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08: Aromaticity: Reactions of Benzene and Substituted Benzenes
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8.04: The Nomenclature of Monosubstituted Benzenes
Benzene, C6H6, is an organic aromatic compound with many interesting properties. Unlike aliphatic (straight chain carbons) or other cyclic organic compounds, the structure of benzene (3 conjugated π bonds) allows benzene and its derived products to be useful in fields such as health, laboratory, and other applications such as rubber synthesis.
Introduction
Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (Figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections Aromaticity and Hückel's Rule.
Figure 1. Top-view of camphor, along with its monoterpene unit. Notice how camphor lacks the benzene ring to be "aromatic".
Many aromatic compounds are however, sweet/pleasant smelling. Eugenol, for example, is extracted from essential oils of cloves and it releases a spicy, clove-like aroma used in perfumes. In addition, it is also used in dentistry as an analgesic.
Figure 2. Eugenol, an aromatic compound extracted from clove essential oils. Used in perfumes and as an analgesic. The benzene ring is labeled in red in the eugenol molecule.
Is it cyclohexane or is it benzene?
Due to the similarity between benzene and cyclohexane, the two is often confused with each other in beginning organic chemistry students.
Figure 3. Structure comparison between cyclohexane and benzene
If you were to count the number of carbons and hydrogens in cyclohexane, you will notice that its molecular formula is C6H12. Since the carbons in the cyclohexane ring is fully saturated with hydrogens (carbon is bound to 2 hydrogens and 2 adjacent carbons), no double bonds are formed in the cyclic ring. In contrast, benzene is only saturated with one hydrogen per carbon, leading to its molecular formula of C6H6. In order to stabilize this structure, 3 conjugated π (double) bonds are formed in the benzene ring in order for carbon to have four adjacent bonds.
In other words, cyclohexane is not the same as benzene! These two compounds have different molecular formulas and their chemical and physical properties are not the same. The hydrogenation technique can be used by chemists to convert from benzene to cyclohexane by saturating the benzene ring with missing hydrogens.
A special catalyst is required to hydrogenate benzene rings due to its unusual stability and configuration. Normal catalytic hydrogenation techniques will not hydrogenate benzene and yield any meaningful products.
What about Resonance?
Benzene can be drawn a number of different ways. This is because benzene's conjugated pi electrons freely resonate within the cyclic ring, thus resulting in its two resonance forms.
Figure 4. The Figure to the left shows the two resonance forms of benzene. The delocalized electrons are moved from one carbon to the next, thus providing stabilization energy. Ring structures stabilized by the movement of delocalized electrons are sometimes referred to as arenes.
As the electrons in the benzene ring can resonate within the ring at a fairly high rate, a simplified notation is often used to designate the two different resonance forms. This notation is shown above, with the initial three pi bonds (#1, #2) replaced with an inner ring circle (#3). Alternatively, the circle within the benzene ring can also be dashed to show the same resonance forms (#4).
The Formation of the Phenyl Group and its Derivatives
The phenyl group can be formed by taking benzene, and removing a hydrogen from it. The resulting molecular formula for the fragment is C6H5. NOTE: Although the molecular formula of the phenyl group is C6H5, the phenyl group would always have something attached to where the hydrogen was removed. Thus, the formula is often written as Ph-R, where Ph refers to the Phenyl group, and R refers to the R group attached to where the hydrogen was removed.
Figure 5. Figure demonstrating the removal of hydrogen to form the phenyl group.
Different R groups on the phenyl group allows different benzene derivatives to be formed. Phenol, Ph-OH, or C6H5OH, for example, is formed when an alcohol (-OH) group displaces a hydrogen atom on the benzene ring. Benzene, for this very same reason, can be formed from the phenyl group by reattaching the hydrogen back its place of removal. Thus benzene, similar to phenol, can be abbreviated Ph-H, or C6H6.
Figure 7: Epigallocatechin gallate (EGCG), an antioxidant found in green teas and its extracts, is famous for its potential health benefits. The molecule is a type of catechin, which is composed of multiple phenol (labeled in red) units (polyphenols - see polycyclic aromatics). Since catechins are usually found in plant extracts, they are often referred as plant polyphenolic antioxidants.
As you can see above, these are only some of the many possibilities of the benzene derived products that have special uses in human health and other industrial fields.
Nomenclature of Benzene Derived Compounds
Unlike aliphatic organics, nomenclature of benzene-derived compounds can be confusing because a single aromatic compound can have multiple possible names (such as common and systematic names) be associated with its structure. In these sections, we will analyze some of the ways these compounds can be named.
Simple Benzene Naming
Some common substituents, like NO2, Br, and Cl, can be named this way when it is attached to a phenyl group. Long chain carbons attached can also be named this way. The general format for this kind of naming is:
(positions of substituents (if >1)- + # (di, tri, ...) + substituent)n + benzene.
For example, chlorine (Cl) attached to a phenyl group would be named chlorobenzene (chloro + benzene). Since there is only one substituent on the benzene ring, we do not have to indicate its position on the benzene ring (as it can freely rotate around and you would end up getting the same compound.)
Figure 8. Example of simple benzene naming with chlorine and NO2 as substituents.
Figure 9. More complicated simple benzene naming examples - Note that standard nomenclature priority rules are applied here, causing the numbering of carbons to switch. See Nomenclature of Organic Compounds for a review on naming and priority rules.
Ortho-, Meta-, Para- (OMP) Nomenclature for Disubstituted Benzenes
Instead of using numbers to indicate substituents on a benzene ring, ortho- (o-), meta- (m-), or para (p-) can be used in place of positional markers when there are two substituents on the benzene ring (disubstituted benzenes). They are defined as the following:
• ortho- (o-): 1,2- (next to each other in a benzene ring)
• meta- (m): 1,3- (separated by one carbon in a benzene ring)
• para- (p): 1,4- (across from each other in a benzene ring)
Using the same example above in Figure 9a (1,3-dichlorobenzene), we can use the ortho-, meta-, para- nomenclature to transform the chemical name into m-dichlorobenzene, as shown in the Figure below.
Figure 10. Transformation of 1,3-dichlorobenzene into m-dichlorobenzene.
Here are some other examples of ortho-, meta-, para- nomenclature used in context:
However, the substituents used in ortho-, meta-, para- nomenclature do not have to be the same. For example, we can use chlorine and a nitro group as substituents in the benzene ring.
In conclusion, these can be pieced together into a summary diagram, as shown below:
Base Name Nomenclature
In addition to simple benzene naming and OMP nomenclature, benzene derived compounds are also sometimes used as bases. The concept of a base is similar to the nomenclature of aliphatic and cyclic compounds, where the parent for the organic compound is used as a base (a name for its chemical name. For example, the following compounds have the base names hexane and cyclohexane, respectively. See Nomenclature of Organic Compounds for a review on naming organic compounds.
Benzene, similar to these compounds shown above, also has base names from its derived compounds. Phenol (C6H5OH), as introduced previously in this article, for example, serves as a base when other substituents are attached to it. This is best illustrated in the diagram below.
Figure 14. An example showing phenol as a base in its chemical name. Note how benzene no longer serves as a base when an OH group is added to the benzene ring.
Alternatively, we can use the numbering system to indicate this compound. When the numbering system is used, the carbon where the substituent is attached on the base will be given the first priority and named as carbon #1 (C1). The normal priority rules then apply in the nomenclature process (give the rest of the substituents the lowest numbering as you could).
Figure 15. The naming process for 2-chlorophenol (o-chlorophenol). Note that 2-chlorophenol = o-chlorophenol.
Below is a list of commonly seen benzene-derived compounds. Some of these mono-substituted compounds (labeled in red and green), such as phenol or toluene, can be used in place of benzene for the chemical's base name.
Figure 16. Common benzene derived compounds with various substituents.
Common vs. Systematic (IUPAC) Nomenclature
According to the indexing preferences of the Chemical Abstracts, phenol, benzaldehyde, and benzoic acid (labeled in red in Figure 16) are some of the common names that are retained in the IUPAC (systematic) nomenclature. Other names such as toluene, styrene, naphthalene, or phenanthrene can also be seen in the IUPAC system in the same way. While the use of other common names are usually acceptable in IUPAC, their use are discouraged in the nomenclature of compounds.
Nomenclature for compounds which has such discouraged names will be named by the simple benzene naming system. An example of this would include toluene derivatives like TNT. (Note that toluene by itself is retained by the IUPAC nomenclature, but its derivatives, which contains additional substituents on the benzene ring, might be excluded from the convention). For this reason, the common chemical name 2,4,6-trinitrotoluene, or TNT, as shown in Figure 17, would not be advisable under the IUPAC (systematic) nomenclature.
In order to correctly name TNT under the IUPAC system, the simple benzene naming system should be used:
Figure 18. Systematic (IUPAC) name of 2,4,6-trinitrotoluene (common name), or TNT. Note that the methyl group is individually named due to the exclusion of toluene from the IUPAC nomenclature.
Figure 19. The common name 2,4-dibromophenol, is shared by the IUPAC systematic nomenclature. Only substituents phenol, benzoic acid, and benzaldehyde share this commonality.
Since the IUPAC nomenclature primarily rely on the simple benzene naming system for the nomenclature of different benzene derived compounds, the OMP (ortho-, meta-, para-) system is not accepted in the IUPAC nomenclature. For this reason, the OMP system will yield common names that can be converted to systematic names by using the same method as above. For example, o-Xylene from the OMP system can be named 1,2-dimethylbenzene by using simple benzene naming (IUPAC standard).
The Phenyl and Benzyl Groups
The Phenyl Group
As mentioned previously, the phenyl group (Ph-R, C6H5-R) can be formed by removing a hydrogen from benzene and attaching a substituent to where the hydrogen was removed. To this phenomenon, we can name compounds formed this way by applying this rule: (phenyl + substituent). For example, a chlorine attached in this manner would be named phenyl chloride, and a bromine attached in this manner would be named phenyl bromide. (See below diagram)
Figure 20. Naming of Phenyl Chloride and Phenyl Bromide
While compounds like these are usually named by simple benzene type naming (chlorobenzene and bromobenzene), the phenyl group naming is usually applied to benzene rings where a substituent with six or more carbons is attached, such as in the diagram below.
Figure 21. Diagram of 2-phenyloctane.
Although the diagram above might be a little daunting to understand at first, it is not as difficult as it seems after careful analysis of the structure is made. By looking for the longest chain in the compound, it should be clear that the longest chain is eight (8) carbons long (octane, as shown in green) and that a benzene ring is attached to the second position of this longest chain (labeled in red). As this rule suggests that the benzene ring will act as a function group (a substituent) whenever a substituent of more than six (6) carbons is attached to it, the name "benzene" is changed to phenyl and is used the same way as any other substituents, such as methyl, ethyl, or bromo. Putting it all together, the name can be derived as: 2-phenyloctane (phenyl is attached at the second position of the longest carbon chain, octane).
The Benzyl Group
The benzyl group (abbv. Bn), similar to the phenyl group, is formed by manipulating the benzene ring. In the case of the benzyl group, it is formed by taking the phenyl group and adding a CH2 group to where the hydrogen was removed. Its molecular fragment can be written as C6H5CH2-R, PhCH2-R, or Bn-R. Nomenclature of benzyl group based compounds are very similar to the phenyl group compounds. For example, a chlorine attached to a benzyl group would simply be called benzyl chloride, whereas an OH group attached to a benzyl group would simply be called benzyl alcohol.
Figure 22. Benzyl Group Nomenclature
Additionally, other substituents can attach on the benzene ring in the presence of the benzyl group. An example of this can be seen in the Figure below:
Figure 23. Nomenclature of 2,4-difluorobenzyl chloride. Similar to the base name nomenclatures system, the carbon in which th base substitutent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible.
Similar to the base name nomenclature system, the carbon in which the base substituent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible. Under this consideration, the above compound can be named: 2,4-difluorobenzyl chloride.
Commonly Named Benzene Compounds Nomenclature Summary Flowchart
Summary Flowchart (Figure 24). Summary of nomenclature rules used in commonly benzene derived compounds.
As benzene derived compounds can be extremely complex, only compounds covered in this article and other commonly named compounds can be named using this flowchart.
Determination of Common and Systematic Names using Flowchart
To demonstrate how this flowchart can be used to name TNT in its common and systematic (IUPAC) name, a replica of the flowchart with the appropriate flow paths are shown below:
Practice Problems
Q1) (True/False) The compound above contains a benzene ring and thus is aromatic.
Q2) Benzene unusual stability is caused by how many conjugated pi bonds in its cyclic ring? ____
Q3) Menthol, a topical analgesic used in many ointments for the relief of pain, releases a peppermint aroma upon exposure to the air. Based on this conclusion, can you imply that a benzene ring is present in its chemical structure? Why or why not?
Q4)
Q5) At normal conditions, benzene has ___ resonance structures.
Q6) Which of the following name(s) is/are correct for the following compound?
a) nitrohydride benzene
b) phenylamine
c) phenylamide
d) aniline
e) nitrogenhydrogen benzene
f) All of the above is correct
Q7) Convert 1,4-dimethylbenzene into its common name.
Q8) TNT's common name is: ______________________________
Q9) Name the following compound using OMP nomenclature:
Q10) Draw the structure of 2,4-dinitrotoluene.
Q11) Name the following compound:
Q12) Which of the following is the correct name for the following compound?
a) 3,4-difluorobenzyl bromide
b) 1,2-difluorobenzyl bromide
c) 4,5-difluorobenzyl bromide
d) 1,2-difluoroethyl bromide
e) 5,6-difluoroethyl bromide
f) 4,5-difluoroethyl bromide
Q13) (True/False) Benzyl chloride can be abbreviated Bz-Cl.
Q14) Benzoic Acid has what R group attached to its phenyl functional group?
Q15) (True/False) A single aromatic compound can have multiple names indicating its structure.
Q16) List the corresponding positions for the OMP system (o-, m-, p-).
Q17) A scientist has conducted an experiment on an unknown compound. He was able to determine that the unknown compound contains a cyclic ring in its structure as well as an alcohol (-OH) group attached to the ring. What is the unknown compound?
a) Cyclohexanol
b) Cyclicheptanol
c) Phenol
d) Methanol
e) Bleach
f) Cannot determine from the above information
Q18) Which of the following statements is false for the compound, phenol?
a) Phenol is a benzene derived compound.
b) Phenol can be made by attaching an -OH group to a phenyl group.
c) Phenol is highly toxic to the body even in small doses.
d) Phenol can be used as a catalyst in the hydrogenation of benzene into cyclohexane.
e) Phenol is used as an antiseptic in minute doses.
f) Phenol is amongst one of the three common names retained in the IUPAC nomenclature.
Answer Key to Practice Questions
Q1) False, this compound does not contain a benzene ring in its structure.
Q2) 3
Q3) No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (Figure 1)
Q4) No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds.
Q5) 2
Q6) b, d
Q7) p-Xylene
Q8) 2,4,6-trinitrotoluene
Q9) p-chloronitrobenzene
Q10)
Q11) 4-phenylheptane
Q12) a
Q13) False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl.
Q14) COOH
Q15) True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene.
Q16) Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4
Q17) The correct answer is f). We cannot determine what structure this is since the question does not tell us what kind of cyclic ring the -OH group is attached on. Just as cyclohexane can be cyclic, benzene and cycloheptane can also be cyclic.
Q18) d
• David Lam
8.11: The Friedel-Crafts Alkylation of Benzene
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and chloromethane in the presence of an aluminium chloride catalyst. Any other chloroalkane would work similarly.
The electrophilic substitution reaction between benzene and chloromethane
Alkylation means substituting an alkyl group into something - in this case into a benzene ring. A hydrogen on the ring is replaced by a group like methyl or ethyl and so on.
Benzene is treated with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminum chloride as a catalyst. On this page, we will look at substituting a methyl group, but any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene - once known as toluene.
$C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl$
or better:
The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl3 over the top of the arrow.
The formation of the electrophile
The electrophile is CH3+. It is formed by reaction between the chloromethane and the aluminum chloride catalyst.
$CH_3Cl + AlCl_3 \rightarrow ^+CH_3 + AlCl_4^-$
The electrophilic substitution mechanism
Stage one
Stage two
The hydrogen is removed by the $AlCl_4^-$ ion wh ich was formed at the same time as the $CH_3^+$ ele ctrophile . The aluminum chloride catalyst is re-generated in this second stage.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/08%3A_Aromaticity%3A_Reactions_of_Benzene_and_Substituted_Benzenes/8.01_The_Two_Criteria_for_Aromaticity.txt |
Benzene, C6H6, is an organic aromatic compound with many interesting properties. Unlike aliphatic (straight chain carbons) or other cyclic organic compounds, the structure of benzene (3 conjugated π bonds) allows benzene and its derived products to be useful in fields such as health, laboratory, and other applications such as rubber synthesis.
Introduction
Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (Figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections Aromaticity and Hückel's Rule.
Figure 1. Top-view of camphor, along with its monoterpene unit. Notice how camphor lacks the benzene ring to be "aromatic".
Many aromatic compounds are however, sweet/pleasant smelling. Eugenol, for example, is extracted from essential oils of cloves and it releases a spicy, clove-like aroma used in perfumes. In addition, it is also used in dentistry as an analgesic.
Figure 2. Eugenol, an aromatic compound extracted from clove essential oils. Used in perfumes and as an analgesic. The benzene ring is labeled in red in the eugenol molecule.
Is it cyclohexane or is it benzene?
Due to the similarity between benzene and cyclohexane, the two is often confused with each other in beginning organic chemistry students.
Figure 3. Structure comparison between cyclohexane and benzene
If you were to count the number of carbons and hydrogens in cyclohexane, you will notice that its molecular formula is C6H12. Since the carbons in the cyclohexane ring is fully saturated with hydrogens (carbon is bound to 2 hydrogens and 2 adjacent carbons), no double bonds are formed in the cyclic ring. In contrast, benzene is only saturated with one hydrogen per carbon, leading to its molecular formula of C6H6. In order to stabilize this structure, 3 conjugated π (double) bonds are formed in the benzene ring in order for carbon to have four adjacent bonds.
In other words, cyclohexane is not the same as benzene! These two compounds have different molecular formulas and their chemical and physical properties are not the same. The hydrogenation technique can be used by chemists to convert from benzene to cyclohexane by saturating the benzene ring with missing hydrogens.
A special catalyst is required to hydrogenate benzene rings due to its unusual stability and configuration. Normal catalytic hydrogenation techniques will not hydrogenate benzene and yield any meaningful products.
What about Resonance?
Benzene can be drawn a number of different ways. This is because benzene's conjugated pi electrons freely resonate within the cyclic ring, thus resulting in its two resonance forms.
Figure 4. The Figure to the left shows the two resonance forms of benzene. The delocalized electrons are moved from one carbon to the next, thus providing stabilization energy. Ring structures stabilized by the movement of delocalized electrons are sometimes referred to as arenes.
As the electrons in the benzene ring can resonate within the ring at a fairly high rate, a simplified notation is often used to designate the two different resonance forms. This notation is shown above, with the initial three pi bonds (#1, #2) replaced with an inner ring circle (#3). Alternatively, the circle within the benzene ring can also be dashed to show the same resonance forms (#4).
The Formation of the Phenyl Group and its Derivatives
The phenyl group can be formed by taking benzene, and removing a hydrogen from it. The resulting molecular formula for the fragment is C6H5. NOTE: Although the molecular formula of the phenyl group is C6H5, the phenyl group would always have something attached to where the hydrogen was removed. Thus, the formula is often written as Ph-R, where Ph refers to the Phenyl group, and R refers to the R group attached to where the hydrogen was removed.
Figure 5. Figure demonstrating the removal of hydrogen to form the phenyl group.
Different R groups on the phenyl group allows different benzene derivatives to be formed. Phenol, Ph-OH, or C6H5OH, for example, is formed when an alcohol (-OH) group displaces a hydrogen atom on the benzene ring. Benzene, for this very same reason, can be formed from the phenyl group by reattaching the hydrogen back its place of removal. Thus benzene, similar to phenol, can be abbreviated Ph-H, or C6H6.
Figure 7: Epigallocatechin gallate (EGCG), an antioxidant found in green teas and its extracts, is famous for its potential health benefits. The molecule is a type of catechin, which is composed of multiple phenol (labeled in red) units (polyphenols - see polycyclic aromatics). Since catechins are usually found in plant extracts, they are often referred as plant polyphenolic antioxidants.
As you can see above, these are only some of the many possibilities of the benzene derived products that have special uses in human health and other industrial fields.
Nomenclature of Benzene Derived Compounds
Unlike aliphatic organics, nomenclature of benzene-derived compounds can be confusing because a single aromatic compound can have multiple possible names (such as common and systematic names) be associated with its structure. In these sections, we will analyze some of the ways these compounds can be named.
Simple Benzene Naming
Some common substituents, like NO2, Br, and Cl, can be named this way when it is attached to a phenyl group. Long chain carbons attached can also be named this way. The general format for this kind of naming is:
(positions of substituents (if >1)- + # (di, tri, ...) + substituent)n + benzene.
For example, chlorine (Cl) attached to a phenyl group would be named chlorobenzene (chloro + benzene). Since there is only one substituent on the benzene ring, we do not have to indicate its position on the benzene ring (as it can freely rotate around and you would end up getting the same compound.)
Figure 8. Example of simple benzene naming with chlorine and NO2 as substituents.
Figure 9. More complicated simple benzene naming examples - Note that standard nomenclature priority rules are applied here, causing the numbering of carbons to switch. See Nomenclature of Organic Compounds for a review on naming and priority rules.
Ortho-, Meta-, Para- (OMP) Nomenclature for Disubstituted Benzenes
Instead of using numbers to indicate substituents on a benzene ring, ortho- (o-), meta- (m-), or para (p-) can be used in place of positional markers when there are two substituents on the benzene ring (disubstituted benzenes). They are defined as the following:
• ortho- (o-): 1,2- (next to each other in a benzene ring)
• meta- (m): 1,3- (separated by one carbon in a benzene ring)
• para- (p): 1,4- (across from each other in a benzene ring)
Using the same example above in Figure 9a (1,3-dichlorobenzene), we can use the ortho-, meta-, para- nomenclature to transform the chemical name into m-dichlorobenzene, as shown in the Figure below.
Figure 10. Transformation of 1,3-dichlorobenzene into m-dichlorobenzene.
Here are some other examples of ortho-, meta-, para- nomenclature used in context:
However, the substituents used in ortho-, meta-, para- nomenclature do not have to be the same. For example, we can use chlorine and a nitro group as substituents in the benzene ring.
In conclusion, these can be pieced together into a summary diagram, as shown below:
Base Name Nomenclature
In addition to simple benzene naming and OMP nomenclature, benzene derived compounds are also sometimes used as bases. The concept of a base is similar to the nomenclature of aliphatic and cyclic compounds, where the parent for the organic compound is used as a base (a name for its chemical name. For example, the following compounds have the base names hexane and cyclohexane, respectively. See Nomenclature of Organic Compounds for a review on naming organic compounds.
Benzene, similar to these compounds shown above, also has base names from its derived compounds. Phenol (C6H5OH), as introduced previously in this article, for example, serves as a base when other substituents are attached to it. This is best illustrated in the diagram below.
Figure 14. An example showing phenol as a base in its chemical name. Note how benzene no longer serves as a base when an OH group is added to the benzene ring.
Alternatively, we can use the numbering system to indicate this compound. When the numbering system is used, the carbon where the substituent is attached on the base will be given the first priority and named as carbon #1 (C1). The normal priority rules then apply in the nomenclature process (give the rest of the substituents the lowest numbering as you could).
Figure 15. The naming process for 2-chlorophenol (o-chlorophenol). Note that 2-chlorophenol = o-chlorophenol.
Below is a list of commonly seen benzene-derived compounds. Some of these mono-substituted compounds (labeled in red and green), such as phenol or toluene, can be used in place of benzene for the chemical's base name.
Figure 16. Common benzene derived compounds with various substituents.
Common vs. Systematic (IUPAC) Nomenclature
According to the indexing preferences of the Chemical Abstracts, phenol, benzaldehyde, and benzoic acid (labeled in red in Figure 16) are some of the common names that are retained in the IUPAC (systematic) nomenclature. Other names such as toluene, styrene, naphthalene, or phenanthrene can also be seen in the IUPAC system in the same way. While the use of other common names are usually acceptable in IUPAC, their use are discouraged in the nomenclature of compounds.
Nomenclature for compounds which has such discouraged names will be named by the simple benzene naming system. An example of this would include toluene derivatives like TNT. (Note that toluene by itself is retained by the IUPAC nomenclature, but its derivatives, which contains additional substituents on the benzene ring, might be excluded from the convention). For this reason, the common chemical name 2,4,6-trinitrotoluene, or TNT, as shown in Figure 17, would not be advisable under the IUPAC (systematic) nomenclature.
In order to correctly name TNT under the IUPAC system, the simple benzene naming system should be used:
Figure 18. Systematic (IUPAC) name of 2,4,6-trinitrotoluene (common name), or TNT. Note that the methyl group is individually named due to the exclusion of toluene from the IUPAC nomenclature.
Figure 19. The common name 2,4-dibromophenol, is shared by the IUPAC systematic nomenclature. Only substituents phenol, benzoic acid, and benzaldehyde share this commonality.
Since the IUPAC nomenclature primarily rely on the simple benzene naming system for the nomenclature of different benzene derived compounds, the OMP (ortho-, meta-, para-) system is not accepted in the IUPAC nomenclature. For this reason, the OMP system will yield common names that can be converted to systematic names by using the same method as above. For example, o-Xylene from the OMP system can be named 1,2-dimethylbenzene by using simple benzene naming (IUPAC standard).
The Phenyl and Benzyl Groups
The Phenyl Group
As mentioned previously, the phenyl group (Ph-R, C6H5-R) can be formed by removing a hydrogen from benzene and attaching a substituent to where the hydrogen was removed. To this phenomenon, we can name compounds formed this way by applying this rule: (phenyl + substituent). For example, a chlorine attached in this manner would be named phenyl chloride, and a bromine attached in this manner would be named phenyl bromide. (See below diagram)
Figure 20. Naming of Phenyl Chloride and Phenyl Bromide
While compounds like these are usually named by simple benzene type naming (chlorobenzene and bromobenzene), the phenyl group naming is usually applied to benzene rings where a substituent with six or more carbons is attached, such as in the diagram below.
Figure 21. Diagram of 2-phenyloctane.
Although the diagram above might be a little daunting to understand at first, it is not as difficult as it seems after careful analysis of the structure is made. By looking for the longest chain in the compound, it should be clear that the longest chain is eight (8) carbons long (octane, as shown in green) and that a benzene ring is attached to the second position of this longest chain (labeled in red). As this rule suggests that the benzene ring will act as a function group (a substituent) whenever a substituent of more than six (6) carbons is attached to it, the name "benzene" is changed to phenyl and is used the same way as any other substituents, such as methyl, ethyl, or bromo. Putting it all together, the name can be derived as: 2-phenyloctane (phenyl is attached at the second position of the longest carbon chain, octane).
The Benzyl Group
The benzyl group (abbv. Bn), similar to the phenyl group, is formed by manipulating the benzene ring. In the case of the benzyl group, it is formed by taking the phenyl group and adding a CH2 group to where the hydrogen was removed. Its molecular fragment can be written as C6H5CH2-R, PhCH2-R, or Bn-R. Nomenclature of benzyl group based compounds are very similar to the phenyl group compounds. For example, a chlorine attached to a benzyl group would simply be called benzyl chloride, whereas an OH group attached to a benzyl group would simply be called benzyl alcohol.
Figure 22. Benzyl Group Nomenclature
Additionally, other substituents can attach on the benzene ring in the presence of the benzyl group. An example of this can be seen in the Figure below:
Figure 23. Nomenclature of 2,4-difluorobenzyl chloride. Similar to the base name nomenclatures system, the carbon in which th base substitutent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible.
Similar to the base name nomenclature system, the carbon in which the base substituent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible. Under this consideration, the above compound can be named: 2,4-difluorobenzyl chloride.
Commonly Named Benzene Compounds Nomenclature Summary Flowchart
Summary Flowchart (Figure 24). Summary of nomenclature rules used in commonly benzene derived compounds.
As benzene derived compounds can be extremely complex, only compounds covered in this article and other commonly named compounds can be named using this flowchart.
Determination of Common and Systematic Names using Flowchart
To demonstrate how this flowchart can be used to name TNT in its common and systematic (IUPAC) name, a replica of the flowchart with the appropriate flow paths are shown below:
Practice Problems
Q1) (True/False) The compound above contains a benzene ring and thus is aromatic.
Q2) Benzene unusual stability is caused by how many conjugated pi bonds in its cyclic ring? ____
Q3) Menthol, a topical analgesic used in many ointments for the relief of pain, releases a peppermint aroma upon exposure to the air. Based on this conclusion, can you imply that a benzene ring is present in its chemical structure? Why or why not?
Q4)
Q5) At normal conditions, benzene has ___ resonance structures.
Q6) Which of the following name(s) is/are correct for the following compound?
a) nitrohydride benzene
b) phenylamine
c) phenylamide
d) aniline
e) nitrogenhydrogen benzene
f) All of the above is correct
Q7) Convert 1,4-dimethylbenzene into its common name.
Q8) TNT's common name is: ______________________________
Q9) Name the following compound using OMP nomenclature:
Q10) Draw the structure of 2,4-dinitrotoluene.
Q11) Name the following compound:
Q12) Which of the following is the correct name for the following compound?
a) 3,4-difluorobenzyl bromide
b) 1,2-difluorobenzyl bromide
c) 4,5-difluorobenzyl bromide
d) 1,2-difluoroethyl bromide
e) 5,6-difluoroethyl bromide
f) 4,5-difluoroethyl bromide
Q13) (True/False) Benzyl chloride can be abbreviated Bz-Cl.
Q14) Benzoic Acid has what R group attached to its phenyl functional group?
Q15) (True/False) A single aromatic compound can have multiple names indicating its structure.
Q16) List the corresponding positions for the OMP system (o-, m-, p-).
Q17) A scientist has conducted an experiment on an unknown compound. He was able to determine that the unknown compound contains a cyclic ring in its structure as well as an alcohol (-OH) group attached to the ring. What is the unknown compound?
a) Cyclohexanol
b) Cyclicheptanol
c) Phenol
d) Methanol
e) Bleach
f) Cannot determine from the above information
Q18) Which of the following statements is false for the compound, phenol?
a) Phenol is a benzene derived compound.
b) Phenol can be made by attaching an -OH group to a phenyl group.
c) Phenol is highly toxic to the body even in small doses.
d) Phenol can be used as a catalyst in the hydrogenation of benzene into cyclohexane.
e) Phenol is used as an antiseptic in minute doses.
f) Phenol is amongst one of the three common names retained in the IUPAC nomenclature.
Answer Key to Practice Questions
Q1) False, this compound does not contain a benzene ring in its structure.
Q2) 3
Q3) No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (Figure 1)
Q4) No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds.
Q5) 2
Q6) b, d
Q7) p-Xylene
Q8) 2,4,6-trinitrotoluene
Q9) p-chloronitrobenzene
Q10)
Q11) 4-phenylheptane
Q12) a
Q13) False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl.
Q14) COOH
Q15) True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene.
Q16) Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4
Q17) The correct answer is f). We cannot determine what structure this is since the question does not tell us what kind of cyclic ring the -OH group is attached on. Just as cyclohexane can be cyclic, benzene and cycloheptane can also be cyclic.
Q18) d
• David Lam
8.17: The Effect of Substituents on pKa
Substitution of the hydroxyl hydrogen atom is even more facile with phenols, which are roughly a million times more acidic than equivalent alcohols. This phenolic acidity is further enhanced by electron-withdrawing substituents ortho and para to the hydroxyl group, as displayed in the following diagram. The alcohol cyclohexanol is shown for reference at the top left. It is noteworthy that the influence of a nitro substituent is over ten times stronger in the para-location than it is meta, despite the fact that the latter position is closer to the hydroxyl group. Furthermore additional nitro groups have an additive influence if they are positioned in ortho or para locations. The trinitro compound shown at the lower right is a very strong acid called picric acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. Formulas illustrating this electron delocalization will be displayed when the "Resonance Structures" button beneath the previous diagram is clicked. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/08%3A_Aromaticity%3A_Reactions_of_Benzene_and_Substituted_Benzenes/8.13%3A_The_Nomenclature_of_Disubstituted_and_Polysubstituted_Benzenes.txt |
Homework Problems
Homework Solutions
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09: Substitution and Elimination Reactions of Alkyl Halides
The high reactivity of alkyl halides can be explained in terms of the nature of C — X bond which is highly polarized covalent bond due to large difference in the electronegativities of carbon and halogen atoms.
9.02: The Mechanism For an SN2 Reaction
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
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. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
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. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
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.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
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. 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. 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.
The SN1 mechanism
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
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
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
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
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.
Exercise
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.
Influence of the solvent in an 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, but it 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 transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.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 polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the 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.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.01%3A_How_Alkyl_Halides_React.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
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. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
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. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
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.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
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. 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. 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.
The SN1 mechanism
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
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
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
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
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.
Exercise
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.
Influence of the solvent in an 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, but it 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 transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.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 polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the 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.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.03%3A_Factors_That_Affect_%28S_N2%29_Reactions.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
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. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
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. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
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.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
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. 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. 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.
The SN1 mechanism
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
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
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
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
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.
Exercise
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.
Influence of the solvent in an 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, but it 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 transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.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 polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the 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.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.03%3A_Factors_That_Affect_(S_N2)_Reactions.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
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. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
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. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
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.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
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. 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. 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.
The SN1 mechanism
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
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
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
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
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.
Exercise
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.
Influence of the solvent in an 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, but it 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 transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.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 polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the 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.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.04%3A_The_Mechanism_for_an_%28S_N1%29_Reaction.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
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. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
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. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
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.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
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. 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. 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.
The SN1 mechanism
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
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
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
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
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.
Exercise
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.
Influence of the solvent in an 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, but it 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 transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.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 polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the 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.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.04%3A_The_Mechanism_for_an_(S_N1)_Reaction.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
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. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
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. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
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.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
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. 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. 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.
The SN1 mechanism
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
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
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
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
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.
Exercise
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.
Influence of the solvent in an 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, but it 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 transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.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 polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the 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.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.05%3A_Factors_That_Affect_%28S_N1%29_Reactions.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
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. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
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. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
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.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
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. 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. 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.
The SN1 mechanism
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
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
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
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
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.
Exercise
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.
Influence of the solvent in an 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, but it 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 transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.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 polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the 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.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.05%3A_Factors_That_Affect_(S_N1)_Reactions.txt |
8.5A: What makes a good leaving group?
In our general discussion of nucleophilic substitution reactions, we have until now been designating the leaving group simply as “X". As you may imagine, however, the nature of the leaving group is an important consideration: if the C-X bond does not break, the new bond between the nucleophile and electrophilic carbon cannot form, regardless of whether the substitution is SN1 or SN2.
When the C-X bond breaks in a nucleophilic substitution, the pair of electrons in the bond goes with the leaving group. In this way, the leaving group is analogous to the conjugate base in a Brønsted-Lowry acid-base reaction. When we were evaluating the strength of acids in chapter 7, what we were really doing was evaluating the stability of the conjugate base that resulted from the proton transfer. All of the concepts that we used to evaluate the stability of conjugate bases we can use again to evaluate leaving groups – essentially they are one and the same. In other words, the trends in basicity are parallel to the trends in leaving group potential - the weaker the base, the better the leaving group. Just as with conjugate bases, the most important question regarding leaving groups is this: when a leaving group leaves and takes a pair of electrons with it, how well is the extra electron density stabilized?
In laboratory synthesis reactions, halides often act as leaving groups. Iodide, which is the least basic of the four main halides, is also the best leaving group – it is the most stable as a negative ion. Fluoride is the least effective leaving group among the halides, because fluoride anion is the most basic.
Exercise 8.14: Predict the structures of A and B in the following reaction:
Solution
8.5D: Predicting SN1 vs. SN2 mechanisms; competition between nucleophilic substitution and elimination reactions
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 protic solvent is used – so the SN2 mechanism is heavily favored. The reaction is expected to proceed with inversion of configuration.
Template:ExampleStart
Exercise 8.15: Determine whether each substitution reaction shown below is likely to proceed by an SN1 or SN2 mechanism.
Solution
Template:ExampleEnd
In all of our discussion so far about nucleophilic substitutions, we have ignored another important possibility. In many cases, including the two examples above, substitution reactions compete with a type of reaction known as elimination. Consider, for example, the two courses that a reaction could take when tert-butyl bromide reacts with water:
We begin with formation of the carbocation intermediate. In pathway ‘a’, water acts as a nucleophile – this is, of course, the familiar SN1 reaction. However, a water molecule encountering the carbocation intermediate could alternatively act as a base rather than as a nucleophile, plucking a proton from one of the methyl carbons and causing the formation of a new carbon-carbon p bond. This alternative pathway is called an elimination reaction, and in fact with the conditions above, both the substitution and the elimination pathways will occur in competition with each other.
We will have lots more to say about elimination reactions in chapter 14, focusing on biochemical eliminations but also thinking about the competition between substitution and elimination that occurs with many nonenzymatic reactions.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.06%3A_Comparing_the_SN2_and_SN1_reaction_of_Alkyl_Halides.txt |
Certain types of reactions that are likely to occur with alkyl halides. In describing these, it is useful to designate the halogen-bearing carbon as alpha and the carbon atom(s) adjacent to it as beta, as noted in the first four equations shown below. Replacement or substitution of the halogen on the α-carbon (colored maroon) by a nucleophilic reagent is a commonly observed reaction, as shown in equations 1, 2, 5, 6 & 7 below.
Also, since the electrophilic character introduced by the halogen extends to the β-carbons, and since nucleophiles are also bases, the possibility of base induced H-X elimination must also be considered, as illustrated by equation 3. Finally, there are some combinations of alkyl halides and nucleophiles that fail to show any reaction over a 24 hour period, such as the example in equation 4. For consistency, alkyl bromides have been used in these examples. Similar reactions occur when alkyl chlorides or iodides are used, but the speed of the reactions and the exact distribution of products will change.
In order to understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation.
One conclusion, relating the structure of the R-group to possible products, should be immediately obvious. If R- has no beta-hydrogens an elimination reaction is not possible, unless a structural rearrangement occurs first. The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds.
It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Using the general reaction shown above as our reference, we can identify the following variables and observables.
Variables
R change α-carbon from 1º to 2º to 3º
if the α-carbon is a chiral center, set as (R) or (S)
X change from Cl to Br to I (F is relatively unreactive)
Nu: change from anion to neutral; change basicity; change polarizability
Solvent polar vs. non-polar; protic vs. non-protic
Observables
Products substitution, elimination, no reaction.
Stereospecificity if the α-carbon is a chiral center what happens to its configuration?
Reaction Rate measure as a function of reactant concentration.
When several reaction variables may be changed, it is important to isolate the effects of each during the course of study. In other words: only one variable should be changed at a time, the others being held as constant as possible. For example, we can examine the effect of changing the halogen substituent from Cl to Br to I, using ethyl as a common R–group, cyanide anion as a common nucleophile, and ethanol as a common solvent. We would find a common substitution product, C2H5–CN, in all cases, but the speed or rate of the reaction would increase in the order: Cl < Br < I. This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.07%3A_Elimination_Reaction_of_Alkyl_Halides.txt |
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, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant.In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination.
The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Anionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH ——> CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.09%3A_Comparing_the_E2_and_E1_Reactions_of_Alkyl_Halides.txt |
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9.12: Solvent Effects
Nucleophilicity
Recall the definitions of electrophile and nucleophile:
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 forming a covalent bond to an electrophile (or Lewis acid).
If we use a common alkyl halide, such as methyl bromide, and a common solvent, ethanol, we can examine the rate at which various nucleophiles substitute the methyl carbon. Nucleophilicity is thereby related to the relative rate of substitution reactions at the halogen-bearing carbon atom of the reference alkyl halide. The most reactive nucleophiles are said to be more nucleophilic than less reactive members of the group. The nucleophilicities of some common Nu:(–) reactants vary as shown in the following
Nucleophilicity: CH3CO2(–) < Cl(–) < Br(–) < N3(–) < CH3O(–) < CN(–) ≈ SCN(–) < I(–) < CH3S(–)
The reactivity range encompassed by these reagents is over 5,000 fold, thiolate being the most reactive. Note that by using methyl bromide as the reference substrate, the complication of competing elimination reactions is avoided. The nucleophiles used in this study were all anions, but this is not a necessary requirement for these substitution reactions. Indeed reactions 6 & 7, presented at the beginning of this section, are examples of neutral nucleophiles participating in substitution reactions. The cumulative results of studies of this kind has led to useful empirical rules pertaining to nucleophilicity:
1. (i) For a given element, negatively charged species are more nucleophilic (and basic) than are equivalent neutral species.
2. (ii) For a given period of the periodic table, nucleophilicity (and basicity) decreases on moving from left to right.
3. (iii) For a given group of the periodic table, nucleophilicity increases from top to bottom (i.e. with increasing size), although there is a solvent dependence due to hydrogen bonding. Basicity varies in the opposite manner.
Solvent Effects
Solvation of nucleophilic anions markedly influences their reactivity. The nucleophilicities cited above were obtained from reactions in methanol solution. Polar, protic solvents such as water and alcohols solvate anions by hydrogen bonding interactions, as shown in the diagram below.
These solvated species are more stable and less reactive than the unsolvated "naked" anions. Polar, aprotic solvents such as DMSO (dimethyl sulfoxide), DMF (dimethylformamide) and acetonitrile do not solvate anions nearly as well as methanol, but provide good solvation of the accompanying cations. Consequently, most of the nucleophiles discussed here react more rapidly in solutions prepared from these solvents. These solvent effects are more pronounced for small basic anions than for large weakly basic anions. Thus, for reaction in DMSO solution we observe the following reactivity order:
Nucleophilicity: I(–) < SCN(–) < Br(–) < Cl(–) ≈ N3(–) < CH3CO2 (–) < CN(–) ≈ CH3S(–) < CH3O(–)
Note that this order is roughly the order of increasing basicity.
9.14: Biological Methylating Reagents
9.1A: 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:
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).
Exercise 9.1: SAM is formed by a nucleophilic substitution reaction between methionine and adenosine triphosphate (ATP). Propose a mechanism for this reaction.
Solution
9.1B: Synthetic parallel – the Williamson ether synthesis
Synthetic organic chemists often use reactions in the laboratory that are conceptually very similar to the SAM-dependent methylation reactions described above. In what is referred to as "O-methylation", an alcohol is first deprotonated by a strong base, typically sodium hydride (this is essentially an irreversible deprotonation, as the hydrogen gas produced can easily be removed from the reaction vessel).
The alkoxide ion, a powerful nucleophile, is then allowed to attack the electrophilic carbon in iodomethane, displacing iodide in an SN2 reaction. Recall (section 7.3A) that iodide ion is the least basic - and thus the best leaving group - among the halogens commonly used in the synthetic lab. CH3I is one of the most commonly used lab reagents for methyl transfer reactions, and is the lab equivalent of SAM. In the synthesis of a modified analog of the signaling molecule myo-inisitol triphosphate, an alcohol group was methylated using sodium hydride and iodomethane (Carbohydrate Research 2004, 339, 51):
Methylation of amines (N-methylation) by iodomethane is also possible. A recent study concerning the optimization of peptide synthesis methods involved the following reaction (J. Org. Chem. 2000 65, 2309):
Template:ExampleStart
Exercise 9.2: Notice that in the N-methylation reaction a weak base (potassium carbonate) is used, rather than the very strong base (sodium hydride) that was used in the o-methylation reaction. Why is this weak base sufficient for N-methylation but not for o-methylation? In your answer, draw a step-by step mechanism for both reactions, and think carefully about pKa values and where in each mechanism the deprotonation step occurs.
Solution
The methylation of an alcohol by iodomethane is just one example of what is generally referred to as the Williamson ether synthesis, in which alcohols are first deprotonated by a strong base and then allowed to attack an alkyl halide electrophile. Below we see the condensation between ethanol and benzyl bromide:
It is important to note that this reaction does not work well at all if secondary or tertiary alkyl halides are used: remember that the alkoxide ion is a strong base as well as a nucleophile, and elimination will compete with nucleophilic substitution (Section 8.5D).
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.10%3A_Does_an_Alkyl_Halide_Undergo_SN2_E2_Reactions_or_SN1_Reactions.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
10: Reactions of Alcohols Amines Ethers and Epoxides
Alcohols are one of the most important functional groups in organic chemistry. Alcohols are a good source of reagents for synthesis reactions. The ability to identify alcohols is important especially when looking at IR and NMR spectra. The alcohol signal is very easy to spot on IR graphs, because they have a strong signal near the 3200 cm-1 region.
Introduction
The following is list of some common primary alcohols based on the IUPAC naming system.
Name
Molecular Formula
Methanol (methyl alcohol)
CH3OH
Ethanol (ethyl alcohol)
C2H5OH
Propanol
C3H7OH
Butanol
C4H9OH
Pentanol
C5H11OH
Hexanol
C6H13OH
Heptanol
C7H15OH
Octanol
C8H17OH
Rules for naming the alcohols
1. Find the longest chain containing the hydroxy group (OH). If there is a chain with more carbons than the one containing the OH group it will be named as a subsitutent.
2. Place the OH on the lowest possible number for the chain. With the exception of carbonyl groups such as ketones and aldehydes, the alcohol or hydroxy groups have first priority for naming.
3. When naming a cyclic structure, the -OH is assumed to be on the first carbon unless the carbonyl group is present, in which case the later will get priority at the first carbon.
4. When multiple -OH groups are on the cyclic structure, number the carbons on which the -OH groups reside.
5. Remove the final e from the parent alkane chain and add -ol. When multiple alcohols are present use di, tri, et.c before the ol, after the parent name. ex. 2,3-hexandiol. If a carbonyl group is present, the -OH group is named with the prefix "hydroxy," with the carbonyl group attached to the parent chain name so that it ends with -al or -one.
Examples
Ethane: CH3CH3 ----->Ethanol: (the alcohol found in beer, wine and other consumed sprits)
Secondary alcohol: 2-propanol
Other functional groups on an alcohol: 3-bromo-2-pentanol
Cyclic alcohol (two -OH groups): cyclohexan-1,4-diol
Other functional group on the cyclic structure: 3-hexeneol (the alkene is in bold and indicated by numbering the carbon closest to the alcohol)
A complex alcohol: 4-ethyl-3hexanol (the parent chain is in red and the substituent is in blue)
Problems
Name the following alchols:
1.
Contributors
• Abhiram Kondajji (UCD)
10.02: Substitution Reactions of Alcohols
When alcohols react with a hydrogen halide, a substitution occurs, producing an alkyl halide and water:
Scope of Reaction
• 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).
The reaction is acid catalyzed. Alcohols react with the strongly acidic hydrogen halides HCl, HBr, and HI, but they do not react with nonacidic NaCl, NaBr, or NaI. Primary and secondary alcohols can be converted to alkyl chlorides and bromides by allowing them to react with a mixture of a sodium halide and sulfuric acid:
Mechanisms of the Reactions of Alcohols with HX
Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation in an \(S_N1\) reaction with the protonated alcohol acting as the substrate.
The \(S_N1\) mechanism is illustrated by the reaction tert-butyl alcohol and aqueous hydrochloric acid (\(H_3O^+\), \(Cl^-\) ). The first two steps in this \(S_n1\) substitution mechanism are protonation of the alcohol to form an oxonium ion. Although the oxonium ion is formed by protonation of the alcohol, it can also be viewed as a Lewis acid-base complex between the cation (\(R^+\)) and \(H_2O\). Protonation of the alcohol converts a poor leaving group (OH-) to a good leaving group (\)H_2O\_), which makes the dissociation step of the \(S_N1\) mechanism more favorable.
In step 3, the carbocation reacts with a nucleophile (a halide ion) to complete the substitution.
When we convert an alcohol to an alkyl halide, we perform the reaction in the presence of acid and in the presence of halide ions and not at elevated temperatures. Halide ions are good nucleophiles (they are much stronger nucleophiles than water), and because halide ions are present in a high concentration, most of the carbocations react with an electron pair of a halide ion to form a more stable species, the alkyl halide product. The overall result is an \(S_n1\) reaction.
Primary Alcohols
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 themselves. Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion:
We can see now why the reactions of alcohols with hydrogen halides are acid-promoted.
The role of acid catalysis
Acid protonates the alcohol hydroxyl group, making it a good leaving group. However, other strong Lewis acids can be used instead of hydrohalic acids. Because the chloride ion is a weaker nucleophile than bromide or iodide ions, hydrogen chloride does not react with primary or secondary alcohols unless zinc chloride or a similar Lewis acid is added to the reaction mixture as well. Zinc chloride, a good Lewis acid, forms a complex with the alcohol through association with an unshared pair of electrons on the oxygen atom. This enhances the hydroxyl’s leaving group potential sufficiently so that chloride can displace it.
Rearrangement
As we might expect, many reactions of alcohols with hydrogen halides, particularly those in which carbocations are formed, are accompanied by rearrangements. The general rule is that if rearrangement CAN OCCUR (to form more stable or equally stable cations), it will! In these reactions, mixtures of products can be formed. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Amines_Ethers_and_Epoxides/10.01%3A_Nomenclature_of_Alcohols.txt |
The discussion of alkyl halide reactions noted that 2º and 3º-alkyl halides experience rapid E2 elimination when treated with strong bases such as hydroxide and alkoxides. Alcohols do not undergo such base-induced elimination reactions and are, in fact, often used as solvents for such reactions. This is yet another example of how leaving-group stability influences the rate of a reaction. When an alcohol is treated with sodium hydroxide, the following acid-base equilibrium occurs. Most alcohols are slightly weaker acids than water, so the left side is favored.
R–O–H + Na(+) OH(–) R–O(–) Na(+) + H–OH
The elimination of water from an alcohol is called dehydration. Recalling that water is a much better leaving group than hydroxide ion, it is sensible to use acid-catalysis rather than base-catalysis in such reactions. Four examples of this useful technique are shown below. Note that hydrohalic acids (HX) are not normally used as catalysts because their conjugate bases are good nucleophiles and may create substitution products. The conjugate bases of sulfuric and phosphoric acids are not good nucleophiles, and do not participate in substitution under typical conditions.
The first two examples (in the top row) are typical, and the more facile elimination of the 3º-alcohol suggests the predominant E1 character for the reaction. This agrees with the tendency of branched 1º and 2º-alcohols to yield rearrangement products, as shown in the last example. The last two reactions also demonstrate that the Zaitsev rule applies to alcohol dehydrations, as well as to alkyl halide eliminations. Therefore, the more highly-substituted double bond isomer is favored among the products.
It should be noted that the acid-catalyzed dehydrations discussed here are the reverse of the acid-catalyzed hydration reactions of alkenes. Indeed, for these types of reversible reactions, the laws of thermodynamics require that the mechanism in both directions proceed by the same reaction path. This is known as the principle of microscopic reversibility. To illustrate, the following diagram lists the three steps in each transformation. The dehydration reaction is shown by the blue arrows; the hydration reaction by magenta arrows. The intermediates in these reactions are common to both, and common transition states are involved. This can be seen clearly in the energy diagrams depicted by clicking the button beneath the equations.
Base-induced E2 eliminations of alcohols may be achieved from sulfonate ester derivatives. This has the advantage of avoiding strong acids, which may cause molecular rearrangement or double bond migration in some cases. Because 3º-sulfonate derivatives are sometimes unstable, this procedure is best used with 1º and 2º-mesylates or tosylates. Application of this reaction sequence is shown here for 2-butanol. The Zaitsev rule favors the formation of 2-butene (cis + trans) over 1-butene.
CH3CH2CH(CH3)–OH CH3SO2Cl
CH3CH2CH(CH3)–OSO2CH3 C2H5O(–)Na(+)
CH3CH=CHCH3 + CH3CH2CH=CH2 + CH3SO2O(–) Na(+) + C2H5OH
The E2 elimination of 3º-alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorous oxychloride (POCl3) in pyridine. This procedure is also effective with hindered 2º-alcohols, but for unhindered and 1º-alcohols, an SN2 chloride ion substitution of the chlorophosphate intermediate competes with elimination. Examples of these and related reactions are given in the following figure. The first equation shows the dehydration of a 3º-alcohol. The predominance of the non-Zaitsev product (less substituted double bond) is presumed due to steric hindrance of the methylene group hydrogen atoms, which interferes with the approach of the base at that site. The second example shows two elimination procedures applied to the same 2º-alcohol. The first uses the single step POCl3 method, which works well in this case because SN2 substitution is impeded by steric hindrance. The second method is another example in which an intermediate sulfonate ester confers halogen-like reactivity on an alcohol. In every case, the anionic leaving group is the conjugate base of a strong acid.
Pyrolytic syn-Eliminations
Ester derivatives of alcohols may undergo unimolecular syn-elimination upon heating. To see examples of these, Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Amines_Ethers_and_Epoxides/10.03%3A_Elimination_Reactions_of_Alcohols%3A_Dehydration.txt |
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, then the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is as follows:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Primary alcohols
Primary alcohols can be oxidized to either aldehydes or carboxylic acids, depending on the reaction conditions. In the case of the formation of carboxylic acids, the alcohol is first oxidized to an aldehyde, which is then oxidized further to the acid.
An aldehyde is obtained if an excess amount of the alcohol is used, and the aldehyde is distilled off as soon as it forms. An excess of the alcohol means that there is not enough oxidizing agent present to carry out the second stage, and removing the aldehyde as soon as it is formed means that it is not present to be oxidized anyway!
If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, $CH_3CHO$. The full equation for this reaction is fairly complicated, and you need to understand the electron-half-equations in order to work it out.
$3CH_3CH_2OH + Cr_2O_7^{2-} + 8H^+ \rightarrow 3CH_3CHO + 2Cr^{3+} + 7H_2O$
In organic chemistry, simplified versions are often used that concentrate on what is happening to the organic substances. To do that, oxygen from an oxidizing agent is represented as $[O]$. That would produce the much simpler equation:
It also helps in remembering what happens. You can draw simple structures to show the relationship between the primary alcohol and the aldehyde formed.
Full oxidation to carboxylic acids
An excess of the oxidizing agent must be used, and the aldehyde formed as the half-way product should remain in the mixture. The alcohol is heated under reflux with an excess of the oxidizing agent. When the reaction is complete, the carboxylic acid is distilled off. The full equation for the oxidation of ethanol to ethanoic acid is as follows:
$3CH_3CH_2OH + 2Cr_2O_7^{2-} + 16H+ \rightarrow 3CH_3COOH + 4Cr^{3+} + 11H_2O$
The more typical simplified version looks like this:
$CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O$
Alternatively, you could write separate equations for the two stages of the reaction - the formation of ethanal and then its subsequent oxidation.
$CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O$
$CH_3CHO + [O] \rightarrow CH_3COOH$
This is what is happening in the second stage:
Secondary alcohols
Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulfuric acid, propanone is formed. Changing the reaction conditions makes no difference to the product. Folloiwng is the simple version of the equation, showing the relationship between the structures:
If you look back at the second stage of the primary alcohol reaction, you will see that an oxygen inserted between the carbon and the hydrogen in the aldehyde group to produce the carboxylic acid. In this case, there is no such hydrogen - and the reaction has nowhere further to go.
Tertiary alcohols
Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever. If you look at what is happening with primary and secondary alcohols, you will see that the oxidizing agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom is attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.
You need to be able to remove those two particular hydrogen atoms in order to set up the carbon-oxygen double bond.
Using these reactions as a test for the different types of alcohols
First, the presence of an alcohol must be confirmed by testing for the -OH group. The liquid would need to be verified as neutral, free of water and that it reacted with solid phosphorus(V) chloride to produce a burst of acidic steamy hydrogen chloride fumes. A few drops of the alcohol would be added to a test tube containing potassium dichromate(VI) solution acidified with dilute sulfuric acid. The tube would be warmed in a hot water bath.
Determining the tertiary alcohol
In the case of a primary or secondary alcohol, the orange solution turns green. With a tertiary alcohol, there is no color change. After heating, the following colors are observed:
Distinguishing between the primary and secondary alcohols
A sufficient amount of the aldehyde (from oxidation of a primary alcohol) or ketone (from a secondary alcohol) must be produced to be able to test them. There are various reactions that aldehydes undergo that ketones do not. These include the reactions with Tollens' reagent, Fehling's solution and Benedict's solution, and these reactions are covered on a separate page.
These tests can be difficult to carry out, and the results are not always as clear-cut as the books say. A much simpler but fairly reliable test is to use Schiff's reagent. Schiff's reagent is a fuchsin dye decolorized by passing sulfur dioxide through it. In the presence of even small amounts of an aldehyde, it turns bright magenta.
It must, however, be used absolutely cold, because ketones react with it very slowly to give the same color. If you heat it, obviously the change is faster - and potentially confusing. While you are warming the reaction mixture in the hot water bath, you can pass any vapors produced through some Schiff's reagent.
• If the Schiff's reagent quickly becomes magenta, then you are producing an aldehyde from a primary alcohol.
• If there is no color change in the Schiff's reagent, or only a trace of pink color within a minute or so, then you are not producing an aldehyde; therefore, no primary alcohol is present.
Because of the color change to the acidified potassium dichromate(VI) solution, you must, therefore, have a secondary alcohol. You should check the result as soon as the potassium dichromate(VI) solution turns green - if you leave it too long, the Schiff's reagent might start to change color in the secondary alcohol case as well. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Amines_Ethers_and_Epoxides/10.04%3A_Oxidation_of_Alcohols.txt |
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
10.06: Nomenclature of Ethers
Ethers are compounds having two alkyl or aryl groups bonded to an oxygen atom, as in the formula R1–O–R2. The ether functional group does not have a characteristic IUPAC nomenclature suffix, so it is necessary to designate it as a substituent. To do so the common alkoxy substituents are given names derived from their alkyl component (below):
Alkyl Group Name Alkoxy Group Name
CH3 Methyl CH3O– Methoxy
CH3CH2 Ethyl CH3CH2O– Ethoxy
(CH3)2CH– Isopropyl (CH3)2CHO– Isopropoxy
(CH3)3C– tert-Butyl (CH3)3CO– tert-Butoxy
C6H5 Phenyl C6H5O– Phenoxy
Ethers can be named by naming each of the two carbon groups as a separate word followed by a space and the word ether. The -OR group can also be named as a substituent using the group name, alkox
Example \(1\)
CH3-CH2-O-CH3 is called ethyl methyl ether or methoxyethane.
The smaller, shorter alkyl group becomes the alkoxy substituent. The larger, longer alkyl group side becomes the alkane base name. Each alkyl group on each side of the oxygen is numbered separately. The numbering priority is given to the carbon closest to the oxgen. The alkoxy side (shorter side) has an "-oxy" ending with its corresponding alkyl group. For example, CH3CH2CH2CH2CH2-O-CH2CH2CH3 is 1-propoxypentane. If there is cis or trans stereochemistry, the same rule still applies.
Examples \(2\)
• \(CH_3CH_2OCH_2CH_3\), diethyl ether (sometimes referred to as just ether)
• \(CH_3OCH_2CH_2OCH_3\), ethylene glycol dimethyl ether (glyme).
Exercises \(2\)
Try to name the following compounds using these conventions:
Try to draw structures for the following compounds:
• 2-pentyl 1-propyl ether J
• 1-(2-propoxy)cyclopentene J
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers".
• anisole (try naming anisole by the other two conventions. J )
• oxirane
1,2-epoxyethane, ethylene oxide, dimethylene oxide, oxacyclopropane,
• furan (this compound is aromatic)
tetrahydrofuran
oxacyclopentane, 1,4-epoxybutane, tetramethylene oxide,
• dioxane
1,4-dioxacyclohexane
Exercise \(2\)
Try to draw structures for the following compounds-
• 3-bromoanisole J
• 2-methyloxirane J
• 3-ethylfuran J
Heterocycles
In cyclic ethers (heterocycles), one or more carbons are replaced with oxygen. Often, it's called heteroatoms, when carbon is replaced by an oxygen or any atom other than carbon or hydrogen. In this case, the stem is called the oxacycloalkane, where the prefix "oxa-" is an indicator of the replacement of the carbon by an oxygen in the ring. These compounds are numbered starting at the oxygen and continues around the ring. For example,
If a substituent is an alcohol, the alcohol has higher priority. However, if a substituent is a halide, ether has higher priority. If there is both an alcohol group and a halide, alcohol has higher priority. The numbering begins with the end that is closest to the higher priority substituent. There are ethers that are contain multiple ether groups that are called cyclic polyethers or crown ethers. These are also named using the IUPAC system.
Sulfides
Sulfur analogs of ethers (R–S–R') are called sulfides, e.g., (CH3)3C–S–CH3 is tert-butyl methyl sulfide. Sulfides are chemically more reactive than ethers, reflecting the greater nucleophilicity of sulfur relative to oxygen.
Problems
Name the following ethers:
(Answers to problems above: 1. diethyl ether; 2. 2-ethoxy-2-methyl-1-propane; 3. cis-1-ethoxy-2-methoxycyclopentane; 4. 1-ethoxy-1-methylcyclohexane; 5. oxacyclopropane; 6. 2,2-Dimethyloxacyclopropane)
Exercise \(3\)
Answer
A one-eyed one-horned flying propyl people ether
10.07: Nucleophilic Substitution Reactions of Ethers
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
10.08: Nucleophilic Substitution Reactions of Epoxides
8.6A: Epoxide structure
Epoxides (also known as oxiranes) are three-membered ring structures in which one of the vertices is an oxygen and the other two are carbons.
The carbons in an epoxide group are very reactive electrophiles, due in large part to the fact that substantial ring strain is relieved when the ring opens upon nucleophilic attack.
Epoxides are very important intermediates in laboratory organic synthesis, and are also found as intermediate products in some biosynthetic pathways. The compound (3S)-2,3-oxidosqualene, for example, is an important intermediate in the biosynthesis of cholesterol (we’ll see the epoxide ring-opening step in chapter 15):
Both in the laboratory and in the cell, epoxides are usually formed by the oxidation of an alkene. These reactions will be discussed in detail in chapter 16.
Exercise 8.17: Predict the major product(s) of the ring opening reaction that occurs when the epoxide shown below is treated with:
a) ethanol and a small amount of sodium hydroxide
b) ethanol and a small amount of sulfuric acid
Hint: be sure to consider both regiochemistry and stereochemistry!
Solution
Template:ExampleEnd
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Amines_Ethers_and_Epoxides/10.05%3A_Amines_Do_Not_Undergo_Substitution_or_Elimination_Reactions.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
11: Carbonyl Compounds I: Reactions of Carboxylic Acids and Carboxylic Derivatives
The IUPAC system of nomenclature assigns a characteristic suffix to these classes. The –e ending is removed from the name of the parent chain and is replaced -anoic acid. Since a carboxylic acid group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name.
Many carboxylic acids are called by the common names. These names were chosen by chemists to usually describe a source of where the compound is found. In common names of aldehydes, carbon atoms near the carboxyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
Formula Common Name Source IUPAC Name Melting Point Boiling Point
HCO2H formic acid ants (L. formica) methanoic acid 8.4 ºC 101 ºC
CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid 16.6 ºC 118 ºC
CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid -20.8 ºC 141 ºC
CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid -5.5 ºC 164 ºC
CH3(CH2)3CO2H valeric acid valerian root pentanoic acid -34.5 ºC 186 ºC
CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid -4.0 ºC 205 ºC
CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid -7.5 ºC 223 ºC
CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid 16.3 ºC 239 ºC
CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid 12.0 ºC 253 ºC
CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid 31.0 ºC 219 ºC
Example (Common Names Are in Red)
Naming carboxyl groups added to a ring
When a carboxyl group is added to a ring the suffix -carboxylic acid is added to the name of the cyclic compound. The ring carbon attached to the carboxyl group is given the #1 location number.
Naming carboxylates
Salts of carboxylic acids are named by writing the name of the cation followed by the name of the acid with the –ic acid ending replaced by an –ate ending. This is true for both the IUPAC and Common nomenclature systems.
Naming carboxylic acids which contain other functional groups
Carboxylic acids are given the highest nomenclature priority by the IUPAC system. This means that the carboxyl group is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of molecules containing carboxylic acid and alcohol functional groups the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
In the case of molecules containing a carboxylic acid and aldehydes and/or ketones functional groups the carbonyl is named as a "Oxo" substituent.
In the case of molecules containing a carboxylic acid an amine functional group the amine is named as an "amino" substituent.
When carboxylic acids are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an –enoic acid ending to indicate the presence of an alkene and carboxylic acid)
Remember that the carboxylic acid has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Naming dicarboxylic acids
For dicarboxylic acids the location numbers for both carboxyl groups are omitted because both functional groups are expected to occupy the ends of the parent chain. The ending –dioic acid is added to the end of the parent chain. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/11%3A_Carbonyl_Compounds_I%3A_Reactions_of_Carboxylic_Acids_and_Carboxylic_Derivatives/11.01%3A_The_Nomenclature_of_Carboxylic_Acids_and_Carboxylic_Acid_Derivat.txt |
Structure of the carboxyl acid group
Carboxylic acids are organic compounds which incorporate a carboxyl functional group, CO2H. The name carboxyl comes from the fact that a carbonyl and a hydroxyl group are attached to the same carbon.
The carbon and oxygen in the carbonyl are both sp2 hybridized which give a carbonyl group a basic trigonal shape. The hydroxyl oxygen is also sp2 hybridized which allows one of its lone pair electrons to conjugate with the pi system of the carbonyl group. This make the carboxyl group planar an can represented with the following resonance structure.
Carboxylic acids are named such because they can donate a hydrogen to produce a carboxylate ion. The factors which affect the acidity of carboxylic acids will be discussed later.
Physical Properties of Some Carboxylic Acids
The table at the beginning of this page gave the melting and boiling points for a homologous group of carboxylic acids having from one to ten carbon atoms. The boiling points increased with size in a regular manner, but the melting points did not. Unbranched acids made up of an even number of carbon atoms have melting points higher than the odd numbered homologs having one more or one less carbon. This reflects differences in intermolecular attractive forces in the crystalline state. In the table of fatty acids we see that the presence of a cis-double bond significantly lowers the melting point of a compound. Thus, palmitoleic acid melts over 60º lower than palmitic acid, and similar decreases occur for the C18 and C20 compounds. Again, changes in crystal packing and intermolecular forces are responsible.
The factors that influence the relative boiling points and water solubilities of various types of compounds were discussed earlier. In general, dipolar attractive forces between molecules act to increase the boiling point of a given compound, with hydrogen bonds being an extreme example. Hydrogen bonding is also a major factor in the water solubility of covalent compounds To refresh your understanding of these principles Click Here. The following table lists a few examples of these properties for some similar sized polar compounds (the non-polar hydrocarbon hexane is provided for comparison).
Physical Properties of Some Organic Compounds
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)4OH 1-pentanol 88 138 ºC slightly soluble
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3(CH2)2CONH2 butanamide 87 216 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 165 ºC very soluble
CH3(CH2)4NH2 1-aminobutane 87 103 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 140 ºC slightly soluble
CH3(CH2)4CH3 hexane 86 69 ºC insoluble
The first five entries all have oxygen functional groups, and the relatively high boiling points of the first two is clearly due to hydrogen bonding. Carboxylic acids have exceptionally high boiling points, due in large part to dimeric associations involving two hydrogen bonds. A structural formula for the dimer of acetic acid is shown here. When the mouse pointer passes over the drawing, an electron cloud diagram will appear. The high boiling points of the amides and nitriles are due in large part to strong dipole attractions, supplemented in some cases by hydrogen bonding.
Acidity of Carboxylic Acids
The pKa 's of some typical carboxylic acids are listed in the following table. When we compare these values with those of comparable alcohols, such as ethanol (pKa = 16) and 2-methyl-2-propanol (pKa = 19), it is clear that carboxylic acids are stronger acids by over ten powers of ten! Furthermore, electronegative substituents near the carboxyl group act to increase the acidity.
Compound
pKa
Compound
pKa
HCO2H 3.75 CH3CH2CH2CO2H 4.82
CH3CO2H 4.74 ClCH2CH2CH2CO2H 4.53
FCH2CO2H 2.65 CH3CHClCH2CO2H 4.05
ClCH2CO2H 2.85 CH3CH2CHClCO2H 2.89
BrCH2CO2H 2.90 C6H5CO2H 4.20
ICH2CO2H 3.10 p-O2NC6H4CO2H 3.45
Cl3CCO2H 0.77 p-CH3OC6H4CO2H 4.45
Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton? To answer this question we must return to the nature of acid-base equilibria and the definition of pKa , illustrated by the general equations given below. These relationships were described in an previous section of this text.
We know that an equilibrium favors the thermodynamically more stable side, and that the magnitude of the equilibrium constant reflects the energy difference between the components of each side. In an acid base equilibrium the equilibrium always favors the weaker acid and base (these are the more stable components). Water is the standard base used for pKa measurements; consequently, anything that stabilizes the conjugate base (A:(–)) of an acid will necessarily make that acid (H–A) stronger and shift the equilibrium to the right. Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater than that of the neutral function, as shown in the following diagram. In the carboxylate anion the two contributing structures have equal weight in the hybrid, and the C–O bonds are of equal length (between a double and a single bond). This stabilization leads to a markedly increased acidity, as illustrated by the energy diagram displayed by clicking the "Toggle Display" button.
The resonance effect described here is undoubtedly the major contributor to the exceptional acidity of carboxylic acids. However, inductive effects also play a role. For example, alcohols have pKa's of 16 or greater but their acidity is increased by electron withdrawing substituents on the alkyl group. The following diagram illustrates this factor for several simple inorganic and organic compounds (row #1), and shows how inductive electron withdrawal may also increase the acidity of carboxylic acids (rows #2 & 3). The acidic hydrogen is colored red in all examples.
Water is less acidic than hydrogen peroxide because hydrogen is less electronegative than oxygen, and the covalent bond joining these atoms is polarized in the manner shown. Alcohols are slightly less acidic than water, due to the poor electronegativity of carbon, but chloral hydrate, Cl3CCH(OH)2, and 2,2,2,-trifluoroethanol are significantly more acidic than water, due to inductive electron withdrawal by the electronegative halogens (and the second oxygen in chloral hydrate). In the case of carboxylic acids, if the electrophilic character of the carbonyl carbon is decreased the acidity of the carboxylic acid will also decrease. Similarly, an increase in its electrophilicity will increase the acidity of the acid. Acetic acid is ten times weaker an acid than formic acid (first two entries in the second row), confirming the electron donating character of an alkyl group relative to hydrogen, as noted earlier in a discussion of carbocation stability. Electronegative substituents increase acidity by inductive electron withdrawal. As expected, the higher the electronegativity of the substituent the greater the increase in acidity (F > Cl > Br > I), and the closer the substituent is to the carboxyl group the greater is its effect (isomers in the 3rd row). Substituents also influence the acidity of benzoic acid derivatives, but resonance effects compete with inductive effects. The methoxy group is electron donating and the nitro group is electron withdrawing (last three entries in the table of pKa values).
For additional information about substituent effects on the acidity of carboxylic acids Click Here
Vinylagous Acids
Compounds in which an enolic hydroxyl group is conjugated with a carbonyl group also show enhanced acidity. To see examples of such compounds Click Here
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/11%3A_Carbonyl_Compounds_I%3A_Reactions_of_Carboxylic_Acids_and_Carboxylic_Derivatives/11.02%3A_The_Structures_of_Carboxylic_Acids_and_Carboxylic_Acid_Derivativ.txt |
Structure of the carboxyl acid group
Carboxylic acids are organic compounds which incorporate a carboxyl functional group, CO2H. The name carboxyl comes from the fact that a carbonyl and a hydroxyl group are attached to the same carbon.
The carbon and oxygen in the carbonyl are both sp2 hybridized which give a carbonyl group a basic trigonal shape. The hydroxyl oxygen is also sp2 hybridized which allows one of its lone pair electrons to conjugate with the pi system of the carbonyl group. This make the carboxyl group planar an can represented with the following resonance structure.
Carboxylic acids are named such because they can donate a hydrogen to produce a carboxylate ion. The factors which affect the acidity of carboxylic acids will be discussed later.
Physical Properties of Some Carboxylic Acids
The table at the beginning of this page gave the melting and boiling points for a homologous group of carboxylic acids having from one to ten carbon atoms. The boiling points increased with size in a regular manner, but the melting points did not. Unbranched acids made up of an even number of carbon atoms have melting points higher than the odd numbered homologs having one more or one less carbon. This reflects differences in intermolecular attractive forces in the crystalline state. In the table of fatty acids we see that the presence of a cis-double bond significantly lowers the melting point of a compound. Thus, palmitoleic acid melts over 60º lower than palmitic acid, and similar decreases occur for the C18 and C20 compounds. Again, changes in crystal packing and intermolecular forces are responsible.
The factors that influence the relative boiling points and water solubilities of various types of compounds were discussed earlier. In general, dipolar attractive forces between molecules act to increase the boiling point of a given compound, with hydrogen bonds being an extreme example. Hydrogen bonding is also a major factor in the water solubility of covalent compounds To refresh your understanding of these principles Click Here. The following table lists a few examples of these properties for some similar sized polar compounds (the non-polar hydrocarbon hexane is provided for comparison).
Physical Properties of Some Organic Compounds
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)4OH 1-pentanol 88 138 ºC slightly soluble
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3(CH2)2CONH2 butanamide 87 216 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 165 ºC very soluble
CH3(CH2)4NH2 1-aminobutane 87 103 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 140 ºC slightly soluble
CH3(CH2)4CH3 hexane 86 69 ºC insoluble
The first five entries all have oxygen functional groups, and the relatively high boiling points of the first two is clearly due to hydrogen bonding. Carboxylic acids have exceptionally high boiling points, due in large part to dimeric associations involving two hydrogen bonds. A structural formula for the dimer of acetic acid is shown here. When the mouse pointer passes over the drawing, an electron cloud diagram will appear. The high boiling points of the amides and nitriles are due in large part to strong dipole attractions, supplemented in some cases by hydrogen bonding.
Acidity of Carboxylic Acids
The pKa 's of some typical carboxylic acids are listed in the following table. When we compare these values with those of comparable alcohols, such as ethanol (pKa = 16) and 2-methyl-2-propanol (pKa = 19), it is clear that carboxylic acids are stronger acids by over ten powers of ten! Furthermore, electronegative substituents near the carboxyl group act to increase the acidity.
Compound
pKa
Compound
pKa
HCO2H 3.75 CH3CH2CH2CO2H 4.82
CH3CO2H 4.74 ClCH2CH2CH2CO2H 4.53
FCH2CO2H 2.65 CH3CHClCH2CO2H 4.05
ClCH2CO2H 2.85 CH3CH2CHClCO2H 2.89
BrCH2CO2H 2.90 C6H5CO2H 4.20
ICH2CO2H 3.10 p-O2NC6H4CO2H 3.45
Cl3CCO2H 0.77 p-CH3OC6H4CO2H 4.45
Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton? To answer this question we must return to the nature of acid-base equilibria and the definition of pKa , illustrated by the general equations given below. These relationships were described in an previous section of this text.
We know that an equilibrium favors the thermodynamically more stable side, and that the magnitude of the equilibrium constant reflects the energy difference between the components of each side. In an acid base equilibrium the equilibrium always favors the weaker acid and base (these are the more stable components). Water is the standard base used for pKa measurements; consequently, anything that stabilizes the conjugate base (A:(–)) of an acid will necessarily make that acid (H–A) stronger and shift the equilibrium to the right. Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater than that of the neutral function, as shown in the following diagram. In the carboxylate anion the two contributing structures have equal weight in the hybrid, and the C–O bonds are of equal length (between a double and a single bond). This stabilization leads to a markedly increased acidity, as illustrated by the energy diagram displayed by clicking the "Toggle Display" button.
The resonance effect described here is undoubtedly the major contributor to the exceptional acidity of carboxylic acids. However, inductive effects also play a role. For example, alcohols have pKa's of 16 or greater but their acidity is increased by electron withdrawing substituents on the alkyl group. The following diagram illustrates this factor for several simple inorganic and organic compounds (row #1), and shows how inductive electron withdrawal may also increase the acidity of carboxylic acids (rows #2 & 3). The acidic hydrogen is colored red in all examples.
Water is less acidic than hydrogen peroxide because hydrogen is less electronegative than oxygen, and the covalent bond joining these atoms is polarized in the manner shown. Alcohols are slightly less acidic than water, due to the poor electronegativity of carbon, but chloral hydrate, Cl3CCH(OH)2, and 2,2,2,-trifluoroethanol are significantly more acidic than water, due to inductive electron withdrawal by the electronegative halogens (and the second oxygen in chloral hydrate). In the case of carboxylic acids, if the electrophilic character of the carbonyl carbon is decreased the acidity of the carboxylic acid will also decrease. Similarly, an increase in its electrophilicity will increase the acidity of the acid. Acetic acid is ten times weaker an acid than formic acid (first two entries in the second row), confirming the electron donating character of an alkyl group relative to hydrogen, as noted earlier in a discussion of carbocation stability. Electronegative substituents increase acidity by inductive electron withdrawal. As expected, the higher the electronegativity of the substituent the greater the increase in acidity (F > Cl > Br > I), and the closer the substituent is to the carboxyl group the greater is its effect (isomers in the 3rd row). Substituents also influence the acidity of benzoic acid derivatives, but resonance effects compete with inductive effects. The methoxy group is electron donating and the nitro group is electron withdrawing (last three entries in the table of pKa values).
For additional information about substituent effects on the acidity of carboxylic acids Click Here
Vinylagous Acids
Compounds in which an enolic hydroxyl group is conjugated with a carbonyl group also show enhanced acidity. To see examples of such compounds Click Here
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/11%3A_Carbonyl_Compounds_I%3A_Reactions_of_Carboxylic_Acids_and_Carboxylic_Derivatives/11.03%3A_The_Physical_Properties_of_Carbonyl_Compounds.txt |
Carboxylic acids are widespread in nature, often combined with other functional groups. Simple alkyl carboxylic acids, composed of four to ten carbon atoms, are liquids or low melting solids having very unpleasant odors. The fatty acids are important components of the biomolecules known as lipids, especially fats and oils. As shown in the following table, these long-chain carboxylic acids are usually referred to by their common names, which in most cases reflect their sources. A mnemonic phrase for the C10 to C20 natural fatty acids capric, lauric, myristic, palmitic, stearic and arachidic is: "Curly, Larry & Moe Perform Silly Antics" (note that the names of the three stooges are in alphabetical order).
Table: \(1\): Saturated Fatty Acids
Formula Common Name Melting Point
CH3(CH2)10CO2H lauric acid 45 ºC
CH3(CH2)12CO2H myristic acid 55 ºC
CH3(CH2)14CO2H palmitic acid 63 ºC
CH3(CH2)16CO2H stearic acid 69 ºC
CH3(CH2)18CO2H arachidic acid 76 ºC
Interestingly, the molecules of most natural fatty acids have an even number of carbon atoms. Analogous compounds composed of odd numbers of carbon atoms are perfectly stable and have been made synthetically. Since nature makes these long-chain acids by linking together acetate units, it is not surprising that the carbon atoms composing the natural products are multiples of two. The double bonds in the unsaturated compounds listed on the right are all cis (or Z).
Table \(2\): Unsaturated Fatty Acids
Formula Common Name Melting Point
CH3(CH2)5CH=CH(CH2)7CO2H palmitoleic acid 0 ºC
CH3(CH2)7CH=CH(CH2)7CO2H oleic acid 13 ºC
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H linoleic acid -5 ºC
CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7CO2H linolenic acid -11 ºC
CH3(CH2)4(CH=CHCH2)4(CH2)2CO2H arachidonic acid -49 ºC
The following formulas are examples of other naturally occurring carboxylic acids. The molecular structures range from simple to complex, often incorporate a variety of other functional groups, and many are chiral.
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
11.05: How Carboxylic Acids and Carboxylic Acids Compounds React
Introduction
Carboxylic acid derivatives are a group of functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during nucleophile substitution reactions. Although there are many types of carboxylic acid derivatives known we will be focusing on just four: Acid halides, Acid anhydrides, Esters, and Amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain a carbonyl their chemistry is distinctly different because they do not contain a suitable leaving group. Once the tetrahedral intermediate is formed aldehydes and ketones cannot reform the carbonyl. Because of this aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdrawn electron density from the carbonyl, thereby, increasing its electrophilicity.
Contributors
Prof. Steven Farmer (Sonoma State University)
11.06: Relative Reactivities of Carboxylic Acids and Carboxylic Acid De
Introduction
Carboxylic acid derivatives are a group of functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during nucleophile substitution reactions. Although there are many types of carboxylic acid derivatives known we will be focusing on just four: Acid halides, Acid anhydrides, Esters, and Amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain a carbonyl their chemistry is distinctly different because they do not contain a suitable leaving group. Once the tetrahedral intermediate is formed aldehydes and ketones cannot reform the carbonyl. Because of this aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdrawn electron density from the carbonyl, thereby, increasing its electrophilicity.
Contributors
Prof. Steven Farmer (Sonoma State University)
11.09: Acid-Catalyzed Ester Hydrolysis
Introduction
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a catalytic amount of acid.
Example 1:
Mechanism
1) Protonation of the Carbonyl
2) Nucleophilic attack by water
3) Proton transfer
4) Leaving group removal
Contributors
Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/11%3A_Carbonyl_Compounds_I%3A_Reactions_of_Carboxylic_Acids_and_Carboxylic_Derivatives/11.04%3A_Carboxylic_Acids_and_Carboxylic_Acid_Derivatives_found_in_Nature.txt |
This page describes ways of hydrolyzing esters - splitting them into carboxylic acids (or their salts) and alcohols by the action of water, dilute acid or dilute alkali. It starts by looking at the hydrolysis of simple esters like ethyl ethanoate, and goes on to look at hydrolyzing bigger, more complicated ones to make soap.
Technically, hydrolysis is a reaction with water. That is exactly what happens when esters are hydrolyzed by water or by dilute acids such as dilute hydrochloric acid. The alkaline hydrolysis of esters actually involves reaction with hydroxide ions, but the overall result is so similar that it is lumped together with the other two.
Hydrolysis using water or dilute acid
The reaction with pure water is so slow that it is never used. The reaction is catalyzed by dilute acid, and so the ester is heated under reflux with a dilute acid like dilute hydrochloric acid or dilute sulfuric acid. Here are two simple examples of hydrolysis using an acid catalyst.
First, hydrolyzing ethyl ethanoate:
. . . and then hydrolyzing methyl propanoate:
Notice that the reactions are reversible. To make the hydrolysis as complete as possible, you would have to use an excess of water. The water comes from the dilute acid, and so you would mix the ester with an excess of dilute acid.
Hydrolysis using dilute alkali
This is the usual way of hydrolyzing esters. The ester is heated under reflux with a dilute alkali like sodium hydroxide solution. There are two advantages of doing this rather than using a dilute acid. The reactions are one-way rather than reversible, and the products are easier to separate. Taking the same esters as above, but using sodium hydroxide solution rather than a dilute acid:
First, hydrolyzing ethyl ethanoate using sodium hydroxide solution:
and then hydrolyzing methyl propanoate in the same way:
Notice that you get the sodium salt formed rather than the carboxylic acid itself. This mixture is relatively easy to separate. Provided you use an excess of sodium hydroxide solution, there will not be any ester left - so you don't have to worry about that. The alcohol formed can be distilled off. That's easy!
If you want the acid rather than its salt, all you have to do is to add an excess of a strong acid like dilute hydrochloric acid or dilute sulfuric acid to the solution left after the first distillation. If you do this, the mixture is flooded with hydrogen ions. These are picked up by the ethanoate ions (or propanoate ions or whatever) present in the salts to make ethanoic acid (or propanoic acid, etc). Because these are weak acids, once they combine with the hydrogen ions, they tend to stay combined. The carboxylic acid can now be distilled off.
Hydrolyzing complicated esters to make soap
This next bit deals with the alkaline hydrolysis (using sodium hydroxide solution) of the big esters found in animal and vegetable fats and oils. If the large esters present in animal or vegetable fats and oils are heated with concentrated sodium hydroxide solution exactly the same reaction happens as with the simple esters. A salt of a carboxylic acid is formed - in this case, the sodium salt of a big acid such as octadecanoic acid (stearic acid). These salts are the important ingredients of soap - the ones that do the cleaning. An alcohol is also produced - in this case, the more complicated alcohol, propane-1,2,3-triol (glycerol).
Because of its relationship with soap making, the alkaline hydrolysis of esters is sometimes known as saponification.
11.12: Reactions of Amides
Amides are reasonably reactive, usually via an attack on the carbonyl breaking the carbonyl double bond and forming a tetrahedral intermediate. Thiols, hydroxyls and amines are all known to serve as nucleophiles. Owing to their resonance stabilization, amides are less reactive under physiological conditions than esters.
11.13: Acid-Catalyzed Amide Hydrolysis
Technically, hydrolysis is a reaction with water. That is exactly what happens when amides are hydrolyzed in the presence of dilute acids such as dilute hydrochloric acid. The acid acts as a catalyst for the reaction between the amide and water. The alkaline hydrolysis of amides actually involves reaction with hydroxide ions, but the result is similar enough that it is still classed as hydrolysis.
Hydrolysis under acidic conditions
Taking ethanamide as a typical amide. If ethanamide is heated with a dilute acid (such as dilute hydrochloric acid), ethanoic acid is formed together with ammonium ions. So, if you were using hydrochloric acid, the final solution would contain ammonium chloride and ethanoic acid.
$CH_3CONH_2 + H_2O + HCl \ rightarrow CH_3COOH + NH_4^+Cl^-$
Hydrolysis under alkaline conditions
Also, if ethanamide is heated with sodium hydroxide solution, ammonia gas is given off and you are left with a solution containing sodium ethanoate.
$CH_3CONH_2 + NaOH \rightarrow CH_3COONa + NH_3$
Using alkaline hydrolysis to test for an amide
If you add sodium hydroxide solution to an unknown organic compound, and it gives off ammonia on heating (but not immediately in the cold), then it is an amide. You can recognize the ammonia by smell and because it turns red litmus paper blue.
The possible confusion using this test is with ammonium salts. Ammonium salts also produce ammonia with sodium hydroxide solution, but in this case there is always enough ammonia produced in the cold for the smell to be immediately obvious. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/11%3A_Carbonyl_Compounds_I%3A_Reactions_of_Carboxylic_Acids_and_Carboxylic_Derivatives/11.10%3A_Soaps_Detergents_and_Micelles.txt |
Applications of Condensation Reactions to Synthesis
The construction of complex molecules by a series of suitable reactions carried out from simple starting compounds is called synthesis. Synthesis is not only of immense practical importance (aspirin and nylon are two examples of commercially valuable synthetic compounds), but it also allows us to prepare novel molecules with which to test our understanding of structure and reactivity. Three challenges must be met in devising a synthesis for a specific compound:
1. The carbon atom framework or skeleton that is found in the desired compound (the target) must be assembled.
2. The functional groups that characterize the target compound must be introduced or transformed from other groups at appropriate locations.
3. If centers of stereoisomerism are present, they must be fixed in a proper manner.
Recognition of these tasks does not imply that they are independent of each other, or should be approached and solved separately. A successful plan or strategy for a synthesis must correlate each step with all these goals, so that an efficient and practical solution to making the target molecule is achieved. Nevertheless, it is useful to classify the various reactions we have studied with respect to their ability to (i) enlarge or expand a given structure, (ii) transform or relocate existing functional groups, and (iii) do both of these in a stereoselective fashion. The organization of this text by functional group behavior partially satisfies the second point, and the following discussion focuses on the first.
1. Carbon-Carbon Bond Formation
A useful assortment of carbon-carbon bond forming reactions have been described in this and earlier chapters. These include:
(1) Friedel-Crafts alkylation and acylation.
(2) Diels-Alder cycloaddition.
(3) addition of organometallic reagents to aldehydes, ketones & carboxylic acid derivatives.
(4) alkylation of acetylide anions.
(5) alkylation of enolate anions.
(6) Claisen and aldol condensations.
With the exception of Friedel-Crafts alkylation these reactions all give products having one or more functional groups at or adjacent to the bonding sites. As a result, subsequent functional group introduction or modification may be carried out in a relatively straightforward manner. This will be illustrated for aldol and Claisen condensations in the following section.
2. Modification of Condensation Products
A. Reactions of Aldol Products
The aldol reaction produces beta-hydroxyaldehydes or ketones, and a number of subsequent reactions may be carried out with these products. As shown in the following diagram, they may be (i) reduced to 1,3-diols, (ii) a 2º-hydroxyl group may be oxidized to a carbonyl group, (iii) acid or base catalyzed beta-dehydration may produce an unsaturated aldehyde or ketone, and (iv) organometallic reagents may be added to the carbonyl group (assuming the hydroxyl group is protected as an ether or a second equivalent of reagent is used).
B. Reactions of Claisen Products
The Claisen condensation produces beta-ketoesters. These products may then be modified or enhanced by further reactions. Among these, the following diagram illustrates (i) partial reduction of the ketone with NaBH4, (ii) complete reduction to a 1,3-diol by LiAlH4, (iii) enolate anion alkylation, and (iv) ester hydrolysis followed by thermal decarboxylation of the resulting beta-ketoacid.
C. Synthesis Examples
To illustrate how the reaction sequences described above may be used to prepare a variety of different compounds, five examples are provided here. The first is a typical aldol reaction followed by reduction to a 1,3-diol (2-ethyl-1,3-hexanediol). In the second example, the absence of alpha-hydrogens on the aldehyde favors the mixed condensation, and conjugation of the double bond facilitates dehydration. The doubly-activated methylene group of malonic and acetoacetic acids or esters makes them good donors in any condensation, as is demonstrated by the third aldol-like reaction. A concerted dehydrative-decarboxylation (shown by the magenta arrows) leads to the unsaturated carboxylic acid product. Amine bases are often used as catalysts for aldol reactions, as in equations #2 & 3. The fourth reaction demonstrates that the conjugate base of the beta-ketoester products from Claisen or Dieckmann condensation may be alkylated directly. Thermal decarboxylation of the resulting beta-ketoacid gives a mono-alkylated cyclic ketone. Finally, both acidic methylene hydrogens in malonic ester or ethyl acetoacetate may be substituted, and the irreversible nature of such alkylations permits strained rings to be formed. In this case thermal decarboxylation of a substituted malonic acid generates a carboxylic acid. In all these examples the remaining functional groups could be used for additional synthetic operations.
Vinylagous Reactions
A large family of vinylagous reactions, related to the condensations, acylations and alkylations described here, increase the bond forming options available to the synthetic chemist. To learn more about these versatile reactions Click Here
Some Exercises
If you understand the previous discussion of reactions useful in synthesis you should try the following problems. Some of them are complex so don't be concerned if you don't solve them all immediately. Analyze each problem carefully, and try to learn from it. The solutions will be displayed by clicking the answer button under the diagram.
The following problems ask you to devise a synthesis for a given target molecule. The first two problems make use of the common starting materials, diethyl malonate and ethyl acetoacetate. The third problem leaves the choice of materials open. The nature of the target molecule suggests that an aldol condensation might be useful. The fourth problem must be solved by using diethyl succinate as the only reagent. Finally, other reactants composed of no more than five carbon atoms may be used in the last problem.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
11.15: Nitriles
Nitriles can be converted to carboxylic acid with heating in sulfuric acid. During the reaction an amide intermediate is formed.
Contributors
Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/11%3A_Carbonyl_Compounds_I%3A_Reactions_of_Carboxylic_Acids_and_Carboxylic_Derivatives/11.14%3A_The_Synthesis_of_Carboxylic_Acid_Derivatives.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
12: Carbonyl Compounds II: Reactions of Aldehydes and Ketones More Reactions of Carboxylic Acid Derivatives
Aldehydes and ketones contain the carbonyl group. Aldehydes are considered the most important functional group. They are often called the formyl or methanoyl group. Aldehydes derive their name from the dehydration of alcohols. Aldehydes contain the carbonyl group bonded to at least one hydrogen atom. Ketones contain the carbonyl group bonded to two carbon atoms.
Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen, alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde. If neither is hydrogen, the compound is a ketone.
Naming Aldehydes
The IUPAC system of nomenclature assigns a characteristic suffix -al to aldehydes. For example, H2C=O is methanal, more commonly called formaldehyde. Since an aldehyde carbonyl group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. There are several simple carbonyl containing compounds which have common names which are retained by IUPAC.
Also, there is a common method for naming aldehydes and ketones. For aldehydes common parent chain names, similar to those used for carboxylic acids, are used and the suffix –aldehyde is added to the end. In common names of aldehydes, carbon atoms near the carbonyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
If the aldehyde moiety (-CHO) is attached to a ring the suffix –carbaldehyde is added to the name of the ring. The carbon attached to this moiety will get the #1 location number in naming the ring.
Summary of Aldehyde Nomenclature rules
1. Aldehydes take their name from their parent alkane chains. The -e is removed from the end and is replaced with -al.
2. The aldehyde funtional group is given the #1 numbering location and this number is not included in the name.
3. For the common name of aldehydes start with the common parent chain name and add the suffix -aldehyde. Substituent positions are shown with Greek letters.
4. When the -CHO functional group is attached to a ring the suffix -carbaldehyde is added, and the carbon attached to that group is C1.
Example 1
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Aldehyde Common Names to Memorize
There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
Naming Ketones
The IUPAC system of nomenclature assigns a characteristic suffix of -one to ketones. A ketone carbonyl function may be located anywhere within a chain or ring, and its position is usually given by a location number. Chain numbering normally starts from the end nearest the carbonyl group. Very simple ketones, such as propanone and phenylethanone do not require a locator number, since there is only one possible site for a ketone carbonyl function. The common names for ketones are formed by naming both alkyl groups attached to the carbonyl then adding the suffix -ketone. The attached alkyl groups are arranged in the name alphabetically.
Summary of Ketone Nomenclature rules
1. Ketones take their name from their parent alkane chains. The ending -e is removed and replaced with -one.
2. The common name for ketones are simply the substituent groups listed alphabetically + ketone.
3. Some common ketones are known by their generic names. Such as the fact that propanone is commonly referred to as acetone.
Example 2
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Ketone Common Names to Memorize
There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
Naming Aldehydes and Ketones in the Same Molecule
As with many molecules with two or more functional groups, one is given priority while the other is named as a substituent. Because aldehydes have a higher priority than ketones, molecules which contain both functional groups are named as aldehydes and the ketone is named as an "oxo" substituent. It is not necessary to give the aldehyde functional group a location number, however, it is usually necessary to give a location number to the ketone.
Naming Dialdehydes and Diketones
For dialdehydes the location numbers for both carbonyls are omitted because the aldehyde functional groups are expected to occupy the ends of the parent chain. The ending –dial is added to the end of the parent chain name.
Example 4
For diketones both carbonyls require a location number. The ending -dione or -dial is added to the end of the parent chain.
Naming Cyclic Ketones and Diketones
In cyclic ketones the carbonyl group is assigned location position #1, and this number is not included in the name, unless more than one carbonyl group is present. The rest of the ring is numbered to give substituents the lowest possible location numbers. Remember the prefix cyclo is included before the parent chain name to indicate that it is in a ring. As with other ketones the –e ending is replaced with the –one to indicate the presence of a ketone.
With cycloalkanes which contain two ketones both carbonyls need to be given a location numbers. Also, an –e is not removed from the end, but the suffix –dione is added.
Naming Carbonyls and Hydroxyls in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains an alcohol functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of alcohols the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
Naming Carbonyls and Alkenes in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains analkene functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. When carbonyls are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an -en ending to indicate the presence of an alkene)-(the location number of the carbonyl if a ketone is present)-(either an –one or and -anal ending).
Remember that the carbonyl has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Aldehydes and Ketones as Fragments
• Alkanoyl is the common name of the fragment, though the older naming, acyl, is still widely used.
• Formyl is the common name of the fragment.
• Acety is the common name of the CH3-C=O- fragment.
Additional Examples of Carbonyl Nomenclature
1) Please give the IUPAC name for each compound:
Answers for Question 1
1. 3,4-dimethylhexanal
2. 5-bromo-2-pentanone
3. 2,4-hexanedione
4. cis-3-pentenal (or (Z)-3-pentenal)
5. 6-methyl-5-hepten-3-one
6. 3-hydroxy-2,4-pentanedione
7. 1,2-cyclobutanedione
8. 2-methyl-propanedial
9. 3-methyl-5-oxo-hexanal
10. cis-2,3-dihydroxycyclohexanone
11. 3-bromo-2-methylcyclopentanecarboaldehyde
12. 3-bromo-2-methylpropanal
2) Please give the structure corresponding to each name:
A) butanal
B) 2-hydroxycyclopentanone
C) 2,3-pentanedione
D) 1,3-cyclohexanedione
E) 4-hydoxy-3-methyl-2-butanone
F) (E) 3-methyl-2-hepten-4-one
G) 3-oxobutanal
H) cis-3-bromocyclohexanecarboaldehyde
I) butanedial
J) trans-2-methyl-3-hexenal
Answers to question 2: | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/12%3A_Carbonyl_Compounds_II%3A_Reactions_of_Aldehydes_and_Ketones__More_Reactions_of_Carboxylic_Acid_Derivatives/12.01%3A_The_Nomenclature_of_Aldehydes_and_Keto.txt |
The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
12.03: How Aldehydes and Ketones React
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 soverlaps 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’ pbond 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.
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 (which we will learn about in section 13.3) 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.
. . . on to the next section . . .
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/12%3A_Carbonyl_Compounds_II%3A_Reactions_of_Aldehydes_and_Ketones__More_Reactions_of_Carboxylic_Acid_Derivatives/12.02%3A_The_Relative_Reactivities_of_Carbonyl_.txt |
A Grignard reagent has a formula RMgX where X is a halogen, and R is an alkyl or aryl (based on a benzene ring) group. For the purposes of this page, we shall take R to be an alkyl group (e.g., CH3CH2MgBr). Grignard reagents are made by adding the halogenoalkane to small bits of magnesium in a flask containing ethoxyethane (commonly called diethyl ether or just "ether"). The flask is fitted with a reflux condenser, and the mixture is warmed over a water bath for 20 - 30 minutes.
Everything must be perfectly dry because Grignard reagents react with water. Any reactions using the Grignard reagent are carried out with the mixture produced from this reaction; you cannot separate it out in any way. Any excess Grignard reagent must be quenched before disposal.
Reactions of Grignard reagents with aldehydes and ketones
These are reactions of the carbon-oxygen double bond, and so aldehydes and ketones react in exactly the same way - all that changes are the groups that happen to be attached to the carbon-oxygen double bond. It is much easier to understand what is going on by looking closely at the general case (using "R" groups rather than specific groups) - and then slotting in the various real groups as and when you need to. The "R" groups can be either hydrogen or alkyl in any combination.
In the first stage, the Grignard reagent adds across the carbon-oxygen double bond:
Dilute acid is then added to this to hydrolyse it.
An alcohol is formed. One of the key uses of Grignard reagents is the ability to make complicated alcohols easily. What sort of alcohol you get depends on the carbonyl compound you started with - in other words, what R and R' are.
Example \(1\): Reaction with Methanal
Methanal is the simplest possible aldehyde with hydrogen as both R groups.
Assuming that you are starting with CH3CH2MgBr and using the general equation above, the alcohol you get always has the form:
Since both R groups are hydrogen atoms, the final product will be:
A primary alcohol is formed. A primary alcohol has only one alkyl group attached to the carbon atom with the -OH group on it. You could obviously get a different primary alcohol if you started from a different Grignard reagent.
Example \(2\): Reaction with Aldehyde
The next biggest aldehyde is ethanal with one of the R groups is hydrogen and the other CH3.
Again, think about how that relates to the general case. The alcohol formed is:
So this time the final product has one CH3 group and one hydrogen attached:
A secondary alcohol has two alkyl groups (the same or different) attached to the carbon with the -OH group on it. You could change the nature of the final secondary alcohol by either:
• changing the nature of the Grignard reagent - which would change the CH3CH2 group into some other alkyl group;
• changing the nature of the aldehyde - which would change the CH3 group into some other alkyl group.
Example \(3\): Reactions with Propanone
Ketones have two alkyl groups attached to the carbon-oxygen double bond. The simplest one is propanone.
This time when you replace the R groups in the general formula for the alcohol produced you get a tertiary alcohol.
A tertiary alcohol has three alkyl groups attached to the carbon with the -OH attached. The alkyl groups can be any combination of same or different. You could ring the changes on the product by:
• changing the nature of the Grignard reagent - which would change the CH3CH2 group into some other alkyl group;
• changing the nature of the ketone - which would change the CH3 groups into whatever other alkyl groups you choose to have in the original ketone. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/12%3A_Carbonyl_Compounds_II%3A_Reactions_of_Aldehydes_and_Ketones__More_Reactions_of_Carboxylic_Acid_Derivatives/12.04%3A_Gringard_Reagents.txt |
Despite the fearsome names, the structures of the two reducing agents are very simple. In each case, there are four hydrogens ("tetrahydido") around either aluminium or boron in a negative ion (shown by the "ate" ending). The "(III)" shows the oxidation state of the aluminium or boron, and is often left out because these elements only ever show the +3 oxidation state in their compounds.
The formulae of the two compounds are \(LiAlH_4\) and \(NaBH_4\). Their structures are:
In each of the negative ions, one of the bonds is a co-ordinate covalent (dative covalent) bond using the lone pair on a hydride ion (H-) to form a bond with an empty orbital on the aluminium or boron.
The reduction of an aldehyde
You get exactly the same organic product whether you use lithium tetrahydridoaluminate or sodium tetrahydridoborate. For example, with ethanal you get ethanol:
Notice that this is a simplified equation where [H] means "hydrogen from a reducing agent". In general terms, reduction of an aldehyde leads to a primary alcohol.
The reduction of a Ketone
Again the product is the same whichever of the two reducing agents you use. For example, with propanone you get propan-2-ol:
Reduction of a ketone leads to a secondary alcohol.
Using lithium tetrahydridoaluminate (lithium aluminium hydride)
Lithium tetrahydridoaluminate is much more reactive than sodium tetrahydridoborate. It reacts violently with water and alcohols, and so any reaction must exclude these common solvents. The reactions are usually carried out in solution in a carefully dried ether such as ethoxyethane (diethyl ether). The reaction happens at room temperature, and takes place in two separate stages.
In the first stage, a salt is formed containing a complex aluminium ion. The following equations show what happens if you start with a general aldehyde or ketone. R and R' can be any combination of hydrogen or alkyl groups.
The product is then treated with a dilute acid (such as dilute sulfuric acid or dilute hydrochloric acid) to release the alcohol from the complex ion.
The alcohol formed can be recovered from the mixture by fractional distillation.
Using sodium tetrahydridoborate (sodium borohydride)
Sodium tetrahydridoborate is a more gentle (and therefore safer) reagent than lithium tetrahydridoaluminate. It can be used in solution in alcohols or even solution in water - provided the solution is alkaline. Solid sodium tetrahydridoborate is added to a solution of the aldehyde or ketone in an alcohol such as methanol, ethanol or propan-2-ol. Depending on which recipe you read, it is either heated under reflux or left for some time around room temperature. This almost certainly varies depending on the nature of the aldehyde or ketone.
At the end of this time, a complex similar to the previous one is formed.
In the second stage of the reaction, water is added and the mixture is boiled to release the alcohol from the complex.
Again, the alcohol formed can be recovered from the mixture by fractional distillation.
12.07: The Reactions of Aldehydes and Ketones
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Converting reactants to products simply
Mechanism of imine formation
1) Nucleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Reversibility of imine forming reactions
Imines can be hydrolyzed back to the corresponding primary amine under acidic conditons.
Reactions involving other reagents of the type Y-NH2
Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH2 have been studied, and found to give stable products (R2C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. Hydrazones are used as part of the Wolff-Kishner reduction and will be discussed in more detail in another module.
With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. It should be noted that although semicarbazide has two amino groups (–NH2) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group.
Problems
1)Please draw the products of the following reactions.
2) Please draw the structure of the reactant needed to produce the indicated product.
3) Please draw the products of the following reactions.
1)
2)
3)
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/12%3A_Carbonyl_Compounds_II%3A_Reactions_of_Aldehydes_and_Ketones__More_Reactions_of_Carboxylic_Acid_Derivatives/12.06%3A_The_Reactions_of_Carbonyl_Compounds_wi.txt |
It has been demonstrated that water, in the presence of an acid or a base, adds rapidly to the carbonyl function of aldehydes and ketones establishing a reversible equilibrium with a hydrate (geminal-diol or gem-diol). The word germinal or gem comes from the Latin word for twin, geminus.
Reversibility of the Reaction
Isolation of gem-diols is difficult because the reaction is reversibly. Removal of the water during a reaction can cause the conversion of a gem-diol back to the corresponding carbonyl.
Factors Affecting the Gem-diol Equilibrium
In most cases the resulting gem-diol is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde where the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition. Thus, a solution of formaldehyde in water (formalin) is almost exclusively the hydrate, or polymers of the hydrate. The addition of electron donating alkyl groups stabilized the partial positive charge on the carbonyl carbon and decreases the amount of gem-diol product at equilibrium. Because of this ketones tend to form less than 1% of the hydrate at equilibrium. Likewise, the addition of strong electron-withdrawing groups destabilizes the carbonyl and tends to form stable gem-diols. Two examples of this are chloral, and 1,2,3-indantrione. It should be noted that chloral hydrate is a sedative and has been added to alcoholic beverages to make a “Knock-out” drink also called a Mickey Finn. Also, ninhydrin is commonly used by forensic investigators to resolve finger prints.
Mechanism of Gem-diol Formation
The mechanism is catalyzed by the addition of an acid or base. Note! This may speed up the reaction but is has not effect on the equilibriums discussed above. Basic conditions speed up the reaction because hydroxide is a better nucleophilic than water. Acidic conditions speed up the reaction because the protonated carbonyl is more electrophilic.
Under Basic conditions
1) Nucleophilic attack by hydroxide
2) Protonation of the alkoxide
Under Acidic conditions
1) Protonation of the carbonyl
2) Nucleophilic attack by water
3) Deprotonation
Problems
1) Draw the expected products of the following reactions.
2) Of the following pairs of molecules which would you expect to form a larger percentage of gem-diol at equilibrium? Please explain your answer.
3) Would you expect the following molecule to form appreciable amount of gem-diol in water? Please explain your answer.
Answers
1)
2) The compound on the left would. Fluorine is more electronegative than bromine and would remove more electron density from the carbonyl carbon. This would destabilize the carbonyl allowing for more gem-diol to form.
3) Although ketones tend to not form gem-diols this compound exists almost entirely in the gem-diol form when placed in water. Ketones tend to not form gem-diols because of the stabilizing effect of the electron donating alkyl group. However, in this case the electron donating effects of alkyl group is dominated by the presence of six highly electronegative fluorines.
12.09: Reactions of Aldehydes and Ketones wit
In this organic chemistry topic, we shall see how alcohols (R-OH) add to carbonyl groups. Carbonyl groups are characterized by a carbon-oxygen double bond. The two main functional groups that consist of this carbon-oxygen double bond are Aldehydes and Ketones.
Introduction
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (hemi, Greek, half). This reaction can continue by adding another alcohol to form an acetal. Hemiacetals and acetals are important functional groups because they appear in sugars.
To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile; and second, the water produced with the acetal must be removed from the reaction by a process such as a molecular sieves or a Dean-Stark trap. The latter is important, since acetal formation is reversible. Indeed, once pure hemiacetal or acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid and an excess of water.
Formation of Acetals
Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. It is important to note that a hemiacetal is formed as an intermediate during the formation of an acetal.
Mechanism for Hemiacetal and Acetal Formation
The mechanism shown here applies to both acetal and hemiacetal formation
1) Protonation of the carbonyl
2) Nucleophilic attack by the alcohol
3) Deprotonation to form a hemiacetal
4) Protonation of the alcohol
5) Removal of water
6) Nucleophilic attack by the alcohol
7) Deprotonation by water
Formation of Cyclic Hemiacetal and Acetals
Molecules which have an alcohol and a carbonyl can undergo an intramolecular reaction to form a cyclic hemiacetal.
Intramolecular Hemiacetal formation is common in sugar chemistry. For example, the common sugar glucose exists in the cylcic manner more than 99% of the time in a mixture of aqueous solution.
Carbonyls reacting with diol produce a cyclic acetal. A common diol used to form cyclic acetals is ethylene glycol.
Acetals as Protecting Groups
The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general. Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents. If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.
In the following example we would like a Grignard reagent to react with the ester and not the ketone. This cannot be done without a protecting group because Grignard reagents react with esters and ketones. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/12%3A_Carbonyl_Compounds_II%3A_Reactions_of_Aldehydes_and_Ketones__More_Reactions_of_Carboxylic_Acid_Derivatives/12.08%3A_The_Reactions_of_Aldehydes_and_Ketones.txt |
One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls. A resonance description of this transmission is shown below.
From this formula it should be clear that nucleophiles may attack either at the carbonyl carbon, as for any aldehyde, ketone or carboxylic acid derivative, or at the beta-carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition.
Basic reaction of 1,2 addition
Here the nucleophile adds to the carbon which is in the one position. The hydrogen adds to the oxygen which is in the two position.
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Going from reactant to products simplified
1,2 vs. 1,4 addition
Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents, both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions.
If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained.
Water
Alcohols
Thiols
1o Amines
2o Amines
HBr
Cyanides
Gilman Reagents
Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls
Organocuprate reagents are made from the reaction of organolithium reagents and $CuI$
$2 RLi + CuI \rightarrow R_2CuLi + LiI$
This acts as a source of R:-
$2 CH_3Li + CuI \rightarrow (CH_3)_2CuLi + LiI$
Example
Nucleophiles which add 1,2 to α, β unsaturated carbonyls
Metal Hydrides
Grignard Reagents
Organolithium Reagents
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/12%3A_Carbonyl_Compounds_II%3A_Reactions_of_Aldehydes_and_Ketones__More_Reactions_of_Carboxylic_Acid_Derivatives/12.10%3A__Nucleophilic_Addition_to__-_Unsaturat.txt |
Organophosphorus ylides react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Preparation of Phosphorus Ylides
It has been noted that dipolar phosphorus compounds are stabilized by p-d bonding. This bonding stabilization extends to carbanions adjacent to phosphonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. An ylide is defined as a compound with opposite charges on adjacent atoms both of which have complete octets. For the Wittig reaction discussed below an organophosphorus ylide, also called Wittig reagents, will be used. The ability of phosphorus to hold more than eight valence electrons allows for a resonance structure to be drawn forming a double bonded structure.
The stabilization of the carbanion provided by the phosphorus causes an increase in acidity (pKa ~35). Very strong bases, such as butyl lithium, are required for complete formation of ylides.
The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes phosphorous ylides to hydrocarbons and phosphine oxides, as shown.
Although many ylides are commercially available it is often necessary to create them synthetically. Ylides can be synthesized from an alkyl halide and a trialkyl phosphine. Typically triphenyl phosphine is used to synthesize ylides. Because a SN2 reaction is used in the ylide synthesis methyl and primary halides perform the best. Secondary halides can also be used but the yields are generally lower. This should be considered when planning out a synthesis which involves a synthesized Wittig reagent.
1) SN2 reaction
2) Deprotonation
The Wittig Reaction
The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Going from reactants to products simplified
Mechanism of the Wittig reaction
Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible.
1) Nucleophillic attack on the carbonyl
2) Formation of a 4 membered ring
3) Formation of the alkene
Limitation of the Wittig reaction
If possible both E and Z isomer of the double bond will be formed. This should be considered when planning a synthesis involving a Wittig Reaction.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please draw the structure of the oxaphosphetane which is made during the mechanism of the reaction given that produces product C.
4) Please draw the structure of the betaine which is made during the mechanism of the reaction given that produces product D.
5) Please give a detailed mechanism and the final product of this reaction
6) It has been shown that reacting and epoxide with triphenylphosphine forms an alkene. Please propose a mechanism for this reaction. Review the section on epoxide reactions if you need help.
Answers
1)
2)
3)
4)
5)
Nucleophillic attack on the carbonyl
Formation of a 4 membered ring
Formation of the alkene
6) Nucleophillic attack on the epoxide
Formation of a 4 membered ring
Formation of the alkene
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
18.14 Stereochemistry of Nucleophilic Additi
Notice that in the course of the nucleophilic addition pictured above, the hybridization of the carbonyl carbon changes from sp2 to sp3, meaning that the bond geometry changes from trigonal planar to tetrahedral. It is also important to note that if the starting carbonyl is asymmetric (in other words, if the two R groups are not equivalent), then a new stereocenter has been created. The configuration of the new stereocenter depends upon which side of the carbonyl plane the nucleophile attacks from.
If the reaction is catalyzed by an enzyme, the stereochemistry of addition is tightly controlled, and leads to one specific stereoisomer - this is because the nucleophilic and electrophilic substrates are bound in a specific positions within the active site, so that attack must occur specifically from one side. If, however, the reaction occurs uncatalyzed in solution, then either side of the carbonyl is equally likely to be attacked, and the result will be a 50:50 racemic mixture.
This is the rule for most nonenzymatic reactions, but as with most rules, there are exceptions. If, for example, the geometry of the carbonyl-containing molecule is constrained in such a way that approach by the nucleophile is less hindered from one side, a 50:50 racemic mixture will not necessarily result. Consider camphor, the distinctive-smelling compound found in many cosmetics and skin creams.
Upon inspection it is clear that topside attack and bottom side attack by a nucleophile are nonequivalent in terms of steric hindrance. A relatively simple experiment shows that, when the incoming nucleophile is a hydride ion from the common synthetic reducing agent sodium borohydride (a reaction type we will study in a later chapter), the product of bottom side attack predominates by a ratio of about 6 to 1 (see section 16.4D for more details on this experiment). We can infer from this result that approach from the bottom (si) face of the carbonyl in camphor is less hindered.
. . . on to the next section . . .
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/12%3A_Carbonyl_Compounds_II%3A_Reactions_of_Aldehydes_and_Ketones__More_Reactions_of_Carboxylic_Acid_Derivatives/18.13____The_Wittig_Reaction_Forms_an_Alkene.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
13: Carbonyl Compounds III: Reactions at the - Carbon
As previously seen, aldehydes and ketones react with 2o amines to reversibly form enamines.
Example
Reversible
Enamines act as nucleophiles in a fashion similar to enolates. Because of this enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps:
1. Formation of the enamine,
2. Reaction with an eletrophile to form an iminium salt,
3. Hydrolysis of the iminium salt to reform the aldehyde or ketone.
Some of the advantages of using an enamine over and enolate are enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. These reactions are generally known as the Stork enamine reaction after Gilbert Stork of Columbia University who originated the work.
Typically we use the following 2o amines for enamine reactions
Alkylation of an Enamine
Enamined undergo an SN2 reaction with reactive alkyl halides to give the iminium salt. The iminium salt can be hydrolyzed back into the carbonyl.
Step 1: Formation of an enamine
Step 2: SN2 Alkylation
Step 3: Reform the carbonyl by hydrolysis
All three steps together:
Acylation of Enamines
Enamine can react with acid halides to form β-dicarbonyls
1) Formation of the enamine
2) Nucleophilic attack
3) Leaving group removal
4) Reform the carbonyl by hydrolysis
All three steps together:
Michael Addition using Enamines
Enamines, like other weak bases, add 1,4 to enones. The end product is a 1,5 dicarbonyl compound.
19.11 Alkylation of the -Carbon: The Michael Reaction
Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942).
Robinson Annulation
Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred.
19.12 An Aldol Addition Forms -Hydroxaldehydes or -Hydroxyketones
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as Grignard reagents. For this reaction to occur at least one of the reactants must have α hydrogens.
Aldol Reaction Mechanism
A three step mechanism:
Step 1: Enolate formation
Step 2: Nucleophilic attack by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of Aldol products to synthesize α, β unsaturated carbonyl (enones)
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this $\beta$-elimination reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed Condensations. Hence, the following examples are properly referred to as aldol condensations. Overall the general reaction involves a dehydration of an aldol product to form an alkene:
Figure: General reaction for an aldol condensation
Going from reactants to products simply
Figure: The aldol condensatio example
Aldol Condensation Mechanism
1) Form enolate
2) Form enone
When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl.
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of $\alpha$ hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred.
As with other aldol reaction the addition of heat causes an aldol condensation to occur.
Mixed Aldol Reaction and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Example 4: Mixed Aldol Reactions
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens.
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
AACH2CHO + BCH2CHO + NaOH → AA + BB + AB + BA
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.10____Alkylation_and_Acylation_of_the_-Carbon_Using_an_Enamine_Intermediate.txt |
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as Grignard reagents. For this reaction to occur at least one of the reactants must have α hydrogens.
Aldol Reaction Mechanism
A three step mechanism:
Step 1: Enolate formation
Step 2: Nucleophilic attack by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of Aldol products to synthesize α, β unsaturated carbonyl (enones)
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this $\beta$-elimination reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed Condensations. Hence, the following examples are properly referred to as aldol condensations. Overall the general reaction involves a dehydration of an aldol product to form an alkene:
Figure: General reaction for an aldol condensation
Going from reactants to products simply
Figure: The aldol condensatio example
Aldol Condensation Mechanism
1) Form enolate
2) Form enone
When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl.
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of $\alpha$ hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred.
As with other aldol reaction the addition of heat causes an aldol condensation to occur.
Mixed Aldol Reaction and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Example 4: Mixed Aldol Reactions
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens.
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
AACH2CHO + BCH2CHO + NaOH → AA + BB + AB + BA
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.13____Dehydration_of_Aldol_Addition_Products_Form__-Unsaturated_Aldehydes_and_Ketones.txt |
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as Grignard reagents. For this reaction to occur at least one of the reactants must have α hydrogens.
Aldol Reaction Mechanism
A three step mechanism:
Step 1: Enolate formation
Step 2: Nucleophilic attack by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of Aldol products to synthesize α, β unsaturated carbonyl (enones)
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this $\beta$-elimination reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed Condensations. Hence, the following examples are properly referred to as aldol condensations. Overall the general reaction involves a dehydration of an aldol product to form an alkene:
Figure: General reaction for an aldol condensation
Going from reactants to products simply
Figure: The aldol condensatio example
Aldol Condensation Mechanism
1) Form enolate
2) Form enone
When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl.
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of $\alpha$ hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred.
As with other aldol reaction the addition of heat causes an aldol condensation to occur.
Mixed Aldol Reaction and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Example 4: Mixed Aldol Reactions
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens.
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
AACH2CHO + BCH2CHO + NaOH → AA + BB + AB + BA
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction.
19.15 A Claisen Condensation Forms a -Keto Ester
Because esters can contain $\alpha$ hydrogens they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed; the product is a β-keto ester. A major difference with the aldol reaction is the fact that hydroxide cannot be used as a base because it could possibly react with the ester. Instead, an alkoxide version of the alcohol used to synthesize the ester is used to prevent transesterification side products.
Claisen Condensation
Basic reaction
Going from reactants to products simply
Claisen Condensation Mechanism
1) Enolate formation
2) Nucleophilic attack
3) Removal of leaving group
Dieckmann Condensation
A diester can undergo an intramolecular reaction called a Dieckmann condensation.
Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures. Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens.
19.16 Other Crossen Condensations
Because esters can contain $\alpha$ hydrogens they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed; the product is a β-keto ester. A major difference with the aldol reaction is the fact that hydroxide cannot be used as a base because it could possibly react with the ester. Instead, an alkoxide version of the alcohol used to synthesize the ester is used to prevent transesterification side products.
Claisen Condensation
Basic reaction
Going from reactants to products simply
Claisen Condensation Mechanism
1) Enolate formation
2) Nucleophilic attack
3) Removal of leaving group
Dieckmann Condensation
A diester can undergo an intramolecular reaction called a Dieckmann condensation.
Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures. Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.14____The_Crossed_Aldol_Addition.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the Robinson annulation reaction.
2. identify the cyclic product formed when a 1,5‑diketone is treated with base.
3. identify the carbonyl compounds and any other reagents needed to synthesize a given cyclic compound by a series of reactions, one of which is a Robinson annulation.
Key Terms
• Robinson annulation reaction
Study Notes
The building of an alicyclic compound from acyclic starting materials can present an interesting challenge to the synthetic organic chemist. One route by which such a synthesis can be achieved is through the use of the Robinson annulation reaction. This reaction, which may at first look very complex, can be readily understood once you realize that it simply involves a Michael reaction followed by an intramolecular aldol condensation. Both of these steps involve attack by enolate anions.
As in some of the other syntheses that you have studied, when you are simply given the structure of a compound and asked how it could have been prepared, it can be difficult to recognize which reactions might have been used. In this instance, keep in mind that you have studied relatively few reactions which have resulted in the formation of an alicyclic compound. Thus, when asked how such a compound might be prepared, you should keep the possibility of using a Robinson annulation reaction in mind.
The Robinson Annulation
Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred.
The nucleophilic enolate donor is typically an enolate ion or enamine of a cyclic ketone, β-keto ester or β-diketone. The electrophilic acceptor is usually an α, β-unsaturated ketone. In the example below, 2-methyl-1,3-cyclohexanedione is deprotonated to form an enolate which affects a Michael reaction addition on 3-buten-2-one forming a C-C bond. The product contains a 1,5-diketone fragment which can undergo an intramolecular aldol condensation. This occurs through creation of a new enolate at the methyl ketone which undergoes an intramolecular aldol reaction. A new C-C bond is formed to one of the ring carbonyls, creating a new six-membered ring. In the last step, the resulting alcohol is eliminated to form a α, β-unsaturated ketone as the final Robinson annulation product. Because the Robinson Annulation involves an aldol reaction, a full equivalent of base is required.
Example
Worked Example
Draw the product of the following Robinson Annulation.
Answer
Analysis: When considering the product of a Robinson annulation it is usually best to consider each reaction individually. Use the steps discussed in Section 23.10 to convert the starting materials into the product of a Michael reaction, then into the product of an intramolecular aldol condensation. If the starting materials are drawn in a skeletal structure it may be helpful to convert to a condensed formula to better keep track of carbons and hydrogens.
Formation of the Michael reaction product
Formation of the intramolecular Aldol condensation product
Planning a Synthesis Using a Robinson Annulation
The presence of a cyclic, six-membered α, β-unsaturated ketone in a target molecule suggest that a Robinson Annulation may be utilized in its synthesis. The key bond cleavages are the C=C bond of the α, β-unsaturated ketone and the C-C bond between carbons in the a and g on the opposite alkyl chain of the ketone.
Worked Example
What would be the starting materials used if the following molecule was made using a Robinson annulation?
Answer
Analysis
Solution
It is necessary to use sodium ethoxide as the reaction base due to the presence of an ester.
Exercise \(1\)
1) Provide products of the following Robinson annulations.
a)
b)
c)
2) What would be the starting materials required to make the following molecule using a Robinson annulation?
Answers
1)
a)
b)
c)
2) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.18____The_Robinson_Annulation.txt |
Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2.
• reaction 1:
reaction 2:
reaction 3:
reaction 4:
A more direct method to convert 3 into 4 is the reaction of 3 with the enolate ion (5) of ethyl acetate followed by hydrolysis of the resultant ester.
However, the generation of 5 from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions.
Malonic ester synthesis provides a more convenient alternative to convert 3 to 4.
Malonic ester synthesis can be adapted to synthesize compounds that have the general structural formula 6.
R3, R4 = identical or different alkyl groups
eg:
reaction 1:
reaction 2:
reaction 1 (repeat):
reaction 2 (repeat):
reaction 3:
reaction 4:
Malonic Ester Synthesis
Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester.
Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides.
After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step.
Mechanism
1) Saponification
2) Decarboxylation
3) Tautomerization
All of the steps together form the Malonic ester synthesis.
$RX \rightarrow RCH_2CO_2H$
Example
19.1 The Acidity of an - Hydrogen
Alkyl hydrogen atoms bonded to a carbon atom in a a (alpha) position relative to a carbonyl group display unusual acidity. While the pKa values for alkyl C-H bonds is typically on the order of 40-50, pKa values for these alpha hydrogens is more on the order of 19-20. This can most easily be explained by resonance stabilization of the product carbanion, as illustrated in the diagram below.
In the presence of a proton source, the product can either revert back into the starting ketone or aldehyde or can form a new product, the enol. The equilibrium reaction between the ketone or aldehyde and the enol form is commonly referred to as "keto-enol tautomerism". The ketone or aldehyde is generally strongly favored in this reaction.
Because carbonyl groups are sp2 hybridized the carbon and oxygen both have unhybridized p orbitals which can overlap to form the C=O $\pi$ bond.
The presence of these overlapping p orbitals gives $\alpha$ hydrogens (Hydrogens on carbons adjacent to carbonyls) special properties. In particular, $\alpha$ hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the $\pi$ orbitals of the carbonyl. The effect of the carbonyl is seen when comparing the pKa for the $\alpha$ hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25) to the pKa of an alkane (~50).
Of the two resonance structures of the enolate ion the one which places the negative charge on the oxygen is the most stable. This is because the negative change will be better stabilized by the greater electronegativity of the oxygen.
Keto-enol Tautomerism
Because of the acidity of α hydrogens carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom 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.
Contributors
Prof. Steven Farmer (Sonoma State University)
• Clarke Earley (Department of Chemistry, Kent State University Stark Campus) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.19____Carboxylic_Acids_with_a_Carbonyl_Group_at_the_3-Position_can_be_Decarboxylated.txt |
Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2.
• reaction 1:
reaction 2:
reaction 3:
reaction 4:
A more direct method to convert 3 into 4 is the reaction of 3 with the enolate ion (5) of ethyl acetate followed by hydrolysis of the resultant ester.
However, the generation of 5 from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions.
Malonic ester synthesis provides a more convenient alternative to convert 3 to 4.
Malonic ester synthesis can be adapted to synthesize compounds that have the general structural formula 6.
R3, R4 = identical or different alkyl groups
eg:
reaction 1:
reaction 2:
reaction 1 (repeat):
reaction 2 (repeat):
reaction 3:
reaction 4:
Malonic Ester Synthesis
Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester.
Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides.
After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step.
Mechanism
1) Saponification
2) Decarboxylation
3) Tautomerization
All of the steps together form the Malonic ester synthesis.
$RX \rightarrow RCH_2CO_2H$
Example
19.21 The Acetoacetic Ester Synthesis: A Way to Synthesize a Methyl Ketone
Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2.
• reaction 1:
reaction 2:
reaction 3:
reaction 4:
A more direct method to convert 3 into 4 is the reaction of 3 with the enolate ion (5) of ethyl acetate followed by hydrolysis of the resultant ester.
However, the generation of 5 from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions.
Malonic ester synthesis provides a more convenient alternative to convert 3 to 4.
Malonic ester synthesis can be adapted to synthesize compounds that have the general structural formula 6.
R3, R4 = identical or different alkyl groups
eg:
reaction 1:
reaction 2:
reaction 1 (repeat):
reaction 2 (repeat):
reaction 3:
reaction 4:
Malonic Ester Synthesis
Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester.
Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides.
After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step.
Mechanism
1) Saponification
2) Decarboxylation
3) Tautomerization
All of the steps together form the Malonic ester synthesis.
$RX \rightarrow RCH_2CO_2H$
Example | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.20____The_Malonic_Ester_Synthesis%3A_A_Way_to_Synthesize_a_Carboxylic_Acid.txt |
We come now to one of the most important mechanism types in metabolism: the carbon-carbon bond-forming 'aldol' condensation reaction.
13.3A: The general mechanism for an aldol reaction
Consider the potential pathways available to a reactive enolate intermediate once the alpha-proton has been abstracted. The oxygen, which bears most of the negative charge, could act as a base, and the result would be an enol.
Alternatively, the enolate carbon, which bears a degree of negative charge, could act as a base, which is simply the reverse of the initial deprotonation step that formed the enolate in the first place.
In both of these cases, the electron-poor species attacked by the enolate is an acidic proton. What if the electron-poor species - the electrophile - is not a proton but a carbonyl carbon? In other words, what if the enolate acts not as a base but rather as a nucleophile in a carbonyl addition (chapter 11) reaction? One possibility is that the enolate oxygen could be the nucleophile in a hemiketal-forming reaction:
Although you can find examples of this type of reaction in biochemistry, it is far more common for the enolate carbon to be the nucleophile, attacking an electrophilic carbonyl to form a new carbon-carbon bond.
Historically, the first examples of this mechanism type to be studied involved the non-enzymatic reaction of an aldehyde with itself (a so-called 'self-condensation' reaction, where 'condensation' means the formation of one larger molecule from two smaller ones).
Because the resulting product contained both an aldehyde and an alcohol functional group, the reaction was termed an 'aldol condensation', a name that has become standard for reactions of this type, whether or not an aldehyde is involved.
The enzymes that catalyze aldol reactions are called, not surprisingly, 'aldolases'.
13.3B: Typical aldolase reactions - three variations on a theme
The first step in an aldolase reaction is the deprotonation of an alpha-carbon to generate a nucleophilic carbanion. Nature has evolved several distinct strategies to stabilize the intermediate that results. Some aldolases use a metal ion to stabilize the negative charge on an enolate intermediate, while others catalyze reactions that proceed through neutral Schiff base or enol intermediates.
Let's examine first a reaction catalyzed by a so-called 'Class II' aldolase, in which a metal cation - generally Zn2+ - bound in the active site serves to stabilize the negative charge on an enolate intermediate. Fructose 1,6-bisphosphate aldolase is an enzyme that participates in both the glycolytic (sugar burning) and gluconeogenesis (sugar building) biochemical pathways. For now, we will concentrate on its role in the gluconeogenesis pathway, but we will see it again later in its glycolytic role. The reaction catalyzed by fructose 1,6-bisphosphate aldolase is a condensation between two 3-carbon sugars, glyceraldehyde-3-phosphate (GAP) and dihydroxyacetone phosphate (DHAP), forming a six-carbon product (which leads, after three more enzymatic steps, to glucose).
In the first step of the condensation, an alpha-carbon on DHAP is deprotonated, leading to an enolate intermediate. The strategy used to stabilize this key intermediate is to coordinate the negatively-charged enolate oxygen to an enzyme-bound zinc cation.
Next, the deprotonated a-carbon attacks the carbonyl carbon of GAP in a nucleophilic addition reaction, and protonation of the resulting alcohol leads directly to the fructose 1,6-bisphosphate product.
As with many other nucleophilic carbonyl addition reactions, a new stereocenter is created in this reaction, as a planar, achiral carbonyl group is converted to a tetrahedral, chiral alcohol. The enzyme-catalyzed reaction, not surprisingly, is completely stereospecific: the DHAP substrate is positioned in the active site so as to attack the re (front)face of the GAP carbonyl group, leading to the R configuration at the new stereocenter.
Interestingly, it appears that in bacteria, the fructose bisphosphate aldolase enzyme evolved separately from the corresponding enzyme in plants and animals. In plants and animals, the same aldol condensation reaction is carried out by a significantly different mechanism, in which the key intermediate is not a zinc-stabilized enolate but an enamine. The nucleophilic substrate (DHAP) is first linked to the enzyme through the formation of an imine (also known as a Schiff base, section 11.6) with a lysine residue in the active site.
The alpha-proton is then abstracted by an active site base to form an enamine.
In the next step, the alpha-carbon attacks the carbonyl carbon of GAP, and the new carbon-carbon bond is formed. In order to release the product from the enzyme active site and free the enzyme to catalyze another reaction, the imine is hydrolyzed back to a ketone group.
There are many more examples of 'Class I' aldolase reactions in which the key intermediate is a lysine-linked imine. Many bacteria are able to incorporate formaldehyde, a toxic compound, into carbohydrate metabolism by condensing it with ribulose monophosphate. The reaction proceeds through imine and enamine intermediates.
Template:ExampleStart
Exercise 13.5:
a) Propose a complete mechanism for the condensation reaction shown above.
b) Propose a complete mechanism for the conversion of hexulose-6-phosphate (formed from the condensation of ribulose-5-phosphate and formaldehyde) into fructose-6-phosphate.
Solution
Template:ExampleEnd
Along with aldehydes and ketones, esters and thioesters can act as the nucleophilic partner in aldol condensations. In the first step of the citric acid (Krebs) cycle, acetyl CoA condenses with oxaloacetate to form (S)-citryl CoA.
Notice that in this aldol reaction, the nucleophilic intermediate is stabilized by protonation, rather than by formation of an imine (as in the Class I aldolases) or by a metal ion (as in the Class II aldolases).
13.3C: Going backwards: the retroaldol reaction
Although aldol reactions play a very important role in the formation of new carbon-carbon bonds in metabolic pathways, it is important to emphasize that they are also highly reversible: in most cases, the energy level of starting compounds and products are very close. This means that, depending on metabolic conditions, aldolases can also catalyze retro-aldol reactions (the reverse of aldol condensations, in which carbon-carbon bonds are broken). Recall that fructose 1,6-bisphosphate aldolase (section 13.3B) is active in the direction of sugar breakdown (glycolysis) as well as sugar synthesis (gluconeogenesis). In the glycolytic direction, the enzyme catalyzes - either by zinc cation or by imine/enamine mechanisms, depending on the organism - the retro-aldol cleavage of fructose bisphosphate into DHAP and GAP.
The mechanism is the exact reverse of the condensation reaction. Shown below is the mechanism for a Zn2+ - dependent (Type II) retroaldol cleavage. Notice that in the retroaldol reaction, the enolate intermediate is the leaving group, rather than the nucleophile.
The key thing to keep in mind when looking for a possible retro-aldol mechanism is that, when the carbon-carbon bond breaks, the electrons must have some place to go, where they will be stabilized by resonance. Generally, this means that there must be a carbonyl or imine group on the next carbon. If there is no adjacent carbonyl or imine group, the carbon-carbon bond is not free to break.
Here are two more examples of retro-aldol reactions. Bacterial carbohydrate metabolism involves this reversible, class I retro-aldol cleavage: (Proc. Natl. Acad. Sci 2001, 98, 3679).
Template:ExampleStart
Exercise 13.6: Draw the structure of the enamine intermediate in the retroaldol reaction shown above.
Solution
Template:ExampleEnd
Another interesting example is the retro-aldol cleavage of indole-3-glycerol phosphate, a step in the biosynthesis of tryptophan.
Look carefully at this reaction - how is the leaving group stabilized? There is an imine group involved, but no participation by an enzymatic lysine. The imine is 'built into' the starting compound, available from the initial tautomerization of the cyclic enamine group in indole-3-glycerol phosphate.
Template:ExampleStart
Exercise 13.7: Draw the reverse (aldol condensation) direction of the reaction above.
Solution
Template:ExampleEnd
13.3D: Going both ways: transaldolase
An enzyme called transaldolase, which is part of the 'pentose phosphate pathway' of carbohydrate metabolism, catalyzes an interesting combination of class I aldol and retro-aldol reactions. The overall reaction, which can proceed in either direction depending on metabolic requirements, converts 3- and 7-carbon sugars into 6- and 4-carbon sugars. Essentially, a 3-carbon unit breaks off from a ketone sugar (ketose) and then is condensed directly with an aldehyde sugar (aldose).
Let's follow the progress of the reaction in the left-to-right direction as depicted above. Because this is a class I aldolase, the first step is the formation of an imine linkage between the ketone carbon of fructose-6-phosphate (F6P) and a lysine group from the enzyme. The enzyme-substrate adduct then undergoes a retro-aldol step to free glyceralde-3-phosphate (GAP), which leaves the active site.
The second substrate, erythrose 4-phosphate (E4P), enters the active site, and an aldol condensation occurs between E4P and the 3-carbon fragment remaining from the cleavage of fructose-6-phosphate.
The final step is hydrolysis of the imine and subsequent dissociation of sedoheptulose 7-phosphate from the active site.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.23____Reactions_at_the_-Carbon_in_Biological_Systems.txt |
Alkyl hydrogen atoms bonded to a carbon atom in a a (alpha) position relative to a carbonyl group display unusual acidity. While the pKa values for alkyl C-H bonds is typically on the order of 40-50, pKa values for these alpha hydrogens is more on the order of 19-20. This can most easily be explained by resonance stabilization of the product carbanion, as illustrated in the diagram below.
In the presence of a proton source, the product can either revert back into the starting ketone or aldehyde or can form a new product, the enol. The equilibrium reaction between the ketone or aldehyde and the enol form is commonly referred to as "keto-enol tautomerism". The ketone or aldehyde is generally strongly favored in this reaction.
Because carbonyl groups are sp2 hybridized the carbon and oxygen both have unhybridized p orbitals which can overlap to form the C=O $\pi$ bond.
The presence of these overlapping p orbitals gives $\alpha$ hydrogens (Hydrogens on carbons adjacent to carbonyls) special properties. In particular, $\alpha$ hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the $\pi$ orbitals of the carbonyl. The effect of the carbonyl is seen when comparing the pKa for the $\alpha$ hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25) to the pKa of an alkane (~50).
Of the two resonance structures of the enolate ion the one which places the negative charge on the oxygen is the most stable. This is because the negative change will be better stabilized by the greater electronegativity of the oxygen.
Keto-enol Tautomerism
Because of the acidity of α hydrogens carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom 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.
Contributors
Prof. Steven Farmer (Sonoma State University)
• Clarke Earley (Department of Chemistry, Kent State University Stark Campus)
19.3 Keto-Enol Interconversion
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and LiN[CH(CH3)2]2 (lithium diisopropylamide, LDA, pKa 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced.
Examples
If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide.
NaOCH2CH3 = Na+ -OCH2CH3 = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom 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.
However under acidic and basic conditions the equilibrium can be shifted to the right
Mechanism for Enol Formation
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.2____Keto-Enol_Tautomers.txt |
Enolates can act as a nucleophile in Sn2 type reactions. The alpha alkylation reaction involves an α hydrogen being replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as Sn2 reactions previously discussed. Good leaving groups like chloride, bromide, iodide, tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylation’s occurring.
Mechanism
Stpe 1: Enolate formation
Step 2: Sn2 attack
Example \(2\)
Please write the structure of the product for the following reactions.
19.5 Halogenation of the - Carbon and Aldehydes and Ketones
A carbonyl containing compound with $\alpha$ hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl2, Br2 or I2 can be used as the halogens.
General reaction
Acid Catalyzed Mechanism
Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen.
Step 1: Protonation of the carbonyl
Step 2: Enol formation
Step 3: SN2 attack
Step 4: Deprotonation
Base Catalyzed Mechanism
Under basic conditions the enolate forms and then reacts with the halogen. Note! This is base promoted and not base catalyzed because an entire equivalent of base is required.
Step 1: Enolate formation
Step 2: SN2 attack
Overreaction during base promoted α halogenation
The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used.
The Haloform Reaction
Methyl ketones typically undergo halogenation three times to give a trihalo ketone due to the increased reactivity of the halogenated product as discussed above. This trihalomethyl group is an effective leaving group due to the three electron withdrawing halogens and can be cleaved by a hydroxide anion to effect the haloform reaction. The product of this reaction is a carboxylate and a haloform molecule (CHCl3, CHBr3, CHI3). Overall the haloform reaction represents an effective method for the conversion of methyl ketones to carboxylic acids. Typically, this reaction is performed using iodine because the subsequent iodoform (CHI3) is a bright yellow precipitate which is easily filtered off.
General reaction
Mechanism
1) Formation of the trihalo species
2) Nulceophilic attack on the carbonyl carbon
3) Removal of the leaving group
4) Deprotonation
Exercise $1$
Please draw the products of the following reactions
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.4____How_Enolate_Ions_and_Enols_React.txt |
Carbon is one of the most common elements on earth, and greatly influences everyday life. Common molecules containing carbon include carbon dioxide (CO2) and methane (CH4). Many scientists in a variety of fields study of carbon: biologists investigating the origins of life; oceanographers measuring the acidification of the oceans; and engineers developing diamond film tools. This article details the periodic properties of the carbon family and briefly discusses of the individual properties of carbon, silicon, germanium, tin, lead, and flerovium.
• Group 14: General Chemistry
Covers the Group 4 (IUPAC: Group 14) chemistry (carbon, silicon, germanium, tin and lead) and specifically the trend from non-metal to metal as you go down the group, and the increasing tendency towards an oxidation state of +2. Also a certain amount of chemistry of the chlorides and oxides.
• Group 14: General Properties and Reactions
Carbon is one of the most common elements on earth, and greatly influences everyday life. This article details the periodic properties of the carbon family and briefly discusses of the individual properties of carbon, silicon, germanium, tin, lead, and flerovium.
• Chemistry of Carbon (Z=6)
Organic chemistry involves structures and reactions of mainly carbon and hydrogen. Inorganic chemistry deal with interactions of all other pure elements besides carbon, amongst geo/biochemistry. So where does inorganic chemistry of carbon fit in? The inorganic chemistry of carbon also known as inorganic carbon chemistry, is the chemistry of carbon that does not fall within the organic chemistry zone.
• Chemistry of Silicon (Z=14)
Silicon, the second most abundant element on earth, is an essential part of the mineral world. Its stable tetrahedral configuration makes it incredibly versatile and is used in various way in our every day lives. Found in everything from spaceships to synthetic body parts, silicon can be found all around us, and sometimes even in us.
• Chemistry of Germanium (Z=32)
Germanium, categorized as a metalloid in group 14, the Carbon family, has five naturally occurring isotopes. Germanium, abundant in the Earth's crust has been said to improve the immune system of cancer patients. It is also used in transistors, but its most important use is in fiber-optic systems and infrared optics.
• Chemistry of Tin (Z=50)
Mentioned in the Hebrew scriptures, tin is of ancient origins. Tin is an element in Group 14 (The carbon family) and has mainly metallic properties. Tin has atomic number 50 and an atomic mass of 118.710 atomic mass units. Tin, or Sn (from the Latin name Stannum) has been known since ancient times, although it could only be obtained by extraction from its ore. Tin shares chemical similarities with germanium and lead. Tin mining began in Australia in 1872 and today Tin is used extensively.
• Chemistry of Lead (Z=82)
Although lead is not very common in the earth's crust, what is there is readily available and easy to refine. Its chief use today is in lead-acid storage batteries such as those used in automobiles. In pure form it is too soft to be used for much else. Lead has a blue-white color when first cut but quickly dulls on exposure to air, forming Pb2O, one of the few lead(I) compounds. Most stable lead compounds contain lead in oxidation states of +2 or +4.
• Chemistry of Flerovium (Z=114)
The synthesis of element 114 was reported in January of 1999 by scientists from the Joint Institute for Nuclear Research in Dubna (near Moscow) and Lawrence Livermore National Laboratory (in California). In an experiment lasting more than 40 days Russian scientists bombarded a film of Pu-244 supplied by Livermore scientists with a beam of Ca-48. One atom of element 114 was detected with a half-life of more than 30 seconds. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.6____Halogenation_of_the_-_Carbon_of_Carboxylic_Acids%3A_The_Hell-Volhard-Zelinski_Reaction.txt |
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and LiN[CH(CH3)2]2 (lithium diisopropylamide, LDA, pKa 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced.
Examples
If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide.
NaOCH2CH3 = Na+ -OCH2CH3 = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom 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.
However under acidic and basic conditions the equilibrium can be shifted to the right
Mechanism for Enol Formation
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation
19.9 Alkylating the -Carbon of Carbonyl Compounds
Enolates can act as a nucleophile in Sn2 type reactions. The alpha alkylation reaction involves an α hydrogen being replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as Sn2 reactions previously discussed. Good leaving groups like chloride, bromide, iodide, tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylation’s occurring.
Mechanism
Stpe 1: Enolate formation
Step 2: Sn2 attack
Example \(2\)
Please write the structure of the product for the following reactions.
14: Determing the Structure of Organic Compounds
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/13%3A_Carbonyl_Compounds_III%3A_Reactions_at_the_-_Carbon/19.8____Using_LDA_to_Form_an_Enolate_Ion.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
15: The Organic Chemistry of Carbohydrates
If a monosaccharide has a carbonyl function on one of the inner atoms of the carbon chain it is classified as a ketose. Dihydroxyacetone may not be a sugar, but it is included as the ketose analog of glyceraldehyde. The carbonyl group is commonly found at C-2, as illustrated by the following examples (chiral centers are colored red). As expected, the carbonyl function of a ketose may be reduced by sodium borohydride, usually to a mixture of epimeric products. D-Fructose, the sweetest of the common natural sugars, is for example reduced to a mixture of D-glucitol (sorbitol) and D-mannitol, named after the aldohexoses from which they may also be obtained by analogous reduction. Mannitol is itself a common natural carbohydrate.
Although the ketoses are distinct isomers of the aldose monosaccharides, the chemistry of both classes is linked due to their facile interconversion in the presence of acid or base catalysts. This interconversion, and the corresponding epimerization at sites alpha to the carbonyl functions, occurs by way of an enediol tautomeric intermediate.
Because of base-catalyzed isomerizations of this kind, the Tollens' reagent is not useful for distinguishing aldoses from ketoses or for specific oxidation of aldoses to the corresponding aldonic acids. Oxidation by HOBr is preferred for the latter conversion.
22.05: The Reactions of Monosaccharides in Basic Solutions
Enols, enolates, and enamines are reactive intermediates in many isomerization reactions. Isomerizations can involve either the interconversion of constitutional isomers, in which bond connectivity is altered, or of stereoisomers, where the stereochemical configuration is changed. Enzymes which interconvert constitutional isomers are called isomerases, while those which catalyze the interconversion of enantiomers and epimers are called racemases and epimerases, respectively.
13.2A: Carbonyl isomerization
Two important chemical steps in the glycolytic pathway, catalyzed by the enzymes phosphoglucose isomerase and triose phosphate isomerase, involve successive keto-enol tautomerization steps. In both reactions, the location of a carbonyl group on a sugar molecule is shifted back and forth by a single carbon, as ketones are converted to aldehydes and back again - this is a conversion between two constitutional isomers.
Let's look first at the triosephosphate isomerase reaction, in the ketone to aldehyde direction. The ketone species, dihydroxyacetone phosphate (DHAP) is first converted to its enol tautomer with the assistance of an enzymatic acid/base pair (actually, this particular intermediate is known as an 'ene-diol' rather than an enol, because there are hydroxyl groups on both sides of the carbon-carbon double bond). The initial proton donor is positioned in the active site near the carbonyl carbon, and significantly lowers the pKa of the alpha-proton.
The second step, leading to glyceraldehyde phosphate (GAP), is simply another tautomerization, this time in the reverse direction. However, because there happens to be a hydroxyl group on C1, the carbonyl can form here as well as at C2. Notice that DHAP is achiral while GAP is chiral, and that a new chiral center is introduced at C2. The catalytic base abstracts the pro-R proton from behind the plane of the page, then gives the same proton back to C2, again from behind the plane of the page.
In the phosphoglucose isomerase reaction, glucose-6-phosphate (an aldehyde sugar) and fructose-6-phosphate (a ketone sugar) are interconverted in a very similar fashion.
The enzyme ribose-5-phosphate isomerase (EC 5.3.1.6), which is active in both the Calvin cycle and the pentose phosphate pathway, catalyzes an analogous aldehyde-to-ketone isomerization between two five-carbon sugars.
Template:exampleStart
Exercise 13.3: draw the ene-diol intermediate in the phosphoglucose isomerase reaction above.
Solution
Template:exampleEnd
13.2B: Stereoisomerization at chiral carbons
The enolate form of a carbonyl is a common intermediate in a type of stereoisomerization reaction known as an epimerization. Recall from section 3.7A that the term 'epimer' refers to a pair of diastereomers that differ at a single stereocenter. The five-carbon sugar phosphates ribulose-5-phosphate and xylulose-5-phosphate, for example, are epimers: they are identical except for their stereochemistry at C3. A reaction in the Calvin cycle, which plants use to incorporate, or 'fix', the carbon atom from carbon dioxide into sugars, involves the interconversion of these two sugar phosphate epimers - in other words, inversion of configuration at C3.
The reaction is simply a deprotonation at C3 to form an enolate, followed by reprotonation to return to the ketone.
The key to epimerization is that deprotonation and reprotonation occur at opposite sides of the planar, sp2-hybridized intermediate. In the epimerase reaction shown here, a coordinated pair of aspartate residues is located at opposite sides of the enzyme's active site. Asp1, which is ionized at the start of the catalytic cycle, abstracts the C3 proton from the front side of the plane of the page, converting C3 from a chiral tetrahedral center to an achiral, planar group. When reprotonation occurs, it is Asp2, on the back side of the plane of the page, which donates the proton, resulting in inversion of stereochemistry at C3 (J. Molec. Biol. 2003, 326, 127-135).
Notice another thing about this epimerization reaction: the key intermediate is an enolate anion, stabilized by coordination to a zinc cation (a Lewis acid). Contrast this with the phosphoglucase isomerase reaction (section 13.2A), where stabilization of the key intermediate was achieved by proton donation to form a short-lived enol species.
There are many more examples of reactions in which the stereochemistry of an a-carbon in a carbonyl compound is inverted. The short polypeptides that form part of the peptidoglycan cell wall structure in bacteria contain some D-amino acids, which have the opposite stereochemistry at the a-carbon compared to the L-amino acids that are found almost exclusively in proteins. Enzymes called amino acid racemases catalyze the conversion between the L and D forms of these amino acids. While many amino acid racemases depend on the coenzyme pyridoxal phosphate to stabilize the key carbanion intermediate (we'll study these reactions in section 14.4B), others proceed through an enolate intermediate. Glutamate racemase, which converts L-glutamate to D-glutamate, is an example (Biochemistry 2001, 40, 6199). Although many mechanistic details are as yet unknown, it has been demonstrated that, at least for some of the cofactor-independent racemases, a pair of cysteine residues serves as the catalytic acid/base pair.
Template:exampleStart
Exercise 13.4: Propose a likely mechanism for glutamate racemase, showing stereochemistry throughout.
Solution
Template:exampleEnd
13.2C: Alkene isomerization in the degradation of unsaturated fatty acids
The oxidation pathway for unsaturated fatty acids (in other words, fatty acids whose hydrocarbon chains contain one or more double bonds) involves the 'shuffling' of the position of carbon-carbon double bonds in the chain. This is accomplished by enoyl CoA isomerase, through an enolate intermediate. Here, the cis double bond between C3 and C4 is isomerized to a trans double bond between C2 and C3.
The catalytic acid and base groups are glutamate and/or aspartate residues (J. Biol Chem 2001, 276, 13622).
In section 15.3A, we will see a different example of an alkene isomerization reaction mechanism in which the key intermediate is a carbocation, rather than an enolate.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.04%3A_The_Configurations_of_Ketoses.txt |
Overview
Emil Fischer made use of several key reactions in the course of his carbohydrate studies. These are described here, together with the information that each delivers.
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Derivatives of HOCH2(CHOH)nCHO
HOBr Oxidation
——>
HOCH2(CHOH)nCO2H
an Aldonic Acid
HNO3 Oxidation
——>
H2OC(CHOH)nCO2H
an Aldaric Acid
NaBH4 Reduction
——>
HOCH2(CHOH)nCH2OH
an Alditol
Osazone Formation
1.
2.
The osazone reaction was developed and used by Emil Fischer to identify aldose sugars differing in configuration only at the alpha-carbon. The upper equation shows the general form of the osazone reaction, which effects an alpha-carbon oxidation with formation of a bis-phenylhydrazone, known as an osazone. Application of the osazone reaction to D-glucose and D-mannose demonstrates that these compounds differ in configuration only at C-2.
Chain Shortening and Lengthening
1.
2.
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Fischer's train of logic in assigning the configuration of D-glucose
1. Ribose and arabinose (two well known pentoses) both gave erythrose on Ruff degradation. As expected, Kiliani-Fischer synthesis applied to erythrose gave a mixture of ribose and arabinose.
2. Oxidation of erythrose gave an achiral (optically inactive) aldaric acid. This defines the configuration of erythrose.
3. Oxidation of ribose gave an achiral (optically inactive) aldaric acid. This defines the configuration of both ribose and arabinose.
4. Ruff shortening of glucose gave arabinose, and Kiliani-Fischer synthesis applied to arabinose gave a mixture of glucose and mannose.
5. Glucose and mannose are therefore epimers at C-2, a fact confirmed by the common product from their osazone reactions.
6. A pair of structures for these epimers can be written, but which is glucose and which is mannose?
To determine which of these epimers was glucose, Fischer made use of the inherent C2 symmetry in the four-carbon dissymmetric core of one epimer (B). This is shown in the following diagram by a red dot where the symmetry axis passes through the projection formula. Because of this symmetry, if the aldehyde and 1º-alcohol functions at the ends of the chain are exchanged, epimer B would be unchanged; whereas A would be converted to a different compound.
Fischer looked for and discovered a second aldohexose that represented the end group exchange for the epimer lacking the latent C2 symmetry (A). This compound was L-(+)-gulose, and its exchange relationship to D-(+)-glucose was demonstrated by oxidation to a common aldaric acid product. The remaining epimer is therefore mannose.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.06%3A_The_Oxidation-Reduction_Reactions_of_Monosaccharides.txt |
Overview
Emil Fischer made use of several key reactions in the course of his carbohydrate studies. These are described here, together with the information that each delivers.
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Derivatives of HOCH2(CHOH)nCHO
HOBr Oxidation
——>
HOCH2(CHOH)nCO2H
an Aldonic Acid
HNO3 Oxidation
——>
H2OC(CHOH)nCO2H
an Aldaric Acid
NaBH4 Reduction
——>
HOCH2(CHOH)nCH2OH
an Alditol
Osazone Formation
1.
2.
The osazone reaction was developed and used by Emil Fischer to identify aldose sugars differing in configuration only at the alpha-carbon. The upper equation shows the general form of the osazone reaction, which effects an alpha-carbon oxidation with formation of a bis-phenylhydrazone, known as an osazone. Application of the osazone reaction to D-glucose and D-mannose demonstrates that these compounds differ in configuration only at C-2.
Chain Shortening and Lengthening
1.
2.
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Fischer's train of logic in assigning the configuration of D-glucose
1. Ribose and arabinose (two well known pentoses) both gave erythrose on Ruff degradation. As expected, Kiliani-Fischer synthesis applied to erythrose gave a mixture of ribose and arabinose.
2. Oxidation of erythrose gave an achiral (optically inactive) aldaric acid. This defines the configuration of erythrose.
3. Oxidation of ribose gave an achiral (optically inactive) aldaric acid. This defines the configuration of both ribose and arabinose.
4. Ruff shortening of glucose gave arabinose, and Kiliani-Fischer synthesis applied to arabinose gave a mixture of glucose and mannose.
5. Glucose and mannose are therefore epimers at C-2, a fact confirmed by the common product from their osazone reactions.
6. A pair of structures for these epimers can be written, but which is glucose and which is mannose?
To determine which of these epimers was glucose, Fischer made use of the inherent C2 symmetry in the four-carbon dissymmetric core of one epimer (B). This is shown in the following diagram by a red dot where the symmetry axis passes through the projection formula. Because of this symmetry, if the aldehyde and 1º-alcohol functions at the ends of the chain are exchanged, epimer B would be unchanged; whereas A would be converted to a different compound.
Fischer looked for and discovered a second aldohexose that represented the end group exchange for the epimer lacking the latent C2 symmetry (A). This compound was L-(+)-gulose, and its exchange relationship to D-(+)-glucose was demonstrated by oxidation to a common aldaric acid product. The remaining epimer is therefore mannose.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.07%3A_Monosaccharides_Form_Crystalline_Osazones.txt |
Overview
Emil Fischer made use of several key reactions in the course of his carbohydrate studies. These are described here, together with the information that each delivers.
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Derivatives of HOCH2(CHOH)nCHO
HOBr Oxidation
——>
HOCH2(CHOH)nCO2H
an Aldonic Acid
HNO3 Oxidation
——>
H2OC(CHOH)nCO2H
an Aldaric Acid
NaBH4 Reduction
——>
HOCH2(CHOH)nCH2OH
an Alditol
Osazone Formation
1.
2.
The osazone reaction was developed and used by Emil Fischer to identify aldose sugars differing in configuration only at the alpha-carbon. The upper equation shows the general form of the osazone reaction, which effects an alpha-carbon oxidation with formation of a bis-phenylhydrazone, known as an osazone. Application of the osazone reaction to D-glucose and D-mannose demonstrates that these compounds differ in configuration only at C-2.
Chain Shortening and Lengthening
1.
2.
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Fischer's train of logic in assigning the configuration of D-glucose
1. Ribose and arabinose (two well known pentoses) both gave erythrose on Ruff degradation. As expected, Kiliani-Fischer synthesis applied to erythrose gave a mixture of ribose and arabinose.
2. Oxidation of erythrose gave an achiral (optically inactive) aldaric acid. This defines the configuration of erythrose.
3. Oxidation of ribose gave an achiral (optically inactive) aldaric acid. This defines the configuration of both ribose and arabinose.
4. Ruff shortening of glucose gave arabinose, and Kiliani-Fischer synthesis applied to arabinose gave a mixture of glucose and mannose.
5. Glucose and mannose are therefore epimers at C-2, a fact confirmed by the common product from their osazone reactions.
6. A pair of structures for these epimers can be written, but which is glucose and which is mannose?
To determine which of these epimers was glucose, Fischer made use of the inherent C2 symmetry in the four-carbon dissymmetric core of one epimer (B). This is shown in the following diagram by a red dot where the symmetry axis passes through the projection formula. Because of this symmetry, if the aldehyde and 1º-alcohol functions at the ends of the chain are exchanged, epimer B would be unchanged; whereas A would be converted to a different compound.
Fischer looked for and discovered a second aldohexose that represented the end group exchange for the epimer lacking the latent C2 symmetry (A). This compound was L-(+)-gulose, and its exchange relationship to D-(+)-glucose was demonstrated by oxidation to a common aldaric acid product. The remaining epimer is therefore mannose.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.08%3A_Lengthening_the_Chain%3A_The_Kiliani-Fischer_Synthesis.txt |
The ability to shorten (degrade) an aldose chain by one carbon was an importanat tool in the structure elucidation of carbohydrates. This was commonly accomplished by the Ruff procedure. An interesting alternative technique, known as the Wohl degradation has also been used. The following equation illustrates the application of this procedure to the aldopentose, arabinose. Based on your knowledge of carbonyl chemistry, and considering that the Wohl degradation is in essence the reverse of the Kiliani-Fischer synthesis, you should be able to draw structural formulas for intermediates A and B. Using the zig-zag chain tool, draw suitable formulas in the drawing window on the left without specifying substituent configurations. When you are finished, check your answer by pressing the appropriate Check Intermediatebutton. A help screen is available for the drawing window.
22.10 The Stereochemistry of Glucose: The Fischer Proof
The four chiral centers in glucose indicate there may be as many as sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. One of the first tasks faced by Fischer was to devise a method of representing the configuration of each chiral center in an unambiguous manner. To this end, he invented a simple technique for drawing chains of chiral centers, that we now call the Fischer projection formula.
At the time Fischer undertook the glucose project it was not possible to establish the absolute configuration of an enantiomer. Consequently, Fischer made an arbitrary choice for (+)-glucose and established a network of related aldose configurations that he called the D-family. The mirror images of these configurations were then designated the L-family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red.
The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Fischer as the D / L designator site. If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family. A left directed hydroxyl group (the mirror image) then represented the L-family. Fischer's initial assignment of the D-configuration had a 50:50 chance of being right, but all his subsequent conclusions concerning the relative configurations of various aldoses were soundly based. In 1951 x-ray fluorescence studies of (+)-tartaric acid, carried out in the Netherlands by Johannes Martin Bijvoet (pronounced "buy foot"), proved that Fischer's choice was correct.
It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L). It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.09%3A_Shortening_the_Chain%3A_The_Wohl_Degradation.txt |
As noted above, the preferred structural form of many monosaccharides may be that of a cyclic hemiacetal. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed furanose (five-membered) or pyranose (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown on the right.
Ribose, an important aldopentose, commonly adopts a furanose structure, as shown in the following illustration. By convention for the D-family, the five-membered furanose ring is drawn in an edgewise projection with the ring oxygen positioned away from the viewer. The anomeric carbon atom (colored red here) is placed on the right. The upper bond to this carbon is defined as beta, the lower bond then is alpha.
The cyclic pyranose forms of various monosaccharides are often drawn in a flat projection known as a Haworth formula, after the British chemist, Norman Haworth. As with the furanose ring, the anomeric carbon is placed on the right with the ring oxygen to the back of the edgewise view. In the D-family, the alpha and beta bonds have the same orientation defined for the furanose ring (beta is up & alpha is down). These Haworth formulas are convenient for displaying stereochemical relationships, but do not represent the true shape of the molecules. We know that these molecules are actually puckered in a fashion we call a chair conformation. Examples of four typical pyranose structures are shown below, both as Section 3.8: Fischer and Haworth projections and as the more representative chair conformers. The anomeric carbons are colored red.
The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features. Aldolhexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. The formation of acetal derivatives illustrates how subtle changes may alter this selectivity. A pyranose structure for D-glucose is drawn in the rose-shaded box on the left. Acetal derivatives have been prepared by acid-catalyzed reactions with benzaldehyde and acetone. As a rule, benzaldehyde forms six-membered cyclic acetals, whereas acetone prefers to form five-membered acetals. The top equation shows the formation and some reactions of the 4,6-O-benzylidene acetal, a commonly employed protective group. A methyl glycoside derivative of this compound (see below) leaves the C-2 and C-3 hydroxyl groups exposed to reactions such as the periodic acid cleavage, shown as the last step. The formation of an isopropylidene acetal at C-1 and C-2, center structure, leaves the C-3 hydroxyl as the only unprotected function. Selective oxidation to a ketone is then possible. Finally, direct di-O-isopropylidene derivatization of glucose by reaction with excess acetone results in a change to a furanose structure in which the C-3 hydroxyl is again unprotected. However, the same reaction with D-galactose, shown in the blue-shaded box, produces a pyranose product in which the C-6 hydroxyl is unprotected. Both derivatives do not react with Tollens' reagent. This difference in behavior is attributed to the cis-orientation of the C-3 and C-4 hydroxyl groups in galactose, which permits formation of a less strained five-membered cyclic acetal, compared with the trans-C-3 and C-4 hydroxyl groups in glucose. Derivatizations of this kind permit selective reactions to be conducted at different locations in these highly functionalized molecules.
22.13 Formation of Glycosides
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.11____Monosaccharides_Form_Cyclic_Hemiacetals.txt |
When the alcohol component of a glycoside is provided by a hydroxyl function on another monosaccharide, the compound is called a disaccharide. Four examples of disaccharides composed of two glucose units are shown in the following diagram. The individual glucopyranose rings are labeled A and B, and the glycoside bonding is circled in light blue. Notice that the glycoside bond may be alpha, as in maltose and trehalose, or beta as in cellobiose and gentiobiose. Acid-catalyzed hydrolysis of these disaccharides yields glucose as the only product. Enzyme-catalyzed hydrolysis is selective for a specific glycoside bond, so an alpha-glycosidase cleaves maltose and trehalose to glucose, but does not cleave cellobiose or gentiobiose. A beta-glycosidase has the opposite activity.
In order to draw a representative structure for cellobiose, one of the glucopyranose rings must be rotated by 180º, but this feature is often omitted in favor of retaining the usual perspective for the individual rings. The bonding between the glucopyranose rings in cellobiose and maltose is from the anomeric carbon in ring A to the C-4 hydroxyl group on ring B. This leaves the anomeric carbon in ring B free, so cellobiose and maltose both may assume alpha and beta anomers at that site (the beta form is shown in the diagram). Gentiobiose has a beta-glycoside link, originating at C-1 in ring A and terminating at C-6 in ring B. Its alpha-anomer is drawn in the diagram. Because cellobiose, maltose and gentiobiose are hemiacetals they are all reducing sugars (oxidized by Tollen's reagent). Trehalose, a disaccharide found in certain mushrooms, is a bis-acetal, and is therefore a non-reducing sugar. A systematic nomenclature for disaccharides exists, but as the following examples illustrate, these are often lengthy.
• Cellobiose : 4-O-β-D-Glucopyranosyl-D-glucose (the beta-anomer is drawn)
• Maltose : 4-O-α-D-Glucopyranosyl-D-glucose (the beta-anomer is drawn)
• Gentiobiose : 6-O-β-D-Glucopyranosyl-D-glucose (the alpha-anomer is drawn)
• Trehalose : α-D-Glucopyranosyl-α-D-glucopyranoside
Although all the disaccharides shown here are made up of two glucopyranose rings, their properties differ in interesting ways. Maltose, sometimes called malt sugar, comes from the hydrolysis of starch. It is about one third as sweet as cane sugar (sucrose), is easily digested by humans, and is fermented by yeast. Cellobiose is obtained by the hydrolysis of cellulose. It has virtually no taste, is indigestible by humans, and is not fermented by yeast. Some bacteria have beta-glucosidase enzymes that hydrolyze the glycosidic bonds in cellobiose and cellulose. The presence of such bacteria in the digestive tracts of cows and termites permits these animals to use cellulose as a food. Finally, it may be noted that trehalose has a distinctly sweet taste, but gentiobiose is bitter.
Disaccharides made up of other sugars are known, but glucose is often one of the components. Two important examples of such mixed disaccharides are displayed above. Lactose, also known as milk sugar, is a galactose-glucose compound joined as a beta-glycoside. It is a reducing sugar because of the hemiacetal function remaining in the glucose moiety. Many adults, particularly those from regions where milk is not a dietary staple, have a metabolic intolerance for lactose. Infants have a digestive enzyme which cleaves the beta-glycoside bond in lactose, but production of this enzyme stops with weaning. Cheese is less subject to the lactose intolerance problem, since most of the lactose is removed with the whey. Sucrose, or cane sugar, is our most commonly used sweetening agent. It is a non-reducing disaccharide composed of glucose and fructose joined at the anomeric carbon of each by glycoside bonds (one alpha and one beta). In the formula shown here the fructose ring has been rotated 180º from its conventional perspective.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.16____Disaccharides.txt |
As the name implies, polysaccharides are large high-molecular weight molecules constructed by joining monosaccharide units together by glycosidic bonds. They are sometimes called glycans. The most important compounds in this class, cellulose, starch and glycogen are all polymers of glucose. This is easily demonstrated by acid-catalyzed hydrolysis to the monosaccharide. Since partial hydrolysis of cellulose gives varying amounts of cellobiose, we conclude the glucose units in this macromolecule are joined by beta-glycoside bonds between C-1 and C-4 sites of adjacent sugars. Partial hydrolysis of starch and glycogen produces the disaccharide maltose together with low molecular weight dextrans, polysaccharides in which glucose molecules are joined by alpha-glycoside links between C-1 and C-6, as well as the alpha C-1 to C-4 links found in maltose. Polysaccharides built from other monosaccharides (e.g. mannose, galactose, xylose and arabinose) are also known, but will not be discussed here.
Over half of the total organic carbon in the earth's biosphere is in cellulose. Cotton fibers are essentially pure cellulose, and the wood of bushes and trees is about 50% cellulose. As a polymer of glucose, cellulose has the formula (C6H10O5)n where n ranges from 500 to 5,000, depending on the source of the polymer. The glucose units in cellulose are linked in a linear fashion, as shown in the drawing below. The beta-glycoside bonds permit these chains to stretch out, and this conformation is stabilized by intramolecular hydrogen bonds. A parallel orientation of adjacent chains is also favored by intermolecular hydrogen bonds. Although an individual hydrogen bond is relatively weak, many such bonds acting together can impart great stability to certain conformations of large molecules. Most animals cannot digest cellulose as a food, and in the diets of humans this part of our vegetable intake functions as roughage and is eliminated largely unchanged. Some animals (the cow and termites, for example) harbor intestinal microorganisms that breakdown cellulose into monosaccharide nutrients by the use of beta-glycosidase enzymes.
Cellulose is commonly accompanied by a lower molecular weight, branched, amorphous polymer called hemicellulose. In contrast to cellulose, hemicellulose is structurally weak and is easily hydrolyzed by dilute acid or base. Also, many enzymes catalyze its hydrolysis. Hemicelluloses are composed of many D-pentose sugars, with xylose being the major component. Mannose and mannuronic acid are often present, as well as galactose and galacturonic acid.
Starch is a polymer of glucose, found in roots, rhizomes, seeds, stems, tubers and corms of plants, as microscopic granules having characteristic shapes and sizes. Most animals, including humans, depend on these plant starches for nourishment. The structure of starch is more complex than that of cellulose. The intact granules are insoluble in cold water, but grinding or swelling them in warm water causes them to burst.
The released starch consists of two fractions. About 20% is a water soluble material called amylose. Molecules of amylose are linear chains of several thousand glucose units joined by alpha C-1 to C-4 glycoside bonds. Amylose solutions are actually dispersions of hydrated helical micelles. The majority of the starch is a much higher molecular weight substance, consisting of nearly a million glucose units, and called amylopectin. Molecules of amylopectin are branched networks built from C-1 to C-4 and C-1 to C-6 glycoside links, and are essentially water insoluble. Representative structural formulas for amylose and amylopectin are shown above. The branching in this diagram is exaggerated, since on average, branches only occur every twenty five glucose units.
Hydrolysis of starch, usually by enzymatic reactions, produces a syrupy liquid consisting largely of glucose. When cornstarch is the feedstock, this product is known as corn syrup. It is widely used to soften texture, add volume, prohibit crystallization and enhance the flavor of foods. Glycogen is the glucose storage polymer used by animals. It has a structure similar to amylopectin, but is even more highly branched (about every tenth glucose unit). The degree of branching in these polysaccharides may be measured by enzymatic or chemical analysis.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.17____Polysaccharides.txt |
Cotton, probably the most useful natural fiber, is nearly pure cellulose. The manufacture of textiles from cotton involves physical manipulation of the raw material by carding, combing and spinning selected fibers. For fabrics the best cotton has long fibers, and short fibers or cotton dust are removed. Crude cellulose is also available from wood pulp by dissolving the lignan matrix surrounding it. These less desirable cellulose sources are widely used for making paper.
In order to expand the ways in which cellulose can be put to practical use, chemists have devised techniques for preparing solutions of cellulose derivatives that can be spun into fibers, spread into a film or cast in various solid forms. A key factor in these transformations are the three free hydroxyl groups on each glucose unit in the cellulose chain, --[C6H7O(OH)3]n--. Esterification of these functions leads to polymeric products having very different properties compared with cellulose itself.
• Cellulose Nitrate, first prepared over 150 years ago by treating cellulose with nitric acid, is the earliest synthetic polymer to see general use. The fully nitrated compound, --[C6H7O(ONO2)3]n--, called guncotton, is explosively flammable and is a component of smokeless powder. Partially nitrated cellulose is called pyroxylin. Pyroxylin is soluble in ether and at one time was used for photographic film and lacquers. The high flammability of pyroxylin caused many tragic cinema fires during its period of use. Furthermore, slow hydrolysis of pyroxylin yields nitric acid, a process that contributes to the deterioration of early motion picture films in storage.
• Cellulose Acetate, --[C6H7O(OAc)3]n--, is less flammable than pyroxylin, and has replaced it in most applications. It is prepared by reaction of cellulose with acetic anhydride and an acid catalyst. The properties of the product vary with the degree of acetylation. Some chain shortening occurs unavoidably in the preparations. An acetone solution of cellulose acetate may be forced through a spinneret to generate filaments, called acetate rayon, that can be woven into fabrics.
• Viscose Rayon, is prepared by formation of an alkali soluble xanthate derivative that can be spun into a fiber that reforms the cellulose polymer by acid quenching. The following general equation illustrates these transformations. The product fiber is called viscose rayon.
Many complex polysaccharides are found in nature. The galactomannans, consisting of a mannose backbone with galactose side groups, are an interesting and useful example. A (1-4)-linked beta-D-mannose chain is adorned with 1-6-linked alpha-D-galactose units, as shown in the diagram below. The ratio of galactose to mannose usually ranges from 1:2 to 1:4. An important source of this substance is the guar bean, grown principally in northwestern India, and Pakistan. Altough guar protein is not of nutritional value to humans (guar means 'cow food' in Hindi), the bean is important as a source of guar gum, a galactomannan which forms a gel in water. The food industry uses this material as a stabilizer in ice cream, cream cheese and salad dressings.
Recently, guar gum has becomes an essential ingredient for mining oil and natural gas in a process called hydraulic fracturing. The demand for guar has increased to such a degree that the poor farmers in northwestern India have had their lives transformed by unexpected wealth.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/15%3A_The_Organic_Chemistry_of_Carbohydrates/22.18____Some_Naturally_Occurring_Products_Derived_from_Carbohydrates.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
16: The Organic Chemistry of Amino Acids Peptides and Proteins
Intermolecular forces are the attractive or repulsive forces between molecules. They are separated into two groups; short range and long range forces. Short range forces happen when the centers of the molecules are separated by three angstroms (10-8 cm) or less. Short range forces tend to be repulsive, where the long range forces that act outside the three angstroms range are attractive. Long range forces are also known as Van der Waals forces. They are responsible for surface tension, friction, viscosity and differences between actual behavior of gases and that predicted by the ideal gas law. Intermolecular forces are responsible for most properties of all the phases. The viscosity, diffusion, and surface tension are examples of physical properties of liquids that depend on intermolecular forces. Vapor pressure, critical point, and boiling point are examples of properties of gases. Melting and sublimation are examples of properties of solids that depend on intermolecular forces.
• Hydrogen Bonding
A hydrogen bond is a special type of dipole-dipole attraction which occurs when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. These bonds are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds.
• Hydrophobic Interactions
Hydrophobic interactions describe the relations between water and hydrophobes (low water-soluble molecules). Hydrophobes are nonpolar molecules and usually have a long chain of carbons that do not interact with water molecules. The mixing of fat and water is a good example of this particular interaction. The common misconception is that water and fat doesn’t mix because the Van der Waals forces that are acting upon both water and fat molecules are too weak.
• Multipole Expansion
A multipole expansion is a series expansion of the effect produced by a given system in terms of an expansion parameter which becomes small as the distance away from the system increases. Therefore, the leading one or terms in a multipole expansion are generally the strongest. The first-order behavior of the system at large distances can therefore be obtained from the first terms of this series, which is generally much easier to compute than the general solution.
• Overview of Intermolecular Forces
Intermolecular forces are forces between molecules. Depending on its strength, intermolecular forces cause the forming of three physical states: solid, liquid and gas. The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces.
• Specific Interactions
Intermolecular forces are forces of attraction or repulsion which act between neighboring particles (atoms, molecules or ions). They are weak compared to the intramolecular forces, which keep a molecule together (e.g., covalent and ionic bonding).
• Van der Waals Forces
Van der Waals forces' is a general term used to define the attraction of intermolecular forces between molecules. There are two kinds of Van der Waals forces: weak London Dispersion Forces and stronger dipole-dipole forces. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.01%3A_Classification_and_Nomenclature_of_Amino_Acids.txt |
Proteins, from the Greek proteios, meaning first, are a class of organic compounds which are present in and vital to every living cell. In the form of skin, hair, callus, cartilage, muscles, tendons and ligaments, proteins hold together, protect, and provide structure to the body of a multi-celled organism. In the form of enzymes, hormones, antibodies, and globulins, they catalyze, regulate, and protect the body chemistry. In the form of hemoglobin, myoglobin and various lipoproteins, they effect the transport of oxygen and other substances within an organism.
Contributors and Attributions
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
16.03: The Acid-Base Properties of Amino Acids
Proteins, from the Greek proteios, meaning first, are a class of organic compounds which are present in and vital to every living cell. In the form of skin, hair, callus, cartilage, muscles, tendons and ligaments, proteins hold together, protect, and provide structure to the body of a multi-celled organism. In the form of enzymes, hormones, antibodies, and globulins, they catalyze, regulate, and protect the body chemistry. In the form of hemoglobin, myoglobin and various lipoproteins, they effect the transport of oxygen and other substances within an organism.
Contributors and Attributions
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
16.04: The Isoelectric Point
Proteins, from the Greek proteios, meaning first, are a class of organic compounds which are present in and vital to every living cell. In the form of skin, hair, callus, cartilage, muscles, tendons and ligaments, proteins hold together, protect, and provide structure to the body of a multi-celled organism. In the form of enzymes, hormones, antibodies, and globulins, they catalyze, regulate, and protect the body chemistry. In the form of hemoglobin, myoglobin and various lipoproteins, they effect the transport of oxygen and other substances within an organism.
Contributors and Attributions
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
16.05: Separating Amino Acids
Proteins, from the Greek proteios, meaning first, are a class of organic compounds which are present in and vital to every living cell. In the form of skin, hair, callus, cartilage, muscles, tendons and ligaments, proteins hold together, protect, and provide structure to the body of a multi-celled organism. In the form of enzymes, hormones, antibodies, and globulins, they catalyze, regulate, and protect the body chemistry. In the form of hemoglobin, myoglobin and various lipoproteins, they effect the transport of oxygen and other substances within an organism.
Contributors and Attributions
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
16.06: Peptide Bond and Disulfide Bonds
Proteins, from the Greek proteios, meaning first, are a class of organic compounds which are present in and vital to every living cell. In the form of skin, hair, callus, cartilage, muscles, tendons and ligaments, proteins hold together, protect, and provide structure to the body of a multi-celled organism. In the form of enzymes, hormones, antibodies, and globulins, they catalyze, regulate, and protect the body chemistry. In the form of hemoglobin, myoglobin and various lipoproteins, they effect the transport of oxygen and other substances within an organism.
Contributors and Attributions
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
16.07: The Strategy of Peptide Bond Synthesis: N-Protection and C-Activation
Proteins, from the Greek proteios, meaning first, are a class of organic compounds which are present in and vital to every living cell. In the form of skin, hair, callus, cartilage, muscles, tendons and ligaments, proteins hold together, protect, and provide structure to the body of a multi-celled organism. In the form of enzymes, hormones, antibodies, and globulins, they catalyze, regulate, and protect the body chemistry. In the form of hemoglobin, myoglobin and various lipoproteins, they effect the transport of oxygen and other substances within an organism.
Contributors and Attributions
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
16.08: An Introduction to Protein Structure
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
• Protein Folding
• Secondary Structure: α-Helices
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil.
• Secondary Structure: β-Pleated Sheet
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement or in anti-parallel arrangement. Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain.
• Secondary Structure: α-Pleated Sheet
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction.
• The Structure of Proteins
This page explains how amino acids combine to make proteins and what is meant by the primary, secondary and tertiary structures of proteins. Quaternary structure isn't covered. It only applies to proteins consisting of more than one polypeptide chain.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY-SA 3.0; Zephyris). | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.02%3A_The_Configuration_of_the_Amino_Acids.txt |
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
• Protein Folding
• Secondary Structure: α-Helices
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil.
• Secondary Structure: β-Pleated Sheet
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement or in anti-parallel arrangement. Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain.
• Secondary Structure: α-Pleated Sheet
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction.
• The Structure of Proteins
This page explains how amino acids combine to make proteins and what is meant by the primary, secondary and tertiary structures of proteins. Quaternary structure isn't covered. It only applies to proteins consisting of more than one polypeptide chain.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY-SA 3.0; Zephyris).
16.10 The Secondary Structure of Proteins
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
• Protein Folding
• Secondary Structure: α-Helices
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil.
• Secondary Structure: β-Pleated Sheet
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement or in anti-parallel arrangement. Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain.
• Secondary Structure: α-Pleated Sheet
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction.
• The Structure of Proteins
This page explains how amino acids combine to make proteins and what is meant by the primary, secondary and tertiary structures of proteins. Quaternary structure isn't covered. It only applies to proteins consisting of more than one polypeptide chain.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY-SA 3.0; Zephyris). | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.09%3A_Determining_the_Primary_Structure_of_a_Polypeptide_or_Protein.txt |
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
• Protein Folding
• Secondary Structure: α-Helices
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil.
• Secondary Structure: β-Pleated Sheet
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement or in anti-parallel arrangement. Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain.
• Secondary Structure: α-Pleated Sheet
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction.
• The Structure of Proteins
This page explains how amino acids combine to make proteins and what is meant by the primary, secondary and tertiary structures of proteins. Quaternary structure isn't covered. It only applies to proteins consisting of more than one polypeptide chain.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY-SA 3.0; Zephyris).
16.12: The Quaternary Structure of Proteins
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
• Protein Folding
• Secondary Structure: α-Helices
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil.
• Secondary Structure: β-Pleated Sheet
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement or in anti-parallel arrangement. Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain.
• Secondary Structure: α-Pleated Sheet
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction.
• The Structure of Proteins
This page explains how amino acids combine to make proteins and what is meant by the primary, secondary and tertiary structures of proteins. Quaternary structure isn't covered. It only applies to proteins consisting of more than one polypeptide chain.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY-SA 3.0; Zephyris). | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.11%3A_The_Tertiary_Structure_of_Proteins.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
17: How Enzymes Catalyze Reactions The Organic Chemisty of Vitamins
Nicotinamide is from the niacin vitamin. The NAD+ coenzyme is involved with many types of oxidation reactions where alcohols are converted to ketones or aldehydes. It is also involved in the first enzyme complex 1 of the electron transport chain.The structure for the coenzyme, NAD+, Nicotinamide Adenine Dinucleotide is shown in Figure $1$.
Role of NAD+
One role of $NAD^+$ is to initiate the electron transport chain by the reaction with an organic metabolite (intermediate in metabolic reactions). This is an oxidation reaction where 2 hydrogen atoms (or 2 hydrogen ions and 2 electrons) are removed from the organic metabolite. (The organic metabolites are usually from the citric acid cycle and the oxidation of fatty acids--details in following pages.) The reaction can be represented simply where M = any metabolite.
$MH_2 + NAD^+ \rightarrow NADH + H^+ + M: + \text{energy}$
One hydrogen is removed with 2 electrons as a hydride ion ($H^-$) while the other is removed as the positive ion ($H+$). Usually the metabolite is some type of alcohol which is oxidized to a ketone.
Figure $1$
Alcohol Dehydrogenase
The NAD+ is represented as cyan in Figure $2$. The alcohol is represented by the space filling red, gray, and white atoms. The reaction is to convert the alcohol, ethanol, into ethanal, an aldehyde.
$CH_3CH_2OH + NAD^+ \rightarrow CH_3CH=O + NADH + H^+$
This is an oxidation reaction and results in the removal of two hydrogen ions and two electrons which are added to the NAD+, converting it to NADH and H+. This is the first reaction in the metabolism of alcohol. The active site of ADH has two binding regions. The coenzyme binding site, where NAD+ binds, and the substrate binding site, where the alcohol binds. Most of the binding site for the NAD+ is hydrophobic as represented in green. Three key amino acids involved in the catalytic oxidation of alcohols to aldehydes and ketones. They are ser-48, phe 140, and phe 93.
Figure $2$: Active site of Alcohol Dehydrogenase
17.06: Vitamin (B 2)
The structure shown on the left is for FAD and is similar to NAD+ in that it contains a vitamin-riboflavin, adenine, ribose, and phosphates. As shown it is the diphosphate, but is also used as the monophosphate (FMN).
Introduction
In the form of FMN it is involved in the first enzyme complex 1 of the electron transport chain. A FMN (flavin adenine mononucleotide) as an oxidizing agent is used to react with NADH for the second step in the electron transport chain. The simplified reaction is:
NADH + H+ + FMN → FMNH2 + NAD+
Red.Ag. Ox.Ag.
Note the fact that the two hydrogens and 2e- are "passed along" from NADH to FFMN. Also note that NAD+ as a product is back to its original state as an oxidizing agent ready to begin the cycle again. The FMN has now been converted to the reducing agent and is the starting point for the third step.
Coenzyme Q or Ubiquinone
Ubiquinone: As its name suggests, is very widely distributed in nature. There are some differences in the length of the isoprene unit (in bracket on left) side chain in various species. All the natural forms of CoQ are insoluble in water, but soluble in membrane lipids where they function as a mobile electron carrier in the electron transport chain. The long hydrocarbon chain gives the non-polar property to the molecule.
CoQ acts as a bridge between enzyme complex 1 and 3 or between complex 2 and 3. Electrons are transferred from NADH along with two hydrogens to the double bond oxygens in the benzene ring. These in turn convert to alcohol groups. The electrons are then passed along to the cytochromes in enzyme complex 3.
Coenzyme A
Although not used in the electron transport chain, Coenzyme A is a major cofactor which is used to transfer a two carbon unit commonly referred to as the acetyl group. The structure has many common features with NAD+ and FAD in that it has the diphosphate, ribose, and adenine. In addition it has a vitamin called pantothenic acid, and finally terminated by a thiol group. The thiol (-SH) is the sulfur analog of an alcohol (-OH). The acetyl group (CH3C=O) is attached to the sulfur of the CoA through a thiol ester type bond. Acetyl CoA is important in the breakdown of fatty acids and is a starting point in the citric acid cycle.
17.06: Vitamin (B 2)
The structure shown on the left is for FAD and is similar to NAD+ in that it contains a vitamin-riboflavin, adenine, ribose, and phosphates. As shown it is the diphosphate, but is also used as the monophosphate (FMN).
Introduction
In the form of FMN it is involved in the first enzyme complex 1 of the electron transport chain. A FMN (flavin adenine mononucleotide) as an oxidizing agent is used to react with NADH for the second step in the electron transport chain. The simplified reaction is:
NADH + H+ + FMN → FMNH2 + NAD+
Red.Ag. Ox.Ag.
Note the fact that the two hydrogens and 2e- are "passed along" from NADH to FFMN. Also note that NAD+ as a product is back to its original state as an oxidizing agent ready to begin the cycle again. The FMN has now been converted to the reducing agent and is the starting point for the third step.
Coenzyme Q or Ubiquinone
Ubiquinone: As its name suggests, is very widely distributed in nature. There are some differences in the length of the isoprene unit (in bracket on left) side chain in various species. All the natural forms of CoQ are insoluble in water, but soluble in membrane lipids where they function as a mobile electron carrier in the electron transport chain. The long hydrocarbon chain gives the non-polar property to the molecule.
CoQ acts as a bridge between enzyme complex 1 and 3 or between complex 2 and 3. Electrons are transferred from NADH along with two hydrogens to the double bond oxygens in the benzene ring. These in turn convert to alcohol groups. The electrons are then passed along to the cytochromes in enzyme complex 3.
Coenzyme A
Although not used in the electron transport chain, Coenzyme A is a major cofactor which is used to transfer a two carbon unit commonly referred to as the acetyl group. The structure has many common features with NAD+ and FAD in that it has the diphosphate, ribose, and adenine. In addition it has a vitamin called pantothenic acid, and finally terminated by a thiol group. The thiol (-SH) is the sulfur analog of an alcohol (-OH). The acetyl group (CH3C=O) is attached to the sulfur of the CoA through a thiol ester type bond. Acetyl CoA is important in the breakdown of fatty acids and is a starting point in the citric acid cycle. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.05%3A_Niacin%3A_The_Vitamin_Needed_for_Many_Redox_Reactions.txt |
Solutions to exercises
Thiamine diphosphate (TPP) is another very important coenzyme which, like PLP, acts as an electron sink to stabilize key carbanion intermediates. The important part of the TPP molecule from a catalytic standpoint is its thiazole ring.
14.5A: The benzoin condensation reaction
In a very simple experiment that can performed in an undergraduate organic chemistry lab, benzaldehyde (a liquid compound at room temperature that is used as an artificial cherry flavoring) self-condenses to form benzoin (a crystalline solid) when stirred in ethanol with a catalytic amount of sodium hydroxide and thiamine.
The laboratory synthesis of benzoin is interesting because it mimics the mechanism of TPP-dependant enzymatic reactions in biological systems. This reaction is not catalyzed by an enzyme – rather, the thiamine molecule acts on its own, playing a similar catalytic role to that played by its diphosphate ester cousin (TPP) in enzymatic reactions.
Look carefully at the connectivity of the starting compounds and product in the benzoin condensation. Essentially what has happened is that the carbonyl carbon of one benzaldehyde molecule has somehow been turned into a nucleophile, and has attacked a second benzaldehyde molecule in a nucleophilic carbonyl addition reaction.
This probably seems quite strange - we know that carbonyl carbons are good electrophiles, but how can they be nucleophilic? For the aldehyde carbon to be a nucleophile, it would have to be deprotonated, and become a carbonyl anion. But aldehyde protons are not at all acidic!
The thiamine catalyst is the key: it allows the formation of what is essentially the equivalent of a nucleophilic benzaldehyde carbanion. Let's follow the benzoin condensation reaction mechanism through step-by-step, and see how thiamine accomplishes this task.
The important part of the thiamine molecule is the thiazole ring (look again at the structure of thiamine diphosphate on the previous page), thus we will draw thiamine (and later, thiamine diphosphate) using R groups to depict the unreactive parts of the molecule. The first step of the benzoin condensation is deprotonation of thiamine by hydroxide.
It may surprise you to learn that this proton is acidic. The reason for its acidity lies partly in the ability of the neighboring sulfur atom to accept, in its open d orbitals, some of the excess electron density of the conjugate base. Another reason is that the positive charge on the nitrogen helps to stabilize the negative charge on the conjugate base. The deprotonated thiazole is called an ylide, which is a species with adjacent positively and negatively charged atoms.
The negatively charged carbon on the thiazole ylide next attacks the carbonyl of the first benzaldehyde molecule in a nucleophilic carbonyl addition (from here on out, thiamine will be colored green in order to help you to focus on the chemistry going on with the substrate).
In the resulting molecule, what used to be the aldehyde hydrogen is now acidic - this is so because, when the proton is abstracted by hydroxide, the negative charge on the conjugate base is stabilized by resonance with the positively-charged thiazole ring of thiamine. This is the function of thiamine: it acts as an electron sink, accepting electron density so as to allow for the formation of what amounts to a carbonyl anion.
Now the first benzaldehyde molecule, assisted by thiamine, can finally act as a nucleophile, attacking the carbonyl of a second benzaldehyde (step 3).
Once this is accomplished, the thiamine ylide can be kicked off as the original carbonyl on the first benzaldehyde re-forms (step 4). The thiamine ylide is now free to catalyze another reaction.
14.5B: The transketolase reaction
Now let's look at a biochemical reaction carried out by an enzyme called transketolase, with the assistance of thiamine diphosphate. Transketolase is one of a series of enzymes (along with ribulose-5-phosphate-3-epimerase, which we considered in section 13.2B) in the 'Calvin cycle' of carbon fixation in plants. Animals and bacteria also use transketolase in sugar metabolism. In the transketolase reaction, a 2-carbon unit (in the smaller, red box) is transferred between two sugar molecules:
First, let's consider how this reaction might hypothetically proceed without the assistance of the TPP cofactor. The breaking off of the 2-carbon unit from fructose-6-phosphate might be depicted as a kind of retro-aldol event:
This would lead to glyceraldehyde-3-phosphate, the first product, plus an intermediate in the form of a carbonyl anion.
Continuing to think hypothetically, this strange 2-carbon ionic intermediate could attack the glyceraldehyde-3-phosphate carbonyl, resulting in sedoheptulose-7-phosphate, the second product.
Everything in this hypothetical scenario is fine, chemically speaking, with one major exception: the carbonyl anion intermediate. It would be an extremely high energy, unlikely species. But we have seen something like this before, haven't we, in the non-enzymatic benzoin condensation reaction above! This is exactly the kind of carbanion that thiamine (this time in its diphosphate form, TPP) makes possible. The carbonyl anion is generated in different ways in the two reactions: in the benzoin condensation it is the result of the deprotonation of benzaldehyde, while in the transketolase it results from a retro-aldol-like cleavage. But that doesn't really matter - what does matter is that in each case, a thiazole ring is present to modify the starting carbonyl group and act as an electron sink, allowing it to take on a negative charge. Here, then, is the real (as opposed to hypothetical) transketolase reaction, with the role of TPP revealed.
Make sure that you can follow the electron movement throughout the mechanism, that you can see how TPP acts as an electron sink cofactor, and that you clearly recognize the mechanistic parallels to the benzoin condensation.
14.5C: Pyruvate decarboxylase
The thiamine diphosphate coenzyme also assists in the decarboxylation of an acyl group, such as in this reaction catalyzed by pyruvate decarboxylase (this is a key reaction in the fermentation of glucose to ethanol by yeast):
In this example, the TPP-stabilized carbonyl carbanion simply acts as a base rather than as a nucleophile, abstracting a proton from an enzymatic acid to form acetaldehyde.
Example 14.7
Propose a mechanism for the reaction catalyzed by pyruvate decarboxylase.
14.5D: Synthetic parallel - carbonyl nucleophiles via dithiane anions
Nature uses thiamine to generate the equivalent of a nucleophilic carbonyl anion, but with the specific exception of the benzoin condensation, a chemist working in an organic synthesis laboratory before the mid-1970's had no equivalent procedure. In 1975, E. J. Corey and Dieter Seebach reported that they had developed a method to accomplish reactions such as the following, in which aldehyde carbons act as nucleophiles in SN2 and carbonyl addition reactions (J. Org. Chem. 1975, 40, 231).
In this technique, the aldehyde is first converted to a cyclic thioketal using 1,3-propanedithiol and acid catalyst - this is the same as the method used to 'protect' aldehydes as cyclic acetals, discussed in section 11.4B, except that a dithiol is used instead of a diol.
The original aldehyde proton is now somewhat acidic, because the empty d orbitals on the two adjacent sulfur atoms are able to delocalize excess electron density of the conjugate base. A strong base, such as an organolithium compound (section 13.6B) will deprotonate the cyclic thioacetal, which can then act as a nucleophile, attacking an alkyl halide or a carbonyl electrophile (the latter case is illustrated below):
The thioacetal can then be hydrolyzed back to an aldehyde group, a process that is facilitated by the use of methyl iodide.
Example 14.8
Show a mechanism for the hydrolysis of a cyclic thioacetal, in the presence of catalytic acid and methyl iodide. Propose a role for methyl iodide in this reaction.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.07%3A_Vitamin_%28B_1%29.txt |
Solutions to exercises
Thiamine diphosphate (TPP) is another very important coenzyme which, like PLP, acts as an electron sink to stabilize key carbanion intermediates. The important part of the TPP molecule from a catalytic standpoint is its thiazole ring.
14.5A: The benzoin condensation reaction
In a very simple experiment that can performed in an undergraduate organic chemistry lab, benzaldehyde (a liquid compound at room temperature that is used as an artificial cherry flavoring) self-condenses to form benzoin (a crystalline solid) when stirred in ethanol with a catalytic amount of sodium hydroxide and thiamine.
The laboratory synthesis of benzoin is interesting because it mimics the mechanism of TPP-dependant enzymatic reactions in biological systems. This reaction is not catalyzed by an enzyme – rather, the thiamine molecule acts on its own, playing a similar catalytic role to that played by its diphosphate ester cousin (TPP) in enzymatic reactions.
Look carefully at the connectivity of the starting compounds and product in the benzoin condensation. Essentially what has happened is that the carbonyl carbon of one benzaldehyde molecule has somehow been turned into a nucleophile, and has attacked a second benzaldehyde molecule in a nucleophilic carbonyl addition reaction.
This probably seems quite strange - we know that carbonyl carbons are good electrophiles, but how can they be nucleophilic? For the aldehyde carbon to be a nucleophile, it would have to be deprotonated, and become a carbonyl anion. But aldehyde protons are not at all acidic!
The thiamine catalyst is the key: it allows the formation of what is essentially the equivalent of a nucleophilic benzaldehyde carbanion. Let's follow the benzoin condensation reaction mechanism through step-by-step, and see how thiamine accomplishes this task.
The important part of the thiamine molecule is the thiazole ring (look again at the structure of thiamine diphosphate on the previous page), thus we will draw thiamine (and later, thiamine diphosphate) using R groups to depict the unreactive parts of the molecule. The first step of the benzoin condensation is deprotonation of thiamine by hydroxide.
It may surprise you to learn that this proton is acidic. The reason for its acidity lies partly in the ability of the neighboring sulfur atom to accept, in its open d orbitals, some of the excess electron density of the conjugate base. Another reason is that the positive charge on the nitrogen helps to stabilize the negative charge on the conjugate base. The deprotonated thiazole is called an ylide, which is a species with adjacent positively and negatively charged atoms.
The negatively charged carbon on the thiazole ylide next attacks the carbonyl of the first benzaldehyde molecule in a nucleophilic carbonyl addition (from here on out, thiamine will be colored green in order to help you to focus on the chemistry going on with the substrate).
In the resulting molecule, what used to be the aldehyde hydrogen is now acidic - this is so because, when the proton is abstracted by hydroxide, the negative charge on the conjugate base is stabilized by resonance with the positively-charged thiazole ring of thiamine. This is the function of thiamine: it acts as an electron sink, accepting electron density so as to allow for the formation of what amounts to a carbonyl anion.
Now the first benzaldehyde molecule, assisted by thiamine, can finally act as a nucleophile, attacking the carbonyl of a second benzaldehyde (step 3).
Once this is accomplished, the thiamine ylide can be kicked off as the original carbonyl on the first benzaldehyde re-forms (step 4). The thiamine ylide is now free to catalyze another reaction.
14.5B: The transketolase reaction
Now let's look at a biochemical reaction carried out by an enzyme called transketolase, with the assistance of thiamine diphosphate. Transketolase is one of a series of enzymes (along with ribulose-5-phosphate-3-epimerase, which we considered in section 13.2B) in the 'Calvin cycle' of carbon fixation in plants. Animals and bacteria also use transketolase in sugar metabolism. In the transketolase reaction, a 2-carbon unit (in the smaller, red box) is transferred between two sugar molecules:
First, let's consider how this reaction might hypothetically proceed without the assistance of the TPP cofactor. The breaking off of the 2-carbon unit from fructose-6-phosphate might be depicted as a kind of retro-aldol event:
This would lead to glyceraldehyde-3-phosphate, the first product, plus an intermediate in the form of a carbonyl anion.
Continuing to think hypothetically, this strange 2-carbon ionic intermediate could attack the glyceraldehyde-3-phosphate carbonyl, resulting in sedoheptulose-7-phosphate, the second product.
Everything in this hypothetical scenario is fine, chemically speaking, with one major exception: the carbonyl anion intermediate. It would be an extremely high energy, unlikely species. But we have seen something like this before, haven't we, in the non-enzymatic benzoin condensation reaction above! This is exactly the kind of carbanion that thiamine (this time in its diphosphate form, TPP) makes possible. The carbonyl anion is generated in different ways in the two reactions: in the benzoin condensation it is the result of the deprotonation of benzaldehyde, while in the transketolase it results from a retro-aldol-like cleavage. But that doesn't really matter - what does matter is that in each case, a thiazole ring is present to modify the starting carbonyl group and act as an electron sink, allowing it to take on a negative charge. Here, then, is the real (as opposed to hypothetical) transketolase reaction, with the role of TPP revealed.
Make sure that you can follow the electron movement throughout the mechanism, that you can see how TPP acts as an electron sink cofactor, and that you clearly recognize the mechanistic parallels to the benzoin condensation.
14.5C: Pyruvate decarboxylase
The thiamine diphosphate coenzyme also assists in the decarboxylation of an acyl group, such as in this reaction catalyzed by pyruvate decarboxylase (this is a key reaction in the fermentation of glucose to ethanol by yeast):
In this example, the TPP-stabilized carbonyl carbanion simply acts as a base rather than as a nucleophile, abstracting a proton from an enzymatic acid to form acetaldehyde.
Example 14.7
Propose a mechanism for the reaction catalyzed by pyruvate decarboxylase.
14.5D: Synthetic parallel - carbonyl nucleophiles via dithiane anions
Nature uses thiamine to generate the equivalent of a nucleophilic carbonyl anion, but with the specific exception of the benzoin condensation, a chemist working in an organic synthesis laboratory before the mid-1970's had no equivalent procedure. In 1975, E. J. Corey and Dieter Seebach reported that they had developed a method to accomplish reactions such as the following, in which aldehyde carbons act as nucleophiles in SN2 and carbonyl addition reactions (J. Org. Chem. 1975, 40, 231).
In this technique, the aldehyde is first converted to a cyclic thioketal using 1,3-propanedithiol and acid catalyst - this is the same as the method used to 'protect' aldehydes as cyclic acetals, discussed in section 11.4B, except that a dithiol is used instead of a diol.
The original aldehyde proton is now somewhat acidic, because the empty d orbitals on the two adjacent sulfur atoms are able to delocalize excess electron density of the conjugate base. A strong base, such as an organolithium compound (section 13.6B) will deprotonate the cyclic thioacetal, which can then act as a nucleophile, attacking an alkyl halide or a carbonyl electrophile (the latter case is illustrated below):
The thioacetal can then be hydrolyzed back to an aldehyde group, a process that is facilitated by the use of methyl iodide.
Example 14.8
Show a mechanism for the hydrolysis of a cyclic thioacetal, in the presence of catalytic acid and methyl iodide. Propose a role for methyl iodide in this reaction.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.07%3A_Vitamin_(B_1).txt |
13.5A: The metabolic context of carboxylation and decarboxylation
Some of the most important carbon-carbon bond-forming and bond-breaking processes in biological chemistry involve the gain or loss, by an organic molecule, of a single carbon in the form of CO2. You undoubtedly have seen this chemical equation before in an introductory biology or chemistry class:
6CO2 + 6H2O + energy → C6H12O6 + 6O2
This of course represents the photosynthetic process, by which plants (and some bacteria) harness energy from sunlight to build glucose from individual carbon dioxide molecules. The key chemical step in which carbon dioxide is 'fixed' (in other words, condensed with an existing organic molecule) is called a carboxylation reaction. It is catalyzed by the enzyme ribulose 1,5-bisphosphate carboxylase, commonly known as Rubisco, in the 'Calvin cycle' of carbon fixation.
The reverse chemical equation is also probably familiar to you:
C6H12O6 + 6O2 → 6CO2 + 6H2O + energy
This equation expresses what happens in respiration: the oxidative breakdown of glucose to form carbon dioxide, water, and energy. In the course of this transformation, each of the carbon atoms of glucose is eventually converted to individual CO2 molecules. The actual chemical step by which a carbon atom, in the form of carbon dioxide, breaks off from a larger organic molecule is called a decarboxylation reaction. The key decarboxylation steps in the conversion of glucose to carbon dioxide occur in the citric acid (Krebs) cycle and the pentose phosphate pathway.
Let's now look at the organic mechanisms of some carboxylation and decarboxylation reactions.
13.5B: The carboxylation mechanism of Rubisco, the carbon fixing enzyme
Carboxylation reactions are essentially just aldol condensations, except that the carbonyl electrophile is CO2 rather than a ketone or aldehyde. The mechanism for Rubisco, the key carbon-fixing enzyme in plants and photosynthetic bacteria (and the most abundant enzyme on earth!), is shown below. Magnesium ion plays a key charge-stabilizing role throughout the reaction. Step 1, not surprisingly, is deprotonation of an alpha-carbon to form an enolate.
Step 2 is simply an intramolecular proton transfer, which has the effect of creating a different enolate intermediate and making C2 into the nucleophile for an aldol-like attack on CO2 (step 3). Carbon dioxide has now been 'fixed' into organic form - it has become a carboxylate group on a six-carbon sugar derivative.
To follow the Rubisco mechanism through to its endpoint:
Step 4 is a retro-Claisen mechanism, with a water nucleophile and enolate leaving group. After protonation of this enolate, we are left with two molecules of 3-phosphoglycerate, which are incorporated into the 'gluconeogenesis' pathway of glucose synthesis.
Example 13.8
Propose a complete mechanism for the retro-Claisen reaction in the figure above.
13.5C: Decarboxylation
Mechanistically, a decarboxylation has parallels to retro-aldol cleavage reactions:
Just as in retro-aldol reactions, the electrons from the broken carbon-carbon bond have to have some place to go - they must, in other words, be stabilized - for the decarboxylation step to take place. Quite often, the electrons are stabilized by the formation of an enolate, as is the case in the general mechanism pictured above. This of course means that a carbonyl group must be positioned beta to (i.e. two carbons down from) the carboxylate carbon. If there is no stable place for the electrons in the carbon-carbon bond to go, then a decarboxylation is very unlikely.
Be especially careful, when drawing decarboxylation mechanisms, to resist the temptation to treat the CO2 molecule as the leaving group:
This is not what a decarboxylation looks like! In a decarboxylation step, it is the organic part of the molecule that is, in fact, the leaving group, 'pushed off' by the electrons on the carboxylate.
Below are two important key b-carbonyl decarboxylation steps in glucose metabolism, each representing a point at which a carbon derived from glucose is released as CO2.
Example 13.9
Exercise 13.9: Draw mechanistic arrows showing the first step in each of the reactions shown above.
The reaction catalyzed by acetoacetate decarboxylase relies on an imine group, rather than a carbonyl, to stabilize the carbanion intermediate. This is a mechanistic parallel to a type II retro-aldolase reaction.
Example 13.10
Exercise 13.10: Draw a mechanism for the decarboxylation step in the reaction above.
There are some examples of decarboxylation reactions in which the carbanion intermediate goes on to form a new carbon-carbon bond, rather than becoming protonated as in the example we have seen so far. This reaction in the fatty acid biosynthetic pathway is a decarboxylation followed by a Claisen condensation. A thiol group on an 'acyl carrier protein' (section 12.3D) is part of the thioester functional group in this reaction.
An interesting variation on decarboxylation in the synthesis of tyrosine in bacteria is shown below.
Template:ExampleStart
Example 13.11
a) Draw a mechanism for the reaction above.
b) Draw hypothetical decarboxylation mechanisms showing the formation of alternate products A and B from the same starting compound:
c) How would you describe in words the relationship between the actual product and hypothetical products A and B? Which of the three would you expect to be most thermodynamically stable, and why?
In chapter 14, we will see decarboxylation reactions occurring in several other mechanistic contexts.
13.5D: Biotin is a CO2-carrying coenzyme
Recall from section 6.5B that many enzymes are dependent upon the assistance of coenzymes, which are small (relative to protein) organic molecules that bind - covalently or non-covalently - in an enzyme's active site and help it to catalyze its reaction. S-adenosylmethionine (SAM, section 9.1A) and ATP (section 10.2) are two examples that we have encountered so far, and we will see several more in the chapters ahead. Although Rubisco (described in part B of this section) is an exception, most enzymes that catalyze carboxylation reactions are dependent upon a coenzyme called biotin, which serves as a temporary carrier of carbon dioxide.
Pyruvate carboxylase, the enzyme catalyzing the first step of the gluconeogensis pathway, is a good example of a biotin-dependent carboxylation reaction. Notice that the CO2 in this reaction is derived from bicarbonate, unlike the Rubisco reaction in which CO2 is 'fixed' directly from the atmosphere.
Biotin is covalently attached to the enzyme through an amide linkage to an active site lysine.
The exact mechanism by which biotin-dependent carboxylation reactions operate is still not completely understood, however the following is a likely picture. First, the bicarbonate ion is phosphorylated by ATP (step 1, see section 10.2), and thus is activated for decarboxylation, which generates free CO2 (step 2).
Biotin's job is to hold on to the carbon dioxide molecule until pyruvate comes into the active site. Carboxylation of biotin involves deprotonation of the amide nitrogen to form an enolate-like intermediate (step 3 - amides have a pKa of approximately 17, and this is lowered by the presence of an active site acid near the oxygen). This step is followed by attack of the nucleophilic nitrogen on carbon dioxide to form carboxybiotinylated enzyme (step 4).
When a pyruvate molecule binds, rearrangement of the active site architecture causes the previous step to go in reverse (step 5), freeing the CO2 and generating a biotin base to deprotonate the alpha-carbon of pyruvate so that it can condense, in an aldol-like fashion, with CO2 to form oxaloacetate (steps 6-7).
If you have studied some biochemistry, you may have heard about biotin in a somewhat different context that what is discussed in this section. A protein called avidin, found in abundance in egg white, binds non-covalently and extremely tightly to biotin (in fact, avidin-biotin is the tightest protein-ligand binding pair known to science). Biochemists often make use of this property by covalently linking a biomolecule of interest to biotin. The 'biotinylated' species can then be easily isolated from a complex mixture by running the mixture through an 'affinity column' that is coated with avidin.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.08%3A_Vitamin_%28H%29.txt |
13.5A: The metabolic context of carboxylation and decarboxylation
Some of the most important carbon-carbon bond-forming and bond-breaking processes in biological chemistry involve the gain or loss, by an organic molecule, of a single carbon in the form of CO2. You undoubtedly have seen this chemical equation before in an introductory biology or chemistry class:
6CO2 + 6H2O + energy → C6H12O6 + 6O2
This of course represents the photosynthetic process, by which plants (and some bacteria) harness energy from sunlight to build glucose from individual carbon dioxide molecules. The key chemical step in which carbon dioxide is 'fixed' (in other words, condensed with an existing organic molecule) is called a carboxylation reaction. It is catalyzed by the enzyme ribulose 1,5-bisphosphate carboxylase, commonly known as Rubisco, in the 'Calvin cycle' of carbon fixation.
The reverse chemical equation is also probably familiar to you:
C6H12O6 + 6O2 → 6CO2 + 6H2O + energy
This equation expresses what happens in respiration: the oxidative breakdown of glucose to form carbon dioxide, water, and energy. In the course of this transformation, each of the carbon atoms of glucose is eventually converted to individual CO2 molecules. The actual chemical step by which a carbon atom, in the form of carbon dioxide, breaks off from a larger organic molecule is called a decarboxylation reaction. The key decarboxylation steps in the conversion of glucose to carbon dioxide occur in the citric acid (Krebs) cycle and the pentose phosphate pathway.
Let's now look at the organic mechanisms of some carboxylation and decarboxylation reactions.
13.5B: The carboxylation mechanism of Rubisco, the carbon fixing enzyme
Carboxylation reactions are essentially just aldol condensations, except that the carbonyl electrophile is CO2 rather than a ketone or aldehyde. The mechanism for Rubisco, the key carbon-fixing enzyme in plants and photosynthetic bacteria (and the most abundant enzyme on earth!), is shown below. Magnesium ion plays a key charge-stabilizing role throughout the reaction. Step 1, not surprisingly, is deprotonation of an alpha-carbon to form an enolate.
Step 2 is simply an intramolecular proton transfer, which has the effect of creating a different enolate intermediate and making C2 into the nucleophile for an aldol-like attack on CO2 (step 3). Carbon dioxide has now been 'fixed' into organic form - it has become a carboxylate group on a six-carbon sugar derivative.
To follow the Rubisco mechanism through to its endpoint:
Step 4 is a retro-Claisen mechanism, with a water nucleophile and enolate leaving group. After protonation of this enolate, we are left with two molecules of 3-phosphoglycerate, which are incorporated into the 'gluconeogenesis' pathway of glucose synthesis.
Example 13.8
Propose a complete mechanism for the retro-Claisen reaction in the figure above.
13.5C: Decarboxylation
Mechanistically, a decarboxylation has parallels to retro-aldol cleavage reactions:
Just as in retro-aldol reactions, the electrons from the broken carbon-carbon bond have to have some place to go - they must, in other words, be stabilized - for the decarboxylation step to take place. Quite often, the electrons are stabilized by the formation of an enolate, as is the case in the general mechanism pictured above. This of course means that a carbonyl group must be positioned beta to (i.e. two carbons down from) the carboxylate carbon. If there is no stable place for the electrons in the carbon-carbon bond to go, then a decarboxylation is very unlikely.
Be especially careful, when drawing decarboxylation mechanisms, to resist the temptation to treat the CO2 molecule as the leaving group:
This is not what a decarboxylation looks like! In a decarboxylation step, it is the organic part of the molecule that is, in fact, the leaving group, 'pushed off' by the electrons on the carboxylate.
Below are two important key b-carbonyl decarboxylation steps in glucose metabolism, each representing a point at which a carbon derived from glucose is released as CO2.
Example 13.9
Exercise 13.9: Draw mechanistic arrows showing the first step in each of the reactions shown above.
The reaction catalyzed by acetoacetate decarboxylase relies on an imine group, rather than a carbonyl, to stabilize the carbanion intermediate. This is a mechanistic parallel to a type II retro-aldolase reaction.
Example 13.10
Exercise 13.10: Draw a mechanism for the decarboxylation step in the reaction above.
There are some examples of decarboxylation reactions in which the carbanion intermediate goes on to form a new carbon-carbon bond, rather than becoming protonated as in the example we have seen so far. This reaction in the fatty acid biosynthetic pathway is a decarboxylation followed by a Claisen condensation. A thiol group on an 'acyl carrier protein' (section 12.3D) is part of the thioester functional group in this reaction.
An interesting variation on decarboxylation in the synthesis of tyrosine in bacteria is shown below.
Template:ExampleStart
Example 13.11
a) Draw a mechanism for the reaction above.
b) Draw hypothetical decarboxylation mechanisms showing the formation of alternate products A and B from the same starting compound:
c) How would you describe in words the relationship between the actual product and hypothetical products A and B? Which of the three would you expect to be most thermodynamically stable, and why?
In chapter 14, we will see decarboxylation reactions occurring in several other mechanistic contexts.
13.5D: Biotin is a CO2-carrying coenzyme
Recall from section 6.5B that many enzymes are dependent upon the assistance of coenzymes, which are small (relative to protein) organic molecules that bind - covalently or non-covalently - in an enzyme's active site and help it to catalyze its reaction. S-adenosylmethionine (SAM, section 9.1A) and ATP (section 10.2) are two examples that we have encountered so far, and we will see several more in the chapters ahead. Although Rubisco (described in part B of this section) is an exception, most enzymes that catalyze carboxylation reactions are dependent upon a coenzyme called biotin, which serves as a temporary carrier of carbon dioxide.
Pyruvate carboxylase, the enzyme catalyzing the first step of the gluconeogensis pathway, is a good example of a biotin-dependent carboxylation reaction. Notice that the CO2 in this reaction is derived from bicarbonate, unlike the Rubisco reaction in which CO2 is 'fixed' directly from the atmosphere.
Biotin is covalently attached to the enzyme through an amide linkage to an active site lysine.
The exact mechanism by which biotin-dependent carboxylation reactions operate is still not completely understood, however the following is a likely picture. First, the bicarbonate ion is phosphorylated by ATP (step 1, see section 10.2), and thus is activated for decarboxylation, which generates free CO2 (step 2).
Biotin's job is to hold on to the carbon dioxide molecule until pyruvate comes into the active site. Carboxylation of biotin involves deprotonation of the amide nitrogen to form an enolate-like intermediate (step 3 - amides have a pKa of approximately 17, and this is lowered by the presence of an active site acid near the oxygen). This step is followed by attack of the nucleophilic nitrogen on carbon dioxide to form carboxybiotinylated enzyme (step 4).
When a pyruvate molecule binds, rearrangement of the active site architecture causes the previous step to go in reverse (step 5), freeing the CO2 and generating a biotin base to deprotonate the alpha-carbon of pyruvate so that it can condense, in an aldol-like fashion, with CO2 to form oxaloacetate (steps 6-7).
If you have studied some biochemistry, you may have heard about biotin in a somewhat different context that what is discussed in this section. A protein called avidin, found in abundance in egg white, binds non-covalently and extremely tightly to biotin (in fact, avidin-biotin is the tightest protein-ligand binding pair known to science). Biochemists often make use of this property by covalently linking a biomolecule of interest to biotin. The 'biotinylated' species can then be easily isolated from a complex mixture by running the mixture through an 'affinity column' that is coated with avidin.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.08%3A_Vitamin_(H).txt |
Solutions to exercises
In section 6.5, we were introduced to the concept of coenzymes - small organic molecules that assist enzymes in doing a particular chemical job. We have seen many specific examples since then: SAM as a methyl group donor, ATP as a phosphoryl group donor, Coenzyme A as an activator for acyl transfer reactions, and most recently glutathione as a 'nucleophile for hire' in the cis/trans isomerization described in section 14.2A.
In this section and the next, we will consider two very important coenzymes that help to stabilize carbanion intermediates in many diverse enzymatic reactions. They may be familiar to you already if you occasionally look at the nutritional information on food packages: thiamine diphosphate (TPP) is the diphosphate ester of thiamine (otherwise known as vitamin B1), and pyridoxal phosphate (PLP), is a derivative of vitamin B6.
Unlike plants and many bacteria, humans do not have the biosynthetic enzymes to make these molecules, so we must get them in our diet.
Many enzymes involved in the metabolism of amino acids make use of PLP as a coenzyme. The basic function of PLP is to act as an 'electron sink', the meaning of which will become clear as we examine in detail the mechanisms of some PLP-dependant reactions. Most of these reactions will be familiar - they include amino acid racemizations, eliminations, additions, and retro-aldol or retro-Claisen cleavages. What will be new is the use of PLP to stabilize the carbanion intermediates in each of these reactions.
14.4A: PLP and the Schiff base linkage to lysine
In the first step of virtually all PLP-dependant reactions, the aldehyde group of the coenzyme forms an imine (Schiff base) linkage with a lysine side chain on the enzyme. (In order to help you to focus on the chemistry happening with the amino acid substrates, the PLP molecule in subsequent figures is colored green - it will be very helpful to refer to color figures).
Generally, the next step (step 2 in the figure above) is an imine exchange, as the amine nitrogen on the amino acid substrate replaces the enzyme lysine nitrogen in the imine linkage. This substrate-coenzyme adduct is stabilized by a favorable hydrogen bond between the phenol of PLP and the imine nitrogen.
14.4B: PLP-dependent amino acid racemases
With these preliminaries accomplished, the real PLP chemistry is ready to start. Let's look first at the reaction catalyzed by PLP-dependent alanine racemase. In section 13.2B we saw an example of a PLP-independent amino acid racemase reaction, in which the negatively-charged intermediate was simply the enolate form of a carboxylate.
Many other amino acid racemases, however, require the participation of PLP. Once it is linked to PLP in the active site, the a-proton of alanine can be abstracted by an active site base (step 3 below). The negative charge on the carbanion intermediate can, of course, be delocalized to the carboxylate group of alanine. The PLP coenzyme, however, provides a much expanded network of conjugated π-bonds over which the electron density can be delocalized all the way down to the PLP nitrogen.
This is what we mean when we say that the job of PLP is to act as an ‘electron sink’: the coenzyme is extremely efficient at absorbing, or delocalizing, the excess electron density on the deprotonated a-carbon of the reaction intermediate. PLP is helping the enzyme to increase the acidity of the a-hydrogen by stabilizing the conjugate base. A PLP-stabilized carbanion intermediate is commonly referred to as a quinonoid intermediate.
In the remaining steps of the alanine racemase reaction, reprotonation occurs on the opposite side of the substrate (step 4), leading to the D-amino acid.
All that remains is another imine exchange (step 5), which frees the D-alanine product and re-attaches PLP to the enzymatic lysine side-chain.
14.4C: PLP-dependent decarboxylation
In the alanine racemase reaction above, PLP assisted in breaking the Cα-H bond of the amino acid. Some PLP-dependant enzymes can catalyze the breaking of the bond between Cα and the carboxylate carbon in an amino acid by stabilizing the resulting carbanion intermediate: these are simply decarboxylation reactions. One of the final steps in the synthesis of lysine is a PLP-dependent decarboxylation.
14.4D: PLP-dependent retro-aldol and retro-Claisen reactions
Still other PLP-dependant enzymes catalyze the breaking of the bond between the alpha-carbon and the first carbon on the amino acid side chain - this is referred to as a Cα-Cβ cleavage, and it actually is a variation on the retro-aldol reaction (section 13.3C). When serine is degraded, it is first converted to glycine by serine hydroxymethyltransferase.
The retro-aldol step occurs from the PLP-serine Schiff base adduct, as the electrons from the broken bond are stabilized by the electron-sink network of the coenzyme. Just like in a standard retroaldol step, these electrons then go on to form a new carbon-hydrogen bond, leading to a PLP-glycine adduct which is freed by imine exchange.
Note that, in this reaction just as in the racemase reaction described previously, the key intermediate is a PLP-stabilized carbanion, or quinonoid.
What happens to the formaldehyde produced in this reaction? Recall from section 11.4D that it is incorporated into the coenzyme tetrahydrofolate to form 5,10-methylenetetrahydrofolate.
PLP also assists in retro-Claisen cleavage reactions (section 13.4B), such as this step in the degradation of threonine:
14.4E: PLP-dependent transamination reactions
One of the most important steps in the degradation of amino acids is elimination of the amino nitrogen atom in the form of urea, which is excreted in the urine. While the metabolic details of this 'urea cycle' are outside of the scope of this text, what is important to understand is that nitrogen atoms from many of the amino acids must be shuttled first to glutamate before being processed for elimination:
The reaction in which the nitrogen group from an amino acid is transferred to alpha-ketoglutarate is accomplished by PLP-dependent enzymes called transaminases. Once again, the first step is abstraction of the alpha-proton from the PLP-amino acid adduct.
However, in the transaminase reactions this initial deprotonation step is immediately followed by a reprotonation at what was originally the aldehyde carbon of PLP (step 2 above), which results in a new carbon-nitrogen double bond (i.e., an imine) between the a-carbon and the nitrogen of the original amino acid. This imine is then hydrolyzed (step 3 above) - this is the step where the nitrogen is removed from the amino acid to form an alpha-keto acid, which can be degraded further.
The coenzyme, which now carries an amine group and is therefore called pyridoxamine phospate (PMP), next transfers the amine group to alpha-ketoglutarate (to form glutamate) through an exact reversal of the whole process we have just seen.
In the overall transaminase reaction, the PLP coenzyme not only provides an electron sink, it also serves as a temporary 'storage space' for a nitrogen atom as it is passed from one amino acid to another.
Example 14.5
Show a complete, step-by-step mechanism for the second half of the transaminase reaction (transfer of the amine group from PLP to a-ketoglutarate to form glutamate).
Transaminase reactions also function in the biosynthesis direction for alanine, aspartate, and glutamate. Alanine, for example, is synthesized from pyruvate by the transfer of an amino group from glutamate.
14.4F: PLP-dependent beta-elimination and beta-substitution reactions
Two more reaction types in the PLP toolbox are beta-eliminations and beta-substitutions. Serine dehydratase catalyzes a beta-elimination of the hydroxyl group of serine, leading eventually to pyruvate.
After forming the normal imine linkage with PLP, the a-proton of serine is abstracted by a basic active-site lysine residue (step 1), and the coenzyme stabilizes the conjugate base.
After elimination of water (step 2), the dehydrated substrate is freed from the PLP coenzyme (step 3) by Schiff base transfer, tautomerizes to the imine form (step 4), and then is hydrolyzed to pyruvate (step 5).
A beta-substitution is simply beta-elimination followed directly by the reverse reaction (Michael addition) with a different nucleophile:
As with virtually all PLP-dependent reactions, the coenzyme serves to stabilize the carbanion intermediates.
In many bacteria, the synthesis of cysteine from serine relies upon a PLP-dependent beta-substitution step. In this pathway, serine is first acetylated by acetyl-CoA (an acyl transfer reaction).
The acetylated serine forms an imine linkage with PLP, then undergoes an elimination (steps 1-2 below) in which the acetyl group is expelled (acetyl is, of course, a much weaker base / better leaving group than a hydroxide - thus the function of the initial serine acetylation step).
A sulfhydryl ion (SH-) then attacks in a Michael addition (steps 3-4), with the intermediate stabilized again by the electron-sink property of PLP. Finally, the cysteine product is released from PLP (step 5) via an imine exchange reaction with an active site lysine.
14.4G: PLP-dependent gamma-elimination and gamma-substitution reactions
The electron sink capability of PLP allows some enzymes to catalyze eliminations at the gamma (γ) carbon of an amino acid side chain, rather than at the beta-carbon:
The secret to understanding the mechanism of a gamma-elimination is that PLP acts as an electron sink twice - it absorbs the excess electron density from not one but two proton abstractions. As an example, let's look at the cystathionine gamma-lyase reaction, which is part of the methionine degradation pathway.
Cystathionine first links to PLP in the normal way, using the amino group that is furthest away from the sulfur atom. In a familiar step, the alpha-proton is then abstracted by an enzymatic base (step 1), and the electron density is absorbed by PLP.
Next comes the new part - before anything happens to the electron density from the first proton abstraction, a second proton, this time from the beta-position on the side chain, is abstracted, forming an enamine intermediate (step 2) This is made possible by the acidic phenolic proton on the pyridoxal ring of PLP. It is this second proton abstraction that is part of a recognizable elimination reaction, as the thiol group (which is actually a cysteine amino acid) leaves and a new pi-bond forms between Cβ and Cγ(step 3). This pi-bond is short-lived, however, as the electron density from the first proton abstraction, which has been 'hiding' in PLP all this time, finally bounces back and protonates Cγ (step 4) With the usual imine transfer to an enzymatic lysine, the final product, 2-aminocrotonate, is released in step 5.
Another related reaction is the PLP-dependant gamma-substitution, which starts off like a gamma-elimination, then reverses itself with the addition of a different nucleophile.
In the synthesis of methionine, cystathionine (the starting compound in the previous gamma-elimination!) is obtained from O-succinyl homoserine and cysteine in a PLP-dependent gamma-substitution.
Example 14.6
Draw a complete mechanism for the cystathionine synthase reaction pictured above.
14.4H: Altering the course of a PLP reaction through site-directed mutagenesis
We have seen how PLP-dependent enzymes catalyze a group of reaction types - racemizations, retroaldols, transamininations, and eliminations - which, despite their apparent diversity, are all characterized by a critical carbanion intermediate that is stabilized by the electron sink of the PLP coenzyme. Given the similarities in the chemistry, it would be reasonable to propose that the active site architecture of these enzymes might also be quite close. This idea was nicely illustrated by an experiment in which site-directed mutagenesis on a single active site amino acid of PLP-dependent alanine racemase was sufficient to turn it into a retro-aldolase when provided with a suitable alternate substrate (J. Am. Chem. Soc. 2003, 125, 10158).
The catalytic base that abstracts the alpha-proton in the alanine racemase reaction is a tyrosine, assisted by a nearby histidine (see figure below). When researchers mutated the tyrosine to an alanine, and substituted beta-hydroxytyrosine for the alanine substrate, a retro-aldol reaction was catalyzed with remarkable efficiency.
Notice what has happened here: the basic histidine, with no tyrosine to deprotonate because of the mutation, instead abstracts a proton from the beta-hydroxyl group of the new substrate, setting up a retroaldol cleavage. What the researchers did, essentially, was to swap a beta-hydroxy amino acid substrate (capable of undergoing retroaldol cleavage) for the normal alanine substrate, then reposition the active site basic group so that a different acidic proton was abstracted. That was all it took to change a racemase into a retroaldolase, because the necessary electron sink system was all left in place. Why did they use the unnatural amino acid beta-hydroxy tyrosine for the retro-aldol substrate rather than threonine, which also has a beta-hydroxyl group and is closer in structure to alanine? The researchers figured that the phenyl ring of beta-hydroxy tyrosine would fit nicely in the space left empty due to the removal of the enzymatic tyrosine from the active site. To reiterate: the point of this experiment was to demonstrate how similar the chemistry being catalyzed by different PLP-dependent enzymes really is - and, as a corollary, how similar the active sites are. Even though one would not normally consider a racemization to be closely related to a retroaldol cleavage, the mechanistic themes are very close, as evidenced by researchers' ability to change the reaction product by making a single active site mutation.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.09%3A_Vitamin_%28B_6%29.txt |
Solutions to exercises
In section 6.5, we were introduced to the concept of coenzymes - small organic molecules that assist enzymes in doing a particular chemical job. We have seen many specific examples since then: SAM as a methyl group donor, ATP as a phosphoryl group donor, Coenzyme A as an activator for acyl transfer reactions, and most recently glutathione as a 'nucleophile for hire' in the cis/trans isomerization described in section 14.2A.
In this section and the next, we will consider two very important coenzymes that help to stabilize carbanion intermediates in many diverse enzymatic reactions. They may be familiar to you already if you occasionally look at the nutritional information on food packages: thiamine diphosphate (TPP) is the diphosphate ester of thiamine (otherwise known as vitamin B1), and pyridoxal phosphate (PLP), is a derivative of vitamin B6.
Unlike plants and many bacteria, humans do not have the biosynthetic enzymes to make these molecules, so we must get them in our diet.
Many enzymes involved in the metabolism of amino acids make use of PLP as a coenzyme. The basic function of PLP is to act as an 'electron sink', the meaning of which will become clear as we examine in detail the mechanisms of some PLP-dependant reactions. Most of these reactions will be familiar - they include amino acid racemizations, eliminations, additions, and retro-aldol or retro-Claisen cleavages. What will be new is the use of PLP to stabilize the carbanion intermediates in each of these reactions.
14.4A: PLP and the Schiff base linkage to lysine
In the first step of virtually all PLP-dependant reactions, the aldehyde group of the coenzyme forms an imine (Schiff base) linkage with a lysine side chain on the enzyme. (In order to help you to focus on the chemistry happening with the amino acid substrates, the PLP molecule in subsequent figures is colored green - it will be very helpful to refer to color figures).
Generally, the next step (step 2 in the figure above) is an imine exchange, as the amine nitrogen on the amino acid substrate replaces the enzyme lysine nitrogen in the imine linkage. This substrate-coenzyme adduct is stabilized by a favorable hydrogen bond between the phenol of PLP and the imine nitrogen.
14.4B: PLP-dependent amino acid racemases
With these preliminaries accomplished, the real PLP chemistry is ready to start. Let's look first at the reaction catalyzed by PLP-dependent alanine racemase. In section 13.2B we saw an example of a PLP-independent amino acid racemase reaction, in which the negatively-charged intermediate was simply the enolate form of a carboxylate.
Many other amino acid racemases, however, require the participation of PLP. Once it is linked to PLP in the active site, the a-proton of alanine can be abstracted by an active site base (step 3 below). The negative charge on the carbanion intermediate can, of course, be delocalized to the carboxylate group of alanine. The PLP coenzyme, however, provides a much expanded network of conjugated π-bonds over which the electron density can be delocalized all the way down to the PLP nitrogen.
This is what we mean when we say that the job of PLP is to act as an ‘electron sink’: the coenzyme is extremely efficient at absorbing, or delocalizing, the excess electron density on the deprotonated a-carbon of the reaction intermediate. PLP is helping the enzyme to increase the acidity of the a-hydrogen by stabilizing the conjugate base. A PLP-stabilized carbanion intermediate is commonly referred to as a quinonoid intermediate.
In the remaining steps of the alanine racemase reaction, reprotonation occurs on the opposite side of the substrate (step 4), leading to the D-amino acid.
All that remains is another imine exchange (step 5), which frees the D-alanine product and re-attaches PLP to the enzymatic lysine side-chain.
14.4C: PLP-dependent decarboxylation
In the alanine racemase reaction above, PLP assisted in breaking the Cα-H bond of the amino acid. Some PLP-dependant enzymes can catalyze the breaking of the bond between Cα and the carboxylate carbon in an amino acid by stabilizing the resulting carbanion intermediate: these are simply decarboxylation reactions. One of the final steps in the synthesis of lysine is a PLP-dependent decarboxylation.
14.4D: PLP-dependent retro-aldol and retro-Claisen reactions
Still other PLP-dependant enzymes catalyze the breaking of the bond between the alpha-carbon and the first carbon on the amino acid side chain - this is referred to as a Cα-Cβ cleavage, and it actually is a variation on the retro-aldol reaction (section 13.3C). When serine is degraded, it is first converted to glycine by serine hydroxymethyltransferase.
The retro-aldol step occurs from the PLP-serine Schiff base adduct, as the electrons from the broken bond are stabilized by the electron-sink network of the coenzyme. Just like in a standard retroaldol step, these electrons then go on to form a new carbon-hydrogen bond, leading to a PLP-glycine adduct which is freed by imine exchange.
Note that, in this reaction just as in the racemase reaction described previously, the key intermediate is a PLP-stabilized carbanion, or quinonoid.
What happens to the formaldehyde produced in this reaction? Recall from section 11.4D that it is incorporated into the coenzyme tetrahydrofolate to form 5,10-methylenetetrahydrofolate.
PLP also assists in retro-Claisen cleavage reactions (section 13.4B), such as this step in the degradation of threonine:
14.4E: PLP-dependent transamination reactions
One of the most important steps in the degradation of amino acids is elimination of the amino nitrogen atom in the form of urea, which is excreted in the urine. While the metabolic details of this 'urea cycle' are outside of the scope of this text, what is important to understand is that nitrogen atoms from many of the amino acids must be shuttled first to glutamate before being processed for elimination:
The reaction in which the nitrogen group from an amino acid is transferred to alpha-ketoglutarate is accomplished by PLP-dependent enzymes called transaminases. Once again, the first step is abstraction of the alpha-proton from the PLP-amino acid adduct.
However, in the transaminase reactions this initial deprotonation step is immediately followed by a reprotonation at what was originally the aldehyde carbon of PLP (step 2 above), which results in a new carbon-nitrogen double bond (i.e., an imine) between the a-carbon and the nitrogen of the original amino acid. This imine is then hydrolyzed (step 3 above) - this is the step where the nitrogen is removed from the amino acid to form an alpha-keto acid, which can be degraded further.
The coenzyme, which now carries an amine group and is therefore called pyridoxamine phospate (PMP), next transfers the amine group to alpha-ketoglutarate (to form glutamate) through an exact reversal of the whole process we have just seen.
In the overall transaminase reaction, the PLP coenzyme not only provides an electron sink, it also serves as a temporary 'storage space' for a nitrogen atom as it is passed from one amino acid to another.
Example 14.5
Show a complete, step-by-step mechanism for the second half of the transaminase reaction (transfer of the amine group from PLP to a-ketoglutarate to form glutamate).
Transaminase reactions also function in the biosynthesis direction for alanine, aspartate, and glutamate. Alanine, for example, is synthesized from pyruvate by the transfer of an amino group from glutamate.
14.4F: PLP-dependent beta-elimination and beta-substitution reactions
Two more reaction types in the PLP toolbox are beta-eliminations and beta-substitutions. Serine dehydratase catalyzes a beta-elimination of the hydroxyl group of serine, leading eventually to pyruvate.
After forming the normal imine linkage with PLP, the a-proton of serine is abstracted by a basic active-site lysine residue (step 1), and the coenzyme stabilizes the conjugate base.
After elimination of water (step 2), the dehydrated substrate is freed from the PLP coenzyme (step 3) by Schiff base transfer, tautomerizes to the imine form (step 4), and then is hydrolyzed to pyruvate (step 5).
A beta-substitution is simply beta-elimination followed directly by the reverse reaction (Michael addition) with a different nucleophile:
As with virtually all PLP-dependent reactions, the coenzyme serves to stabilize the carbanion intermediates.
In many bacteria, the synthesis of cysteine from serine relies upon a PLP-dependent beta-substitution step. In this pathway, serine is first acetylated by acetyl-CoA (an acyl transfer reaction).
The acetylated serine forms an imine linkage with PLP, then undergoes an elimination (steps 1-2 below) in which the acetyl group is expelled (acetyl is, of course, a much weaker base / better leaving group than a hydroxide - thus the function of the initial serine acetylation step).
A sulfhydryl ion (SH-) then attacks in a Michael addition (steps 3-4), with the intermediate stabilized again by the electron-sink property of PLP. Finally, the cysteine product is released from PLP (step 5) via an imine exchange reaction with an active site lysine.
14.4G: PLP-dependent gamma-elimination and gamma-substitution reactions
The electron sink capability of PLP allows some enzymes to catalyze eliminations at the gamma (γ) carbon of an amino acid side chain, rather than at the beta-carbon:
The secret to understanding the mechanism of a gamma-elimination is that PLP acts as an electron sink twice - it absorbs the excess electron density from not one but two proton abstractions. As an example, let's look at the cystathionine gamma-lyase reaction, which is part of the methionine degradation pathway.
Cystathionine first links to PLP in the normal way, using the amino group that is furthest away from the sulfur atom. In a familiar step, the alpha-proton is then abstracted by an enzymatic base (step 1), and the electron density is absorbed by PLP.
Next comes the new part - before anything happens to the electron density from the first proton abstraction, a second proton, this time from the beta-position on the side chain, is abstracted, forming an enamine intermediate (step 2) This is made possible by the acidic phenolic proton on the pyridoxal ring of PLP. It is this second proton abstraction that is part of a recognizable elimination reaction, as the thiol group (which is actually a cysteine amino acid) leaves and a new pi-bond forms between Cβ and Cγ(step 3). This pi-bond is short-lived, however, as the electron density from the first proton abstraction, which has been 'hiding' in PLP all this time, finally bounces back and protonates Cγ (step 4) With the usual imine transfer to an enzymatic lysine, the final product, 2-aminocrotonate, is released in step 5.
Another related reaction is the PLP-dependant gamma-substitution, which starts off like a gamma-elimination, then reverses itself with the addition of a different nucleophile.
In the synthesis of methionine, cystathionine (the starting compound in the previous gamma-elimination!) is obtained from O-succinyl homoserine and cysteine in a PLP-dependent gamma-substitution.
Example 14.6
Draw a complete mechanism for the cystathionine synthase reaction pictured above.
14.4H: Altering the course of a PLP reaction through site-directed mutagenesis
We have seen how PLP-dependent enzymes catalyze a group of reaction types - racemizations, retroaldols, transamininations, and eliminations - which, despite their apparent diversity, are all characterized by a critical carbanion intermediate that is stabilized by the electron sink of the PLP coenzyme. Given the similarities in the chemistry, it would be reasonable to propose that the active site architecture of these enzymes might also be quite close. This idea was nicely illustrated by an experiment in which site-directed mutagenesis on a single active site amino acid of PLP-dependent alanine racemase was sufficient to turn it into a retro-aldolase when provided with a suitable alternate substrate (J. Am. Chem. Soc. 2003, 125, 10158).
The catalytic base that abstracts the alpha-proton in the alanine racemase reaction is a tyrosine, assisted by a nearby histidine (see figure below). When researchers mutated the tyrosine to an alanine, and substituted beta-hydroxytyrosine for the alanine substrate, a retro-aldol reaction was catalyzed with remarkable efficiency.
Notice what has happened here: the basic histidine, with no tyrosine to deprotonate because of the mutation, instead abstracts a proton from the beta-hydroxyl group of the new substrate, setting up a retroaldol cleavage. What the researchers did, essentially, was to swap a beta-hydroxy amino acid substrate (capable of undergoing retroaldol cleavage) for the normal alanine substrate, then reposition the active site basic group so that a different acidic proton was abstracted. That was all it took to change a racemase into a retroaldolase, because the necessary electron sink system was all left in place. Why did they use the unnatural amino acid beta-hydroxy tyrosine for the retro-aldol substrate rather than threonine, which also has a beta-hydroxyl group and is closer in structure to alanine? The researchers figured that the phenyl ring of beta-hydroxy tyrosine would fit nicely in the space left empty due to the removal of the enzymatic tyrosine from the active site. To reiterate: the point of this experiment was to demonstrate how similar the chemistry being catalyzed by different PLP-dependent enzymes really is - and, as a corollary, how similar the active sites are. Even though one would not normally consider a racemization to be closely related to a retroaldol cleavage, the mechanistic themes are very close, as evidenced by researchers' ability to change the reaction product by making a single active site mutation.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.09%3A_Vitamin_(B_6).txt |
Cobalamin, or Vitamin B12, is the largest and the most complex out of all the types of Vitamins. The discovery of Cobalamin was made as scientists were seeking to find a cure for pernicious anemia, an anemic disease caused by an absence of intrinsic factor in the stomach. Cobalamin was studied, purified, and collected into small red crystals, and its crystallize structure was determined during an X ray analysis experiment conducted by Scientist Hodkin. A molecule structure of Cobalamin is simple, yet contains a lot of different varieties and complexes as shown in Figure \(1\). The examination of the vitamin’s molecular structure helps scientists to have a better understanding of how the body utilizes Vitamin B12 into building red blood cells and preventing pernicious anemia syndromes.
The metalloenzyme structure of Cobalamin presents a corrin ring with Cobalt, the only metal in the molecule, positioned right in the center of the structure by four coordinated bonds of nitrogen from four pyrrole groups. These four subunit groups are separated evenly on the same plane, directly across from each other. They are also connected to each other by a C-CH3 methylene link on the other sides, by a C-H on one side and by two pyrroles directly coming together. Together, they form a perfect corrin ring as shown in figure 2. The fifth ligand connected to Cobalt is a nitrogen coming from the 5,6-dimethhylbenzimidazole. It presents itself as an axial running straight down from the cobalt right under the corrin ring. This benzimidazole is also connected to a five carbon sugar, which eventually attaches itself to a phosphate group, and then straps back to the rest of the structure. Since the axial is stretched all the way down, the bonding between the Cobalt and the 5,6-dimethylbenzimidazole is weak and can sometimes be replaced by related molecules such as a 5-hydrozyl-benzimidazole, an adenine, or any other similar group. In the sixth position above the Corrin ring, the active site of Cobalt can directly connect to several different types of ligands. It can connect to CN to form a Cyanocobalami, to a Methyl group to form a methylcobalamin, to a 5’-deoxy adenosy group to form an adenosylcobalamin, and OH, Hydroxycobalamin. Cobalt is always ready to oxidize from 1+ change into 2+ and 3+ in order to match up with these R groups that are connected to it. For example, Hydroxocobalamin contains cobalt that has a 3+ charge while Methyladenosyl contains a cobalt that has a 1+ Charge.
The point group configuration of Cobalamin is C4v. In order to determine this symmetry, one must see that the structure is able to rotate itself four times and will eventually arrive back to its original position. Furthermore, there are no sigma h plane and no perpendicular C2 axe. However, since there are sigma v planes that cut the molecules into even parts, it is clear to determine that the structure of Cobalamin is a C4v. With Cobalt being the center metal of the molecule, Cobalamin carried a distorted octahedral configuration. The axial that connects Cobalt to the 5,6 dimethyl benzimidazole is stretched all the way down to the bottom. Its distance is several times longer than the distance from the Cobalt and the attached R group above it. This sometimes can also be referred to as a tetragonal structure. The whole shape overall is similar to an octahedral, but the two axial groups are different and separated into uneven distances. Since there is only one metalloenzyme center in the system, the point group and configuration just mentioned is also assigned to the structure as a whole. Since the metallocoenzyme structure is stretched out, it is quite weakly coordinated and can be break apart or replaced with other groups as mentioned above.
Scientists have shown that both IR and Raman Spectroscopy were used to determine the structure of the molecule. This is determined by observing the character tables of point group C4v, the point group symmetry of Cobalamin. On the IR side, one can see that there are groups such as drz, (x, y), (rz, ry). On the other hand, on the Raman side, there are groups such as x square +y square, z square, x square – y square, xy, xz, yz. The Raman side indicated that there were stretching modes in the molecule and relates back to the stretching of the 5,6 dimethyl benzimidazole axial that connected directly below the Cobalt metal. The stretching can be seen in Figure 3.
Cobalamin enzymes can catalyze a few different types of reactions. One of them is the reaction of Intramolecular rearrangements. During this rearrangement coenzyme is exchanged to the two groups attached to adjacent carbon atoms. Another reaction involves transferring the methyl group in certain methylation reactions, such as the conversion of homocysteine to methionine, biosysnthesis of choline and thymine etc. These interactions can bring beneficial values to the biological bodies.
Cobalamin has many beneficial effects in regard to biological existences. They play a role to maintain healthy body system and help to aid the production of the body’s genetic materials. Cyanocobalamin, one type of cobalamin, works to generate the forming of red blood cells and heal many different damages in the nervous system. Cobalamin also serves as a vital role in the metabolism of fatty acids essential for the maintainence of myelin. Studies have shown that people with Vitamin B12 deficiency will reveal irregular destruction of the myeline shealth, which leads to parlysis and death. Some of the other symptoms of the lack of cobalamin are poor growth, megaloblastic bone marrow, Gi tract changes, Leucoopenia and hyper-segmented nutrophills, degenerative changes in spinal cord and nervous system and excretion of methyl malonic acid and homocystin in urine.
Throughout the years, Vitamin B12 has shown to be essential for the functioning of the nervous system and the production of red blood cell. A study conducted by researchers at the National Institutes of Health, Trinity College Dublin, suggested that a deficiency in Vitamin B12 might increase the risk of neural tubes defect in children (Miller). Therefore, by studying the structure and function of Cobalamin, scientists can experiment and form Vitamin B12 in their laboratories and serve the community as a whole. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.10%3A_Vitamin_%28B_12%29.txt |
Cobalamin, or Vitamin B12, is the largest and the most complex out of all the types of Vitamins. The discovery of Cobalamin was made as scientists were seeking to find a cure for pernicious anemia, an anemic disease caused by an absence of intrinsic factor in the stomach. Cobalamin was studied, purified, and collected into small red crystals, and its crystallize structure was determined during an X ray analysis experiment conducted by Scientist Hodkin. A molecule structure of Cobalamin is simple, yet contains a lot of different varieties and complexes as shown in Figure \(1\). The examination of the vitamin’s molecular structure helps scientists to have a better understanding of how the body utilizes Vitamin B12 into building red blood cells and preventing pernicious anemia syndromes.
The metalloenzyme structure of Cobalamin presents a corrin ring with Cobalt, the only metal in the molecule, positioned right in the center of the structure by four coordinated bonds of nitrogen from four pyrrole groups. These four subunit groups are separated evenly on the same plane, directly across from each other. They are also connected to each other by a C-CH3 methylene link on the other sides, by a C-H on one side and by two pyrroles directly coming together. Together, they form a perfect corrin ring as shown in figure 2. The fifth ligand connected to Cobalt is a nitrogen coming from the 5,6-dimethhylbenzimidazole. It presents itself as an axial running straight down from the cobalt right under the corrin ring. This benzimidazole is also connected to a five carbon sugar, which eventually attaches itself to a phosphate group, and then straps back to the rest of the structure. Since the axial is stretched all the way down, the bonding between the Cobalt and the 5,6-dimethylbenzimidazole is weak and can sometimes be replaced by related molecules such as a 5-hydrozyl-benzimidazole, an adenine, or any other similar group. In the sixth position above the Corrin ring, the active site of Cobalt can directly connect to several different types of ligands. It can connect to CN to form a Cyanocobalami, to a Methyl group to form a methylcobalamin, to a 5’-deoxy adenosy group to form an adenosylcobalamin, and OH, Hydroxycobalamin. Cobalt is always ready to oxidize from 1+ change into 2+ and 3+ in order to match up with these R groups that are connected to it. For example, Hydroxocobalamin contains cobalt that has a 3+ charge while Methyladenosyl contains a cobalt that has a 1+ Charge.
The point group configuration of Cobalamin is C4v. In order to determine this symmetry, one must see that the structure is able to rotate itself four times and will eventually arrive back to its original position. Furthermore, there are no sigma h plane and no perpendicular C2 axe. However, since there are sigma v planes that cut the molecules into even parts, it is clear to determine that the structure of Cobalamin is a C4v. With Cobalt being the center metal of the molecule, Cobalamin carried a distorted octahedral configuration. The axial that connects Cobalt to the 5,6 dimethyl benzimidazole is stretched all the way down to the bottom. Its distance is several times longer than the distance from the Cobalt and the attached R group above it. This sometimes can also be referred to as a tetragonal structure. The whole shape overall is similar to an octahedral, but the two axial groups are different and separated into uneven distances. Since there is only one metalloenzyme center in the system, the point group and configuration just mentioned is also assigned to the structure as a whole. Since the metallocoenzyme structure is stretched out, it is quite weakly coordinated and can be break apart or replaced with other groups as mentioned above.
Scientists have shown that both IR and Raman Spectroscopy were used to determine the structure of the molecule. This is determined by observing the character tables of point group C4v, the point group symmetry of Cobalamin. On the IR side, one can see that there are groups such as drz, (x, y), (rz, ry). On the other hand, on the Raman side, there are groups such as x square +y square, z square, x square – y square, xy, xz, yz. The Raman side indicated that there were stretching modes in the molecule and relates back to the stretching of the 5,6 dimethyl benzimidazole axial that connected directly below the Cobalt metal. The stretching can be seen in Figure 3.
Cobalamin enzymes can catalyze a few different types of reactions. One of them is the reaction of Intramolecular rearrangements. During this rearrangement coenzyme is exchanged to the two groups attached to adjacent carbon atoms. Another reaction involves transferring the methyl group in certain methylation reactions, such as the conversion of homocysteine to methionine, biosysnthesis of choline and thymine etc. These interactions can bring beneficial values to the biological bodies.
Cobalamin has many beneficial effects in regard to biological existences. They play a role to maintain healthy body system and help to aid the production of the body’s genetic materials. Cyanocobalamin, one type of cobalamin, works to generate the forming of red blood cells and heal many different damages in the nervous system. Cobalamin also serves as a vital role in the metabolism of fatty acids essential for the maintainence of myelin. Studies have shown that people with Vitamin B12 deficiency will reveal irregular destruction of the myeline shealth, which leads to parlysis and death. Some of the other symptoms of the lack of cobalamin are poor growth, megaloblastic bone marrow, Gi tract changes, Leucoopenia and hyper-segmented nutrophills, degenerative changes in spinal cord and nervous system and excretion of methyl malonic acid and homocystin in urine.
Throughout the years, Vitamin B12 has shown to be essential for the functioning of the nervous system and the production of red blood cell. A study conducted by researchers at the National Institutes of Health, Trinity College Dublin, suggested that a deficiency in Vitamin B12 might increase the risk of neural tubes defect in children (Miller). Therefore, by studying the structure and function of Cobalamin, scientists can experiment and form Vitamin B12 in their laboratories and serve the community as a whole. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.10%3A_Vitamin_(B_12).txt |
The use of food by organisms is termed nutrition. Vitamins and minerals necessary for biochemical processes. There are three general categories of food: (1) Essential fiber which are non-digestible polysaccharide material, essential for normal functioning of animal digestive systems (i.e. colon), (2) Energy-yielding nutrients which are protein, carbohydrate and lipid and (3) Micronutrients.
Protein
Animals are unable to synthesize certain amino acids (humans can only make 10 of the 20 common amino acids). The amino acids that an animal is unable to synthesize must be obtained from the diet (i.e. by consuming plants or microorganisms), and these amino acids are termed "essential amino acids".
Excess dietary protein becomes a source of metabolic energy
• Glucogenic amino acids: can be converted into glucose
• Ketogenic amino acids: can be converted into fatty acids or keto acids
• If plenty of fats and carbohydrates are available, then the glucogenic and ketogenic amino acids from excess dietary protein are converted to triacylglycerol and stored as fat
Protein is an important source of nitrogen in the diet. Protein within the body is constantly turning over (i.e. being degraded and resynthesized). Furthermore, there is a general demand for protein synthesis when an organism is growing. The Nitrogen Balance refers to the relationship between the supply and demand for nitrogen (i.e. protein) within an organism.
• A positive nitrogen balance means that the organism is taking in more protein that it needs for growth or turnover
• A negative nitrogen balance means that the organism is not getting enough protein for its normal turnover, or growth. This would represent a nutritional deficiency of protein
Carbohydrate
Carbohydrates are also an essential structural component of nucleic acids, nucleotides, glycoproteins and glycolipids. However, the principle role of carbohydrate in the diet is production of metabolic energy.
• Simple sugars are metabolized in the glycolytic pathway to release energy
• Complex carbohydrates are degraded into simple sugars, which then enter the glycolytic pathway
• Metabolism can make use of a wide variety of sugars for energy production. However, the brain relies solely on glucose for an energy source
• When dietary carbohydrate exceeds the supply needed for energy requirements, it is converted to glycogen and triacylglycerols for storage
• When dietary carbohydrate is in short supply, ketone bodies are formed from acetate units to provide fuel for the brain
Lipids
Fatty acids and triacylglycerols can be used as fuel by many tissues in the human body. Phospholipids are essential components of all biological membranes
• Excess dietary fat is stored as triacylglycerols in adipose tissue
• A deficiency of dietary fat is problematic because some fatty acids cannot be synthesized by the human body, and must be obtained through diet. These are termed
• essential fatty acids
• The human body cannot synthesize linoleic, linolenic or arachidonic fatty acids. These are key components of biological membranes, and arachidonic acid is a precursor of prostaglandins (an important class of hormones). These are therefore considered essential fatty acids
Fiber
"Dietary fiber" refers to molecules that cannot be broken down by enzymes in the human body.
• Cellulose (polysaccharide component of plant cell walls). Required for proper function of colon.
• Lignins (plant polymer of aromatic ring structures). Absorbs organic molecules in the digestive system (binds cholesterol).
Vitamins and minerals
Vitamins are essential nutrients that are required in the diet because they cannot be synthesized by human metabolic enzymes. Often, only trace levels are required, but a shortage can result in disease or death.
• A common categorization of human vitamins is whether they are
• water soluble or fat soluble compounds.
Coenzymes are low molecular weight molecules that provide unique chemical functionalities for certain enzyme/coenzyme complexes.
• Coenzymes may act as carriers of specific functional groups (e.g. methyl or acyl groups)
• They can provide chemically reactive groups that the common 20 amino acid side chains cannot provide
• Coenzymes are
• usually modified in the course of a reaction, and subsequently chemically regenerated back to their useful active form. Thus, this recycling of coenzymes means that only small concentrations are required.
• All the
• water soluble vitamins (with the exception of vitamin C) are coenzymes or precursors of coenzymes.
Summary of water soluble and fat soluble vitamins:
Common name
Chemical name
Related cofactor(s)
Water Soluble Vitamins
Vitamin B1
Thiamine
Thiamine pyrophosphate
Vitamin B2
Riboflavin
Flavin adenine dinucleotide (FAD)
Flavin mononucleotide (FMN)
Vitamin B6
Pyridoxal, pyridoxine, pyridoxamine
Pyridoxal phosphate
Vitamin B12
Cobalamin
5'-deoxyadenosylcobalamin
Methylcobalamin
Niacin
Nicotinic acid
Nicotinamide adenine dinucleotide (NAD+)
Nicotinamide adenine dinucleotide phosphate (NADP+)
Vitamin B3
Pantothenic acid
Coenzyme A
Biotin
Biotin-lysine conjugates (biocytin)
Lipoic acid
Lipoyl-lysine conjugates (lipoamide)
Folic acid
Tetrahydrofolate
Vitamin C
L-ascorbate
Fat Soluble Vitamins
Vitamin A
Retinol
Vitamin D2
Ergocalciferol
Vitamin D3
Cholecalciferol
Vitamin E
a-Tocopherol
Vitamin K
Vitamin B1: Thiamine and thiamine pyrophosphate
Thiamine is the precursor of thiamine pyrophosphate (TPP):
• TPP is a coenzyme
• It is a coenzyme for certain enzymes involved in carbohydrate metabolism
• It catalyzes the synthesis or cleavage of bonds to carbonyl carbons
Niacin (Nicotinic Acid) and nicotinamide coenzymes
Nicotinamide is an essential part of two important coenzymes: nicotinamide adenine dinucleotide (NAD+) and nicotinamide adenine dinucleotide phosphate (NADP+).
• The reduced forms of these coenzymes are NADH and NADPH
• The coenzymes participate in
• redox reactions via the direct transfer of hydride (H-) ions either to or from the cofactor and a substrate. The hydride transfer carries two electrons along with it (a proton transfer in acid/base catalysis carries no electrons)
• The hydride transfer involves the C4 carbon of the nicotinamide ring. The quaternary amine of the nicotinamide ring acts as an electron sink to promote acceptance of a hydride ion, or to facilitate leaving of a hydride ion.
• Enzymes that are involved in such redox reactions are called
• dehydrogenases
• The nucleotide part of the molecule does not enter into any chemistry, but
• is important for recognition and binding to enzymes that will use FMN or FAD as a cofactor.
Vitamin B2: Riboflavin
Riboflavin is a constituent of riboflavin 5'-phosphate (flavin mononucleotide, or FMN) and flavin adenine dinucleotide (FAD). The nucleotide part of the molecule does not enter into any chemistry, but is important for recognition and binding to enzymes that will use FMN or FAD as a cofactor.
The isoalloxazine ring is the core structure of the different flavin molecules. It is yellow in color and the word "flavin" is derived from the latin word for yellow, flavus.
• Flavin coenzymes can exists in three different redox states, and each state has a different color (the reduced form is colorless)
• Flavin molecules can participate in both
• one- and two- electron transfer reactions
Vitamin B3: pantothenic acid and coenzyme A
Pantothenic acid is a component of coenzyme A (CoA). The two main functions of CoA are:
a-hydrogen of the acyl group for removal as a proton
1. Activation of acyl groups (R-COX) for transfer to nucleophilic acceptors
2. Activation of the
Both of these functions involve the reactive sulfhydryl group through the formation of thioester linkages with acyl groups
• The 4-phosphopantetheine part of CoA is also used in the same way in
• acyl carrier proteins (ACP's) involved in fatty acid biosynthesis
Vitamin B6: Pyridoxine and pyridoxal phosphate
The biologically active form of vitamin B6 is pyridoxal-5-phosphate (PLP), however, the nutritional requirements can be met by either pyridoxine, pyridoxal or pyridoxol.
PLP participates in a wide variety of reactions involving amino acids, including:
• Transamination
• a- and b-decarboxylation
• b- and g- elimination (not to be confused with painful elimination)
• Racemization
• Aldol reactions
These involve bonds to the amino acid Ca as well as side chain carbons. The wide variety of reactions is due to the ability of PLP to form stable Schiff base adducts with a-amino groups of amino acids:
• In PLP-dependent enzymes, the PLP is present in a Schiff base linkage with the
• e-amino group of an acitve site lysine
• Rearrangement to a Schiff base with the arriving amino acid substrate is a
• transaldiminization reaction
Vitamin B12: Cyanocobalamin
Vitamin B12 is not made by any animal or plant, it is produced by only a few species of bacteria. Once in the food chain, vitamin B12 is obtained by animals by eating other animals, but plants are sadly deficient. Therefore, herbivorous animals (and vegetarians) can suffer a deficiency. The structure contains a cobalt ion, coordinated within a corrin ring structure:
Vitamin B12 (cyanocobalamin) is converted in the body into two coenzymes:
1. 5'-deoxyadenosylcobalamin (the predominant form)
2. Methylcobalamin
Vitamin B12 coenzymes participate in three types of reactions:
1. Intramolecular rearrangements
2. Reductions of ribonucleotides to deoxyribonucleotides in certain bacteria
3. Methyl group transfers (these use methylcobalamin for this purpose)
Vitamin C (L-Ascorbate)
L-Ascorbate is a reducing sugar (has a reactive ene-diol structure) that is involved in the following biochemical processes:
• Hydroxylation of proline and lysine residues in collagen. Without these post-translational modifications the triple helix of collagen is unstable and connective tissue loses its integrity. This is the problem in the disease known as
• scurvy.
• Mobilization of iron, stimulation of immune system, anti-oxidant for scavenging of reactive free-radicals.
Almost all animals can synthesize vitamin C (its in the pathway of carbohydrate synthesis). Humans and great apes have suffered a mutation in the last enzyme in the pathway of synthesis for L-ascorbate (mutation occurred about 10-40 million years ago). Since that time, all great apes (of which humans are a member) must get L-ascorbate from their diet (fresh fruits and vegetable contain an abundance). Thus, for the great apes L-ascorbate is a "vitamin" (another way of looking at it is that all great apes suffer an in-born error in metabolism). Humans still have the gene for the enzyme to make vitamin C. However, it has suffered a couple of deletions that introduce a frame shift mutation, in addition to numerous point mutations.
Biotin
Biotin acts as a mobile carboxyl group carrier in a variety of enzymatic carboxylation reactions.
• Synthesized by intestinal bacteria (finaly, they do something for you)
• Biotin is bound covalently to the enzyme as a prosthetic group via an
• e-amino group of a lysine residue in the enzyme
• This biotin-lysine conjugated amino acid is termed a "biocytin" residue
• The lysine side-chain acts as a flexible "tether" for the biotin, and this flexibility allows the transfer of carboxylate groups within the enzyme
• It is the carrier for the most oxidized form of carbon
• - CO2 (using bicarbonate as the carboxylating agent). The carbon dioxide binds as a carboxy group to one of the ring nitrogens in the biotin
Lipoic Acid
Lipoic acid contains two sulfur atoms that can exist as a disulfide bonded pair, or as two free sulfhydrils. Conversion between the two forms involves a redox reaction (the two free sulfhydrils represent the reduced form). Lipoic acid is typically found covalently attached to a lysine side chain in enzymes that use it as a cofactor, as a lipoamide complex.
• Lipoic acid is an
• acyl group carrier (R-CO-X)
• It also functions to transfer electrons during oxidation and decarboxylation of
• a-keto acids
• Pyruvate dehydrogenase and
• a-ketoglutarate dehydrogenase use lipoic acid as a cofactor
• Its not clear whether a dietary deficiency of lipoic acid contributes to a disease state, so its not technically considered a vitamin
Folic Acid
Folic acid derivatives (i.e. "folates") are acceptors and donors of one-carbon units for all oxidation levels of carbon (except for the most oxidized form - CO2. See biotin above).
• The active coenzyme form is tetrahydrofolate (THF). Folate is reduced to THF by the action of tetrahydrofolate reductase.
• Three different oxidation states of carbon can bind to THF. These are oxidation states of -2 (methanol group), 0 (formaldehyde group) and +2 (formate group)
• These groups are attached to the THF molecule at either the N5 or N10 atom positions
• The biosynthetic pathways of
• methionine, homocysteine, purines, and thymine rely one one-carbon units being provided by THF.
Vitamin A: Retinol
Vitamin A occurs as an ester (Retinyl ester), aldehyde (Retinal) or acidic form (Retinoic acid).
• It is a fat soluble vitamin, and is synthesized from isoprene building blocks
• Obtained directly from an animal diet, or synthesized from
• b-carotene provided by plants
• It is essential to vision. Retinol transported to the eyes is oxidized by retinol dehydrogenase to produce trans-retinal. Trans-retinal is converted to 11-cis-retinal by retinal isomerase. The aldehyde group of retinal forms a Schiff base with a lysin of the protein opsin, to form rhodopsin (the light-sensitive pigment of vision).
• Vitamin A is essential for various biological processes - including fetal development and sperm development. But excessive vitamin A is toxic.
Vitamin D: Ergocalciferol (D2) and cholecalciferol (D3)
Cholecalciferol is produced in the skin of animals by the action of U.V. light on the precursor molecule 7-dehydrocholesterol.
• Light energy induces bond-breakage (between carbons 9 and 10) and formation of previtamin D3. Spontaneous isomerization produces D3
• Ergocalciferol is produced by the action of U.V. light on the plant sterol ergosterol.
• Since humans can produce D3 from 7-dehydrocholesterol, vitamin D3 is technically not really a vitamin
• Cholecalciferol is really a prohormone. Derivatives of this compound
• regulate calcium and phosphate metabolism.
• Inadequate intestinal absorption of calcium and phosphate can result in demineralization of bones, and the disease Rickets.
Vitamin E: Tocopherol
a-Tocopherol is a potent antioxidant, however, molecular details of its function are not clearly understood.
• Fatty acids in membranes are susceptible to oxidative damage
• Vitamin E is fat soluble and may protect membrane fatty acids from oxidation
• A deficiency of vitamin E results in red blood cells that are susceptible to oxidative damage
• Retinal damage in premature infants, due to supplemental oxygen, may be preventable by administering vitamin E
Vitamin K: Napthoquinone
Vitamin K is essential to the blood-clotting process. Vitamin K is required for the post-translational modification to produce g-carboxy glutamic acid from glutamic acid. Such modified residues can bind Ca2+, which is an essential part of the process in the clotting cascade.
g-carboxy glutamic acids in their structure
• Prothrombin ("factor II"), and factors VII, IX and X are serine proteases that participate in a protease activation cascade that is involved in blood coagulation, and have
Contributors and Attributions
• Dr Michael Blaber | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.11%3A_Folic_Acid.txt |
The essential dietary substances called vitamins are commonly classified as "water soluble" or "fat soluble". Water soluble vitamins, such as vitamin C, are rapidly eliminated from the body and their dietary levels need to be relatively high. The recommended daily allotment (RDA) of vitamin C is 100 mg, and amounts as large as 2 to 3 g are taken by many people without adverse effects. The lipid soluble vitamins, shown in the diagram below, are not as easily eliminated and may accumulate to toxic levels if consumed in large quantity. The RDA for these vitamins are:
• Vitamin A 800 μg ( upper limit ca. 3000 μg)
• Vitamin D 5 to 10 μg ( upper limit ca. 2000 μg)
• Vitamin E 15 mg ( upper limit ca. 1 g)
• Vitamin K 110 μg ( upper limit not specified)
From this data it is clear that vitamins A and D, while essential to good health in proper amounts, can be very toxic. Vitamin D, for example, is used as a rat poison, and in equal weight is more than 100 times as poisonous as sodium cyanide.
From the structures shown here, it should be clear that these compounds have more than a solubility connection with lipids. Vitamins A is a terpene, and vitamins E and K have long terpene chains attached to an aromatic moiety. The structure of vitamin D can be described as a steroid in which ring B is cut open and the remaining three rings remain unchanged. The precursors of vitamins A and D have been identified as the tetraterpene beta-carotene and the steroid ergosterol, respectively.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
17.12: Vitamin (K)
The essential dietary substances called vitamins are commonly classified as "water soluble" or "fat soluble". Water soluble vitamins, such as vitamin C, are rapidly eliminated from the body and their dietary levels need to be relatively high. The recommended daily allotment (RDA) of vitamin C is 100 mg, and amounts as large as 2 to 3 g are taken by many people without adverse effects. The lipid soluble vitamins, shown in the diagram below, are not as easily eliminated and may accumulate to toxic levels if consumed in large quantity. The RDA for these vitamins are:
• Vitamin A 800 μg ( upper limit ca. 3000 μg)
• Vitamin D 5 to 10 μg ( upper limit ca. 2000 μg)
• Vitamin E 15 mg ( upper limit ca. 1 g)
• Vitamin K 110 μg ( upper limit not specified)
From this data it is clear that vitamins A and D, while essential to good health in proper amounts, can be very toxic. Vitamin D, for example, is used as a rat poison, and in equal weight is more than 100 times as poisonous as sodium cyanide.
From the structures shown here, it should be clear that these compounds have more than a solubility connection with lipids. Vitamins A is a terpene, and vitamins E and K have long terpene chains attached to an aromatic moiety. The structure of vitamin D can be described as a steroid in which ring B is cut open and the remaining three rings remain unchanged. The precursors of vitamins A and D have been identified as the tetraterpene beta-carotene and the steroid ergosterol, respectively.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/17%3A_How_Enzymes_Catalyze_Reactions_The_Organic_Chemisty_of_Vitamins/17.12%3A_Vitamin_%28K%29.txt |
Homework Problems
Homework Solutions
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18: The Organic Chemistry of Metabolic Pathways
10.1A: Nomenclature and abbreviations
Phosphoryl groups are derivatives of phosphoric acid, a strong acid that is commonly used in the laboratory. The fully deprotonated conjugate base of phosphoric acid is called a phosphate ion, or inorganic phosphate (often abbreviated 'Pi'). When two phosphate groups are linked to each other, the linkage is referred to as a 'phosphate anhydride', and the ion is called 'inorganic pyrophosphate' (abbreviation PPi).
When a phosphate ion is attached to a carbon atom on an organic molecule, the chemical linkage is referred to as a phosphate ester, and the whole species is called an organic monophosphate. Glucose-6-phosphate is an example.
If an organic molecule is linked to two or three phosphate groups, the resulting species are called organic diphosphates and organic triphosphates.
Isopententyl diphosphate and adenosine triphosphate (ATP) are good examples:
Oxygen atoms in phosphate groups are referred to either 'bridging' and 'non-bridging', depending on their position. An organic diphosphate has two bridging and five non-bridging oxygens.
When a single phosphate is linked to two organic groups, the term 'phosphate diester' is used. The backbone of DNA is composed of phosphate diesters.
The term 'phosphoryl group' is a general way to refer to all of the phosphate-based groups mentioned in the paragraphs above.
Recall (section 1.4A) that phosphate groups on organic structures are sometimes abbreviated simply as 'P', a convention that we will use throughout this text. For example, glucose-6-phosphate and isopentenyl diphosphate are often depicted as shown below. Notice that the 'P' abbreviation includes the oxygen atoms and negative charges associated with the phosphate groups.
10.1B: Acid constants and protonation states
Phosphoric acid is triprotic, meaning that it has three acidic hydrogens available to donate, with pKa values of 2.1, 7.2, and 12.3, respectively. These acid constant values tell us that, at the physiological pH of approximately 7.3, all phosphoric acid species in solution will have donated at least one proton, and more than half will have donated two, meaning that the average charge on the phosphate ion is slightly higher than -1.5. Organic monophosphates, diphosphates, and triphosphates are predominantly deprotonated at pH 7.3, meaning that they carry charges of slightly less than -2, -3, and -4, respectively. By convention, they are usually drawn in their fully deprotonated states.
10.1C: Bonding in phosphines and phosphates
Looking at the location of phosphorus on the periodic table, you might expect it to bond and react in a fashion similar to nitrogen, which is located just above it in the fifth column. Indeed, phosphines - phosphorus analogs of amines - are commonly used in the organic laboratory.
Just like in amines, the central phosphorus atom in a phosphine is sp3 hybridized, with a lone pair of electrons occupying one of the four sp3 hybrid orbitals. In the case of phosphines, however, the sp3 orbitals are hybrids of a single 3s orbital and three 3p orbitals, rather than 2s and 2p orbitals as in amines.
In biological molecules, the most important form of phosphorus is not phosphine, but the phosphoryl group. The four oxygen substituents in phosphoryl groups are arranged about the central phosphorus atom with tetrahedral geometry, however there are a total of five bonds to phosphorus - four s igmabonds and one delocalized pi bond. The reason that phosphorus can break the 'octet rule' is that it is on the third row of the periodic table, and thus has d orbitals available for bonding. In the hybrid orbital picture for phosphate ion (PO4 3-), a single 3s and three 3p orbitals combine to form four sp3 hybrid orbitals with tetrahedral geometry. Four of the five valance electrons on phosphorus occupy sp3 orbitals, and the fifth occupies an unhybridized d orbital.
The phosphorus is thus able to form five bonding interactions, rather than three as is phosphine. The four sp3 orbitals on phosphorus are each able to overlap with an sp2 orbital on an oxygen atom (forming a tetrahedral framework), while a delocalized fifth bond (a pi bond) is formed by side-by-side overlap of the d orbital on phosphorus with 2p orbitals on the oxygens.
The -3 charge on a fully deprotonated phosphate ion is spread evenly over the four oxygens, and each phosphorus-oxygen bond can be considered have some double bond character.
In phosphate esters, diesters, and anhydrides the pi bonding is delocalized primarily over the non-bridging bonds, while the bridging bonds have mainly single-bond character. In a phosphate diester, for example, the two non-bridging oxygens share a -1 charge, as illustrated by the two major resonance contributors. In the resonance contributors in which the bridging oxygens are shown as double bonds, there is an additional separation of charge - thus these contributors are minor and make a relatively unimportant contribution to the overall bonding picture.
Template:examplestart
Exercise 10.1: Draw all of the resonance structures showing the delocalization of charge on a (fully deprotonated) organic monophosphate. If a 'bond order' of 1.0 is a single bond, and a bond order of 2.0 is a double bond, what is the approximate bond order of bridging and non-bridging P-O bonds?
Solution
Template:exampleEnd
Throughout this chapter, phosphoryl groups are usually drawn without attempting to show tetrahedral geometry, and π bonds and negative charges will often be shown as localized to one position. This is done for the sake of simplification - however it is important always to remember that the phosphoryl group is really tetrahedral, and that the pi electrons are delocalized over the non-bridging bonds.
10.1D: Phosphoryl transfer reactions - the general picture
In a phosphoryl transfer reaction, a phosphoryl group is transferred from a phosphoryl group donor molecule (designated 'R2' in the figure below) to a phosphoryl group acceptor (designated 'R1'). The figure below illustrates the transfer of a single phosphate group from R2 to R1:
An alcohol, for example, can be transformed into an organic monophosphate by accepting one of the phosphate groups of ATP (we'll learn more about ATP soon).
An organic monophosphate is converted back to an alcohol when it transfers its phosphate group to a water molecule:
In the course of this chapter, we will see many other types of phosphoryl transfer reactions, but they all can be described by essentially the same mechanism.
Let's look more closely at the phosphoryl transfer reaction mechanism, using as an example the transfer of a phosphate group from adenosine triphosphate (ATP) to the C6 hydroxyl group of glucose, a reaction catalyzed by glucose kinase.
To simplify things in the mechanistic discussion to come, we'll abbreviate this reaction as:
One very important aspect of biological phosphoryl transfer reactions is that the electrophilicity of the phosphorus atom is enhanced by the Lewis acid (electron-accepting) effect of one or more magnesium ions. Phosphoryl transfer-catalyzing enzymes bind Mg2+ ions in such a way that they can interact with non-bridging phosphoryl oxygens on the substrate.
The positively charged metal ions stabilize negative charge on the oxygen atoms, which has the effect of increasing the dipole moment of the phosphorus-oxygen bond. The phosphorus therefore has a larger partial positive charge, which makes it a better electrophile.
A phosphoryl transfer reaction is very much like a SN2 reaction at a carbon. Just like in an SN2 reaction, the nucleophile in a phosphoryl transfer approaches the electrophilic center from the backside, opposite the leaving group. As the nucleophile gets closer and the leaving group begins its departure, the bonding geometry at the phosphorus atom changes from tetrahedral to trigonal bipyramidal at the transition state. As the phosphorus-nucleophile bond gets shorter and the phosphorus-leaving group bond grows longer, the bonding picture around the phosphorus atom returns to its original tetrahedral state, but the stereochemical configuration has been 'flipped', or inverted.
In the trigonal bipyramidal transition state, the five substituents are not equivalent: the three non-bridging oxygens are said to be equatorial (forming the base of a trigonal bipyramid), while the nucleophile and the leaving group are said to be apical (occupying the tips of the two pyramids).
Although stereochemical inversion in phosphoryl transfer reactions is predicted by theory, the fact that phosphoryl groups are achiral made it impossible to observe the phenomenon directly until 1978, when a group of researchers was able to synthesize organic phosphate esters in which stable oxygen isotopes 17O and 18O were specifically incorporated. This created a chiral phosphate center.
Subsequent experiments with phosphoryl transfer-catalyzing enzymes confirmed that these reactions proceed with stereochemical inversion. (Nature 1978 275, 564; Ann Rev Biochem 1980 49, 877).
10.1E: Phosphoryl transfer reactions - concerted, addition-elimination, or dissociative?
In the above discussion, the phosphoryl transfer reaction mechanism was depicted as passing through a concerted SN2-like transition state, with both apical bonds in some stage of breaking or forming at the top of the 'energy hill':
This is not the only mechanism that has been proposed for these reactions - in fact, two other possible mechanisms have been suggested.
In an alternative two-step mechanistic model, the nucleophile could attack first, forming a pentavalent, trigonal bipyramidal intermediate, (as apposed to a pentavalent transition state). The reaction is completed when the leaving group is expelled. The intermediate species would occupy an energy valley between the two transition states.
This is often referred to as an 'addition-elimination' mechanism - the nucleophile adds to the phosphate first, forming a pentavalent intermediate, and then the leaving group is eliminated.
A pentavalent intermediate is not possible for an SN2 reaction at a carbon center, because carbon, as a second-row element, does not have any d orbitals and cannot form five bonds. Phosphorus, on the other hand, is a third-row element and is quite capable of forming more than four bonds. Phosphorus pentachloride, after all, is a stable compound that has five bonds to chlorine arranged in trigonal bipyramidal geometry around the central phosphorus.
The phosphorus atom in PCl5 (and in the hypothetical pentavalent intermediate pictured above) is considered to be sp3d hybridized:
There is a third possibility: the reaction could proceed in a dissociative, SN1-like manner. In this model, the phosphorus-leaving group bond breaks first, resulting in a 'metaphosphate' intermediate. This intermediate, which corresponds to the carbocation intermediate in an SN1 reaction, is then attacked by the nucleophile to form the reaction product.
So what is the actual mechanism for a phosphoryl transfer reaction - concerted, addition-elimination, or dissociative? Chemists love to investigate and argue about questions like this! Just like with the SN1/SN2 argument discussed in the previous chapter, it really boils down to one question. Which happens first, bond-forming or bond-breaking - or do these two events occur at the same time? From the evidence accumulated to date, it appears that enzymatic phosphoryl transfer reactions may occur by all three mechanisms - and often somewhere in between - depending on the nature of the nucleophile, the electrophile, and the leaving group, as well as on the active-site architecture of the enzyme catalyzing the reaction. Although it is thought that many phosphoryl transfer reactions, both enzymatic and non-enzymatic, proceed with some degree of dissociative (SN1-like) character, there is not yet a clear understanding of exactly what happens between starting compound and product. Considering the importance of phosphoryl transfer reactions in metabolic pathways, this area is clearly a very promising one for further investigation. (FASEB J. 1995 9, 1585; Trends Biochem Sci. 2004 29, 495).
For the sake of simplicity and clarity, phosphoryl transfers in this text will generally be depicted as concerted, SN2-like reactions, in two dimensions, with a localized double bond. but you should be always keep in mind the existence of a pentavalent, trigonal bipyramidal transition state/intermediate. Also, be aware that in other books and articles these reactions may be drawn somewhat differently.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/18%3A_The_Organic_Chemistry_of_Metabolic_Pathways/18.2%3A_ATP_and_Phosphoryl_Transfer_Reactions.txt |
Learning Objectives
• To describe the reactions needed to completely oxidize a fatty acid to carbon dioxide and water.
Like glucose, the fatty acids released in the digestion of triglycerides and other lipids are broken down in a series of sequential reactions accompanied by the gradual release of usable energy. Some of these reactions are oxidative and require nicotinamide adenine dinucleotide (NAD+) and flavin adenine dinucleotide (FAD). The enzymes that participate in fatty acid catabolism are located in the mitochondria, along with the enzymes of the citric acid cycle, the electron transport chain, and oxidative phosphorylation. This localization of enzymes in the mitochondria is of the utmost importance because it facilitates efficient utilization of energy stored in fatty acids and other molecules.
Fatty acid oxidation is initiated on the outer mitochondrial membrane. There the fatty acids, which like carbohydrates are relatively inert, must first be activated by conversion to an energy-rich fatty acid derivative of coenzyme A called fatty acyl-coenzyme A (CoA). The activation is catalyzed by acyl-CoA synthetase. For each molecule of fatty acid activated, one molecule of coenzyme A and one molecule of adenosine triphosphate (ATP) are used, equaling a net utilization of the two high-energy bonds in one ATP molecule (which is therefore converted to adenosine monophosphate [AMP] rather than adenosine diphosphate [ADP]):
The fatty acyl-CoA diffuses to the inner mitochondrial membrane, where it combines with a carrier molecule known as carnitine in a reaction catalyzed by carnitine acyltransferase. The acyl-carnitine derivative is transported into the mitochondrial matrix and converted back to the fatty acyl-CoA.
Steps in the β-Oxidation of Fatty Acids
Further oxidation of the fatty acyl-CoA occurs in the mitochondrial matrix via a sequence of four reactions known collectively as β-oxidation because the β-carbon undergoes successive oxidations in the progressive removal of two carbon atoms from the carboxyl end of the fatty acyl-CoA (Figure $1$).
The first step in the catabolism of fatty acids is the formation of an alkene in an oxidation reaction catalyzed by acyl-CoA dehydrogenase. In this reaction, the coenzyme FAD accepts two hydrogen atoms from the acyl-CoA, one from the α-carbon and one from the β-carbon, forming reduced flavin adenine dinucleotide (FADH2).
Next, the trans-alkene is hydrated to form a secondary alcohol in a reaction catalyzed by enoyl-CoA hydratase. The enzyme forms only the L-isomer.
The secondary alcohol is then oxidized to a ketone by β-hydroxyacyl-CoA dehydrogenase, with NAD+ acting as the oxidizing agent. The reoxidation of each molecule of NADH to NAD+ by the electron transport chain furnishes 2.5–3 molecules of ATP.
The final reaction is cleavage of the β-ketoacyl-CoA by a molecule of coenzyme A. The products are acetyl-CoA and a fatty acyl-CoA that has been shortened by two carbon atoms. The reaction is catalyzed by thiolase.
The shortened fatty acyl-CoA is then degraded by repetitions of these four steps, each time releasing a molecule of acetyl-CoA. The overall equation for the β-oxidation of palmitoyl-CoA (16 carbon atoms) is as follows:
Because each shortened fatty acyl-CoA cycles back to the beginning of the pathway, β-oxidation is sometimes referred to as the fatty acid spiral.
The fate of the acetyl-CoA obtained from fatty acid oxidation depends on the needs of an organism. It may enter the citric acid cycle and be oxidized to produce energy, it may be used for the formation of water-soluble derivatives known as ketone bodies, or it may serve as the starting material for the synthesis of fatty acids. For more information about the citric acid cycle, see Section 20.4.
Looking Closer: Ketone Bodies
In the liver, most of the acetyl-CoA obtained from fatty acid oxidation is oxidized by the citric acid cycle. However, some of the acetyl-CoA is used to synthesize a group of compounds known as ketone bodies: acetoacetate, β-hydroxybutyrate, and acetone. Two acetyl-CoA molecules combine, in a reversal of the final step of β-oxidation, to produce acetoacetyl-CoA. The acetoacetyl-CoA reacts with another molecule of acetyl-CoA and water to form β-hydroxy-β-methylglutaryl-CoA, which is then cleaved to acetoacetate and acetyl-CoA. Most of the acetoacetate is reduced to β-hydroxybutyrate, while a small amount is decarboxylated to carbon dioxide and acetone.
The acetoacetate and β-hydroxybutyrate synthesized by the liver are released into the blood for use as a metabolic fuel (to be converted back to acetyl-CoA) by other tissues, particularly the kidney and the heart. Thus, during prolonged starvation, ketone bodies provide about 70% of the energy requirements of the brain. Under normal conditions, the kidneys excrete about 20 mg of ketone bodies each day, and the blood levels are maintained at about 1 mg of ketone bodies per 100 mL of blood.
In starvation, diabetes mellitus, and certain other physiological conditions in which cells do not receive sufficient amounts of carbohydrate, the rate of fatty acid oxidation increases to provide energy. This leads to an increase in the concentration of acetyl-CoA. The increased acetyl-CoA cannot be oxidized by the citric acid cycle because of a decrease in the concentration of oxaloacetate, which is diverted to glucose synthesis. In response, the rate of ketone body formation in the liver increases further, to a level much higher than can be used by other tissues. The excess ketone bodies accumulate in the blood and the urine, a condition referred to as ketosis. When the acetone in the blood reaches the lungs, its volatility causes it to be expelled in the breath. The sweet smell of acetone, a characteristic of ketosis, is frequently noticed on the breath of severely diabetic patients.
Because two of the three kinds of ketone bodies are weak acids, their presence in the blood in excessive amounts overwhelms the blood buffers and causes a marked decrease in blood pH (to 6.9 from a normal value of 7.4). This decrease in pH leads to a serious condition known as acidosis. One of the effects of acidosis is a decrease in the ability of hemoglobin to transport oxygen in the blood. In moderate to severe acidosis, breathing becomes labored and very painful. The body also loses fluids and becomes dehydrated as the kidneys attempt to get rid of the acids by eliminating large quantities of water. The lowered oxygen supply and dehydration lead to depression; even mild acidosis leads to lethargy, loss of appetite, and a generally run-down feeling. Untreated patients may go into a coma. At that point, prompt treatment is necessary if the person’s life is to be saved.
ATP Yield from Fatty Acid Oxidation
The amount of ATP obtained from fatty acid oxidation depends on the size of the fatty acid being oxidized. For our purposes here. we’ll study palmitic acid, a saturated fatty acid with 16 carbon atoms, as a typical fatty acid in the human diet. Calculating its energy yield provides a model for determining the ATP yield of all other fatty acids.
The breakdown by an organism of 1 mol of palmitic acid requires 1 mol of ATP (for activation) and forms 8 mol of acetyl-CoA. Recall from Table 20.4.1 that each mole of acetyl-CoA metabolized by the citric acid cycle yields 10 mol of ATP. The complete degradation of 1 mol of palmitic acid requires the β-oxidation reactions to be repeated seven times. Thus, 7 mol of NADH and 7 mol of FADH2 are produced. Reoxidation of these compounds through respiration yields 2.5–3 and 1.5–2 mol of ATP, respectively. The energy calculations can be summarized as follows:
ATP Yield from Fatty Acid Oxidation
1 mol of ATP is split to AMP and 2Pi −2 ATP
8 mol of acetyl-CoA formed (8 × 12) 96 ATP
7 mol of FADH2 formed (7 × 2) 14 ATP
7 mol of NADH formed (7 × 3) 21 ATP
Total 129 ATP
The number of times β-oxidation is repeated for a fatty acid containing n carbon atoms is n/2 – 1 because the final turn yields two acetyl-CoA molecules.
The combustion of 1 mol of palmitic acid releases a considerable amount of energy:
$C_{16}H_{32}O_2 + 23O_2 → 16CO_2 + 16H_2O + 2,340\; kcal \nonumber$
The percentage of this energy that is conserved by the cell in the form of ATP is as follows:
$\mathrm{\dfrac{energy\: conserved}{total\: energy\: available}\times100=\dfrac{(129\: ATP)(7.4\: kcal/ATP)}{2,340\: kcal}\times100=41\%} \nonumber$
The efficiency of fatty acid metabolism is comparable to that of carbohydrate metabolism, which we calculated to be 42%. For more information about the efficiency of fatty acid metabolism, see II of Carbohydrate Catabolism" data-cke-saved-href="/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20:_Energy_Metabolism/20.05:_Stage_II_of_Carbohydrate_Catabolism" href="/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20:_Energy_Metabolism/20.05:_Stage_II_of_Carbohydrate_Catabolism" data-quail-id="91">Section 20.5.
The oxidation of fatty acids produces large quantities of water. This water, which sustains migratory birds and animals (such as the camel) for long periods of time.
Summary
• Fatty acids, obtained from the breakdown of triglycerides and other lipids, are oxidized through a series of reactions known as β-oxidation.
• In each round of β-oxidation, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH2 are produced.
• The acetyl-CoA, NADH, and FADH2 are used in the citric acid cycle, the electron transport chain, and oxidative phosphorylation to produce ATP. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/18%3A_The_Organic_Chemistry_of_Metabolic_Pathways/18.3%3A_The_Catabolism_of_Fats.txt |
Learning Objectives
• Describe the function of glycolysis and identify its major products.
• Describe how the presence or absence of oxygen determines what happens to the pyruvate and the NADH that are produced in glycolysis.
• Determine the amount of ATP produced by the oxidation of glucose in the presence and absence of oxygen.
In stage II of catabolism, the metabolic pathway known as glycolysis converts glucose into two molecules of pyruvate (a three-carbon compound with three carbon atoms) with the corresponding production of adenosine triphosphate (ATP). The individual reactions in glycolysis were determined during the first part of the 20th century. It was the first metabolic pathway to be elucidated, in part because the participating enzymes are found in soluble form in the cell and are readily isolated and purified. The pathway is structured so that the product of one enzyme-catalyzed reaction becomes the substrate of the next. The transfer of intermediates from one enzyme to the next occurs by diffusion.
Steps in Glycolysis
The 10 reactions of glycolysis, summarized in Figures $1$ and $2$, can be divided into two phases. In the first 5 reactions—phase I—glucose is broken down into two molecules of glyceraldehyde 3-phosphate. In the last five reactions—phase II—each glyceraldehyde 3-phosphate is converted into pyruvate, and ATP is generated. Notice that all the intermediates in glycolysis are phosphorylated and contain either six or three carbon atoms.
• When glucose enters a cell, it is immediately phosphorylated to form glucose 6-phosphate, in the first reaction of phase I. The phosphate donor in this reaction is ATP, and the enzyme—which requires magnesium ions for its activity—is hexokinase. In this reaction, ATP is being used rather than being synthesized. The presence of such a reaction in a catabolic pathway that is supposed to generate energy may surprise you. However, in addition to activating the glucose molecule, this initial reaction is essentially irreversible, an added benefit that keeps the overall process moving in the right direction. Furthermore, the addition of the negatively charged phosphate group prevents the intermediates formed in glycolysis from diffusing through the cell membrane, as neutral molecules such as glucose can do.
• In the next reaction, phosphoglucose isomerase catalyzes the isomerization of glucose 6-phosphate to fructose 6-phosphate. This reaction is important because it creates a primary alcohol, which can be readily phosphorylated.
• The subsequent phosphorylation of fructose 6-phosphate to form fructose 1,6-bisphosphate is catalyzed by phosphofructokinase, which requires magnesium ions for activity. ATP is again the phosphate donor.
• Fructose 1,6-bisphosphate is enzymatically cleaved by aldolase to form two triose phosphates: dihydroxyacetone phosphate and glyceraldehyde 3-phosphate.
• Isomerization of dihydroxyacetone phosphate into a second molecule of glyceraldehyde 3-phosphate is the final step in phase I. The enzyme catalyzing this reaction is triose phosphate isomerase.
When a molecule contains two phosphate groups on different carbon atoms, the convention is to use the prefix bis. When the two phosphate groups are bonded to each other on the same carbon atom (for example, adenosine diphosphate [ADP]), the prefix is di.
In steps 4 and 5, aldolase and triose phosphate isomerase effectively convert one molecule of fructose 1,6-bisphosphate into two molecules of glyceraldehyde 3-phosphate. Thus, phase I of glycolysis requires energy in the form of two molecules of ATP and releases none of the energy stored in glucose.
In the initial step of phase II (Figure $2$), glyceraldehyde 3-phosphate is both oxidized and phosphorylated in a reaction catalyzed by glyceraldehyde-3-phosphate dehydrogenase, an enzyme that requires nicotinamide adenine dinucleotide (NAD+) as the oxidizing agent and inorganic phosphate as the phosphate donor. In the reaction, NAD+ is reduced to reduced nicotinamide adenine dinucleotide (NADH), and 1,3-bisphosphoglycerate (BPG) is formed.
• BPG has a high-energy phosphate bond (Table $1$) joining a phosphate group to C1. This phosphate group is now transferred directly to a molecule of ADP, thus forming ATP and 3-phosphoglycerate. The enzyme that catalyzes the reaction is phosphoglycerate kinase, which, like all other kinases, requires magnesium ions to function. This is the first reaction to produce ATP in the pathway. Because the ATP is formed by a direct transfer of a phosphate group from a metabolite to ADP—that is, from one substrate to another—the process is referred to as substrate-level phosphorylation, to distinguish it from the oxidative phosphorylation discussed in Section 20.4.
• In the next reaction, the phosphate group on 3-phosphoglycerate is transferred from the OH group of C3 to the OH group of C2, forming 2-phosphoglycerate in a reaction catalyzed by phosphoglyceromutase.
• A dehydration reaction, catalyzed by enolase, forms phosphoenolpyruvate (PEP), another compound possessing a high-energy phosphate group.
• The final step is irreversible and is the second reaction in which substrate-level phosphorylation occurs. The phosphate group of PEP is transferred to ADP, with one molecule of ATP being produced per molecule of PEP. The reaction is catalyzed by pyruvate kinase, which requires both magnesium and potassium ions to be active.
Table $1$: Maximum Yield of ATP from the Complete Oxidation of 1 Mol of Glucose
Reaction Comments Yield of ATP (moles)
glucose → glucose 6-phosphate consumes 1 mol ATP −1
fructose 6-phosphate → fructose 1,6-bisphosphate consumes 1 mol ATP −1
glyceraldehyde 3-phosphate → BPG produces 2 mol of cytoplasmic NADH
BPG → 3-phosphoglycerate produces 2 mol ATP +2
phosphoenolpyruvate → pyruvate produces 2 mol ATP +2
pyruvate → acetyl-CoA + CO2 produces 2 mol NADH
isocitrate → α-ketoglutarate + CO2 produces 2 mol NADH
α-ketoglutarate → succinyl-CoA + CO2 produces 2 mol NADH
succinyl-CoA → succinate produces 2 mol GTP +2
succinate → fumarate produces 2 mol FADH2
malate → oxaloacetate produces 2 mol NADH
2 cytoplasmic NADH from glycolysis yields 2–3 mol ATP per NADH (depending on tissue) +4 to +6
2 NADH from the oxidation of pyruvate yields 3 mol ATP per NADH +6
2 FADH2 from the citric acid cycle yields 2 ATP per FADH2 +4
3 NADH from the citric acid cycle yields 3 ATP per NADH +18
Net yield of ATP: +36 to +38
In phase II, two molecules of glyceraldehyde 3-phosphate are converted to two molecules of pyruvate, along with the production of four molecules of ATP and two molecules of NADH.
To Your Health: Diabetes
Although medical science has made significant progress against diabetes , it continues to be a major health threat. Some of the serious complications of diabetes are as follows:
• It is the leading cause of lower limb amputations in the United States.
• It is the leading cause of blindness in adults over age 20.
• It is the leading cause of kidney failure.
• It increases the risk of having a heart attack or stroke by two to four times.
Because a person with diabetes is unable to use glucose properly, excessive quantities accumulate in the blood and the urine. Other characteristic symptoms are constant hunger, weight loss, extreme thirst, and frequent urination because the kidneys excrete large amounts of water in an attempt to remove excess sugar from the blood.
There are two types of diabetes. In immune-mediated diabetes, insufficient amounts of insulin are produced. This type of diabetes develops early in life and is also known as Type 1 diabetes, as well as insulin-dependent or juvenile-onset diabetes. Symptoms are rapidly reversed by the administration of insulin, and Type 1 diabetics can lead active lives provided they receive insulin as needed. Because insulin is a protein that is readily digested in the small intestine, it cannot be taken orally and must be injected at least once a day.
In Type 1 diabetes, insulin-producing cells of the pancreas are destroyed by the body’s immune system. Researchers are still trying to find out why. Meanwhile, they have developed a simple blood test capable of predicting who will develop Type 1 diabetes several years before the disease becomes apparent. The blood test reveals the presence of antibodies that destroy the body’s insulin-producing cells.
Type 2 diabetes, also known as noninsulin-dependent or adult-onset diabetes, is by far the more common, representing about 95% of diagnosed diabetic cases. (This translates to about 16 million Americans.) Type 2 diabetics usually produce sufficient amounts of insulin, but either the insulin-producing cells in the pancreas do not release enough of it, or it is not used properly because of defective insulin receptors or a lack of insulin receptors on the target cells. In many of these people, the disease can be controlled with a combination of diet and exercise alone. For some people who are overweight, losing weight is sufficient to bring their blood sugar level into the normal range, after which medication is not required if they exercise regularly and eat wisely.
Those who require medication may use oral antidiabetic drugs that stimulate the islet cells to secrete insulin. First-generation antidiabetic drugs stimulated the release of insulin. Newer second-generation drugs, such as glyburide, do as well, but they also increase the sensitivity of cell receptors to insulin. Some individuals with Type 2 diabetes do not produce enough insulin and thus do not respond to these oral medications; they must use insulin. In both Type 1 and Type 2 diabetes, the blood sugar level must be carefully monitored and adjustments made in diet or medication to keep the level as normal as possible (70–120 mg/dL).
Metabolism of Pyruvate
The presence or absence of oxygen determines the fates of the pyruvate and the NADH produced in glycolysis. When plenty of oxygen is available, pyruvate is completely oxidized to carbon dioxide, with the release of much greater amounts of ATP through the combined actions of the citric acid cycle, the electron transport chain, and oxidative phosphorylation. However, in the absence of oxygen (that is, under anaerobic conditions), the fate of pyruvate is different in different organisms. In vertebrates, pyruvate is converted to lactate, while other organisms, such as yeast, convert pyruvate to ethanol and carbon dioxide. These possible fates of pyruvate are summarized in Figure $2$. The conversion to lactate or ethanol under anaerobic conditions allows for the reoxidation of NADH to NAD+ in the absence of oxygen.
ATP Yield from Glycolysis
The net energy yield from anaerobic glucose metabolism can readily be calculated in moles of ATP. In the initial phosphorylation of glucose (step 1), 1 mol of ATP is expended, along with another in the phosphorylation of fructose 6-phosphate (step 3). In step 7, 2 mol of BPG (recall that 2 mol of 1,3-BPG are formed for each mole of glucose) are converted to 2 mol of 3-phosphoglycerate, and 2 mol of ATP are produced. In step 10, 2 mol of pyruvate and 2 mol of ATP are formed per mole of glucose.
For every mole of glucose degraded, 2 mol of ATP are initially consumed and 4 mol of ATP are ultimately produced. The net production of ATP is thus 2 mol for each mole of glucose converted to lactate or ethanol. If 7.4 kcal of energy is conserved per mole of ATP produced, and the total amount of energy that can theoretically be obtained from the complete oxidation of 1 mol of glucose is 670 kcal (as stated in the chapter introduction), the energy conserved in the anaerobic catabolism of glucose to two molecules of lactate (or ethanol) is as follows:
$\mathrm{\dfrac{2\times 7.4\: kcal}{670\: kcal}\times100=2.2\%} \nonumber$
Thus anaerobic cells extract only a very small fraction of the total energy of the glucose molecule.
Contrast this result with the amount of energy obtained when glucose is completely oxidized to carbon dioxide and water through glycolysis, the citric acid cycle, the electron transport chain, and oxidative phosphorylation as summarized in Table $1$. Note the indication in the table that a variable amount of ATP is synthesized, depending on the tissue, from the NADH formed in the cytoplasm during glycolysis. This is because NADH is not transported into the inner mitochondrial membrane where the enzymes for the electron transport chain are located. Instead, brain and muscle cells use a transport mechanism that passes electrons from the cytoplasmic NADH through the membrane to flavin adenine dinucleotide (FAD) molecules inside the mitochondria, forming reduced flavin adenine dinucleotide (FADH2), which then feeds the electrons into the electron transport chain. This route lowers the yield of ATP to 1.5–2 molecules of ATP, rather than the usual 2.5–3 molecules. A more efficient transport system is found in liver, heart, and kidney cells where the formation of one cytoplasmic NADH molecule results in the formation of one mitochondrial NADH molecule, which leads to the formation of 2.5–3 molecules of ATP.The total amount of energy conserved in the aerobic catabolism of glucose in the liver is as follows:
$\mathrm{\dfrac{38\times7.4\: kcal}{670\: kcal}\times100=42\%} \nonumber$
Conservation of 42% of the total energy released compares favorably with the efficiency of any machine. In comparison, automobiles are only about 20%–25% efficient in using the energy released by the combustion of gasoline.
As indicated earlier, the 58% of released energy that is not conserved enters the surroundings (that is, the cell) as heat that helps to maintain body temperature. If we are exercising strenuously and our metabolism speeds up to provide the energy needed for muscle contraction, more heat is produced. We begin to perspire to dissipate some of that heat. As the perspiration evaporates, the excess heat is carried away from the body by the departing water vapor.
Summary
• The monosaccharide glucose is broken down through a series of enzyme-catalyzed reactions known as glycolysis.
• For each molecule of glucose that is broken down, two molecules of pyruvate, two molecules of ATP, and two molecules of NADH are produced.
• In the absence of oxygen, pyruvate is converted to lactate, and NADH is reoxidized to NAD+. In the presence of oxygen, pyruvate is converted to acetyl-CoA and then enters the citric acid cycle.
• More ATP can be formed from the breakdown of glucose when oxygen is present. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/18%3A_The_Organic_Chemistry_of_Metabolic_Pathways/18.4%3A_The_Catabolism_of_Carbohydrates.txt |
Learning Objectives
• To describe how excess amino acids are degraded.
The liver is the principal site of amino acid metabolism, but other tissues, such as the kidney, the small intestine, muscles, and adipose tissue, take part. Generally, the first step in the breakdown of amino acids is the separation of the amino group from the carbon skeleton, usually by a transamination reaction. The carbon skeletons resulting from the deaminated amino acids are used to form either glucose or fats, or they are converted to a metabolic intermediate that can be oxidized by the citric acid cycle. The latter alternative, amino acid catabolism, is more likely to occur when glucose levels are low—for example, when a person is fasting or starving.
Transamination
Transamination is an exchange of functional groups between any amino acid (except lysine, proline, and threonine) and an α-keto acid. The amino group is usually transferred to the keto carbon atom of pyruvate, oxaloacetate, or α-ketoglutarate, converting the α-keto acid to alanine, aspartate, or glutamate, respectively. Transamination reactions are catalyzed by specific transaminases (also called aminotransferases), which require pyridoxal phosphate as a coenzyme.
In an α-keto acid, the carbonyl or keto group is located on the carbon atom adjacent to the carboxyl group of the acid.
Oxidative Deamination
In the breakdown of amino acids for energy, the final acceptor of the α-amino group is α-ketoglutarate, forming glutamate. Glutamate can then undergooxidative deamination, in which it loses its amino group as an ammonium (NH4+) ion and is oxidized back to α-ketoglutarate (ready to accept another amino group):
This reaction occurs primarily in liver mitochondria. Most of the NH4+ ion formed by oxidative deamination of glutamate is converted to urea and excreted in the urine in a series of reactions known as the urea cycle.
The synthesis of glutamate occurs in animal cells by reversing the reaction catalyzed by glutamate dehydrogenase. For this reaction nicotinamide adenine dinucleotide phosphate (NADPH) acts as the reducing agent. The synthesis of glutamate is significant because it is one of the few reactions in animals that can incorporate inorganic nitrogen (NH4+) into an α-keto acid to form an amino acid. The amino group can then be passed on through transamination reactions, to produce other amino acids from the appropriate α-keto acids.
The Fate of the Carbon Skeleton
Any amino acid can be converted into an intermediate of the citric acid cycle. Once the amino group is removed, usually by transamination, the α-keto acid that remains is catabolized by a pathway unique to that acid and consisting of one or more reactions. For example, phenylalanine undergoes a series of six reactions before it splits into fumarate and acetoacetate. Fumarate is an intermediate in the citric acid cycle, while acetoacetate must be converted to acetoacetyl-coenzyme A (CoA) and then to acetyl-CoA before it enters the citric acid cycle.
Those amino acids that can form any of the intermediates of carbohydrate metabolism can subsequently be converted to glucose via a metabolic pathway known as gluconeogenesis. These amino acids are called glucogenic amino acids. Amino acids that are converted to acetoacetyl-CoA or acetyl-CoA, which can be used for the synthesis of ketone bodies but not glucose, are called ketogenic amino acids. Some amino acids fall into both categories. Leucine and lysine are the only amino acids that are exclusively ketogenic. Figure \(2\) summarizes the ultimate fates of the carbon skeletons of the 20 amino acids.
Career Focus: Exercise Physiologist
An exercise physiologist works with individuals who have or wish to prevent developing a wide variety of chronic diseases, such as diabetes, in which exercise has been shown to be beneficial. Each individual must be referred by a licensed physician. An exercise physiologist works in a variety of settings, such as a hospital or in a wellness program at a commercial business, to design and monitor individual exercise plans. A registered clinical exercise physiologist must have an undergraduate degree in exercise physiology or a related degree. Some job opportunities require a master’s degree in exercise physiology or a related degree.
Summary
Generally the first step in the breakdown of amino acids is the removal of the amino group, usually through a reaction known as transamination. The carbon skeletons of the amino acids undergo further reactions to form compounds that can either be used for the synthesis of glucose or the synthesis of ketone bodies.
18.7: The Citric Acid Cycle
Sections/problems listed with an asterisk (*) do not discuss the exact reaction indicated, but do discuss a closely related reaction.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/18%3A_The_Organic_Chemistry_of_Metabolic_Pathways/18.6%3A_The_Catabolism_of_Proteins.txt |
10.2A: ATP - the principle phosphoryl group donor
Enzymes called kinases catalyze the transfer of phosphoryl groups to organic molecules. The source of the phosphoryl group in most phosphorylation reactions is a molecule called adenosine triphosphate, abbreviated ATP.
Notice that there are essentially three parts to the molecule: an adenine nucleoside base, a five-carbon sugar (ribose), and a triphosphate group. The three phosphates are designated by Greek letters a, b, and g. Adenosine diphosphate (ADP) and adenosine monophosphate (AMP) are also important players in the reactions of this chapter.
You will see ATP, ADP, and AMP abbreviated in many different ways in this text and throughout the biochemical literature. For example, the three structures below are all abbreviated depictions of ATP:
The following exercise will give you some practice in recognizing different abbreviations for biological molecules that contain phosphate groups.
Template:ExampleStart
Exercise 10.2: Below are a number of representations, labeled A-S, of molecules that contain phosphoryl groups. Different abbreviations are used. Arrange A-S into groups of drawings that depict the same molecule, using different abbreviations (or no abbreviation at all).
You are probably familiar with the physiological role of ATP from your biology classes - it is commonly called 'the cell's energy currency'. What this means is that ATP stores - for a very short time - some of the energy derived from the oxidation of fuel molecules like carbohydrates or fats (in plants and photosynthetic bacteria, the energy comes from sunlight). The energy in ATP is stored in the two high-energy phosphate anhydride linkages.
When we speak of the energy of an ATP molecule being 'spent', what we mean is that a phosphoryl group is being transferred from ATP to some other acceptor molecule, making the acceptor molecule more reactive. For example, in many phosphoryl transfer reactions (such as the phosphorylation of glucose, which we used as an example in section 10.1D) the gamma (γ) phosphate of ATP is transferred to an organic acceptor, releasing ADP.
In other reactions, the base, ribose, and the alpha phosphate is transferred to the organic molecule to form an organic-AMP adduct, while inorganic pyrophosphate (PPi) is released.
Occasionally, the beta and gamma phosphate groups are transferred together, with the release of AMP.
In all of these reactions, a relatively stable organic molecule is being transformed into a higher energy phosphorylated product. This activated product can then go on to react in ways that its more stable, non-phosphorylated counterpart could not - phosphoryl groups, as we know, are much better leaving groups in nucleophilic substitution reactions than the hydroxyl group of alcohols. Even though the conversion of a lower-energy starting compound into a higher energy product is, by itself, a thermodynamically uphill process, the overall phosphoryl transfer reaction is thermodynamically downhill, because the conversion of ATP into ADP or AMP - the breaking of a phosphate anhydride bond - releases a great deal of energy. In other words, the energy stored in the phosphate anhydride bond of ATP has been 'spent' to create an activated (higher energy) molecule. When AMP or ADP is converted back to ATP, energy from fuel molecules (or from sunlight) is required to re-form the high energy anhydride bond (this process is the subject of discussion later in this section)
The explanation for why the phosphate anhydrides linkages in ATP are so energetic lies primarily in the concept of charge separation. Recall from section 10.1 that ATP, at the physiological pH of ~7, is almost completely ionized with a total charge of close to -4. When one of the two anhydride bonds is broken, the negative charges on the phosphate groups are able to separate, eliminating some of the same-charge repulsion that existed in ATP. One way to picture this is as a coil springing open.
Another reason has to do with the energy of solvation by water. When the gamma phosphate of ATP is transferred to an alcohol, for example, surrounding water molecules are able to form more hydrogen-bonding interactions with the products (ADP and the organic phosphate) than was possible with ATP and alcohol. These additional solvation interactions stabilize the products of the phosphorylation reaction relative to the starting compounds.
You will learn more about the thermodynamic role of ATP in metabolic pathways if you take a class in biochemistry - what is most important to understand at this point is that, because of the energy stored in its phosphate anhydride bonds, ATP is a powerful phosphoryl group donor, and is used as such in many important biochemical reactions. Some examples are discussed in the remainder of this section.
10.2B: Monophosphorylation of alcohols
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in section 10.1D).
+ B-H
The glucose kinase reaction is the first step in glycolysis, a metabolic pathway in which the 6-carbon sugar glucose is broken down into two 3-carbon fragments called pyruvate. The third step of glycolysis also a kinase reaction: this time, it is the hydroxyl group on carbon #1 of fructose-6-phosphate that is phosphorylated (step 2 of glycolysis is the isomerization of glucose-6-phosphate to fructose-6-phosphate, a reaction we will study in section 13.2A) Once again, ATP is the phosphate donor in the fructose-6-phosphate kinase reaction:
Now, when the 6-carbon sugar breaks into two 3-carbon pieces, each piece has its own phosphate group (the carbon-carbon bond-breaking step is a reaction that we will learn about in section 13.3C).
The biological activity of many proteins is regulated by means of a very similar phosphorylation reaction catalyzed by protein kinases. In these reactions, the side chain hydroxyl groups on serine, threonine, and tyrosine residues of certain proteins are modified with the gamma phosphate from ATP.
Notice the new " ATP in, ADP out" notation used in this figure, showing that ATP is converted to ADP in the course of the reaction. From here on, we will frequently use this shorthand convention to indicate when common molecules such as ATP, water, or phosphate are participants in a reaction, either as reactants or products.
Template:ExampleStart
Exercise 10.3: draw a mechanism for the kinase reaction above, using appropriate abbreviations for ATP and ADP.
Solution
Template:ExampleEnd
The conversion of a neutral hydroxyl group to a charged phosphate represents a very dramatic change in the local architecture of the protein, and thus it may behave very differently when phosphorylated, in terms of its overall conformation and ability to bind to small molecules or other proteins. A protein's biological function, whatever that may be, may be turned 'off' until the phosphorylation of a specific serine, threonine, or tyrosine serves as an activating 'on' switch (or vice-versa). In order to reverse the process and 'flip the switch' again, the phosphate group must be removed by a phosphatase enzyme, a reaction which we will examine later in this chapter (section 10.3).
Template:ExampleStart
Exercise 10.4: Threonine kinase catalyzes the phosphorylation of the side chain hydroxyl group of threonine residues in proteins. Draw the structure, including the configuration of all stereocenters, of a phosphothreonine residue. Explain how you can predict the stereochemistry of the side chain.
Solution
Template:ExampleEnd
10.2C: Diphosphorylation of alcohols
We have just seen how alcohol groups can be converted to monophosphates using ATP as the phosphate donor. In some biochemical pathways, the next step is the addition of a second phosphate group to form a diphosphate. In the early stages of the biosynthesis of ‘isoprenoid’ compounds such as cholesterol, for example, two phosphates are added sequentially to a primary alcohol group on an intermediate compound called mevalonate. The first phosphorylation is essentially the same as the reactions described in part B of this section.
In the second phosphorylation reaction, the gamma phosphate of a second ATP molecule is transferred to an oxygen atom on the first phosphate, forming a new phosphate anhydride linkage.
Another example of a diphosphorylation reaction takes place in a single step, rather than sequentially. Phosphoribosyl diphosphate (PRPP) is a very important intermediate compound in the biosynthesis of nucleotides and some amino acids, and is the product of the diphosphorylation of ribose-5-phosphate.
In this reaction, two phosphate groups (the beta and the gamma) are transferred together from ATP to a hydroxyl on ribose-5-phosphate. Notice that the beta phosphorus of ATP is the electrophile in this case, rather then the alpha phosphorus. Consequentially, the reaction results in the conversion on one ATP to one AMP, rather than two ATPs to two ADPs.
Are we getting more for our 'ATP money' in this one-step diphosphorylation? Not really - in order to convert the energy-poor AMP back up to the energy-rich ATP, the cell first has to transfer a phosphate from a second ATP molecule in a reaction catalyzed by an enzyme called adenylate kinase.
So in the end, the diphosphorylation reaction still costs the cell two ATP-to-ADP conversions.
It is worth noting that both of the diphosphate groups produced in these reactions end up as leaving groups in subsequent nucleophilic substitution reactions. Mevalonate diphosphate is eventually converted to isoprenyl diphosphate, the substrate for protein prenyltransferase (section 9.3). PRPP is the starting point for the biosynthesis of both pyrimidine (C and U/T) and purine (G, and A) nucleotides. The SN1 displacement of pyrophosphate in pyrimidine biosynthesis is shown below.
10.2D: Phosphorylation of carboxylates
Thus far we have seen alcohol oxygens and phosphate oxygens acting as nucleophilic accepting groups in phosphoryl transfer reactions. Consider next the first step of the reaction catalyzed by the enzyme glutamine synthase:
Once again in this reaction, the gamma-phosphate of ATP is transferred to an oxygen acceptor - however in this case the acceptor is a carboxylate oxygen, and the product is an acyl phosphate. As we shall see in chapter 12, acyl phosphates are commonly referred to as 'activated carboxylates', and are primed to undergo reactions called 'nucleophilic acyl substitutions'.
Template:ExampleStart
Exercise 10.5: Draw a mechanism for the acyl-phosphate-forming reaction above, using appropriate abbreviations for ATP and ADP.
Solution
Template:ExampleEnd
10.2E: Generation of nucleotide phosphates
Activation of the side chain carboxylate of aspartate is somewhat different from the parallel activation of glutamate shown above. While the carboxylate group in glutamate accepts a simple phosphate group from ATP, the carboxylate in aspartate attacks the alpha-phosphate of ATP, displacing inorganic pyrophosphate and accepting an entire AMP group.
The resulting 'acyl adenosine phosphate', which is technically a phosphate diester, is another form of 'activated carboxylate' that we will learn more about in chapter 12.
For some interesting variations on the phosphoryl transfer reaction, consider the early steps of isoprenoid biosynthesis in bacteria (this is a completely different pathway than that mentioned in section 10.2C, which is operative in animals). In the first step, the oxygen of a monophosphate ester attacks the a phosphate of CTP (not ATP!) to expel inorganic pyrophosphate.
In step 2, a second hydroxyl group is phosphorylated in the normal way by an ATP-dependent kinase, and in step 3 that phosphate proceeds to attack the first of the two phosphates on the nucleotide diphosphate diester, expelling CMP and forming a cyclic diphosphate. Several more steps lead to the formation of isopentenyl diphosphate, the building block molecule for all isoprenoid compounds.
10.2F: Regeneration of ATP from ADP
Throughout this section we have seen reactions in which the energy contained in an ATP anhydride bond is 'spent', and ADP or AMP is formed as a result. In order to regenerate ATP, a phosphate group must be transferred to ADP, which is of course a thermodynamically uphill reaction requiring the input of energy from the breakdown of fuel molecules or, in the case of plants, from sunlight. By far the most important source of ATP regeneration is the enzyme ATP synthase, which catalyzes the direct condensation between inorganic phosphate and ADP.
Despite the apparent simplicity of the chemistry going on here, ATP synthase is an extremely large, complex, and fascinating enzyme, with multiple protein subunits and a intricate 'molecular motor' design. The reaction must be 'driven' uphill by using the energy from a proton gradient that is set up across the inner mitochondrial membrane. You will learn much more about this amazing biochemical machine if you take a course in biochemistry.
Two other reactions in the gycolytic (sugar breakdown) pathway also result in the generation of ATP from ADP, but these are minor physiological sources of ATP compared to ATP synthase. Phosphoglycerate kinase (named according to the reverse of the reaction shown below) transfers a phosphate from an acyl phosphate to ADP.
Note that here the leaving group is a carboxylate group.
Pyruvate kinase (again the name refers to the reverse reaction) catalyzes a less familiar-looking phosphate transfer.
We will revisit this reaction in section 13.1A.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/18%3A_The_Organic_Chemistry_of_Metabolic_Pathways/18.8%3A_Oxidative_Phosphorylation.txt |
Anabolic reactions are those that lead to the synthesis of biomolecules. In contrast to the catabolic reactions just discussed (glycolysis, TCA cycle and electron transport/oxidative phosphorylation) which lead to the oxidative degradation of carbohydrates and fatty acids and energy release, anabolic reactions lead to the synthesis of more complex biomolecules including biopolymers (glycogen, proteins, nucleic acids) and complex lipids. Many biosynthetic reactions, including those for fatty acid synthesis, are reductive and hence require reducing agents. Reductive biosynthesis and complex polymer formation require energy input, usually in the form of ATP whose exergonic cleavage is coupled to endergonic biosynthesis.
Cells have evolved interesting mechanism so as not to have oxidative degradation reactions (which release energy) proceed at the same time and in the same cell as reductive biosynthesis (which requires energy input). Consider this scenario. You dive into a liver cell and find palmitic acid, a 16C fatty acid. From where did it come? Was it just synthesized by the liver cell or did it just enter the cell from a distant location such as adipocytes (fat cells). Should it be oxidized, which should happen if there is a demand for energy production by the cell, or should the liver cell export it, perhaps to adipocytes, which might happen if there is an excess of energy storage molecules? Cells have devised many ways to distinguish these opposing needs. One is by using a slightly different pool of redox reagents for anabolic and catabolic reactions. Oxidative degradation reactions typically use the redox pair NAD+/NADH (or FAD/FADH2) while reductive biosynthesis often uses phosphorylated variants of NAD+, NADP+/NADPH. In addition, cells often carry out competing reactions in different cellular compartments. Fatty acid oxidation of our example molecule (palmitic acid) occurs in the mitochondrial matrix, while reductive fatty acid synthesis occurs in the cytoplasm of the cell. Fatty acids entering the cell destined for oxidative degradation are transported into the mitochondria by the carnitine transport system. This transport system is inhibited under conditions when fatty acid synthesis is favored. We will discuss the regulation of metabolic pathways in a subsequent section. One of the main methods, as we will see, is to activate or inhibit key enzymes in the pathways under a given set of cellular conditions. The key enzyme in fatty acid synthesis, acetyl-CoA carboxylase, is inhibited when cellular conditions require fatty acid oxidation.
The following examples give short descriptions of anabolic pathways. Compare them to the catabolic pathways from the previous section.
• Glucose synthesis, better known as Gluconeogenesis: In glycolysis, glucose (C6H12O6), a 6C molecule, is converted to two, 3C molecules (pyruvate) in an oxidative process that requires NAD+ and makes two net ATP molecules. In a few organs, most predominately in the liver, the reverse pathway can take place. The liver does this to provide glucose to the brain when the body is deficient in circulating glucose, for example, under fasting and starving conditions. (The liver under these conditions can get its energy from oxidation of fatty acids). The reactions in gluconeogenesis are the same reactions in glycolysis but run in reverse, with the exception of three glycolytic steps which are essentially irreversible. These three steps have bypass enzymes in the gluconeogenesis pathway. Although the synthesis of glucose is a reductive pathway, it uses NADH instead of NADPH as the redundant as the same enzyme used in glycolysis is simply run in reverse. Gluconeogenesis, which also occurs in the cortex of the kidney, is more than just a simple reversal of glycolysis, however. It can be thought of as the net synthesis of glucose from non-carbohydrate precursors. Pyruvate, as seen in the section on catabolism, can be formed from protein degradation to glucogenic amino acids which can be converted to pyruvate. It can also be formed from triacylglycerides from the 3C molecule glycerol formed and released from adipocytes after hydrolysis of three fatty acids from triacylglycerides. However, in humans, glucose can not be made in net fashion from fatty acids. Fatty acids can be converted to acetyl-CoA by fatty acid oxidation. The resulting acetyl-CoA can not form pyruvate since the enzyme that catalyzes the formation for acetyl-CoA from pyruvate, pyruvate dehydrogenase, is irreversible and there is no bypass reaction known. The acetyl-CoA can enter the TCA cycle but since the pathway is cyclic and proceeds in one direction, it can not form in net fashion oxaloacetate. Although oxaloacetate can be remove from the TCA cycle and be use to form phosphoenolpyuvate, a glycolytic intermediate, one acetyl-CoA condenses with one oxaloacetate to form citrate which leads back to one oxaloacetate. Hence fatty acids can not be converted to glucose and other sugars in a net fashion.
• Pentose Phosphate Shunt: This two-part pathway doesn't appear to start as a reductive biosynthetic pathway as the first part is the oxidative conversion of a glycolytic intermediate, glucose-6-phosphate, to ribulose-5-phosphate. The next, nonoxidative branch leads to the formation of ribose-5-phosphate, a key biosynthetic intermediate in nucleic acid synthesis as well as erthyrose-4-phosphate used for biosynthesis of aromatic amino acids . The oxidative branch is important in reductive biosynthesis as it is a major source of the reductant NADPH used in biosynthetic reactions.
• Fatty acid and isoprenoid/sterol biosynthesis: Acetyl-CoA is the source of carbon atoms for the synthesis of more complex lipids such as fatty acids, isoprenoids, and sterols. When energy needs in a cell are not high, citrate, the condensation product of oxaloacetate and acetyl-CoA in the TCA cycle, builds up in the mitochondrial matrix. It is then transported by the citrate transporter (an inner mitochondrial membrane protein) to the cytoplasm, where it is cleaved back to oxaloacetate and acetyl-CoA by the cytoplasmic enzyme citrate lyase. The oxaloacetate is returned to the mitochondria by conversion first to malate (reduction reaction using NADH), which can move back into the mitochondria through the malate transporter, or further conversion to pyruate, using the cytosolic malic enzyme, which uses NADP+ to oxidize malate to pyruvate which then enters the mitochondria. The acetyl-CoA formed in the cytoplasm can then be used in reductive biosynthesis using NADPH as the reductant to form fatty acids, isoprenoids, and sterols. The NADPH for the reduction comes from the oxidative branch of the pentose phosphate pathway and from the reaction catalyzed by malic enzyme. The liver cells can still run the glycolytic pathway as the NADH/NAD+ ratio is low in the cytoplasm while NADPH/NADP+ ratio is high.
Now its time to see how the various pathways fit together to form an integrated set of pathways. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/18%3A_The_Organic_Chemistry_of_Metabolic_Pathways/18.9%3A_Anabolism.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
19: The Organic Chemistry of Lipids
Fatty acids are merely carboxylic acids with long hydrocarbon chains. The hydrocarbon chain length may vary from 10-30 carbons (most usual is 12-18). The non-polar hydrocarbon alkane chain is an important counter balance to the polar acid functional group. In acids with only a few carbons, the acid functional group dominates and gives the whole molecule a polar character. However, in fatty acids, the non-polar hydrocarbon chain gives the molecule a non-polar character.
• Hydrogenation of Unsaturated Fats and Trans Fat
Saturated fats have a chain like structure which allows them to stack very well forming a solid at room temperature. Unsaturated fats are not linear due to double bonded carbons which results in a different molecular shape because the sp2 carbons are trigonal planar. This causes the fat molecules to poorly stack resulting in fats that are liquid at room temperature. Unsaturated fats can be converted to saturated fats via hydrogenation reactions.
• Introduction to Fatty Acids
Fatty acids are carboxylic acids with long hydrocarbon chains. There are two groups of fatty acids--saturated and unsaturated. Recall that the term unsaturated refers to the presence of one or more double bonds between carbons as in alkenes. A saturated fatty acid has all bonding positions between carbons occupied by hydrogens. The melting points for the saturated fatty acids follow the boiling point principle observed previously.
• Prostaglandins
Prostaglandins were first discovered and isolated from human semen in the 1930s by Ulf von Euler of Sweden. Thinking they had come from the prostate gland, he named them prostaglandins. It has since been determined that they exist and are synthesized in virtually every cell of the body. Prostaglandins, are like hormones in that they act as chemical messengers, but do not move to other sites, but work right within the cells where they are synthesized.
Thumbnail: A ball-and-stick diagram of arachidonic acid. (Public Domain; SubDural12). Arachidonic acid is a polyunsaturated fatty acid present in the phospholipids of membranes of the body's cells, and is abundant in the brain, muscles, and liver.
19.2: Waxes are High-Molecular-Weight Esters
Waxes are esters of fatty acids with long chain monohydric alcohols (one hydroxyl group). Natural waxes are often mixtures of such esters, and may also contain hydrocarbons. The formulas for three well known waxes are given below, with the carboxylic acid moiety colored red and the alcohol colored blue.
Spermaceti Beeswax Carnuba wax
CH3(CH2)14CO2-(CH2)15CH3 CH3(CH2)24CO2-(CH2)29CH3 CH3(CH2)30CO2-(CH2)33CH3
Waxes are widely distributed in nature. The leaves and fruits of many plants have waxy coatings, which may protect them from dehydration and small predators. The feathers of birds and the fur of some animals have similar coatings which serve as a water repellent. Carnuba wax is valued for its toughness and water resistance.
19.3: Fats and Oils
Triglycerides are esters of fatty acids and a trifunctional alcohol - glycerol (IUPAC name is 1,2,3-propantriol). The properties of fats and oils follow the same general principles as already described for the fatty acids. The important properties to be considered are: melting points and degree of unsaturation from component fatty acids.
Introduction
Since glycerol has three alcohol functional groups, three fatty acids must react to make three ester functional groups. The three fatty acids may or may not be identical. In fact, three different fatty acids may be present. The synthesis of a triglyceride is another application of the ester synthesis reaction. To write the structure of the triglyceride you must know the structure of glycerol and be given or look up the structure of the fatty acid in the table.
Table \(1\):The common fats and oils including fatty acid content are listed below.
Fat or Oil Saturated Unsaturated
Palmitic Stearic Oleic Linoleic Other
Animal Origin
Butter 29 9 27 4 31
Lard 30 18 41 6 5
Beef 32 25 38 3 2
Vegetable Origin
Corn oil 10 4 34 48 4
Soybean 7 3 25 56 9
Peanut 7 5 60 21 7
Olive 6 4 83 7 -
Synthesis of a Triglyceride
Since glycerol, (IUPAC name is 1,2,3-propantriol), has three alcohol functional groups, three fatty acids must react to make three ester functional groups. The three fatty acids may or may not be identical. In fact, three different fatty acids may be present. nThe synthesis of a triglyceride is another application of the ester synthesis reaction. To write the structure of the triglyceride you must know the structure of glycerol and be given or look up the structure of the fatty acid in Table \(1\) - find lauric acid.
Glycerol
The simplified reaction reveals the process of breaking some bonds and forming the ester and the by product, water. Refer to the graphic on the left for the synthesis of trilauroylglycerol. First, the -OH (red) bond on the acid is broken and the -H (red) bond on the alcohol is also broken. Both join to make HOH, a water molecule. Secondly, the oxygen of the alcohol forms a bond (green) to the acid at the carbon with the double bond oxygen. This forms the ester functional group. This process is carried out three times to make three ester groups and three water molecules.
Structure of a Triglyceride
As you can see from the graphic on the left, the actual molecular model of the triglyceride does not look at all like the line drawing. The reason for this difference lies in the concepts of molecular geometry. Trilauroylglycerol. All of the above factors contribute to the apparent "T" shape of the molecule.
Problems
Quiz: Which acid (short chain or fatty) would most likely be soluble in water?
... in hexane?
1 Practice writing out a triglyceride of stearic acid. Again look up the formula of stearic acid and use the structure of glycerol.
2. Write down your answers. Then check the answers from the drop down menu.
What is the molecular geometry of all three carbons in glycerol (look at model above)?
What is the molecular geometry of the carbon at the center of the ester group?
What is the molecular geometry of the single bond oxygen?
• Snell, Foster D. "Soap and glycerol." J. Chem. Educ. 1942, 19, 172.
19.4: Phospholipids Are Components of Membranes
Phospholipids are the main constituents of cell membranes. They resemble the triglycerides in being ester or amide derivatives of glycerol or sphingosine with fatty acids and phosphoric acid. The phosphate moiety of the resulting phosphatidic acid is further esterified with ethanolamine, choline or serine in the phospholipid itself. The following diagram shows the structures of some of these components. Clicking on the diagram will change it to display structures for two representative phospholipids. Note that the fatty acid components (R & R') may be saturated or unsaturated.
As ionic amphiphiles, phospholipids aggregate or self-assemble when mixed with water, but in a different manner than the soaps and detergents. Because of the two pendant alkyl chains present in phospholipids and the unusual mixed charges in their head groups, micelle formation is unfavorable relative to a bilayer structure. If a phospholipid is smeared over a small hole in a thin piece of plastic immersed in water, a stable planar bilayer of phospholipid molecules is created at the hole. As shown in the following diagram, the polar head groups on the faces of the bilayer contact water, and the hydrophobic alkyl chains form a nonpolar interior. The phospholipid molecules can move about in their half the bilayer, but there is a significant energy barrier preventing migration to the other side of the bilayer.
This bilayer membrane structure is also found in aggregate structures called liposomes. Liposomes are microscopic vesicles consisting of an aqueous core enclosed in one or more phospholipid layers. They are formed when phospholipids are vigorously mixed with water. Unlike micelles, liposomes have both aqueous interiors and exteriors.
A cell may be considered a very complex liposome. The bilayer membrane that separates the interior of a cell from the surrounding fluids is largely composed of phospholipids, but it incorporates many other components, such as cholesterol, that contribute to its structural integrity. Protein channels that permit the transport of various kinds of chemical species in and out of the cell are also important components of cell membranes.
The interior of a cell contains a variety of structures (organelles) that conduct chemical operations vital to the cells existence. Molecules bonded to the surfaces of cells serve to identify specific cells and facilitate interaction with external chemical entities. The sphingomyelins are also membrane lipids. They are the major component of the myelin sheath surrounding nerve fibers. Multiple Sclerosis is a devastating disease in which the myelin sheath is lost, causing eventual paralysis. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/19%3A_The_Organic_Chemistry_of_Lipids/19.1%3A_Fatty_Acids.txt |
Compounds classified as terpenes constitute what is arguably the largest and most diverse class of natural products. A majority of these compounds are found only in plants, but some of the larger and more complex terpenes (e.g. squalene & lanosterol) occur in animals. Terpenes incorporating most of the common functional groups are known, so this does not provide a useful means of classification. Instead, the number and structural organization of carbons is a definitive characteristic. Terpenes may be considered to be made up of isoprene (more accurately isopentane) units, an empirical feature known as the isoprene rule. Because of this, terpenes usually have \(5n\) carbon atoms (\(n\) is an integer), and are subdivided as follows:
Classification Isoprene Units Carbon Atoms
monoterpenes 2 C10
sesquiterpenes 3 C15
diterpenes 4 C20
sesterterpenes 5 C25
triterpenes 6 C30
Isoprene itself, a C5H8 gaseous hydrocarbon, is emitted by the leaves of various plants as a natural byproduct of plant metabolism. Next to methane it is the most common volatile organic compound found in the atmosphere. Examples of C10 and higher terpenes, representing the four most common classes are shown in the following diagrams. Most terpenes may be structurally dissected into isopentane segments. How this is done can be seen in the diagram directly below.
Figure: Monoterpenes and diterpenes
The isopentane units in most of these terpenes are easy to discern, and are defined by the shaded areas. In the case of the monoterpene camphor, the units overlap to such a degree it is easier to distinguish them by coloring the carbon chains. This is also done for alpha-pinene. In the case of the triterpene lanosterol we see an interesting deviation from the isoprene rule. This thirty carbon compound is clearly a terpene, and four of the six isopentane units can be identified. However, the ten carbons in center of the molecule cannot be dissected in this manner. Evidence exists that the two methyl groups circled in magenta and light blue have moved from their original isoprenoid locations (marked by small circles of the same color) to their present location. This rearrangement is described in the biosynthesis section. Similar alkyl group rearrangements account for other terpenes that do not strictly follow the isoprene rule.
Figure: Triterpenes
Polymeric isoprenoid hydrocarbons have also been identified. Rubber is undoubtedly the best known and most widely used compound of this kind. It occurs as a colloidal suspension called latex in a number of plants, ranging from the dandelion to the rubber tree (Hevea brasiliensis). Rubber is a polyene, and exhibits all the expected reactions of the C=C function. Bromine, hydrogen chloride and hydrogen all add with a stoichiometry of one molar equivalent per isoprene unit. Ozonolysis of rubber generates a mixture of levulinic acid ( \(CH_3COCH_2CH_2CO_2H\) ) and the corresponding aldehyde. Pyrolysis of rubber produces the diene isoprene along with other products.
The double bonds in rubber all have a Z-configuration, which causes this macromolecule to adopt a kinked or coiled conformation. This is reflected in the physical properties of rubber. Despite its high molecular weight (about one million), crude latex rubber is a soft, sticky, elastic substance. Chemical modification of this material is normal for commercial applications. Gutta-percha (structure above) is a naturally occurring E-isomer of rubber. Here the hydrocarbon chains adopt a uniform zig-zag or rod like conformation, which produces a more rigid and tough substance. Uses of gutta-percha include electrical insulation and the covering of golf balls. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/19%3A_The_Organic_Chemistry_of_Lipids/19.5%3A_Terpenes_Contain_Carbon_Atoms_in_Multiples_of_Five.txt |
Compounds classified as terpenes constitute what is arguably the largest and most diverse class of natural products. A majority of these compounds are found only in plants, but some of the larger and more complex terpenes (e.g. squalene & lanosterol) occur in animals. Terpenes incorporating most of the common functional groups are known, so this does not provide a useful means of classification. Instead, the number and structural organization of carbons is a definitive characteristic. Terpenes may be considered to be made up of isoprene (more accurately isopentane) units, an empirical feature known as the isoprene rule. Because of this, terpenes usually have \(5n\) carbon atoms (\(n\) is an integer), and are subdivided as follows:
Classification Isoprene Units Carbon Atoms
monoterpenes 2 C10
sesquiterpenes 3 C15
diterpenes 4 C20
sesterterpenes 5 C25
triterpenes 6 C30
Isoprene itself, a C5H8 gaseous hydrocarbon, is emitted by the leaves of various plants as a natural byproduct of plant metabolism. Next to methane it is the most common volatile organic compound found in the atmosphere. Examples of C10 and higher terpenes, representing the four most common classes are shown in the following diagrams. Most terpenes may be structurally dissected into isopentane segments. How this is done can be seen in the diagram directly below.
Figure: Monoterpenes and diterpenes
The isopentane units in most of these terpenes are easy to discern, and are defined by the shaded areas. In the case of the monoterpene camphor, the units overlap to such a degree it is easier to distinguish them by coloring the carbon chains. This is also done for alpha-pinene. In the case of the triterpene lanosterol we see an interesting deviation from the isoprene rule. This thirty carbon compound is clearly a terpene, and four of the six isopentane units can be identified. However, the ten carbons in center of the molecule cannot be dissected in this manner. Evidence exists that the two methyl groups circled in magenta and light blue have moved from their original isoprenoid locations (marked by small circles of the same color) to their present location. This rearrangement is described in the biosynthesis section. Similar alkyl group rearrangements account for other terpenes that do not strictly follow the isoprene rule.
Figure: Triterpenes
Polymeric isoprenoid hydrocarbons have also been identified. Rubber is undoubtedly the best known and most widely used compound of this kind. It occurs as a colloidal suspension called latex in a number of plants, ranging from the dandelion to the rubber tree (Hevea brasiliensis). Rubber is a polyene, and exhibits all the expected reactions of the C=C function. Bromine, hydrogen chloride and hydrogen all add with a stoichiometry of one molar equivalent per isoprene unit. Ozonolysis of rubber generates a mixture of levulinic acid ( \(CH_3COCH_2CH_2CO_2H\) ) and the corresponding aldehyde. Pyrolysis of rubber produces the diene isoprene along with other products.
The double bonds in rubber all have a Z-configuration, which causes this macromolecule to adopt a kinked or coiled conformation. This is reflected in the physical properties of rubber. Despite its high molecular weight (about one million), crude latex rubber is a soft, sticky, elastic substance. Chemical modification of this material is normal for commercial applications. Gutta-percha (structure above) is a naturally occurring E-isomer of rubber. Here the hydrocarbon chains adopt a uniform zig-zag or rod like conformation, which produces a more rigid and tough substance. Uses of gutta-percha include electrical insulation and the covering of golf balls. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/19%3A_The_Organic_Chemistry_of_Lipids/19.6%3A_How_Terpenes_Are_Biosynthesized.txt |
The important class of lipids called steroids are actually metabolic derivatives of terpenes, but they are customarily treated as a separate group. Steroids may be recognized by their tetracyclic skeleton, consisting of three fused six-membered and one five-membered ring, as shown in the diagram to the right. The four rings are designated A, B, C & D as noted, and the peculiar numbering of the ring carbon atoms (shown in red) is the result of an earlier misassignment of the structure. The substituents designated by R are often alkyl groups, but may also have functionality. The R group at the A:B ring fusion is most commonly methyl or hydrogen, that at the C:D fusion is usually methyl. The substituent at C-17 varies considerably, and is usually larger than methyl if it is not a functional group. The most common locations of functional groups are C-3, C-4, C-7, C-11, C-12 & C-17. Ring A is sometimes aromatic. Since a number of tetracyclic triterpenes also have this tetracyclic structure, it cannot be considered a unique identifier.
Steroids are widely distributed in animals, where they are associated with a number of physiological processes. Examples of some important steroids are shown in the following diagram. Norethindrone is a synthetic steroid, all the other examples occur naturally. A common strategy in pharmaceutical chemistry is to take a natural compound, having certain desired biological properties together with undesired side effects, and to modify its structure to enhance the desired characteristics and diminish the undesired. This is sometimes accomplished by trial and error.
The generic steroid structure drawn above has seven chiral stereocenters (carbons 5, 8, 9, 10, 13, 14 & 17), which means that it may have as many as 128 stereoisomers. With the exception of C-5, natural steroids generally have a single common configuration. This is shown in the last of the toggled displays, along with the preferred conformations of the rings.
Chemical studies of the steroids were very important to our present understanding of the configurations and conformations of six-membered rings. Substituent groups at different sites on the tetracyclic skeleton will have axial or equatorial orientations that are fixed because of the rigid structure of the trans-fused rings. This fixed orientation influences chemical reactivity, largely due to the greater steric hindrance of axial groups versus their equatorial isomers. Thus an equatorial hydroxyl group is esterified more rapidly than its axial isomer.
It is instructive to examine a simple bicyclic system as a model for the fused rings of the steroid molecule. Decalin, short for decahydronaphthalene, exists as cis and trans isomers at the ring fusion carbon atoms. Planar representations of these isomers are drawn at the top of the following diagram, with corresponding conformational formulas displayed underneath. The numbering shown for the ring carbons follows IUPAC rules, and is different from the unusual numbering used for steroids. For purposes of discussion, the left ring is labeled A (colored blue) and the right ring B (colored red). In the conformational drawings the ring fusion and the angular hydrogens are black.
The trans-isomer is the easiest to describe because the fusion of the A & B rings creates a rigid, roughly planar, structure made up of two chair conformations. Each chair is fused to the other by equatorial bonds, leaving the angular hydrogens (Ha) axial to both rings. Note that the bonds directed above the plane of the two rings alternate from axial to equatorial and back if we proceed around the rings from C-1 to C-10 in numerical order. The bonds directed below the rings also alternate in a complementary fashion.
Conformational descriptions of cis- decalin are complicated by the fact that two energetically equivalent fusions of chair cyclohexanes are possible, and are in rapid equilibrium as the rings flip from one chair conformation to the other. In each of these all chair conformations the rings are fused by one axial and one equatorial bond, and the overall structure is bent at the ring fusion. In the conformer on the left, the red ring (B) is attached to the blue ring (A) by an axial bond to C-1 and an equatorial bond to C-6 (these terms refer to ring A substituents). In the conformer on the right, the carbon bond to C-1 is equatorial and the bond to C-6 is axial. Each of the angular hydrogens (Hae or Hea) is oriented axial to one of the rings and equatorial to the other. This relationship reverses when double ring flipping converts one cis-conformer into the other.
Cis-decalin is less stable than trans-decalin by about 2.7 kcal/mol (from heats of combustion and heats of isomerization data). This is due to steric crowding (hindrance) of the axial hydrogens in the concave region of both cis-conformers, as may be seen in the model display activated by the following button. This difference is roughly three times the energy of a gauche butane conformer relative to its anti conformer. Indeed three gauche butane interactions may be identified in each of the cis-decalin conformations, as will be displayed by clicking on the above conformational diagram. These gauche interactions are also shown in the model.
Steroids in which rings A and B are fused cis, such as the example on the right, do not have the sameconformational mobility exhibited by cis-decalin. The fusion of ring C to ring B in a trans configuration prevents ring B from undergoing a conformational flip to another chair form. If this were to occur, ring C would have to be attached to ring B by two adjacent axial bonds directed 180º apart. This is too great a distance to be bridged by the four carbon atoms making up ring C. Consequently, the steroid molecule is locked in the all chair conformation shown here. Of course, all these steroids and decalins may have one or more six-membered rings in a boat conformation. However the high energy of boat conformers relative to chairs would make such structures minor components in the overall ensemble of conformations available to these molecules.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
19.8: Synthetic Steroids
A steroid hormone is a steroid that acts as a hormone. Steroid hormones can be grouped into five groups by the receptors to which they bind: glucocorticoids, mineralocorticoids, androgens, estrogens, and progestogens. Vitamin D derivatives are a sixth closely related hormone system with homologous receptors. They have some of the characteristics of true steroids as receptor ligands, but lack the planar fused four ring system of true steroids.
Introduction
Steroid hormones help control metabolism, inflammation, immune functions, salt and water balance, development of sexual characteristics, and the ability to withstand illness and injury. The term steroid describes both hormones produced by the body and artificially produced medications that duplicate the action for the naturally occurring steroids.
A variety of synthetic steroids and sterols have also been contrived. Most are steroids, but some non-steroidal molecules can interact with the steroid receptors because of a similarity of shape. Some synthetic steroids are weaker, and some much stronger, than the natural steroids whose receptors they activate.[6]
Some examples of synthetic steroid hormones:
Some steroid antagonists:
Synthesis
The natural steroid hormones are generally synthesized from cholesterol in the gonads and adrenal glands. These forms of hormones are lipids. They can pass through the cell membrane[5] as they are fat-soluble, and then bind to steroid hormone receptors which may be nuclear or cytosolic depending on the steroid hormone, to bring about changes within the cell. Steroid hormones are generally carried in the blood bound to specific carrier proteins such as sex hormone-binding globulin or corticosteroid-binding globulin. Further conversions and catabolism occurs in the liver, in other "peripheral" tissues, and in the target tissues.
Figure 1: Enzymes, their cellular location, substrates and products in human steroidogenesis. Shown also is the major classes of steroid hormones: progestagens, mineralocorticoids, glucocorticoids, androgens and estrogens. However, they partly overlap, e.g. mineralocorticoids and glucocorticoids. White circles indicate changes in molecular structure compared with precursors. For more information on interpretation of molecular structures, see Wikipedia:structural formula. HSD: Hydroxysteroid dehydrogenase References: Walter F. Boron, Emile L. Boulpaep (2003) Medical Physiology: A Cellular And Molecular Approach, Elsevier/Saunders, pp. page 1,300 ISBN: 1-4160-2328-3. Figure used with permission from Häggström M, Richfield D (2014). "Diagram of the pathways of human steroidogenesis". Wikiversity Journal of Medicine 1 (1).
• Wikipedia | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/19%3A_The_Organic_Chemistry_of_Lipids/19.7%3A_Steriods_Are_Chemical_Messengers.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
20: The Chemistry of Nucleic Acids
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
20.02: Nucleic Acids Are Composed of Nucleotide Subunits
Intermolecular forces are the attractive or repulsive forces between molecules. They are separated into two groups; short range and long range forces. Short range forces happen when the centers of the molecules are separated by three angstroms (10-8 cm) or less. Short range forces tend to be repulsive, where the long range forces that act outside the three angstroms range are attractive. Long range forces are also known as Van der Waals forces. They are responsible for surface tension, friction, viscosity and differences between actual behavior of gases and that predicted by the ideal gas law. Intermolecular forces are responsible for most properties of all the phases. The viscosity, diffusion, and surface tension are examples of physical properties of liquids that depend on intermolecular forces. Vapor pressure, critical point, and boiling point are examples of properties of gases. Melting and sublimation are examples of properties of solids that depend on intermolecular forces.
• Hydrogen Bonding
A hydrogen bond is a special type of dipole-dipole attraction which occurs when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. These bonds are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds.
• Hydrophobic Interactions
Hydrophobic interactions describe the relations between water and hydrophobes (low water-soluble molecules). Hydrophobes are nonpolar molecules and usually have a long chain of carbons that do not interact with water molecules. The mixing of fat and water is a good example of this particular interaction. The common misconception is that water and fat doesn’t mix because the Van der Waals forces that are acting upon both water and fat molecules are too weak.
• Multipole Expansion
A multipole expansion is a series expansion of the effect produced by a given system in terms of an expansion parameter which becomes small as the distance away from the system increases. Therefore, the leading one or terms in a multipole expansion are generally the strongest. The first-order behavior of the system at large distances can therefore be obtained from the first terms of this series, which is generally much easier to compute than the general solution.
• Overview of Intermolecular Forces
Intermolecular forces are forces between molecules. Depending on its strength, intermolecular forces cause the forming of three physical states: solid, liquid and gas. The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces.
• Specific Interactions
Intermolecular forces are forces of attraction or repulsion which act between neighboring particles (atoms, molecules or ions). They are weak compared to the intramolecular forces, which keep a molecule together (e.g., covalent and ionic bonding).
• Van der Waals Forces
Van der Waals forces' is a general term used to define the attraction of intermolecular forces between molecules. There are two kinds of Van der Waals forces: weak London Dispersion Forces and stronger dipole-dipole forces. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/20%3A_The_Chemistry_of_Nucleic_Acids/20.01%3A_Nucleosides_and_Nucleotides.txt |
Ensemble properties result from being or relating to the greater in number of atoms in a sample. This are in contrast to atomic or molecular properties.
• Capillary Action
Capillary action can be defined as the ascension of liquids through slim tube, cylinder or permeable substance due to adhesive and cohesive forces interacting between the liquid and the surface. When intermolecular bonding of a liquid itself is substantially inferior to a substances’ surface it is interacting, capillarity occurs. Also, the diameter of the container as well as the gravitational forces will determine amount of liquid raised.
• Cohesive and Adhesive Forces
Cohesive and adhesive forces are associated with bulk (or macroscopic) properties and hence the terms are not applicable to discussion of atomic and molecular properties. When a liquid comes into contact with a surface (such as the walls of a graduated cylinder or a tabletop), both cohesive and adhesive forces will act on it. These forces govern the shape which the liquid takes on.
• Contact Angles
Contact angle is one of the common ways to measure the wettability of a surface or material. Wetting refers to the study of how a liquid deposited on a solid (or liquid) substrate spreads out or the ability of liquids to form boundary surfaces with solid states. The wetting, as mentioned before is determined by measuring the contact angle, which the liquid forms in contact with the solids or liquids. The wetting tendency is larger, the smaller the contact angle or the surface tension is.
• Surface Tension
Surface tension is the energy, or work, required to increase the surface area of a liquid due to intermolecular forces. Since these intermolecular forces vary depending on the nature of the liquid (e.g. water vs. gasoline) or solutes in the liquid (e.g. surfactants like detergent), each solution exhibits differing surface tension properties.
• Unusual Properties of Water
With 70% of our earth being ocean water and 65% of our bodies being water, it is hard to not be aware of how important it is in our lives. There are 3 different forms of water, or H2O: solid (ice), liquid (water), and gas (steam). Because water seems so ubiquitous, many people are unaware of the unusual and unique properties of water, including:
• Vapor Pressure
Pressure is the average force that material (gas, liquid or solid) exert upon the surface, e.g. walls of a container or other confining boundary. Vapor pressure or equilibrium vapor pressure is the pressure of a vapor in thermodynamic equilibrium with its condensed phases in a closed container. All liquids and solids have a tendency to evaporate or sublime into a gaseous form and all gases have a tendency to condense back to their liquid or solid form.
• Viscosity
Viscosity is another type of bulk property defined as a liquid’s resistance to flow. When the intermolecular forces of attraction are strong within a liquid, there is a larger viscosity. An example of this phenomenon is imagining a race between two liquids down a windshield. Which would you expect to roll down the windshield faster honey or water? Obviously from experience one would expect water to easily speed right past the honey, a fact that reveals honey has a much higher viscocity than wate
• Wetting Agents
A substance is referred to as a wetting agent if it lowers the surface tension of a liquid and thus allows it to spread more easily.
Thumbnail: A water drop on a lotus leaf surface showing contact angles of approximately 147°. (Public Domain; Na2jojon).
20.04: The Biosynthesis of DNA is Called Replication
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
20.05: DNA and Heredity
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/20%3A_The_Chemistry_of_Nucleic_Acids/20.03%3A_Why_DNA_Does_Not_Have_A_2-_OH_Group.txt |
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
20.07: There Are Three Kinds of RNA
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
20.08: The Biosynthesis of Proteins Is Called Translation
Contributors and Attributions
• Nathalie Interiano
20.09: Why DNA Contains Thymine Instead of Uracil
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
20.10: How the Base Sequence of DNA Is Determined
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
20.11: The Polymerase Chain Reaction (PCR)
Contributors and Attributions
• Nathalie Interiano
20.11: The Polymerase Chain Reaction (PCR)
Contributors and Attributions
• Nathalie Interiano
20.12: Genetic Engineering
Contributors and Attributions
• Nathalie Interiano | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/20%3A_The_Chemistry_of_Nucleic_Acids/20.06%3A_The_Biosynthesis_of_RNA_is_Called_Transcription.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
21: The Organic Chemistry of Drugs: Discovery and Design
A very broad definition of a drug would include "all chemicals other than food that affect living processes." If the affect helps the body, the drug is a medicine. However, if a drug causes a harmful effect on the body, the drug is a poison. The same chemical can be a medicine and a poison depending on conditions of use and the person using it. Another definition would be "medicinal agents used for diagnosis, prevention, treatment of symptoms, and cure of diseases." Contraceptives would be outside of this definition unless pregnancy were considered a disease.
• Anti-Cancer Drugs I
• Anti-Cancer Drugs II
• Antidepressants
Antidepressant drugs are used to restore mentally depressed patients to an improved mental status. Depression results from a deficiency of norepinephrine at receptors in the brain. Mechanisms that increase their effective concentration at the receptor sites should alleviate depression.
• Barbiturates and Benzodiazepines
Barbiturates are central nervous system depressants and are similar, in many ways, to the depressant effects of alcohol.
• Drugs Acting Upon the Central Nervous System
Chemical influences are capable of producing a myriad of effects on the activity and function of the central nervous system. Since our knowledge of different regions of brain function and the neurotransmitters in the brain is limited, the explanations for the mechanisms of drug action may be vague. The known neurotransmitters are: acetylcholine which is involved with memory and learning; norepinephrine which is involved with mania-depression and emotions.
• Drug Activity
Page notifications Off Barbiturates and Benzodiazepines Drug Receptor Interactions picture_as_pdf Batch Donate Table of contents A very broad definition of a drug would include "all chemicals other than food that affect living processes." If the affect helps the body, the drug is a medicine. However, if a drug causes a harmful effect on the body, the drug is a poison. The same chemical can be a medicine and a poison depending on conditions of use and the person using it.
• Drug Receptor Interactions
Drugs interact with receptor sites localized in macromolecules which have protein-like properties and specific three dimensional shapes. A minimum three point attachment of a drug to a receptor site is required. In most cases a rather specific chemical structure is required for the receptor site and a complementary drug structure. Slight changes in the molecular structure of the drug may drastically change specificity.
• Enzyme Inhibition
Although activation of enzymes may be exploited therapeutically, most effects are produced by enzyme inhibition.
• Hallucinogenic Drugs
Hallucinogenic agents, also called psychomimetic agents, are capable of producing hallucinations, sensory illusions and bizarre thoughts. The primary effect of these compounds is to consistently alter thought and sensory perceptions.
• Local Anesthetics
Local anesthetics are agents that reversibly block the generation and conduction of nerve impulses along a nerve fiber. They depress impulses from sensory nerves of the skin, surfaces of mucosa, and muscles to the central nervous system. These agents are widely used in surgery, dentistry, and ophthalmology to block transmission of impulses in peripheral nerve endings.
• Misc Antibiotics
Antibiotics are specific chemical substances derived from or produced by living organisms that are capable of inhibiting the life processes of other organisms.
• Narcotic Analgesic Drugs
Narcotic agents are potent analgesics which are effective for the relief of severe pain. Analgesics are selective central nervous system depressants used to relieve pain. The term analgesic means "without pain". Even in therapeutic doses, narcotic analgesics can cause respiratory depression, nausea, and drowsiness.
• Penicillin
The penicillins were the first antibiotics discovered as natural products from the mold Penicillium.
• Sulfa Drugs
Sulfonamides are synthetic antimicrobial agents with a wide spectrum encompassing most gram-positive and many gram-negative organisms. These drugs were the first efficient treatment to be employed systematically for the prevention and cure of bacterial infections.
Thumbnail: Ritalin SR 20 mg, a brand-name sustained-release formulation of methylphenidate. (CC SA-BY 3.0; Sponge). | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/21%3A_The_Organic_Chemistry_of_Drugs%3A_Discovery_and_Design/21.01%3A_Naming_Drugs.txt |
A very broad definition of a drug would include "all chemicals other than food that affect living processes." If the affect helps the body, the drug is a medicine. However, if a drug causes a harmful effect on the body, the drug is a poison. The same chemical can be a medicine and a poison depending on conditions of use and the person using it. Another definition would be "medicinal agents used for diagnosis, prevention, treatment of symptoms, and cure of diseases." Contraceptives would be outside of this definition unless pregnancy were considered a disease.
• Anti-Cancer Drugs I
• Anti-Cancer Drugs II
• Antidepressants
Antidepressant drugs are used to restore mentally depressed patients to an improved mental status. Depression results from a deficiency of norepinephrine at receptors in the brain. Mechanisms that increase their effective concentration at the receptor sites should alleviate depression.
• Barbiturates and Benzodiazepines
Barbiturates are central nervous system depressants and are similar, in many ways, to the depressant effects of alcohol.
• Drugs Acting Upon the Central Nervous System
Chemical influences are capable of producing a myriad of effects on the activity and function of the central nervous system. Since our knowledge of different regions of brain function and the neurotransmitters in the brain is limited, the explanations for the mechanisms of drug action may be vague. The known neurotransmitters are: acetylcholine which is involved with memory and learning; norepinephrine which is involved with mania-depression and emotions.
• Drug Activity
Page notifications Off Barbiturates and Benzodiazepines Drug Receptor Interactions picture_as_pdf Batch Donate Table of contents A very broad definition of a drug would include "all chemicals other than food that affect living processes." If the affect helps the body, the drug is a medicine. However, if a drug causes a harmful effect on the body, the drug is a poison. The same chemical can be a medicine and a poison depending on conditions of use and the person using it.
• Drug Receptor Interactions
Drugs interact with receptor sites localized in macromolecules which have protein-like properties and specific three dimensional shapes. A minimum three point attachment of a drug to a receptor site is required. In most cases a rather specific chemical structure is required for the receptor site and a complementary drug structure. Slight changes in the molecular structure of the drug may drastically change specificity.
• Enzyme Inhibition
Although activation of enzymes may be exploited therapeutically, most effects are produced by enzyme inhibition.
• Hallucinogenic Drugs
Hallucinogenic agents, also called psychomimetic agents, are capable of producing hallucinations, sensory illusions and bizarre thoughts. The primary effect of these compounds is to consistently alter thought and sensory perceptions.
• Local Anesthetics
Local anesthetics are agents that reversibly block the generation and conduction of nerve impulses along a nerve fiber. They depress impulses from sensory nerves of the skin, surfaces of mucosa, and muscles to the central nervous system. These agents are widely used in surgery, dentistry, and ophthalmology to block transmission of impulses in peripheral nerve endings.
• Misc Antibiotics
Antibiotics are specific chemical substances derived from or produced by living organisms that are capable of inhibiting the life processes of other organisms.
• Narcotic Analgesic Drugs
Narcotic agents are potent analgesics which are effective for the relief of severe pain. Analgesics are selective central nervous system depressants used to relieve pain. The term analgesic means "without pain". Even in therapeutic doses, narcotic analgesics can cause respiratory depression, nausea, and drowsiness.
• Penicillin
The penicillins were the first antibiotics discovered as natural products from the mold Penicillium.
• Sulfa Drugs
Sulfonamides are synthetic antimicrobial agents with a wide spectrum encompassing most gram-positive and many gram-negative organisms. These drugs were the first efficient treatment to be employed systematically for the prevention and cure of bacterial infections.
Thumbnail: Ritalin SR 20 mg, a brand-name sustained-release formulation of methylphenidate. (CC SA-BY 3.0; Sponge). | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/21%3A_The_Organic_Chemistry_of_Drugs%3A_Discovery_and_Design/21.02%3A_Lead_Compounds.txt |
Unlike other drugs which act in the region of the synapse, local anesthetics are agents that reversibly block the generation and conduction of nerve impulses along a nerve fiber. They depress impulses from sensory nerves of the skin, surfaces of mucosa, and muscles to the central nervous system. These agents are widely used in surgery, dentistry, and ophthalmology to block transmission of impulses in peripheral nerve endings.
Introduction
Most local anesthetics can be represented by the following general formula. In both the official chemical name and the proprietary name, a local anesthetic drug can be recognized by the "-caine" ending. The ester linkage can also be an amide linkage. The most recent research indicates that the local anesthetic binds to a phospholipid in the nerve membrane and inhibits the ability of the phospholipid to bind Ca+2 ions.
Practically all of the free-base forms of the drugs are liquids. For this reason most of these drugs are used as salts (chloride, sulfate, etc.) which are water soluble, odorless, and crystalline solids. As esters these drugs are easily hydrolyzed with consequent loss of activity. The amide form of the drug is more stable and resistant to hydrolysis.
Benzocaine and Lidocaine
Two local anesthtics are shown below.
Contributors
Narcotic agents are potent analgesics which are effective for the relief of severe pain. Analgesics are selective central nervous system depressants used to relieve pain. The term analgesic means "without pain". Even in therapeutic doses, narcotic analgesics can cause respiratory depression, nausea, and drowsiness. Long term administration produces tolerance, psychic, and physical dependence called addiction.
Introduction
Narcotic agents may be classified into four categories:
1. Morphine and codeine - natural alkaloids of opium.
2. Synthetic derivatives of morphine such as heroin.
3. Synthetic agents which resemble the morphine structure.
4. Narcotic antagonists which are used as antidotes for overdoses of narcotic analgesics.
The main pharmacological action of analgesics is on the cerebrum and medulla of the central nervous system. Another effect is on the smooth muscle and glandular secretions of the respiratory and gastro-intestinal tract. The precise mechanism of action is unknown although the narcotics appear to interact with specific receptor sites to interfere with pain impulses.
Receptor Site
A schematic for an analgesic receptor site may look as shown in the graphic below with morphine. Three areas are needed: a flat areas to accommodate a flat nonpolar aromatic ring, a cavity to accept another series of rings perpendicular, and an anionic site for polar interaction of the amine group.
Enkephalins
Recently investigators have discovered two compounds in the brain called enkephalins which resemble morphine in structure. Each one is a peptide composed of 5 amino acids and differ only in the last amino acid. The peptide sequences are: tyr-gly-gly-phe-leu and tyr-gly-gly-phe-met. Molecular models show that the structures of the enkephalins has some similarities with morphine. The main feature in common appears to be the aromatic ring with the -OH group attached (tyr). Methadone and other similar analgesics have 2 aromatic rings which would be similar to the enkephalins (tyr and phe).
Analgesics may relieve pain by preventing the release of acetylcholine. Enkephalin molecules are released from a nerve cell and bind to analgesic receptor sites on the nerve cell sending the impulse. The binding of enkephalin or morphine-like drugs changes the shape of the nerve sending the impulse in such a fashion as to prevent the cell from releasing acetylcholine. As a result, the pain impulse cannot be transmitted and the brain does not preceive pain.
Morphine and Codeine
Morphine exerts a narcotic action manifested by analgesia, drowsiness, changes in mood, and mental clouding. The major medical action of morphine sought in the CNS is analgesia. Opiates suppress the "cough center" which is also located in the brain stem, the medulla. Such an action is thought to underlie the use of opiate narcotics as cough suppressants. Codeine appears to be particularly effective in this action and is widely used for this purpose.
Narcotic analgesics cause an addictive physical dependence. If the drug is discontinued, withdrawal symptoms are experienced. Although the reasons for addiction and withdrawal symptoms are not completely known, recent experiments have provided some information. A nucleotide known as cyclicadenosine monophosphate (cAMP) is synthesized with the aid of the enzyme adenylate cyclase. Enkephalin and morphine-like drugs inhibit this enzyme and thus decrease the amount of cAMP in the cells. In order to compensate for the decreased cAMP, the cells synthesize more enzyme in an attempt to produce more cAMP. Since more enzyme has been produced, more morphine is required as an inhibitor to keep the cAMP at a low level. This cycle repeats itself causing an increase in the tolerance level and increasing the amounts of morphine required. If morphine is suddenly withheld, withdrawal symptoms are probably caused by a high concentration of cAMP since the synthesizing enzyme, adenylate cyclase, is no longer being inhibited.
Morphine and codeine are contained in opium from the poppy (Papaver Somniterum) plant found in Turkey, Mexico, Southeast Asia, China, and India. This plant is 3-4 feet tall with 5-8 egg shaped capsules on top. Ten days after the poppy blooms in June, incisions are made in the capsules permitting a milky fluid to ooze out. The following day the gummy mass (now brown) is carefully scraped off and pressed into cakes of raw opium to dry.
Opium contains over 20 compounds but only morphine (10%) and codeine (0.5%) are of any importance. Morphine is extracted from the opium and isolated in a relatively pure form. Since codeine is in such low concentration, it is synthesized from morphine by an ether-type methylation of an alcohol group. Codeine has only a fraction of the potency compared to morphine. It is used with aspirin and as a cough suppressant.
Heroin
Heroin is synthesized from morphine by a relatively simple esterification reaction of two alcohol (phenol) groups with acetic anhydride (equivalent to acetic acid). Heroin is much more potent than morphine but without the respiratory depression effect. A possible reason may be that heroin passes the blood-brain barrier much more rapidly than morphine. Once in the brain, the heroin is hydrolyzed to morphine which is responsible for its activity.
Synthetic narcotic analgesics include meperidine and methadone. Meperidine is the most common subsitute for morphine. It exerts several pharmacological effects: analgesic, local anesthetic, and mild antihistamine. This multiple activity may be explained by its structural resemblance to morphine, atropine, and histamine.
Methadone is more active and more toxic than morphine. It can be used for the relief of may types of pain. In addition it is used as a narcotic substitute in addiction treatment because it prevents morphine abstinence syndrome. Methadone was synthesized by German chemists during Wold War II when the United States and our allies cut off their opium supply. And it is difficult to fight a war without analgesics so the Germans went to work and synthesized a number of medications in use today, including demerol and darvon which is structurally simular to methadone. And before we go further lets clear up another myth. Methadone, or dolophine was not named after Adolf Hitler. The "dol" in dolophine comes from the latin root "dolor." The female name Dolores is derived from it and the term dol is used in pain research to measure pain e.g., one dol is 1 unit of pain.
Even methadone, which looks strikingly different from other opioid agonists, has steric forces which produce a configuration that closely resembles that of other opiates. See the graphic on the left and the top graphic on this page. In other words, steric forces bend the molecule of methadone into the correct configuration to fit into the opiate receptor. When you take methadone it first must be metabolized in the liver to a product that your body can use. Excess methadone is also stored in the liver and blood stream and this is how methadone works its 'time release trick' and last for 24 hours or more. Once in the blood stream metabolized methadone is slowly passed to the brain when it is needed to fill opiate receptors. Methadone is the only effective treatment for heroin addiction. It works to smooth the ups and down of heroin craving and allows the person to function normally.
Narcotic Antagonists
Narcotic Antagonists prevent or abolish excessive respiratory depression caused by the administration of morphine or related compounds. They act by competing for the same analgesic receptor sites. They are structurally related to morphine with the exception of the group attached to nitrogen.
Nalorphine precipitates withdrawal symptoms and produces behavioral disturbances in addition to the antogonism action. Naloxane is a pure antagonist with no morphine like effects. It blocks the euphoric effect of heroin when given before heroin. Naltrexone became clinically available in 1985 as a new narcotic antagonist. Its actions resemble those of naloxone, but naltrexone is well is well absorbed orally and is long acting, necessitating only a dose of 50 to 100 mg. Therefore, it is useful in narcotic treatment programs where it is desired to maintain an individual on chronic therapy with a narcotic antagonist. In individuals taking naltrexone, subsequent injection of an opiate will produce little or no effect. Naltrexone appears to be particularly effective for the treatment of narcotic dependence in addicts who have more to gain by being drug-free rather than drug dependant
Contributors and Attributions
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
• Poppies image from: leda.lycaeum.org | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/21%3A_The_Organic_Chemistry_of_Drugs%3A_Discovery_and_Design/21.03%3A_Molecular_Modification.txt |
Ensemble properties result from being or relating to the greater in number of atoms in a sample. This are in contrast to atomic or molecular properties.
• Capillary Action
Capillary action can be defined as the ascension of liquids through slim tube, cylinder or permeable substance due to adhesive and cohesive forces interacting between the liquid and the surface. When intermolecular bonding of a liquid itself is substantially inferior to a substances’ surface it is interacting, capillarity occurs. Also, the diameter of the container as well as the gravitational forces will determine amount of liquid raised.
• Cohesive and Adhesive Forces
Cohesive and adhesive forces are associated with bulk (or macroscopic) properties and hence the terms are not applicable to discussion of atomic and molecular properties. When a liquid comes into contact with a surface (such as the walls of a graduated cylinder or a tabletop), both cohesive and adhesive forces will act on it. These forces govern the shape which the liquid takes on.
• Contact Angles
Contact angle is one of the common ways to measure the wettability of a surface or material. Wetting refers to the study of how a liquid deposited on a solid (or liquid) substrate spreads out or the ability of liquids to form boundary surfaces with solid states. The wetting, as mentioned before is determined by measuring the contact angle, which the liquid forms in contact with the solids or liquids. The wetting tendency is larger, the smaller the contact angle or the surface tension is.
• Surface Tension
Surface tension is the energy, or work, required to increase the surface area of a liquid due to intermolecular forces. Since these intermolecular forces vary depending on the nature of the liquid (e.g. water vs. gasoline) or solutes in the liquid (e.g. surfactants like detergent), each solution exhibits differing surface tension properties.
• Unusual Properties of Water
With 70% of our earth being ocean water and 65% of our bodies being water, it is hard to not be aware of how important it is in our lives. There are 3 different forms of water, or H2O: solid (ice), liquid (water), and gas (steam). Because water seems so ubiquitous, many people are unaware of the unusual and unique properties of water, including:
• Vapor Pressure
Pressure is the average force that material (gas, liquid or solid) exert upon the surface, e.g. walls of a container or other confining boundary. Vapor pressure or equilibrium vapor pressure is the pressure of a vapor in thermodynamic equilibrium with its condensed phases in a closed container. All liquids and solids have a tendency to evaporate or sublime into a gaseous form and all gases have a tendency to condense back to their liquid or solid form.
• Viscosity
Viscosity is another type of bulk property defined as a liquid’s resistance to flow. When the intermolecular forces of attraction are strong within a liquid, there is a larger viscosity. An example of this phenomenon is imagining a race between two liquids down a windshield. Which would you expect to roll down the windshield faster honey or water? Obviously from experience one would expect water to easily speed right past the honey, a fact that reveals honey has a much higher viscocity than wate
• Wetting Agents
A substance is referred to as a wetting agent if it lowers the surface tension of a liquid and thus allows it to spread more easily.
Thumbnail: A water drop on a lotus leaf surface showing contact angles of approximately 147°. (Public Domain; Na2jojon). | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/21%3A_The_Organic_Chemistry_of_Drugs%3A_Discovery_and_Design/21.04%3A_Random_Screening.txt |
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
21.06: Receptors
The vast majority of drugs show a remarkably high correlation of structure and specificity to produce pharmacological effects. Experimental evidence indicates that drugs interact with receptor sites localized in macromolecules which have protein-like properties and specific three dimensional shapes. A minimum three point attachment of a drug to a receptor site is required. In most cases a rather specific chemical structure is required for the receptor site and a complementary drug structure. Slight changes in the molecular structure of the drug may drastically change specificity.
Introduction
Several chemical forces may result in a temporary binding of the drug to the receptor. Essentially any bond could be involved with the drug-receptor interaction. Covalent bonds would be very tight and practically irreversible. Since by definition the drug-receptor interaction is reversible, covalent bond formation is rather rare except in a rather toxic situation. Since many drugs contain acid or amine functional groups which are ionized at physiological pH, ionic bonds are formed by the attraction of opposite charges in the receptor site.
Polar-polar interactions as in hydrogen bonding are a further extension of the attraction of opposite charges. The drug-receptor reaction is essentially an exchange of the hydrogen bond between a drug molecule, surrounding water, and the receptor site.
Finally hydrophobic bonds are formed between non-polar hydrocarbon groups on the drug and those in the receptor site. These bonds are not very specific but the interactions do occur to exclude water molecules. Repulsive forces which decrease the stability of the drug-receptor interaction include repulsion of like charges and steric hindrance. Steric hindrance refers to certain 3-dimensional features where repulsion occurs between electron clouds, inflexible chemical bonds, or bulky alkyl groups.
Drug Interaction with Receptor Site
• A neurotransmitter has a specific shape to fit into a receptor site and cause a pharmacological response such as a nerve impulse being sent. The neurotransmitter is similar to a substrate in an enzyme interaction. After attachment to a receptor site, a drug may either initiate a response or prevent a response from occurring. A drug must be a close "mimic" of the neurotransmitter.
• An agonist is a drug which produces a stimulation type response. The agonist is a very close mimic and "fits" with the receptor site and is thus able to initiate a response.
• An antagonist drug interacts with the receptor site and blocks or depresses the normal response for that receptor because it only partially fits the receptor site and can not produce an effect. However, it does block the site preventing any other agonist or the normal neurotransmitter from interacting with the receptor site. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/21%3A_The_Organic_Chemistry_of_Drugs%3A_Discovery_and_Design/21.05%3A_Serendipity_in_Drug_Development.txt |
Learning Objectives
• Explain what an enzyme inhibitor is.
• Distinguish between reversible and irreversible inhibitors.
• Distinguish between competitive and noncompetitive inhibitors.
Previously, we noted that enzymes are inactivated at high temperatures and by changes in pH. These are nonspecific factors that would inactivate any enzyme. The activity of enzymes can also be regulated by more specific inhibitors. Many compounds are poisons because they bind covalently to particular enzymes or kinds of enzymes and inactivate them (Table \(1\)).
Table \(1\): Poisons as Enzyme Inhibitors
Poison Formula Example of Enzyme Inhibited Action
arsenate \(\ce{AsO4^{3−}}\) glyceraldehyde 3-phosphate dehydrogenase substitutes for phosphate
iodoacetate \(\ce{ICH2COO^{−}}\) triose phosphate dehydrogenase binds to cysteine \(\ce{SH}\) group
diisopropylfluoro-phosphate (DIFP; a nerve poison) acetylcholinesterase binds to serine \(\ce{OH}\) group
Irreversible Inhibition: Poisons
An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. The inhibitor-enzyme bond is so strong that the inhibition cannot be reversed by the addition of excess substrate. The nerve gases, especially Diisopropyl fluorophosphate (DIFP), irreversibly inhibit biological systems by forming an enzyme-inhibitor complex with a specific OH group of serine situated at the active sites of certain enzymes. The peptidases trypsin and chymotrypsin contain serine groups at the active site and are inhibited by DIFP.
Reversible Inhibition
A reversible inhibitor inactivates an enzyme through noncovalent, more easily reversed, interactions. Unlike an irreversible inhibitor, a reversible inhibitor can dissociate from the enzyme. Reversible inhibitors include competitive inhibitors and noncompetitive inhibitors. (There are additional types of reversible inhibitors.) A competitive inhibitor is any compound that bears a structural resemblance to a particular substrate and thus competes with that substrate for binding at the active site of an enzyme. The inhibitor is not acted on by the enzyme but does prevent the substrate from approaching the active site.
The degree to which a competitive inhibitor interferes with an enzyme’s activity depends on the relative concentrations of the substrate and the inhibitor. If the inhibitor is present in relatively large quantities, it will initially block most of the active sites. But because the binding is reversible, some substrate molecules will eventually bind to the active site and be converted to product. Increasing the substrate concentration promotes displacement of the inhibitor from the active site. Competitive inhibition can be completely reversed by adding substrate so that it reaches a much higher concentration than that of the inhibitor.
Studies of competitive inhibition have provided helpful information about certain enzyme-substrate complexes and the interactions of specific groups at the active sites. As a result, pharmaceutical companies have synthesized drugs that competitively inhibit metabolic processes in bacteria and certain cancer cells. Many drugs are competitive inhibitors of specific enzymes.
A classic example of competitive inhibition is the effect of malonate on the enzyme activity of succinate dehydrogenase (Figure \(1\)). Malonate and succinate are the anions of dicarboxylic acids and contain three and four carbon atoms, respectively. The malonate molecule binds to the active site because the spacing of its carboxyl groups is not greatly different from that of succinate. However, no catalytic reaction occurs because malonate does not have a CH2CH2 group to convert to CH=CH. This reaction will also be discussed in connection with the Krebs cycle and energy production.
To Your Health: Penicillin
Chemotherapy is the strategic use of chemicals (that is, drugs) to destroy infectious microorganisms or cancer cells without causing excessive damage to the other, healthy cells of the host. From bacteria to humans, the metabolic pathways of all living organisms are quite similar, so the search for safe and effective chemotherapeutic agents is a formidable task. Many well-established chemotherapeutic drugs function by inhibiting a critical enzyme in the cells of the invading organism.
An antibiotic is a compound that kills bacteria; it may come from a natural source such as molds or be synthesized with a structure analogous to a naturally occurring antibacterial compound. Antibiotics constitute no well-defined class of chemically related substances, but many of them work by effectively inhibiting a variety of enzymes essential to bacterial growth.
Penicillin, one of the most widely used antibiotics in the world, was fortuitously discovered by Alexander Fleming in 1928, when he noticed antibacterial properties in a mold growing on a bacterial culture plate. In 1938, Ernst Chain and Howard Florey began an intensive effort to isolate penicillin from the mold and study its properties. The large quantities of penicillin needed for this research became available through development of a corn-based nutrient medium that the mold loved and through the discovery of a higher-yielding strain of mold at a United States Department of Agriculture research center near Peoria, Illinois. Even so, it was not until 1944 that large quantities of penicillin were being produced and made available for the treatment of bacterial infections.
Penicillin functions by interfering with the synthesis of cell walls of reproducing bacteria. It does so by inhibiting an enzyme—transpeptidase—that catalyzes the last step in bacterial cell-wall biosynthesis. The defective walls cause bacterial cells to burst. Human cells are not affected because they have cell membranes, not cell walls.
Several naturally occurring penicillins have been isolated. They are distinguished by different R groups connected to a common structure: a four-member cyclic amide (called a lactam ring) fused to a five-member ring. The addition of appropriate organic compounds to the culture medium leads to the production of the different kinds of penicillin.
The penicillins are effective against gram-positive bacteria (bacteria capable of being stained by Gram’s stain) and a few gram-negative bacteria (including the intestinal bacterium Escherichia coli). They are effective in the treatment of diphtheria, gonorrhea, pneumonia, syphilis, many pus infections, and certain types of boils. Penicillin G was the earliest penicillin to be used on a wide scale. However, it cannot be administered orally because it is quite unstable; the acidic pH of the stomach converts it to an inactive derivative. The major oral penicillins—penicillin V, ampicillin, and amoxicillin—on the other hand, are acid stable.
Some strains of bacteria become resistant to penicillin through a mutation that allows them to synthesize an enzyme—penicillinase—that breaks the antibiotic down (by cleavage of the amide linkage in the lactam ring). To combat these strains, scientists have synthesized penicillin analogs (such as methicillin) that are not inactivated by penicillinase.
Some people (perhaps 5% of the population) are allergic to penicillin and therefore must be treated with other antibiotics. Their allergic reaction can be so severe that a fatal coma may occur if penicillin is inadvertently administered to them. Fortunately, several other antibiotics have been discovered. Most, including aureomycin and streptomycin, are the products of microbial synthesis. Others, such as the semisynthetic penicillins and tetracyclines, are made by chemical modifications of antibiotics; and some, like chloramphenicol, are manufactured entirely by chemical synthesis. They are as effective as penicillin in destroying infectious microorganisms. Many of these antibiotics exert their effects by blocking protein synthesis in microorganisms.
Initially, antibiotics were considered miracle drugs, substantially reducing the number of deaths from blood poisoning, pneumonia, and other infectious diseases. Some seven decades ago, a person with a major infection almost always died. Today, such deaths are rare. Seven decades ago, pneumonia was a dreaded killer of people of all ages. Today, it kills only the very old or those ill from other causes. Antibiotics have indeed worked miracles in our time, but even miracle drugs have limitations. Not long after the drugs were first used, disease organisms began to develop strains resistant to them. In a race to stay ahead of resistant bacterial strains, scientists continue to seek new antibiotics. The penicillins have now been partially displaced by related compounds, such as the cephalosporins and vancomycin. Unfortunately, some strains of bacteria have already shown resistance to these antibiotics.
Some reversible inhibitors are noncompetitive. A noncompetitive inhibitor can combine with either the free enzyme or the enzyme-substrate complex because its binding site on the enzyme is distinct from the active site. Binding of this kind of inhibitor alters the three-dimensional conformation of the enzyme, changing the configuration of the active site with one of two results. Either the enzyme-substrate complex does not form at its normal rate, or, once formed, it does not yield products at the normal rate. Because the inhibitor does not structurally resemble the substrate, the addition of excess substrate does not reverse the inhibitory effect.
Feedback inhibition is a normal biochemical process that makes use of noncompetitive inhibitors to control some enzymatic activity. In this process, the final product inhibits the enzyme that catalyzes the first step in a series of reactions. Feedback inhibition is used to regulate the synthesis of many amino acids. For example, bacteria synthesize isoleucine from threonine in a series of five enzyme-catalyzed steps. As the concentration of isoleucine increases, some of it binds as a noncompetitive inhibitor to the first enzyme of the series (threonine deaminase), thus bringing about a decrease in the amount of isoleucine being formed (Figure \(2\)).
Summary
An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. A reversible inhibitor inactivates an enzyme through noncovalent, reversible interactions. A competitive inhibitor competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site.
21.09: Antiviral Drugs
Sulfonamides are synthetic antimicrobial agents with a wide spectrum encompassing most gram-positive and many gram-negative organisms. These drugs were the first efficient treatment to be employed systematically for the prevention and cure of bacterial infections.
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
21.10: Economics of Drugs (Governmental Regulations)
Contributors and Attributions
• Nathalie Interiano
21.10: Economics of Drugs (Governmental Regulations)
Contributors and Attributions
• Nathalie Interiano | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/21%3A_The_Organic_Chemistry_of_Drugs%3A_Discovery_and_Design/21.07%3A_Drug_Resistance.txt |
• Basic Electronic Structure of Atoms
This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies.
• Case Study: Electron Configuration of Mn vs. Cu
• d-orbital Occupation and Electronic Configurations
• Electronic Configurations Intro
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to electro
• Electronic Structure and Orbitals
• Hund's Rules
Hund's rule states that: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin). When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible.
• Pauli Exclusion Principle
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one is assigned an up-spin ( +1/2), the other must be down-spin (-1/2).
• Spin Pairing Energy
Spin pairing energy refers to the energy associated with paired electrons sharing one orbital and its effect on the molecules surrounding it. Electron pairing determining the direction of spin depends on several laws founded by chemists over the years such as Hund's law, the Aufbau principle, and Pauli's exclusion principle. An overview of the different types laws associated with the electron pairing rules.
• Spin Quantum Number
The Spin Quantum Number (s) is a value (of 1/2) that describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -). This vector is called the magnetic spin quantum number (ms).
• The Aufbau Process
The Aufbau model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons.
• The Octet Rule
The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds.
• The Order of Filling 3d and 4s Orbitals
This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. The way that the order of filling of orbitals is normally taught gives an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.01%3A_The_Structure_of_an_Atom.txt |
• Basic Electronic Structure of Atoms
This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies.
• Case Study: Electron Configuration of Mn vs. Cu
• d-orbital Occupation and Electronic Configurations
• Electronic Configurations Intro
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to electro
• Electronic Structure and Orbitals
• Hund's Rules
Hund's rule states that: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin). When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible.
• Pauli Exclusion Principle
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one is assigned an up-spin ( +1/2), the other must be down-spin (-1/2).
• Spin Pairing Energy
Spin pairing energy refers to the energy associated with paired electrons sharing one orbital and its effect on the molecules surrounding it. Electron pairing determining the direction of spin depends on several laws founded by chemists over the years such as Hund's law, the Aufbau principle, and Pauli's exclusion principle. An overview of the different types laws associated with the electron pairing rules.
• Spin Quantum Number
The Spin Quantum Number (s) is a value (of 1/2) that describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -). This vector is called the magnetic spin quantum number (ms).
• The Aufbau Process
The Aufbau model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons.
• The Octet Rule
The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds.
• The Order of Filling 3d and 4s Orbitals
This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. The way that the order of filling of orbitals is normally taught gives an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.02%3A_How_Electrons_in_an_Atom_are_Distributed.txt |
There are many types of chemical bonds and forces that bind molecules together. The two most basic types of bonds are characterized as either ionic or covalent. In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.
Introduction
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion. Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal.
Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. In ionic bonding, more than 1 electron can be donated or received to satisfy the octet rule. The charges on the anion and cation correspond to the number of electrons donated or received. In ionic bonds, the net charge of the compound must be zero.
This sodium molecule donates the lone electron in its valence orbital in order to achieve octet configuration. This creates a positively charged cation due to the loss of electron.
This chlorine atom receives one electron to achieve its octet configuration, which creates a negatively charged anion.
The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of the electrostatic attraction between the particles. At the ideal interatomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ion.
Example \(1\): Chloride Salts
In this example, the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.
In this example, the magnesium atom is donating both of its valence electrons to chlorine atoms. Each chlorine atom can only accept 1 electron before it can achieve its noble gas configuration; therefore, 2 atoms of chlorine are required to accept the 2 electrons donated by the magnesium. Notice that the net charge of the compound is 0.
Covalent Bonding
Covalent bonding is the sharing of electrons between atoms. This type of bonding occurs between two atoms of the same element or of elements close to each other in the periodic table. This bonding occurs primarily between nonmetals; however, it can also be observed between nonmetals and metals.
If atoms have similar electronegativities (the same affinity for electrons), covalent bonds are most likely to occur. Because both atoms have the same affinity for electrons and neither has a tendency to donate them, they share electrons in order to achieve octet configuration and become more stable. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur. For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet. To form ionic bonds, Carbon molecules must either gain or lose 4 electrons. This is highly unfavorable; therefore, carbon molecules share their 4 valence electrons through single, double, and triple bonds so that each atom can achieve noble gas configurations. Covalent bonds include interactions of the sigma and pi orbitals; therefore, covalent bonds lead to formation of single, double, triple, and quadruple bonds.
Example \(2\): \(PCl_3\)
In this example, a phosphorous atom is sharing its three unpaired electrons with three chlorine atoms. In the end product, all four of these molecules have 8 valence electrons and satisfy the octet rule.
Bonding in Organic Chemistry
Ionic and covalent bonds are the two extremes of bonding. Polar covalent is the intermediate type of bonding between the two extremes. Some ionic bonds contain covalent characteristics and some covalent bonds are partially ionic. For example, most carbon-based compounds are covalently bonded but can also be partially ionic. Polarity is a measure of the separation of charge in a compound. A compound's polarity is dependent on the symmetry of the compound and on differences in electronegativity between atoms. Polarity occurs when the electron pushing elements, found on the left side of the periodic table, exchanges electrons with the electron pulling elements, on the right side of the table. This creates a spectrum of polarity, with ionic (polar) at one extreme, covalent (nonpolar) at another, and polar covalent in the middle.
Both of these bonds are important in organic chemistry. Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life.
Problems
1. Are these compounds ionic or covalent?
2. In the following reactions, indicate whether the reactants and products are ionic or covalently bonded.
a)
b) Clarification: What is the nature of the bond between sodium and amide? What kind of bond forms between the anion carbon chain and sodium?
c)
Solutions
• 1) From left to right: Covalent, Ionic, Ionic, Covalent, Covalent, Covalent, Ionic.
• 2a) All products and reactants are ionic.
• 2b) From left to right: Covalent, Ionic, Ionic, Covalent, Ionic, Covalent, Covalent, Ionic.
• 2c) All products and reactants are covalent. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.03%3A_Ionic_and_Covalent_Bonds.txt |
Objectives
After completing this section, you should be able to
1. describe the physical significance of an orbital.
2. list the atomic orbitals from 1s to 3d in order of increasing energy.
3. sketch the shapes of s and p orbitals.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• nodal plane
• node
• orbital
• quantum mechanics
• wave function
For a refresher on quantum numbers view this text
Atomic Orbitals
An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of having an electron.
One way of representing electron probability distributions was illustrated in Figure 6.5.2 for the 1s orbital of hydrogen. Because Ψ2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ2 versus distance from the nucleus (r) is a plot of the probability density. The 1s orbital is spherically symmetrical, so the probability of finding a 1s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at r = 0 (at the nucleus) and decreases steadily with increasing distance. At very large values of r, the electron probability density is very small but not zero.
In contrast, we can calculate the radial probability (the probability of finding a 1s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r1, r2, r3,…, rx − 1, rx. In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (part (a) in Figure \(1\)), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (part (b) in Figure \(1\)), so the density of dots is greatest for the smallest spherical shells in part (a) in Figure \(1\). In contrast, the surface area of each spherical shell is equal to 4πr2, which increases very rapidly with increasing r (part (c) in Figure \(1\)). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (part (d) in Figure \(1\)). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (part (d) in Figure \(1\)).
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the quantum mechanical Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
Figure \(2\) compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (part (c) in Figure \(2\)). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability.
s Orbitals
Three things happen to s orbitals as n increases (Figure \(2\)):
1. They become larger, extending farther from the nucleus.
2. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude.
3. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density, as was shown for the hydrogen 1s, 2s, and 3s orbitals in part (b) in Figure \(2\). Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
p Orbitals
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals.
The electron probability distribution for one of the hydrogen 2p orbitals is shown in Figure \(3\). Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a 2pz orbital. As shown in Figure \(4\), the other two 2p orbitals have identical shapes, but they lie along the x axis (2px) and y axis (2py), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative.
Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in Figure \(4\).
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.05%3A_Atomic_Orbitals.txt |
Hybridization was introduced to explain molecular structure when the valence bond theory failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109o, 120o, or 180o. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles.
Introduction
Carbon is a perfect example showing the value of hybrid orbitals. Carbon's ground state configuration is:
According to Valence Bond Theory, carbon should form two covalent bonds, resulting in a CH2, because it has two unpaired electrons in its electronic configuration.However, experiments have shown that \(CH_2\) is highly reactive and cannot exist outside of a reaction. Therefore, this does not explain how CH4 can exist. To form four bonds the configuration of carbon must have four unpaired electrons.
One way CH4 can be explained is, the 2s and the 3 2p orbitals combine to make four, equal energy sp3 hybrid orbitals. That would give us the following configuration:
Now that carbon has four unpaired electrons it can have four equal energy bonds. The hybridization of orbitals is favored because hybridized orbitals are more directional which leads to greater overlap when forming bonds, therefore the bonds formed are stronger. This results in more stable compounds when hybridization occurs.
The next section will explain the various types of hybridization and how each type helps explain the structure of certain molecules.
sp3 hybridization
sp3 hybridization can explain the tetrahedral structure of molecules. In it, the 2s orbitals and all three of the 2p orbitals hybridize to form four sp3 orbitals, each consisting of 75% p character and 25% s character. The frontal lobes align themselves in the manner shown below. In this structure, electron repulsion is minimized.
Energy changes occurring in hybridization
Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5o from each other. This 109.5o arrangement gives tetrahedral geometry (Figure 4).
Example: sp3 Hybridization in Methane
Because carbon plays such a significant role in organic chemistry, we will be using it as an example here. Carbon's 2s and all three of its 2p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane. The resulting shape is tetrahedral, since that minimizes electron repulsion.
Hybridization
Lone Pairs: Remember to take into account lone pairs of electrons. These lone pairs cannot double bond so they are placed in their own hybrid orbital. This is why H2O is tetrahedral. We can also build sp3d and sp3d2 hybrid orbitals if we go beyond s and p subshells.
sp2 hybridization
sp2 hybridization can explain the trigonal planar structure of molecules. In it, the 2s orbitals and two of the 2p orbitals hybridize to form three sp orbitals, each consisting of 67% p and 33% s character. The frontal lobes align themselves in the trigonal planar structure, pointing to the corners of a triangle in order to minimize electron repulsion and to improve overlap. The remaining p orbital remains unchanged and is perpendicular to the plane of the three sp2 orbitals.
Energy changes occurring in hybridization
Hybridization of an s orbital with two p orbitals (px and py) results in three sp2 hybrid orbitals that are oriented at 120o angle to each other (Figure 3). Sp2 hybridization results in trigonal geometry.
Example: sp2 Hybridization in Aluminum Trihydride
In aluminum trihydride, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals that align themselves in the trigonal planar structure. The three Al sp2 orbitals bond with with 1s orbitals from the three hydrogens through sp2-s orbital overlap.
Example: sp2 Hybridization in Ethene
Similar hybridization occurs in each carbon of ethene. For each carbon, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals. These hybridized orbitals align themselves in the trigonal planar structure. For each carbon, two of these sp orbitals bond with two 1s hydrogen orbitals through s-sp orbital overlap. The remaining sp2 orbitals on each carbon are bonded with each other, forming a bond between each carbon through sp2-sp2 orbital overlap. This leaves us with the two p orbitals on each carbon that have a single carbon in them. These orbitals form a ? bonds through p-p orbital overlap, creating a double bond between the two carbons. Because a double bond was created, the overall structure of the ethene compound is linear. However, the structure of each molecule in ethene, the two carbons, is still trigonal planar.
sp Hybridization
sp Hybridization can explain the linear structure in molecules. In it, the 2s orbital and one of the 2p orbitals hybridize to form two sp orbitals, each consisting of 50% s and 50% p character. The front lobes face away from each other and form a straight line leaving a 180° angle between the two orbitals. This formation minimizes electron repulsion. Because only one p orbital was used, we are left with two unaltered 2p orbitals that the atom can use. These p orbitals are at right angles to one another and to the line formed by the two sp orbitals.
Energy changes occurring in hybridization
Figure 1: Notice how the energy of the electrons lowers when hybridized.
These p orbitals come into play in compounds such as ethyne where they form two addition? bonds, resulting in in a triple bond. This only happens when two atoms, such as two carbons, both have two p orbitals that each contain an electron. An sp hybrid orbital results when an s orbital is combined with p orbital (Figure 2). We will get two sp hybrid orbitals since we started with two orbitals (s and p). sp hybridization results in a pair of directional sp hybrid orbitals pointed in opposite directions. These hybridized orbitals result in higher electron density in the bonding region for a sigma bond toward the left of the atom and for another sigma bond toward the right. In addition, sp hybridization provides linear geometry with a bond angle of 180o.
Example: sp Hybridization in Magnesium Hydride
In magnesium hydride, the 3s orbital and one of the 3p orbitals from magnesium hybridize to form two sp orbitals. The two frontal lobes of the sp orbitals face away from each other forming a straight line leading to a linear structure. These two sp orbitals bond with the two 1s orbitals of the two hydrogen atoms through sp-s orbital overlap.
Hybridization
Example: sp Hybridization in Ethyne
The hybridization in ethyne is similar to the hybridization in magnesium hydride. For each carbon, the 2s orbital hybridizes with one of the 2p orbitals to form two sp hybridized orbitals. The frontal lobes of these orbitals face away from each other forming a straight line. The first bond consists of sp-sp orbital overlap between the two carbons. Another two bonds consist of s-sp orbital overlap between the sp hybridized orbitals of the carbons and the 1s orbitals of the hydrogens. This leaves us with two p orbitals on each carbon that have a single carbon in them. This allows for the formation of two ? bonds through p-p orbital overlap. The linear shape, or 180° angle, is formed because electron repulsion is minimized the greatest in this position.
Hybridization
Problems
Using the Lewis Structures, try to figure out the hybridization (sp, sp2, sp3) of the indicated atom and indicate the atom's shape.
1. The carbon.
2. The oxygen.
3. The carbon on the right.
Answers
1. sp2- Trigonal Planar
The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp2.
2. sp3 - Tetrahedral
Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp3 orbital. That makes 4 orbitals, aka sp3.
3. sp - Linear
The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp.
An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it. It's a lot easier to figure out the hybridization this way.
Contributors
• Harpreet Chima (UCD), Farah Yasmeen | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.07%3A_How_Single_Bonds_Are_Formed_in_Organic_Compounds.txt |
Hybridization was introduced to explain molecular structure when the valence bond theory failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109o, 120o, or 180o. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles.
Introduction
Carbon is a perfect example showing the value of hybrid orbitals. Carbon's ground state configuration is:
According to Valence Bond Theory, carbon should form two covalent bonds, resulting in a CH2, because it has two unpaired electrons in its electronic configuration.However, experiments have shown that \(CH_2\) is highly reactive and cannot exist outside of a reaction. Therefore, this does not explain how CH4 can exist. To form four bonds the configuration of carbon must have four unpaired electrons.
One way CH4 can be explained is, the 2s and the 3 2p orbitals combine to make four, equal energy sp3 hybrid orbitals. That would give us the following configuration:
Now that carbon has four unpaired electrons it can have four equal energy bonds. The hybridization of orbitals is favored because hybridized orbitals are more directional which leads to greater overlap when forming bonds, therefore the bonds formed are stronger. This results in more stable compounds when hybridization occurs.
The next section will explain the various types of hybridization and how each type helps explain the structure of certain molecules.
sp3 hybridization
sp3 hybridization can explain the tetrahedral structure of molecules. In it, the 2s orbitals and all three of the 2p orbitals hybridize to form four sp3 orbitals, each consisting of 75% p character and 25% s character. The frontal lobes align themselves in the manner shown below. In this structure, electron repulsion is minimized.
Energy changes occurring in hybridization
Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5o from each other. This 109.5o arrangement gives tetrahedral geometry (Figure 4).
Example: sp3 Hybridization in Methane
Because carbon plays such a significant role in organic chemistry, we will be using it as an example here. Carbon's 2s and all three of its 2p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane. The resulting shape is tetrahedral, since that minimizes electron repulsion.
Hybridization
Lone Pairs: Remember to take into account lone pairs of electrons. These lone pairs cannot double bond so they are placed in their own hybrid orbital. This is why H2O is tetrahedral. We can also build sp3d and sp3d2 hybrid orbitals if we go beyond s and p subshells.
sp2 hybridization
sp2 hybridization can explain the trigonal planar structure of molecules. In it, the 2s orbitals and two of the 2p orbitals hybridize to form three sp orbitals, each consisting of 67% p and 33% s character. The frontal lobes align themselves in the trigonal planar structure, pointing to the corners of a triangle in order to minimize electron repulsion and to improve overlap. The remaining p orbital remains unchanged and is perpendicular to the plane of the three sp2 orbitals.
Energy changes occurring in hybridization
Hybridization of an s orbital with two p orbitals (px and py) results in three sp2 hybrid orbitals that are oriented at 120o angle to each other (Figure 3). Sp2 hybridization results in trigonal geometry.
Example: sp2 Hybridization in Aluminum Trihydride
In aluminum trihydride, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals that align themselves in the trigonal planar structure. The three Al sp2 orbitals bond with with 1s orbitals from the three hydrogens through sp2-s orbital overlap.
Example: sp2 Hybridization in Ethene
Similar hybridization occurs in each carbon of ethene. For each carbon, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals. These hybridized orbitals align themselves in the trigonal planar structure. For each carbon, two of these sp orbitals bond with two 1s hydrogen orbitals through s-sp orbital overlap. The remaining sp2 orbitals on each carbon are bonded with each other, forming a bond between each carbon through sp2-sp2 orbital overlap. This leaves us with the two p orbitals on each carbon that have a single carbon in them. These orbitals form a ? bonds through p-p orbital overlap, creating a double bond between the two carbons. Because a double bond was created, the overall structure of the ethene compound is linear. However, the structure of each molecule in ethene, the two carbons, is still trigonal planar.
sp Hybridization
sp Hybridization can explain the linear structure in molecules. In it, the 2s orbital and one of the 2p orbitals hybridize to form two sp orbitals, each consisting of 50% s and 50% p character. The front lobes face away from each other and form a straight line leaving a 180° angle between the two orbitals. This formation minimizes electron repulsion. Because only one p orbital was used, we are left with two unaltered 2p orbitals that the atom can use. These p orbitals are at right angles to one another and to the line formed by the two sp orbitals.
Energy changes occurring in hybridization
Figure 1: Notice how the energy of the electrons lowers when hybridized.
These p orbitals come into play in compounds such as ethyne where they form two addition? bonds, resulting in in a triple bond. This only happens when two atoms, such as two carbons, both have two p orbitals that each contain an electron. An sp hybrid orbital results when an s orbital is combined with p orbital (Figure 2). We will get two sp hybrid orbitals since we started with two orbitals (s and p). sp hybridization results in a pair of directional sp hybrid orbitals pointed in opposite directions. These hybridized orbitals result in higher electron density in the bonding region for a sigma bond toward the left of the atom and for another sigma bond toward the right. In addition, sp hybridization provides linear geometry with a bond angle of 180o.
Example: sp Hybridization in Magnesium Hydride
In magnesium hydride, the 3s orbital and one of the 3p orbitals from magnesium hybridize to form two sp orbitals. The two frontal lobes of the sp orbitals face away from each other forming a straight line leading to a linear structure. These two sp orbitals bond with the two 1s orbitals of the two hydrogen atoms through sp-s orbital overlap.
Hybridization
Example: sp Hybridization in Ethyne
The hybridization in ethyne is similar to the hybridization in magnesium hydride. For each carbon, the 2s orbital hybridizes with one of the 2p orbitals to form two sp hybridized orbitals. The frontal lobes of these orbitals face away from each other forming a straight line. The first bond consists of sp-sp orbital overlap between the two carbons. Another two bonds consist of s-sp orbital overlap between the sp hybridized orbitals of the carbons and the 1s orbitals of the hydrogens. This leaves us with two p orbitals on each carbon that have a single carbon in them. This allows for the formation of two ? bonds through p-p orbital overlap. The linear shape, or 180° angle, is formed because electron repulsion is minimized the greatest in this position.
Hybridization
Problems
Using the Lewis Structures, try to figure out the hybridization (sp, sp2, sp3) of the indicated atom and indicate the atom's shape.
1. The carbon.
2. The oxygen.
3. The carbon on the right.
Answers
1. sp2- Trigonal Planar
The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp2.
2. sp3 - Tetrahedral
Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp3 orbital. That makes 4 orbitals, aka sp3.
3. sp - Linear
The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp.
An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it. It's a lot easier to figure out the hybridization this way.
Contributors
• Harpreet Chima (UCD), Farah Yasmeen | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.09%3A_How_a_Triple_Bond_is_Formed-_The_Bonds_in_Ethyne.txt |
Solutions to exercises
17.1A: 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.
Template:ExampleStart
Exercise 17.1: Draw the structure of a benzylic radical compound, and then draw a resonance form showing how the radical is stabilized.
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With enough resonance stabilization, radicals can be made that are quite unreactive. One example of an inert organic radical structure is shown below.
In this molecule, the already extensive resonance stabilization is further enhanced by the ability of the chlorine atoms to shield the radical center from external reagents. The radical is, in some sense, inside a protective 'cage'.
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Exercise 17.2: Draw a resonance contributor of the structure above in which the unpaired electron is located on a chlorine atom.
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17.1B: The diradical character of triplet oxygen
You may be surprised to learn that molecular oxygen (O2) often reacts like a radical species - or more accurately, like a diradical, with two separate unpaired valence electrons. This puzzling phenomenon is best explained by molecular orbital theory (you may want to go back to chapter 2 at this point to review basic MO theory).
In molecular orbital energy diagram form, the configuration of O2 looks like this:
When the molecular orbitals of O2 are filled up with electrons according to Hund's rule (section 1.1C) , the HOMOs (Highest Occupied Molecular Orbitals) are the two antibonding π*2p orbitals, each holding a single, unpaired electron. This electron configuration, which describes oxygen in its lowest energy (ground) state, is referred to as the triplet state - the oxygen in the air around you istriplet oxygen. A higher energy state, in which the two highest energy electrons are paired in the same π*2p orbital, is called the singlet state of oxygen.
It is the last two unpaired electrons in triplet oxygen that react in radical-like fashion. For this reason, triplet O2 is often depicted by a Lewis structure as a diradical, with a single covalent bond, four lone pairs, and two unpaired electrons. The excited singlet state is often depicted as a doubly-bonded molecule.
These Lewis-dot depictions, while only approximations and unsatisfactory in many respects, can nonetheless be helpful in illustrating how O2 reacts. In any event, the molecular orbital picture of O2 is not simply an academic exercise: the diradical model for triplet oxygen is key to understanding its reactivity in many enzymatic and non-enzymatic contexts. We will see one example of triplet oxygen reactivity in section 17.2D.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.10%3A_Bonding_in_the_Methyl_Cation_the_Methyl_Radical_and_the_Methyl_Anion.txt |
Learning Objectives
• Identify three special properties of water that make it unusual for a molecule of its size, and explain how these result from hydrogen bonding.
• Explain what is meant by hydrogen bonding and the molecular structural features that bring it about.
• Describe the "structure", such as it is, of liquid water.
• Sketch out structural examples of hydrogen bonding in three small molecules other than H2O.
• Describe the roles of hydrogen bonding in proteins and in DNA.
Most students of chemistry quickly learn to relate the structure of a molecule to its general properties. Thus we generally expect small molecules to form gases or liquids, and large ones to exist as solids under ordinary conditions. And then we come to H2O, and are shocked to find that many of the predictions are way off, and that water (and by implication, life itself) should not even exist on our planet! In this section we will learn why this tiny combination of three nuclei and ten electrons possesses special properties that make it unique among the more than 15 million chemical species we presently know.
In water, each hydrogen nucleus is covalently bound to the central oxygen atom by a pair of electrons that are shared between them. In H2O, only two of the six outer-shell electrons of oxygen are used for this purpose, leaving four electrons which are organized into two non-bonding pairs. The four electron pairs surrounding the oxygen tend to arrange themselves as far from each other as possible in order to minimize repulsions between these clouds of negative charge. This would ordinarily result in a tetrahedral geometry in which the angle between electron pairs (and therefore the H-O-H bond angle) is 109.5°. However, because the two non-bonding pairs remain closer to the oxygen atom, these exert a stronger repulsion against the two covalent bonding pairs, effectively pushing the two hydrogen atoms closer together. The result is a distorted tetrahedral arrangement in which the H—O—H angle is 104.5°.
Water's large dipole moment leads to hydrogen bonding
The H2O molecule is electrically neutral, but the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an electric dipole, represented by the arrow at the bottom; you can think of this dipole as the electrical "image" of a water molecule.
Opposite charges attract, so it is not surprising that the negative end of one water molecule will tend to orient itself so as to be close to the positive end of another molecule that happens to be nearby. The strength of this dipole-dipole attraction is less than that of a normal chemical bond, and so it is completely overwhelmed by ordinary thermal motions in the gas phase. However, when the H2O molecules are crowded together in the liquid, these attractive forces exert a very noticeable effect, which we call (somewhat misleadingly) hydrogen bonding. And at temperatures low enough to turn off the disruptive effects of thermal motions, water freezes into ice in which the hydrogen bonds form a rigid and stable network.
Notice that the hydrogen bond (shown by the dashed green line) is somewhat longer than the covalent O—H bond. It is also much weaker, about 23 kJ mol–1 compared to the O–H covalent bond strength of 492 kJ mol–1.
Forty-one anomalies of water" — some of them rather esoteric.
Water has long been known to exhibit many physical properties that distinguish it from other small molecules of comparable mass. Although chemists refer to these as the "anomalous" properties of water, they are by no means mysterious; all are entirely predictable consequences of the way the size and nuclear charge of the oxygen atom conspire to distort the electronic charge clouds of the atoms of other elements when these are chemically bonded to the oxygen.
Boiling point
The most apparent peculiarity of water is its very high boiling point for such a light molecule. Liquid methane CH4 (molecular weight 16) boils at –161°C. As you can see from this diagram, extrapolation of the boiling points of the various Group 16 hydrogen compounds to H2O suggests that this substance should be a gas under normal conditions.
Surface Tension
Compared to most other liquids, water also has a high surface tension. Have you ever watched an insect walk across the surface of a pond? The water strider takes advantage of the fact that the water surface acts like an elastic film that resists deformation when a small weight is placed on it. (If you are careful, you can also "float" a small paper clip or steel staple on the surface of water in a cup.) This is all due to the surface tension of the water. A molecule within the bulk of a liquid experiences attractions to neighboring molecules in all directions, but since these average out to zero, there is no net force on the molecule. For a molecule that finds itself at the surface, the situation is quite different; it experiences forces only sideways and downward, and this is what creates the stretched-membrane effect.
The distinction between molecules located at the surface and those deep inside is especially prominent in H2O, owing to the strong hydrogen-bonding forces. The difference between the forces experienced by a molecule at the surface and one in the bulk liquid gives rise to the liquid's surface tension. This drawing highlights two H2O molecules, one at the surface, and the other in the bulk of the liquid. The surface molecule is attracted to its neighbors below and to either side, but there are no attractions pointing in the 180° solid angle angle above the surface. As a consequence, a molecule at the surface will tend to be drawn into the bulk of the liquid. But since there must always be some surface, the overall effect is to minimize the surface area of a liquid.
The geometric shape that has the smallest ratio of surface area to volume is the sphere, so very small quantities of liquids tend to form spherical drops. As the drops get bigger, their weight deforms them into the typical tear shape.
Ice floats on water
The most energetically favorable configuration of H2O molecules is one in which each molecule is hydrogen-bonded to four neighboring molecules. Owing to the thermal motions described above, this ideal is never achieved in the liquid, but when water freezes to ice, the molecules settle into exactly this kind of an arrangement in the ice crystal. This arrangement requires that the molecules be somewhat farther apart then would otherwise be the case; as a consequence, ice, in which hydrogen bonding is at its maximum, has a more open structure, and thus a lower density than water.
Here are three-dimensional views of a typical local structure of water (left) and ice (right.) Notice the greater openness of the ice structure which is necessary to ensure the strongest degree of hydrogen bonding in a uniform, extended crystal lattice. The more crowded and jumbled arrangement in liquid water can be sustained only by the greater amount of thermal energy available above the freezing point.
When ice melts, the more vigorous thermal motion disrupts much of the hydrogen-bonded structure, allowing the molecules to pack more closely. Water is thus one of the very few substances whose solid form has a lower density than the liquid at the freezing point. Localized clusters of hydrogen bonds still remain, however; these are continually breaking and reforming as the thermal motions jiggle and shove the individual molecules. As the temperature of the water is raised above freezing, the extent and lifetimes of these clusters diminish, so the density of the water increases.
At higher temperatures, another effect, common to all substances, begins to dominate: as the temperature increases, so does the amplitude of thermal motions. This more vigorous jostling causes the average distance between the molecules to increase, reducing the density of the liquid; this is ordinary thermal expansion.
Because the two competing effects (hydrogen bonding at low temperatures and thermal expansion at higher temperatures) both lead to a decrease in density, it follows that there must be some temperature at which the density of water passes through a maximum. This temperature is 4° C; this is the temperature of the water you will find at the bottom of an ice-covered lake in which this most dense of all water has displaced the colder water and pushed it nearer to the surface.
Structure of Liquid Water
The nature of liquid water and how the H2O molecules within it are organized and interact are questions that have attracted the interest of chemists for many years. There is probably no liquid that has received more intensive study, and there is now a huge literature on this subject. The following facts are well established:
• H2O molecules attract each other through the special type of dipole-dipole interaction known as hydrogen bonding
• a hydrogen-bonded cluster in which four H2Os are located at the corners of an imaginary tetrahedron is an especially favorable (low-potential energy) configuration, but...
• the molecules undergo rapid thermal motions on a time scale of picoseconds (10–12 second), so the lifetime of any specific clustered configuration will be fleetingly brief.
A variety of techniques including infrared absorption, neutron scattering, and nuclear magnetic resonance have been used to probe the microscopic structure of water. The information garnered from these experiments and from theoretical calculations has led to the development of around twenty "models" that attempt to explain the structure and behavior of water. More recently, computer simulations of various kinds have been employed to explore how well these models are able to predict the observed physical properties of water.
This work has led to a gradual refinement of our views about the structure of liquid water, but it has not produced any definitive answer. There are several reasons for this, but the principal one is that the very concept of "structure" (and of water "clusters") depends on both the time frame and volume under consideration. Thus, questions of the following kinds are still open:
• How do you distinguish the members of a "cluster" from adjacent molecules that are not in that cluster?
• Since individual hydrogen bonds are continually breaking and re-forming on a picosecond time scale, do water clusters have any meaningful existence over longer periods of time? In other words, clusters are transient, whereas "structure" implies a molecular arrangement that is more enduring. Can we then legitimately use the term "clusters" in describing the structure of water?
• The possible locations of neighboring molecules around a given H2O are limited by energetic and geometric considerations, thus giving rise to a certain amount of "structure" within any small volume element. It is not clear, however, to what extent these structures interact as the size of the volume element is enlarged. And as mentioned above, to what extent are these structures maintained for periods longer than a few picoseconds?
In the 1950's it was assumed that liquid water consists of a mixture of hydrogen-bonded clusters (H2O)n in which n can have a variety of values, but little evidence for the existence of such aggregates was ever found. The present view, supported by computer-modeling and spectroscopy, is that on a very short time scale, water is more like a "gel" consisting of a single, huge hydrogen-bonded cluster. On a 10–12-10–9 sec time scale, rotations and other thermal motions cause individual hydrogen bonds to break and re-form in new configurations, inducing ever-changing local discontinuities whose extent and influence depends on the temperature and pressure.
Ice
Ice, like all solids, has a well-defined structure; each water molecule is surrounded by four neighboring H2Os. two of these are hydrogen-bonded to the oxygen atom on the central H2O molecule, and each of the two hydrogen atoms is similarly bonded to another neighboring H2O.
Ice forms crystals having a hexagonal lattice structure, which in their full development would tend to form hexagonal prisms very similar to those sometimes seen in quartz. This does occasionally happen, and anyone who has done much winter mountaineering has likely seen needle-shaped prisms of ice crystals floating in the air. Under most conditions, however, the snowflake crystals we see are flattened into the beautiful fractal-like hexagonal structures that are commonly observed.
Snowflakes
The H2O molecules that make up the top and bottom plane faces of the prism are packed very closely and linked (through hydrogen bonding) to the molecules inside. In contrast to this, the molecules that make up the sides of the prism, and especially those at the hexagonal corners, are much more exposed, so that atmospheric H2O molecules that come into contact with most places on the crystal surface attach very loosely and migrate along it until they are able to form hydrogen-bonded attachments to these corners, thus becoming part of the solid and extending the structure along these six directions. This process perpetuates itself as the new extensions themselves acquire a hexagonal structure.
Why is ice slippery?
At temperatures as low as 200 K, the surface of ice is highly disordered and water-like. As the temperature approaches the freezing point, this region of disorder extends farther down from the surface and acts as a lubricant.
The illustration is taken from from an article in the April 7, 2008 issue of C&EN honoring the physical chemist Gabor Somorjai who pioneered modern methods of studying surfaces.
"Pure" water
To a chemist, the term "pure" has meaning only in the context of a particular application or process. The distilled or de-ionized water we use in the laboratory contains dissolved atmospheric gases and occasionally some silica, but their small amounts and relative inertness make these impurities insignificant for most purposes. When water of the highest obtainable purity is required for certain types of exacting measurements, it is commonly filtered, de-ionized, and triple-vacuum distilled. But even this "chemically pure" water is a mixture of isotopic species: there are two stable isotopes of both hydrogen (H1 and H2, the latter often denoted by D) and oxygen (O16 and O18) which give rise to combinations such as H2O18, HDO16, etc., all of which are readily identifiable in the infrared spectra of water vapor. And to top this off, the two hydrogen atoms in water contain protons whose magnetic moments can be parallel or antiparallel, giving rise to ortho- and para-water, respectively. The two forms are normally present in a o/p ratio of 3:1.
The amount of the rare isotopes of oxygen and hydrogen in water varies enough from place to place that it is now possible to determine the age and source of a particular water sample with some precision. These differences are reflected in the H and O isotopic profiles of organisms. Thus the isotopic analysis of human hair can be a useful tool for crime investigations and anthropology research.
More about hydrogen bonding
Hydrogen bonds form when the electron cloud of a hydrogen atom that is attached to one of the more electronegative atoms is distorted by that atom, leaving a partial positive charge on the hydrogen. Owing to the very small size of the hydrogen atom, the density of this partial charge is large enough to allow it to interact with the lone-pair electrons on a nearby electronegative atom. Although hydrogen bonding is commonly described as a form of dipole-dipole attraction, it is now clear that it involves a certain measure of electron-sharing (between the external non-bonding electrons and the hydrogen) as well, so these bonds possess some covalent character.
Hydrogen bonds are longer than ordinary covalent bonds, and they are also weaker. The experimental evidence for hydrogen bonding usually comes from X-ray diffraction studies on solids that reveal shorter-than-normal distances between hydrogen and other atoms.
Hydrogen bonding in small molecules
The following examples show something of the wide scope of hydrogen bonding in molecules.
Ammonia (mp –78, bp –33°C) is hydrogen-bonded in the liquid and solid states.
Hydrogen bonding is responsible for ammonia's remarkably high solubility in water.
Many organic (carboxylic) acids form hydrogen-bonded dimers in the solid state.
Here the hydrogen bond acceptor is the π electron cloud of a benzene ring. This type of interaction is important in maintaining the shape of proteins.
Hydrogen fluoride (mp –92, bp 33°C) is another common substance that is strongly hydrogen-bonded in its condensed phases.
The bifluoride ion (for which no proper Lewis structure can be written) can be regarded as a complex ion held together by the strongest hydrogen bond known: about 155 kJ mol–1.
"As slow as molasses in the winter!" Multiple hydroxyl groups provide lots of opportunities for hydrogen bonding and lead to the high viscosities of substances such as glycerine and sugar syrups.
Hydrogen bonding in biopolymers
Hydrogen bonding plays an essential role in natural polymers of biological origin in two ways:
• Hydrogen bonding between adjacent polymer chains (intermolecular bonding);
• Hydrogen bonding between different parts of the same chain (intramolecular bonding;
• Hydrogen bonding of water molecules to –OH groups on the polymer chain ("bound water") that helps maintain the shape of the polymer.
The examples that follow are representative of several types of biopolymers.
Cellulose
Cellulose is a linear polymer of glucose (see above), containing 300 to over 10,000 units, depending on the source. As the principal structural component of plants (along with lignin in trees), cellulose is the most abundant organic substance on the earth. The role of hydrogen bonding is to cross-link individual molecules to build up sheets as shown here. These sheets than stack up in a staggered array held together by van der Waals forces. Further hydrogen-bonding of adjacent stacks bundles them together into a stronger and more rigid structure.
Proteins
These polymers made from amino acids R—CH(NH2)COOH depend on intramolecular hydrogen bonding to maintain their shape (secondary and tertiary structure) which is essential for their important function as biological catalysts (enzymes). Hydrogen-bonded water molecules embedded in the protein are also important for their structural integrity.
The principal hydrogen bonding in proteins is between the -N—H groups of the "amino" parts with the -C=O groups of the "acid" parts. These interactions give rise to the two major types of the secondary structure which refers to the arrangement of the amino acid polymer chain:
[images]
beta-sheet
Although carbon is not usually considered particularly electronegative, C—H----X hydrogen bonds are also now known to be significant in proteins.
DNA (Deoxyribonucleic acid)
Who you are is totally dependent on hydrogen bonds! DNA, as you probably know, is the most famous of the biopolymers owing to its central role in defining the structure and function of all living organisms. Each strand of DNA is built from a sequence of four different nucleotide monomers consisting of a deoxyribose sugar, phosphate groups, and a nitrogenous base conventionally identified by the letters A,T, C and G. DNA itself consists of two of these polynucleotide chains that are coiled around a common axis in a configuration something like the protein alpha helix depicted above. The sugar-and-phosphate backbones are on the outside so that the nucleotide bases are on the inside and facing each other. The two strands are held together by hydrogen bonds that link a nitrogen atom of a nucleotide in one chain with a nitrogen or oxygen on the nucleotide that is across from it on the other chain.
Efficient hydrogen bonding within this configuration can only occur between the pairs A-T and C-G, so these two complementary pairs constitute the "alphabet" that encodes the genetic information that gets transcribed whenever new protein molecules are built. Water molecules, hydrogen-bonded to the outer parts of the DNA helix, help stabilize it. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.11%3A_The_Bonds_in_Water.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
What is the first step that you must do in order to draw the electron dot formula for ammonia, NH3?
The first step in drawing the electron dot formula for ammonia is to determine the number of bonding electrons for each of the atoms. The arabic numeral above the element's column in the periodic table gives you that number.
What are the numbers of bonding electrons for the nitrogen atom and the hydrogen atoms?
The nitrogen atom has 5 electrons and each of the three hydrogen atoms has 1 electron. The total number of electrons for ammonia will therefore be 8 electrons.
Draw a hydrogen atom next to a nitrogen atom and place a pair of electrons between these two atoms.
Now draw a second hydrogen atom next to the nitrogen atom and place a second pair of electrons between these atoms.
Now draw a third hydrogen atom next to the nitrogen atom and place a third pair of electrons between these atoms.
You have now used 6 of the 8 electrons. Place the remaining two electrons on the nitrogen atom. Remember that each hydrogen atom should not have more than 2 electrons and the nitrogen atom should have a total of 8 electrons.
The electron dot formula for ammonia is now complete. Note that the nitrogen atom now has three bonds. This is the normal number of bonds for a nitrogen atom when it does not have a charge.
Click here to see an animated gif of this process...
1.13: The Bond in a Hydrogen Halide
This page discusses the acidity of the hydrogen halides: hydrogen fluoride, hydrogen chloride, hydrogen bromide and hydrogen iodide. It begins by describing their physical properties and synthesis and then explains what happens when they react with water to make acids such as hydrofluoric acid and hydrochloric acid.
Physical properties
The hydrogen halides are colorless gases at room temperature, producing steamy fumes in moist air. The boiling points of these compounds are shown in the figure below:
Hydrogen fluoride has an abnormally high boiling point for a molecule of its size (293 K or 20°C), and can condense under cool conditions. This is due to the fact that hydrogen fluoride can form hydrogen bonds. Because fluorine is the most electronegative of all the elements, the fluorine-hydrogen bond is highly polarized. The hydrogen atom carries a high partial positive charge (δ+); the fluorine is fairly negatively charged (δ-).
In addition, each fluorine atom has 3 lone pairs of electrons. Fluorine's outer electrons are at the n=2 level, and the lone pairs represent small, highly charged regions of space. Hydrogen bonds form between the δ+ hydrogen on one HF molecule and a lone pair on the fluorine of another one.The figure below illustrates this association:
The other hydrogen halides do not form hydrogen bonds because the larger halogens are not as electronegative as fluorine; therefore, the bonds are less polar. In addition, their lone pairs are at higher energy levels, so the halogen does not carry such an intensely concentrated negative charge; therefore, other hydrogen atoms are not attracted as strongly.
Making hydrogen halides
There are several ways of synthesizing hydrogen halides; the method considered here is the reaction between an ionic halide, like sodium chloride, and an acid like concentrated phosphoric(V) acid, H3PO4, or concentrated sulfuric acid.
Making hydrogen chloride
When concentrated sulfuric acid is added to sodium chloride under cold conditions, the acid donates a proton to a chloride ion, forming hydrogen chloride. In the gas phase, it immediately escapes from the system.
$Cl^- + H_2SO_4 \rightarrow HCl + HSO_4^- \nonumber$
The full equation for the reaction is:
$NaCl + H_2SO_4 \rightarrow HCl + NaHSO_4 \nonumber$
Sodium bisulfate is also formed in the reaction. Concentrated phosphoric(V) acid reacts similarly, according to the following equation:
$Cl^- + H_3PO_4 \rightarrow HCl + H_2PO_4^- \nonumber$
The full ionic equation showing the formation of the salt, sodium biphosphate(V), is given below:
$NaCl + H_3PO_4 \rightarrow HCl + NaH_2PO_4 \nonumber$
Making other hydrogen halides
All hydrogen halides can be formed by the same method, using concentrated phosphoric(V) acid.
Concentrated sulfuric acid, however, behaves differently. Hydrogen fluoride can be made with sulfuric acid, but hydrogen bromide and hydrogen iodide cannot.
The problem is that concentrated sulfuric acid is an oxidizing agent, and as well as producing hydrogen bromide or hydrogen iodide, some of the halide ions are oxidized to bromine or iodine. Phosphoric acid does not have this ability because it is not an oxidant.
The acidity of the hydrogen halides
Hydrogen chloride as an acid
By the Bronsted-Lowry definition of an acid as a proton donor, hydrogen chloride is an acid because it transfers protons to other species. Consider its reaction with water.
Hydrogen chloride gas is soluble in water; its solvated form is hydrochloric acid. Hydrogen chloride fumes in moist air are caused by hydrogen chloride reacting with water vapor in the air to produce a cloud of concentrated hydrochloric acid.
A proton is donated from the hydrogen chloride to one of the lone pairs on a water molecule.
A coordinate (dative covalent) bond is formed between the oxygen and the transferred proton.
The equation for the reaction is the following:
$H_2O + HCl \rightarrow H_3O^+ + Cl^- \nonumber$
The H3O+ ion is the hydroxonium ion (also known as the hydronium ion or the oxonium ion). This is the normal form of protons in water; sometimes it is shortened to the proton form, H+(aq), for brevity.
When hydrogen chloride dissolves in water (to produce hydrochloric acid), almost all the hydrogen chloride molecules react in this way. Hydrochloric acid is therefore a strong acid. An acid is strong if it is fully ionized in solution.
Hydrobromic acid and hydriodic acid as strong acids
Hydrogen bromide and hydrogen iodide dissolve in (and react with) water in exactly the same way as hydrogen chloride does. Hydrogen bromide forms hydrobromic acid; hydrogen iodide gives hydriodic acid. Both of these are also strong acids.
Hydrofluoric acid as an exception
By contrast, although hydrogen fluoride dissolves freely in water, hydrofluoric acid is only a weak acid; it is similar in strength to organic acids like methanoic acid. The complicated reason for this is discussed below.
The bond enthalpy of the H-F bond
Because the fluorine atom is so small, the bond enthalpy (bond energy) of the hydrogen-fluorine bond is very high. The chart below gives values for all the hydrogen-halogen bond enthalpies:
bond enthalpy
(kJ mol-1)
H-F +562
H-Cl +431
H-Br +366
H-I +299
In order for ions to form when the hydrogen fluoride reacts with water, the H-F bond must be broken. It would seem reasonable to say that the relative reluctance of hydrogen fluoride to react with water is due to the large amount of energy needed to break that bond, but this explanation does not hold.
The energetics of the process from HX(g) to X-(aq)
The energetics of this sequence are of interest:
All of these terms are involved in the overall enthalpy change as you convert HX(g) into its ions in water.
However, the terms involving the hydrogen are the same for every hydrogen halide. Only the values for the red terms in the diagram need be considered. The values are tabulated below:
bond enthalpy of HX
(kJ mol-1)
electron affinity of X
(kJ mol-1)
hydration enthalpy of X-
(kJ mol-1)
sum of these
(kJ mol-1)
HF +562 -328 -506 -272
HCl +431 -349 -364 -282
HBr +366 -324 -335 -293
HI +299 -295 -293 -289
There is virtually no difference in the total HF and HCl values.
The large bond enthalpy of the H-F bond is offset by the large hydration enthalpy of the fluoride ion. There is a very strong attraction between the very small fluoride ion and the water molecules. This releases a lot of heat (the hydration enthalpy) when the fluoride ion becomes wrapped in water molecules.
Other attractions in the system
The energy terms considered previously have concerned HX molecules in the gas phase. To reach a more correct explanation, the molecules must first be considered as unreacted aqueous HX molecules. The equation for this is given below:
The equation is incorporated into an improved energy cycle as follows:
Unfortunately, values for the first step in the reaction are not readily available. However, in each case, the initial separation of the HX from water molecules is endothermic. Energy is required to break the intermolecular attractions between the HX molecules and water.
That energy is much greater for hydrogen fluoride because it forms hydrogen bonds with water. The other hydrogen halides experience only the weaker van der Waals dispersion forces or dipole-dipole attractions.
The overall enthalpy changes (including all the stages in the energy cycle) for the reactions are given in the table below:
$HX(aq) + H_2O (l) \rightarrow H_3O^+ (aq) + X^- (aq) \nonumber$
enthalpy change
(kJ mol-1)
HF -13
HCl -59
HBr -63
H-I -57
The enthalpy change for HF is much smaller in magnitude than that for the other three hydrogen halides, but it is still negative exothermic change. Therefore, more information is needed to explain why HF is a weak acid.
Entropy and free energy considerations
The free energy change, not the enthalpy change, determines the extent and direction of a reaction.
Free energy change is calculated from the enthalpy change, the temperature of the reaction and the entropy change during the reaction.
For simplicity, entropy can be thought of as a measure of the amount of disorder in a system. Entropy is given the symbol S. If a system becomes more disordered, then its entropy increases. If it becomes more ordered, its entropy decreases.
The key equation is given below:
In simple terms, for a reaction to happen, the free energy change must be negative. But more accurately, the free energy change can be used to calculate a value for the equilibrium constant for a reaction using the following expression:
The term Ka is the equilibrium constant for the reaction below:
$HX(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + X^-(aq) \nonumber$
The values for TΔS (needed to calculate ΔG) for the four reactions at a temperature of 298 K are tabulated below:
TS
(kJ mol-1)
HF -29
HCl -13
HBr -4
H-I +4
Notice that at the top of the group, the systems become more ordered when the HX reacts with the water. The entropy of the system (the amount of disorder) decreases, particularly for the hydrogen fluoride.
The reason for this is that the very strong attraction between H3O+ and F-(aq) imposes a lot of order on the system, as does the attraction between the water molecules and the various ions present. These attractions are each greatest for the small fluoride ions.
The total effect on the free energy change, and therefore the value of the equilibrium constant, can now be considered. These values are calculated in the following table:
H
(kJ mol-1)
TS
(kJ mol-1)
G
(kJ mol-1)
Ka
(mol dm-3)
HF -13 -29 +16 1.6 x 10-3
HCl -59 -13 -46 1.2 x 108
HBr -63 -4 -59 2.2 x 1010
HI -57 +4 -61 5.0 x 1010
The values for these estimated equilibrium constants for HCl, HBr and HI are so high that the reaction can be considered "one-way". The ionization is virtually 100% complete. These are all strong acids, increasing in strength down the group.
By contrast, the estimated Ka for hydrofluoric acid is small. Hydrofluoric acid only ionizes to a limited extent in water. Therefore, it is a weak acid.
The estimated value for HF in the table can be compared to the experimental value:
• Experimental value: 5.6 x 10-4 mol dm-3
• Estimated value: 1.6 x 10-3 mol dm-3
These values differ by an order of magnitude, but because of the logarithmic relationship between the free energy and the equilibrium constant, a very small change in ΔG has a very large effect on Ka.
To have the values in close agreement, ΔG would have to increase from +16 to +18.5 kJ mol-1. Given the uncertainty in the values used to calculate ΔG, the difference between the calculated value and the experimental value could easily fall within this range.
Summary: Why is hydrofluoric acid a weak acid?
The two main factors are:
• Entropy decreases dramatically when the hydrogen fluoride reacts with water. This is particularly noticeable with hydrogen fluoride because the attraction of the small fluoride ions produced imposes significant order on the surrounding water molecules and nearby hydronium ions. The effect decreases with larger halide ions.
• Very strong hydrogen bonding exists between the hydrogen fluoride molecules and water molecules. This costs a large amount of energy to break. This effect does not occur in the other hydrogen halides.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.12%3A_The_Bonds_in_Ammonia_and_in_the_Ammonium_Ion.txt |
Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. Bond order and bond length indicate the type and strength of covalent bonds between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased.
Introduction
Chemistry deals with the way in which subatomic particles bond together to form atoms. Chemistry also focuses on the way in which atoms bond together to form molecules. In the atomic structure, electrons surround the atomic nucleus in regions called orbitals. Each orbital shell can hold a certain number of electrons. When the nearest orbital shell is full, new electrons start to gather in the next orbital shell out from the nucleus, and continue until that shell is also full. The collection of electrons continues in ever widening orbital shells as larger atoms have more electrons than smaller atoms. When two atoms bond to form a molecule, their electrons bond them together by mixing into openings in each others' orbital shells. As with the collection of electrons by the atom, the formation of bonds by the molecule starts at the nearest available orbital shell opening and expand outward.
Bond Order
Bond order is the number of bonding pairs of electrons between two atoms. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. To determine the bond order between two covalently bonded atoms, follow these steps:
1. Draw the Lewis structure.
2. Determine the type of bonds between the two atoms.
• 0: No bond
• 1: Single bond
• 2: double bond
• 3: triple bond
If the bond order is zero, the molecule cannot form. The higher bond orders indicate greater stability for the new molecule. In molecules that have resonance bonding, the bond order does not need to be an integer.
Example $1$: $CN^-$
Determine the bond order for cyanide, CN-.
Solution
1) Draw the Lewis structure.
2) Determine the type of bond between the two atoms.
Because there are 3 dashes, the bond is a triple bond. A triple bond corresponds to a bond order of 3.
Example $2$: $H_2$
Determine the bond order for hydrogen gas, H2.
Solution
1) Draw the Lewis structure.
2) Determine the type of bond between the two atoms.
There is only one pair of shared electrons (or dash), indicating is a single bond, with a bond order of 1.
Polyatomic molecules
If there are more than two atoms in the molecule, follow these steps to determine the bond order:
1. Draw the Lewis structure.
2. Count the total number of bonds.
3. Count the number of bond groups between individual atoms.
4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.
Example $3$: $NO_3^-$
Determine the bond order for nitrate, $NO_3^-$.
Solution
1) Draw the Lewis structure.
2) Count the total number of bonds.
4
The total number of bonds is 4.
3) Count the number of bond groups between individual atoms.
3
The number of bond groups between individual atoms is 3.
4) Divide the number of bonds between individual atoms by the total number of bonds.
$\dfrac{4}{3}= 1.33$
The bond order is 1.33
Example $4$: $NO^+_2$
Determine the bond order for nitronium ion: $NO_2^+$.
Solution
1) Draw the Lewis Structure.
2) Count the total number of bonds.
4
The total number of bonds is 4.
3) Count the number of bond groups between individual atoms.
2
The number of bond groups between atoms is 2.
4) Divide the bond groups between individual atoms by the total number of bonds.
$\frac{4}{2} = 2$
The bond order is 2.
A high bond order indicates more attraction between electrons. A higher bond order also means that the atoms are held together more tightly. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together more loosely. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability.
Bond order increases across a period and decreases down a group.
Bond Length
Bond length is defined as the distance between the centers of two covalently bonded atoms. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. Bond length is reported in picometers. Therefore, bond length increases in the following order: triple bond < double bond < single bond.
To find the bond length, follow these steps:
1. Draw the Lewis structure.
2. Look up the chart below for the radii for the corresponding bond.
3. Find the sum of the two radii.
4
Determine the carbon-to-chlorine bond length in CCl4.
Solution
Using Table A3, a C single bond has a length of 75 picometers and that a Cl single bond has a length of 99 picometers. When added together, the bond length of a C-Cl bond is approximately 174 picometers.
2
Determine the carbon-oxygen bond length in CO2.
Solution
Using Table A3, we see that a C double bond has a length of 67 picometers and that an O double bond has a length of 57 picometers. When added together, the bond length of a C=O bond is approximately 124 picometers.
Because the bond length is proportional to the atomic radius, the bond length trends in the periodic table follow the same trends as atomic radii: bond length decreases across a period and increases down a group.
Problems
1. What is the bond order of $O_2$?
2. What is the bond order of $NO_3^-$?
3. What is the carbon-nitrogen bond length in $HCN$?
4. Is the carbon-to-oxygen bond length greater in $CO_2$ or $CO$?
5. What is the nitrogen-fluoride bond length in $NF_3$?
Solutions
1. First, write the Lewis structure for $O_2$.
There is a double bond between the two oxygen atoms; therefore, the bond order of the molecule is 2.
2. The Lewis structure for NO3- is given below:
To find the bond order of this molecule, take the average of the bond orders. N=O has a bond order of two, and both N-O bonds have a bond order of one. Adding these together and dividing by the number of bonds (3) reveals that the bond order of nitrate is 1.33.
3. To find the carbon-nitrogen bond length in HCN, draw the Lewis structure of HCN.
The bond between carbon and nitrogen is a triple bond, and a triple bond between carbon and nitrogen has a bond length of approximately 60 + 54 =114 pm.
4. From the Lewis structures for CO2 and CO, there is a double bond between the carbon and oxygen in CO2 and a triple bond between the carbon and oxygen in CO.
Referring to the table above, a double bond between carbon and oxygen has a bond length of approximately 67 + 57 = 124 pm and a triple bond between carbon and oxygen has a bond length of approximately 60 + 53 =113 pm. Therefore, the bond length is greater in CO2.
Another method makes use of the fact that the more electron bonds between the atoms, the tighter the electrons are pulling the atoms together. Therefore, the bond length is greater in CO2.
5. To find the nitrogen-to-fluorine bond length in NF3, draw the Lewis structure.
The bond between fluorine and nitrogen is a single bond. From the table above, a single bond between fluorine and nitrogen has a bond length of approximately 64 + 71 =135 pm.
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1.16: An Introduction to Acids and Bases
An acid–base reaction is a chemical reaction that occurs between an acid and a base. Several theoretical frameworks provide alternative conceptions of the reaction mechanisms and their application in solving related problems; these are called acid–base theories, for example, Brønsted–Lowry acid–base theory. Their importance becomes apparent in analyzing acid–base reactions for gaseous or liquid species, or when acid or base character may be somewhat less apparent.
Thumbnail: Warning picture used with dangerous acids and dangerous bases. Bases are the opposites of acids. (Public Domain). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.14%3A_Summary-_Hybridization_Bond_Lengths_Bond_Strengths_and_Bond_Angles.txt |
Learning Objectives
• To define the pH scale as a measure of acidity of a solution
• Tell the origin and the logic of using the pH scale.
• Apply the same strategy for representing other types of quantities such as pKa, pKb, pKw.
Auto-Ionization of Water
Because of its amphoteric nature (i.e., acts as both an acid or a base), water does not always remain as $H_2O$ molecules. In fact, two water molecules react to form hydronium and hydroxide ions:
$\ce{ 2 H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^{−} (aq)} \label{1}$
This is also called the self-ionization of water. The concentration of $H_3O^+$ and $OH^-$ are equal in pure water because of the 1:1 stoichiometric ratio of Equation $\ref{1}$. The molarity of H3O+ and OH- in water are also both $1.0 \times 10^{-7} \,M$ at 25° C. Therefore, a constant of water ($K_w$) is created to show the equilibrium condition for the self-ionization of water. The product of the molarity of hydronium and hydroxide ion is always $1.0 \times 10^{-14}$ (at room temperature).
$K_w= [H_3O^+][OH^-] = 1.0 \times 10^{-14} \label{2}$
Equation $\ref{2}$ also applies to all aqueous solutions. However, $K_w$ does change at different temperatures, which affects the pH range discussed below.
$H^+$ and $H_3O^+$ is often used interchangeably to represent the hydrated proton, commonly call the hydronium ion.
Equation \ref{1} can also be written as
$H_2O \rightleftharpoons H^+ + OH^- \label{3}$
As expected for any equilibrium, the reaction can be shifted to the reactants or products:
• If an acid ($H^+$) is added to the water, the equilibrium shifts to the left and the $OH^-$ ion concentration decreases
• If base ( $OH^-$) is added to water, the equilibrium shifts to left and the $H^+$ concentration decreases.
pH and pOH
Because the constant of water, Kw is $1.0 \times 10^{-14}$ (at 25° C), the $pK_w$ is 14, the constant of water determines the range of the pH scale. To understand what the pKw is, it is important to understand first what the "p" means in pOH and pH. The addition of the "p" reflects the negative of the logarithm, $-\log$. Therefore, the pH is the negative logarithm of the molarity of H, the pOH is the negative logarithm of the molarity of $\ce{OH^-}$, and the $pK_w$ is the negative logarithm of the constant of water:
\begin{align} pH &= -\log [H^+] \label{4a} \[4pt] pOH &= -\log [OH^-] \label{4b} \[4pt] pK_w &= -\log [K_w] \label{4c} \end{align}
At room temperature,
$K_w =1.0 \times 10^{-14} \label{4d}$
So
\begin{align} pK_w &=-\log [1.0 \times 10^{-14}] \label{4e} \[4pt] &=14 \end{align}
Using the properties of logarithms, Equation $\ref{4e}$ can be rewritten as
$10^{-pK_w}=10^{-14}. \label{4f}$
The equation also shows that each increasing unit on the scale decreases by the factor of ten on the concentration of $\ce{H^{+}}$. Combining Equations \ref{4a} - \ref{4c} and \ref{4e} results in this important relationship:
$pK_w= pH + pOH = 14 \label{5b}$
Equation \ref{5b} is correct only at room temperature since changing the temperature will change $K_w$.
The pH scale is logarithmic, meaning that an increase or decrease of an integer value changes the concentration by a tenfold. For example, a pH of 3 is ten times more acidic than a pH of 4. Likewise, a pH of 3 is one hundred times more acidic than a pH of 5. Similarly a pH of 11 is ten times more basic than a pH of 10.
Properties of the pH Scale
From the simple definition of pH in Equation \ref{4a}, the following properties can be identified:
• This scale is convenient to use, because it converts some odd expressions such as $1.23 \times 10^{-4}$ into a single number of 3.91.
• This scale covers a very large range of $\ce{[H+]}$, from 0.1 to 10-14. When $\ce{[H+]}$ is high, we usually do not use the pH value, but simply the $\ce{[H+]}$. For example, when $\mathrm{[H^+] = 1.0}$, pH = 0. We seldom say the pH is 0, and that is why you consider pH = 0 such an odd expression. A pH = -0.30 is equivalent to a $\ce{[H+]}$ of 2.0 M. Negative pH values are only for academic exercises. Using the concentrations directly conveys a better sense than the pH scales.
• The pH scale expands the division between zero and 1 in a linear scale or a compact scale into a large scale for comparison purposes. In mathematics, you learned that there are infinite values between 0 and 1, or between 0 and 0.1, or between 0 and 0.01 or between 0 and any small value. Using a log scale certainly converts infinite small quantities into infinite large quantities.
• The non-linearity of the pH scale in terms of $\ce{[H+]}$ is easily illustrated by looking at the corresponding values for pH between 0.1 and 0.9 as follows:
pH 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
[H+] 1 0.79 0.63 0.50 0.40 0.32 0.25 0.20 0.16 0.13
• Because the negative log of $\ce{[H+]}$ is used in the pH scale, the pH scale usually has positive values. Furthermore, the larger the pH, the smaller the $\ce{[H+]}$.
The Effective Range of the pH Scale
It is common that the pH scale is argued to range from 0-14 or perhaps 1-14, but neither is correct. The pH range does not have an upper nor lower bound, since as defined above, the pH is an indication of concentration of H+. For example, at a pH of zero the hydronium ion concentration is one molar, while at pH 14 the hydroxide ion concentration is one molar. Typically the concentrations of H+ in water in most solutions fall between a range of 1 M (pH=0) and 10-14 M (pH=14). Hence a range of 0 to 14 provides sensible (but not absolute) "bookends" for the scale. One can go somewhat below zero and somewhat above 14 in water, because the concentrations of hydronium ions or hydroxide ions can exceed one molar. Figure $1$ depicts the pH scale with common solutions and where they are on the scale.
Quick Interpretation
• If pH >7, the solution is basic. The pOH should be looked in the perspective of OH- instead of H+. Whenever the value of pOH is less than 7, then it is considered basic. And therefore there are more OH- than H+ in the solution.
• At pH 7, the substance or solution is at neutral and means that the concentration of H+ and OH- ion is the same.
• If pH < 7, the solution is acidic. There are more H+ than OH- in an acidic solution.
• The pH scale does not have an upper nor lower bound.
Example $1$
If the concentration of $NaOH$ in a solution is $2.5 \times 10^{-4}\; M$, what is the concentration of $H_3O^+$?
Solution
We can assume room temperature, so
$1.0 \times 10^{-14} = [H_3O^+][OH^-] \nonumber$
to find the concentration of H3O+, solve for the [H3O+].
$\dfrac{1.0 \times 10^{-14}}{[OH^-]} = [H_3O^+]$
$\dfrac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}} = [H_3O^+] = 4.0 \times 10^{-11}\; M$
Example $2$
1. Find the pH of a solution of 0.002 M of HCl.
2. Find the pH of a solution of 0.00005 M NaOH.
Solution
1. The equation for pH is -log [H+]
$[H^+]= 2.0 \times 10^{-3}\; M \nonumber$
$pH = -\log [2.0 \times 10^{-3}] = 2.70 \nonumber$
1. The equation for pOH is -log [OH-]
$[OH^-]= 5.0 \times 10^{-5}\; M \nonumber$
$pOH = -\log [5.0 \times 10^{-5}] = 4.30 \nonumber$
$pK_w = pH + pOH \nonumber$
and
$pH = pK_w - pOH \nonumber$
then
$pH = 14 - 4.30 = 9.70 \nonumber$
Example $3$: Soil
If moist soil has a pH of 7.84, what is the H+ concentration of the soil solution?
Solution
$pH = -\log [H^+] \nonumber$
$7.84 = -\log [H^+] \nonumber$
$[H^+] = 1.45 \times 10^{-8} M \nonumber$
Hint
Place -7.84 in your calculator and take the antilog (often inverse log or 10x) = 1.45 x 10-8M
Proper Definition of pH
The pH scale was originally introduced by the Danish biochemist S.P.L. Sørenson in 1909 using the symbol pH. The letter p is derived from the German word potenz meaning power or exponent of, in this case, 10. In 1909, S.P.L. Sørenson published a paper in Biochem Z in which he discussed the effect of H+ ions on the activity of enzymes. In the paper, he invented the term pH (purported to mean pondus hydrogenii in Latin) to describe this effect and defined it as the $-\log[H^+]$. In 1924, Sørenson realized that the pH of a solution is a function of the "activity" of the H+ ion and not the concentration. Thus, he published a second paper on the subject. A better definition would be
$pH = -\log\,a\{\ce{H^{+}}\}$
where $a\{H^+\}$ denotes the activity (an effective concentration) of the H+ ions. The activity of an ion is a function of many variables of which concentration is one.
• Concentration is abbreviated by using square brackets, e.g., $[H_3O^+]$ is the concentration of hydronium ion in solution.
• Activity is abbreviated by using "a" with curly brackets, e.g., $a\{H_3O^+\}$ is the activity of hydronium ions in solution
Because of the difficulty in accurately measuring the activity of the $\ce{H^{+}}$ ion for most solutions the International Union of Pure and Applied Chemistry (IUPAC) and the National Bureau of Standards (NBS) has defined pH as the reading on a pH meter that has been standardized against standard buffers. The following equation is used to calculate the pH of all solutions:
\begin{align} pH &= \dfrac{F(E-E_{standard})}{RT\;\ln 10} + pH_{standard} \label{6a} \[4pt] &= \dfrac{5039.879 (E-E_{standard})}{T} + pH_{standard} \label{6b} \end{align}
with
• $R$ is the ideal gas constant,
• $F$ is the Faraday's constant, and
• $T$ is absolute temperature (in K)
The activity of the H+ ion is determined as accurately as possible for the standard solutions used. The identity of these solutions vary from one authority to another, but all give the same values of pH to ± 0.005 pH unit. The historical definition of pH is correct for those solutions that are so dilute and so pure the H+ ions are not influenced by anything but the solvent molecules (usually water).
When measuring pH, [H+] is in units of moles of H+ per liter of solution. This is a reasonably accurate definition at low concentrations (the dilute limit) of H+. At very high concentrations (10 M hydrochloric acid or sodium hydroxide, for example,) a significant fraction of the ions will be associated into neutral pairs such as H+Cl, thus reducing the concentration of “available” ions to a smaller value which we will call the effective concentration. It is the effective concentration of H+ and OH that determines the pH and pOH. The pH scale as shown above is called sometimes "concentration pH scale" as opposed to the "thermodynamic pH scale". The main difference between both scales is that in thermodynamic pH scale one is interested not in H+concentration, but in H+activity. What a person measures in the solution is just activity, not the concentration. Thus it is thermodynamic pH scale that describes real solutions, not the concentration one.
For solutions in which ion concentrations don't exceed 0.1 M, the formulas pH = –log [H+] and pOH = –log[OH] are generally reliable, but don't expect a 10.0 M solution of a strong acid to have a pH of exactly –1.00! However, this definition is only an approximation (albeit very good under most situations) of the proper definition of pH, which depends on the activity of the hydrogen ion:
$pH= -\log a\{H^+\} \approx -\log [H^+] \label{7}$
The activity is a measure of the "effective concentration" of a substance, is often related to the true concentration via an activity coefficient, $\gamma$:
$a{H^+}=\gamma [H^+] \label{8}$
Calculating the activity coefficient requires detailed theories of how charged species interact in solution at high concentrations (e.g., the Debye-Hückel Theory). In most solutions the pH differs from the -log[H+ ] in the first decimal point. The following table gives experimentally determined pH values for a series of HCl solutions of increasing concentration at 25 °C.
Table $1$: HCl Solutions with corresponding pH values. Data taken from Christopher G. McCarty and Ed Vitz, Journal of Chemical Education, 83(5), 752 (2006) and G.N. Lewis, M. Randall, K. Pitzer, D.F. Brewer, Thermodynamics (McGraw-Hill: New York, 1961; pp. 233-34).
Molar Concentration of $HCl$ pH defined as Concentration Experimentally Determined pH Relative Deviation
0.00050 3.30 3.31 0.3%
0.0100 2 2.04 1.9%
0.100 1 1.10 9%
0.40 0.39 0.52 25%
7.6 -0.88 -1.85 52%
While the pH scale formally measures the activity of hydrogen ions in a substance or solution, it is typically approximated as the concentration of hydrogen ions; this approximation is applicable only under low concentrations.
Living Systems
Molecules that make up or are produced by living organisms usually function within a narrow pH range (near neutral) and a narrow temperature range (body temperature). Many biological solutions, such as blood, have a pH near neutral. pH influences the structure and the function of many enzymes (protein catalysts) in living systems. Many of these enzymes have narrow ranges of pH activity. Cellular pH is so important that death may occur within hours if a person becomes acidotic (having increased acidity in the blood). As one can see pH is critical to life, biochemistry, and important chemical reactions. Common examples of how pH plays a very important role in our daily lives are given below:
• Water in swimming pool is maintained by checking its pH. Acidic or basic chemicals can be added if the water becomes too acidic or too basic.
• Whenever we get a heartburn, more acid build up in the stomach and causes pain. We needs to take antacid tablets (a base) to neutralize excess acid in the stomach.
• The pH of blood is slightly basic. A fluctuation in the pH of the blood can cause in serious harm to vital organs in the body.
• Certain diseases are diagnosed only by checking the pH of blood and urine.
• Certain crops thrive better at certain pH range.
• Enzymes activate at a certain pH in our body.
Table $2$: pH in Living Systems
Compartment pH
Gastric Acid 1
Lysosomes 4.5
Granules of Chromaffin Cells 5.5
Human Skin 5.5
Urine 6
Neutral H2O at 37 °C 6.81
Cytosol 7.2
Cerebrospinal Fluid 7.3
Blood 7.43-7.45
Mitochondrial Matrix 7.5
Pancreas Secretions 8.1
Problems
1. In a solution of $2.4 \times 10^{-3} M$ of HI, find the concentration of $OH^-$.
2. Determine the pH of a solution that is 0.0035 M HCl.
3. Determine the [H3O+] of a solution with a pH = 5.65
4. If the pOH of NH3, ammonia, in water is 4.74. What is the pH?
5. Pepsin, a digestive enzyme in our stomach, has a pH of 1.5. Find the concentration of OH- in the stomach.
Solutions
1. We use the dissociation of water equation to find [OH-].
Kw = [H3O+][OH-] = 1.0 X 10-14
Solve for [OH-]
[OH-] = (1.0 X 10-14)/ [H3O+]
Plug in the molarity of HI and solve for OH-.
[OH-] = (1.0 X 10-14)/ [2.4 X 10-3] = 4.17 X 10-12 M.
2. pH = -log[H3O+]
Plug the molarity of the HCl in and solve for pH.
pH = -log[0.0035] = 2.46
3. pH = -log[H3O+]
Plug in the pH and solve for [H3O+]
5.65 = -log[H3O+]
Move the negative sign to the pH. -5.65 = log[H3O+]
10-5.65=10log[H3O+] = 2.24 X 10-6 M
4. pH + pOH = 14
Solve for pH.
14 - pOH = pH
14 - 4.74 = pH = 9.26
5. There are several ways to do this problem.
Answer 1.
pH + pOH = 14
Solve for pOH.
pOH = 14 - pH
pOH = 14 - 1.5 = 12.5
When the pOH is solved, solve for the concentration by using log.
pOH = -log[OH-]
12.5 = -log[OH-]
-12.5 = log[OH-]
10-12.5 = 10log[OH-] = 3.16 X 10-13 M.
Answer 2.
pH = -log[H+]
Plug in the pH and solve for the molarity of H+ of pepsin.
1.5 = -log[H+]
-1.5 = log[H+]
10-1.5 = 10log[H+] = [H+]= 0.032
Use the concentration of H+ to solve for the concentration of OH-.
[H+][OH-] = 1.0 X 10-14
Plug in the [H+] and solve for [OH-].
[OH-] = (1.0 X 10-14)/[H3O+]
[OH-] = (1.0 X 10-14)/(0.032) = 3.125 X 10-14 M | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.17%3A_pka_and_pH.txt |
This page explains the acidity of simple organic acids and looks at the factors which affect their relative strengths.
Organic acids as weak acids
For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water.
An acid in solution sets up this equilibrium:
A hydroxonium ion is formed together with the anion (negative ion) from the acid.
This equilibrium is sometimes simplified by leaving out the water to emphasise the ionisation of the acid.
If you write it like this, you must include the state symbols - "(aq)". Writing H+(aq) implies that the hydrogen ion is attached to a water molecule as H3O+. Hydrogen ions are always attached to something during chemical reactions.
The organic acids are weak in the sense that this ionisation is very incomplete. At any one time, most of the acid will be present in the solution as un-ionised molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionised. The position of equilibrium therefore lies well to the left.
Comparing the strengths of weak acids
The strengths of weak acids are measured on the pKa scale. The smaller the number on this scale, the stronger the acid is.
Three of the compounds we shall be looking at, together with their pKa values are:
Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pKa of 10.00, and ethanol is so weak with a pKa of about 16 that it hardly counts as acidic at all!
Why are these acids acidic?
In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X":
So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid strengths?
Differences in acid strengths between carboxylic acids, phenols and alcohols
Two of the factors which influence the ionization of an acid are:
• the strength of the bond being broken,
• the stability of the ions being formed.
In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar. The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case.
Example \(1\): Ethanoic Acid
Ethanoic acid has the structure:
The acidic hydrogen is the one attached to the oxygen. When ethanoic acid ionises it forms the ethanoate ion, CH3COO-.
You might reasonably suppose that the structure of the ethanoate ion was as below, but measurements of bond lengths show that the two carbon-oxygen bonds are identical and somewhere in length between a single and a double bond.
To understand why this is, you have to look in some detail at the bonding in the ethanoate ion. Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond. Pi bonds are made by sideways overlap between p orbitals on the carbon and the oxygen.
In an ethanoate ion, one of the lone pairs on the negative oxygen ends up almost parallel to these p orbitals, and overlaps with them
This leads to a delocalised pi system over the whole of the -COO- group, rather like that in benzene.
All the oxygen lone pairs have been left out of this diagram to avoid confusion. Because the oxygens are more electronegative than the carbon, the delocalised system is heavily distorted so that the electrons spend much more time in the region of the oxygen atoms.
So where is the negative charge in all this? It has been spread around over the whole of the -COO- group, but with the greatest chance of finding it in the region of the two oxygen atoms. Ethanoate ions can be drawn simply as:
The dotted line represents the delocalisation. The negative charge is written centrally on that end of the molecule to show that it isn't localised on one of the oxygen atoms. The more you can spread charge around, the more stable an ion becomes. In this case, if you delocalise the negative charge over several atoms, it is going to be much less attractive to hydrogen ions - and so you are less likely to re-form the ethanoic acid.
Example \(2\): Phenol
Phenols have an -OH group attached directly to a benzene ring. Phenol itself is the simplest of these with nothing else attached to the ring apart from the -OH group.
When the hydrogen-oxygen bond in phenol breaks, you get a phenoxide ion, C6H5O-.
Delocalisation also occurs in this ion. This time, one of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalisation which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localised on the oxygen, but is spread out around the whole ion.
Why then is phenol a much weaker acid than ethanoic acid?
Think about the ethanoate ion again. If there wasn't any delocalisation, the charge would all be on one of the oxygen atoms, like this:
But the delocalisation spreads this charge over the whole of the COO group. Because oxygen is more electronegative than carbon, you can think of most of the charge being shared between the two oxygens (shown by the heavy red shading in this diagram).
If there wasn't any delocalisation, one of the oxygens would have a full charge which would be very attractive towards hydrogen ions. With delocalisation, that charge is spread over two oxygen atoms, and neither will be as attractive to a hydrogen ion as if one of the oxygens carried the whole charge.
That means that the ethanoate ion won't take up a hydrogen ion as easily as it would if there wasn't any delocalisation. Because some of it stays ionised, the formation of the hydrogen ions means that it is acidic.
In the phenoxide ion, the single oxygen atom is still the most electronegative thing present, and the delocalised system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge.
What delocalisation there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent.
However, the delocalisation hasn't shared the charge around very effectively. There is still lots of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic.
Example \(3\): Ethanol
Ethanol, CH3CH2OH, is so weakly acidic that you would hardly count it as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen ion, an ethoxide ion is formed:
This has nothing at all going for it. There is no way of delocalising the negative charge, which remains firmly on the oxygen atom. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form.
Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all.
Variations in acid strengths between different carboxylic acids
You might think that all carboxylic acids would have the same strength because each depends on the delocalization of the negative charge around the -COO- group to make the anion more stable, and so more reluctant to re-combine with a hydrogen ion.
In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple carboxylic acids:
pKa
HCOOH 3.75
CH3COOH 4.76
CH3CH2COOH 4.87
CH3CH2CH2COOH 4.82
Remember that the higher the value for pKa, the weaker the acid is.
Why is ethanoic acid weaker than methanoic acid? It again depends on the stability of the anions formed - on how much it is possible to delocalise the negative charge. The less the charge is delocalised, the less stable the ion, and the weaker the acid.
The methanoate ion (from methanoic acid) is:
The only difference between this and the ethanoate ion is the presence of the CH3 group in the ethanoate.
But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the -COO- group. Any build-up of charge will make the ion less stable, and more attractive to hydrogen ions.
Ethanoic acid is therefore weaker than methanoic acid, because it will re-form more easily from its ions.
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of propanoic acid and butanoic acid are very similar to ethanoic acid. The acids can be strengthened by pulling charge away from the -COO- end. You can do this by attaching electronegative atoms like chlorine to the chain.
As the next table shows, the more chlorines you can attach the better:
pKa
CH3COOH 4.76
CH2ClCOOH 2.86
CHCl2COOH 1.29
CCl3COOH 0.65
Trichloroethanoic acid is quite a strong acid.
Attaching different halogens also makes a difference. Fluorine is the most electronegative and so you would expect it to be most successful at pulling charge away from the -COO- end and so strengthening the acid.
pKa
CH2FCOOH 2.66
CH2ClCOOH 2.86
CH2BrCOOH 2.90
CH2ICOOH 3.17
The effect is there, but isn't as great as you might expect.
Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO- end. Here is what happens if you move a chlorine atom along the chain in butanoic acid.
pKa
CH3CH2CH2COOH 4.82
CH3CH2CHClCOOH 2.84
CH3CHClCH2COOH 4.06
CH2ClCH2CH2COOH 4.52
The chlorine is effective at withdrawing charge when it is next-door to the -COO- group, and much less so as it gets even one carbon further away.
This page explains why simple organic bases are basic and looks at the factors which affect their relative strengths. For A'level purposes, all the bases we are concerned with are primary amines - compounds in which one of the hydrogens in an ammonia molecule, NH3, is replaced either by an alkyl group or a benzene ring.
Ammonia as a weak base
All of the compounds we are concerned with are derived from ammonia and so we'll start by looking at the reason for its basic properties. For the purposes of this topic, we are going to take the definition of a base as "a substance which combines with hydrogen ions (protons)". We are going to get a measure of this by looking at how easily the bases take hydrogen ions from water molecules when they are in solution in water.
Ammonia in solution sets up this equilibrium:
An ammonium ion is formed together with hydroxide ions. Because the ammonia is only a weak base, it doesn't hang on to the extra hydrogen ion very effectively and so the reaction is reversible. At any one time, about 99% of the ammonia is present as unreacted molecules. The position of equilibrium lies well to the left.
The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom. That combination of extra negativity and active lone pair attracts the new hydrogen from the water.
Comparing the strengths of weak bases
The strengths of weak bases are measured on the pKb scale. The smaller the number on this scale, the stronger the base is. Three of the compounds we shall be looking at, together with their pKb values are:
Remember - the smaller the number the stronger the base. Comparing the other two to ammonia, you will see that methylamine is a stronger base, whereas phenylamine is very much weaker.
• Methylamine is typical of aliphatic primary amines - where the -NH2 group is attached to a carbon chain. All aliphatic primary amines are stronger bases than ammonia.
• Phenylamine is typical of aromatic primary amines - where the -NH2 group is attached directly to a benzene ring. These are very much weaker bases than ammonia.
Explaining the differences in base strengths
Two of the factors which influence the strength of a base are:
• the ease with which the lone pair picks up a hydrogen ion,
• the stability of the ions being formed.
Why are aliphatic primary amines stronger bases than ammonia?
Methylamine
Methylamine has the structure:
The only difference between this and ammonia is the presence of the CH3 group in the methylamine. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the nitrogen atom. That extra negativity around the nitrogen makes the lone pair even more attractive towards hydrogen ions.
Making the nitrogen more negative helps the lone pair to pick up a hydrogen ion. What about the effect on the positive methylammonium ion formed? Is this more stable than a simple ammonium ion? Compare the methylammonium ion with an ammonium ion:
In the methylammonium ion, the positive charge is spread around the ion by the "electron-pushing" effect of the methyl group. The more you can spread charge around, the more stable an ion becomes. In the ammonium ion there isn't any way of spreading the charge.
To summarize:
• The nitrogen is more negative in methylamine than in ammonia, and so it picks up a hydrogen ion more readily.
• The ion formed from methylamine is more stable than the one formed from ammonia, and so is less likely to shed the hydrogen ion again.
Taken together, these mean that methylamine is a stronger base than ammonia.
The other aliphatic primary amines
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of the other aliphatic primary amines are very similar to methylamine. For example:
pKb
CH3NH2 3.36
CH3CH2NH2 3.27
CH3CH2CH2NH2 3.16
CH3CH2CH2CH2NH2 3.39
Why are aromatic primary amines much weaker bases than ammonia?
An aromatic primary amine is one in which the -NH2 group is attached directly to a benzene ring. The only one you are likely to come across is phenylamine. Phenylamine has the structure:
The lone pair on the nitrogen touches the delocalized ring electrons . . .
. . . and becomes delocalized with them:
That means that the lone pair is no longer fully available to combine with hydrogen ions. The nitrogen is still the most electronegative atom in the molecule, and so the delocalized electrons will be attracted towards it, but the intensity of charge around the nitrogen is nothing like what it is in, say, an ammonia molecule.
The other problem is that if the lone pair is used to join to a hydrogen ion, it is no longer available to contribute to the delocalisation. That means that the delocalization would have to be disrupted if the phenylamine acts as a base. Delocalization makes molecules more stable, and so disrupting the delocalization costs energy and will not happen easily.
Taken together - the lack of intense charge around the nitrogen, and the need to break some delocalization - this means that phenylamine is a very weak base indeed. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.18%3A_Organic_Acids_and_Bases.txt |
/*<![CDATA[*/ MathJax.Hub.Config({ TeX: { equationNumbers: { autoNumber: "all", formatNumber: function (n) { return 3.6 + '.' + n } }, Macros: { PageIndex: ["{3.6. #1}",1] } } }); /*]]>*/A: Defining the acidity constant
You are no doubt aware that some acids are stronger than others. The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the acidity constant, abbreviated Ka. The common base chosen for comparison is water.
We will consider acetic acid as our first example. If we make a dilute solution of acetic acid in water, an acid-base reaction occurs between the acid (proton donor) and water (proton acceptor).
Acetic acid is a weak acid, so the equilibrium favors reactants over products - it is thermodynamically 'uphill', as indicated in the figure above by the relative length of the forward and reverse reaction arrows, and in the reaction coordinate diagram below in which products are higher energy than reactants.
As you know, the equilibrium constant Keq is defined as:
Every equilibrium constant expressions is actually a ratio of the activities of all species involved in the reaction. To avoid the use of activities, and to simplify experimental measurements, the equilibrium constant of concentrations approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the activities of solids, pure liquids, and solvents are not approximated with their molarities. Instead these activities are defined to have a value equal to 1 (one).
Thus, if we acknowledge that the activity of water in a dilute solution is approximated with the value of unity (1), we can divide by 1 to get the common form of the expression for Ka, the acid constant for acetic acid:
$K_a = K_{eq} = \dfrac{a_{H_3O^+}· a_{CH_3COO^-}}{a_{CH_3COOH} · a_{H_2O}} ≈ \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH](1)} = \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$
In fact, for an dilute aqueous solution, the activity of water is approximated with the value of 1, so the generic dissociation constant for a given acid HA or HB+ is expressed as:
The value of Ka for acetic acid is 1.75 x 10-5 - much less than 1, indicating there is much more acetic acid in solution at equilibrium than acetate and hydronium ions.
Conversely, sulfuric acid, with a Ka of approximately 109, or hydrochloric acid, with a Ka of approximately 107, both undergo essentially complete dissociation in water: they are very strong acids.
A number like 1.75 x 10- 5 is not very easy either to say, remember, or visualize, so chemists usually use a more convenient term to express relative acidity. The pKa value of an acid is simply the log (base 10) of its Ka value.
pKa = -log Ka Ka = 10-pKa
Doing the math, we find that the pKa of acetic acid is 4.8. The pKa of sulfuric acid is -10, and of hydrochloric acid is -7. The use of pKa values allows us to express the relative acidity of common compounds and functional groups on a numerical scale of about –10 (for a very strong acid) to 50 (for a compound that is not acidic at all). The lower the pKa value, the stronger the acid.
The ionizable (proton donating or accepting) functional groups relevant to biological organic chemistry generally have pKa values ranging from about 5 to about 20. The most important of these are summarized below, with very rough pKa values for the conjugate acid forms. More acidic groups with pKa values near zero are also included for reference.
Approximate pKa values to know
hydronium ion (H3O+) : 0
protonated alcohol: 0
protonated carbonyl: 0
carboxylic acids: 5
protonated imines: 7
protonated amines: 10
phenols: 10
thiols: 10
water: 14
alcohols: 15-18
alpha-carbon acids*: 20
*alpha-carbon acids will be discussed later in this chapter
You are strongly recommended to commit these rough values to memory now - then if you need a more precise value, you can always look it up in a pKa table.
Caution! pKa is not the same as pH!
It is important to realize that pKa is not the same thing as pH: the former is an inherent property of a compound or functional group, while the latter is a measure of hydronium ion concentration in a given aqueous solution:
pH = -log [H3O+]
Knowing pKa values not only allows us to compare acid strength, it also allows us to compare base strength. The key idea to remember is this: the stronger the conjugate acid, the weaker the conjugate base. We can determine that hydroxide ion is a stronger base than ammonia (NH3), because ammonium ion (NH4+, pKa = 9.2) is a stronger acid than water (pKa = 14.0).
Exercise 7.2.1
Which is the stronger base, CH3O- or CH3S-? Acetate ion or ammonia? Hydroxide ion or acetate ion?
Solution
Let's put our understanding of the pKa concept to use in the context of a more complex molecule. For example, what is the pKa of the compound below?
We need to evaluate the potential acidity of four different types of protons on the molecule, and find the most acidic one. The aromatic protons are not all acidic - their pKa is about 45. The amine group is also not acidic, its pKa is about 35. (Remember, uncharged amines are basic: it is positively-charged protonated amines, with pKa values around 10, that are weakly acidic.) The alcohol proton has a pKa of about 15, and the phenol proton has a pKa of about 10: thus, the most acidic group on the molecule above is the phenol. (Be sure that you can recognize the difference between a phenol and an alcohol - remember, in a phenol the OH group is bound directly to the aromatic ring). If this molecule were to react with one molar equivalent of a strong base such as sodium hydroxide, it is the phenol proton which would be donated to form a phenolate anion.
Exercise 7.2.2
Identify the most acidic functional group on each of the molecules below, and give its approximate pKa.
Solution
B: Using pKa values to predict reaction equilibria
By definition, the pKa value tells us the extent to which an acid will react with water as the base, but by extension we can also calculate the equilibrium constant for a reaction between any acid-base pair. Mathematically, it can be shown that:
Keq = 10ΔpKa
. . . where ΔpKa = (pKa of product acid minus pKa of reactant acid).
Consider a reaction between methylamine and acetic acid:
The first step is to identify the acid species on either side of the equation, and look up or estimate their pKa values. On the left side, the acid is of course acetic acid while on the right side the acid is methyl ammonium ion (in other words, methyl ammonium ion is the acid in the reaction going from right to left). We can look up the precise pKa values in table 7 (at the back of the book), but we already know (because we have this information memorized, right?!) that the pKa of acetic acids is about 5, and methyl ammonium is about 10. More precise values are 4.8 and 10.6, respectively.
Without performing any calculations at all, you should be able to see that this equilibrium lies far to the right-hand side: acetic acid has a lower pKa, meaning it is a stronger acid than methyl ammonium, and thus it wants to give up its proton more than methyl ammonium does. Doing the math, we see that
Keq = 10ΔpKa = 10(10.6 – 4.8) = 105.8 = 6.3 x 105
So Keq is a very large number (much greater than 1) and the equilibrium for the reaction between acetic acid and methylamine lies far to the right-hand side of the equation, just as we had predicted. This also tells us that the reaction has a negative Gibbs free energy change, and is thermodynamically favorable.
If you had just wanted to quickly approximate the value of Keq without benefit of precise pKa information or a calculator, you could have approximated pKa ~ 5 (for the carboxylic acid) and pKa ~10 (for the ammonium ion) and calculated in your head that the equilibrium constant should be somewhere in the order of 105.
Exercise 7.2.3
Show the products of the following acid-base reactions, and roughly estimate the value of Keq.
Solution
C: Organic molecules in buffered solution: the Henderson-Hasselbalch equation
The environment inside a living cell, where most biochemical reactions take place, is an aqueous buffer with pH ~ 7. Recall from your General Chemistry course that a buffer is a solution of a weak acid and its conjugate base. The key equation for working with buffers is the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation:
The equation tells us that if our buffer is an equimolar solution of a weak acid and its conjugate base, the pH of the buffer will equal the pKa of the acid (because the log of 1 is equal to zero). If there is more of the acid form than the base, then of course the pH of the buffer is lower than the pKa of the acid.
Exercise 7.2.4
What is the pH of an aqueous buffer solution that is 30 mM in acetic acid and 40 mM in sodium acetate? The pKa of acetic acid is 4.8.
Solution
The Henderson-Hasselbalch equation is particularly useful when we want to think about the protonation state of different biomolecule functional groups in a pH 7 buffer. When we do this, we are always assuming that the concentration of the biomolecule is small compared to the concentration of the buffer components. (The actual composition of physiological buffer is complex, but it is primarily based on phosphoric and carbonic acids).
Imagine an aspartic acid residue located on the surface of a protein in a human cell. Being on the surface, the side chain is in full contact with the pH 7 buffer surrounding the protein. In what state is the side chain functional group: the protonated state (a carboxylic acid) or the deprotonated state (a carboxylate ion)? Using the Henderson-Hasselbalch equation, we fill in our values for the pH of the buffer and a rough pKa approximation of pKa = 5 for the carboxylic acid functional group. Doing the math, we find that the ratio of carboxylate to carboxylic acid is about 100 to 1: the carboxylic acid is almost completely ionized (in the deprotonated state) inside the cell. This result extends to all other carboxylic acid groups you might find on natural biomolecules or drug molecules: in the physiological environment, carboxylic acids are almost completely deprotonated.
Now, let's use the equation again, this time for an amine functional group, such as the side chain of a lysine residue: inside a cell, are we likely to see a neutral amine (R-NH2) or an ammonium cation (R-NH3+?) Using the equation with pH = 7 (for the biological buffer) and pKa = 10 (for the ammonium group), we find that the ratio of neutral amine to ammonium cation is about 1 to 100: the group is close to completely protonated inside the cell, so we will see R-NH3+, not R-NH2.
We can do the same rough calculation for other common functional groups found in biomolecules. It will be very helpful going forward to commit the following to memory:
At physiological pH:
Carboxylic acids are deprotonated (in the carboxylate anion form)
Amines are protonated (in the ammonium cation form)
Thiols, phenols, alcohols, and amides are uncharged
Imines are a mixture of the protonated (cationic) and deprotonated (neutral) states
We will talk about the physiological protonation state of phosphate groups in chapter 9.
Exercise 7.2.5
The molecule below is not drawn in a reasonable protonation state for pH 7. Redraw it in with the functional groups in the correct protonation state.
Solution
While we are most interested in the state of molecules at pH 7, the Henderson-Hasselbalch equation can of course be used to determine the protonation state of functional groups in solutions buffered to other pH levels. The exercises below provide some practice in this type of calculation.
Exercise 7.2.6
What is the ratio of acetate ion to neutral acetic acid when a small amount of acetic acid (pKa = 4.8) is dissolved in a buffer of pH 2.8? pH 3.8? pH 4.8? pH 5.8? pH 6.8?
Solution
Exercise 7.2.7
Would you expect phenol to be soluble in an aqueous solution buffered to pH 2? pH 7? pH 12? Explain your answer.
Solution
Exercise 7.2.8
Exercise 7.2.8: What is the approximate net charge on a tetrapeptide Cys-Asp-Lys-Glu in pH 7 buffer?
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.19%3A_How_to_Predict_the_Outcome_of_an_Acid-Base_Reaction.txt |
Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind why one compound is more acidic or more basic than another. This is a big step: we are, for the first time, taking our knowledge of organic structure and applying it to a question of organic reactivity. Many of the ideas that we’ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).
Horizontal periodic trend in acidity and basicity
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively. Remember that electronegativity also increases as we move from left to right along a row of the periodic table, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.
The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.
Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic). Conversely, ethanol is the strongest acid, and ethane the weakest acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the haloacids and halides: basicity, like electronegativity, increases as we move up the column.
Vertical periodic trend in acidity and basicity
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ over a larger area.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, we are applying the concept only to the influence of atomic radius on base strength. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid. HI, with a pKa of about -9, is almost as strong as sulfuric acid.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the alcohol group on the serine side chain is on the order of 17.
Remember the concept of 'driving force' that we learned about in chapter 6? Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy. Let's see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:
HCl + F- HF + Cl-
We know that HCl (pKa -7) is a stronger acid than HF (pKa 3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a 'driving force' pushes reactant to product.
What explains this driving force? Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side). This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.
What about total bond energy, the other factor in driving force? If you consult a table of bond energies, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the reactant side: 565 kJ/mol vs 427 kJ/mol, respectively). This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.
B: Resonance effects
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.
Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are strikingly different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the stability of the conjugate bases, remembering that a more stable (weaker) conjugate base corresponds to a stronger acid.
In both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or ‘locked’ on the single oxygen – it has nowhere else to go. This makes the ethoxide ion much less stable.
Recall the important general statement that we made a little earlier: 'Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.' Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a fator of 1012 between the Ka values for the two molecules!)
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.
Whereas the lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi bonding system. The lone pair on an amine nitrogen, by contrast, is not so comfortable - it is not part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.
If an amide group is protonated, it will be at the oxygen rather than the nitrogen.
Exercise 7.3.1
a) Draw the Lewis structure of nitric acid, HNO3.
b) Nitric acid is a strong acid - it has a pKa of -1.4. Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.
Solution
Exercise 7.3.2
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Exercise 7.3.3
(challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1 - the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to two oxygen atoms. Which if the four OH protons on the molecule is most acidic? Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom. Hint - try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.
Solution
C: Inductive effects
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent can be referred to as an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced pKa-lowered effect than chlorine substituents.
In addition, the inductive takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise 7.3.4
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.22%3A_How_Substituents_Affect_the_Strength_of_an_Acid.txt |
Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind why one compound is more acidic or more basic than another. This is a big step: we are, for the first time, taking our knowledge of organic structure and applying it to a question of organic reactivity. Many of the ideas that we’ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).
Horizontal periodic trend in acidity and basicity
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively. Remember that electronegativity also increases as we move from left to right along a row of the periodic table, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.
The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.
Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic). Conversely, ethanol is the strongest acid, and ethane the weakest acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the haloacids and halides: basicity, like electronegativity, increases as we move up the column.
Vertical periodic trend in acidity and basicity
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ over a larger area.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, we are applying the concept only to the influence of atomic radius on base strength. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid. HI, with a pKa of about -9, is almost as strong as sulfuric acid.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the alcohol group on the serine side chain is on the order of 17.
Remember the concept of 'driving force' that we learned about in chapter 6? Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy. Let's see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:
HCl + F- HF + Cl-
We know that HCl (pKa -7) is a stronger acid than HF (pKa 3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a 'driving force' pushes reactant to product.
What explains this driving force? Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side). This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.
What about total bond energy, the other factor in driving force? If you consult a table of bond energies, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the reactant side: 565 kJ/mol vs 427 kJ/mol, respectively). This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.
B: Resonance effects
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.
Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are strikingly different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the stability of the conjugate bases, remembering that a more stable (weaker) conjugate base corresponds to a stronger acid.
In both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or ‘locked’ on the single oxygen – it has nowhere else to go. This makes the ethoxide ion much less stable.
Recall the important general statement that we made a little earlier: 'Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.' Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a fator of 1012 between the Ka values for the two molecules!)
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.
Whereas the lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi bonding system. The lone pair on an amine nitrogen, by contrast, is not so comfortable - it is not part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.
If an amide group is protonated, it will be at the oxygen rather than the nitrogen.
Exercise 7.3.1
a) Draw the Lewis structure of nitric acid, HNO3.
b) Nitric acid is a strong acid - it has a pKa of -1.4. Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.
Solution
Exercise 7.3.2
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Exercise 7.3.3
(challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1 - the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to two oxygen atoms. Which if the four OH protons on the molecule is most acidic? Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom. Hint - try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.
Solution
C: Inductive effects
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent can be referred to as an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced pKa-lowered effect than chlorine substituents.
In addition, the inductive takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise 7.3.4
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.23%3A_An_Introduction_to_Delocalized_Electrons.txt |
Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind why one compound is more acidic or more basic than another. This is a big step: we are, for the first time, taking our knowledge of organic structure and applying it to a question of organic reactivity. Many of the ideas that we’ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).
Horizontal periodic trend in acidity and basicity
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively. Remember that electronegativity also increases as we move from left to right along a row of the periodic table, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.
The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.
Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic). Conversely, ethanol is the strongest acid, and ethane the weakest acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the haloacids and halides: basicity, like electronegativity, increases as we move up the column.
Vertical periodic trend in acidity and basicity
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ over a larger area.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, we are applying the concept only to the influence of atomic radius on base strength. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid. HI, with a pKa of about -9, is almost as strong as sulfuric acid.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the alcohol group on the serine side chain is on the order of 17.
Remember the concept of 'driving force' that we learned about in chapter 6? Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy. Let's see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:
HCl + F- HF + Cl-
We know that HCl (pKa -7) is a stronger acid than HF (pKa 3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a 'driving force' pushes reactant to product.
What explains this driving force? Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side). This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.
What about total bond energy, the other factor in driving force? If you consult a table of bond energies, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the reactant side: 565 kJ/mol vs 427 kJ/mol, respectively). This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.
B: Resonance effects
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.
Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are strikingly different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the stability of the conjugate bases, remembering that a more stable (weaker) conjugate base corresponds to a stronger acid.
In both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or ‘locked’ on the single oxygen – it has nowhere else to go. This makes the ethoxide ion much less stable.
Recall the important general statement that we made a little earlier: 'Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.' Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a fator of 1012 between the Ka values for the two molecules!)
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.
Whereas the lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi bonding system. The lone pair on an amine nitrogen, by contrast, is not so comfortable - it is not part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.
If an amide group is protonated, it will be at the oxygen rather than the nitrogen.
Exercise 7.3.1
a) Draw the Lewis structure of nitric acid, HNO3.
b) Nitric acid is a strong acid - it has a pKa of -1.4. Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.
Solution
Exercise 7.3.2
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Exercise 7.3.3
(challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1 - the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to two oxygen atoms. Which if the four OH protons on the molecule is most acidic? Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom. Hint - try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.
Solution
C: Inductive effects
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent can be referred to as an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced pKa-lowered effect than chlorine substituents.
In addition, the inductive takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise 7.3.4
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.24%3A_A_Summary_of_the_Factors_that_Determine_Acid_Strength.txt |
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