notice the similarity between the behaviour of alkenes and that of alkynes. In the laboratory, you will observe that alkynes readily decolourize a solution of bromine in dichloromethane. Section 9.7 describes a test that allows you to distinguish between a terminal alkyne (i.e., one in which the triple bond occurs between the last two carbons in the chain) and nonterminal alkynes and alkenes.
The Alkyne Triple Bond
As discussed in Section 1-9, the carbon-carbon triple bonds of alkynes are created by the the overlap of orbitals on two sp hybridized carbon atoms. The molecule acetylene (HCCH) is said to contain three sigma bonds and two pi bonds. The C-C sigma bond of acetylene is formed by the overlap an sp hybrid orbital from each of the carbon atoms. The two C-H sigma bonds are formed by the overlap of the second sp orbital on each carbon atom with a 1s orbital from a hydrogen. Each carbon atom still has two half-filled P orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds. These two perpendicular pairs of p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds). The electrostatic potential map of acetylene shows that the pi electrons of the triple bond form a negative belt (shown in red) around the center of the molecule.
Acetylene is linear, as predicted by VSEPR, with all four atoms lying in a straight line and both H-C-C bond angles being 180o. The triple bond in acetylene is the shortest (120 pm) and the strongest (964 kJ/mol) carbon-carbon bond known.
Electrophilic Addition of HX to Alkynes
Alkynes undergo electrophilic addition in much the same manner as alkenes, however, the presence to two pi bonds allows for the possibly of the addition happening twice. The addition of one equivalent of hydrogen chloride or hydrogen bromide converts alkynes to haloalkenes. The addition of two or more equivalents of HCl or HBr converts alkynes to geminal dihalides through an haloalkene intermediate. These additions are regioselective and follow Markovnikov's rule. The double bonds formed during the reaction with internal alkynes tend to have Z stereochemistry although not always.
HBr Addition to an Asymmetrical Internal Alkyne
The addition of HX to an asymmetrical internal alkyne tend to make a mixture of isomers as products.
Mechanism
The mechanism for the electrophilic addition of HX to an alkyne is analogous to the HX addition to an alkene. The presence of two pi bonds in the alkyne allows for the addition of HX to occur twice. The addition of H+ to the alkyne forms a vinyl cation will preferably form on the more substituted side of the alkyne following Markovnikov's rule. The subsequent addition of Br- forms a haloalkene which undergoes electrophilic addition to a second H+. The carbocation form will preferably form on the carbon attached to the halogen already in place. The carbocation is stabilized by the halogen through the creation of a resonance structure which obeys the octet rule. This stabilizing effect ensures that a geminal-dihalide is the sole product and no vicinal-dihalide is formed.
Electrophilic Addition of X2 to Alkynes
Alkynes undergo the same type of electrophilic addition with chloride and bromine as alkenes. However, with alkynes the halogen addition can occur once or twice depending on the molar equivalents of halogen used in the reaction. If one molar equivalent of halogen is used, a dihaloalkene is formed. The anti addition of the reaction mechanism causes the halogens to be trans in the resulting alkene. The addition of two or more molar equivalents of halogen converts the alkyne to a tetrahaloalkane through a dihaloalkene intermediate.
Mechanism
The alkyne undergoes electrophilic addition with bromine to form a bromonium ion in a three-membered ring. The ejected bromide ion performs an SN2 reaction with the bromonium ion causing the ring to open and the bromines in the resulting alkene to be in a trans configuration. The process is repeated with a second pi bond creating a tetrahaloalkane as a product.
Relative Reactivity of Alkynes and Alkenes to Electrophilic Reagents
When the addition reactions of electrophilic reagents, such as strong Brønsted acids and halogens, to alkynes are studied there is a curious paradox. The reactions of alkynes are even more exothermic than the additions to alkenes, and yet the rate of addition to alkynes is slower by a factor of 100 to 1000. This concept is shown in the reaction of one equivalent of bromine with 1-penten-4-yne to produce 4,5-dibromopent-1-yne as the chief product.
Why are the reactions of alkynes with electrophilic reagents more sluggish than the corresponding reactions of alkenes? Typically, addition reactions to alkynes are more exothermic than additions to alkenes, and there would seem to be a higher π-electron density about the triple bond ( two π-bonds versus one ). Two factors are significant in explaining this apparent paradox. First, although there are more π-electrons associated with the triple bond, the sp-hybridized alkyne carbons are more electronegative than the sp2-hybridized alkene carbons. The alkyne carbons exert a strong attraction for their π-electrons, which are consequently bound more tightly to the functional group than are the π-electrons of a double bond. This is seen in the ionization potentials of ethylene and acetylene. Remember an ionization potential is the minimum energy required to remove an electron from a molecule of a compound. Since the initial interaction between an electrophile and an alkene or alkyne involves the donation of electrons, the relatively slower reactions of alkynes becomes understandable.
Acetylene HC≡CH + Energy → [HC≡CH •(+) + e(–) ΔH = +264 kcal/mole
Ethylene H2C=CH2 + Energy → [H2C=CH2] •(+) + e(–) ΔH = +244 kcal/mole
A second factor is presumed to be the stability of the carbocation intermediate generated by sigma-bonding of a proton or other electrophile to one of the triple bond carbon atoms. When comparing the mechanism for the addition of HBr to an alkene and an alkyne, the alkyne reaction creates a vinyl carbocation which is less stable than the alkyl carbocation made during the alkene reaction.
Indeed, we can modify our earlier ordering of carbocation stability to include these vinyl cations in the manner shown below.
Substitution Methyl 1°-Vinyl 2°-Vinyl 1°-Allyl
Stability CH3(+) RCH=CH(+) < RCH2(+) RCH=CR(+) < R2CH(+) CH2=CH-CH2(+) < R3C(+)
Application of the Hammond postulate indicates that the activation energy for the generation of a vinyl cation intermediate would be higher than that for a lower energy intermediate. Thus, electrophilic reactions with alkenes have a lower activation energy and process faster than the corresponding reaction with an alkyne. This is illustrated for alkenes versus alkynes by the following energy diagrams. Despite these differences, electrophilic additions to alkynes have emerged as exceptionally useful synthetic transforms.
Electrophilic Addition of HX with Peroxides to Alkynes
When 1 equivalent of HBr is reacted with alkynes in the presence of peroxides and Anti-Markovnikov addition occurs. The use of peroxides causes the reaction to occur via a free radical mechanism. The bromine adds to the less substituted alkyne carbon while the hydrogen adds to the more substituted creating a haloalkene. Typically, H and Br are added in a syn or anti manner creating a mixture of products.
Exercise \(1\)
Draw the structure and give the IUPAC name of the product formed in each of the reactions listed below:
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.03%3A_Reactions_of_Alkynes_-_Addition_of_HX_and_X.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the reaction of water with an alkyne in the presence of sulfuric acid and mercury(II) sulfate.
2. describe keto-enol tautomerism.
3. predict the structure of the ketone formed when a given alkyne reacts with sulfuric acid in the presence of mercury(II) sulfate.
4. identify the reagents needed to convert a given alkyne to a given ketone.
5. identify the alkyne needed to prepare a given ketone by hydration of the triple bond.
6. write an equation for the reaction of an alkyne with borane.
7. write the equation for the reaction of a vinylic borane with basic hydrogen peroxide or hot acetic acid.
8. identify the reagents, the alkyne, or both, needed to prepare a given ketone or a given cis alkene through a vinylic borane intermediate.
9. identify the ketone produced when a given alkyne is reacted with borane followed by basic hydrogen peroxide.
10. identify the cis alkene produced when a given alkyne is reacted with borane followed by hot acetic acid.
11. explain why it is necessary to use a bulky, sterically hindered borane when preparing vinylic boranes from terminal alkynes.
12. predict the product formed when the vinylic borane produced from a terminal alkyne is treated with basic hydrogen peroxide.
13. identify the alkyne needed to prepare a given aldehyde by a vinylic borane.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• enol
• keto-enol tautomeric equilibrium
• tautomerism
• tautomers
Study Notes
Rapid interconversion between tautomers is called tautomerism; however, as the two tautomers are in equilibrium, the term tautomeric equilibrium may be used. This section demonstrates the equilibrium between a ketone and an enol; hence, the term keto-enol tautomeric equilibrium is appropriate. The term “enol” indicates the presence of a carbon-carbon double bond and a hydroxyl (i.e., alcohol) group. Later in the course, you will see the importance of keto-enol tautomerism in discussions of the reactions of ketones, carbohydrates and nucleic acids.
It is important to note that tautomerism is not restricted to keto-enol systems. Other examples include imine-enamine tautomerism
and nitroso-oxime tautomerism
However, at the moment you need only concern yourself with keto-enol tautomerism.
Notice how hydroboration complements hydration in the chemistry of both alkenes and alkynes.
Mercury(II)-Catalyzed Hydration of Alkynes
As with alkenes, hydration (addition of water) of alkynes requires a strong acid, usually sulfuric acid, and is facilitated by the mercuric ion (Hg2+). However, the hydration of alkynes gives ketone products while the hydration of alkenes gives alcohol products. Notice that the addition of oxygen in both reactions follows Markovnikov rule.
During the hydration of an alkyne, the initial product is an enol intermediate (a compound having a hydroxyl substituent attached to a double-bond), which immediately rearranges to the more stable ketone through a process called enol-keto tautomerization.
Tautomers are defined as rapidly inter-converted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The keto and enol tautomers are in equilibrium with each other and with few exceptions the keto tautomer is more thermodynamically stable and therefor favored by the equilibrium. This mechanism for tautomerization will be discussed in greater detail in Section 22-1.
General Reaction
For terminal alkynes, the addition of water follows the Markovnikov rule, and the final product is a methyl ketone. For internal alkynes the addition of water is not regioselective. Hydration of symmetrical internal alkynes produces a single ketone product. However, hydration of asymmetrical alkynes, (i.e. if R & R' are not the same ) produces two isomeric ketone products.
Mechanism
The mechanism starts with the eletrophilic addition of the mercuric ion (Hg2+) to the alkyne producing a mercury-containing vinylic carbocation intermediate. Nucleophilic attack of water on the vinylic carbocation forms a C-O bond to produce a protonated enol. Deprotonation of the enol by water then produces a organomurcury enol. The murcury is substituted with H+ to produced a neutral enol and regenerate the Hg2+ catalyst. The enol is converted to the ketone product through keto-enol tautomerization the mechanism of which is provided in Section 22-1.
1) Electrophilic addition of Hg2+
2) Nucleophilic attack by water
3) Deprotonation
4) Substitution
5) Tautomerization
Hydroboration–Oxidation of Alkynes
The hydroboration-oxidation of alkynes is analagous to the reaction with alkenes. However, where alkenes for alcohol products, alkynes for aldehyde or ketone products. In both cases the addition is anti-Markovnikov and an oxygen is placed on the less alkyl substituted carbon. With the hydroboration of an alkyne the presences of a second pi bond allows the initial product to under tautomerization to become the final aldehyde product.
Alkynes have two pi bonds both of which are capable of reacting with borane (BH3). To limit the reactivity to only one alkyne pi bond, a dialkyl borane reagent (R2BH) is used. Replacing two of the hydrogens on the borane with alkyl groups also creates steric hindrance which enhances the anti-Markovnikov regioselective of the reaction. Disiamylborane (Sia2BH) and 9-borabicyclo[3.3.1]nonane (9-BBN) are two common reagents for this hydroboration reaction. The oxidation reagents (a basic hydrogen peroxide solution) are the same for both alkenes and alkynes.
General Reaction
The hydroboration of terminal alkynes produces aldehyde products while internal alkynes produce ketone products. The hydroboration of symmetrical alkynes produces one ketone product and asymmetrical alkynes produce a mixture of product ketones.
Mechanism
The mechanism starts with the electrophilic addition of the B-H bond of the borane. The hydrogen atom and the borane all on the same side of the alkyne creating a syn addition configuration in the alkene product. Also, the addition is anti-Markovnikov regioselective which means the borane adds to the less substituted carbon of the alkyne and the hydrogen atom adds to the more substituted. The oxidative work-up replaces the borane with a hydroxy group (-OH) creating an enol intermediate. The enol immediately tautomerizes to the product aldehyde for terminal alkynes and the product ketone for internal alkynes.
Comparison of Mercury(II)-Catalyzed Hydration and Hydroboration–Oxidation of Alkynes
These two reactions are complementary for the reaction of a terminal alkyne because the produce distinctly different products. The mercury(II) catalyzed hydration of a terminal alkyne produces a methyl ketone, while the hydroboration-oxidation produces an aldehyde.
For internal alkynes, the regioslectivity of these reactions are rendered ineffective. The reactions are redundant in that they both produce the same ketone products.
Exercise \(1\)
1) Draw the structure of the product formed when each of the substances below is treated with H2O/H2SO4 in the presence of HgSO4.
a)
b)
2) Draw the structure of the keto form of the compound shown below. Which form would you expect to be the most stable?
3) What alkyne would you start with to gain the following products?
4) What alkyne would you start with to gain the following product?
5) Draw the product(s) of the following reactions:
Answer
Answers:
1)
a)
b)
2.
The keto form should be the most stable.
3)
4)
5)
For internal alkynes, there is no difference in the reaction products. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.04%3A_Hydration_of_Alkynes.txt |
Objectives
After completing this section, you should be able to
1. write equations for the catalytic hydrogenation of alkynes to alkanes and cis alkenes.
2. identify the reagent and catalyst required to produce a given alkane or cis alkene from a given alkyne.
3. identify the product formed from the reaction of a given alkyne with hydrogen and a specified catalyst.
4. identify the alkyne that must be used to produce a given alkane or cis alkene by catalytic hydrogenation.
5. write the equation for the reduction of an alkyne with an alkali metal and liquid ammonia.
6. predict the structure of the product formed when a given alkyne is reduced with an alkali metal and liquid ammonia.
7. identify the alkyne that must be used to produce a given alkene by reduction with an alkali metal and ammonia.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anion radical
• Lindlar catalyst
Study Notes
The Lindlar catalyst allows a chemist to reduce a triple bond in the presence of a double bond.
Thus
but
Hydrogenation and the Relative Stability of Hydrocarbons
Like alkenes, alkynes readily undergo catalytic hydrogenation partially to cis- or trans- alkenes or fully to alkanes depending on the reaction employed.
The catalytic addition of hydrogen to 2-butyne provides heat of reaction data that reflect the relative thermodynamic stabilities of these hydrocarbons, as shown above. From the heats of hydrogenation, shown in blue in units of kcal/mole, it would appear that alkynes are thermodynamically less stable than alkenes to a greater degree than alkenes are less stable than alkanes. The standard bond energies for carbon-carbon bonds confirm this conclusion. Thus, a double bond is stronger than a single bond, but not twice as strong. The difference ( 63 kcal/mole ) may be regarded as the strength of the π-bond component. Similarly, a triple bond is stronger than a double bond, but not 50% stronger. Here the difference ( 54 kcal/mole ) may be taken as the strength of the second π-bond. The 9 kcal/mole weakening of this second π-bond is reflected in the heat of hydrogenation numbers ( 36.7 - 28.3 = 8.4 ).
Overview of Reduction of Alkynes
Alkynes can undergo reductive hydrogenation reactions similar to alkenes. With the presence of two pi bonds within alkynes, the reduction reactions can be partial to form an alkene or complete to form an alkane. Since partial reduction of an alkyne produces an alkene, the stereochemistry provided by the reaction's mechanism determines whether a cis- or trans- alkene is formed. The three most significant alkyne reduction reactions are summarized below:
Catalytic Hydrogenation of an Alkyne
Much like alkenes, alkynes can be fully hydrogenated into alkanes with the help of a platinum, palladium, or nickel catalyst. Because the reaction is catalyzed on the surface of the metal, it is common for these catalysts to dispersed on carbon (Pd/C) or finely dispersed as nickel (Raney-Ni). The presence of two pi bonds in the alkyne cause two equivalents of H2 to be added during the reaction. During the reaction an alkene intermediate is form but not isolated.
Hydrogenation of an Alkyne to a Cis-Alkene
For catalytic hydrogenation, the Pt, Pd, or Ni catalysts are so effective in promoting addition of hydrogen to both double and triple carbon-carbon bonds that the alkene intermediate formed by hydrogen addition to an alkyne cannot be isolated. A less efficient catalyst, Lindlar's catalyst permits alkynes to be converted to alkenes without further reduction to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis-alkene because the hydrogenation reaction is occurring on the surface of the metal. Both hydrogen atoms are added to the same side of the alkyne as shown in the syn-addition mechanism for hydrogenation of alkenes in the previous chapter.
Lindlar's catalyst is prepared by deactivating (or poisoning) a conventional palladium catalyst. Lindlar’s catalyst has three components: palladium-calcium carbonate, lead acetate, and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane.
Hydrogenation of an Alkyne to a Trans-Alkene
The anti-addition of hydrogen to an alkyne pi bond occurs when reacted with sodium or lithium metal dissolved in ammonia. This reaction, also called dissolving metal reduction, involves radicals in its mechanism and produces a trans-alkene as it product.
Mechanism
Sodium metal is a powerful reducing agent due to the presence of a 3s1 electron in its valence. Sodium metal easily gives up this electron to become Na+. The mechanism start with a sodium atom donating an electron to the alkyne creating an intermediate with a negative charge and an unpaired electron called a radical anion. Next the amine solvent protonates the anion to create a vinyl radical. A second sodium atom then donates an electron to pair the radical converting it to a vinyl anion. This vinyl anion intermediate rapid interconverts between cis and trans conformations and determines the stereochemistry of the reaction. The trans-vinyl anion is more stable due to reduced steric crowding and is preferentially formed. Finally, the protonation of the trans-vinyl anion creates the trans-alkene product.
1) Electron Donation
2) Protonation
3) Electron Donation
4) Protonation
Exercise
Using any alkyne how would you prepare the following compounds: pentane, trans-4-methyl-2-pentene, cis-4-methyl-2-pentene.
Answer
9.06: Oxidative Cleavage of Alkynes
Objectives
After completing this section, you should be able to
1. write an equation to represent the oxidative cleavage of an alkyne with potassium permanganate or ozone.
2. identify the products that result from the oxidative cleavage of a given alkyne.
3. identify the reagents needed to carry out the oxidative cleavage of an alkyne.
4. use the results of an oxidative cleavage to determine the identity of an alkyne of unknown structure.
Study Notes
Compare the oxidative cleavage of alkynes with the oxidative cleavage of alkenes, discussed in Section 8.8.
Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenes, the reaction conditions can be gentler. For example, alkynes form vicinal dicarbonyls in neutral permanganate solution.
Gentle Alkyne Oxidation
Strong Alkyne Oxidation - Oxidative Cleavage
During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products. Because at least one of the reaction products is a carboxylic acid, it is important to consider the acid-base chemistry of the product in the reaction solution. Carboxylic acids are deprotonated in basic solutions to carboxylates. A second reaction step is required to protonate the carboxylate to the neutral form of the carboxylic acid.
Exercise
1. Draw the bond-line structures for the product(s) of the following reactions.
Answer
1.
Oxidative cleavage of alkynes produces carboxylic acids and/or carbon dioxide.
Aldehydes are not produced. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.05%3A_Reduction_of_Alkynes.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the reaction that occurs between a terminal alkyne and a strong base, such as sodamide, NaNH2.
2. rank a given list of compounds, including water, acetylene and ammonia, in order of increasing or decreasing acidity.
3. rank a given list of hydrocarbons, such as acetylene, ethylene and ethane, in order of increasing or decreasing acidity.
4. describe a general method for determining which of two given compounds is the stronger acid.
5. provide an acceptable explanation of why terminal alkynes are more acidic than alkanes or alkenes.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• acetylide anion
• acidity order
Study Notes
An acetylide anion is an anion formed by removing the proton from the end carbon of a terminal alkyne:
An acidity order is a list of compounds arranged in order of increasing or decreasing acidity.
The general ideas discussed in this section should already be familiar to you from your previous exposure to chemistry and from the review in Section 2.8. A slightly different account of why terminal alkynes are stronger acids than are alkenes or alkanes is given below. However, the argument is still based on the differences between sp-, sp2- and sp3-hybrid orbitals.
The carbons of a triple bond are sp-hybridized. An sp‑hybrid orbital has a 50% s character and a 50% p character, whereas an sp2‑hybrid orbital is 33% s and 67% p, and an sp3‑hybrid orbital is 25% s and 75% p. The greater the s character of the orbital, the closer the electrons are to the nucleus. Thus in a C(sp)\$\ce{-}\$H bond, the bonding electrons are closer to the carbon nucleus than they are in a C(sp2)\$\ce{-}\$H bond. In other words, compared to a C(sp2)\$\ce{-}\$H bond (or a C(sp3)\$\ce{-}\$H bond), a C(sp)\$\ce{-}\$H bond is very slightly polar: Cδ\$\ce{-}\$Hδ+. This slight polarity makes it easier for a base to remove a proton from a terminal alkyne than from a less polar or non-polar alkene or alkane.
As you will appreciate, the reaction between sodium amide and a terminal alkyne is an acid-base reaction. The sodium acetylide product is, of course, a salt. Terminal alkynes can also form salts with certain heavy-metal cations, notably silver(I) and copper(I). In the laboratory component of this course, you will use the formation of an insoluble silver acetylide as a method for distinguishing terminal alkynes from alkenes and non-terminal alkynes:
Metal acetylides are explosive when dry. They should be destroyed while still wet by warming with dilute nitric acid:
Acidity of Terminal Alkynes: Formation of Acetylide Anions
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the terminal proton leads to the formation of an acetylide anion, RC≡C:.
As discussed in Section 2.10, acidity typically increases with the stability of the corresponding conjugate base. The origin of the enhanced acidity of terminal alkynes can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The hybridization of an orbital affects its electronegativity. Within a shell, the s orbitals occupy the region closer to the nucleus than the p orbitals. Therefore, the spherical s orbitals are more electronegative than the lobed p orbitals. The relative electronegativity of hybridized orbitals increases as the percent s character increases and follows the order sp > sp2 > sp3. This trend indicates the sp hybridized orbitals of the acetylide anion are more electronegative and better able to stabilize a negative charge than sp2 or sp3 hybridized orbitals. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The table below shows how orbital hybridization compares with the identity of the atom when predicting relative acidity. Remember that as the pKa of a compound decreases its acidity increases.
Table 9.7.1: Akynes
Compound Conjugate Base Hybridization "s Character" pKa C-H BDE (kJ/mol)
CH3CH3 CH3CH2 sp3 25% 50 410
CH2CH2 CH2CH sp2 33% 44 473
HCCH HCC sp 50% 25 523
Acetylene, with a pKa of 25 is shown to be much more acidic than ethylene (pKa = 44) or ethane (pKa = 50). Consequently, acetylide anions can be readily formed by deprotonation of a terminal alkynes with a sufficiently strong base. The amide anion (NH2-), in the form of sodium amide (NaNH2) is commonly used for the formation of acetylide anions.
Exercise 9.7.1
Given that the pKa of water is 14.00, would you expect hydroxide ion to be capable of removing a proton from each of the substances listed below? Justify your answers, briefly.
1. ethanol (pKa = 16)
2. acetic acid (pKa = 4.72)
3. acetylene (pKa = 25)
Answer
1.
a. No, not very well. The pKa of ethanol is greater than that of water, thus the equilibrium lies to the left rather than to the right.Add texts here. Do not delete this text first.
b. Yes, very well. There is a difference of 11 pKa units between the pKa of water and the pKa of acetic acid. The equilibrium lies well to the right.
c. No, hardly at all. The hydroxide ion is too weak a base to remove a proton from acetylene. The equilibrium lies well to the left. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.07%3A_Alkyne_Acidity_-_Formation_of_Acetylide_Anions.txt |
Objectives
After completing this section, you should be able to
1. write an equation to describe the reaction of an acetylide ion with an alkyl halide.
2. discuss the importance of the reaction between acetylide ions and alkyl halides as a method of extending a carbon chain.
3. identify the alkyne (and hence the acetylide ion) and the alkyl halide needed to synthesize a given alkyne.
4. determine whether or not the reaction of an acetylide ion with a given alkyl halide will result in substitution or elimination, and draw the structure of the product formed in either case.
Key Terms
Make certain that you can define, and use in context, the key term below.
• alkylation
Study Notes
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.
The alkyl halide used in this reaction must be primary. Thus, if you were asked for a suitable synthesis of 2,2-dimethyl-3-hexyne, you would choose to attack iodoethane with the anion of 3,3- dimethyl-1-butyne
rather than to attack 2-iodo-2-methylpropane with the anion of 1-butyne.
The reasons will be made clear in Chapter 11.
Nucleophilic Substitution Reactions of Acetylides
The presence of lone pair electrons and a negative charge on a carbon, makes acetylide anions are strong bases and strong nucleophiles. Therefore, acetylide anions can attack electrophiles such as alkyl halides to cause a substitution reaction. These substitution reactions will be discussed in detail in Chapter 11.
Mechanism
The C-X bonds in 1o alkyl halides are polarized due to the high electronegativity of the halogen. The electrons of the C-X sigma bond are shifted towards the halogen giving it a partial negative charge. This also causes electrons to be shifted away from the carbon giving it a partial positive and making it electrophilic. During this reaction, the lone pair electrons on the acetylide anion attack the electrophilic carbon in the 1o alkyl halide forming a new C-C bond. The formation of this new bond causes the expulsion of the halogen as what is called a leaving group. Overall, this reaction forms a C-C bond and converts a terminal alkyne into a internal alkyne. Because a new alkyl group is added to the alkyne during this reaction, it is commonly called an alkylation.
This substitution reaction is often coupled with the acetylide formation, discussed in the previous section, and shown as a single reaction.
Example \(1\)
Terminal alkynes can be generated through the reaction of acetylene and a 1o alkyl halide.
Example \(2\)
Because the acetylide anion is a very strong base, this substitution reaction is most efficient with methyl or primary halides. Secondary, tertiary, or even bulky primary halogens will give alkenes by the E2 elimination mechanism discussed in Section 11.10. An example of this effect is seen in the reaction of bromocyclopentane with a propyne anion. The reaction produces the elimination product cyclopentene rather than the substitution product 1-propynylcyclopentane.
Nucleophilic Addition of Acetylides to Carbonyls
Acetylide anions also add to the electrophilic carbon in aldehydes and ketones to form alkoxides, which, upon protonation, give propargyl alcohols. With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers. These types of reaction will be discussed in more detail in Chapter 19.
Exercise \(1\)
1) The pKa of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by sodium amide, as shown below.
2) Give the possible reactants which could form the following molecules by an alkylation.
3) Propose a synthetic route to produce 2-pentene from propyne and an alkyl halide.
4) Using acetylene as the starting material, show how you would synthesize the following compounds
a)
b) but-2-yne
c)
d)
5) Show how you would accomplish the following synthetic transformation.
Answer
1) Assuming the pKa of pent-1-yne is about 25, then the difference in pKas is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 1010.
2)
3)
4)
a)
b)
c)
d)
5) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.08%3A_Alkylation_of_Acetylide_Anions.txt |
Objective
After completing this section, you should be able to design a multistep synthesis to prepare a given product from a given starting material, using any of the reactions introduced in the textbook up to this point.
Study Notes
You should have noticed that some of the assigned problems have required that you string together a number of organic reactions to convert one organic compound to another when there is no single reaction to achieve this goal. Such a string of reactions is called an “organic synthesis.” One of the major objectives of this course is to assist you in designing such syntheses. To achieve this objective, you will need to have all of the reactions described in the course available in your memory. You will need to recall some reactions much more frequently than others, and the only way to master this objective is to practise. The examples given in this chapter will be relatively simple, but you will soon see that you can devise some quite sophisticated syntheses using a limited number of basic reactions.
Introduction
The study of organic chemistry introduces students to a wide range of interrelated reactions. Alkenes, for example, may be converted to structurally similar alkanes, alcohols, alkyl halides, epoxides, glycols and boranes; cleaved to smaller aldehydes, ketones and carboxylic acids; and enlarged by carbocation and radical additions as well as cycloadditions. Most of these reactions are shown in the Alkene Reaction Map below. All of these products may be subsequently transformed into a host of new compounds incorporating a wide variety of functional groups. Consequently, the logical conception of a multi-step synthesis for the construction of a designated compound from a specified starting material becomes one of the most challenging problems that may be posed. Functional group reaction maps like the one below for alkenes can be helpful in designing multi-step syntheses. It can be helpful to build and design your own reaction maps for each functional group studied.
Alkene Reaction Map
Please note: The reagents for each chemical transformation have been intentionally omitted so that this map can be used as a study tool. The answers are provided at the end of this section as part of the exercises.
Simple Multi-Step Syntheses
A one or two step sequence of simple reactions is not that difficult to deduce. For example, the synthesis of meso-3,4-hexanediol from 3-hexyne can occur by more than one multi-step pathway.
One approach would be to reduce the alkyne to cis or trans-3-hexene before undertaking glycol formation. Permanaganate or osmium tetroxide hydroxylation of cis-3-hexene would form the desired meso isomer.
From trans-3-hexene, it would be necessary to first epoxidize the alkene with a peracid followed by ring opening with acidic or basic hydrolysis.
Longer multi-step syntheses require careful analysis and thought, since many options need to be considered. Like an expert chess player evaluating the long range pros and cons of potential moves, the chemist must appraise the potential success of various possible reaction paths, focusing on the scope and limitations constraining each of the individual reactions being employed. The skill is acquired by practice, experience, and often trial and error.
Thinking it Through with 3 Examples
The following three examples illustrate strategies for developing multi-step syntheses from the reactions studied in the first ten chapters of this text. It is helpful to systematically look for structural changes beginning with the carbon chain and brainstorm relevant functional group conversion reactions. Retro-synthesis is the approach of working backwards from the product to the starting material.
In the first example, we are asked to synthesize 1-butanol from acetylene.
The carbon chain doubles in size indicating an acetylide SN2 reaction with an alkyl halide. Primary alcohol formation from an anti-Markovnikov alkene hydration reaction (hydroboration-oxidation) is more likely than a substitution reaction. Applying retro-synthesis, we work backwards from the alcohol to the alkene to the alkyne from an acetylide reaction that initially builds the carbon chain.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation identified from the retro-synthesis. The ethylbromide must also be derived from acetylene so multiple reaction pathways are combined as shown below.
In the second example, we are asked to synthesize 1,2-dibromobutane from acetylene.
Once again there is an increase in the carbon chain length indicating an acetylide SN2 reaction with an alkyl halide similar to the first example. The hydrohalogenation can be subtle to discern because the hydrogen atoms are not shown in bond-line structures. Comparing the chemical formulas of 1-butyne with 1,2-dibromobutane, there is a difference of two H atoms and two Br atoms indicating hydrohalogenation and not halogenation. The addition of both bromine atoms to the same carbon atom also supports the idea that hydrohalogenation occurs on an alkyne and not an alkene. The formation of the geminal dihalide also indicates hydrohalogenation instead of halogenation because halogenation produces vicinal dihalides. With this insight, the retro-synthesis indicates the following series of chemical transformations.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation.
In the third example, we are asked to produce 6-oxoheptanal from methylcyclohexane.
Counting the carbons, the starting material and product both contain seven carbon atoms and there is a cleavage reaction of an alkene under reductive conditions. One important missing aspect of this reaction is a good leaving group (LG). Alkanes are chemically quite boring. We can burn them as fuel or perform free-radical halogenation to create alkyl halides with excellent leaving groups. With these observations, the following retro-synthesis is reasonable.
Retro-Synthesis
Working forwards, we specify the reagents needed for each reaction. For the initial free-radical halogenation of the alkane, we have the option of chlorine (Cl2) or bromine (Br2). Because methylcyclohexane has several different classifications of carbons, the selectivity of Br2 is more important than the faster reactivity of Cl2. A strong base with heat can be used for the second step to follow an E2 mechanism and form 1-methylcyclohexene. The aldehyde group on the final product indicates gentle oxidative cleavage by any of several reaction pathways. These reactions can be combined in to the following multi-step synthesis.
Reaction Maps to Build Functional Group Conversion Mastery
After working through the examples above, we can see how important it is to memorize all of the functional group reactions studied in the first ten chapters. We can apply the knowledge of these reactions to the wisdom of multi-step syntheses.
Please note: The reagents for each chemical transformation have been intentionally omitted so that these maps can be used as a study tools. The answers are provided at the end of this section as part of the exercises.
Alkyne Reaction Map
Exercise
1) Starting at 3-hexyne predict synthetic routes to achieve:
a) trans-3-hexene
b) 3,4-dibromohexane
c) 3-hexanol.
2) Starting with acetylene and any alkyl halides propose a synthesis to make
a) pentanal
b) hexane.
3) Show how you would accomplish the following synthetic transformations.
Answer
1)
2)
a)
b)
3)
a)
b)
9.S: Alkynes - An Introduction to Organic Synthesis (Summary)
Concepts & Vocabulary
9.1 Naming Alkynes
• Follow IUPAC rules in naming alkynes.
9.2 Preparation of Alkynes - Elimination Reactions of Dihalides
• Vicinal describes two groups on adjacent carbon atoms.
• Geminal describes two groups on the same carbon atom.
• Alkynes can be prepared by two successive eliminations of HX from either vicinal or geminal dihalides.
9.3 Reactions of Alkynes - Addition of HX and X2
• Alkynes undergo addition reactions similarly to alkenes yielding Markovnikov products.
9.4 Hydration of Alkynes
• Enols have a hydroxyl group bonded to a sp2 hybrid carbon (double-bonded carbon).
• Enols are usually not stable and undergo keto-enol tautomerization to form a ketone or aldehyde.
• Hydration of alkynes leads to an enol product which then rapidly tautomerizes into a ketone or aldehyde.
9.5 Reduction of Alkynes
• Alkynes can be hydrogenated with hydrogen gas and strong catalysts to yield alkanes.
• Alkynes can be hydrogenated with hydrogen gas and Lindlar's catalyst to yield Z alkenes.
• Alkynes can be hydrogenated with sodium metal and liquid ammonia to yield E alkenes.
9.6 Oxidative Cleavage of Alkynes
• Oxidative cleavage of internal alkynes forms two molecules of carboxylic acids.
• Oxidative cleavage of terminal alkynes forms one molecule of carbon dioxide and one carboxylic acid.
9.7 Alkyne Acidity - Formation of Acetylide Anions
• Terminal alkynes are relatively acidic compared to alkene and alkane carbon-hydrogen bonds.
• Deprotonation of a terminal alkyne forms an acetylide ion, which is a good nucleophile.
9.8 Alkylation of Acetylide Anions
• Acetylide ions can be alkylated by adding to alkyl halides and carbonyl compounds.
9.9 An Introduction to Organic Synthesis
• Desired products cannot always be made from available starting materials through one reaction. Formation of these materials may require multiple reactions completed in sequence. This type of reaction sequence is termed synthesis.
Skills to Master
• Skill 9.1 Use IUPAC rules to accurately name alkynes.
• Skill 9.2 Draw elimination mechanisms to form alkynes.
• Skill 9.3 Draw addition mechanisms to alkynes incorporating carbocation intermediates.
• Skill 9.4 Draw addition mechanisms to alkynes incorporating halonium intermediates.
• Skill 9.5 Describe relative stability of enols to ketones and aldehydes.
• Skill 9.6 Draw keto-enol tautomerism mechanism.
• Skill 9.7 Draw products that differentiate between multiple reduction reactions of alkynes.
• Skill 9.8 Draw products of oxidative cleavage of alkynes.
• Skill 9.9 Draw mechanism for deprotonation of terminal alkynes.
• Skill 9.10 Compare acidity of terminal alkynes with other organic compounds.
• Skill 9.11 Draw reaction mechanisms using acetylide ions as nucleophiles.
• Skill 9.12 Describe schemes to accomplish synthesis of organic products given a starting material.
Summary of Reactions
Preparation of Alkynes
Reactions of Alkynes | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.09%3A_An_Introduction_to_Organic_Synthesis.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.
• 10.0: Introduction to Organohalides
Many organic compounds are closely related to the alkanes. Alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion.
• 10.1: Names and Properties of Alkyl Halides
Alkyl halides are also known as haloalkanes. This page explains what they are and discusses their physical properties. alkyl halides are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine).
• 10.2: Preparing Alkyl Halides from Alkanes - Radical Halogenation
Alkanes (the simplest of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it functionalizes alkanes which opens a gateway to further chemical reactions.
• 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
This page looks at reactions in which the -OH group in an alcohol is replaced by a halogen such as chlorine or bromine. It includes a simple test for an -OH group using phosphorus(V) chloride.
• 10.6: Reactions of Alkyl Halides - Grignard Reagents
The organomagnesium compounds formed by the reaction of an alkyl or aryl halide with magnesium are called Grignard reagents. As you will see throughout the remainder of this course, Grignard reagents can be used to synthesize a wide range of organic compounds and are extremely useful to the organic chemist.
• 10.7: Organometallic Coupling Reactions
• 10.8: Oxidation and Reduction in Organic Chemistry
In organic chemistry, redox reactions look a little different. Electrons in an organic redox reaction often are transferred in the form of a hydride ion - a proton and two electrons. Because they occur in conjunction with the transfer of a proton, these are commonly referred to as hydrogenation and dehydrogenation reactions: a hydride plus a proton adds up to a hydrogen (H2) molecule. Be careful - do not confuse the terms hydrogenation and dehydrogenation with hydration and dehydration.
• 10.S: Organohalides (Summary)
10: Organohalides
Objectives
After completing this section, you should be able to
1. list the industrial uses of some important halogenated hydrocarbons including 1,1,1-trichloroethane, tetrafluoroethylene and dichlorodifluoromethane.
2. outline, briefly, how the chemistry of vinyl halides and aryl halides differs from that of the alkyl halides discussed.
Study Notes
There are several different types of halogen-substituted organic compounds, including aryl halides, acyl halides, vinyl halides and alkynyl halides. The primary focus of this chapter is on alkyl halides.
Freons™, also called fluorocarbons or chlorofluorocarbons, have been a source of concern to environmentalists since 1974, when Frank S. Rowland and Mario J. Molina suggested that these substances might be contributing to the destruction of Earth’s ozone layer. The stratospheric ozone layer filters out much of the ultraviolet radiation from the sun’s rays. It is believed that extensive depletion of this layer, and the consequent increase in the amount of ultraviolet radiation reaching Earth, could result in the destruction of certain crops, in climate modification, and in an increase in the incidence of skin cancer. In recent years, the manufacture and use of freons has declined sharply as the general public has become more aware of the problems that might be caused by these substances.
Note: “Freon” is a DuPont trademark.
Related to the freons are the halons—now used in some fire extinguishers, particularly in areas where foams or dry-chemical extinguishers cannot be used (e.g., in and around computers). If you examine such extinguishers, you will find that the halon is identified by a number; for example, halon 1301 or halon 1211. The first number represents the number of carbon atoms present, the second is the number of fluorines, the third is the number of chlorines and the fourth is the number of bromines.
Thus the halons given as examples above have the following structures:
You need not remember the names of the various freons and halons, but you should be prepared to name them by the IUPAC system according to the rules developed in the next section.
Many organic compounds are closely related to the alkanes. Alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion.
The reactions of alkanes with halogens produce halogenated hydrocarbons, compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms:
The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane). A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH4) can react with chlorine (Cl2), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH3CH3) are listed in Table 10.1, along with some of their uses.
Table 10.1: Some Halogenated Hydrocarbons
Formula Common Name IUPAC Name Some Important Uses
Derived from CH4
CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber
CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent
CHCl3 chloroform trichloromethane industrial solvent
CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use)
CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems
CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics
CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant
Derived from CH3CH3
CH3CH2Cl ethyl chloride chloroethane local anesthetic
ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber
CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastics
To Your Health: Halogenated Hydrocarbons
Once widely used in consumer products, many chlorinated hydrocarbons are suspected carcinogens (cancer-causing substances) and also are known to cause severe liver damage. An example is carbon tetrachloride (CCl4), once used as a dry-cleaning solvent and in fire extinguishers but no longer recommended for either use. Even in small amounts, its vapor can cause serious illness if exposure is prolonged. Moreover, it reacts with water at high temperatures to form deadly phosgene (COCl2) gas, which makes the use of CCl4 in fire extinguishers particularly dangerous.
Ethyl chloride, in contrast, is used as an external local anesthetic. When sprayed on the skin, it evaporates quickly, cooling the area enough to make it insensitive to pain. It can also be used as an emergency general anesthetic.
Bromine-containing compounds are widely used in fire extinguishers and as fire retardants on clothing and other materials. Because they too are toxic and have adverse effects on the environment, scientists are engaged in designing safer substitutes for them, as for many other halogenated compounds.
Reactivity of Halides
Alkyl halides have an sp3 carbon atom with a halogen attached, this is also true for allylic halides and benzylic halides. These types of halides are all reactive toward most substitution and elimination reactions. Allyl and benzyl halides tend to form carbocations more easily due to resonance stabilization.
However, halogens bonded to sp2 carbon atoms are not typically reactive. Examples of this are vinyl and aryl halides. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.00%3A_Introduction_to_Organohalides.txt |
Objectives
After completing this section, you should be able to
1. write the IUPAC name of a halogenated aliphatic hydrocarbon, given its Kekulé, condensed or shorthand structure.
2. draw the Kekulé, condensed or shorthand structure of a halogenated aliphatic hydrocarbon, given it IUPAC name.
3. write the IUPAC name and draw the Kekulé, condensed or shorthand structure of a simple alkyl halide, given a systematic, non-IUPAC name (e.g., sec-butyl iodide).
4. arrange a given series of carbon-halogen bonds in order of increasing or decreasing length and strength.
Study Notes
This section contains little that is new. If you mastered the IUPAC nomenclature of alkanes, you should have little difficulty in naming alkyl halides. Notice that when a group such as CH2Br must be regarded as a substituent, rather than as part of the main chain, we may use terms such as bromomethyl.
You will find it easier to understand the reactions of the alkyl halides if you keep the polarity of the C-X bond fixed permanently in your mind (see ”The Polar C-X Bond” shown in the reading below).
Alkyl halides are also known as haloalkanes. This page explains what they are and discusses their physical properties. alkyl halides are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). For example:
Halide Designations
Alkyl halides fall into different classes depending on how many alkyl groups are attached to the carbon which holds the halogen. There are some chemical differences between the various types. When there are no alkyl groups attached to the carbon holding the halogen, these are considered methyl halides (CH3X).
Primary alkyl halides
In a primary (1°) haloalkane, the carbon which carries the halogen atom is only attached to one other alkyl group. Some examples of primary alkyl halides include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) haloalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different. Examples:
Tertiary alkyl halides
In a tertiary (3°) haloalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Example $1$
Please indicate if the following haloalkanes are methyl, 1o, 2o, or 3o:
a) CH3I
b) CH3CH2Br
c)
d)
e)
f)
a) methyl
b) 1o
c) 2o
d) 3o
e) 1o
f) 2o
Nomenclature of Alkyl Halides
Alkyl halides are systematically named as alkanes (Section 3-4) where the halogen is a substituent on the parent alkane chain. To summarize the rules discussed in detail in Section 3-4, there are three basic steps to naming alkyl halides.
1. Find and name the longest carbon chain and name it as the parent. Remember is an alkene or alkyne is present, the parent chain must contain both carbons of the multiple bond.
2. Number the parent chain consecutively, starting at the end nearest a substituent group. Then assign each substituent a number. Remember the IUPAC system uses prefix to indicate the halogen followed by the suffix -ide. The prefixes are fluoro- for fluorine, chloro- for chlorine, bromo- from bromine, and iodo- for iodine. The name of a halogen is preceded by a number indicating the substituent’s location on the parent chain.
1. If there is an ambiguity in numbering the parent chain, begin on the end which is closer to the substituent which comes first alphabetically.
Common Names of Alkyl Halides
Alkyl halides with simple alkyl groups are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names. The common names of alkyl halides consist of two parts: the name of the alkyl group plus the first syllable of the name of the halogen, with the ending -ide. The names of common alkyl groups are listed in Section 3.3.
Exercise $1$
1) Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
3. CH3CH2I
4. CH3CH2CH2CH2F
2) Give the IUPAC name for each compound.
a)
b)
c)
d)
Answer
1) a) The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
b) The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
c) The alkyl group (CH3CH2–) is a ethyl group, and the halogen is iodide (I). The common name is therefore ethyl iodide. For the IUPAC name, the prefix for Iodide (Iodod) is combined with the name for a two-carbon chain (ethane), preceded by a number identifying the carbon atom to which the I atom is attached, so the IUPAC name is 1-bromoethane
d) The alkyl group (CH3CH2CH2CH2–) is a butyl group, and the halogen is fluorine (F). The common name is therefore butyl fluoride. For the IUPAC name, the prefix for Fluorine (Fluoro) is combined with the name for a four-carbon chain (butane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-fluorobutane.
2) a) The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
b) The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
c) 2-Chloro-3-methylbutane
d) 1-Bromo-2-chloro-4-methylpentane rine (F). The common name is therefore butyl fluoride. For the IUPAC name, the prefix for Fluorine (Fluoro) is combined with the name for a four-carbon chain (butane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-fluorobutane.
There is a fairly large distinction between the structural and physical properties of haloalkanes and the structural and physical properties of alkanes. As mentioned above, the structural differences are due to the replacement of one or more hydrogens with a halogen atom. The differences in physical properties are a result of factors such as electronegativity, bond length, bond strength, and molecular size.
Halogens and the Character of the Carbon-Halogen Bond
As discussed in Section 6. 4, halogens are more electronegative than carbon. This results in a carbon-halogen bond that is polarized with the carbon atom bearing a partial positive charge and the halogen a partial negative charge. This polarity can be distinctly seen when viewing the electrostatic potential map of a methyl halide. Electron density is shown by a red/yellow color which is almost exclusively around the halogen atom. The methyl portion of the compound lacks electron density which is shown by a blue/green color.
The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
Haloalkanes Have Higher Boiling Points than Alkanes
When comparing alkanes and haloalkanes, we will see that haloalkanes have higher boiling points than alkanes containing the same number of carbons. London dispersion forces are the first of two types of forces that contribute to this physical property. You might recall from general chemistry that London dispersion forces increase with molecular surface area. In comparing haloalkanes with alkanes, haloalkanes exhibit an increase in surface area due to the substitution of a halogen for hydrogen. The increase in surface area leads to an increase in London dispersion forces, which then results in a higher boiling point.
Dipole-dipole interaction is the second type of force that contributes to a higher boiling point. As you may recall, this type of interaction is a coulombic attraction between the partial positive and partial negative charges that exist between carbon-halogen bonds on separate haloalkane molecules. Similar to London dispersion forces, dipole-dipole interactions establish a higher boiling point for haloalkanes in comparison to alkanes with the same number of carbons.
The table below illustrates how boiling points are affected by some of these properties. Notice that the boiling point increases when hydrogen is replaced by a halogen, a consequence of the increase in molecular size, as well as an increase in both London dispersion forces and dipole-dipole attractions. The boiling point also increases as a result of increasing the size of the halogen, as well as increasing the size of the carbon chain.
Solubility
Solubility in water
The alkyl halides are at best only slightly soluble in water. For a haloalkane to dissolve in water you have to break attractions between the haloalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Both of these cost energy. Energy is released when new intermolecular forces are generated between the haloalkane molecules and water molecules. These will only be dispersion forces and dipole-dipole interactions. These are not as strong as the original hydrogen bonds in the water, and so not as much energy is released as was used to separate the water molecules. The energetics of the change are sufficiently "unprofitable" that very little dissolves.
Solubility in organic solvents
Alkyl halides tend to dissolve in organic solvents because the new intermolecular attractions have much the same strength as the ones being broken in the separate haloalkane and solvent.
Chemical Reactivity
The pattern in strengths lies in the strength of the bond between the carbon atom and the halogen atom. Previously in this section, it was noted that the trend for bond strength increases from C-I to C-Br to C-Cl with C-F bonds being the strongest. To react with the alkyl halides, the carbon-halogen bond has got to be broken. Because that gets easier as you go from fluoride to chloride to bromide to iodide, the compounds get more reactive in that order. Iodoalkanes are the most reactive and fluoroalkanes are the least. In fact, fluoroalkanes are so unreactive that we will ignore them completely from now on in this section!
Exercise $2$
1) Give the names of the following organohalides:
2) Draw the structures of the following compounds:
a) 2-Chloro-3,3-dimethylpentane
b) 1,1-Dichloro-4-isopropylcyclohexane
c) 3-bromo-3-ethylhexane
Answer
1)
a) 5-ethyl-4-iodo-3methyl-octane
b) 1-bromo-2,3,4-trimethyl-pentane
c) 4-bromo-5-chloro-2-methyl-heptane
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.01%3A_Names_and_Properties_of_Alkyl_Halides.txt |
Objectives
After completing this section, you should be able to
1. explain why the radical halogenation of alkanes is not usually a particularly good method of preparing pure samples of alkyl halides.
2. use $\ce{\sf{C–H}}$ bond energies to account for the fact that in radical chlorinations, the reactivity of hydrogen atoms decreases in the order $\text{tertiary} > \text{secondary} > \text{primary}. \nonumber$
3. predict the approximate ratio of the expected products from the monochlorination of a given alkane.
Study Notes
The following terms are synonymous:
1. methyl hydrogens, primary hydrogens, and 1° hydrogens.
2. methylene hydrogens, secondary hydrogens, and 2° hydrogens.
3. methine hydrogens, tertiary hydrogens, and 3° hydrogens.
Note that in radical chlorination reactions, the reactivity of methine, methylene and methyl hydrogens decreases in the ratio of approximately 5 : 3.5 : 1. This will aid in the prediction of expected products from the monochlorination of a given alkane.
Radical Halogenation
Alkanes (the simplest of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen (Cl2 or Br2) to form a haloalkane. This reaction is very important in organic chemistry because it functionalizes alkanes which opens a gateway to further chemical reactions.
General Reaction
$CH_4 + Cl_2 + energy → CH_3Cl + HCl \nonumber$
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism which is characterized by three steps: initiation, propagation, and termination. Initiation requires an input of energy but after that the reaction is self-sustaining.
Step 1: Initiation
During the initiation step free radicals are created when ultraviolet light or heat causes the X-X halogen bond to undergo homolytic to create two halogen free radicals. It is important to note that this step is not energetically favorable and cannot occur without some external energy input. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical abstracts hydrogen atom from methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step, the methyl radical reacts with more of the chlorine starting material (Cl2). One of the chlorine atoms becomes a radical and the other combines with the methyl radical to form the alkyl halide product.
Step 3: Termination
In the three termination steps of this mechanism, radicals produced in the mechanism an undergo radical coupling to form a sigma bond. These are called termination steps because a free radical is not produced as a product, which prevents the reaction from continuing. Combining the two types of radicals produced can be combined to from three possible products. Two chlorine radicals and couple to form more halogen reactant (Cl2). A chlorine radical and a methyl radical can couple to form more product (CH3Cl). An finally, two methyl radicals can couple to form a side product of ethane (CH3CH3).
This reaction is a poor synthetic method due to the formation of polyhalogenated side products. The desired product occurs when one of the hydrogen atoms in the methane has been replaced by a chlorine atom. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms to produce a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.
Energetics
Why do these reactions occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ΔH that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed} \nonumber$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Chlorination of Other Alkanes
When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.
CH3-CH2-CH3 + Cl2 → 45% CH3-CH2-CH2Cl + 55% CH3-CHCl-CH3
These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3.5:1. Further experiments showed that 3º-hydrogens are about 5 times more toward halogen atoms 1º-hydrogens. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.
(CH3)3CH + Cl2 → 65% (CH3)3CCl + 35% (CH3)2CHCH2Cl
The Relative Reactivity of Hydrogens to
Radical Chlorination
This difference in reactivity can only be attributed to differences in C-H bond dissociation energies. In our previous discussion of bond energy we assumed average values for all bonds of a given kind, but now we see that this is not strictly true. In the case of carbon-hydrogen bonds, there are significant differences, and the specific dissociation energies (energy required to break a bond homolytically) for various kinds of C-H bonds have been measured. These values are given in the following table.
R (in R–H) methyl ethyl i-propyl t-butyl
Bond Dissociation Energy
(kcal/mole)
103 98 95 93
This data shows that a tertiary C-H bond (93 kcal/mole) is easier to break than a secondary (95 kcal/mole) and primary (98 kcal/mole) C-H bond. These bond dissociation energies can be used to estimate the relative stability of the radicals formed after homolytic cleavage. Because a tertiary C-H bond requires less energy to undergo homolytic cleavage than a secondary or primary C-H bond, it can be inferred that a tertiary radical is more stable than secondary or primary.
Relative Stability of Free Radicals
Exercise $1$
Write out the complete mechanism for the chlorination of methane.
Answer
The answer to this problem is actually above in the initiation, propagation and termination descriptions.
Exercise $2$
Explain, in your own words, how the first propagation step can occur without input of energy if it is energetically unfavorable.
Answer
Since the second step in propagation is energetically favorable and fast, it drives the equilibrium toward products, even though the first step is not favorable.
Exercise $3$
Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
Answer
Initiation step requires energy which can be in the form of light or het.
Exercise $4$
Having learned how to calculate the change in enthalpy for the chlorination of methane apply your knowledge and using the table provided below calculate the change in enthalpy for the bromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Answer
To calculate the enthalpy of reaction, you subtract the BDE of the bonds formed from the BDE of the bonds broken.
Bonds broken are C-H and Br-Br.
Bonds formed are H-Br adn C-Br.
Bonds broken - bonds formed = change in enthalpy
(101 kcal/mol + 46 kcal/mol) - (87 kcal/mol + 70 kcal/mol) = change in enthalpy
-10 kcal/mol = change in enthalpy for bromination of ethane.
Exercise $5$
1) Predict the mono-substituted halogenated product(s) of chlorine gas reacting with 2-methylbutane.
2) Predict the relative amount of each mono-brominated product when 3-methylpentane is reacted with Br2. Consider 1°, 2°, 3° hydrogen.
3) For the following compounds, give all possible monochlorinated derivatives.
Answer
1)
2)
.
3) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.02%3A_Preparing_Alkyl_Halides_from__Alkanes_-_Radical_Halogenation.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the bromination of a symmetrical alkene using N-bromosuccinimide.
2. predict the product formed when a given symmetrical alkene is treated with N-bromosuccinimide.
3. identify the reagent, the symmetrical alkene, or both, needed to produce a given allyl halide by allylic bromination.
4. list the following radicals in order of increasing or decreasing stability: allyl, vinyl, primary alkyl, secondary alkyl, tertiary alkyl, methyl.
5. explain the ease of forming an allyl radical, and the difficulty of forming a vinyl radical, in terms of relative \$\ce{\sf{C–H}}\$ bond dissociation energies.
Key Terms
Make certain that you can define, and use in context, the key term below.
Study Notes
We have discussed the electrophilic addition of X2 and HX to alkenes as a route to forming alkyl halides (Sections 7.8 and 8.2). In this section we introduce bromination at the allyic position with N-bromosuccinimide (NBS). Notice that at the moment we are restricting our studies to the allylic bromination of symmetrical alkenes, such as cyclohexene. When we introduce an element of asymmetry, we find that more than one allyl radical can be formed; therefore, we must assess the relative stability of each radical when trying to predict which product will predominate. The method of doing this assessment is described in the next section.
Allylic Bromination
Previously, alkyl halides have been produced though reactions with alkenes. Hydrogen halides (HCl, HBr, and HI) react with alkenes in an electrophilic addition reaction discussed in Section 7-8 to yield alkyl halides as products. Also, Bromine (Br2) and chlorine (Cl2) can react with alkenes to provide dihalogenated products as discussed in Section 8-2.
Another method for preparing alkyl halides from alkenes is with N-bromosuccinimide (NBS) in carbon tetrachloride (CCl4) solution with the presence of light. The reaction specifically causes the substitution of bromine with a hydrogen attached to a carbon adjacent to the double bond - the allylic position.
Mechanism
The allylic bromination with NBS is analagous to the alkane halogenation reaction (Section 10.2) since it also occurs as a radical chain reaction. NBS is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl4), NBS reacts very rapidly with the HBr formed during the reaction mechanism to provide bromine (Br2) which is required for the reaction to continue. Under the correct conditions, NBS provides a constant but very low concentration of Br2 in the reaction mixture. The low concentration of Br2 helps to prevent the formation of unwanted side-products.
The mechanism starts with the formation of a small amount of bromine radical which then abstracts an allylic hydrogen to form an allylic radical and HBr. The HBr can then react with NBS to form the Br2 required for the reaction. The allylic radical then abstracts a bromine atom from Br2 to form the allyl halide product and a bromine radical. The bromine radical produced allows the reaction to continue.
The predominance of allylic substitution over other positions is based on bond dissociation energies. An allylic C-H bond has a strength of about 88 kcal/mol which is much weaker than a typical alkyl C-H bond (98 kcal/mol) or vinylic C-H bond (111 kcal/mol). Therefore, an allylic C-H bond is most likely to form a free radical and react.
Because a allylic C-H bond requires less energy to undergo homolytic cleavage than even a tertiary C-H bond, it can be inferred that a allylic radical is more stable than a tertiary radical. The ordering of stability in radicals can be expanded to include vinylic and allylic radicals. The enhanced stability of allyl radicals can be attributed to resonance stabilization which will be discussed in the next section. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.03%3A_Preparing_Alkyl_Halides_from_Alkenes_-_Allylic_Bromination.txt |
Objectives
After completing this section, you should be able to
1. explain the stability of the allyl radical in terms of resonance.
2. explain the difference between resonance and tautomerism.
3. write an equation for the reaction of an unsymmetrical alkene with N-bromosuccinimide.
4. draw the structure of each of the possible products that could be obtained from the reaction of a given unsymmetrical alkene with N-bromosuccinimide, and predict which product will predominate.
5. explain the formation of more than one product from the reaction of N-bromosuccinimide with a given unsymmetrical alkene.
6. explain the observed product ratio when a given unsymmetrical alkene is treated with N-bromosuccinimide.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• delocalized
• resonance forms
• resonance hybrid
Study Notes
You will have encountered the concept of resonance if you have taken general first-year chemistry course. You should also briefly review Section 2.5.
When we can represent a species by two or more different Lewis or Kekulé structures, neither of which represents the true structure of the species, these structures are referred to as resonance forms. A common example used in general chemistry courses to illustrate the concept of resonance is ozone, O3. The two resonance forms of ozone may be represented as follows:
The concept of resonance is quite important, and will be used frequently throughout the remainder of this course. The guidelines below may assist you in drawing resonance contributors.
1. Resonance occurs whenever a molecule, radical or ion can be represented by two or more structures differing only in the arrangement of electrons (no atoms may be moved).
2. The true structure of a species is a hybrid of the resonance contributors and is more stable (i.e., lower in energy) than any of the contributors.
3. The most important contributors are those containing the most covalent bonds. Another way of saying the same thing is that the most important contributors have the least amount of charge separation.
4. Contributors in which all the atoms (except hydrogen) have a complete octet (i.e., are surrounded by eight electrons) are particularly important.
In the previous section we discussed the allylic bromination of a symmetrical alkene with NBS such as this cyclopentene, which affords one product.
However, with an unsymmetrical alkene and the delocalized unpaired electron forming various allylic resonances, several products are possible. For example, the NBS bromination of 4-methyl-cyclohexene leads to three products.
The geometry and relative stability of carbon radicals
As organic chemists, we are particularly interested in radical intermediates in which the unpaired electron resides on a carbon atom. Experimental evidence indicates that the three bonds in a carbon radical have trigonal planar geometry, and therefore the carbon is considered to be sp2-hybridized with the unpaired electron occupying the perpendicular, unhybridized 2pzorbital. Contrast this picture with carbocation and carbanion intermediates, which are both also trigonal planar but whose 2pz orbitals contain zero or two electrons, respectively.
The trend in the stability of carbon radicals parallels that of carbocations (section 8.4B): tertiary radicals, for example, are more stable than secondary radicals, followed by primary and methyl radicals. This should make intuitive sense, because radicals, like carbocations, can be considered to be electron deficient, and thus are stabilized by the electron-donating effects of nearby alkyl groups.
Benzylic and allylic radicals are more stable than alkyl radicals due to resonance effects - an unpaired electron can be delocalized over a system of conjugated pi bonds. An allylic radical, for example, can be pictured as a system of three parallel 2pz orbitals sharing three electrons. With two resonance forms, the allylic radical is electronically symmetrical. Due to resonance hybrid theory, neither structure is correct, but instead the structure lies somewhere between the two resonance forms. Another way to phrase this is that the unpaired electron is delocalized across all the carbon atoms through the pi system and not localized on one site. The more resonance structures, the more stable the molecule. This is why the allylic radical is more stable than the alkyl radical.
Because the allylic radical is symmetrical, a reaction can occur on either side. Therefore if reacting with bromine, the bromination could occur on either end of the allylic radical. When the allyl radical is symmetrical, this yields the same product. However, if you have an unsymmetrical allyl radical, it would lead to a mixture of products and not necessarily in equal amounts. This is because the intermediate radical is unsymmetrical. The reaction will occur at the less hindered site. An example would be 1-octene as a starting material in a bromination. The products the reaction would yield would be 3-bromo-1-octene and 1-bromo-2-octene.
Further Reactions
The products from allylic bromination reactions can easily be converted into dienes by elimination using a base. If the alkyl chain is long enough, alkenes can be converted to dienes through a two-step process: allylic bromination followed by elimination.
Exercise \(1\)
1) The following reaction shows the major product. Explain why this would be the final product and why the 2° bromo product is not the major product.
2) Predict the products of the following reactions:
Answer
1) The product (A) is a 1° halogen which is more predominant product even though the (B) had a better transition state with a 2° radical. The 1° radical intermediate is not as sterically hindered.
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.04%3A_Stability_of_the_Allyl_Radical_-_Resonance_Revisited.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the conversion of an alcohol to an alkyl halide.
2. list a given series of alcohols in increasing or decreasing order of reactivity with hydrogen halides.
3. identify the alkyl halide formed when a given alcohol reacts with thionyl chloride, phosphorus tribromide, or a hydrogen halide.
4. identify the alcohol which should be used to prepare a given alkyl halide using one of the reagents specified in Objective 3.
5. select the most appropriate reagent for converting a given alcohol to a given alkyl halide.
Study Notes
The use of thionyl chloride for converting alcohols to alkyl chlorides has the added benefit that both of the by-products, sulfur dioxide and hydrogen chloride, are gases. This characteristic simplifies the isolation and purification of the reaction product.
In the laboratory, one can test for the presence of alcohols with Lucas reagent (a mixture of concentrated hydrochloric acid and zinc chloride). Lucas reagent converts alcohols to alkyl chlorides: tertiary alcohols fastest followed by secondary alcohols; primary alcohols do not react to any significant extent. Thus, Lucas reagent can help distinguish among primary, secondary and tertiary alcohols due to going through a substitution reaction.
General Reaction
When alcohols react with an acid halide, a substitution takes place producing an alkyl halide and water:
• The order of reactivity of alcohols is 3° > 2° > 1° > Methyl.
• The order of reactivity of the hydrogen halides is HI > HBr > HCl (HF is generally unreactive).
Tertiary alcohols react reasonably rapidly HCl, HBr, or HI, but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance. For the reactions that do occur, bubbling HX into an alcohol solution yields a haloalkane or alkyl halide. Below the reaction shows, tert-butanol and hydrochloric acid reacting to yield t-butyl chloride and water.
Mechanism
Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation, in an SN1 reaction with the protonated alcohol acting as a leaving group.
The SN1 mechanism is illustrated by the reaction of tert-butyl alcohol and aqueous hydrochloric acid (H3O+, Cl-). The first two steps in this SN1 substitution mechanism are protonation of the alcohol to form an oxonium ion. Protonation of the alcohol converts a poor leaving group (OH-) to a good leaving group H2O which makes the dissociation step of the SN1 mechanism more favorable.
In step 3, the carbocation reacts with a nucleophile (a halide ion) to complete the substitution.
Carbocation rearrangements are extremely common in organic chemistry reactions are are defined as the movement of a carbocation from an unstable state to a more stable state through the use of various structural reorganizational "shifts" within the molecule. Once the carbocation has shifted over to a different carbon, we can say that there is a structural isomer of the initial molecule.
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism. In these reactions the function of the acid is to produce a protonated alcohol. The halide ion then displaces a molecule of water (a good leaving group) from carbon; this produces an alkyl halide:
Again, acid is required. Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols that are not activated (converted to a better leaving group). Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion.
Conversion of Alcohols into Alkyl Halides Using SOCl2 or PBr3
The most common methods for converting 1º- and 2º-alcohols to the corresponding chloro and bromo alkanes (i.e. replacement of the hydroxyl group) are treatments with thionyl chloride (SOCl2) and phosphorus tribromide (PBr3), respectively. These reagents are generally preferred over the use of concentrated HX due to the harsh acidity of these hydrohalic acids and the carbocation rearrangements associated with their use.
Mechanisms
Both of these reagents form an alkyl halide through an SN2 mechanism. The mechanism for both reactions starts by converting the hydroxide of the alcohol into a better leaving group through formation of an intermediate. Thionyl chloride creates an intermediate chlorosulfite (-OSOCl2) compound and phosphorus tribromide makes an intermediate dibromophosphite (-OPBr2) compound. These intermediate compounds can subsequently be eliminated as a leaving group during an SN2 reaction with the corresponding nucleophilic halide ion. Since these reactions proceeds through a backside attack, there is inversion of configuration at the carbon
Thionyl chloride
Notice that during the reaction with thonyl chloride hydrochloric acid (HCl) and sulfur dioxide (SO2) are produced as byproducts.
Phosphorus tribromide
During this reaction HOPBr2 is made as a biproduct.
Stereochemical Considerations
The SN2 reaction with the corresponding nucleophilic halide ion contained in the mechanism of both reactions causes an inversion of configuration at the carbon atom.
Exercise \(1\)
1) Predict the alcohol required for the synthesis of the following halides:
Answer
1) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.05%3A_Preparing_Alkyl_Halides_from_Alcohols.txt |
Objectives
After completing this section, you should be able to
1. write an equation to describe the formation of a Grignard reagent.
2. give examples of Grignard reagents formed from aryl and vinyl halides as well as from alkyl halides.
3. explain the reactivity of Grignard reagents in terms of the polarity of the carbon-magnesium bond.
4. write an equation for the reaction of a Grignard reagent with a proton donor, such as water.
5. predict the product formed from the reaction of a given organohalide with magnesium followed by a proton donor.
6. identify the organohalide, the reagents, or both, needed to prepare a given alkane.
7. describe how a deuterium atom may be introduced at a specific location in an organic molecule through use of a Grignard reagent.
8. describe at least one limitation on the use of Grignard reagents in organic synthesis.
9. write an equation for the direct conversion of an alkyl halide to an alkane using a hydride donor, such as lithium aluminum hydride.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• carbanion
• Grignard reagent
Study Notes
The organomagnesium compounds formed by the reaction of an alkyl or aryl halide with magnesium are called Grignard reagents. As you will see throughout the remainder of this course, Grignard reagents can be used to synthesize a wide range of organic compounds and are extremely useful to the organic chemist.
In the introductory section, we tried to stress that the chemistry of alkyl halides is quite different from that of aryl (or vinyl) halides. However, both alkyl and aryl halides react with magnesium to form Grignard reagents.
The reaction of a Grignard reagent with D2O (“heavy water”) provides a convenient method for introducing a deuterium atom (remember D is equivalent to 2H) into a molecule at a specific location. For example:
Formation of Organometallic Reagents
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, of which the alkali metals are stronger. These same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to a halide anion, and the carbon bonds to the metal which has characteristics similar to a carbanion (R:-). Most alkyl halides can undergo this reaction including: 1o, 2o, 3o, vinyl, and aryl. Halide reactivity in these reactions increases in the order: Cl < Br < I and Fluorides are usually not used.
Many organometallic reagents are commercially available, however, it is often necessary to make then. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination). The alkyl magnesium halides described in the first reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them and received the Nobel prize in 1912 for this work. The products of the second reaction are called Alkyllithium Reagents The other metals mentioned above react in a similar manner, but Grignard and Alkylithium Reagents are most widely used. Although the formulae drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.
Example \(1\)
These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). The Polarity change is evident when looking at the electrostatic maps of the alkyl halide and the Grignard reagent. In the alkyl halide, the electronegative halogen atom holds much of the electron density which gives the carbon a slight positive charge (shown in blue in the electrostatic potential map below) making it electrophilic. In the Grignard reagent, the carbon has a slight negative charge making it nucleophilic (shown in yellow & red).
A suitable solvent must be used for reactions that form organometallic molecules. For alkyl lithium formation pentane or hexane are usually used. Diethyl ether can also be used but the subsequent alkyl lithium reagent must be used immediately after preparation due to an interaction with the solvent. Diethyl ether or tetrahydrofuran (THF) are essential for Grignard reagent formation. Lone pair electrons from two ether molecules form a complex with the magnesium in the Grignard reagent (as pictured below). This complex helps stabilize the organometallic and increases its ability to react.
Organometallic Reagents as Bases
These reagents are very strong bases (pKa's of saturated hydrocarbons range from 42 to 50). Although not usually done with Grignard reagents, organolithium reagents can be used as strong bases. Both Grignard reagents and organolithium reagents react with water to form the corresponding hydrocarbon. This is why so much care is needed to insure dry glassware and solvents when working with organometallic reagents.
In fact, the reactivity of Grignard reagents and organolithium reagents can be exploited to create a new method for the conversion of halogens to the corresponding hydrocarbon (illustrated below). The halogen is converted to an organometallic reagent and then subsequently reacted with water to from an alkane.
Exercise \(1\)
1) How strong of a base would you expect ethyl Grignard to be? Would the following reactions be able to take place?
2) How would you make a deuterated compound from an alkyl halide?
Answer
1) Because hydrocarbons like ethane are very weak acids (pKa = 50), then the corresponding carbanion (CH3CH2-) is expected to be a strong base. Both reactions will occur.
2) By first making a Grignard and then exposing it to heavy water. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.06%3A_Reactions_of_Alkyl_Halides_-_Grignard_Reagents.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the formation of an alkyllithium from an alkyl halide.
2. write an equation for the formation of a lithium dialkylcopper (Gilman) reagent from an alkyllithium and copper(I) iodide.
3. write an equation for the coupling of a lithium dialkylcopper reagent with an alkyl halide (i.e., a Corey-House synthesis).
4. draw the structure of the product formed from a given Corey-House synthesis.
5. identify the reagents needed to convert two given organohalides to a specified hydrocarbon through a Corey-House synthesis.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Corey-House synthesis
• pheromone
Study Notes
A pheromone is a chemical released by members of one species to cause specific behavioural or physiological changes in other members of the same species. Examples include sex pheromones, alarm pheromones and trail pheromones.
The Corey-House synthesis provides us with a method of coupling together two alkyl groups through the formation of a new carbon-carbon bond. The product of such a reaction is an alkane, and this synthetic method gives us a route for the preparation of unsymmetrical alkanes. The method was developed during the late 1960s by E. J. Corey and Herbert House working independently at Harvard University and Massachusetts Institute of Technology, respectively. The overall synthetic route is shown on the next page. Note that R and R′ represent alkyl groups, which can be the same or different, and X represents a halogen (preferably bromine or iodine).
In order to obtain a good yield of alkane, both R′X and RX should be primary alkyl halides. However, the experimental procedure can be modified so that this synthesis can be carried out using a wide range of alkyl, aryl, vinyl, benzyl and allyl halides. A detailed discussion of these modifications is beyond the scope of this course, but you should be aware of possible limitations in the use of the Corey-House synthesis.
Note: In some textbooks, lithium diorganocoppers are referred to as lithium dialkylcoppers, or cuprates.
Gilman Reagents
Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium diorganocopper reagent (also called diorganocuprates), which often is referred to as a Gilman reagent. Remember that organolithium reagents are formed by a reaction of lithium metal with an organohalide. Lithium diorganocopper reagents are considered a source of carbanion-like nucleophiles similar to Grignard and Organolithium reagents. However, the reactivity of lithium diorganocuprate reagents is slightly different and this difference will be exploited in specific situations. Diorganocuprate reagents are made from the reaction of two equivalents of an organolithium reagent and copper (I) iodide (CuI). The created lithium diorganocuprate reagent acts as a source of R:-
Lithium Diorganocopper (Gilman Reagent)
Coupling Reactions with Gilman Reagents
Gilman reagents undergo a coupling reaction with organochlorides, bromides, and iodides to form a carbon-carbon bond. During the reaction one of the alkyl groups from the Gilman reagent replaces the halogen atom in the organohalide. This reaction is useful in organic synthesis because it allows a larger molecule to be built from smaller fragments. Most alkyl halides, including aryl and vinyl halides, are capable of undergoing this reaction.
Mechanism
The mechanism begins with copper in the Gilman reagent being oxidized from Cu+1 to Cu+3. Losing electrons allows the copper atom to act as a nucleophile and undergo an SN2 like substitution reaction with the alkyl halide. The resulting neutral tri-organocopper intermediate undergoes reductive elimination, reducing Cu+3 to Cu+1, and creating the alkyl coupling product.
The Suzuki-Miyaura Reaction
The reactions of diorganocopper reagents with organohalides are related to the processes that occur with other organometallic reagents such as organopalladiums. Suzuki-Miyaura coupling (or Suzuki coupling) is a metal-catalyzed reaction, typically with Pd, between an alkenyl (vinyl), aryl, or alkynyl organoborane (boronic acid or boronic ester, or special cases with aryl trifluoroborane) and halide or triflate under basic conditions. This reaction is used to create carbon-carbon bonds to produce conjugated systems of alkenes, styrenes, or biaryl compounds. Organopalladium reagents are often preferred over diorganocopper reagents, because only a catalytic amount of the organopalladium is necessary rather than a full equivalent which makes the reaction less toxic.
Mechanism
The general catalytic cycle for Suzuki cross coupling involves three fundamental steps: oxidative addition, transmetalation, and reductive elimination as demonstrated in the figure below.
The oxidative addition of aryl halides to the Pd(0) complex gives an intermediate 1, a Pd(II) species. Under the participation of base, an organoborane compound reacts with intermediate 1 in a transmetalation step to afford intermediate 2.
This is followed by reductive elimination to give the desired coupling product and regenerate the original Pd(0) species. Depending on different catalytic systems with various catalysts, ligands, and solvents, there are additional processes in the catalytic cycle, including ligand or solvent association and dissociation.
Exercise \(1\)
1) Starting with alkyl halides containing no more than four carbon atoms, how would you synthesize each of the following alkanes?
a) 2,5-dimethylhexane
b) 2-methylhexane
Answer
1) Notice that in (a), both the alkyl halides are primary. This fact should ensure a good yield of product. In (b) we have the choice of using 2-bromopropane and 1-bromobutane, or 1-bromo-2-methylpropane and 1-bromopropane. We chose the latter as it enables us to use two primary alkyl halides, and hence a simpler procedure.
a)
b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.07%3A_Organometallic_Coupling_Reactions.txt |
Objectives
After completing this section, you should be able to
1. identify organic reactions as being oxidations, reductions, or neither.
2. rank given compounds in order of their oxidation level.
Key Terms
Make certain that you can define, and use in context, the terms below.
• oxidation
• reduction
• heteroatom
General Redox Reactions
General chemistry courses describe oxidation and reduction - when a compound or atom is oxidized it loses electrons, and when it is reduced it gains electrons. Also, oxidation and reduction half reactions occur in pairs: if one species is oxidized, another must be reduced at the same time. Thus, the combination of an oxidation and a reduction half reaction is termed a 'redox reaction.' Most of the redox reactions you have seen previously in general chemistry typically involved the flow of electrons from one metal to another, such as the reaction between copper ion in solution and metallic zinc:
$Cu^{+2}_{(aq)} + Zn_{(s)} → Cu_{(s)} + Zn^{+2}_{(aq)} \nonumber$
$Cu^{+2}_{(aq)} + 2 e{-} → Cu_{(s)} \nonumber$ Reduction
$Zn_{(s)} → Zn^{+2}_{(aq)} + 2 e{-} \nonumber$ Oxidation
In order, to keep track of electrons in organic molecules an oxidation state formalism is used. Oxidation states do not represent the actual charge on an atom, but it will allow the number of electrons being gained or lost by a particular atom to be determined during a reaction.
To calculate the oxidation state of a carbon atom the following rules are used:
1. A C-C bond does not affect the oxidation state of a carbon. So a carbon attached to 4 carbons has an oxidation state of zero.
2. Every C-H bond will decrease the oxidation state of the carbon by 1.
3. Each C-X bond will increase the oxidation state of the carbon by 1. Where X is an electronegative atom, such as nitrogen, oxygen, sulfur, or a halogen.
When looking at the oxidation states of carbon in the common functional groups shown below it can be said that carbon loses electron density as it becomes more oxidized. We'll take a series of single carbon compounds as an example. Methane (CH4) is at the lowest oxidation level of carbon because is has the maximum possible number of bonds to hydrogen. Carbon dioxide (CO2) is at the highest oxidation level because it has the maximum number of bonds to an electronegative atom.
This pattern holds true for the relevant functional groups on organic molecules with two or more carbon atoms:
Organic Redox Reactions
It is important to be able to recognize when an organic molecule is being oxidized or reduced, because this information tells you to look for the participation of a corresponding redox agent that is being reduced or oxidized. If a reaction converts a compound to a higher oxidation level that is an oxidation. If it converts a compound to a lower oxidation level it is a reduction. If the oxidation level of the reactant does not change it is not a redox reaction.
Two examples of organic redox reactions are shown below. The conversion of an alcohol to a ketone is considered an oxidation because the oxidation level of the carbon increases from 0 to +2. This implies that the reaction would require an oxidizing agent. Likewise, the conversion of a ketone to an alcohol is a reduction and would require a reducing agent. The conversion of an alkane to alkene is an oxidation because the oxidation state on both carbons is increasing while the reverse reaction would be a reduction.
Now reaction previously discussed in this textbook can be considered to determine if they are in fact redox reaction. The free radical bromination of methane to bromomethane would be an oxidation because the oxidation level of carbon is raised from -4 to -3.
The electrophilic addition of Br2 to an alkene to provide a 1,2-dibromide is an oxidation because both carbons increase their oxidation level from -2 to -1. However, the electrophilic addion of HBr to an alkene to provide an alkyl halide is not a redox reaction because the overall oxidation state of carbons involved are not changed. One carbon has its oxidation level decreased from -2 to -3 while the other carbon's oxidation level is increased from -2 to -1. Overall, the change in oxidation level cancels out to leave an overall change of oxidcation level in the compound of 0.
You should learn to recognize when a reaction involves a change in oxidation state in an organic reactant . Looking at the following transformation, for example, you should be able to quickly recognize that it is an oxidation: an alcohol functional group is converted to a ketone, which is one step up on the oxidation ladder.
Likewise, this next reaction involves the transformation of a carboxylic acid derivative (a thioester) first to an aldehyde, then to an alcohol: this is a double reduction, as the substrate loses two bonds to heteroatoms and gains two bonds to hydrogens.
An acyl transfer reaction (for example the conversion of an acyl phosphate to an amide) is not considered to be a redox reaction - the oxidation state of the organic molecule is does not change as substrate is converted to product, because a bond to one heteroatom (oxygen) has simply been traded for a bond to another heteroatom (nitrogen).
It is important to be able to recognize when an organic molecule is being oxidized or reduced, because this information tells you to look for the participation of a corresponding redox agent that is being reduced or oxidized- remember, oxidation and reduction always occur in tandem! We will soon learn in detail about the most important biochemical and laboratory redox agents.
Worked Example $1$
1) Rank the following compounds in order of increasing oxidation level:
Answer
The easiest way to solve this problem is to calculate the oxidation level of the carbon in each compound. Remembering that hydrogens decrease the oxidation level by one, electronegative elements increase the oxidation level by one, and carbons do not change the oxidation level, the oxidation level of each carbon can be calculate. The carbon in CH3OH has three bonds to hydrogens and one bond to oxygen so it oxidation level is 3(-1) + 1(+1) = -2. The carbon in HCN has one bond to hydrogens and three bonds to nitrogen so it oxidation level is 1(-1) + 3(+1) = +2. The carbon in CH2NH has two bonds to hydrogens and two bond to nitrogen so it oxidation level is 2(-1) + 2(+1) = 0. The carbon in CCl4 has zero bonds to hydrogens and four bonds to chlorine so it oxidation level is o(-1) + 4(+1) = +4. The compounds now can be listing in the following order of increasing oxidation level.
Exercise $1$
1) In each case state whether the reaction is an oxidation or reduction of the organic compound.
Answer
1)
A – Reduction
B – Oxidation | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.08%3A_Oxidation_and_Reduction_in_Organic_Chemistry.txt |
Concepts & Vocabulary
10.1 Introduction to Organohalides
• Alkyl halides (and allyl and benzyl halides) are more reactive than vinyl and aryl halides.
10.2 Names and Properties of Alkyl Halides
• Reactivity of alkyl halides is often related to the substitution of the carbon atom the halogen is attached to.
• Alkyl halides are categorized by the number of bonds to other alkyl groups (primary, secondary, and tertiary).
• Carbon-halogen bonds are polarized with partial positive charges on carbon and partial negative charges on the halogen.
• Fluorine is the most electronegative of the halogens while iodine is the least electronegative.
• Iodine is the largest of the halogens yielding the longest/weakest bonds to carbon of the halogens.
• Since haloalkanes have dipole-dipole interactions, they have greater intermolecular forces than similar sized alkanes and therefore higher boiling points.
• Alkyl halides are either slightly soluble or insoluble in water, but are soluble in organic solvents.
10.3 Preparing Alkyl Halides from Alkanes - Radical Halogenation
• Halogenation of alkanes is exothermic, so it is energetically favorable.
• Radical chain mechanisms consist of three steps: initiation, propagation and termination.
• Hydrogens on more substituted carbon atoms are more reactive to radical halogenation.
10.4 Preparing Alkyl Halides from Alkenes - Allylic Bromination
• More substituted radicals and radicals with resonance structures are more stable than other radicals.
• Radical substitution can be carried out at the allylic or benzylic carbon by reacting with NBS.
10.5 Stability of the Allyl Radical - Resonance Revisited
• Allyl cations, anions and radicals have resonance structures. To draw these resonance structures non-bonded and pi-bond electrons can be moved.
• Resonance hybrids are used to show the combination of all resonance structures for a molecule or ion.
10.6 Preparing Alkyl Halides form Alcohols
• Alcohols can be reacted with hydrohalogen acids or a mixture of halogen salts and a stronger acid (to form hydrohalogen acids in situ).
• Alcohols will also react with thionyl chloride or with phosphorus halides to from haloalkanes.
10.7 Reactions of Alkyl Halides - Grignard Reactions
• Organometallic reagents can be formed from alkyl halides and reactive metals (such as lithium and magnesium).
• Alkyl magnesium halide compounds are callled Grignard reagents.
• Grignard reagents react as bases where the alkyl group gets protonated and the metal complexes to the conjugate base of the reacting acid.
10.8 Organometallic Coupling Reactions
• Lithium dialkyl copper compounds are called Gilman reagents.
• Gilman reagents have different reactivity from the other organometallics (lithium and Grignard reagents).
• Organometallics can be reacted with alkyl halides to join to alkyl groups (coupling reactions).
10.9 Oxidation and Reduction in Organic Chemistry
• Gaining bonds to hydrogen for organic molecules is reduction.
• Losing bonds to hydrogen for organic molecules is oxidation.
Skills to Master
• Skill 10.1 Differentiate between types of halides (alkyl, allyl, aryl, benzyl, and vinyl).
• Skill 10.2 Differentiate between substitution of alkyl halides (primary, secondary, and tertiary).
• Skill 10.3 Identify relative reactivity of carbon-hydrogen bonds to radical halogenation.
• Skill 10.4 Draw resonance structures for radical compounds.
• Skill 10.5 Draw mechanisms for radical halogenation of alkanes (initiation, propagation and termination).
• Skill 10.6 Calculate the enthalpy change of a reaction using bond dissociation energies of reactants and products.
• Skill 10.7 Determine products for allylic bromination reactions.
• Skill 10.8 Draw resonance structures for allylic and other similar compounds and ions.
• Skill 10.9 Draw products of reactions of alcohols to form alkyl halides.
• Skill 10.10 Write equations to form Grignard reagents from alkyl halides.
• Skill 10.11 Draw reaction products for Grignard reagents acting as bases.
• Skill 10.12 Write equations for the formation of Gilman reagents.
• Skill 10.13 Draw reaction products of organometallic coupling reactions.
• Skill 10.14 Explain oxidation and reduction in organic molecules.
Summary of Reactions
Preparation of Alkyl Halides
Reactions Alkyl Halides | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/10%3A_Organohalides/10.S%3A_Organohalides_%28Summary%29.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
Objective
After completing this section, you should be able to identify substitution and elimination as being the two most important reactions of alkyl halides.
Study Notes
Alkyl halides are electrophiles, which means they can undergo nucleophilic substitution and base-induced elimination reactions. These reaction types offer a large and useful range of reactions for organic synthesis in the laboratory.
The Reactions
Two reactions are shown her with both involving heating a halogenoalkane under reflux with sodium or potassium hydroxide solution. Two different reactions can occur.
Nucleophilic substitution
The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction.
In the example, 2-bromopropane is converted into propan-2-ol.
Elimination
Hydroxide ions are also strong bases, therefore halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide.
In this reaction, the 2-bromopropane has reacted to form an alkene - propene.
Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons.
What decides whether you get substitution or elimination?
The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors.
11.01: The Discovery of Nucleophilic Substitution Reactions
Objectives
After completing this section, you should be able to
1. write an equation to represent the Walden inversion.
2. write a short paragraph describing the Walden inversion.
3. describe, using equations, a series of reactions interconverting two enantiomers of 1-phenyl-2-propanol which led to the conclusion that nucleophilic substitution of primary and secondary alkyl halides proceeds with inversion of configuration.
Study Notes
The IUPAC name for malic acid is 2-hydroxybutanedioic acid. This acid is produced by apples, a fact which seems to have been appreciated by the British novelist Thomas Hardy in The Woodlanders:
Up, upward they crept, a stray beam of the sun alighting every now and then like a star on the blades of the pomace-shovels, which had been converted to steel mirrors by the action of the malic acid.
In 1896, the German chemist Paul Walden discovered that he could interconvert pure enantiomeric (+) and (-) malic acids through a series of reactions. This conversion meant that there was some kind of change in the stereochemistry made during the series of reactions. First, (−)-malic acid was reacted with phosphorus pentachloride (PCl5) to provide (+)-chlorosuccinic acid.
This was reacted with silver(I)oxide (Ag2O) to provide (+)-malic acid. These two combined steps caused an inversion of stereochemistry of (−)-malic acid to (+)-malic acid. The reaction series was then continued to convert (+)-malic acid back into (−)-malic acid by further reaction with PCl5 and Ag2O.
These results were considered astonishing. The fact that (−)-malic acid was converted into (+)-malic acid meant that the configuration of the chiral center has somehow been changed during the reaction series.
These reactions are currently referred to as nucleophilic substitution reactions because each step involves the substitution of one nucleophile by another. These are among the most common and versatile reaction types in organic chemistry.
Further investigations into these reaction were undertaken during the 1920's and 1930's to clarify the mechanism and clarify how the inversion of configurations occur. These reactions involved nucleophilic substitution of an alkyl p-toluenesulfonate (called a tosylate group). For this purpose the tosylate groups act similarly to a halogen substituent. In the series of reactions (+)-1-phenyl-2-propanol is interconverted with (-)-1-phenyl-2-propanol.
Somewhere in this three-step series of reactions the configuration at a chiral center is being inverted. In the first step the tosylate is formed without breaking the C-O bond of the chiral center, which means the configuration is unchanged. Similarly, the cleavage of the ester in step three occurs without breaking the C-O bond of the chiral center, which also means the configuration of the chiral carbon is unaffected. It was determined that the second step where acetate nucleophile undergoes a substitution with tosylate was causing the stereochemical configuration to be inverted.
Exercise \(1\)
1) Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH3CO2-, Show stereochemistry.
Answer
1) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.01%3A_Introduction.txt |
Objectives
After completing this section, you should be able to
1. write an expression relating reaction rate to the concentration of reagents for a second-order reaction.
2. determine the order of a chemical reaction from experimentally obtained rate data.
3. describe the essential features of the SN2 mechanism, and draw a generalized transition state for such a reaction.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bimolecular
• kinetics
• rate coefficient
• rate equation
• reaction rate
• second-order reaction
• SN2
Study Notes
Most of the key terms introduced in this section should already be familiar to you from your previous general chemistry course.
Reaction rate refers to the change in concentration of a reactant or product per unit of time. Using strict SI units, reaction rates are expressed in mol · L−1 · s−1, but in some textbooks you will find this value written as M/s. In general, the reaction rate of a given reaction changes with time, as it is dependent on the concentration of one or more of the reactants.
An equation which shows the relationship between the reaction rate and the concentrations of the reactants is known as the rate equation. All rate equations contain a proportionality constant, usually given the symbol k, which is known as the rate coefficient. Some textbooks refer to this value as the “rate constant,” but this name is a little misleading as it is not a true constant. The rate coefficient of a given reaction depends on such factors as temperature and the nature of the solvent.
SN2 is short for “bimolecular nucleophilic substitution.” You will encounter abbreviations for other types of reactions later in this chapter.
If you are unclear on the point about the inversion of configuration during an SN2 reaction, construct a molecular model of a chiral alkyl halide, the transition state formed when this substance reacts with a nucleophile in an SN2 process, and the product obtained from this reaction.
Brønsted-Lowry acid-base reactions
In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base. Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. We will generalize the three other groups bonded on the electrophilic central carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction.
The SN2 Mechanism
Bimolecular nucleophilic substitution (SN2) reactions are concerted, meaning they are a one step process. This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electrophilic carbon and the halogen. This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. The mechanism starts when lone pair electrons from the nucleophile attacks the electrophilic carbon of the alkyl halide to form a C-Nu sigma bond. Simultaneously, X-C bond is broken when the electrons are pushed onto the leaving group. Overall during this mechanism, a set of lone pair electrons are transferred from the nucleophile to the leaving groups.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way.
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. SN2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans. If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Bimolecular Nucleophilic Substitution Reactions and Kinetics
In the term SN2, (as previously stated) the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane.
If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate.
If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate.
The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics:
SN2 Reactions Are Stereospecific
The SN2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an R enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the S enantiomer.
Conversely, if the substrate is an S enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the R enantiomer.
In conclusion, SN2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans. If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis.
Exercise \(1\)
1) The reaction below follows the SN2 mechanism.
a) Write the rate law for this reaction.
b) Determine the value of the rate coefficient, k, if the initial concentrations are 0.01 M CH3Cl, 0.01 M NaOH, and the initial reaction rate is 6 x 10-10 M/s.
c) Calculate the initial reaction rate if the initial reactant concentrations are changed to 0.02 M CH3Cl and 0.0005 M NaOH.
2) Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH3CO2-, Show stereochemistry.
3) Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
4) Since everything is relative in chemistry, one reaction's nucleophile can be another reaction's leaving group. Some functional groups can only react as a nucleophile or electrophile, while other functional groups can react as either a nucleophile or electrophile depending on the reaction conditions. Classify the following compounds as nucleophile, electrophile, or leaving groups. More than one answer may be possible.
a) bromoethane
b) hydroxide
c) water
d) chlorocyclohexane
e) ethanol
f) bromide
Answer
1) a) rate = k [CH3Cl] [OH-]
b) substitute the data into the rate expression above and apply algebra to solve for k
k = 6 x 10-6 Lmol-1s-1
c) Using the rate law above, substitute the value for k from the previous question along with the new concentrations to determine the new initial rate.
rate = 6 x 10-10 M/s
$\begin{array}{r}2\right)\end{array}$
3)
4)
a) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
b) strong nucleophile
c) weak nucleophile and good leaving group
d) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
e) weak nucleophile, a poor electrophile without clever chemistry (stay tuned for future chapters), good leaving group
f) good nucleophile and a good leaving group | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.02%3A_The_SN2_Reaction.txt |
Objectives
After completing this section, you should be able to
1. discuss the role of steric effects in SN2 reactions.
2. arrange a given series of alkyl halides in order of increasing or decreasing reactivity towards nucleophilic substitution through the SN2 mechanism.
3. suggest a reason why vinyl halides and aryl halides do not undergo SN2 reactions.
4. discuss how the nature of the nucleophile affects the rate of an SN2 reaction.
5. arrange a given series of common nucleophiles (e.g., CN, I, Br Cl, H2O) in order of increasing or decreasing nucleophilicity.
6. discuss how the nature of the leaving group affects the rate of an SN2 reaction.
7. arrange a given series of leaving groups in order of increasing or decreasing ability to leave during an SN2 reaction.
8. discuss the role played by the solvent in an SN2 reaction.
9. give examples of the solvents which are commonly used for SN2 reactions, and identify those that promote a high reaction rate.
10. predict which of two given SN2 reactions will proceed faster, by taking into account the structure of the substrates, the nucleophiles involved, leaving-group ability, solvent effects, or any combination of these factors.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• leaving group
• polar aprotic solvent
• solvation
Study Notes
You may wish to review the discussion of acid-base theory given in Sections 2.7-2.11.
Both aryl and vinylic halides are relatively unreactive in SN2 displacement mechanisms, mostly because during the backside attack of the molecule the incoming nucleophile is sterically hindered by both substituents and electron density from any double bonds present. Also, leaving groups on sp2-hybridized carbons tend to be held tighter than sp3-hybridized carbons.
Solvation may be defined as the interaction between molecules of solvent and particles of solute. The result of solvation is to stabilize (i.e., lower the energy of) the solute particles. Solvents with lone pairs of electrons are good at solvating cations. Protic (i.e., hydroxylic) solvents are able to solvate anions through hydrogen bonding. As water has two lone pairs of electrons and is also protic, it is good at solvating both anions and cations.
Potential Energy Diagram for an SN2 Reaction
The potential energy diagram for an SN2 reaction is shown below. This is a single-step, concerted process so a single transition state is formed. A transition state, unlike a reaction intermediate, is a very short-lived species that cannot be isolated or directly observed. For an SN2 reaction the transition state directly determines the energy of activation which must be overcome for the reaction to occur. Factors which affect the stability of the transition state affect the rate of the SN2 reaction.
For an SN2 reaction, the transition state represents the half way point of the reaction. The C-Nu bond is in the process of forming and is represented by a dashed bond. The X-C bond is in the process of breaking and is also represented by a dash bond. The negative charge of the nucleophile is in the process of being transfered to the leaving groups. This is represented by both the nucleophile and the leaving group having a partial negative charge. The electrophilic carbon and the three 'R' substituents all lie on the same plane. Brackets and a double-dagger are placed around the structure to represent that it is a transition state.
Sterically Hindered Substrates Will Reduce the SN2 Reaction Rate
Steric hindrance about the electrophilic carbon, is one of the most important factors Determining the rate of SN2 reactions. Although the substrate, in the case of nucleophilic substitution of haloalkanes, is considered to be the entire molecule circled below we are most interested in the electrophilic carbon that bears the leaving group.
The SN2 transition state is very crowded. Recall that there are a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents.
SN2 Transition State
If each of the three substituents were hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion created in the planar portion of the transition state thereby increasing the ease at which the nucleophilic substitution reaction occurs. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion created in the planar portion of the transition state. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion in the transition state.
Steric repulsion of SN2 Transition States
How does steric hindrance affect the rate at which an SN2 reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. The increases steric hindrance destabilizes the transition state causing it to become higher in energy. This in turn increases the energy of activation and decreases the reaction rate.
The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked.
Vinyl and aryl halides are unreactive toward SN2 displacement due to extreme steric factors. To perform an SN2 reaction, the incoming nucleophile would have to enter the plane of the C-C double bond and move through the molecule to achieve the backside displacement required.
In addition to alkyl groups being added to the electrophilic carbon, it turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well.
In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction.
If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect.
The Nucleophile
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Some confusion in distinguishing basicity (base strength) and nucleophilicity (nucleophile strength) is inevitable. Since basicity is a less troublesome concept; it is convenient to start with it. Basicity refers to the ability of a base to accept a proton. Basicity may be related to the pKa of the corresponding conjugate acid, as shown below. The strongest bases have the weakest conjugate acids and vice versa. The range of basicities included in the following table is remarkable, covering over fifty powers of ten! In general, as the pKa of the conjugate acid increases the base becomes a stronger nuclophile. This, however, is not always the case.
Base I (–) Cl (–) H2O CH3CO2(–) RS(–) CN(–) RO(–) NH2(–) CH3(–)
Conj. Acid HI HCl H3O(+) CH3CO2H RSH HCN ROH NH3 CH4
pKa -9 -7 0.0 4.8 8 9.1 16 33 48
Nucleophilicity is a more complex property. Any compound, neutral or charged, which has lone pair electrons can act a neuclophile. After a SN2 reaction neutral nucleophiles make a positively charges products and negatively charged nucleophiles make neutral products. In general, negatively charges compound tend to make better nucleophilies.
Specifically, nucleophilicity refers to the rate of substitution reactions at the halogen-bearing carbon atom of a reference alkyl halide, such as CH3-Br. The nucleophilicity of the Nu:(–) reactant in the following substitution reaction varies as shown in the chart below:
Nucleophilicity is a more complex property. It commonly refers to the rate of substitution reactions at the halogen-bearing carbon atom of a reference alkyl halide, such as CH3-Br. Thus the nucleophilicity of the Nu:(–) reactant in the following substitution reaction varies as shown in the chart below:
Nucleophilicity: H2O < CH3CO2(–) < NH3 < Cl(–) < Br(–) < HO(–) < CH3O(–) < I(–) < CN(–) < CH3S(–)
The versatility of the SN2 reaction is shown by the wide variety of functional groups which can be formed.
Nucleophile Products
Formula Name Formula Name Functional Group
CH3S(–) Methanethiolate CH3SCH3 Dimethylsulfide Sulfide
CN(–) Cyanide CH3CN Acetonitrile Nitrile
I(–) Iodide CH3I Iodomethane Alkyl Halide
CH3O(–) Methoxide CH3OCH3 Diethylether Ether
HO(–) Hydroxide CH3OH Methanol Alcohol
Br(–) Bromide CH3Br Bromomethane Alkyl Halide
Cl(–) Chloride CH3Cl Chloromethane Alkyl Halide
NH3 Ammonia CH3NH3(+) Methylammonium ion Ammonium
CH3CO2 (–) Acetate CH3O2CH3 Methyl acetate Ester
H2O Water CH3OH2(+) Methylhydronium ion Protonated Alcohol
Periodic Trends in Nucleophilicity
There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity:
The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond. This horizontal trend also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions.
Recall that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity).
As Size Increases, Basicity Decreases: In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons and act as a nucleophile.
Resonance Effects on Nucleophilicity
Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance.
The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group.
The Leaving Group
As Electronegativity Increases, The Ability of the Leaving Group to Leave Increases
As mentioned previously, if we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basicity decreases, and the ability of the leaving group to leave increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them. The following diagram illustrates this concept, showing -CH3 to be the worst leaving group and F- to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that SN2 reactions of fluoroalkanes are rarely observed.
As Size Increases, The Ability of the Leaving Group to Leave Increases: Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Influence of the Solvent in an SN2 Reaction
The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile, because of strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile. In order for the nucleophile to attack the electrophile, it must break free, at least in part, from its solvent cage. A less powerful nucleophile in turn means a slower SN2 reaction. SN2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability.
Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO).
Below are several polar aprotic solvents that are commonly used in the laboratory:
These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
Summary of Factors in Affecting SN2 Reactions
There are four main factors which affect SN2 reaction:
1) The structure of the alkyl portion of the substrate: SN2 reactions are affected by steric hindrance around the electrophilic carbon. As steric hindrance increases the rate of SN2 reactions decrease. Methyl and primary substrates work well for SN2 reactions, secondary substrates react slowly, and teritary substrates do not undergo SN2 reactions at all.
2) The reactivity of the nucleophile: The rate of SN2 reaction is increased when strong nucleophiles are used. Strong nucleophiles tend to be negatively charged and good bases. Also, being of an increases size tends to increase the nucleopihlicity of an atom.
3) The nature of the leaving group: The rate of SN2 reaction is increased when leaving groups are used in the substrate. Good leaving groups tend to stabilize the electrons gained during an SN2 reaction. High electronegativity, resonance, and an increases size all can stabilize electrons.
4) The solvent: SN2 reaction are severely hindered when a polar protic solvent is used for the reaction. Instead, polar aprotic solvents tend to be used for SN2 reaction.
Example \(1\)
In each pair (A and B) below, which electrophile would be expected to react more rapidly in an SN2 reaction with the thiol group of cysteine as the common nucleophile?
Solution (8.13)
Exercise \(1\)
1) What product(s) do you expect from the reaction of 1-bromopentane with each of the following reagents in an SN2 reaction?
a) KI
b) NaOH
c) CH3C≡C-Li
d) NH3
2)
Which in the following pairs is a better nuceophile?
a) (CH3CH2)2N- or (CH3CH2)2NH
b) (CH3CH2)3N or (CH3CH2)3B
c) H2O or H2S
3) Order the following in increasing reactivity for an SN2 reaction.
CH3CH2Br CH3CH2OTos (CH3CH2)3CCl (CH3CH2)2CHCl
4) Solvents benzene, ether, chloroform are non-polar and not strongly polar solvents. What effects do these solvents have on an SN2 reaction?
5) Predict the products of these nucleophilic substitution reactions, including stereochemistry when appropriate.
a)
b)
d)
d)
6) Predict which compound in each pair would undergo the SN2 reaction faster.
a) or
b) or
c) or
d) or
Answer
1)
2)
a) (CH3CH2)2N- as there is a charge present on the nitrogen.
b) (CH3CH2)3N because a lone pair of electrons is present.
c) H2O as oxygen is more electronegative.
3)
4) They will decrease the reactivity of the reaction.
5)
a)
b)
c) No reaction
d)
6)
a)
b)
c)
d) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.03%3A_Characteristics_of_the_SN2_Reaction.txt |
Objectives
After completing this section, you should be able to
1. write an expression relating reaction rate and reactant concentration for a first-order reaction.
2. compare the kinetics of SN1 and SN2 reactions.
3. identify the rate-limiting step for a reaction, given the reaction energy diagram.
4. sketch a reaction energy diagram for a reaction, given the mechanism and sufficient information to identify the rate-limiting step.
5. write the mechanism of a typical SN1 reaction, and discuss the important features of the mechanism.
6. discuss the stereochemistry of an SN1 reaction, and explain why a racemic mixture is expected when substitution takes place at the chiral carbon atom of an optically pure substrate.
7. explain why unimolecular nucleophilic substitution at the chiral carbon atom of an optically pure substrate does not result in complete racemization.
8. compare the stereochemical consequences of the SN1 mechanism with those of the SN2 mechanism.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• first-order reaction
• rate-limiting step
• SN1
Study Notes
You should recognize that certain compounds (notably tertiary alkyl halides) which react very slowly by the SN2 mechanism can undergo rapid nucleophilic substitution by an alternative mechanism.
The abbreviation SN1 refers to “unimolecular nucleophilic substitution.” In a first-order reaction, the rate of the reaction depends on the concentration of only one of the reactants. Thus, when an alkyl halide reacts by an SN1 mechanism, the rate of reaction is dependent on the concentration of the alkyl halide, but is independent of the concentration of the attacking nucleophile.
Note: In many textbooks the “rate-limiting step” is called the “rate-determining step.”
Racemization problems can be a potential source of confusion. Most students entering an introductory organic chemistry course have a reasonable background in mathematics, and feel comfortable if they have a formula or equation they can use in this type of situation. Thus, we recommend that you consider the following approach.
1. In a given mixture of enantiomers, let x = the fraction of the (+)-enantiomer, and 1 − x = the fraction of the (−)-enantiomer. [Remember that the fraction can be obtained by dividing the percentage by 100%.]
2. The observed [α]D of the mixture is then given by:
Observed [α]D = x ([α]D of the (+)-enantiomer)
+ (1 − x)([α]D of the (−)-enantiomer)
Another source of confusion is the way in which the terms “percentage racemization” and “percentage inversion” are used. You must be clear in your mind that 80% racemized means that we have 40% of the original configuration and 60% (40% + 20%) of the inverted configuration. This point is illustrated in the diagram below.
General SN1 Reaction
When looking at the following substitution reaction it would be expected to be extremely slow. The reaction represents a worst case scenario based off the rules of an SN2 reaction. The substrate is sterically hindered, the nucleophile is relatively weak, and the solvent is polar protic. However, the reaction proceeds quickly and prefers tertiary substrates over methyl, which is opposite to the trend seen in SN2 reactions. It makes sense that this substitution reaction occurs using a different mechanism than SN2.
The SN1 mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism. In the SN1 reaction, the bond between the substrate and the leaving group is broken when the leaving group departs with the pair of electrons that formerly composed the bond.
This results in the formation of a carbocation (from "carbon" and "cation") the word for a positively charged carbon atom. Because the carbocation has only three bonds, it bears a formal charge of +1. Recall that a carbocation is sp2 hybridized, with trigonal planar geometry. The positive charge is contained in an empty, unhybridized p orbital perpendicular to the plane formed by the three sigma bonds. The formation of a carbocation is not energetically favored, so this step in the reaction is the slowest step and determines the overall rate of the reaction. The step which controls the overall rate of a reaction is called the rate-determining step.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry. Since the nucleophile attacks the carbocation only after the leaving group has departed, there is no need for back-side attack. The carbocation and its substituents are all in the same plane, meaning that the nucleophile can attack from either side. As a result, both enantiomers are formed in an the SN1 reaction, leading to a racemic mixture of both enantiomers.
Because SN1 reactions almost exclusively involve neutral nucleophile the product of this second step of the mechanism is often positively charged. The neutral substitution product is usually formed after a third, deprotonation step.
Reaction Coordinate Diagrams and SN1 Mechanism
During the SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states. The following reaction coordinate diagram shows an example of a an SN1 reaction with a negatively charged nucleophile.
There are several important consequences to the unimolecular nature of the rate-determining step in the the SN1 reaction. The first step (dissociation) is the rate determining step, so the rate is controlled only by the loss of the leaving group and does not involve any participation of the nucleophile. Therefore, the rate of the reaction is dependent only on the concentration of the substrate, not on the concentration of the nucleophile.
This can also be shown for the reactions with a neutral nucleophile which includes an extra step of deprotonation. A potential energy diagram for this mechanism shows that each of the two positively-charged intermediates (I1 and I2) can be visualized as a valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the transition states. The first, bond-breaking step is the slowest, rate-determining step - notice it has the highest activation energy and leads to the highest-energy species, I1, the carbocation intermediate. Step 2 is rapid: a new covalent bond forms between a carbocation and a water nucleophile, and no covalent bonds are broken. The third step represents a Bronsted-Lowry proton transfer which is rapid, with low activation energy.
Because the first step of the mechanism is the rate determining step, the first transition state will be the most important. During the first step, a carbon-leaving group bond in a neutral substrate is broken to form a carbocation intermediate and a leaving group anion. The breaking bond is represented as a dashed-partial bond in the transition state. A partial positive charge is shown on the carbon to represent the formation of a carbocation and a partial negative charge is shown on the leaving group to represent the formation of an anion.
SN1 Reaction Kinetics
In the first step of an SN1 mechanism, two charged species are formed from a neutral molecule. This step is much the slower of the three steps, and is therefore rate-determining. In the reaction energy diagram, the activation energy for the first step is higher than that for the second step indicating that the SN1 reaction has first order, unimolecular kinetics because the rate determining step involves one molecule splitting apart, not two molecules colliding. It is important to remember that first order refers to the rate law expression where the generic term substrate is used to describe the alkyl halide.
rate = k [substrate]
Because an SN1 reaction is first order overall the concentration of the nucleophile does not affect the rate. Since the nucleophile is not involved in the rate-limiting first step, the nature of the nucleophile does not affect the rate of an SN1 reaction.
Stereochemical Considerations
We saw that SN2 reactions result in inversion of stereochemical configuration at the carbon center. The difference in mechanisms means the stereochemical outcome of SN1 can be different than SN2. Recall that a carbocation intermediate produced during an SN1 reaction is sp2-hybridized, with an empty p orbital perpendicular to the plane formed by the three sigma bonds:
Because the nucleophile is free to attack from either side of the carbocation electrophile, the reaction leads to a 50:50 mixture of two stereoisomeric products. In other words: the SN1 reaction occurs with both retention or inversion of configuration at the electrophilic carbon, leading to racemization if the carbon is chiral after the substitution.
However, often chemical reactions are more complicated than the mechanisms that represent limiting cases like SN1 and SN2. Experimental data has shown that the inversion product is favored over racemization due to the formation of ion pairs. After dissociation during the rate determining step the carbocation and the leaving group are still electrostatically associated. The leaving group anion is temporarily held in place which provides a shield to one side of the carbocation. If substitution occurs before the leaving group anion fully diffuses away the stereochemical inversion substitution product is created. Only after the leaving group fully diffuses away can racemization occur. Overall, the product of an SN1 reaction shows an excess of the stereochemical inversion product.
For an example, consider the hydrolysis of (S)-3-chloro-3-methylhexane.
Exercise \(1\)
1) Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
2) Give the products of the following SN1 reaction. Show stereochemistry.
3) Predict with compound in each pair will undergo an SN1 reaction more quickly.
a) or
b) or
c) or
d) or
Answer
1) For Reaction A, the rate law is rate = k[CH3I][CH3S-]. Therefore, if the concentration of the nucleophile, CH3S-, is doubled and the concentration of the alkyl halide remains the same, then the reaction rate will double.
For Reaction B, the rate law is rate = k[CH3)3Br]. Therefore, if the concentration of the nucleophile, CH3SH, is doubled and the concentration of the alkyl halide remains the same, then reaction rate stays the same.
2)
3)
a)
b)
c)
d) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.04%3A_The_SN1_Reaction.txt |
Objectives
After completing this section, you should be able to
1. discuss how the structure of the substrate affects the rate of a reaction occurring by the SN1 mechanism.
2. arrange a given list of carbocations (including benzyl and allyl) in order of increasing or decreasing stability.
3. explain the high stability of the allyl and benzyl carbocations.
4. arrange a given series of compounds in order of increasing or decreasing reactivity in SN1 reactions, and discuss this order in terms of the Hammond postulate.
5. discuss how the nature of the leaving group affects the rate of an SN1 reaction, and in particular, explain why SN1 reactions involving alcohols are carried out under acidic conditions.
6. explain why the nature of the nucleophile does not affect the rate of an SN1 reaction.
7. discuss the role played by the solvent in an SN1 reaction, and hence determine whether a given solvent will promote reaction by this mechanism.
8. compare the roles played by the solvent in SN1 and in SN2 reactions.
9. determine which of two SN1 reactions will occur faster, by taking into account factors such as the structure of the substrate and the polarity of the solvent.
10. determine whether a given reaction is most likely to occur by an SN1 or SN2 mechanism, based on factors such as the structure of the substrate, the solvent used, etc.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• benzylic
• dielectric constant
• polarity
The Substrate in SN1Reactions
As discussed in the previous section SN1 reactions follow first order kinetics due to a multi-step mechanism in which the rate-determining step consists of the ionization of the alkyl halide to form a carbocation.
The transition state for the rate determining step shows the transition of an alkyl halide to a carbocation. Because the rate determining step is endothermic, the Hammond postulate says that the transition state more closely resembles the carbocation intermediate. Factors which stabilize this intermediate will lower the energy of activation for the rate determining step and cause the rate of reaction to increase. In general, a more stable carbocation intermediate formed during the reaction allows for a faster the SN1 reaction rate.
In Section 7-9, the stability of alky carbocations was shown to be 3o > 2o > 1o > methyl. Two special cases of resonance-stabilized carbocations, allyl and benzyl, must be considered and added to the list. The delocalization of the positive charge over an extended p orbital system allows for allyl and benzyl carbocations to be exceptionally stable. The resonance hybrid of an allyic cation is made up of two resonance forms while the resonance hybrid of the benzylic carbocation is made up of five.
Benzyl Carbocation
Carbocation
Stability
CH3(+) < CH3CH2(+) < (CH3)2CH(+) CH2=CH-CH2(+) C6H5CH2(+) (CH3)3C(+)
Effects of Leaving Group
Since leaving group removal is involved in the rate-determining step of the SN1 mechanism, reaction rates increase with a good leaving group. A good leaving group can stabilize the electron pair it obtains after the breaking of the C-Leaving Group bond faster. Once the bond breaks, the carbocation is formed and the faster the carbocation is formed, the faster the nucleophile can come in and the faster the reaction will be completed.
Because weak bases tend to strongly hold onto their electrons, they are lower energy molecules and they tend to make good leaving groups. Strong bases, on the other hand, donate electrons readily because they are high energy, reactive molecules. Therefore, they are not typically good leaving groups. As you go from left to right on the periodic table, electron donating ability decreases and thus ability to be a good leaving group increases. Halides are an example of a good leaving group whose leaving-group ability increases as you go down the halogen column of the periodic table.
The two reactions below only vary by the different leaving groups in each reaction. The reaction with a more stable leaving group is significantly faster than the other. This is because the better leaving group leaves faster and thus the reaction can proceed faster.
Under acidic conditions, the -OH group of an alcohol can be converted into a neutral water leaving group through protonation. As discussed in Section 10-5, this occurs during the conversion of a tertiary alcohol to an alkyl halide through reaction with HCl or HBr. Because the -OH group itself is a poor nucleophile, the mechanism starts with protonation to form a hydronium ion. Neutral water is then lost as a leaving group to create the carbocation intermediate which then reacts with the halide ion nucleophile to provide the alkyl halide product. This reaction works best when a tertiary alcohol is used because it produces a stable carbocation intermediate.
The Nucleophile
Since nucleophiles only participate in the fast second step, their relative molar concentrations rather than their nucleophilicities should be the primary product-determining factor. If a nucleophilic solvent such as water is used, its high concentration will assure that alcohols are the major product. While recombination of the halide anion with the carbocation intermediate can occur, it simply reforms the starting compound. Also this is less likely since there are less molecules of the leaving group in solution than there are of the solvent. Note that SN1 reactions in which the nucleophile is also the solvent are commonly called solvolysis reactions. The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:
The strength of the nucleophile does not affect the reaction rate of SN1 because, as described previously, the nucleophile is not involved in the rate-determining step. However, if you have more than one nucleophile competing to bond to the carbocation, the strengths and concentrations of those nucleophiles affects the distribution of products that you will get. For example, if you have (CH3)3CCl reacting in water and formic acid where the water and formic acid are competing nucleophiles, you will get two different products: (CH3)3COH and (CH3)3COCOH. The relative yields of these products depend on the concentrations and relative reactivities of the nucleophiles.
Solvent Effects on the SN1 Reaction
Since the hydrogen atom in a polar protic solvent is highly positively charged, it can interact with the anionic nucleophile which would negatively affect an SN2 reaction which depends on nucleophilic attack during the rate-determining step of the reaction. However, this does not affect an SN1 reaction because the nucleophile is not a part of the rate-determining step. Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the carbocation-like transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for a solvolysis (see table below). The dielectric constant of a solvent roughly provides a measure of the solvent's polarity. A dielectric constant below 15 is usually considered non-polar. Basically, the dielectric constant can be thought of as the solvent's ability to reduce the internal charge of the solvent. So for our purposes, the higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate.
(note that even though acetic acid is a highly polar molecule, it tends to make a dimer with itself greatly reducing its dielectric constant)
Predicting SN1 vs. SN2 mechanisms
When considering whether a nucleophilic substitution is likely to occur via an SN1 or SN2 mechanism, we really need to consider three factors:
1) The electrophile: when the leaving group is attached to a methyl group or a primary carbon, an SN2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an SN1 mechanism is favored.
2) The nucleophile: powerful nucleophiles, especially those with negative charges, favor the SN2 mechanism. Weaker nucleophiles such as water or alcohols favor the SN1 mechanism.
3) The solvent: Polar aprotic solvents favor the SN2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the SN1 mechanism by stabilizing the carbocation intermediate. SN1 reactions are frequently solvolysis reactions.
For example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we’ll assume that methanol is the solvent). Thus we’d confidently predict an SN1 reaction mechanism. Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.
In the reaction below, on the other hand, the electrophile is a secondary alkyl bromide – with these, both SN1 and SN2 mechanisms are possible, depending on the nucleophile and the solvent. In this example, the nucleophile (a thiolate anion) is strong, and a polar aprotic solvent is used – so the SN2 mechanism is heavily favored. The reaction is expected to proceed with inversion of configuration.
Exercise \(1\)
1. Put the following leaving groups in order of decreasing leaving group ability
2. Which solvent would an SN1 reaction occur faster in? H2O or CH3CN
3. What kind of conditions disfavor SN1 reactions?
4. What are the products of the following reaction and does it proceed via SN1 or SN2?
5. How could you change the reactants in the problem 4 to favor the other substitution reaction?
6. Indicate the expected product and list why it occurs through SN1 instead of SN2?
7. Rank the following by increasing reactivity in an SN1 reaction.
8. 3-bromo-1-pentene and 1-bromo-2-pentene undergo SN1 reaction at almost the same rate, but one is a secondary halide while the other is a primary halide. Explain why this is.
9. Label the following reactions as most likely occuring by an SN1 or SN2 mechanism. Suggest why.
10. Give the solvolysis product expected when the compound is heated in methanol.
a)
b)
c)
d)
11. Predict whether each compound below would be more likely to undergo a SN2 or SN1 reaction.
a)
b)
c)
d)
12. Show how each compound may be synthesized using nucleophilic substitution reactions.
a)
b)
c)
d)
e)
f)
g)
Answers
1.
2. An SN1 reaction would occur faster in H2O because it's polar protic and would stailize the carbocation and CH3CN is polar aprotic.
3. Polar aprotic solvents, a weak leaving group and primary substrates disfavor SN1 reactions.
4.
Reaction proceeds via SN1 because a tertiary carbocation was formed, the solvent is polar protic and Br- is a good leaving group.
5. You could change the solvent to something polar aprotic like CH3CN or DMSO and you could use a better base for a nucleophile such as NH2- or OH-.
6.
This reaction occurs via SN1 because Cl- is a good leaving group and the solvent is polar protic. This is an example of a solvolysis reaction because the nucleophile is also the solvent.
7. Consider the stability of the intermediate, the carbocation.
A < D < B < C (most reactive)
8. They have the same intermediates when you look at the resonance forms.
9.
A – SN1 *poor leaving group, protic solvent, secondary cation intermediate
B – SN2 *good leaving group, polar solvent, primary position.
10.
a)
b)
c) No reaction
d)
11.
a) SN2
b) SN2
c) SN1
d) SN1
13.
a)
b)
c)
d)
e)
f)
g) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.05%3A_Characteristics_of_the_SN1_Reaction.txt |
Objective
After completing this section, you should have an appreciation that SN1 and SN2 mechanisms exist and are well-known in biological chemistry.
Leaving Groups in Biochemical Reactions
In biological reactions, we do not often see halides serving as leaving groups (in fact, outside of some marine organisms, halogens are fairly unusual in biological molecules). More common leaving groups in biochemical reactions are phosphates, water, alcohols, and thiols. In many cases, the leaving group is protonated by an acidic group on the enzyme as bond-breaking occurs. For example, hydroxide ion itself seldom acts as a leaving group – it is simply too high in energy (too basic). Rather, the hydroxide oxygen is generally protonated by an enzymatic acid before or during the bond-breaking event, resulting in a (very stable) water leaving group.
More often, however, the hydroxyl group of an alcohol is first converted enzymatically to a phosphate ester in order to create a better leaving group. This phosphate ester can take the form of a simple monophosphate (arrow 1 in the figure below), a diphosphate (arrow 2), or a nucleotide monophosphate (arrow 3).
Due to resonance delocalization of the developing negative charge, phosphates are excellent leaving groups.
Here’s a specific example (from DNA nucleotide biosynthesis) that we will encounter in more detail in section 11.5:
Here, the OH group on ribofuranose is converted to a diphosphate, a much better leaving group. Ammonia is the nucleophile in the second step of this SN1-like reaction.
We will learn much more about phosphates in chapter 10. What is important for now is that in each case, an alcohol has been converted into a much better leaving group, and is now primed for a nucleophilic substitution reaction.
SAM Methyltransferases
Some of the most important examples of SN2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an SN2 reaction from SAM to the amine group on the nucleotide base adenosine:
(Nucleic Acids Res. 2000, 28, 3950).
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline.
Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group.
Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise SN1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is SN2. Does this SN2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase:
Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium (3H, T) and deuterium (2H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an SN2 displacement (J. Biol. Chem. 1980, 255, 9124).
Substitution by Electrophilic Addition/Elimination
The electrophilic double bond isomerization catalyzed by IPP isomerase is a highly reversible reaction, with an equilibrium IPP:DMAPP ratio of about 6:1. In the next step of isoprenoid biosynthesis, the two five-carbon isomers condense to form a 10-carbon isoprenoid product called geranyl diphosphate (GPP).
This is a nice example of an electrophilic addition/elimination mechanism, which we saw in general form in section 15.1:
The first step is ionization of the electrophile - in other words, the leaving group departs and a carbocation intermediate is formed. In this case, the pyrophosphate group on DMAPP is the leaving group, and the electrophilic species is the resulting allylic carbocation.
In the condensation (addition) step, the C3-C4 double bond in IPP attacks the positively-charged C1 of DMAPP, resulting in a new carbon-carbon bond and a second carbocation intermediate, this time at a tertiary carbon. In the elimination phase, proton abstraction leads to re-establishment of a double bond in the GPP product. Notice that the enzyme specifically takes the pro-R proton in this step.
To continue the chain elongation process, another IPP molecule can then condense, in a very similar reaction, with C1 of geranyl diphosphate to form a 15-carbon product called farnesyl diphosphate (FPP).
How do we know that these are indeed SN1-like mechanisms with carbocation intermediates, rather than concerted SN2-like mechanisms? First of all, recall that the question of whether a substitution is dissociative (SN1-like) or associative (SN2-like) is not always clear-cut - it could be somewhere in between, like the protein prenyltransferase reaction (section 9.3). The protein prenyltransferase reaction and the isoprenoid chain elongation reactions are very similar: the electrophile is the same, but in the former the nucleophile is a thiolate, while in the latter the nucleophile is a pi bond.
This difference in the identity of the nucleophilic species would lead one to predict that the chain elongation reaction has more SN1-like character than the protein prenylation reaction. A thiolate is a very powerful nucleophile, and thus is able to push the pyrophosphate leaving group off, implying some degree of SN2 character. The electrons in a pi bond, in contrast, are only weakly nucleophilic, and thus need to be pulled in by a powerful electrophile - ie. a carbocation.
So it makes perfect sense that the chain elongation reaction should more SN1-like than SN2-like. Is this in fact the case? We know how to answer this question experimentally - just run the reaction with fluorinated DMAPP or GPP substrates and observe how much the fluorines slow things down (see section 9.3B).
If the reaction is SN1-like, the electron-withdrawing fluorines should destabilize the allylic carbocation intermediate and thus slow the reaction down considerably. If the mechanism is SN2-like, the fluorine substitutions should not have a noticeable effect, because a carbocation intermediate would not be formed. When this experiment was performed with FPP synthase, the results were dramatic: the presence of a single fluorine slowed down the rate of the reaction by a factor of about 60, while two and three fluorines resulted in a reaction that was 500,000 and 3 million times slower, respectively (J. Am. Chem. Soc. 1981, 103, 3926.) These results strongly suggest indicate the formation of a carbocation intermediate in an SN1-like displacement. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.06%3A_Biological_Substitution_Reactions.txt |
Objective
After completing this section, you should be able to apply Zaitsev’s rule to predict the major product in a base-induced elimination of an unsymmetrical halide.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Zaitsev’s rule
When an alkyl halide is reacted with a nucleophile/Lewis base two major types of reaction can occur. Thus far in this chapter, we have discussed substitution reactions where a nucleophile displaces a leaving group at the electrophilic carbon of a substrate. Alternatively, the nucleophile could act as a Lewis base and cause an elimination reaction by removing a hydrogen adjacent to the leaving group. These reaction are similar and are often in competition with each other.
Introduction
The prefix "regio" indicates the interaction of reactants during bond making and/or bond breaking occurs preferentially by one orientation. If two or more structurally distinct groups of adjacent hydrogens are present in a given reactant, then multiple constitutionally isomeric alkenes may be formed by an elimination. Zaitsev’s rule is an empirical rule used to predict the major products of elimination reactions. It states that in an elimination reaction the major product is the more stable alkene with the more highly substituted double bond. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
By using the strong base hydroxide, we direct these reactions toward elimination (rather than substitution). In both cases there are two different sets of adjacent hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six adjacent 1º hydrogens compared with one 3º-hydrogen. These results point to a strong favoring the more highly substituted product double bond predicted by Zaitsev's Rule.
The E1, E2, and E1cB Reactions
Elimination reaction take place by three common mechanism, E1, E2, and E1cB, all of which break the H-C and X-C bonds at different points of their mechanism. In addition, the different mechanisms will have subtle effects on the reaction products which will be discussed later in this chapter.
E1 Mechanism
This mechanism starts the breaking of the C-X to provide a carbocation intermediate. A base removes a hydrogen adjacent to the original electrophilic carbon. The electrons from the broken H-C bond move to form the pi bond of the alkene. In much the same fashion as the SN1 mechanism, the first step of the mechanism is slow making it the rate determining step. This means that the reaction kinetics are unimolecular and first-order with respect to the substrate.
E2 Mechanism
The E2 mechanism takes place in a single concerted step. The rate at which this mechanism occurs follows second order kinetics, and depends on the concentration of both the base and alkyl halide. The base removes a hydrogen from a carbon adjacent to the leaving group. The electrons of the broken H-C move to form the pi bond of the alkene. In doing this the C-X bond is broken causing the removal of the leaving group.
E1cB Mechanism
The E1cB mechanism starts with the base deprotonating a hydrogen adjacent to the leaving to form a carbanion. In the second step of the mechanism the lone pair electrons of the carbanion move to become the pi bond of the alkene. This causes the C-X bond to break and the leaving group to be removed.
Predicting the Products of an Elimination Reaction
For most elimination reactions, the formation of the product involves the breaking of a C-X bond from the electrophilic carbon, the breaking of a C-H bond from a carbon adjacent to the electrophilic carbon, and the formation of a pi bond between these two carbons. Which elimination mechanism is being followed has little effect on these steps. The limitations of each elimination mechanism will be discussed later in this chapter.
To determining the possible products, it is vital to first identify the electrophilic carbon in the substrate. Next identify all hydrogens on carbons directly adjacent to the electrophilic carbon. Each unique adjacent hydrogen has the possibility of forming a unique elimination product. Break a C-H bond from each unique group of adjacent hydrogens then break the C-X bond. Finally connect the adjacent carbon and the electrophilic carbon with a double bond. Repeat this process for each unique group of adjacent hydrogens. Finally, compare all of the possible elimination products. The product whose double bond has the most alkyl substituents will most likely be the preferred product.
Worked Example \(1\)
What would be the expected products of the following reaction? Which would be expected to be the major product?
Solution
To solve this problem, first find the electrophilic carbon in the starting compound. This carbon is directly attached to the chlorine leaving groups and is shown in blue in the structure below. Next, identify all unique groups of hydrogens on carbons directly adjacent to the electrophilic carbon. In the starting compound, there are two distinct groups of hygrogens which can create a unique elimination product if removed. They are shown as red and green in the structure below.
Create the possible elimination product by breaking a C-H bond from each unique group of adjacent hydrogens then breaking the C-Cl bond. Then connect the adjacent carbon and the electrophilic carbon with a double bond to create an alkene elimiation product. Repeat this process for each unique group of adjacent hydrogens. Because the starting compound in this example has two unique groups of adjacent hydrogens, two elimination products can possibly be made.
Product 1
Product 2
Finally, compare the possible elimination products to determine which has the most alkyl substituents. This product will most likely be the preferred. For this example product 1 has three alkyl substituents and product 2 has only two. This means product 1 will likely be the preferred product of the reaction.
Exercise \(1\)
1) Ignoring the alkene stereochemistry show the elimination product(s) of the following compounds:
2) Predict the major products of the following reactions.
a)
b)
c)
Answer
a)
2)
a)
b)
c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.07%3A_Elimination_Reactions-_Zaitsev%27s_Rule.txt |
Objectives
After completing this section, you should be able to
1. write the mechanism of a typical E2 reaction.
2. sketch the transition state of a typical E2 reaction.
3. discuss the kinetic evidence that supports the proposed E2 mechanism.
4. discuss the stereochemistry of an E2 reaction, and explain why the anti periplanar geometry is preferred.
5. determine the structure of the alkene produced from the E2 reaction of a substrate containing two chiral carbon atoms.
6. describe the deuterium isotope effect, and discuss how it can be used to provide evidence in support of the E2 mechanism.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anti periplanar
• deuterium isotope effect
• E2 reaction
• periplanar
• syn periplanar
Study Notes
An E2 reaction is a bimolecular elimination reaction; thus, two molecules are involved in the rate-limiting step. In this section, we are concerned with E2 reactions involving an alkyl halide and a base.
Use molecular models to assist you to understand the difference between syn periplanar and anti periplanar, and to appreciate why E2 eliminations are stereospecific.
Note that when deuterium is used the kinetic isotope effect (KIE) is referred to as the deuterium isotope effect. A C–H bond is about 5 kJ/mol weaker than a C–D bond. So if the rate-limiting step involves a breaking of this bond as it does at the E2 transition state there will be a substantial difference in reaction rates when comparing deuterated and non-deuterated analogues. Indeed, the reaction of 2-bromopropane with sodium ethoxide (NaOEt) is 6.7 times faster than its deuterated counterpart, providing evidence consistent with an E2 mechanism.
Introduction
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. E2 reactions are typically seen with secondary and tertiary alkyl halides, in the presence of a base (OH-, RO-, R2N-). For a primary halide to undergo an E2 reaction a strong sterically hindered base is usually required. The products of an E2 reaction follow Zaitsev's rule, the most substituted alkene is usually the major product.
E2 Mechanism
The E2 mechanism takes place in a single concerted step. The rate at which this mechanism occurs follows second order kinetics, and depends on the concentration of both the base and alkyl halide. The base removes a hydrogen from a carbon adjacent to the carbon with the leaving group. The electrons of the broken H-C bond move to form the pi bond of the neutral product alkene. In doing this, the C-X bond is broken causing the removal of the leaving group. Overall during this reaction an electron pair is transferred from the base to the leaving group.
E2 Reaction Transition State
Because the E2 reaction is predicted to follow a concerted mechanism only one transition state is expected in its energy coordinate diagram.
As the base begins to remove a hydrogen adjacent to the leaving group an H-B is starting to form and an H-C bond is in the process of breaking. Simultaneously, a C=C pi bond forms and the C-X bond breaks. During this reaction a set of electrons and a negative charge are transferred from the base to the leaving group. This is represented by a partial negative charge being present on both the base and the leaving group in the transition state.
Evidence for the E2 Mechanism and Transition State
Kinetic studies of E2 reactions show that they are second order overall and follow the rate law: rate = k[RX][Base]. This data is consistent with the predicted bimolecular E2 mechanism that includes both the alkyl halide and the base participating in the mechanism's rate-determining step.
Further evidence for the predicted mechanism for the E2 reaction was provided by experiments involving the deuterium isotope effect (DIE). The deuterium isotope effect (DIE) is a phenomenon associated with molecules which have had hydrogen (H) atoms isotopically substituted with deuterium (D) atoms, exhibiting different reaction rates. The bond dissociation energy for C-D bonds is about 5 kJ/mol stronger than the bond dissociation energy of C-H bonds. This difference in energy results in a reduction in the rate of reaction, if the deuterium replacement occurs at a site of bond breaking in the rate determining step of a reaction. So if the rate-limiting step involves a breaking of this bond, as it does at the E2 transition state, there will be a substantial decrease in reaction rates when comparing deuterated and non-deuterated analogues. Indeed, the reaction of 2-bromopropane with sodium ethoxide (NaOEt) is 6.7 times faster than its deuterated counterpart, providing evidence that the C-H bond is broken during the rate determining step of the E2 mechanism. This is consistent with the predicted mechanism of the E2 reaction consisting of a single concerted step. If the rate-determining step was the leaving group bond breaking first (as in an E1 reaction) the deuterium substitution would not have caused an observable change in reaction rate.
Yet another piece of evidence which supports the E2 mechanism comes when considering the stereochemistry of the elimination products. In the transition state of the E2 mechanism there are two carbons rehybridization from sp3 to sp2 while forming a π-bond. For good orbital overlap to occur during this change all the atoms involved in the reaction, the two carbons, the hydrogen, and the leaving group need to all lie in the same plane or periplanar. There are two possible ways this geometry can be achieved: syn-periplanar where the hydrogen and the leaving group are on the same side of the C-C bond and anti-periplanar where the hydrogen and the leaving group are on opposite sides of the C-C bond. To obtain a syn-periplanar geometry the substituents attached to the C-C bond must adopt an energetically unfavorable eclipsed conformation. The energy barrier to syn-orientation is such that syn-elimination is rarely observed in E2 reactions. E2 eliminations typically take place from the anti-periplanar conformation, as this is the most stable conformation due to the substituents attached to the C-C bond being staggered.
When viewing the transition state of the E2 mechanism when in the anti-periplanar conformation, the two sp3 hybridized orbitals making up the C-H and C-X sigma bonds are in the same plane. When these two carbons rehybridization from sp3 to sp2, the p orbitals forming the π-bond will also have good overlap.
The stereochemical consequences of reactant molecules obtaining anti-periplanar geometry prior to E2 reactions has been observed in numerous experiments and provides further evidence of the proposed mechanism. For example, (2S, 3R)-2-bromo-2,3-diphenylbutane only forms (Z)-2,3-diphenyl-2-butene as a product of E2 elimination. To form the Z alkene isomer, the starting material must obtain the anti-periplanar geometry preferred for E2 reactions.
No isomeric E alkene product is formed because the starting materials would need to obtain a staggard, syn-periplanar geometry. The syn-periplanar geometry would be transferred to the transition state making it higher in energy and harder to form. The fact that only the Z alkene isomer forms provides further evidence that the E2 mechanism takes place by a single bimolecular step.
Predicting the Stereochemical Product of E2 Reactions
E2 reactions have the requirements of a leaving group and a hydrogen on a carbon adjacent to the leaving group carbon. For non-ringed compounds, the C-C can usually undergo free rotation to place the C-X and C-H bonds anti-periplaner to each other. Stereochemical considerations usually come into play when both of the carbons in the C-C bond are chiral. When these carbons are achiral but contain the other requirements for an E2 reaction, products lacking stereochemistry are usually produced.
When both of the carbons in the C-C bond are chiral, one carbon will have a leaving group and the other carbon will have a single hydrogen. To consider the stereochemistry of an E2 product, start by creating a Newman projection looking down the C-C bond. Then rotate the C-C bond until the C-X and C-H bonds are in the anti position. The Newman projection now shows the relative positioning of the substituents on the C-C bond as they will appear in the double bond formed. It should be noted that if syn elimination was possible, a positional isomer of the double bond will be created. Consider the elimination products for (2R, 3R)-2-bromo-3-methylpentane.
First, locate hydrogens on carbons adjacent to the leaving group. The compound used in this example has two unique sets of hydrogens.
Because the adjacent hydrogen colored fuchsia is attached to a chiral carbon it require special consideration. Convert the structure into a Newman projection along the C-C bond that will for the alkene elimination product. The C-C bond can be view down either direction and still product the correct product.
Then rotate such the the adjacent H and the leaving group are in the anti position.
This Newman projection shows the relative positioning of the substituents on the double bond formed. Note, none of the Z isomer is formed because it would require syn orientation of the H and Br.
Lastly, consider the products made by other adjacent hydrogens and apply Zaitsev's rule to predict the preferred product.
Determining Cis and Trans Products in Elimination Reaction
Understanding that E2 reactions require an anti-periplanar geometry can explain cis and trans alkene isomers can form as products. This occurs as an important exception to the rule that stereoisomers only form as elimination products when both of the carbons in the C-C bond forming the alkene are chiral. When considering the E2 reaction products of (R)-2-butane, there are two groups of hydrogens on carbons adjacent to the leaving group a CH3 and a CH2.
The carbon in the CH2 group is achiral so both the H's appear to be equivalent. However, when the anti-periplanar geometry requirement is applied each hydrogen will produce its own unique product. Once the reactant is converted into a Newman projection it can be seen that the Br leaving group and the hydrogen colored blue are in the anti orientation and are ready to be eliminated. It should also be noted that the two CH3 groups are also in the anti position which gives them the least amount of steric interaction possible. Because the CH3 groups are on opposite sides of the Newman projection the CH3 groups will be on opposite sides of the alkene thus creating the product trans-2-butene. When the Newman projection is rotated such that the fuchsia colored hydrogen is anti to the Br leaving group, the two CH3 groups are now on the same. This orientation creates the isomer cis-2-butene. However, to obtain this orientation the two CH3 groups must overcome the steric strain associated with a gauche conformation. This makes the transition state leading the cis isomer higher in energy and more difficult to form. Because of this, the trans isomer will be the major product. It is clear that the two hydrogens of the CH2 group are not equivalent despite their not being attached to a chiral carbon.
After considering all the products made during this E2 reaction the preferred product can be determined. The cis and trans isomer both have two alkene substituents while the third product only has one. Of the cis and trans isomer, the trans isomer is expected to be easier to form. Zaitsev's rule says that trans-2-butene will be the major product of this reaction.
Exercise \(1\)
1) What is the product of the following molecule in an E2 reaction? What is the stereochemistry?
Answer
The stereochemistry is (Z) for the reaction. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.08%3A_The_E2_Reaction_and_the_Deuterium_Isotope_Effect.txt |
Objectives
After completing this section, you should be able to
1. identify anti periplanar arrangements of atoms in substituted cyclohexanes.
2. determine which cyclohexane conformation will generate a specific anti periplanar arrangement.
Stereochemical Requirements of the E2 Reaction
E2 elimination reactions of certain isomeric cycloalkyl halides show unusual rates and regioselectivity which can provide important supporting evidence that anti-periplanar is the preferred orientation of reactant species in the E2 transition state. Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. The compounds used here all have six-membered rings, so to achieve the anti-periplanar orientation required for the E2 reaction both the halogen and the adjacent hydrogen much assume an axial conformation.
For the compound, 2-methyl-1-chlorocyclohexane, the cis to the trans isomers have distinctly different reaction rates and form different preferred products. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. For trans-methyl-1-chlorocyclohexane, to obtains a chair conformation which places the chlorine substituent in the axial orientation required for E2 elimination, the methyl substituents is also forced into an axial position. Having both substituents in the axial position makes this chair conformer of the trans-isomer much less stable and presents an energy barrier that must be overcome for an E2 reaction to occur. When cis-2-methyl-1-chlorocyclohexane obtains a chair conformation which places its chlorine substituent in an axial orientation, the methyl substituent is forced into an equatorial orientation. Having one substituent axial and one equatorial makes this chair conformation of the cis-isomer lower in energy and thus easier to form. Consequently, the E2 reaction rates with the trans-isomer are slower than with the cis-isomer.
Furthermore, the product from E2 elimination of the trans-isomer is 3-methylcyclohexene (not predicted by Zaitsev's rule), whereas the cis-isomer gives the 1-methylcyclohexene as the preferred product (as predicted by Zaitsev's rule). The trans-isomer only has one adjacent hydrogen which can obtain an axial orientation along with the chlorine. The adjacent hydrogen which would lead to a 1-methylcyclohexene product cannot obtain the diaxial, anti-periplanar orientation with chlorine so an E2 reaction cannot occur. This makes 3-methylcyclohexene the preferred E2 elimination product of the trans-isomer showing that the anti-periplanar orientation requirement of E2 reactions is more important in determining products for this reaction than Zaitsev's rule.
In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two adjacent axial hydrogens. Removing one hydrogen forms the product, 3-methylcyclohexene, while removing the other hydrogen forms 1-methylcyclohexene. Here Zaitsev's rule determines that the more substituted alkene, 1-methylcyclohexene, is the preferred product.
Worked Exercise\(1\)
Which isomer would be expected to undergo E2 elimination quicker cis or trans-1-bromo-4-tert-butylcyclohexane? Explain your answer.
Answer
In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans-isomer. Because of symmetry, the two axial adjacent hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the bromine atom. To assume a conformation having an axial bromine, the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates in this situation.
Exercise \(1\)
Which of the following compounds will react faster in an E2 reaction; trans-1-bromo-2-isopropylcyclohexane or cis-1-bromo-2-isopropylcyclohexane?
Answer
The cis isomer will react faster than the trans. The cis isomer has two possible perpendicular hydrogen in which it can eliminate from. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.09%3A_The_E2_Reaction_and_Cyclohexane__Conformation.txt |
Objectives
After completing this section, you should be able to
1. write the mechanism for a typical E1 reaction.
2. explain why E1 elimination often accompanies SN1 substitution.
3. write an equation to describe the kinetics of an E1 reaction.
4. discuss the stereochemistry of E1 reactions.
5. account for the lack of a deuterium isotope effect in E1 reactions.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• E1 reaction
• E1cB reaction
Study Notes
The abbreviation E1 stands for “unimolecular elimination”; that is, an E1 reaction is an elimination reaction in which only one species is involved in the rate-limiting step.
Unimolecular Elimination (E1) is a reaction in which the removal of an HX substituent results in the formation of a double bond. It is similar to a unimolecular nucleophilic substitution reaction (SN1) in various ways. One being the formation of a carbocation intermediate. Also, the only rate determining (slow) step is the dissociation of the leaving group to form a carbocation, hence the name unimolecular. Thus, since these two reactions behave similarly, they compete against each other. Many times, both these reactions will occur simultaneously to form different products from a single reaction. However, one can be favored over another through thermodynamic control. Although Elimination entails two types of reactions, E1 and E2, we will focus mainly on E1 reactions with some reference to E2.
General E1 Reaction
An E1 reaction involves the removal of the halogen leaving group followed by the deprotonation of an adjacent hydrogen to produce an alkene product. In order to accomplish this, a Lewis base is required.
Mechanism
This mechanism starts with the spontaneous removal of the leaving group. The leaving group also removes the electrons from the C-Br bond, making the attached carbon a carbocation. In much the same fashion as the SN1 mechanism, the first step of the E1 mechanism is slow thus making it the rate determining step. This makes E1 reaction kinetics unimolecular and first-order with respect to the substrate. Next, a Lewis Base (B-) deprotonates an adjacent hydrogen from the carbocation. The electrons of the C-H bond are donated to the adjacent C-C bond, forming a double bond. Unlike E2 reactions, which require the adjacent proton to be anti to the leaving group, E1 reactions only require a neighboring hydrogen. This is due to the fact that the leaving group has left the molecule to form an achiral, trigonal planar carbocation.
Evidence for the E1 Mechanism
The primary evidence in support of the E1 mechanism is the fact that E1 reaction follow first-order kinetics which is consistent with the reaction mechanism containing a unimolecular dissociation as the rate determining step. Another piece of evidence is seen in the fact that E1 reactions show no deuterium isotope effect. Experiments have shown that there is no difference in reaction rate when deuterated and nondeuterated substrates are used. This is consistent with the E1 mechanism because a C-H bond is not broken in the rate determining step.
Lastly, because the halogen and hydrogen are removed in different steps, the mechanism of E1 reactions predicts that they will not require the anti-periplanar geometry required for E2 reactions. E1 reactions do not have the geometric constraints of E2 reactions discussed in Sections 11-8 and 11-9. In Section 11-9 an example showed that trans-1-Chloro-2-methylcyclohexane was capable of only forming the product 3-methylhexene during an E2 reaction due to the anti-periplanar constraints. When the same substrate is reacted under E1 conditions two elimination products (1-methylhexene and 3-methylhexene) are produced. In the substrate, the hydrogen marked as green cannot obtain anti-periplanar geometry with the Cl leaving group. Under E2 conditions this hydrogen and the Cl cannot be eliminated to form an elimination product. However, under E1 conditions the hydrogen and the Cl eliminate, despite the lack of anti-periplanar geometry, to form 1-methylhexene.
Alkyl Halide Reactivity in E1 Reactions
Due to the fact that E1 reactions create a carbocation intermediate, reactivity of alkyl halides toward E1 reaction mirror those present in SN1 reactions.
As expected, tertiary carbocations are favored over secondary, primary and methyl carbocations. This is due to the phenomena of hyperconjugation, which essentially allows a nearby C-C or C-H bond to interact with the p orbital of the carbon to bring the electrons down to a lower energy state. Thus, this has a stabilizing effect on the molecule as a whole. In general, primary and methyl carbocations do not proceed through the E1 pathway for this reason, unless there is a means of carbocation rearrangement to move the positive charge to a nearby carbon that is more stable. Secondary and Tertiary carbons form more stable carbocations, thus E1 reactions occur quite rapidly at these atoms.
Secondary carbocations can be subject to the E2 reaction pathway, but this generally occurs in the presence of a good/strong base. Adding a weak base to the reaction disfavors E2, essentially pushing towards the E1 pathway. In many instances, solvolysis occurs rather than using a base to deprotonate. This means heat is added to the solution, and the solvent itself deprotonates a hydrogen. The medium can effect the pathway of the reaction as well. Polar protic solvents may be used to hinder nucleophiles, thus disfavoring E2/SN2 from occurring.
The Connection Between SN1 and E1 Reactions
The E1 and SN1 mechanisms both begin with the rate-determing step, the unimolecular removal of a leaving group to form a carbocation intermediate. Sharing the rate-determining step causes alkyl halides to produce both SN1 substitution and E1 elimination products whenever they are reacted with nonbasic nucleophiles in a protic solvent. For example, the hydrolysis of tert-butyl chloride in a solution of water and ethanol gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration.
To produce the alcohol product, water attacks the carbocation as a nucleophile as part of an SN1 reaction. To produce the alkene product, water acts as a base and deprotonates an adjacent hydrogen as part of an E1 reaction. As expected, the mechanism of the two reactions have similar characteristics. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific.
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution product.
2. The cation may transfer a adjacent proton to a base, giving an alkene product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
Comparing E1 and E2 Mechanisms
When considering whether an elimination reaction is likely to occur via an E1 or E2 mechanism, we really need to consider three factors:
1) The base: strong bases favor the E2 mechanism, whereas, E1 mechanisms only require a weak base.
2) The solvent: good ionizing solvents (polar protic) favor the E1 mechanism by stabilizing the carbocation intermediate.
3) The alkyl halide: primary alkyl halides have the only structure useful in distinguishing between the E2 and E1 pathways. Since primary carbocations do not form, only the E2 mechanism is possible.
Ultimately, whether the elimination mechanism is E1 or E2 is not very important, since the product is the same alkene. We need to remember, however, that Zeitzev´s rule always determines the most likely alkene to be formed.
Reaction Parameter E2 E1
alkyl halide structure tertiary > secondary > primary tertiary > secondary >>>> primary
nucleophile high concentration of a strong base weak base
mechanism 1-step 2-step
rate limiting step bimolecular transition state carbocation formation
rate law rate = k[R-X][Base] rate = k[R-X]
solvent not important polar protic
The E1cB Reaction
Although E1 reactions typically involve a carbocation intermediate, the E1cB reaction utilizes a carbanion intermediate. A proton adjacent to a carbonyl group is removed using a strong base. This proton is acidic because the resulting conjugate base anion is stabilized by delocalization on to the carbonyl group. This anion causes the expulsion of an adjacent leaving group to create an alkene which is conjugated with the carbonyl which is called an enone. This reaction is generally utilized when a poor leaving group, such as a hydroxide, is involved. This poor leaving group makes the direct E1 or E2 reactions difficult. This reaction is used later in a reaction called an aldol condensation which will be discussed in Section 23-3.
E1cB Mechanism
The mechanism starts with the base removing a hydrogen to form an alkoxide anion. The alkoxide reforms the carbonyl C=O bond promoting the elimination of alcohol OH as a leaving group which also reforms the base catalyst. Although the base catalyzed elimination of alcohols is rare, it happens in this case in part due to the stability of the conjugated enone product.
1) Deprotonation to form the anion
2) Leaving Group Removal
Note! The double bond always forms in conjugation with the carbonyl.
Exercise \(1\)
1. Predict the dominant elimination mechanism (E1 or E2) for each reaction below. Explain your reasoning.
2) Specify the reaction conditions to favor the indicated elimination mechanism.
Answer
1)
2)
a) strong base, such as hydroxide, an alkoxide, or equivalent
b) water or alcohol or equivalent weak base with heat
c) strong base, such as hydroxide, an alkoxide, or equivalent | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.10%3A_The_E1_and_E1cB_Reactions.txt |
Objective
After completing this section, you should have an appreciation that E1, E2 and E1cB mechanisms exist and are well-known in biological chemistry.
Enzymatic E1 and E2 reactions
While most biochemical b-elimination reactions are of the E1cb type, some enzymatic E2 and E1 reactions are known. Like the enzymatic SN2 and SN1 substitution mechanisms discussed in chapters 8 and 9, the E2 and E1 models represent two possible mechanistic extremes, and actual enzymatic elimination reactions may fall somewhere in between. In an E1/E2 hybrid elimination, for example, Cβ-X bond cleavage may be quite advanced (but not complete) before proton abstraction takes place - this would lead to the build-up of transient partial positive charge on Cβ, but a discreet carbocation intermediate would not form. The extent to which partial positive charge builds up determines whether we refer to the mechanism as 'E1-like' or 'E2-like'.
A reaction in the histidine biosynthetic pathway provides a good example of a biological E1-like elimination step (we're looking specifically here at the first, enol-forming step in the reaction below - the second step is simply a tautomerization from the enol to the ketone product (section 13.1A)).
Notice in this mechanism that an E1cb elimination is not possible - there is no electron-withdrawing group (like a carbonyl) to stabilize the carbanion intermediate that would form if the proton were abstracted first. There is, however, an electron-donating group (the lone pair on a nitrogen) that can stabilize a positively-charged intermediate that forms when the water leaves.
Another good example of a biological E1-like reaction is the elimination of phosphate in the formation of 5-enolpyruvylshikimate-3-phosphate (EPSP), an intermediate in the synthesis of aromatic amino acids.
Experimental evidence indicates that significant positive charge probably builds up on Cβ of the starting compound, implying that C-O bond cleavage is advanced before proton abstraction occurs (notice the parallels to the Cope elimination in the previous section):
The very next step in the aromatic acid biosynthesis pathway is also an elimination, this time a 1,6-conjugated elimination rather than a simple beta-elimination.
An E1-like mechanism (as illustrated above) has been proposed for this step, but other evidence suggests that a free-radical mechanism may be involved.
While most E1 and E2 reactions involve proton abstraction, eliminations can also incorporate a decarboxylation step.
Isopentenyl diphosphate, the 'building block' for all isoprenoid compounds, is formed from a decarboxylation-elimination reaction.
Phenylpyruvate, a precursor in the biosynthesis of phenylalanine, results from a conjugated 1,6 decarboxylation-elimination.
11.13: A Summary of Reactivity - SN1 SN2 E1 E1cB and E2
Objectives
After completing this section, you should be able to
1. determine whether a specified substrate is most likely to undergo an E1, E2, SN1 or SN2 reaction under a given set of conditions.
2. describe the conditions under which a given substrate is most likely to react by a specified mechanism (E1, E2, SN1 or SN2).
Study Notes
This section summarizes much of what has been discussed in the chapter. It focuses on how a given substrate will behave under certain conditions, but does not deal with the stereochemistry of the products.
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, there are multiple variables to be considered, the most important being the substitution of the alkyl halide. SN2 reactions favor alkyl halides with little steric hindrance such as methyl halides and primary halides. In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable such as with tertiary halides, secondary allylic halides, and secondary benzylic halides.
The next most import variable for predicting the outcome of a reaction is the nature of the nucleophilic reactant. Strong nucleophiles favor SN2 substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor E2 elimination. Weak nucleophiles that are also weak bases tend to favor SN1 and E1 reactions.
• Good Nucleophiles Which are Weak Bases: I, Br, SCN, N3, CH3CO2 , RS, CN, Amines, etc.
• Good Nucleophiles Which are Strong Bases: HO, RO.
• Poor Nucleophiles which are Weak Bases: H2O, ROH, RSH.
General Rules For Predicting a Reaction
Consider these in the order listed.
Methyl Alkyl Halides
• An SN2 substitution occurs regardless if a good or poor nucleophile is used.
Primary Alkyl Halides
• An E2 elimination occurs if a strong, sterically hindered base is used.
• An E1cB elimination occurs if a strong base is used and the leaving group is two carbons away from a carbonyl group.
• An SN2 substitution occurs if a good nucleophile is used.
Secondary Alkyl Halides
• An E1cB elimination occurs if a strong base is used and the leaving group is two carbons away from a carbonyl group.
• An E2 elimination occurs if a strong base is used.
• An SN2 reaction occurs if a good nucleophile that is a weak bases is used in a polar aprotic solvent.
• An SN1 reaction along with an E1 reaction occurs if a poor nucleophile that is a weak bases is used in a protic solvent.
Tertiary Alkyl Halides
• An E1cB elimination occurs if a strong base is used and the leaving group is two carbons away from a carbonyl group.
• An E2 elimination occurs if a strong base is used.
• An SN1 reaction along with an E1 reaction occurs if a poor nucleophile that is a weak bases is used in a protic solvent.
Worked Example$1$
1) For the following, please determine what kind of reaction is occurring and predict the product(s).
a)
b)
c)
Answer
a) The substrate is a secondary halide so the product is determined by the nature of the nucleophile used in the reaction. Cyanide (-CN) is a good nucleophile which is a weak base. The fact that the nucleophile is a weak base means that an E2 reaction is not favored. Also, the fact that cyanide is a good nucleophile means that SN2 substitutions are favored over SN1. When a secondary halide is reacted with a good nucleophile which is a weak base, the preferred reaction is SN2. After SN2 substitution the product is a nitrile.
b) The substrate is a secondary halide so the product is determined by the nature of the nucleophile used in the reaction. Methoxide (-OCH3) is a strong base so it could prefer to remove a hydrogen from the substrate. This makes an E2 elimination the preferred reaction and an alkene the product.
c) The substrate is a secondary halide so the product is determined by the nature of the nucleophile used in the reaction. Methanol (HOCH3) is a weak nucleophile that is a weak base. Being a weak base means that methanol is not capable of actively removing a hydrogen to cause an E2 reaction. Also, because methanol is a weak nucleophile is not capable of attacking the substrate and causing an SN2 reaction. For the methanol nucleophile to react the substrate must first eject it leaving group to form the highly reactive carbocation intermediate. Formation of the carbocation is the rate determine step for both the SN1 and E1 reactions so they each form a separate product.
Exercises $1$
1. Identify the dominant reaction mechanism (SN1, SN2, E1, or E2) and predict the major product for the following reactions.
a)
b)
c)
d)
e)
2) Identify the function of the following reagents. The reagents will be a strong/weak nucleophile and/or a strong/weak base.
a) Cl-
b) NaH
c) t-BuO-
d) OH-
e) H2O
f) HS-
g) MeOH
3) Identify which mechanism the following reactions would undergo.
4) Identify all the products of the following reactions and specify the major product.
a)
b)
c)
d)
e)
5) The following reaction yields five different products. Give the mechanisms for how each is formed.
6) Label the following reactions as SN1, SN2, E1, or E2.
Answer
1)
2)
a) Cl- ; strong nucleophile
b) NaH ; strong base
c) t-BuO- ; strong base
d) OH- ; strong nucleophile ; strong base
e) H2O ; weak nucleophile ; weak base
f) HS- ; strong nucleophile
g) MeOH ; weak nucleophile ; weak base
3)
a) E2, SN1
b) SN2, E2
c) SN2
d) SN1, E1
e) E2, SN1
4)
a)
b)
c)
d)
e)
5)
6)
A – SN2
B – E1
C – SN1
D – E2 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.11%3A_Biological_Elimination_Reactions.txt |
Concepts & Vocabulary
11.1 Introduction
• Alkyl halides react as electrophiles and undergo nucleophilic substitution and elimination reactions.
11.2 The Discovery of Nucleophilic Substitution Reactions
• Some nucleophilic substitution reactions invert stereochemistry at the reactive carbon.
11.3 The SN2 Reaction
• Reaction steps with two molecules involved in the rate determining step are called bimolecular.
• A substitution mechanism that has the nucleophile entering at the same time the leaving group leaves, in a concerted step, is called SN2 - substitution nucleophilic bimolecular.
• Concerted substitution mechanisms (SN2) occur via backside attack, which causes inversion of the carbon where the reaction occurs.
• Rates of SN2 reactions depend on concentration of nucleophile and alkyl halide.
11.4 Characteristics of the SN2 Reaction
• SN2 reactions are concerted.
• Sterically hindered substrates reduce SN2 reaction rate.
• A transition state in a reaction mechanism is the highest energy point on a pathway from reactants to an intermediate or products.
• Larger groups (such as alkyl vs. hydrogen) cause greater steric repulsion in SN2 transition states, reducing rates of SN2 reactions.
• Groups that have electron-rich atoms are typically good nucleophiles.
• In general, stronger bases are better nucleophiles.
• Polar aprotic solvents increase rates of SN2 reactions.
• Polar protic solvents decrease rates of SN2 reactions.
• As basicity of leaving groups decreases, their ability to leave increases.
11.5 The SN1 Reaction
• A substitution mechanism that occurs with the leaving group leaving in the first step, creating a carbocation intermediate, followed by the nucleophile entering is called SN1 - substitution nucleophilic unimolecular.
• SN1 reactions occur through a stepwise mechanism.
• The first step (dissociation) of an SN1 mechanism is rate limiting.
• In SN1 reactions the nucleophile is not involved in the rate limiting step, therefore nucleophile strength or concentration do not affect the rate.
• The intermediate for SN1 mechanisms contains a planar carbocation. The nucleophile can then enter from either side of the molecule giving racemic products with no additional stereocenters in the molecule.
11.6 Characteristics of the SN1 Reaction
• Polar solvents increase rates of SN1 reactions.
• Better leaving groups increase rates of SN1 and SN2 reactions.
• Predicting whether a reaction will follow an SN1 or SN2 mechanism requires analysis of:
• Electrophile - primary favor SN2, tertiary (and allyl or benzyl) favor SN1, secondary depends on other factors
• Nucleophile - strong favor SN2, weak favor SN1
• Solvent - polar aprotic favor SN2, polar protic favor SN1
11.7 Biological Substitution Reactions
• When biological substitution reactions occur, the electrophiles are often different though the mechanisms are primarily the same.
11.8 Elimination Reactions - Zaitsev's Rule
• The major product of Elimination reactions is the product with the more substituted double bond. This is known as Zaitsev's rule.
11.9 The E2 Reaction and Deuterium Isotope Effect
• The E2 mechanism is concerted with the base removing a proton and the leaving group leaving at the same time.
• Since E2 mechanisms are concerted, both the base and the electrophile are present in the rate equation.
• E2 reactions require strong bases and polar aprotic solvents.
• Kinetic Isotope Effects can provide evidence for E2 mechanisms since they can show when breaking of the C-H bond is part of the rate-determining step.
11.10 The E2 Reaction and Cyclohexane Conformation
• E2 reactions of cyclic structures show necessity for anti orientation of the proton being removed and the leaving group.
11.11 The E1 and E1cB Reactions
• E1 mechanisms begin with a leaving group leaving which forms a carbocation intermediate, which is then deprotonated in a second step.
• E1 mechanisms are step-wise.
• More substituted electrophiles are more reactive in E1 reactions.
• Zaitsev products are preferred, similarly to E2 reactions.
• E1 and SN1 proceed via the same carbocation intermediate and the same rate-determining step so typically happen concurrently.
• E1cB reactions begin with deprotonation (usually resulting in a resonance stabilized carbanion), followed by loss of the leaving group in the second step.
11.12 Biological Elimination Reactions
• There are many important examples of biological elimination reactions.
11.13 A Summary of Reactivity - SN1, SN2, E1, E1CcB, and E2
Skills to Master
• Skill 11.1 Draw SN1/SN2 mechanisms showing appropriate stereochemistry.
• Skill 11.2 Explain when SN1/SN2 mechanisms are likely to occur.
• Skill 11.3 Describe/draw the intermediate for an SN1 mechanism and transition state(s) for SN1/SN2 mechanisms.
• Skill 11.4 Write out rate laws for SN1/SN2 mechanisms.
• Skill 11.5 Differentiate between which mechanism is more likely between SN1/SN2.
• Skill 11.6 Draw reaction coordinate diagrams for SN1/SN2 mechanisms.
• Skill 11.7 Explain how the electrophile, nucleophile, leaving group, and solvent affect SN1/SN2 mechanisms.
• Skill 11.8 Recognize use of nucleophilic substitution and elimination reactions in biological systems.
• Skill 11.9 Draw E1/E2 mechanisms showing appropriate stereochemistry.
• Skill 11.10 Explain when E1/E2 mechanisms are likely to occur.
• Skill 11.11Describe/draw the intermediate for an E1 mechanism and transition state(s) for E1/E2mechanisms.
• Skill 11.12 Write out rate laws for E1/E2 mechanisms.
• Skill 11.13 Differentiate between which mechanism is more likely between E1/E2.
• Skill 11.14 Draw reaction coordinate diagrams for E1/E2 mechanisms.
• Skill 11.15 Explain how kinetic isotope effects can be used to support or refute a proposed mechanism.
• Skill 11.16 Draw an E1cB mechanism and explain when it is a viable option.
• Skill 11.17 Differentiate between which mechanism is more likely between SN1/SN2 and E1/E2.
Memorization Tasks (MT)
MT 11.1 Memorize the order of good leaving groups.
MT 11.2 Memorize which solvents are polar protic and polar aprotic.
MT 11.3 Memorize the stability order of carbocations.
Summary of Reactions
Nucleophilic Substitutions
Eliminations | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.S%3A_Reactions_of_Alkyl_Halides_-_Nucleophilic_Substitutions_and_Eliminations_%28Summary%.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
Objective
After completing this section, you should be able to recognize the various spectroscopic techniques used to identify and characterize organic compounds.
Key Terms
Make certain that you can define, and use in context, the key term below.
• spectroscopy
Study Notes
The term spectroscopy is used to describe a number of techniques used by chemists to obtain information about the structure and bonding of chemical compounds. Four types of spectroscopy are described in the course:
1. mass spectroscopy (also called mass spectrometry, Chapter 12).
2. infrared spectroscopy (often simply called IR, Chapter 12).
3. nuclear magnetic resonance spectroscopy (usually referred to as NMR, Chapter 13).
4. ultraviolet spectroscopy (abbreviated UV, Chapter 14).
Of these four techniques, we shall spend the least time on ultraviolet spectroscopy, as it is much less powerful than the other three. If you do any reading on chemistry outside of the course materials, you will almost certainly see references to other spectroscopic techniques, such as Raman spectroscopy, electron spin resonance (ESR) spectroscopy, and atomic absorption (AA) spectroscopy. Even a description of these techniques and the information they can provide is beyond the scope of this course.
12.01: Mass Spectrometry of Small Molecules - Magnetic-Sector Instruments
Objectives
After completing this section, you should be able to
1. describe, briefly, how a mass spectrometer works.
2. sketch a simple diagram to show the essential features of a mass spectrometer.
3. identify peaks in a simple mass spectrum, and explain how they arise.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• base peak
• parent peak (molecular ion peak)
• cation radical
• relative abundance
• mass spectrometer
• mass spectroscopy
• mass spectrum
• molecular ion (M+·)
• mass-to-charge ratio (m/z)
Study Notes
You may remember from general first-year chemistry how mass spectroscopy has been used to establish the atomic mass and abundance of isotopes.
Mass spectrometers are large and expensive, and usually operated only by fully trained personnel, so you may not have the opportunity to use such an instrument as part of this course. Research chemists often rely quite heavily on mass spectra to assist them in the identification of compounds, and you will be required to interpret simple mass spectra both in assignments and on examinations. Note that in most attempts to identify an unknown compound, chemists do not rely exclusively on the results obtained from a single spectroscopic technique. A combination of chemical and physical properties and spectral evidence is usually employed.
The Mass Spectrometer
In order to measure the characteristics of individual molecules, a mass spectrometer converts them to ions so that they can be moved about and manipulated by external electric and magnetic fields. The three essential functions of a mass spectrometer, and the associated components, are:
1. A small sample is ionized, usually to cations by loss of an electron. The Ion Source
2. The ions are sorted and separated according to their mass and charge. The Mass Analyzer
3. The separated ions are then measured, and the results displayed on a chart. The Detector
Because ions are very reactive and short-lived, their formation and manipulation must be conducted in a vacuum. Atmospheric pressure is around 760 torr (mm of mercury). The pressure under which ions may be handled is roughly 10-5 to 10-8 torr (less than a billionth of an atmosphere). Each of the three tasks listed above may be accomplished in different ways. In one common procedure, ionization is effected by a high energy beam of electrons, and ion separation is achieved by accelerating and focusing the ions in a beam, which is then bent by an external magnetic field. The ions are then detected electronically and the resulting information is stored and analyzed in a computer. A mass spectrometer operating in this fashion is outlined in the following diagram. The heart of the spectrometer is the ion source. Here molecules of the sample (black dots) are bombarded by electrons (light blue lines) issuing from a heated filament. This is called an EI (electron-ionization) source. Gases and volatile liquid samples are allowed to leak into the ion source from a reservoir (as shown). Non-volatile solids and liquids may be introduced directly. Cations formed by the electron bombardment (red dots) are pushed away by a charged repellor plate (anions are attracted to it), and accelerated toward other electrodes, having slits through which the ions pass as a beam. Some of these ions fragment into smaller cations and neutral fragments. A perpendicular magnetic field deflects the ion beam in an arc whose radius is inversely proportional to the mass of each ion. Lighter ions are deflected more than heavier ions. By varying the strength of the magnetic field, ions of different mass can be focused progressively on a detector fixed at the end of a curved tube (also under a high vacuum).
When a high energy electron collides with a molecule it often ionizes it by knocking away one of the molecular electrons (either bonding or non-bonding). This leaves behind a molecular ion (colored red in the following diagram). Residual energy from the collision may cause the molecular ion to fragment into neutral pieces (colored green) and smaller fragment ions (colored pink and orange). The molecular ion is a radical cation, but the fragment ions may either be radical cations (pink) or carbocations (orange), depending on the nature of the neutral fragment. An animated display of this ionization process will appear if you click on the ion source of the mass spectrometer diagram.
Below is typical output for an electron-ionization MS experiment (MS data below is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z, which is the mass to charge ratio (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak corresponds to a fragment with m/z = 43 - . The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.00%3A_Introduction.txt |
Objectives
After completing this section, you should be able to
1. suggest possible molecular formulas for a compound, given the m/z value for the molecular ion, or a mass spectrum from which this value can be obtained.
2. predict the relative heights of the M+·, (M + 1)+·, etc., peaks in the mass spectrum of a compound, given the natural abundance of the isotopes of carbon and the other elements present in the compound.
3. interpret the fragmentation pattern of the mass spectrum of a relatively simple, known compound (e.g., hexane).
4. use the fragmentation pattern in a given mass spectrum to assist in the identification of a relatively simple, unknown compound (e.g., an unknown alkane).
Study Notes
When interpreting fragmentation patterns, you may find it helpful to know that, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break. You might wish to refer to the table of bond dissociation energies when attempting problems involving the interpretation of mass spectra.
This page looks at how fragmentation patterns are formed when organic molecules are fed into a mass spectrometer, and how you can get information from the mass spectrum.
The Origin of Fragmentation Patterns
When the vaporized organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion - or sometimes the parent ion and is often given the symbol M+ or . The dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process.
The molecular ions are energetically unstable, and some of them will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is an uncharged free radical.
The uncharged free radical will not produce a line on the mass spectrum. Only charged particles will be accelerated, deflected and detected by the mass spectrometer. These uncharged particles will simply get lost in the machine - eventually, they get removed by the vacuum pump.
The ion, X+, will travel through the mass spectrometer just like any other positive ion - and will produce a line on the stick diagram. All sorts of fragmentations of the original molecular ion are possible - and that means that you will get a whole host of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this:
Note
The pattern of lines in the mass spectrum of an organic compound tells you something quite different from the pattern of lines in the mass spectrum of an element. With an element, each line represents a different isotope of that element. With a compound, each line represents a different fragment produced when the molecular ion breaks up.
In the stick diagram showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion. The tallest line in the stick diagram (in this case at m/z = 43) is called the base peak. This is usually given an arbitrary height of 100, and the height of everything else is measured relative to this. The base peak is the tallest peak because it represents the commonest fragment ion to be formed - either because there are several ways in which it could be produced during fragmentation of the parent ion, or because it is a particularly stable ion.
Using Fragmentation Patterns
This section will ignore the information you can get from the molecular ion (or ions). That is covered in three other pages which you can get at via the mass spectrometry menu. You will find a link at the bottom of the page.
Example 12.2.1: Pentane
Let's have another look at the mass spectrum for pentane:
What causes the line at m/z = 57?
How many carbon atoms are there in this ion? There cannot be 5 because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to make up a total of 57. How about C4H9+ then?
C4H9+ would be [CH3CH2CH2CH2]+, and this would be produced by the following fragmentation:
The methyl radical produced will simply get lost in the machine.
The line at m/z = 43 can be worked out similarly. If you play around with the numbers, you will find that this corresponds to a break producing a 3-carbon ion:
The line at m/z = 29 is typical of an ethyl ion, [CH3CH2]+:
The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 less than one of the easy lines are often due to loss of one or more hydrogen atoms during the fragmentation process.
Example 12.2.2: Pentan-3-one
This time the base peak (the tallest peak - and so the commonest fragment ion) is at m/z = 57. But this is not produced by the same ion as the same m/z value peak in pentane.
If you remember, the m/z = 57 peak in pentane was produced by [CH3CH2CH2CH2]+. If you look at the structure of pentan-3-one, it's impossible to get that particular fragment from it.
Work along the molecule mentally chopping bits off until you come up with something that adds up to 57. With a small amount of patience, you'll eventually find [CH3CH2CO]+ - which is produced by this fragmentation:
You would get exactly the same products whichever side of the CO group you split the molecular ion. The m/z = 29 peak is produced by the ethyl ion - which once again could be formed by splitting the molecular ion either side of the CO group.
Peak Heights and Stability
The more stable an ion is, the more likely it is to form. The more of a particular sort of ion that's formed, the higher its peak height will be. We'll look at two common examples of this.
Carbocations (carbonium ions)
Summarizing the most important conclusion from the page on carbocations:
Order of stability of carbocations
primary < secondary < tertiary
Applying the logic of this to fragmentation patterns, it means that a split which produces a secondary carbocation is going to be more successful than one producing a primary one. A split producing a tertiary carbocation will be more successful still. Let's look at the mass spectrum of 2-methylbutane. 2-methylbutane is an isomer of pentane - isomers are molecules with the same molecular formula, but a different spatial arrangement of the atoms.
Look first at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This peak in 2-methylbutane is caused by:
The ion formed is a secondary carbocation - it has two alkyl groups attached to the carbon with the positive charge. As such, it is relatively stable. The peak at m/z = 57 is much taller than the corresponding line in pentane. Again a secondary carbocation is formed - this time, by:
You would get the same ion, of course, if the left-hand CH3 group broke off instead of the bottom one as we've drawn it. In these two spectra, this is probably the most dramatic example of the extra stability of a secondary carbocation.
Acylium ions, [RCO]+
Ions with the positive charge on the carbon of a carbonyl group, C=O, are also relatively stable. This is fairly clearly seen in the mass spectra of ketones like pentan-3-one.
The base peak, at m/z=57, is due to the [CH3CH2CO]+ ion. We've already discussed the fragmentation that produces this.
Note
The more stable an ion is, the more likely it is to form. The more of a particular ion that is formed, the higher will be its peak height.
Using mass spectra to distinguish between compounds
Suppose you had to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra.
pentan-2-one CH3COCH2CH2CH3
pentan-3-one CH3CH2COCH2CH3
Each of these is likely to split to produce ions with a positive charge on the CO group. In the pentan-2-one case, there are two different ions like this:
• [CH3CO]+
• [COCH2CH2CH3]+
That would give you strong lines at m/z = 43 and 71. With pentan-3-one, you would only get one ion of this kind:
• [CH3CH2CO]+
In that case, you would get a strong line at 57. You don't need to worry about the other lines in the spectra - the 43, 57 and 71 lines give you plenty of difference between the two. The 43 and 71 lines are missing from the pentan-3-one spectrum, and the 57 line is missing from the pentan-2-one one.
The two mass spectra look like this:
As you've seen, the mass spectrum of even very similar organic compounds will be quite different because of the different fragmentation patterns that can occur. Provided you have a computer data base of mass spectra, any unknown spectrum can be computer analyzed and simply matched against the data base. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.02%3A_Interpreting_Mass_Spectra.txt |
Objective
After completing this section, you should be able to predict the expected fragmentation for common functional groups, such as alcohols, amines, and carbonyl compounds.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alpha (α) cleavage
• McLafferty rearrangement
Much of the utility in electron-ionization MS comes from the fact that the radical cations generated in the electron-bombardment process tend to fragment in predictable ways. Detailed analysis of the typical fragmentation patterns of different functional groups is beyond the scope of this text, but it is worthwhile to see a few representative examples, even if we don’t attempt to understand the exact process by which the fragmentation occurs. We saw, for example, that the base peak in the mass spectrum of acetone is m/z = 43. This is the result of cleavage at the ‘alpha’ position - in other words, at the carbon-carbon bond adjacent to the carbonyl. Alpha cleavage results in the formation of an acylium ion (which accounts for the base peak at m/z = 43) and a methyl radical, which is neutral and therefore not detected.
After the parent peak and the base peak, the next largest peak, at a relative abundance of 23%, is at m/z = 15. This, as you might expect, is the result of formation of a methyl cation, in addition to an acyl radical (which is neutral and not detected).
A common fragmentation pattern for larger carbonyl compounds is called the McLafferty rearrangement:
The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion.
Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectrometrist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample.
Click here for examples of compounds listed by functional group, which demonstrate patterns which can be seen in mass spectra of compounds ionized by electron impact ionization.
Example 12.3.1
The mass spectrum of an aldehyde gives prominent peaks at m/z = 59 (12%, highest value of m/z in the spectrum), 58 (85%), and 29 (100%), as well as others. Propose a structure, and identify the three species whose m/z values were listed.
Solution
12.04: Mass Spectrometry in Biological- Time-of-flight (TOF) Instruments
Objective
This section is intended only to demonstrate that mass spectrometry can be useful for the investigation of some very large molecules present in biological systems.
Mass spectrometry of proteins - applications in proteomics
Mass spectrometry has become in recent years an increasingly important tool in the field of proteomics. Traditionally, protein biochemists tend to study the structure and function of individual proteins. Proteomics researchers, in contrast, want to learn more about how large numbers of proteins in a living system interact with each other, and how they respond to changes in the state of the organism. One very important subfield of proteomics is the search for protein biomarkers for human disease. These can be proteins which are present in greater quantities in a sick person than in a healthy person, and their detection and identification can provide medical researchers with valuable information about possible causes or treatments. Detection in a healthy person of a known biomarker for a disease such as diabetes or cancer could also provide doctors with an early warning that the patient may be especially susceptible, so that preventive measures could be taken to prevent or delay onset of the disease.
New developments in MS technology have made it easier to detect and identify proteins that are present in very small quantities in biological samples. Mass spectrometrists who study proteins often use instrumentation that is somewhat different from the electron-ionization, magnetic deflection system described earlier. When proteins are being analyzed, the object is often to ionize the proteins without causing fragmentation, so 'softer' ionization methods are required. In one such method, called electrospray ionization, the protein sample, in solution, is sprayed into a tube and the molecules are induced by an electric field to pick up extra protons from the solvent. Another common 'soft ionization' method is 'matrix-assisted laser desorption ionization' (MALDI). Here, the protein sample is adsorbed onto a solid matrix, and protonation is achieved with a laser.
Typically, both electrospray ionization and MALDI are used in conjunction with a time-of-flight (TOF) mass analyzer component.
The ionized proteins are accelerated by an electrode through a column, and separation is achieved because lighter ions travel at greater velocity than heavier ions with the same overall charge. In this way, the many proteins in a complex biological sample (such as blood plasma, urine, etc.) can be separated and their individual masses determined very accurately. Modern protein MS is extremely sensitive – very recently, scientists were even able to obtain a mass spectrum of Tyrannosaurus rex protein from fossilized bone! (Science 2007, 316, 277).
In one recent study, MALDI-TOF mass spectrometry was used to compare fluid samples from lung transplant recipients who had suffered from tissue rejection to control samples from recipients who had not suffered rejection. Three peptides (short proteins) were found to be present at elevated levels specifically in the tissue rejection samples. It is hoped that these peptides might serve as biomarkers to identify patients who are at increased risk of rejecting their transplanted lungs (Proteomics 2005, 5, 1705). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.03%3A_Mass_Spectrometry_of_Some_Common_Functional_Groups.txt |
Objectives
After completing this section, you should be able to
1. write a brief paragraph discussing the nature of electromagnetic radiation.
2. write the equations that relate energy to frequency, frequency to wavelength and energy to wavelength, and perform calculations using these relationships.
3. describe, in general terms, how absorption spectra are obtained.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electromagnetic radiation
• electromagnetic spectrum
• hertz (Hz)
• infrared spectroscopy
• photon
• quantum
Study Notes
From your studies in general chemistry or physics, you should be familiar with the idea that electromagnetic radiation is a form of energy that possesses wave character and travels through space at a speed of 3.00 × 108m · s−1. However, such radiation also displays some of the properties of particles, and on occasion it is more convenient to think of electromagnetic radiation as consisting of a stream of particles called photons.
In spectroscopy, the frequency of the electromagnetic radiation being used is usually expressed in hertz (Hz), that is, cycles per second. Note that 1 Hz = s−1.
A quantum is a small, definite quantity of electromagnetic radiation whose energy is directly proportional to its frequency. (The plural is “quanta.”) If you wish, you can read about the properties of electromagnetic radiation and the relationships among wavelength, frequency and energy, or refer to your general chemistry textbook if you still have it.
Note also that in SI units, Planck’s constant is 6.626 × 10−34J · s.
The electromagnetic spectrum
Electromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of electrical and magnetic waves which oscillate on perpendicular planes. Visible light is electromagnetic radiation. So are the gamma rays that are emitted by spent nuclear fuel, the x-rays that a doctor uses to visualize your bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that the army uses in night-vision goggles, the microwaves that you use to heat up your frozen burritos, and the radio-frequency waves that bring music to anybody who is old-fashioned enough to still listen to FM or AM radio.
Just like ocean waves, electromagnetic waves travel in a defined direction. While the speed of ocean waves can vary, however, the speed of electromagnetic waves – commonly referred to as the speed of light – is essentially a constant, approximately 300 million meters per second. This is true whether we are talking about gamma radiation or visible light. Obviously, there is a big difference between these two types of waves – we are surrounded by the latter for more than half of our time on earth, whereas we hopefully never become exposed to the former to any significant degree. The different properties of the various types of electromagnetic radiation are due to differences in their wavelengths, and the corresponding differences in their energies: shorter wavelengths correspond to higher energy.
High-energy radiation (such as gamma- and x-rays) is composed of very short waves – as short as 10-16 meter from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Visible light waves are in the range of 400 – 700 nm (nanometers, or 10-9 m), while radio waves can be several hundred meters in length.
The notion that electromagnetic radiation contains a quantifiable amount of energy can perhaps be better understood if we talk about light as a stream of particles, called photons, rather than as a wave. (Recall the concept known as ‘wave-particle duality’: at the quantum level, wave behavior and particle behavior become indistinguishable, and very small particles have an observable ‘wavelength’). If we describe light as a stream of photons, the energy of a particular wavelength can be expressed as:
$E = \dfrac{hc}{\lambda} \tag{12.5.1}$
where E is energy in kcal/mol, λ (the Greek letter lambda) is wavelength in meters, c is 3.00 x 108 m/s (the speed of light), and h is 6.626 × 10−34 J · s, a number known as Planck’s constant.
Because electromagnetic radiation travels at a constant speed, each wavelength corresponds to a given frequency, which is the number of times per second that a crest passes a given point. Longer waves have lower frequencies, and shorter waves have higher frequencies. Frequency is commonly reported in hertz (Hz), meaning ‘cycles per second’, or ‘waves per second’. The standard unit for frequency is s-1.
When talking about electromagnetic waves, we can refer either to wavelength or to frequency - the two values are interconverted using the simple expression:
$\lambda \nu = c \tag{12.5.2}$
where ν (the Greek letter ‘nu’) is frequency in s-1. Visible red light with a wavelength of 700 nm, for example, has a frequency of 2.84 x 10-19 J per photon or 171 kJ per mole of photons (remember Avogadro’s number = 6.02 × 1023 mol−1). The full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.
Notice in the figure above that visible light takes up just a narrow band of the full spectrum. White light from the sun or a light bulb is a mixture of all of the visible wavelengths. You see the visible region of the electromagnetic spectrum divided into its different wavelengths every time you see a rainbow: violet light has the shortest wavelength, and red light has the longest.
Example 12.5.1
Visible light has a wavelength range of about 400-700 nm. What is the corresponding frequency range? What is the corresponding energy range, in kcal/mol of photons?
Answer
Using λν = c, we first rearrange to ν = c/λ to solve for frequency.
For light with a wavelength of 400 nm, the frequency is 7.50 × 1014 Hz:
In the same way, we calculate that light with a wavelength of 700 nm has a frequency of 4.29 × 1014 Hz.
To calculate corresponding energies using hc/λ. We find for light at 400 nm:
Using the same equation, we find that light at 700 nm corresponds to 171 kJ mol−1.
Molecular spectroscopy – the basic idea
In a spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a low energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are not absorbed by the sample molecule, so they pass on through. A detector on the other side of the sample records which wavelengths were absorbed, and to what extent they were absorbed.
Here is the key to molecular spectroscopy: a given molecule will specifically absorb only those wavelengths which have energies that correspond to the energy difference of the transition that is occurring. Thus, if the transition involves the molecule jumping from ground state A to excited state B, with an energy difference of ΔE, the molecule will specifically absorb radiation with wavelength that corresponds to ΔE, while allowing other wavelengths to pass through unabsorbed.
By observing which wavelengths a molecule absorbs, and to what extent it absorbs them, we can gain information about the nature of the energetic transitions that a molecule is able to undergo, and thus information about its structure.
These generalized ideas may all sound quite confusing at this point, but things will become much clearer as we begin to discuss specific examples. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.05%3A_Spectroscopy_and_the_Electromagnetic_Spectrum.txt |
Objectives
After completing this section, you should be able to
1. identify (by wavelength, wavenumber, or both) the region of the electromagnetic spectrum which is used in infrared (IR) spectroscopy.
2. interconvert between wavelength and wavenumber.
3. discuss, in general terms, the effect that the absorption of infrared radiation can have on a molecule.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• infrared spectrum
• wavenumber (reciprocal centimetres)
Study Notes
Notice that the scale at the bottom of the infrared spectrum for 2-hexanone shown is calibrated in wavenumbers (cm−1). A wavenumber is the reciprocal of a wavelength (1/λ); thus, a wavenumber of 1600 cm−1 corresponds to a wavelength of
$\frac{1}{1600\text{\hspace{0.17em}}{\text{cm}}^{-1}}=6.25×{10}^{-4}\text{cm or}6.25\text{\hspace{0.17em}}\mu \text{\hspace{0.17em}}\text{m}$
Organic chemists find it more convenient to deal with wavenumbers rather than wavelengths when discussing infrared spectra.
You will obtain infrared spectra for a number of the compounds you will synthesize in the laboratory component of this course.
The inverted peaks observed in the spectra correspond to molecular stretching and bending vibrations that only occur at certain quantized frequencies. When infrared radiation matching these frequencies falls on the molecule, the molecule absorbs energy and becomes excited. Eventually the molecule returns to its original (ground) state, and the energy which was absorbed is released as heat.
Infrared Spectroscopy
The full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.
Notice in the figure above that infrared light is lower energy than visible light. The wavelengths of infrared radiation are between 0.8 and 250 μm. The units that are typically used for infrared spectroscopy are wavenumbers (which is cm-1). IR spectroscopy analyzes radiation between 40 to 13,000 cm-1. But what type of excitation is occurring when infrared radiation is absorbed by a molecule?
Covalent bonds in organic molecules are not rigid sticks – rather, they behave more like springs. At room temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. These complex vibrations can be broken down mathematically into individual vibrational modes, a few of which are illustrated below.
The energy of molecular vibration is quantized rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the amplitude of the vibration will increase, but the vibrational frequency will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the infrared region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds.
Let's take 2-hexanone as an example. Picture the carbonyl bond of the ketone group as a spring. This spring is constantly bouncing back and forth, stretching and compressing, pushing the carbon and oxygen atoms further apart and then pulling them together. This is the stretching mode of the carbonyl bond. In the space of one second, the spring 'bounces' back and forth 5.15 x 1013 times - in other words, the ground-state frequency of carbonyl stretching for a the ketone group is about 5.15 x 1013 Hz.
If our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10-6 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state.
The value of ΔE - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state.
With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 1013 to 1014 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side.
Our 5.15 x 1013 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity.
The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light.
The power of infrared spectroscopy arises from the observation that different functional groups have different characteristic absorption frequencies. The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 1013 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 1013 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne.
Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. In order for a vibrational mode to absorb infrared light, it must result in a periodic change in the dipole moment of the molecule. Such vibrations are said to be infrared active. In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls.
Some kinds of vibrations are infrared inactive. The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules do absorb IR light).
Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone.
There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called wavenumber (cm-1), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘% transmittance’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you! | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.06%3A_Infrared_Spectroscopy.txt |
Objectives
After completing this section, you should be able to
1. describe how the so-called “fingerprint region” of an infrared spectrum can assist in the identification of an unknown compound.
2. identify the functional group or groups present in a compound, given a list of the most prominent absorptions in the infrared spectrum and a table of characteristic absorption frequencies.
3. identify the broad regions of the infrared spectrum in which occur absorptions caused by
1. \(\ce{\sf{N-H}}\), \(\ce{\sf{C-H}}\), and \(\ce{\sf{O-H}}\)
2. \(\ce{\sf{C#C}}\) and \(\ce{\sf{C#N}}\)
3. \(\ce{\sf{C=O}}\), \(\ce{\sf{C=N}}\$, and \$\ce{\sf{C=C}}\)
Key Terms
Make certain that you can define, and use in context, the key term below.
• fingerprint region
Study Notes
When answering assignment questions, you may use this IR table to find the characteristic infrared absorptions of the various functional groups. However, you should be able to indicate in broad terms where certain characteristic absorptions occur. You can achieve this objective by memorizing the following table.
Region of Spectrum (cm−1) Absorption
2500-4000 \$\ce{\sf{N−H}}\$, \$\ce{\sf{O−H}}\$, \$\ce{\sf{C−H}}\$
2000-2500 \$\ce{\sf{C#C}}\$, \$\ce{\sf{C#N}}\$
1500-2000 \$\ce{\sf{C=O}}\$, \$\ce{\sf{C=N}}\$, \$\ce{\sf{C=C}}\$
below 1500 Fingerprint region
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
The key absorption peak in this spectrum is that from the carbonyl double bond, at 1716 cm-1 (corresponding to a wavelength of 5.86 mm, a frequency of 5.15 x 1013 Hz, and a ΔE value of 4.91 kcal/mol). Notice how strong this peak is, relative to the others on the spectrum: a strong peak in the 1650-1750 cm-1 region is a dead giveaway for the presence of a carbonyl group. Within that range, carboxylic acids, esters, ketones, and aldehydes tend to absorb in the shorter wavelength end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the longer wavelength end (1650-1700 cm-1).
The jagged peak at approximately 2900-3000 cm-1 is characteristic of tetrahedral carbon-hydrogen bonds. This peak is not terribly useful, as just about every organic molecule that you will have occasion to analyze has these bonds. Nevertheless, it can serve as a familiar reference point to orient yourself in a spectrum.
You will notice that there are many additional peaks in this spectrum in the longer-wavelength 400 -1400 cm-1 region. This part of the spectrum is called the fingerprint region. While it is usually very difficult to pick out any specific functional group identifications from this region, it does, nevertheless, contain valuable information. The reason for this is suggested by the name: just like a human fingerprint, the pattern of absorbance peaks in the fingerprint region is unique to every molecule, meaning that the data from an unknown sample can be compared to the IR spectra of known standards in order to make a positive identification. In the mid-1990's, for example, several paintings were identified as forgeries because scientists were able to identify the IR footprint region of red and yellow pigment compounds that would not have been available to the artist who supposedly created the painting (for more details see Chemical and Engineering News, Sept 10, 2007, p. 28).
Now, let’s take a look at the IR spectrum for 1-hexanol.
As you can see, the carbonyl peak is gone, and in its place is a very broad ‘mountain’ centered at about 3400 cm-1. This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules.
In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm-1.
We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance.
The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm-1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens.
Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm-1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen.
It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in this table.
As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol.
More examples of IR spectra
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum with one of the isomers in the row above it. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.07%3A_Interpreting_Infrared_Spectra.txt |
Objective
After completing this section, you should be able to use an infrared spectrum to determine the presence of functional groups, such as alcohols, amines and carbonyl groups, in an unknown compound, given a list of infrared absorption frequencies.
Study Notes
In Chapter 12.7 you should have learned, in broad terms, where a few key absorptions occur. Otherwise, to find the characteristic infrared absorptions of the various functional groups, refer to this IR table.
Spectral Interpretation by Application of Group Frequencies
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• C–H stretch from 3000–2850 cm-1
• C–H bend or scissoring from 1470-1450 cm-1
• C–H rock, methyl from 1370-1350 cm-1
• C–H rock, methyl, seen only in long chain alkanes, from 725-720 cm-1
Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• C=C stretch from 1680-1640 cm-1
• =C–H stretch from 3100-3000 cm-1
• =C–H bend from 1000-650 cm-1
Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• –C≡C– stretch from 2260-2100 cm-1
• –C≡C–H: C–H stretch from 3330-3270 cm-1
• –C≡C–H: C–H bend from 700-610 cm-1
The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• C–H stretch from 3100-3000 cm-1
• overtones, weak, from 2000-1665 cm-1
• C–C stretch (in-ring) from 1600-1585 cm-1
• C–C stretch (in-ring) from 1500-1400 cm-1
• C–H "oop" from 900-675 cm-1
Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• O–H stretch, hydrogen bonded 3500-3200 cm-1
• C–O stretch 1260-1050 cm-1 (s)
Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• C=O stretch - aliphatic ketones 1715 cm-1
• α, β -unsaturated ketones 1685-1666 cm-1
Figure 8. shows the spectrum of 2-butanone. This is a saturated ketone, and the C=O band appears at 1715.
If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• H–C=O stretch 2830-2695 cm-1
• C=O stretch:
• aliphatic aldehydes 1740-1720 cm-1
• α, β -unsaturated aldehydes 1710-1685 cm-1
Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• C=O stretch
• aliphatic from 1750-1735 cm-1
• α, β -unsaturated from 1730-1715 cm-1
• C–O stretch from 1300-1000 cm-1
Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• O–H stretch from 3300-2500 cm-1
• C=O stretch from 1760-1690 cm-1
• C–O stretch from 1320-1210 cm-1
• O–H bend from 1440-1395 and 950-910 cm-1
Figure 11. shows the spectrum of hexanoic acid.
Organic Nitrogen Compounds
• N–O asymmetric stretch from 1550-1475 cm-1
• N–O symmetric stretch from 1360-1290 cm-1
Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• C–H wag (-CH2X) from 1300-1150 cm-1
• C–X stretches (general) from 850-515 cm-1
• C–Cl stretch 850-550 cm-1
• C–Br stretch 690-515 cm-1
The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation table is linked on bottom of page to find other assigned IR peaks.
Exercises
Exercise \(1\)
Caffeine has a mass of 194.19 amu, determined by mass spectrometry, and contains C, N, H, O. What is a molecular formula for this molecule?
Answer
C8H10N4O2
C = 12 × 8 = 96
N = 14 × 4 = 56
H = 1 × 10 = 10
O = 2 × 16 = 32
96+56+10+32 = 194 g/mol
Exercise \(2\)
The following are the spectra for 2-methyl-2-hexene and 2-heptene, which spectra belongs to the correct molecule. Explain.
A:
B:
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
Answer
The (A) spectrum is 2-methyl-2-hexene and the (B) spectrum is 2-heptene. Looking at (A) the peak at 68 m/z is the fractioned molecule with just the tri-substituted alkene present. While (B) has a strong peak around the 56 m/z, which in this case is the di-substituted alkene left behind from the linear heptene.
Exercise \(3\)
What are the masses of all the components in the following fragmentations?
Answer
Exercise \(4\)
Which of the following frequencies/wavelengths are higher energy
A. λ = 2.0x10-6 m or λ = 3.0x10-9 m
B. υ = 3.0x109 Hz or υ = 3.0x10-6 Hz
Answer
A. λ = 3.0x10-9 m
B. υ = 3.0x109 Hz
Exercise \(5\)
Calculate the energies for the following;
A. Gamma Ray λ = 4.0x10-11 m
B. X-Ray λ = 4.0x10-9 m
C. UV light υ = 5.0x1015 Hz
D. Infrared Radiation λ = 3.0x10-5 m
E. Microwave Radiation υ = 3.0x1011 Hz
Answer
A. 4.965x10-15 J
B. 4.965x10-17 J
C. 3.31x10-18 J
D. 6.62x10-21 J
E. 1.99x10-22 J
Exercise \(6\)
What functional groups give the following signals in an IR spectrum?
A) 1700 cm-1
B) 1550 cm-1
C) 1700 cm-1 and 2510-3000 cm-1
Answer
Exercise \(7\)
How can you distinguish the following pairs of compounds through IR analysis?
A) CH3OH (Methanol) and CH3CH2OCH2CH3 (Diethylether)
B) Cyclopentane and 1-pentene.
C)
Answer
A) A OH peak will be present around 3300 cm-1 for methanol and will be absent in the ether.
B) 1-pentene will have a alkene peak around 1650 cm-1 for the C=C and there will be another peak around 3100 cm-1 for the sp2 C-H group on the alkene
C) Cannot distinguish these two isomers. They both have the same functional groups and therefore would have the same peaks on an IR spectra.
Exercise \(8\)
The following spectra is for the accompanying compound. What are the peaks that you can I identify in the spectrum?
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
Answer
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1610 C=C
(There is also an aromatic undertone region between 2000-1600 which describes the substitution on the phenyl ring.)
Exercise \(9\)
What absorptions would the following compounds have in an IR spectra?
Answer
A)
Frequency (cm-1) Functional Group
2900-3000 C-C-H, C=C-H
1710 C=O
1610 C=C
1100 C-O
B)
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1710 C=O
C)
Frequency (cm-1) Functional Group
3300 (broad) O-H
2900-3000 C-C-H, C=C-H
2000-1800 Aromatic Overtones
1710 C=O
1610 C=C | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.08%3A_Infrared_Spectra_of_Some_Common_Functional_Groups.txt |
Concepts & Vocabulary
12.1 Introduction
• Spectroscopy describes several techniques used by chemists to understand chemical structures and bonds.
12.2 Mass Spectrometry of Small Molecules - Magnetic Sector Instruments
• Mass spectrometers consist of an ion source, mass analyzer and dectector.
• There are several common ion sources including electron ionization and chemical ionization.
• Upon ionization, a molecular ion is formed (the molecule after losing a single electron) which will break into smaller pieces (fragments).
• Fragments that are charged will appear in the mass spectrum and are helpful in identifying the parent molecule.
• The most abundant ion in a mass spectrum is called the base peak.
• The ion with the same mass as the parent molecule is called the molecular ion.
• Isotopes of carbon and hydrogen lead to common M+1 peaks.
• The x-axis of a mass spectrum is m/z - the mass to charge ratio, which in practice equals the mass of the ion.
12.3 Interpreting Mass Spectra
• Uncharged particles do not appear in mass spectra.
• The y-axis of a mass spectrum is the relative abundance, with the base peak set at 100 as the most abundant ion.
• Abundance of ions is related to their stability.
12.4 Mass Spectrometry of Some Common Functional Groups
12.5 Mass Spectrometry in Biological - Time-of-flight (TOF) Instruments
12.6 Spectroscopy and the Electromagnetic Spectrum
• Electromagnetic radiation is composed of waves where shorter wavelengths correspond to higher energy radiation.
• Electromagnetic radiation can also be thought of as a stream of particles called photons.
• The electromagnetic spectrum is made up of many types of radiation including infrared, ultraviolet, and visible lights as well as x-rays, gamma rays, microwaves, and radio waves.
• Molecular spectroscopy works by exposing a chemical sample to electromagnetic radiation. It will only absorb radiation with energy that corresponds to some excited state, while all other energies will pass through unabsorbed.
12.7 Infrared Spectroscopy
• When infrared radiation is absorbed, molecules will move to a higher vibrational energy state.
• Examples of molecular vibrations include bending and stretching of bonds. These vibrations can be symmetric or asymmetric.
• In general, more polar bonds have stronger IR absorption.
• IR spectra typically use wavenumbers (cm-1) as units for the x-axis.
• The y-axis for IR spectra is usually % transmittance, with 100% at the top of the spectrum and absorbances looking like valleys (or downward peaks).
12.8 Interpreting Infrared Spectra
• Functional groups have standard regions within the IR spectrum where they absorb.
• The general regions include hydrogen bonding (O-H and N-H), carbon-hydrogen bonds, triple bonds, carbonyls, alkenes, and fingerprint region.
12.9 Infrared Spectra of Some Common Functional Groups
Skills to Master
• Skill 12.1 Determine specific atoms from mass spectra based on molecular ion and M+2 peaks (N, Cl, Br).
• Skill 12.2 Interpret mass spectra fragments - recognizing common fragments.
• Skill 12.3 Interpret infrared spectra to determine functional groups that are present or absent.
Memorization Tasks (MT)
• MT 12.1 Memorize common mass spectra fragments.
• MT 12.2 Memorize common functional group regions in infrared spectroscopy. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/12%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy/12.S%3A_Structure_Determination_-_Mass_Spectrometry_and_Infrared_Spectroscopy_%28Summary%29.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
Objectives
After completing this section, you should be able to
1. discuss the principles of NMR spectroscopy.
2. identify the two magnetic nuclei that are most important to an organic chemist.
Key Terms
Make certain that you can define, and use in context, the key term below.
• resonance
Study Notes
Notice that the word “resonance” has a different meaning when we are discussing nuclear magnetic resonance spectroscopy than it does when discussing molecular structures.
Introduction
Some types of atomic nuclei act as though they spin on their axis similar to the Earth. Since they are positively charged they generate an electromagnetic field just as the Earth does. So, in effect, they will act as tiny bar magnetics. Not all nuclei act this way, but fortunately both 1H and 13C do have nuclear spins and will respond to this technique.
NMR Spectrometer
In the absence of an external magnetic field the direction of the spin of the nuclei will be randomly oriented (see figure below left). However, when a sample of these nuclei is place in an external magnetic field, the nuclear spins will adopt specific orientations much as a compass needle responses to the Earth’s magnetic field and aligns with it. Two possible orientations are possible, with the external field (i.e. parallel to and in the same direction as the external field) or against the field (i.e. antiparallel to the external field) - see Figure \(1\).
If the ordered nuclei are now subjected to EM radiation of the proper frequency the nuclei aligned with the field will absorb energy and "spin-flip" to align themselves against the field, a higher energy state. When this spin-flip occurs the nuclei are said to be in "resonance" with the field, hence the name for the technique, Nuclear Magentic Resonance or NMR.
The amount of energy, and hence the exact frequency of EM radiation required for resonance to occur is dependent on both the strength of the magnetic field applied and the type of the nuclei being studied. As the strength of the magnetic field increases the energy difference between the two spin states increases and a higher frequency (more energy) EM radiation needs to be applied to achieve a spin-flip (see image below).
Superconducting magnets can be used to produce very strong magnetic field, on the order of 21 tesla (T). Lower field strengths can also be used, in the range of 4 - 7 T. At these levels the energy required to bring the nuclei into resonance is in the MHz range and corresponds to radio wavelength energies, i.e. at a field strength of 4.7 T 200 MHz bring 1H nuclei into resonance and 50 MHz bring 13C into resonance. This is considerably less energy then is required for IR spectroscopy, ~10-4 kJ/mol versus ~5 - ~50 kJ/mol.
1H and 13C are not unique in their ability to undergo NMR. All nuclei with an odd number of protons (1H, 2H, 14N, 19F, 31P ...) or nuclei with an odd number of neutrons (i.e. 13C) show the magnetic properties required for NMR. Only nuclei with even number of both protons and neutrons (12C and 16O) do not have the required magnetic properties. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.00%3A_Nuclear_Magnetic_Resonance__Spectroscopy.txt |
Objectives
After completing this section, you should be able to
1. explain, in general terms, the origin of shielding effects in NMR spectroscopy.
2. explain the number of peaks occurring in the 1H or 13C NMR spectrum of a simple compound, such as methyl acetate.
3. describe, and sketch a diagram of, a simple NMR spectrometer.
4. explain the difference in time scales of NMR and infrared spectroscopy.
5. predict the number of peaks expected in the 1H or 13C NMR spectrum of a given compound.
Study Notes
Before you go on, make sure that you understand that each signal in the 1H NMR spectrum shown for methyl acetate is due to a different proton environment. The three protons on the same methyl group are equivalent and appear in the spectrum as one signal. However, the two methyl groups are in two different environments (one is more deshielded) and so we see two signals in the whole spectrum (aside from the TMS reference peak).
Methyl acetate has a very simple 1H NMR spectrum, because there is no proton-proton coupling, and therefore no splitting of the signals. In later sections, we discuss splitting patterns in 1H NMR spectra and how they help a chemist determine the structure of organic compounds.
Nuclear precession, spin states, and the resonance condition
When a sample of an organic compound is sitting in a flask on a laboratory benchtop, the magnetic moments of its hydrogen atoms are randomly oriented. When the same sample is placed within the field of a very strong magnet in an NMR instrument (this field is referred to by NMR spectroscopists as the applied field, abbreviated B0 ) each hydrogen will assume one of two possible spin states. In what is referred to as the +½ spin state, the hydrogen's magnetic moment is aligned with the direction of B0, while in the -½ spin state it is aligned opposed to the direction of B0.
Because the +½ spin state is slightly lower in energy, in a large population of organic molecules slightly more than half of the hydrogen atoms will occupy this state, while slightly less than half will occupy the –½ state. The difference in energy between the two spin states increases with increasing strength of B0.This last statement is in italics because it is one of the key ideas in NMR spectroscopy, as we shall soon see.
At this point, we need to look a little more closely at how a proton spins in an applied magnetic field. You may recall playing with spinning tops as a child. When a top slows down a little and the spin axis is no longer completely vertical, it begins to exhibit precessional motion, as the spin axis rotates slowly around the vertical. In the same way, hydrogen atoms spinning in an applied magnetic field also exhibit precessional motion about a vertical axis. It is this axis (which is either parallel or antiparallel to B0) that defines the proton’s magnetic moment. In the figure below, the proton is in the +1/2 spin state.
The frequency of precession (also called the Larmour frequency, abbreviated ωL) is simply the number of times per second that the proton precesses in a complete circle. A proton`s precessional frequency increases with the strength of B0.
If a proton that is precessing in an applied magnetic field is exposed to electromagnetic radiation of a frequency ν that matches its precessional frequency ωL, we have a condition called resonance. In the resonance condition, a proton in the lower-energy +½ spin state (aligned with B0) will transition (flip) to the higher energy –½ spin state (opposed to B0). In doing so, it will absorb radiation at this resonance frequency ν = ωL. This frequency, as you might have already guessed, corresponds to the energy difference between the proton’s two spin states. With the strong magnetic fields generated by the superconducting magnets used in modern NMR instruments, the resonance frequency for protons falls within the radio-wave range, anywhere from 100 MHz to 800 MHz depending on the strength of the magnet.
The basics of an NMR experiment
So far, you may have the impression that all 1H nuclei in a molecule would absorb the same frequency. However, this would be of little use to organic chemists if that were the case. It turns out that not all 1H nuclei absorb the same frequency and this is the same for all other NMR active nuclei. It turns out that chemically nonequivalent protons (or other nuclei) have different resonance frequencies in the same applied magnetic field. Nonequivalent protons are in different chemical environments. This allows NMR spectroscopy to provide us with useful information about the structure of an organic molecule. A full explanation of how a modern NMR instrument functions is beyond the scope of this text, but here is what happens. First, a sample compound (we'll use methyl acetate) is placed inside a very strong applied magnetic field (B0). There are two types of protons in methyl acetate. Ha are bonded to a C that is then bonded to a carbonyl, whereas Hb are bonded to a carbon that is then bonded to an oxygen atom. This difference in bonding leads to different types of environments for Ha and Hb. All the Ha protons are the same since they all have the same type of bonding and will be in the same chemical environment and the same is true for Hb.
The basic arrangement of an NMR spectrometer is displayed below. A sample (in a small glass tube, where the methyl acetate is in solution) is placed between the poles of a strong magnet. A radio frequency generator pulses the sample and excites the nuclei causing a spin-flip. The spin flip is detected by the detector and the signal sent to a computer where it is processed.
In the magnet, all of the protons begin to precess: the Ha protons at precessional frequency ωa, the Hb protons at ωb. At first, the magnetic moments of (slightly more than) half of the protons are aligned with B0, and half (slightly less than half) are aligned against B0. Then, the sample is hit with electromagnetic radiation in the radio frequency range. The two specific frequencies which match ωa and ωb (i.e. the resonance frequencies) cause those Ha and Hb protons which are aligned with B0 to 'flip' so that they are now aligned against B0. In doing so, the protons absorb radiation at the two resonance frequencies. The NMR instrument records which frequencies were absorbed, as well as the intensity of each absorbance.
In most cases, a sample being analyzed by NMR is in solution. If we used a common laboratory solvent (diethyl ether, acetone, dichloromethane, ethanol, water, etc.) to dissolve our NMR sample, however, we run into a problem – there many more solvent protons in solution than there are sample protons, therefore the signals from the sample protons will be overwhelmed. To get around this problem, we use special NMR solvents in which all protons have been replaced by deuterium. Recall that deuterium is NMR-active, but its resonance frequency is very different from that of protons, and thus it is `invisible` in 1H-NMR. Some common NMR solvents are shown below. There are multiple deuterated solvents since molecules have different solubilities, so one molecule may dissolve in deuterated chloroform while others may not.
The Chemical Shift
Let's look at an actual 1H-NMR plot for methyl acetate. Just as in IR and UV-vis spectroscopy, the vertical axis corresponds to intensity of absorbance, the horizontal axis to frequency (typically the vertical axis is not shown in an NMR spectrum).
We see three absorbance signals: two of these correspond to Ha and Hb, while the peak at the far right of the spectrum corresponds to the 12 chemically equivalent protons in tetramethylsilane (TMS), a standard reference compound that was added to our sample.
You may be wondering about a few things at this point - why is TMS necessary, and what is the meaning of the `ppm (δ)` label on the horizontal axis? Shouldn't the frequency units be in Hz? Keep in mind that NMR instruments of many different applied field strengths are used in organic chemistry laboratories, and that the proton's resonance frequency range depends on the strength of the applied field. The spectrum above was generated on an instrument with an applied field of approximately 7.1 Tesla, at which strength protons resonate in the neighborhood of 300 million Hz (chemists refer to this as a 300 MHz instrument). If our colleague in another lab takes the NMR spectrum of the same molecule using an instrument with a 2.4 Tesla magnet, the protons will resonate at around 100 million Hz (so we’d call this a 100 MHz instrument). It would be inconvenient and confusing to always have to convert NMR data according to the field strength of the instrument used. Therefore, chemists report resonance frequencies not as absolute values in Hz, but rather as values relative to a common standard, generally the signal generated by the protons in TMS. This is where the ppm – parts per million – term comes in. Regardless of the magnetic field strength of the instrument being used, the resonance frequency of the 12 equivalent protons in TMS is defined as a zero point. The resonance frequencies of protons in the sample molecule are then reported in terms of how much higher they are, in ppm, relative to the TMS signal (almost all protons in organic molecules have a higher resonance frequency than those in TMS, for reasons we shall explore quite soon).
The two proton groups in our methyl acetate sample are recorded as resonating at frequencies 2.05 and 3.67 ppm higher than TMS. One-millionth (1.0 ppm) of 300 MHz is 300 Hz. Thus 2.05 ppm, on this instrument, corresponds to 615 Hz, and 3.67 ppm corresponds to 1101 Hz. If the TMS protons observed by our 7.1 Tesla instrument resonate at exactly 300,000,000 Hz, this means that the protons in our ethyl acetate samples are resonating at 300,000,615 and 300,001,101 Hz, respectively. Likewise, if the TMS protons in our colleague's 2.4 Tesla instrument resonate at exactly 100 MHz, the methyl acetate protons in her sample resonate at 100,000,205 and 100,000,367 Hz (on the 100 MHz instrument, 1.0 ppm corresponds to 100 Hz). The absolute frequency values in each case are not very useful – they will vary according to the instrument used – but the difference in resonance frequency from the TMS standard, expressed in parts per million, should be the same regardless of the instrument.
Expressed this way, the resonance frequency for a given proton in a molecule is called its chemical shift. A frequently used symbolic designation for chemical shift in ppm is the lower-case Greek letter delta (δ). Most protons in organic compounds have chemical shift values between 0 and 12 ppm from TMS, although values below zero and above 12 are occasionally observed. By convention, the left-hand side of an NMR spectrum (higher chemical shift) is called downfield, and the right-hand direction is called upfield.
In our methyl acetate example we included for illustrative purposes a small amount of TMS standard directly in the sample, as was the common procedure for determining the zero point with older NMR instruments．That practice is generally no longer necessary, as modern NMR instruments are designed to use the deuterium signal from the solvent as a standard reference point, then to extrapolate the 0 ppm baseline that corresponds to the TMS proton signal (in an applied field of 7.1 Tesla, the deuterium atom in CDCl3 resonates at 32 MHz, compared to 300 MHz for the protons in TMS). In the remaining NMR spectra that we will see in this text we will not see an actual TMS signal, but we can always assume that the 0 ppm point corresponds to where the TMS protons would resonate if they were present.
Example
A proton has a chemical shift (relative to TMS) of 4.56 ppm.
1. a) What is its chemical shift, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
2. b) What is its resonance frequency, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
(Assume that in these instruments, the TMS protons resonate at exactly 300 or 200 MHz, respectively)
Solution
Diamagnetic shielding and deshielding
We come now to the question of why nonequivalent protons have different chemical shifts. The chemical shift of a given proton is determined primarily by its immediate electronic environment. Consider the methane molecule (CH4), in which the protons have a chemical shift of 0.23 ppm. The valence electrons around the methyl carbon, when subjected to B0, are induced to circulate and thus generate their own very small magnetic field that opposes B0. This induced field, to a small but significant degree, shields the nearby protons from experiencing the full force of B0, an effect known as local diamagnetic shielding. The methane protons therefore do not experience the full force of B0 - what they experience is called Beff, or the effective field, which is slightly weaker than B0.
Therefore, their resonance frequency is slightly lower than what it would be if they did not have electrons nearby to shield them.
Now consider methyl fluoride, CH3F, in which the protons have a chemical shift of 4.26 ppm, significantly higher than that of methane. This is caused by something called the deshielding effect. Because fluorine is more electronegative than carbon, it pulls valence electrons away from the carbon, effectively decreasing the electron density around each of the protons. For the protons, lower electron density means less diamagnetic shielding, which in turn means a greater overall exposure to B0, a stronger Beff, and a higher resonance frequency. Put another way, the fluorine, by pulling electron density away from the protons, is deshielding them, leaving them more exposed to B0. As the electronegativity of the substituent increases, so does the extent of deshielding, and so does the chemical shift. This is evident when we look at the chemical shifts of methane and three halomethane compounds (remember that electronegativity increases as we move up a column in the periodic table).
To a large extent, then, we can predict trends in chemical shift by considering how much deshielding is taking place near a proton. The chemical shift of trichloromethane is, as expected, higher than that of dichloromethane, which is in turn higher than that of chloromethane.
The deshielding effect of an electronegative substituent diminishes sharply with increasing distance:
The presence of an electronegative oxygen, nitrogen, sulfur, or sp2-hybridized carbon also tends to shift the NMR signals of nearby protons slightly downfield:
Table 2 lists typical chemical shift values for protons in different chemical environments.
Armed with this information, we can finally assign the two peaks in the the 1H-NMR spectrum of methyl acetate that we saw a few pages back. The signal at 3.65 ppm corresponds to the methyl ester protons (Hb), which are deshielded by the adjacent oxygen atom. The upfield signal at 2.05 ppm corresponds to the acetate protons (Ha), which is deshielded - but to a lesser extent - by the adjacent carbonyl group.
Finally, a note on the use of TMS as a standard in NMR spectroscopy: one of the main reasons why the TMS proton signal was chosen as a zero-point is that the TMS protons are highly shielded: silicon is slightly less electronegative than carbon, and therefore donates some additional shielding electron density. Very few organic molecules contain protons with chemical shifts that are negative relative to TMS. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.01%3A_The_Nature_of_NMR_Absorptions.txt |
Objectives
After completing this section, you should be able to
1. describe the delta scale used in NMR spectroscopy.
2. perform calculations based on the relationship between the delta value (in ppm), the observed chemical shift (in Hz), and the operating frequency of an NMR spectrometer (in Hz).
Key Terms
Make certain that you can define, and use in context, the key terms below.
• chemical shift
• delta scale
• upfield/downfield
Study Notes
Although the calculations described in this section will help you understand the principles of NMR, it is the actual delta values, not the calculations, which are of greatest importance to the beginning organic chemist. Thus, we shall try to focus on the interpretation of NMR spectra, not the mathematical aspects of the technique.
In Section 13.9 we discuss 1H NMR chemical shifts in more detail. Although you will eventually be expected to associate the approximate region of a 1H NMR spectrum with a particular type of proton, you are expected to use a general table of 1H NMR chemical shifts such as the one shown in Section 13.9.
Chemical Shifts
The NMR spectra is displayed as a plot of the applied radio frequency versus the absorption. The applied frequency increases from left to right, thus the left side of the plot is the low field, downfield or deshielded side and the right side of the plot is the high field, upfield or shielded side (see the figure below). The concept of shielding will be explained shortly.
The position on the plot at which the nuclei absorbs is called the chemical shift. Since this has an arbitrary value a standard reference point must be used. The two most common standards are TMS (tetramethylsilane, (Si(CH3)4) which has been assigned a chemical shift of zero, and CDCl3 (deuterochloroform) which has a chemical shift of 7.26 for 1H NMR and 77 for 13C NMR. The scale is commonly expressed as parts per million (ppm) which is independent of the spectrometer frequency. The scale is the delta (δ) scale.
$\delta=\frac{\text { frequency of signal }-\text { frequency of standard }}{\text { spectrometerfrequency }} \times 10^6 \nonumber$
The range at which most NMR absorptions occur is quite narrow. Almost all 1H absorptions occur downfield within 10 ppm of TMS. For 13C NMR almost all absorptions occurs within 220 ppm downfield of the C atom in TMS.
Shielding in NMR
Structural features of the molecule will have an effect on the exact magnitude of the magnetic field experienced by a particular nucleus. This means that H atoms which have different chemical environments will have different chemical shifts. This is what makes NMR so useful for structure determination in organic chemistry. There are three main features that will affect the shielding of the nucleus, electronegativity, magnetic anisotropy of π systems and hydrogen bonding.
Electronegativity
The electrons that surround the nucleus are in motion so they created their own electromagnetic field. This field opposes the the applied magnetic field and so reduces the field experienced by the nucleus. Thus the electrons are said to shield the nucleus. Since the magnetic field experienced at the nucleus defines the energy difference between spin states it also defines what the chemical shift will be for that nucleus. Electron with-drawing groups can decrease the electron density at the nucleus, deshielding the nucleus and result in a larger chemical shift. Compare the data in the table below.
Compound, CH3X CH3F CH3OH CH3Cl CH3Br CH3I CH4 (CH3)4Si
Electronegativity of X 4.0 3.5 3.1 2.8 2.5 2.1 1.8
Chemical shift δ (ppm) 4.26 3.4 3.05 2.68 2.16 0.23 0
As can be seen from the data, as the electronegativity of X increases the chemical shift, δ increases. This is an effect of the halide atom pulling the electron density away from the methyl group. This exposes the nuclei of both the C and H atoms, "deshielding" the nuclei and shifting the peak downfield.
The effects are cumulative so the presence of more electron withdrawing groups will produce a greater deshielding and therefore a larger chemical shift, i.e.
Compound CH4 CH3Cl CH2Cl2 CHCl3
δ (ppm) 0.23 3.05 5.30 7.27
These inductive effects are not only felt by the immediately adjacent atoms, but the deshielding can occur further down the chain, i.e.
NMR signal -CH2-CH2-CH2Br
δ (ppm) 1.25 1.69 3.30
Magnetic Anisotropy: π Electron Effects
The π electrons in a compound, when placed in a magnetic field, will move and generate their own magnetic field. The new magnetic field will have an effect on the shielding of atoms within the field. The best example of this is benzene (see the figure below).
This effect is common for any atoms near a π bond, i.e.
Proton Type Effect Chemical shift (ppm)
C6H5-H highly deshielded 6.5 - 8
C=C-H deshielded 4.5 - 6
C≡C-H shielded* ~2.5
O=C-H very highly deshielded 9 - 10
* the acetylene H is shielded due to its location relative to the π system
Hydrogen Bonding
Protons that are involved in hydrogen bonding (i.e.-OH or -NH) are usually observed over a wide range of chemical shifts. This is due to the deshielding that occurs in the hydrogen bond. Since hydrogen bonds are dynamic, constantly forming, breaking and forming again, there will be a wide range of hydrogen bonds strengths and consequently a wide range of deshielding. This as well as solvation effects, acidity, concentration and temperature make it very difficult to predict the chemical shifts for these atoms.
Experimentally -OH and -NH can be identified by carrying out a simple D2O exchange experiment since these protons are exchangeable.
• run the normal H-NMR experiment on your sample
• add a few drops of D2O
• re-run the H-NMR experiment
• compare the two spectra and look for peaks that have "disappeared" | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.02%3A_The_Chemical_Shift.txt |
Objectives
After completing this section, you should be able to
1. state the approximate chemical shift (δ) for the following types of protons:
1. aromatic.
2. vinylic.
3. those bonded to carbon atoms which are in turn bonded to a highly electronegative element.
4. those bonded to carbons which are next to unsaturated centres.
5. those bonded to carbons which are part of a saturated system.
2. predict the approximate chemical shifts of each of the protons in an organic compound, given its structure and a table of chemical shift correlations.
Study Notes
You should not attempt to memorize the chemical shifts listed in the table of this section, although it is probable that you will need to refer to it quite frequently throughout the remainder of this course. To fulfill Objective 1, above, you should be familiar with the information presented in the figure of chemical shift ranges for organic compounds. If you have an approximate idea of the chemical shifts of some of the most common types of protons, you will find the interpretation of 1H NMR spectra less arduous than it might otherwise be. Notice that we shall not try to understand why aromatic protons are deshielded or why alkynyl protons are not deshielded as much as vinylic protons. These phenomena can be explained, but the focus is on the interpretation of 1H NMR spectra, not on the underlying theory.
1H NMR Chemical Shifts
Chemical shift is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilane [TMS;(CH3)4Si] is generally used for standard to determine chemical shift of compounds: δTMS=0ppm. In other words, frequencies for chemicals are measured for a 1H nucleus of a sample from the 1H or resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane. 1H chemical shift play a role in identifying many functional groups. Figure 1. indicates important example to figure out the functional groups.
Chemical shift values are in parts per million (ppm) relative to tetramethylsilane.
Hydrogen type Chemical shift (ppm)
RCH3 0.9 - 1.0
RCH2R 1.2 - 1.7
R3CH 1.5 – 2.0
2.0 – 2.3
1.5 – 1.8
RNH2 1 - 3
ArCH3 2.2 – 2.4
2.3 – 3.0
ROCH3 3.7 – 3.9
3.7 – 3.9
ROH 1 - 5
3.7 – 6.5
5 - 9
ArH 6.0 – 8.7
9.5 – 10.0
10 - 13
13.04: Integration of H NMR Absorptions- Proton Counting
Objectives
After completing this section, you should be able to
1. explain what information can be obtained from an integrated 1H NMR spectrum, and use this information in the interpretation of such a spectrum.
2. use an integrated 1H NMR spectrum to determine the ratio of the different types of protons present in an organic compound.
Study Notes
The concept of peak integration is that the area of a given peak in a 1H NMR spectrum is proportional to the number of (equivalent) protons giving rise to the peak. Thus, a peak which is caused by a single, unique proton has an area which measures one third of the area of a peak resulting from a methyl (CH3) group in the same spectrum.
In practice, we do not have to measure these areas ourselves: it is all done electronically by the spectrometer, and an integration curve is superimposed on the rest of the spectrum. The integration curve appears as a series of steps, with the height of each step being proportional to the area of the corresponding absorption peak, and consequently, to the number of protons responsible for the absorption.
As it can be difficult to decide precisely where to start and stop when measuring integrations, you should not expect your ratios to be exact whole numbers.
Signal integration
The computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals. This is very useful, because in 1H-NMR spectroscopy the area under a signal is proportional to the number of hydrogens to which the peak corresponds. In the previous example of methyl acetate from Section 13.2, for example, the Ha and Hb peaks would integrate to approximately the same area, because they both correspond to a set of three equivalent protons.
Now, take a look next at the spectrum of para-xylene (IUPAC name 1,4-dimethylbenzene):
This molecule has two sets of protons: the six methyl (Ha) protons and the four aromatic (Hb) protons. When we instruct the instrument to integrate the areas under the two signals, we find that the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm, which is the case with 6 methyl protons and 4 aromatic protons. This (along with the actual chemical shift values, which we'll discuss soon) tells us which set of protons corresponds to which NMR signal.
The integration function can also be used to determine the relative amounts of two or more compounds in a mixed sample. If we have a sample that is a 50:50 (mole/mole) mixture of benzene and acetone, for example, the acetone signal should integrate to the same value as the benzene sample, because both signals represent six equivalent protons. If we have a 50:50 mixture of acetone and cyclopentane, on the other hand, the ratio of the acetone peak area to the cylopentane peak area will be 3:5 (or 6:10), because the cyclopentane signal represents ten protons.
Example \(1\)
You take a 1H-NMR spectrum of a mixed sample of acetone (CH3(CO)CH3) and dichloromethane (CH2Cl2). The integral ratio of the two signals (acetone : dichloromethane) is 2.3 to 1. What is the molar ratio of the two compounds in the sample?
Example \(2\)
You take the 1H-NMR spectrum of a mixed sample of 36% para-xylene and 64% acetone in CDCl3 solvent (structures are shown earlier in this chapter). How many peaks do you expect to see? What is the expected ratio of integration values for these peaks? (set the acetone peak integration equal to 1.0)
Solutions | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.03%3A_Chemical_Shifts_in_H_NMR__Spectroscopy.txt |
Objectives
After completing this section, you should be able to
1. explain the spin-spin splitting pattern observed in the 1H NMR spectrum of a simple organic compound, such as chloroethane or 2-bromopropane.
2. interpret the splitting pattern of a given 1H NMR spectrum.
3. determine the structure of a relatively simple organic compound, given its 1H NMR spectrum and other relevant information.
4. use coupling constants to determine which groups of protons are coupling with one another in a 1H NMR spectrum.
5. predict the splitting pattern which should be observed in the 1H NMR spectrum of a given organic compound.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• coupling constant
• multiplet
• quartet
• triplet
• doublet
Study Notes
From what we have learned about 1H NMR spectra so far, we might predict that the spectrum of 1,1,2-trichloroethane, CHCl2CH2Cl, would consist of two peaks—one, at about 2.5-4.0 δ, expected for CH2-halogen compounds and one shifted downfield because of the presence of an additional electronegative chlorine atom on the second carbon. However, when we look at the spectrum it appears to be much more complex. True, we see absorptions in the regions we predicted, but these absorptions appear as a group of two peaks (a doublet) and a group of three peaks (a triplet). This complication, which may be disturbing to a student who longs for the simple life, is in fact very useful to the organic chemist, and adds greatly to the power of NMR spectroscopy as a tool for the elucidation of chemical structures. The split peaks (multiplets) arise because the magnetic field experienced by the protons of one group is influenced by the spin arrangements of the protons in an adjacent group.
Spin-spin coupling is often one of the more challenging topics for organic chemistry students to master. Remember the n + 1 rule and the associated coupling patterns.
The source of spin-spin coupling
The 1H-NMR spectra that we have seen so far (of methyl acetate and para-xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1H-NMR spectra of most organic molecules contain proton signals that are 'split' into two or more sub-peaks. Rather than being a complication, however, this splitting behavior actually provides us with more information about our sample molecule.
Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, we show enlargements of individual signals so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet. The Hb signal at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet.
The source of signal splitting is a phenomenon called spin-spin coupling, a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2 trichloromethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of Hb will be aligned with B0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B0. The Beff ‘felt’ by Ha is a slightly weaker if Hb is aligned against B0, or slightly stronger if Hb is aligned with B0. In other words, in half of the molecules Ha is shielded by Hb (thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb(and the NMR signal shifted slightly downfield). What would otherwise be a single Ha peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram, as shown below.
Now, let's think about the Hbsignal. The magnetic environment experienced by Hb is influenced by the fields of both neighboring Ha protons, which we will call Ha1 and Ha2. There are four possibilities here, each of which is equally probable. First, the magnetic fields of both Ha1 and Ha2 could be aligned with B0, which would deshield Hb, shifting its NMR signal slightly downfield. Second, both the Ha1 and Ha2 magnetic fields could be aligned opposed to B0, which would shield Hb, shifting its resonance signal slightly upfield. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1opposed to B0 and Ha2 with B0. In each of the last two cases, the shielding effect of one Ha proton would cancel the deshielding effect of the other, and the chemical shift of Hb would be unchanged.
So in the end, the signal for Hb is a triplet, with the middle peak twice as large as the two outer peaks because there are two ways that Ha1 and Ha2 can cancel each other out.
Now, consider the spectrum for ethyl acetate:
We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (Ha) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. The signal at 1.055 ppm for the Hc hydrogens is split into a triplet by the two Hb hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane.
The Hbhydrogens give rise to a quartet signal at 3.915 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three Hc hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns.
Example 13.11.1
1. Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the Hc hydrogens, why the Hb signal in ethyl acetate is split into a quartet.
2. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the Hb quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.)
Solution
By now, you probably have recognized the pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. Thus the two Hb hydrogens in ethyl acetate split the Hc signal into a triplet, and the three Hc hydrogens split the Hb signal into a quartet. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two `neighbors`. When we begin to determine structures of unknown compounds using 1H-NMR spectral data, it will become more apparent how this kind of information can be used.
Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, Ha1 in 1,1,2-trichloroethane is not split by Ha2, and vice-versa.
Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the Ha hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away, too far for coupling to occur.
Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more details about how we quantify coupling interactions is provided in section 5.5B). Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules.
Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules.
Multiplicity in Proton NMR
The number of lines in a peak is always one more (n+1) than the number of hydrogens on the neighboring carbon. This table summarizes coupling patterns that arise when protons have different numbers of neighbors.
# of lines ratio of lines term for peak # of neighbors
1 - singlet 0
2 1:1 doublet 1
3 1:2:1 triplet 2
4 1:3:3:1 quartet 3
5 1:4:6:4:1 quintet 4
6 1:5:10:10:5:1 sextet 5
7 1:6:15:20:15:6:1 septet 6
8 1:7:21:35:35:21:7:1 octet 7
9 1:8:28:56:70:56:28:8:1 nonet 8
Example 13.11.2
How many proton signals would you expect to see in the 1H-NMR spectrum of the structure shown? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Answer
Because of the symmetry in the molecule, there are only four proton signals. Predicted splitting is indicated.
Example 13.11.3
Predict the splitting pattern for the 1H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin).
Solutions
Coupling constants
Chemists quantify the spin-spin coupling effect using something called the coupling constant, which is abbreviated with the capital letter J. The coupling constant is simply the difference, expressed in Hz, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two subpeaks are separated by 6.1 Hz, and thus we write 3Ja-b = 6.1 Hz.
The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between Ha and Hb. Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet. This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength.
When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! The coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Coupling constants between proton sets on neighboring sp3-hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2-hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement.
For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens.
Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values.
Exercise
Note: Remember, chemically equivalent protons do not couple with one another to give spin-spin splitting. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.05%3A_Spin-Spin_Splitting_in_H_NMR__Spectra.txt |
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