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Corey-House Reaction is also called as 'coupling of alkyl halides with organo metallic compounds'. It is a better method than Wurtz reaction. An alkyl halides and a lithium dialkyl copper are reacted to give a higher hydrocarbon
$R'-X + R_2CuLi \rightarrow R-R' + R-Cu + LiX$
(R and R' may be same or different)
Contributors
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Cracking
This page describes what cracking is, and the differences between catalytic cracking and thermal cracking used in the petrochemical industry.
Introduction
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.
There is not any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline).
Catalytic cracking
Modern cracking uses zeolites as the catalyst. These are complex aluminosilicates, and are large lattices of aluminium, silicon and oxygen atoms carrying a negative charge. They are, of course, associated with positive ions such as sodium ions. You may have come across a zeolite if you know about ion exchange resins used in water softeners. The alkane is brought into contact with the catalyst at a temperature of about 500°C and moderately low pressures.
The zeolites used in catalytic cracking are chosen to give high percentages of hydrocarbons with between 5 and 10 carbon atoms - particularly useful for petrol (gasoline). It also produces high proportions of branched alkanes and aromatic hydrocarbons like benzene. The zeolite catalyst has sites which can remove a hydrogen from an alkane together with the two electrons which bound it to the carbon. That leaves the carbon atom with a positive charge. Ions like this are called carbonium ions (or carbocations). Reorganization of these leads to the various products of the reaction.
Thermal cracking
In thermal cracking, high temperatures (typically in the range of 450°C to 750°C) and pressures (up to about 70 atmospheres) are used to break the large hydrocarbons into smaller ones. Thermal cracking gives mixtures of products containing high proportions of hydrocarbons with double bonds - alkenes. Thermal cracking does not go via ionic intermediates like catalytic cracking. Instead, carbon-carbon bonds are broken so that each carbon atom ends up with a single electron. In other words, free radicals are formed.
Reactions of the free radicals lead to the various products.
Contributors
Jim Clark (Chemguide.co.uk)
Wurtz reaction
Wurtz reaction is coupling of haloalkanes using sodium metal in solvent like dry ether
$2R-X + 2Na \rightarrow R-R + 2Na^+X^−$
The reaction consists of a halogen-metal exchange involving the free radical species R• (in a similar fashion to the formation of a Grignard reagent and then the carbon-carbon bond formation in a nucleophilic substitution reaction.)
One electron from the metal is transferred to the halogen to produce a metal halide and an alkyl radical.
$R-X + M → R^{\cdot} + M^+X^−$
The alkyl radical then accepts an electron from another metal atom to form an alkyl anion and the metal becomes cationic. This intermediate has been isolated in a several cases.
$R^{\cdot} + M → R^−M^+$
The nucleophilic carbon of the alkyl anion then displaces the halide in an SN2 reaction, forming a new carbon-carbon covalent bond.
$R^−M^+ + R-X → R-R + M^+X^−$ | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Synthesis_of_Alkanes/Corey-House_Reaction.txt |
The molecular formula of a hydrocarbon provides information about the possible structural types it may represent. For example, consider compounds having the formula \(\ce{C5H8}\). The formula of the five-carbon alkane pentane is \(\ce{C5H12}\) so the difference in hydrogen content is 4. This difference suggests such compounds may have a triple bond, two double bonds, a ring plus a double bond, or two rings. Some examples are shown here, and there are at least fourteen others!
Hence, as with alkanes, a consistent nomenclature system needs to be adopted that can separate the nature of these unsaturated chemicals. The simplest are the alkenes, which are hydrocarbons which have carbon-carbon double bond functional groups and are unsaturated hydrocarbons with the molecular formula is \(\ce{CnH2n}\), which is also the same molecular formula as cycloalkanes.
IUPAC Rules for Alkene Nomenclature
1. The ene suffix (ending) indicates an alkene or cycloalkene.
2. The longest chain chosen for the root name must include both carbon atoms of the double bond.
3. The root chain must be numbered from the end nearest a double bond carbon atom. If the double bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts.
4. The smaller of the two numbers designating the carbon atoms of the double bond is used as the double bond locator.
5. If more than one double bond is present the compound is named as a diene, triene or equivalent prefix indicating the number of double bonds, and each double bond is assigned a locator number.
6. Substituent groups containing double bonds are:
• H2C=CH– Vinyl group
• H2C=CH–CH2 Allyl group
Rule 1
Alkenes are named using the same general naming rules for alkanes, except that the suffix is now -ene. Here is a chart containing the systemic name for the first twenty straight chain alkenes.
Name Molecular formula Name Molecular formula
Ethene C2H4 Undecene C11H22
Propene C3H6 Dodecene C12H24
Butene C4H8 Tridecene C13H26
Pentene C5H10 Tetradecene C14H28
Hexene C6H12 Pentadecene C15H30
Heptene C7H14 Hexadecene C16H32
Octene C8H16 Heptadecene C17H34
Nonene C9H18 Octadecene C18H36
Decene C10H20 Nonadecene C19H38
Did you notice how there is no methene? Because it is impossible for a carbon to have a double bond with nothing.
Rule 2
The parent structure is the longest chain containing both carbon atoms of the double bond. If the alkene contains only one double bond and that double bond is terminal (the double bond is at one end of the molecule or another) then it is not necessary to place any number in front of the name.
• butane: C4H10 (\(\ce{CH3CH2CH2CH3}\))
• butene: C4H8 (\(\ce{CH2=CHCH2CH3}\))
If the double bond is not terminal (if it is on a carbon somewhere in the center of the chain) then the carbons should be numbered in such a way as to give the first of the two double-bonded carbons the lowest possible number, and that number should precede the "ene" suffix with a dash, as shown below.
• correct: pent-2-ene (\(\ce{CH3CH=CHCH2CH3}\))
• incorrect: pent-3-ene (\(\ce{CH3CH2CH=CHCH3}\))
The second one is incorrect because flipping the formula horizontally results in a lower number for the alkene.
Exercise \(1\)
Name the following compounds:
a:
b.
Answer a
1-pentene or pent-1-ene
Answer b
2-ethyl-1-hexene or 2-ethylhex-1-ene
Exercise \(2\)
a: Name the following compound (hint: give the double bond the lowest possible numbers regardless of substituent placement).
b. Draw a structure for 4-methyl-2-pentene.
Answer a
4-methylpent-1-ene
Answer b
Rule 4
If there is more than one double bond in an alkene, all of the bonds should be numbered in the name of the molecule - even terminal double bonds. The numbers should go from lowest to highest, and be separated from one another by a comma. The IUPAC numerical prefixes are used to indicate the number of double bonds.
• octa-2,4-diene: \(\ce{CH3CH=CHCH=CHCH2CH2CH3}\)
• deca-1,5-diene: \(\ce{CH2=CHCH2CH2CH=CHCH2CH2CH2CH3}\)
Note that the numbering of "2-4" above yields a molecule with two double bonds separated by just one single bond. Double bonds in such a condition are called "conjugated", and they represent an enhanced stability of conformation, so they are energetically favored as reactants in many situations and combinations.
Rule 4: Geometric Isomers
Double bonds can exist as geometric isomers and these isomers are designated by using either the cis / trans designation or the more flexible E / Z designation.
cis Isomers have the two largest groups are on the same side of the double bond (left structure above) and trans Isomers have the two largest groups are on opposite sides of the double bond (right structure above).E/Z nomenclature
If there are 3 or 4 non-hydrogen different atoms attached to the alkene then use the E, Z system.
• E (entgegen) means the higher priority groups are opposite one another relative to the double bond.
• Z (zusammen) means the higher priority groups are on the same side relative to the double bond.
E = entgegan ("trans") Z = zusamen ("cis")
Priority of groups is based on the atomic mass of attached atoms (not the size of the group). An atom attached by a multiple bond is counted once for each bond.
fluorine atom > isopropyl group > n-hexyl group
deuterium atom > hydrogen atom
-CH2-CH=CH2 > -CH2CH2CH3
Example \(5\)
Try to name the following compounds using both cis-trans and E/Z conventions:
a:
b:
Answer a
4-methylpent-1-ene
Answer b
Example \(3\)
What is the name of this molecule?
Solution
In this diagram this is a cis conformation. It has both the substituents going upward. This molecule would be called (cis) 5-chloro-3-heptene.)
Trans would look like this
Example \(4\)
What is the name of this molecule?
Solution
In this example it is E-4-chloro-3-heptene. It is E because the Chlorine and the CH2CH3 are the two higher priorities and they are on opposite sides.
vi. A hydroxyl group gets precedence over the double bond. Therefore alkenes containing alchol groups are called alkenols. And the prefix becomes --enol. And this means that now the alcohol gets lowest priority over the alkene.
vii. Lastly remember that alkene substituents are called alkenyl. Suffix --enyl.
Common names
Remove the -ane suffix and add -ylene. There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl. That you should know are...
• vinyl substituent H2C=CH-
• allyl substituent H2C=CH-CH2-
• allene molecule H2C=C=CH2
• isoprene
Endocyclic Alkenes
Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring.
Cyclopentene is an example of an endocyclic double bond.
Methylenecylopentane is an example of an exocyclic double bond.
Exercise \(1\)
Name the following compounds:
a:
b:
Answer a
1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene.
Answer b
1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene.
Exercise \(1\)
Name the following compounds:
a:
b:
Answer a
1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene.
Answer b
1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene.
Exercise \(1\)
Draw structures for the following 2-vinyl-1,3-cyclohexadiene
Answer
Problems
Try to name the following compounds...
1-pentene or pent-1-ene
2-ethyl-1-hexene or 2-ethylhex-1-ene
Try to draw structures for the following compounds...
• 2-pentene
CH3–CH=CH–CH2–CH3
• 3-heptene
CH3–CH2–CH=CH–CH2–CH2–CH3
b. Give the double bond the lowest possible numbers regardless of substituent placement.
• Try to name the following compound...
• Try to draw a structure for the following compound...
4-methyl-2-pentene J
Name the following structures:
v. Draw (Z)-5-Chloro-3-ethly-4-hexen-2-ol.
Answers
I. trans-8-ethyl-3-undecene
II. E-5-bromo-4-chloro-7,7-dimethyl-4-undecene
III. Z-1,2-difluoro-cyclohexene
IV. 4-ethenylcyclohexanol.
V. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Naming_the_Alkenes.txt |
The melting and boiling points of alkenes are dictated by the regularity with which they can pack and the surface area of interaction.
Source: Boundless. “Properties of Alkenes.” Boundless Chemistry. Boundless, 05 Jun. 2015. Retrieved 06 Jun. 2015 from https://www.boundless.com/chemistry/...enes-631-3627/
The melting and boiling points of alkenes are dictated by the regularity with which they can pack and the surface area of interaction.
Source: Boundless. “Properties of Alkenes.” Boundless Chemistry. Boundless, 05 Jun. 2015. Retrieved 06 Jun. 2015 from https://www.boundless.com/chemistry/...enes-631-3627/
Alkenes contains a carbon-carbon double bond. This carbon-carbon double bond changes the physicals properties of alkenes. At room temperatue, alkenes exist in all three phases, solid, liquids, and gases. Melting and boiling points of alkenes are similar to that of alkanes, however, isomers of cis alkenes have lower melting points than that of trans isomers. Alkenes display a weak dipole-dipole interactions due to the electron-attracting sp2carbon.
Properties of Alkenes
This is an introductory page about alkenes such as ethene, propene and the rest. It deals with their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity.
What are alkenes?
Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. The first two are:
ethene C2H4
propene C3H6
You can work out the formula of any of them using: CnH2n The table is limited to the first two, because after that there are isomers which affect the names.
Isomerism in the alkenes
Structural isomerism
All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more different structural formulae that you can draw for each molecular formula.
For example, with C4H8, it isn't too difficult to come up with these three structural isomers:
There is, however, another isomer. But-2-ene also exhibits geometric isomerism.
Geometric (cis-trans) isomerism
The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH3 groups on either end of the molecule locked either on one side of the molecule or opposite each other.
These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides).
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below.
Physical properties of the alkenes
Boiling Points
The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.
In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.
Solubility
Alkenes are virtually insoluble in water, but dissolve in organic solvents.
Chemical Reactivity
Bonding in the alkenes
We just need to look at ethene, because what is true of C=C in ethene will be equally true of C=C in more complicated alkenes. Ethene is often modeled like this:
The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other.
One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons.
In this diagram, the line between the two carbon atoms represents a normal bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. This sort of bond is called a sigma bond.
The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other.
The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond and, because they lie exposed above and below the rest of the molecule, they are relatively open to attack by other things.
The reactions of alkenes
Like any other hydrocarbons, alkenes burn in air or oxygen, but these reactions are unimportant. Alkenes are too valuable to waste in this way. The important reactions all center around the double bond. Typically, the pi bond breaks and the electrons from it are used to join the two carbon atoms to other things. Alkenes undergo addition reactions. For example, using a general molecule X-Y . . .
The rather exposed electrons in the pi bond are particularly open to attack by things which carry some degree of positive charge. These are called electrophiles. If you explore the rest of the alkene menu, you will find lots of examples of this kind.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Background_of_Alkenes.txt |
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of $\pi$ bonds and/or cyclic rings.
Saturated vs. Unsaturated Molecules
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 1-methyoxypentane
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 3-chloro-5-octyne
Calculating Degrees of Unsaturation (DoU)
Degree of Unsaturation (DoU) is also known as Double Bond Equivalent. If the molecular formula is given, plug in the numbers into this formula:
$DoU= \dfrac{2C+2+N-X-H}{2}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
DoU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
3 3 0 0
2 1 0
1 2 0
0 1 1
0 3 0
1 0 1
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.
Problems
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from 1., give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation for the following molecular formulas:
1. (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Using the molecular formulas from 3, are the molecules unsaturated or saturated.
5. Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answers
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
(b.) The molecule can contain any of these combinations (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
Contributors
• Kim Quach (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Degree_of_Unsaturation.txt |
When considering compounds having two or more double bonds in a molecule, it is useful to identify three distinct ways in which these functions may be oriented with respect to each other. First, the double bonds may be separated by one or more sp3-hybridized carbon atoms, as in 1,5-hexadiene. In this circumstance each double bond behaves independently of the other, and we refer to them as isolated. A second relationship has the double bonds connected to each other by a single bond, as in 1,3-hexadiene, and we refer to this arrangement as conjugated. Finally, two double bonds might share a carbon atom, as in 1,2-hexadiene. The central carbon atom in such a system is sp-hybridized, and we call such double bonds cumulated. These three isomers are shown in the following diagram.
Another stereoisomeric factor associated with conjugated dienes exists. Rotation about the single bond joining the two double bonds (colored blue) converts a trans-like s-trans conformation to its s-cis form. The energy barrier to this conformational isomerization is normally low, and the s-trans conformer is often more stable than the s-cis conformer, as shown in the diagram.
These categories are based on more than obvious structural variations. We find significant differences in the chemical properties of dienes depending on their structural type. For example, catalytic hydrogenation converts all the dienes shown here to the alkane hexane, but the heats of reaction (heat of hydrogenation) reflect characteristic differences in their thermodynamic stability. This is illustrated in the diagram below. Taking the heat of hydrogenation of 1-hexene (30.1 kcal/mole) as a reference, we find that the isolated diene, 1,5-hexadiene, as expected, generates double this heat of reaction on conversion to hexane. The cumulated diene, 1,2-hexadiene, has a 6 kcal/mole higher heat of reaction, indicating it is less stable than the isolated diene by this magnitude. On the other hand, conjugation of double bonds seems to stabilize a diene by about 5 kcal/mole. The increase in stability of 2,4-hexadiene over 1,3-hexadiene (both are conjugated) is due to the increased double bond substitution of the former, a factor noted earlier for simple alkenes.
The stabilization of dienes by conjugation is less dramatic than the aromatic stabilization of benzene. Nevertheless, similar resonance and molecular orbital descriptions of conjugation may be written. A resonance description, such as the one shown here, involves charge separation, implying a relatively small degree of stabilization.
CH2=CH-CH=CH2 (+)CH2-CH=CH-CH2:(–)
A molecular orbital model for 1,3-butadiene is shown below. Note that the lobes of the four p-orbital components in each pi-orbital are colored differently and carry a plus or minus sign. This distinction refers to different phases, defined by the mathematical wave equations for such orbitals. Regions in which adjacent orbital lobes undergo a phase change are called nodes. Orbital electron density is zero in such regions. Thus a single p-orbital has a node at the nucleus, and all the pi-orbitals shown here have a nodal plane that is defined by the atoms of the diene. This is the only nodal surface in the lowest energy pi-orbital, π1. Higher energy pi-orbitals have an increasing number of nodes. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Dienes.txt |
In the previous section, we learned about the physical properties of alkenes (Physical Properties of Alkenes). The double bonded carbons in alkene molecules also have an effect of shifts shown in 1H and 13C nuclear magnetic resonance spectr.
Hydrogens near double bonds are deshielded
For background information of 1H NMR, you can refer 1H Nuclear Magnetic Resonance from the last chapter. In 1H NMR spectrum, hydrogen atoms bound to a carbon consisting of a double bond (these hydrogens are called alkenyl hydrogens) are typically found in low field of the NMR spectrum, which is the left side, and the hydrogens are said to be deshielded. The cause for this is due to the movement of the electrons in the pi bond of the carbon-carbon double bond.
Alkenyl hydrogens create an external magnetic field that is perpendicular to the double bond axis and causes the electrons in the pi bond to enter a circular motion (shown in red). The circular motion actually reinforces the external field at the edge of the double bond on both sides of the pi bond but creates a local field (shown in purple and green) that opposes the external field in the center of the double bond. Because of this pulling force within the pi bond across the double bond which reinforces the regions occupied by alkenyl hydrogens, the alkenyl hydrogens are strongly deshielded.
Additionally, alkenyl hydrogens do not have to be all chemical-shift equivalent, and when they aren't, coupling will be observed which is the different peaks in an MNR spectrum.
Cis and trans coupling appear differently on 1H NMR spectrum
Here are a couple of terms to know:
• Vicinal - Coupling between hydrogens on adjacent carbons.
• Geminal - Coupling between nonequivalent hydrogens on the same carbon atom.
Coupling Constants Around a Double Bond
Type of coupling Name Range (Hz) Typical (Hz)
Vicinal, cis 6-14 10
Vicinal, trans 11-18 16
Geminal 0-3 2
None 4-10 6
Allylic 0.5-3.0 2
When alkynel hydrogen atoms are not symmetrically substituted on a double bonded carbon, the hydrogens of a cis and trans isomer will yield a different shift on the NMR spectrum. Because the coupling constant is smaller in a cis isomer than in a trans isomer, the NMR spectrums of the two isomers are different conveying the hydrogens in a cis isomer to be slightly more upfield to-- the right of the spectrum-- and trans hydrogens to be more downfield to the left.
Sometimes coupling will lead to very complicated patterns as a result of the J values that vary widely due to the relationship between the hydrogens involved. When this occurs, information can still be derived to determine the structure of a molecule by looking at the number of signals, the chemical shift of each one, integration, and splitting patterns similarly to identifying alkane NMR.
Alkenyl carbons are deshielded in 13C NMR
For background information on 13C NMR, please refer to 13C Nuclear Magnetic Resonance from the previous chapter. Compared to alkane carbons with one bond, alkene carbons show a relatively low field shift on the 13C NMR spectrum and absorb about 100 ppm lower field. Also, in broad-band decoupled 13C NMR, sp2 carbons absorb as sharp single lines so with these two methods, it is easy to determine the presence of a double bond in 13C NMR spectrum.
Here are the common 13C Chemical Shift Ranges:
Note that the carbon-carbon double bonds are found in the range between 100-170 ppm. Carbon atoms on alkenes that are attatched to another carbon group are found more downfield than carbon alkenes attatched to hydrogens.
Let's try a 1H NMR practice problem with C4H7Cl:
Remember from previous sections that to solve an NMR spectrum with double bonds, we must know the Degrees of Unsaturation. From this, we get degrees of unsaturation= (9-7)/2=1 so there is one pi bond or ring in our molecule. Next we must look at the integration of the NMR spectrums.
• ppm= 1.8 with 3H reveals a CH3attached to an unsaturated functional group
• ppm= 4.0 with integration of 2H is a CH2 most likely attached to the Cl
• ppm = 4.9 and 5.1 are singlets of 1H and must be our two alkene hydrogens
There are three ways to attach the discovery we made, but only one of them are the correct answer. Since the coupling of the two alkene hydrogens are small whereas vicinal hydrogens tend to be large, we conclude that the hydrogens are geminal and appear on the same carbon. This leaves the two other groups to be located on the other alkene carbon.
Practice Problems
1. If an alkene isomer showed coupling at 17 J (Hz), would it be cis or trans?
2. Why are alkenyl hydrogens deshielded?
3. Rank the order of coupling from highest to lowest range:
1. vicinal, cis
2. vicinal, trans
3. geminal
4. True or false: In 13C NMR, alkene carbons show an upfield shift compared to alkane carbons.
5. What are two ways to distinguish an alkene in a 13C NMR spectrum?
Solutions
1. trans because the range of trans coupling in an alkene is 11-18 J (Hz) while cis is 6-14 J (Hz)
2. Alkenyl hydrogens are deshielded due to the movement of the electrons in the pi bond. Alkenyl hydrogens create an external magnetic field that is perpendicular to the double bond axis and causes the electrons in the pi bond to enter a circular motion. The circular motion of the pi bonds reinfornce the external field where the hydrogens are located thus being strongly deshielded and appearing low field in an NMR spectrum.
3. b) vicinal, trans > a) vicinal, cis > c) geminal
4. False. Alkene carbons absorb at about 100 ppm lower field than alkane carbons thus are found low field in a 13C NMR spectrum.
5. Alkenes typically absorb around 122 ppm and appear as sharp lines in 13C NMR spectrums making them easy to distinguish. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Nuclear_Magnetic_Resonance_%28NMR%29_of_Alkenes.txt |
Alkenes contains a carbon-carbon double bond. This carbon-carbon double bond changes the physicals properties of alkenes. At room temperatue, alkenes exist in all three phases, solid, liquids, and gases. Melting and boiling points of alkenes are similar to that of alkanes, however, isomers of cis alkenes have lower melting points than that of trans isomers. Alkenes display a weak dipole-dipole interactions due to the electron-attracting sp2carbon.
Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. The first two are:
• ethene ( \(C_2H_4\) )
• propene (\(C_3H_6 \))
You can work out the formula of any of them using: \(C_nH_{2n}\). The list is limited to the first two, because after that there are isomers which affect the names.
Structural Isomerism
All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more different structural formulae that you can draw for each molecular formula. For example, with C4H8, it isn't too difficult to come up with these three structural isomers:
There is, however, another isomer. But-2-ene also exhibits geometric isomerism.
Geometric (cis-trans) Isomerism
The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH3 groups on either end of the molecule locked either on one side of the molecule or opposite each other. These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides).
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds.
Physical state
Ethene, Propene, and Butene exists as colorless gases. Members of the 5 or more carbons such as Pentene, Hexene, and Heptene are liquid, and members of the 15 carbons or more are solids.
Density
Alkenes are lighter than water and are insoluble in water due to their non-polar characteristics. Alkenes are only soluble in nonpolar solvents.
Solubility
Alkenes are virtually insoluble in water, but dissolve in organic solvents. The reasons for this are exactly the same as for the alkanes.
Boiling Points
The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids. Boiling points of alkenes depends on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.
Table 1: Meting Points and Boiling Points of common Alkenes
Compound Melting Points (°C) Boiling points (°C)
Ethene -169 -104
Propene -185 -47
Trans-2-Butene 0.9
Cis-2-butene 3.7
Trans 1,2-dichlorobutene 155
Cis 1,2-dichlorobutene 152
1-Pentene -165 30
Trans-2-Pentene -135 36
Cis-2-Pentene -180 37
1-Heptene -119 115
3-Octene -101.9 122
3-Nonene -81.4 147
5-Decene -66.3 170
In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.
Melting Points
Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.
Polarity
Chemical structure and fuctional groups can affect the polarity of alkenes compounds. The sp2 carbon is much more electron-withdrawing than the sp3 hybridize orbitals, therefore, creates a weak dipole along the substituent weak alkenly carbon bond. The two individual dipoles together form a net molecular dipole. In trans-subsituted alkenes, the dipole cancel each other out. In cis-subsituted alkenes there is a net dipole, therefore contributing to higher boiling in cis-isomers than trans-isomers.
Problems
1. Compound containing carbon-carbon double bonds are called:
• (a) Alkanes
• (b) Alkynes
• (c) Alkenes
• (d) Alcohols
2. Cis-alkenes exhibit a weak net dipole-dipole interactions:
• True
• False
3. Which has a lower melting poin
• (a) 5-Decene
• (b) 3-Octene
• (c) cis-2-Pentene
• (d) 1-Pentene | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Physical_Properties_of_Alkenes.txt |
Ethene is the formal IUPAC name for H2C=CH2, but it also goes by a common name: Ethylene. The name Ethylene is used because it is like an ethyl group ($CH_2CH_3$) but there is a double bond between the two carbon atoms in it. Ethene has the formula $C_2H_4$ and is the simplest alkene because it has the fewest carbons (two) necessary for a carbon-carbon double bond.
Introduction
Bonding in carbon is covalent, containing either sigma or $\pi$ bonds. Carbon can make single, double, or triple bonds. The number of bonds it makes determines the structure. With four single bonds, carbon has a tetrahedral structure, while with one double bond it's structure is trigonal planar, and with a triple bond it has a linear structure.
A solitary carbon atom has four electrons, two in the 2s orbital, and one in each of the 2$p_x$ and 2$p_y$ orbitals, leaving the $2p_z$ orbital empty. A single carbon atom can make up to four bonds, but by looking at its electron configuration this would not be possible because there are only two electrons available to bond with. The other two are in a lone pair state, making them much less reactive to another electron that is by itself. Well it is, in order to make the four bonds, the carbon atom promotes one of the 2s electrons into the empty $2p_z$ orbital, leaving the carbon with four unpaired electrons allowing it to now form four bonds. The electron is not promoted spontaneously. It becomes promoted when a photon of light with the correct wavelength hits the carbon atom. When this photon hits the carbon atom it gives the atom enough energy to promote one of the lone pair electrons to the $2p_z$ orbital.
Sigma and Pi Bonds
All the bonds in Ethene are covalent, meaning that they are all formed by two adjacent atoms sharing their valence electrons. As opposed to ionic bonds which hold atoms together through the attraction of two ions of opposite charges.
Sigma bonds are created when there is overlap of similar orbitals, orbitals that are aligned along the inter-nuclear axis. Common sigma bonds are $s+s$, $p_z+p_z$ and $s+p_z$, $z$ is the axis of the bond on the xyz-plane of the atom.
$\pi$ bonds are created when there is adequate overlap of similar, adjacent $p$ orbitals, such as $p_x$+$p_x$ and $p_y$+$p_y$. Each p orbital has two lobes, one usually indicated by a + and the other indicated by a - (sometimes one may be shaded while the other is not). This + and - (shaded, not shaded) are only meant to indicate the opposite phase $\phi$ the wave functions, they do not indicate any type of electrical charge. For a $\pi$ bond to form both lobes of the $p$ orbital must overlap, + with + and - with -. When a + lobe overlaps with a - lobe this creates an anti-bonding orbital interaction which is much higher in energy, and therefore not a desirable interaction.
Usually there can be no $\pi$ bonds between two atoms without having at least one sigma bond present first. But there are special cases such as dicarbon ($C_2$) where the central bond is a $\pi$ bond not a sigma bond, but in cases like these the two atoms want to have as much orbital overlap as possible so the bond lengths between the atoms are smaller than what is normally expected.
The $\pi$ bond in ethene is weak compared to the sigma bond between the two carbons. This weakness makes the $\pi$ bond and the overall molecule a site of comparatively high chemical reactivity to an array of different substances. This is due to the high electron density in the $\pi$ bond, and because it is a weak bond with high electron density the $\pi$ bond will easily break in order to form two separate sigma bonds. Sites such as these are referred to as functional groups or functionalities. These groups have characteristic properties and they control the reactivity of the molecule as a whole. How these functional groups and other reactants form various products are an important concept in organic chemistry.
Orbital Bonding in Ethene
Ethene is made up of four 1s1 Hydrogen atoms and two 2s2 2$p_x$1 2$p_y$2 carbon atoms. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight in the valence shell. Having eight valence electrons around carbon gives the atom itself the same electron configuration as neon, a noble gas. Carbon wants to have the same configuration as Neon because when it has eight valence electrons carbon is at its most stable, lowest energy state, it has all of the electrons that it wants, so it is no longer reactive.
Structure of Ethene
Ethene is not a very complicated molecule. It contains two carbon atoms that are double bonded to each other, with each of these atoms also bonded to two Hydrogen atoms.
This forms a total of three bonds to each carbon atom, giving them an $sp^2$ hybridization. Since the carbon atom is forming three sigma bonds instead of the four that it can, it only needs to hybridize three of its outer orbitals, instead of four. It does this by using the $2s$ electron and two of the $2p$ electrons, leaving the other unchanged. This new orbital is called an $sp^2$ hybrid because that's exactly what it is, it is made from one s orbital and two p orbitals.
When atoms are an $sp^2$ hybrid they have a trigonal planar structure. These structures are very similar to a 'peace' sign, there is a central atom with three atoms around it, all on one plane. Trigonal planar molecules have an ideal bond angle of 120° on each side.
The H-C-H bond angle is 117°, which is very close to the ideal 120° of a carbon with $sp^2$ hybridization. The other two angles (H-C=C) are both 121.5°.
Rigidity in Ethene
There is rigidity in the Ethene molecule due to the double-bonded carbons. In Ethane there are two carbons that share a single bond, this allows the two Methyl groups to rotate with respect to each other. These different conformations result in higher and lower energy forms of Ethane. In Ethene there is no free rotation about the carbon-carbon sigma bond. There is no rotation because there is also a $\pi$ bond along with the sigma bond between the two carbons. A $\pi$ bond is only formed when there is adequate overlap between both top and bottom p-orbitals. In order for there to be free rotation the p-orbitals would have to go through a phase where they are 90° from each other, which would break the $\pi$ bond because there would be no overlap. Since the $\pi$ bond is essential to the structure of Ethene it must not break, so there can be not free rotation about the carbon-carbon sigma bond.
Problems
1. Write out the condensed formula for ethene.
2. Write out the Kekulé formula for ethene.
3. Write out the bond-line formula for ethene.
4. What are the distances between carbon and hydrogen atoms when they are bonded? Carbon-carbon single bond? Carbon-carbon double bond?
5. Why is it that the carbons in ethene cannot freely rotate around the carbon-carbon double bond?
Answers
1. H2CCH2
2.
3. -
4. C-H: 1.076 angstroms, C-C: 1.54 angstroms, C=C: 1.330 angstroms
5. the carbons cannot freely rotate about the carbon-carbon double bond because in order to rotate the p-orbitals would have to pass through a 90° point where there would no longer be any overlap, so the $\pi$ bond would have to break for there to be free rotation. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Structure_and_Bonding_in_Ethene-The_Pi_Bond.txt |
Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. Alkenes are relatively stable compounds, but are more reactive than alkanes because of the reactivity of the carbon–carbon π-bond. Most reactions of alkenes involve additions to this π bond, forming new single bonds.
• Addition of Sulfuric acid to Alkenes
The carbon-carbon double bond in alkenes such as ethene react with concentrated sulfuric acid. It includes the conversion of the product into an alcohol.
• Addition Reactions of Dienes
Addition reactions of isolated dienes proceed more or less as expected from the behavior of simple alkenes. Thus, if one molar equivalent of 1,5-hexadiene is treated with one equivalent of bromine a mixture of 5,6-dibromo-1-hexene, 1,2,5,6-tetrabromohexane and unreacted diene is obtained, with the dibromo compound being the major product (about 50%).
• Catalytic Hydrogenation of Alkenes
The double bond of an alkene consists of a sigma (σ) bond and a pi (π) bond. Because the carbon-carbon π bond is relatively weak, it is quite reactive and can be easily broken and reagents can be added to carbon. Reagents are added through the formation of single bonds to carbon in an addition reaction.
• Catalytic Hydrogenation of Alkenes II
Alkene hydrogenation is the syn-addition of hydrogen to an alkene, saturating the bond. The alkene reacts with hydrogen gas in the presence of a metal catalyst which allows the reaction to occur quickly. The energy released in this process, called the heat of hydrogenation, indicates the relative stabily of the double bond in the molecule (see Catalytic Hydrogenation).
• Catalytic Hydrogenation of Alkenes II
The carbon-carbon double bond in alkenes react with hydrogen in the presence of a metal catalyst. This is called hydrogenation. It includes the manufacture of margarine from animal or vegetable fats and oils.
• Diels-Alder Cycloaddition
The unique character of conjugated dienes manifests itself dramatically in the Diels-Alder Cycloaddition Reaction. A cycloaddition reaction is the concerted bonding together of two independent pi-electron systems to form a new ring of atoms. When this occurs, two pi-bonds are converted to two sigma-bonds, the simplest example being the hypothetical combination of two ethene molecules to give cyclobutane. This does not occur under normal conditions, though.
• Electrophilic Addition of Hydrogen Halides
Due to the nature of the π bond, π bonds can act like a nucleophile and undergo addition of electrophiles. The π electrons have a high electron density in the π electron cloud which can be easily polarized (give up the electrons) and can act like a nucleophile (nucleus-loving due to too many electrons; wants to give them away to electrophile). This behavior is very similar to the behavior that the lone pair electrons have on a Lewis base.
• Electrophilic Addition of Hydrogen Halides II
Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.
• Free Radical Reactions of Alkenes
• Hydroboration-Oxidation of Alkenes
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from \(BH_3\) or \(BHR_2\) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene bouble bond. Furthermore, the borane acts as a lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet.
• Oxidation of Alkenes with Potassium Manganate
The carbon-carbon double bonds in alkenes such as ethene react with potassium manganate(VII) solution (potassium permanganate solution).
• Ozonolysis
Ozonolysis is a type of weak oxidative cleavage where we cleave alkenes (double bonds) into either ketones, aldehydes or carboxylic acid using ozone.
• Ozonolysis of Alkenes and Alkynes
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces.
• Reactions of Alkenes with Halogens
This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with halogens such as chlorine, bromine and iodine. This is called halogenation. Reactions where the chlorine or bromine are in solution (for example, "bromine water") are slightly more complicated and are treated separately at the end.
• Stereoselectivity in Addition Reactions to Double Bonds
• Vicinal Syn Dihydroxylation
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
Reactivity of Alkenes
The most common chemical transformation of a carbon-carbon double bond is the addition reaction. A large number of reagents, both inorganic and organic, have been found to add to this functional group, and in this section we shall review many of these reactions.
• Addition of Lewis Acids (Electrophilic Reagents)
• Addition of Strong Brønsted Acids
• Rearrangement of Carbocations
A large number of reagents, both inorganic and organic, have been found to add to this functional group, and in this section we shall review many of these reactions. A majority of these reactions are exothermic, due to the fact that the C-C pi-bond is relatively weak (ca. 63 kcal/mole) compared to the sigma-bonds formed with the atoms or groups of the reagent. Remember, the bond energies of a molecule are the energies required to break (homolytically) all the covalent bonds in the molecule. Consequently, if the bond energies of the product molecules are greater than the bond energies of the reactants, the reaction will be exothermic. The following calculations for the addition of H-Br are typical. Note that by convention exothermic reactions have a negative heat of reaction.
Addition Reactions of Alkenes
The proton is not the only electrophilic species that initiates addition reactions to the double bond. Lewis acids like the halogens, boron hydrides and certain transition metal ions are able to bond to the alkene pi-electrons, and the resulting complexes rearrange or are attacked by nucleophiles to give addition products.
The electrophilic character of the halogens is well known. Although fluorine is uncontrollably reactive, chlorine, bromine and to a lesser degree iodine react selectively with the double bond of alkenes. The addition of chlorine and bromine to alkenes, as shown in the following general equation, proceeds by an initial electrophilic attack on the pi-electrons of the double bond. Iodine adds reversibly to double bonds, but the equilibrium does not normally favor the addition product, so it is not a useful preparative method. Dihalo-compounds in which the halogens are juxtaposed in the manner shown are called vicinal, from the Latin vicinalis, meaning neighboring.
R2C=CR2 + X2 → R2CX-CR2X
Other halogen containing reagents which add to double bonds include hypohalous acids, HOX, and sulfenyl chlorides, RSCl. These reagents are unsymmetrical, so their addition to unsymmetrical double bonds may in principle take place in two ways. In practice, these addition reactions are regioselective, with one of the two possible constitutionally isomeric products being favored. The electrophilic moiety of these reagents is the halogen.
(CH3)2C=CH2 + HOBr → (CH3)2COH-CH2Br
(CH3)2C=CH2 + C6H5SCl → (CH3)2CCl-CH2SC6H5
The regioselectivity of the above reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Thus, bonding of an electrophilic species to the double bond of an alkene should result in preferential formation of the more stable (more highly substituted) carbocation, and this intermediate should then combine rapidly with a nucleophilic species to produce the addition product. This is illustrated by the following equation.
Electrophilic Moiety
To apply this mechanism we need to determine the electrophilic moiety in each of the reagents. By using electronegativity differences we can dissect common addition reagents into electrophilic and nucleophilic moieties, as shown on the right. In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pKa's ca. 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation. The nucleophilic species that bonds to the intermediate carbocation is then hydroxide ion, or more likely water (the usual solvent for these reagents), and the products are called halohydrins. Sulfenyl chlorides add in the opposite manner because the electrophile is a sulfur cation, RS(+), whereas the nucleophilic moiety is chloride anion (chlorine is more electronegative than sulfur).
If you understand this mechanism you should be able to write products for the following reactions:
Oxymercuration and Hydroboration
The addition products formed in reactions of alkenes with mercuric acetate and boron hydrides (compounds shown at the bottom of of the reagent list) are normally not isolated, but instead are converted to alcohols by a substitution reaction. These important synthetic transformations are illustrated for 2-methylpropene by the following equations, in which the electrophilic moiety is colored red and the nucleophile blue. The top reaction sequence illustrates the oxymercuration procedure and the bottom is an example of hydroboration.
The light blue vertical line separates the addition reaction on the left from the substitution on the right. The atoms or groups that have been added to the original double bond are colored orange in the final product. In both cases the overall reaction is the addition of water to the double bond, but the regioselectivity is reversed. The oxymercuration reaction gives the product predicted by Markovnikov's rule; hydroboration on the other hand gives the "anti-Markovnikov" product. Complementary reactions such as these are important because they allow us to direct a molecular transformation whichever way is desired.
Mercury and boron are removed from the organic substrate in the second step of oxymercuration and hydroboration respectively. These reactions are seldom discussed in detail; however, it is worth noting that the mercury moiety is reduced to metallic mercury by borohydride (probably by way of radical intermediates), and boron is oxidized to borate by the alkaline peroxide. Addition of hydroperoxide anion to the electrophilic borane generates a tetra-coordinate boron peroxide, having the general formula R3B-O-OH(-). This undergoes successive intramolecular shifts of alkyl groups from boron to oxygen, accompanied in each event by additional peroxide addition to electron deficient boron. The retention of configuration of the migrating alkyl group is attributed to the intramolecular nature of the rearrangement.
Since the oxymercuration sequence gives the same hydration product as acid-catalyzed addition of water (see Brønsted acid addition), we might question why this two-step procedure is used at all. The reason lies in the milder reaction conditions used for oxymercuration. The strong acid used for direct hydration may not be tolerated by other functional groups, and in some cases may cause molecular rearrangement (see above).
The addition of borane, BH3, requires additional comment. In pure form this reagent is a dimeric gas B2H6, called diborane, but in ether or THF solution it is dissociated into a solvent coordinated monomer, R2O-BH3. Although diborane itself does not react easily with alkene double bonds, H.C. Brown (Purdue, Nobel Prize 1979) discovered that the solvated monomer adds rapidly under mild conditions. Boron and hydrogen have rather similar electronegativities, with hydrogen being slightly greater, so it is not likely there is significant dipolar character to the B-H bond. Since boron is electron deficient (it does not have a valence shell electron octet) the reagent itself is a Lewis acid and can bond to the pi-electrons of a double bond by displacement of the ether moiety from the solvated monomer. As shown in the following equation, this bonding might generate a dipolar intermediate consisting of a negatively-charged boron and a carbocation. Such a species would not be stable and would rearrange to a neutral product by the shift of a hydride to the carbocation center. Indeed, this hydride shift is believed to occur concurrently with the initial bonding to boron, as shown by the transition state drawn below the equation, so the discrete intermediate shown in the equation is not actually formed. Nevertheless, the carbocation stability rule cited above remains a useful way to predict the products from hydroboration reactions. You may correct the top equation by clicking the button on its right. Note that this addition is unique among those we have discussed, in that it is a single-step process. Also, all three hydrogens in borane are potentially reactive, so that the alkyl borane product from the first addition may serve as the hydroboration reagent for two additional alkene molecules. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Addition_Reactions_of_Alkenes/Addition_of_Lewis_Acids_%28Electrophilic_Reagents%29.txt |
The formation of carbocations is sometimes accompanied by a structural rearrangement. Such rearrangements take place by a shift of a neighboring alkyl group or hydrogen, and are favored when the rearranged carbocation is more stable than the initial cation. The addition of HCl to 3,3-dimethyl-1-butene, for example, leads to an unexpected product, 2-chloro-2,3-dimethylbutane, in somewhat greater yield than 3-chloro-2,2-dimethylbutane, the expected Markovnikov product. This surprising result may be explained by a carbocation rearrangement of the initially formed 2º-carbocation to a 3º-carbocation by a 1,2-shift of a methyl group.
Another factor that may induce rearrangement of carbocation intermediates is strain. The addition of HCl to α-pinene, the major hydrocarbon component of turpentine, gives the rearranged product, bornyl chloride, in high yield. As shown in the following equation, this rearrangement converts a 3º-carbocation to a 2º-carbocation, a transformation that is normally unfavorable. However, the rearrangement also expands a strained four-membered ring to a much less-strained five-membered ring, and this relief of strain provides a driving force for the rearrangement. The atom numbers (colored red) for the pinene structure are retained throughout the rearrangement to help orient the viewer. The green numbers in the final product represent the proper numbering of this bicyclic ring system.
The propensity for structural rearrangement shown by certain molecular constitutions, as illustrated above, serves as a useful probe for the intermediacy of carbocations in a reaction. We shall use this test later.
Addition Reactions of Dienes
Addition reactions of isolated dienes proceed more or less as expected from the behavior of simple alkenes. Thus, if one molar equivalent of 1,5-hexadiene is treated with one equivalent of bromine a mixture of 5,6-dibromo-1-hexene, 1,2,5,6-tetrabromohexane and unreacted diene is obtained, with the dibromo compound being the major product (about 50%).
CH2=CH(CH2)2CH=CH2 + Br2 BrCH2CHBr(CH2)2CH=CH2 + BrCH2CHBr(CH2)2CHBrCH2Br + CH2=CH(CH2)2CH=CH2
5,6-dibromo-1-hexene 1,2,5,6-tetrabromohexane 1,5-hexadiene
Similar reactions of conjugated dienes, on the other hand, often give unexpected products. The addition of bromine to 1,3-butadiene is an example. As shown below, a roughly 50:50 mixture of 3,4-dibromo-1-butene (the expected product) and 1,4-dibromo-2-butene (chiefly the E-isomer) is obtained. The latter compound is remarkable in that the remaining double bond is found in a location where there was no double bond in the reactant. This interesting relocation requires an explanation.
CH2=CH-CH=CH2 + Br2 BrCH2CHBr-CH=CH2 + BrCH2CH=CHCH2Br
3,4-dibromo-1-butene 1,4-dibromo-2-butene
The expected addition product from reactions of this kind is the result of 1,2-addition, i.e. bonding to the adjacent carbons of a double bond. The unexpected product comes from 1,4-addition, i.e. bonding at the terminal carbon atoms of a conjugated diene with a shift of the remaining double bond to the 2,3-location. These numbers refer to the four carbons of the conjugated diene and are not IUPAC nomenclature numbers. Product compositions are often temperature dependent, as the addition of HBr to 1,3-butadiene demonstrates.
CH2=CH-CH=CH2 + HBr
reaction temperature
CH3CHBr-CH=CH2 +
1,2 addition yield
CH3CH=CHCH2Br
1,4 addition yield
0 ºC
40 ºC
70%
15%
30%
85%
Bonding of an electrophilic atom or group to one of the end carbon atoms of a conjugated diene (#1) generates an allyl cation intermediate. Such cations are stabilized by charge delocalization, and it is this delocalization that accounts for the 1,4-addition product produced in such addition reactions. As shown in the diagram, the positive charge is distributed over carbons #2 and #4 so it is at these sites that the nucleophilic component bonds. Note that resonance stabilization of the allyl cation is greater than comparable stabilization of 1,3-butadiene, because charge is delocalized in the former, but created and separated in the latter.
An explanation for the temperature influence is shown in the following energy diagram for the addition of HBr to 1,3-butadiene. The initial step in which a proton bonds to carbon #1 is the rate determining step, as indicated by the large activation energy (light gray arrow). The second faster step is the product determining step, and there are two reaction paths (colored blue for 1,2-addition and magenta for 1,4-addition). The 1,2-addition has a smaller activation energy than 1,4-addition, but the 1,4-product is more stable than the 1,2-product. At low temperatures, the products are formed irreversibly and reflect the relative rates of the two competing reactions. This is termed kinetic control. At higher temperatures, equilibrium is established between the products, and the thermodynamically favored 1,4-product dominates. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Addition_Reactions_of_Alkenes/Rearrangement_of_Carbocations.txt |
The carbon-carbon double bond in alkenes such as ethene react with concentrated sulfuric acid. It includes the conversion of the product into an alcohol.
The reaction with ethene
Alkenes react with concentrated sulfuric acid in the cold to produce alkyl hydrogensulfates. Ethene reacts to give ethyl hydrogensulfate.
The structure of the product molecule is sometimes written as CH3CH2HSO4, but the version in the equation is better because it shows how all the atoms are linked up. You may also find it written as CH3CH2OSO3H.
Confused by all this? Don't be! All you need to do is to learn the structure of sulfuric acid. A hydrogen from the sulfuric acid joins on to one of the carbon atoms, and the rest joins on to the other one. Make sure that you can see how the structure of the sulfuric acid relates to the various ways of writing the formula for the product.
The reaction with propene
This is typical of the reaction with unsymmetrical alkenes. An unsymmetrical alkene has different groups at either end of the carbon-carbon double bond. If sulfuric acid adds to an unsymmetrical alkene like propene, there are two possible ways it could add. You could end up with one of two products depending on which carbon atom the hydrogen attaches itself to.
However, in practice, there is only one major product.
This is in line with Markovnikov's Rule which says:
When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.
In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group.
Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant.
Using these Reactions to Make Alcohols
Making ethanol
Ethene is passed into concentrated sulfuric acid to make ethyl hydrogensulphate (as above). The product is diluted with water and then distilled. The water reacts with the ethyl hydrogensulphate to produce ethanol which distils off.
Making propan-2-ol
More complicated alkyl hydrogensulphates react with water in exactly the same way. For example:
Notice that the position of the -OH group is determined by where the HSO4 group was attached. You get propan-2-ol rather than propan-1-ol because of the way the sulfuric acid originally added across the double bond in propene.
These reactions were originally used as a way of manufacturing alcohols from alkenes in the petrochemical industry. These days, alcohols like ethanol or propan-2-ol tend to be manufactured by direct hydration of the alkene because it is cheaper and easier.
Catalytic Hydrogenation
The double bond of an alkene consists of a sigma (σ) bond and a pi (π) bond. Because the carbon-carbon π bond is relatively weak, it is quite reactive and can be easily broken and reagents can be added to carbon. Reagents are added through the formation of single bonds to carbon in an addition reaction.
Introduction
An example of an alkene addition reaction is a process called hydrogenation.In a hydrogenation reaction, two hydrogen atoms are added across the double bond of an alkene, resulting in a saturated alkane. Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability.
Although the hydrogenation of an alkene is a thermodynamically favorable reaction, it will not proceed without the addition of a catalyst.
Common catalysts used are insoluble metals such as palladium in the form Pd-C, platinum in the form PtO2, and nickel in the form Ra-Ni. With the presence of a metal catalyst, the H-H bond in H2 cleaves, and each hydrogen attaches to the metal catalyst surface, forming metal-hydrogen bonds. The metal catalyst also absorbs the alkene onto its surface. A hydrogen atom is then transferred to the alkene, forming a new C-H bond. A second hydrogen atom is transferred forming another C-H bond. At this point, two hydrogens have added to the carbons across the double bond. Because of the physical arrangement of the alkene and the hydrogens on a flat metal catalyst surface, the two hydrogens must add to the same face of the double bond, displaying syn addition.
Common Applications
Hydrogenation reactions are extensively used to create commercial goods.Hydrogenation is used in the food industry to make a large variety of manufactured goods, like spreads and shortenings, from liquid oils. This process also increases the chemical stability of products and yields semi-solid products like margarine. Hydrogenation is also used in coal processing. Solid coal is converted to a liquid through the addition of hydrogen. Liquefying coal makes it available to be used as fuel.
Problems
Complete the following reactions. Provide stereochemistry if necessary.
Contributors
• Jennifer Lew (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Addition_of_Sulfuric_acid_to_Alkenes.txt |
Alkene hydrogenation is the syn-addition of hydrogen to an alkene, saturating the bond. The alkene reacts with hydrogen gas in the presence of a metal catalyst which allows the reaction to occur quickly. The energy released in this process, called the heat of hydrogenation, indicates the relative stabily of the double bond in the molecule (see Catalytic Hydrogenation).
Introduction
The reaction begins with H2 gas and an alkene (a carbon-carbon double bond). The pi bond in the alkene acts as a nucleophile; the two electrons in it form a sigma bond with one of the hydrogen atoms in H2. With the pi bond broken, the other carbon (the one that did not newly receive a hydrogen) is left with a positive formal charge. This is the carbocation intermediate. The remaining (unreacted) hydrogen is now a hydride anion, as it was left with two electrons previously in the H-H sigma bond. Next, the electrons of the negatively charged hydride ion form a bond with the positively charged carbon. This reaction is exothermic. It will occur, but it is very slow without a catalyst.
The Catalyst
A catalyst increases the reaction rate by lowering the activation energy of the reaction. Although the catalyst is not consumed in the reaction, it is required to accelerate the reaction sufficiently to be observed in a reasonable amount of time. Catalysts commonly used in alkene hydrogenation are: platinum, palladium, and nickel. The metal catalyst acts as a surface on which the reaction takes place. This increases the rate by putting the reactants in close proximity to each other, facilitating interactions between them. With this catalyst present, the sigma bond of H2 breaks, and the two hydrogen atoms instead bind to the metal (see #2 in the figure below). The $\pi$ bond of the alkene weakens as it also interacts with the metal (see #3 below).
Since both the reactants are bound to the metal catalyst, the hydrogen atoms can easily add, one at a time, to the previously double-bonded carbons (see #4 and #5 below). The position of both of the reactants bound to the catalyst makes it so the hydrogen atoms are only exposed to one side of the alkene. This explains why the hydrogen atoms add to same side of the molecule, called syn-addition.
Hydrogenation takes place in the presence of a metal catalyst.
Note: The catalyst remains intact and unchanged throughout the reaction.
Heats of Hydrogenation
The stability of alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule. This is a useful tool because heats of hydrogenation can be measured very accurately. The $\Delta H^o$ is usually around -30 kcal/mol for alkenes. Stability is simply a measure of energy. Lower energy molecules are more stable than higher energy molecules. More substituted alkenes are more stable than less substituted ones due to hyperconjugation. They have a lower heat of hydrogenation. The following illustrates stability of alkenes with various substituents:
In disubstituted alkenes, trans isomers are more stable than cis isomers due to steric hindrance. Also, internal alkenes are more stable than terminal ones. See the following isomers of butene:
Trans-2-butene is the most stable because it has the lowest heat of hydrogenation.
Note: In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain.
Problems and Review Questions
1. Bromobutene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product?
2. Cyclohexene reacts with hydrogen gas in the presence of a palladium catalyst. What is the name of the product?
3. What is the stereochemistry of an alkene hydrogenation reaction?
4. When looking at their heats of hydrogenation, is the cis or the trans isomer generally more stable?
5. 2-chloro-4-ethyl-3methylcyclohexene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product?
6.
Answers
1. Bromobutane
2. Cyclohexane
3. Syn-addition
4. Trans
5. 2-chloro-4-ethyl-3methylcyclohexane
6.
Contributors
• Anna Manis (2009) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Catalytic_Hydrogenation_of_Alkenes.txt |
The carbon-carbon double bond in alkenes react with hydrogen in the presence of a metal catalyst. This is called hydrogenation. It includes the manufacture of margarine from animal or vegetable fats and oils.
The hydrogenation of ethene
Ethene reacts with hydrogen in the presence of a finely divided nickel catalyst at a temperature of about 150°C. Ethane is produced.
This is a fairly pointless reaction because ethene is a far more useful compound than ethane! However, what is true of the reaction of the carbon-carbon double bond in ethene is equally true of it in much more complicated cases.
Margarine manufacture
Some margarine is made by hydrogenating carbon-carbon double bonds in animal or vegetable fats and oils. You can recognize the presence of this in foods because the ingredients list will include words showing that it contains "hydrogenated vegetable oils" or "hydrogenated fats". The impression is sometimes given that all margarine is made by hydrogenation - that's simply not true.
Animal and vegetable fats and oils
These are similar molecules, differing in their melting points. If the compound is a solid at room temperature, you usually call it a fat. If it is a liquid, it is often described as an oil. Their melting points are largely determined by the presence of carbon-carbon double bonds in the molecule. The higher the number of carbon-carbon double bonds, the lower the melting point. If there aren't any carbon-carbon double bonds, the substance is said to be saturated. A typical saturated fat might have the structure:
Molecules of this sort are usually solid at room temperature. If there is only one carbon-carbon double bond in each of the hydrocarbon chains, it is called a mono-unsaturated fat (or mono-unsaturated oil, because it is likely to be a liquid at room temperature.) A typical mono-unsaturated oil might be:
If there are two or more carbon-carbon double bonds in each chain, then it is said to be polyunsaturated.
For example:
For simplicity, in all these diagrams, all three hydrocarbon chains in each molecule are the same. That doesn't have to be the case - you can have a mixture of types of chain in the same molecule.
Making margarine
Vegetable oils often contain high proportions of polyunsaturated and mono-unsaturated fats (oils), and as a result are liquids at room temperature. That makes them messy to spread on your bread or toast, and inconvenient for some baking purposes. You can "harden" (raise the melting point of) the oil by hydrogenating it in the presence of a nickel catalyst. Conditions (like the precise temperature, or the length of time the hydrogen is passed through the oil) are carefully controlled so that some, but not necessarily all, of the carbon-carbon double bonds are hydrogenated. This produces a "partially hydrogenated oil" or "partially hydrogenated fat".
You need to hydrogenate enough of the bonds to give the final texture you want. However, there are possible health benefits in eating mono-unsaturated or polyunsaturated fats or oils rather than saturated ones - so you wouldn't want to remove all the carbon-carbon double bonds. The flow diagram below shows the complete hydrogenation of a typical mono-unsaturated oil.
The downside of hydrogenation as a means of hardening fats and oils
There are some probable health risks from eating hydrogenated fats or oils. Consumers are becoming more aware of this, and manufacturers are increasingly finding alternative ways of converting oils into spreadable solids. One of the problems arises from the hydrogenation process. The double bonds in unsaturated fats and oils tend to have the groups around them arranged in the "cis" form.
The relatively high temperatures used in the hydrogenation process tend to flip some of the carbon-carbon double bonds into the "trans" form. If these particular bonds aren't hydrogenated during the process, they will still be present in the final margarine in molecules of trans fats. The consumption of trans fats has been shown to increase cholesterol levels (particularly of the more harmful LDL form) - leading to an increased risk of heart disease. Any process which tends to increase the amount of trans fat in the diet is best avoided. Read food labels, and avoid any food which contains (or is cooked in) hydrogenated oil or hydrogenated fat. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Catalytic_Hydrogenation_of_Alkenes_II.txt |
The highly strained nature of cyclopropane compounds makes them very reactive and interesting synthetic targets. Additionally cyclopropanes are present in numerous biological compounds. One common method of cyclopropane synthesis is the reaction of carbenes with the double bond in alkenes or cycloalkenes. Methylene, H2C, is simplest carbene, and in general carbenes have the formula R2C. Other species that will also react with alkenes to form cyclopropanes but do not follow the formula of carbenes are referred to as carbenoids.
Introduction
Carbenes were once only thought of as short lived intermediates. The reactions of this section only deal with these short lived carbenes which are mostly prepared in situ, in conjunction with the main reaction. However, there do exist so called persistent carbenes. These persistent carbenes are stabilized by a variety of methods often including aromatic rings or transition metals. In general a carbene is neutral and has 6 valence electrons, 2 of which are non bonding. These electrons can either occupy the same sp2 hybridized orbital to form a singlet carbene (with paired electrons), or two different sp2 orbitals to from a triplet carbene (with unpaired electrons). The chemistry of triplet and singlet carbenes is quite different but can be oversimplified to the statement: singlet carbenes usually retain stereochemistry while triplet carbenes do not. The carbenes discussed in this section are singlet and thus retain stereochemistry.
The reactivity of a singlet carbene is concerted and similar to that of electrophilic or nucleophilic addition (although, triplet carbenes react like biradicals, explaining why sterochemistry is not retained). The highly reactive nature of carbenes leads to very fast reactions in which the rate determining step is generally carbene formation.
Preparation of methylene
The preparation of methylene starts with the yellow gas diazomethane, CH2N2. Diazomethane can be exposed to light, heat or copper to facilitate the loss of nitrogen gas and the formation of the simplest carbene methylene. The process is driven by the formation of the nitrogen gas which is a very stable molecule.
Carbene reaction with alkenes
A carbene such as methlyene will react with an alkene which will break the double bond and result with a cyclopropane. The reaction will usually leave stereochemistry of the double bond unchanged. As stated before, carbenes are generally formed along with the main reaction; hence the starting material is diazomethane not methylene.
In the above case cis-2-butene is converted to cis-1,2-dimethylcyclopropane. Likewise, below the trans configuration is maintained.
Additional Types of Carbenes and Carbenoids
In addition to the general carbene with formula R2C there exist a number of other compounds that behave in much the same way as carbenes in the synthesis of cyclopropane. Halogenated carbenes are formed from halomethanes. An example is dicholorcarbene, Cl2C. These halogenated carbenes will form cyclopropanes in the same manner as methylene but with the interesting presence of two halogen atoms in place of the hydrogen atoms.
Carbenoids are substances that form cyclopropanes like carbenes but are not technically carbenes. One common example is the stereospecific Simmon-Smith reaction which utilizes the carbenoid ICH2ZnI. The carbenoid is formed in situ via the mixing of a Zn-Cu couple with CH2I2.Since this reacts thesame as a carbene, the same methods can be applied to determine the product. An example of this is given as problem 5.
Problems
1. Knowing that cycloalkenes react much the same as regular alkenes what would be the expected structure of the product of cyclohexene and diazomethane facilitated by copper metal?
2. What would be the result of a Simmons-Smith reaction that used trans-3-pentene as a reagent?
3. What starting material could be used to form cis-1,2-diethylcyclopropane?
4. What would the following reaction yield?
5. Draw the product of this reaction. What type of reaction is this?
Answers
1. The product will be a bicyclic ring, Bicyclo[4.1.0]heptane.
2. The stereochemistry will be retained making a cyclopropane with trans methyl and ethyl groups. Trans-1-ethyl-2-methylcyclopropane
3. The cis configuration will be maintained from reagent to product so we would want to start with cis-3-hexene. A Simmons Smith reagent, or methylene could be used as the carbene or carbenoid.
4. The halogenated carbene will react the same as methylene yielding, cis-1,1-dichloro-2,3dimethylcyclopropane.
5. This is a Simmons-Smith reaction which uses the carbenoid formed by the CH2I2 and Zu-Cu. The reaction results in the same product as if methylene was used and retains stereospecificity. Iodine metal and the Zn-Cu are not part of the product. The product is trans-1,2-ethyl-methylcyclopropane.
• Paul Tisher | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Diazomethane_Carbenes_and_Cyclopropane_Synthesis.txt |
The unique character of conjugated dienes manifests itself dramatically in the Diels-Alder Cycloaddition Reaction. A cycloaddition reaction is the concerted bonding together of two independent pi-electron systems to form a new ring of atoms. When this occurs, two pi-bonds are converted to two sigma-bonds, the simplest example being the hypothetical combination of two ethene molecules to give cyclobutane. This does not occur under normal conditions, but the cycloaddition of 1,3-butadiene to cyanoethene (acrylonitrile) does, and this is an example of the Diels-Alder reaction. The following diagram illustrates two cycloadditions, and introduces several terms that are useful in discussing reactions of this kind.
In the hypothetical ethylene dimerization on the left, each reactant molecule has a pi-bond (colored orange) occupied by two electrons. The cycloaddition converts these pi-bonds into new sigma-bonds (colored green), and this transformation is then designated a [2+2] cycloaddition, to enumerate the reactant pi-electrons that change their bonding location.
The Diels-Alder reaction is an important and widely used method for making six-membered rings, as shown on the right. The reactants used in such reactions are a conjugated diene, simply referred to as the diene, and a double or triple bond coreactant called the dienophile, because it combines with (has an affinity for) the diene. The Diels-Alder cycloaddition is classified as a [4+2] process because the diene has four pi-electrons that shift position in the reaction and the dienophile has two.
The Diels-Alder reaction is a single step process, so the diene component must adopt a cis-like conformation in order for the end carbon atoms (#1 & #4) to bond simultaneously to the dienophile. Such conformations are called s-cis, the s referring to the single bond connecting the two double bonds. The s-cis and s-trans conformers of 1,3-butadiene are shown in the preceding diagram. For many acyclic dienes the s-trans conformer is more stable than the s-cis conformer (due to steric crowding of the end groups), but the two are generally in rapid equilibrium, permitting the use of all but the most hindered dienes as reactants in Diels-Alder reactions. In its usual form, the diene component is electron rich, and the best dienophiles are electron poor due to electron withdrawing substituents such as CN, C=O & NO2. The initial bonding interaction reflects this electron imbalance, with the two new sigma-bonds being formed simultaneously, but not necessarily at equal rates.
Stereospecificity
We noted earlier that addition reactions of alkenes often exhibited stereoselectivity, in that the reagent elements in some cases added syn and in other cases anti to the the plane of the double bond. Both reactants in the Diels-Alder reaction may demonstrate stereoisomerism, and when they do it is found that the relative configurations of the reactants are preserved in the product (the adduct). The following drawing illustrates this fact for the reaction of 1,3-butadiene with (E)-dicyanoethene. The trans relationship of the cyano groups in the dienophile is preserved in the six-membered ring of the adduct. Likewise, if the terminal carbons of the diene bear substituents, their relative configuration will be retained in the adduct. Using the earlier terminology, we could say that bonding to both the diene and the dienophile is syn. An alternative description, however, refers to the planar nature of both reactants and terms the bonding in each case to be suprafacial (i.e. to or from the same face of each plane). This stereospecificity also confirms the synchronous nature of the 1,4-bonding that takes place.
The essential characteristics of the Diels-Alder cycloaddition reaction may be summarized as follows:
1. The reaction always creates a new six-membered ring. When intramolecular, another ring may also be formed.
2. The diene component must be able to assume a s-cis conformation.
3. Electron withdrawing groups on the dienophile facilitate reaction.
4. Electron donating groups on the diene facilitate reaction.
5. Steric hindrance at the bonding sites may inhibit or prevent reaction.
6. The reaction is stereospecific with respect to substituent configuration in both the dienophile and the diene.
These features are illustrated by the following eight examples, one of which does not give a Diels-Alder cycloaddition. Try to predict the course of each reaction before checking the answers. The formation of a new six-membered ring should be apparent in every case where reaction occurs.
There is no reaction in example D because this diene cannot adopt a s-cis orientation. In examples B, C, F, G & H at least one of the reactants is cyclic so that the product has more than one ring, but the newly formed ring is always six-membered. In example B the the same cyclic compound acts as both the diene colored blue) and the dienophile (colored red). The adduct has three rings, two of which are the five-membered rings present in the reactant, and the third is the new six-membered ring (shaded light yellow). Example C has an alkyne as a dienophile (colored red), so the adduct retains a double bond at that location. This double bond could still serve as a dienophile, but in the present case the diene is sufficiently hindered to retard a second cycloaddition.
The quinone dienophile in reaction F has two dienophilic double bonds. However, the double bond with two methyl substituents is less reactive than the unsubstituted dienophile due in part to the electron donating properties of the methyl groups and in part to steric hindrance. The stereospecificity of the Diels-Alder reaction is demonstrated by examples A, E & H. In A & H the stereogenic centers lie on the dienophile, whereas in E these centers are on the diene. In all cases the configuration of the reactant is preserved in the adduct.
Exo and Endo Bicyclics
Cyclic dienes, such as those in examples B, C & G, give bridged bicyclic adducts for which an additional configurational feature must be designated. As shown in the following diagram, there are two possible configurations for compounds of this kind. If a substituent (colored magenta here) is oriented cis to the longest or more unsaturated bridge (colored blue here), it is said to be endo. When directed trans to the bridge it is exo. When the Diels-Alder reaction forms bridged bicyclic adducts and an unsaturated substituent is located on this bicyclic structure (as in B & G), the chief product is normally the endo isomer "Alder's Endo Rule". Example C does not merit such a nomenclature, since stereoisomeric orientations of the substituent are not possible.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Diels-Alder_Cycloaddition.txt |
Halogens can act as electrophiles to attack a double bond in alkene. Double bond represents a region of electron density and therefore functions as a nucleophile. How is it possible for a halogen to obtain positive charge to be an electrophile?
Introduction
As halogen molecule, for example Br2, approaches a double bond of the alkene, electrons in the double bond repel electrons in bromine molecule causing polarization of the halogen bond. This creates a dipolar moment in the halogen molecule bond. Heterolytic bond cleavage occurs and one of the halogens obtains positive charge and reacts as an electrophile. The reaction of the addition is not regioselective but stereoselective.Stereochemistry of this addition can be explained by the mechanism of the reaction.In the first step electrophilic halogen with a positive charge approaches the double carbon bond and 2 p orbitals of the halogen, bond with two carbon atoms and create a cyclic ion with a halogen as the intermediate step. In the second step, halogen with the negative charge attacks any of the two carbons in the cyclic ion from the back side of the cycle as in the SN2 reaction. Therefore stereochemistry of the product is vicinial dihalides through anti addition.
$\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X}$
Halogens that are commonly used in this type of the reaction are: $Br$ and $Cl$. In thermodynamical terms $I$ is too slow for this reaction because of the size of its atom, and $F$ is too vigorous and explosive.
Solvents that are used for this type of electrophilic halogenation are inert (e.g., CCl4) can be used in this reaction.
Because halogen with negative charge can attack any carbon from the opposite side of the cycle it creates a mixture of steric products.Optically inactive starting material produce optically inactive achiral products (meso) or a racemic mixture.
Electrophilic addition mechanism consists of two steps.
Before constructing the mechanism let us summarize conditions for this reaction. We will use Br2 in our example for halogenation of ethylene.
Nucleophile Double bond in alkene
Electrophile Br2, Cl2
Regiochemistry not relevant
Stereochemistry
ANTI
Step 1: In the first step of the addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a intermediate cycle with the double bond.
Step 2: In the second step, bromide anion attacks any carbon of the bridged bromonium ion from the back side of the cycle. Cycle opens up and two halogens are in the position anti.
Summary
Hallogens can act as electrophiles due to polarizability of their covalent bond.Addition of halogens is stereospecific and produces vicinial dihalides with anti addition.Cis starting material will give mixture of enantiomers and trans produces a meso compound.
Problems
1.What is the mechanism of adding Cl2 to the cyclohexene?
2.A reaction of Br2 molecule in an inert solvent with alkene follows?
a) syn addition
b) anti addition
c) Morkovnikov rule
3)
4)
Key:
1.
2. b
3.enantiomer
4.
Electrophilic Addition of Hydrogen Halides
Due to the nature of the $\pi$ bond, $\pi$ bonds can act like a nucleophile and undergo addition of electrophiles. The $\pi$ electrons have a high electron density in the $\pi$ electron cloud which can be easily polarized (give up the electrons) and can act like a nucleophile (nucleus-loving due to too many electrons; wants to give them away to electrophile). This behavior is very similar to the behavior that the lone pair electrons have on a Lewis base.
Introduction
There are many types of electrophilic addition, but this section will focus on the addition of hydrogen halides.
Reaction
The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both a electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of $\pi$ electrons, attaching it to the molecule (1). This is basically the reverse of the last step in the E1 reaction (deprotonation step). The resulting molecule will have a single carbon- carbon bond with a positive charge on one of them (carbocation). The next step is when the nucleophile (halide) bonds to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant (2). The second step will only occur if a good nucleophile is used.
Mechanism of Electrophilic Addition of Hydrogen Halide to Ethene
Mechanism of Electrophilic Addition of Hydrogen Halide to Propene
All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals).
Relative Rates of Addition of HX and their causes
Name Rate Size of "X" Overlap of Orbitals
HF slowest smallest good
HCl slow small okay
HBr fast large bad
HI fastest largest horrifically bad
• Gaseous hydrogen halides are used and bubbled through the reactant or the hydrogen halide is added in a solvent such as acetic acid to produce decent amounts of the product. It also helps to do a aqueous work up afterward to get a good yield of product. This reaction is usually carried out at lower temperatures.
Regiochemistry
The regiochemistry of this reaction can be explained through Markovnikov's Rule. The basics of this rule states that the proton will add to the less substituted carbon and the halogen goes to the more substituted carbon. This may seem odd that the larger atom is going to the more sterically hindered spot, but this rule is more about the stability of the intermediate. The addition will occur in a way that produces the most stable carbocation after the initial protonation (the intermediate needs to have a stable carbocation). Tertiary and secondary carbocations are the most stable due to hyperconjugation, so the more substituted the carbon is the more stable the carbocation will be.
Examples of regioselective behavior in Electrophilic Addition: Markovnikov's Rule
Since the intermediate of this reaction is a carbocation the carbocation wants to be the most stable it can be. In some case this will mean that the carbocation will rearrange. This is very likely when there are no good nucleophiles present to trap the carbocation in a certain position. The rearrangement of the carbocation depends on many things but generally rearrangements will occur when there are no good nucleophiles present such as in a strong acid.
Stereochemistry
The stereochemistry for this reaction is random. The important thing to understand is the regiochemistry and the Markovnikov's Rule (this will come up a lot!).
Problems
Write out the following reactions and think about the rate of the reaction:
• Ivy Gardner | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Electrophilic_Addition_of_Halogens_to_Alkenes.txt |
Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.
Addition to symmetrical alkenes
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other. For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Conditions
The alkenes react with gaseous hydrogen halides at room temperature. If the alkene is also a gas, you can simply mix the gases. If the alkene is a liquid, you can bubble the hydrogen halide through the liquid. Alkenes will also react with concentrated solutions of the gases in water. A solution of hydrogen chloride in water is, of course, hydrochloric acid. A solution of hydrogen bromide in water is hydrobromic acid - and so on. There are, however, problems with this. The water will also get involved in the reaction and you end up with a mixture of products.
Reaction rates
Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions. When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be. Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the \(pi\) bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this. Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes. The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.
The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Addition to unsymmetrical alkenes
In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition - in other words, which way around the hydrogen and the halogen add across the double bond.
If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product.
This is in line with Markovnikov's Rule
Markovnikov's Rule
When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.
In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant.
A special problem with hydrogen bromide
Unlike the other hydrogen halides, hydrogen bromide can add to a carbon-carbon double bond either way around - depending on the conditions of the reaction. If the hydrogen bromide and alkene are entirely pure, then the hydrogen bromide adds on according to Markovnikov's Rule. For example, with propene you would get 2-bromopropane.
That is exactly the same as the way the other hydrogen halides add.
If the hydrogen bromide and alkene contain traces of organic peroxides
Oxygen from the air tends to react slowly with alkenes to produce some organic peroxides, and so you do not necessarily have to add them separately. This is therefore the reaction that you will tend to get unless you take care to exclude all air from the system. In this case, the addition is the other way around, and you get 1-bromopropane:
This is sometimes described as an anti-Markovnikov addition or as the peroxide effect. Organic peroxides are excellent sources of free radicals. In the presence of these, the hydrogen bromide reacts with alkenes using a different (faster) mechanism. For various reasons, this doesn't happen with the other hydrogen halides. This reaction can also happen in this way in the presence of ultra-violet light of the right wavelength to break the hydrogen-bromine bond into hydrogen and bromine free radicals. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Electrophilic_Addition_of_Hydrogen_Halides_II.txt |
Protons and other electrophiles are not the only reactive species that initiate addition reactions to carbon-carbon double bonds. Curiously, this first became evident as a result of conflicting reports concerning the regioselectivity of HBr additions. As noted earlier, the acid-induced addition of HBr to 1-butene gave predominantly 2-bromobutane, the Markovnikov Rule product. However, in some early experiments in which peroxide contaminated reactants were used, 1-bromobutane was the chief product. Further study showed that an alternative radical chain-reaction, initiated by peroxides, was responsible for the anti-Markovnikov product. This is shown by the following equations.
The weak O–O bond of a peroxide initiator is broken homolytically by thermal or hight energy. The resulting alkoxy radical then abstracts a hydrogen atom from HBr in a strongly exothermic reaction. Once a bromine atom is formed it adds to the π-bond of the alkene in the first step of a chain reaction. This addition is regioselective, giving the more stable carbon radical as an intermediate. The second step is carbon radical abstraction of another hydrogen from HBr, generating the anti-Markovnikov alkyl bromide and a new bromine atom. Each of the steps in this chain reaction is exothermic, so once started the process continues until radicals are lost to termination events.
This free radical chain addition competes very favorably with the slower ionic addition of HBr described earlier, especially in non-polar solvents. It is important to note, however, that HBr is unique in this respect. The radical addition process is unfavorable for HCl and HI because one of the chain steps becomes endothermic (the second for HCl & the first for HI).
Other radical addition reactions to alkenes have been observed, one example being the peroxide induced addition of carbon tetrachloride shown in the following equation
RCH=CH2 + CCl4 (peroxide initiator) > RCHClCH2CCl3
The best known and most important use of free radical addition to alkenes is probably polymerization. Since the addition of carbon radicals to double bonds is energetically favorable, concentrated solutions of alkenes are prone to radical-initiated polymerization, as illustrated for propene by the following equation. The blue colored R-group represents an initiating radical species or a growing polymer chain; the propene monomers are colored maroon. The addition always occurs so that the more stable radical intermediate is formed.
RCH2(CH3)CH· + CH3CH=CH2 > RCH2(CH3)CH-CH2(CH3)CH· + CH3CH=CH2 > RCH2(CH3)CHCH2(CH3)CH-CH2(CH3)CH· > etc.
Allylic Substitution
We noted earlier that benzylic and allylic sites are exceptionally reactive in free radical halogenation reactions. Since carbon-carbon double bonds add chlorine and bromine in liquid phase solutions, radical substitution reactions by these halogens are often carried out at elevated temperature in the gas phase (first equation below). Formation of the ionic π-complexes that are intermediates in halogen addition is unfavorable in the absence of polar solvents, and entropy generally favors substitution over addition.
The brominating reagent, N-bromosuccinimide (NBS), has proven useful for achieving allylic or benzylic substitution in CCl4 solution at temperatures below its boiling point (77 ºC). One such application is shown in the second equation.
The predominance of allylic substitution over addition in the NBS reaction is interesting. The N–Br bond is undoubtedly weak (probably less than 50 kcal/mol) so bromine atom abstraction by radicals should be very favorable. The resulting succinimyl radical might then establish a chain reaction by removing an allylic hydrogen from the alkene. One problem with this mechanism is that NBS is very insoluble in CCl4, about 0.006 mole / liter at reflux. Although it is possible that the allylic bromination occurs at a solid-liquid interface, evidence for another pathway has been obtained. In the non-polar solvent used for these reactions, very low concentrations of bromine may be generated from NBS. This would serve as a source of bromine atoms, which would abstract allylic hydrogens irreversibly (an exothermic reaction) in competition with reversible addition to the double bond. The HBr produced in this way is known to react with NBS, giving a new bromine molecule and succinimide, as shown here. Ionic addition of bromine to the double bond would be very slow in these circumstances (CCl4 is a nonpolar solvent).
HBr + (CH2CO)2NBr → Br2 + (CH2CO)2NH
This mechanism is essentially the same as that for the free radical halogenation of alkanes, with NBS serving as a source of very low concentrations of bromine. Unsymmetrical allylic radicals will react to give two regioisomers. Thus, 1-octene on bromination with NBS yields a mixture of 3-bromo-1-octene (ca. 18%) and 1-bromo-2-octene (82%) - both cis and trans isomers.
RCH2CH=CH2 + (CH2CO)2NBr → RCHBrCH=CH2 + RCH=CHCH2Br + (CH2CO)2N | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Free_Radical_Reactions_of_Alkenes/Addition_of_Radicals_to_Alkenes.txt |
Anti-Markovnikov rule describes the regiochemistry where the substituent is bonded to a less substituted carbon, rather than the more substitued carbon. This process is quite unusual, as carboncations which are commonly formed during alkene, or alkyne reactions tend to favor the more substitued carbon. This is because substituted carbocation allow more hyperconjugation and indution to happen, making the carbocation more stable.
Introduction
This process was first explained by Morris Selig Karasch in his paper: 'The Addition of Hydrogen Bromide to Allyl Bromide' in 1933.1 Examples of Anti-Markovnikov includes Hydroboration-Oxidation and Radical Addition of HBr. A free radical is any chemical substance with unpaired electron. The more substituents the carbon is connected to, the more substituted is that carbon. For example: Tertiary carbon (most substituted), Secondary carbon (medium substituted), primary carbon (least substituted)
Anti-Markovnikov Radical Addition of Haloalkane can ONLY happen to HBr and there MUST be presence of Hydrogen Peroxide (H2O2). Hydrogen Peroxide is essential for this process, as it is the chemical which starts off the chain reaction in the initiation step. HI and HCl cannot be used in radical reactions, because in their radical reaction one of the radical reaction steps: Initiation is Endothermic, as recalled from Chem 118A, this means the reaction is unfavorable. To demonstrate the anti-Markovnikov regiochemistry, I will use 2-Methylprop-1-ene as an example below:
Initiation Steps
Hydrogen Peroxide is an unstable molecule, if we heat it, or shine it with sunlight, two free radicals of OH will be formed. These OH radicals will go on and attack HBr, which will take the Hydrogen and create a Bromine radical. Hydrogen radical do not form as they tend to be extremely unstable with only one electron, thus bromine radical which is more stable will be readily formed.
Propagation Steps
The Bromine Radical will go on and attack the LESS SUBSTITUTED carbon of the alkene. This is because after the bromine radical attacked the alkene a carbon radical will be formed. A carbon radical is more stable when it is at a more substituted carbon due to induction and hyperconjugation. Thus, the radical will be formed at the more substituted carbon, while the bromine is bonded to the less substituted carbon. After a carbon radical is formed, it will go on and attack the hydrogen of a HBr, which a bromine radical will be formed again.
Termination Steps
There are also Termination Steps, but we do not concern about the termination steps as they are just the radicals combining to create waste products. For example two bromine radical combined to give bromine. This radical addition of bromine to alkene by radical addition reaction will go on until all the alkene turns into bromoalkane, and this process will take some time to finish.
Problems
Please give the product(s) of the reactions below:
1. $\ce{CH_3-C(CH3)=CH-CH_3 + HBr + H_2O_2 \rightarrow}$
2. $\ce{CH_3C(CH_3)=CH-CH_3 + HI + H_2O_2 \rightarrow }$
3. $\ce{CH_3C(CH_3)=CH-CH_3 + HCl + H_2O_2 \rightarrow }$
4. $\ce{CH_3CH=CH-CH3 + HBr + H_2O_2 \rightarrow }$
5. $\ce{CH_3C(CH_3)=CH-CH_3 + HBr \rightarrow }$
Answers
1. CH3-CH(CH3)-CHBr-CH3 (Anti-Markovnikov)
2. CH3-C(CH3)I-CH2-CH3 (Markovnikov)
3. CH3-C(CH3)Cl-CH2-CH3 (Markovnikov)
4. CH3-CHBr-CH-CH3 or CH3-CH-CHBr-CH3 (Both molecules are the same)
5. CH3-C(CH3)Br-CH2-CH3 (Markovnikov)
Contributors
• Kelvin Kan (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Free_Radical_Reactions_of_Alkenes/Radical_Additions%3A_Anti-Markovnikov_Product_Formation.txt |
When halogens are in the presence of unsaturated molecules such as alkenes, the expected reaction is addition to the double bond carbons resulting in a vicinal dihalide (halogens on adjacent carbons). However, when the halogen concentration is low enough, alkenes containing allylic hydrogens undergo substitution at the allylic position rather than addition at the double bond. The product is an allylic halide (halogen on carbon next to double bond carbons), which is acquired through a radical chain mechanism.
Why Substitution of Allylic Hydrogens?
As the table below shows, the dissociation energy for the allylic C-H bond is lower than the dissociation energies for the C-H bonds at the vinylic and alkylic positions. This is because the radical formed when the allylic hydrogen is removed is resonance-stabilized. Hence, given that the halogen concentration is low, substitution at the allylic position is favored over competing reactions. However, when the halogen concentration is high, addition at the double bond is favored because a polar reaction outcompetes the radical chain reaction.
Radical Allylic Bromination (Wohl-Ziegler Reaction)
Preparation of Bromine (low concentration)
NBS (N-bromosuccinimide) is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl4), NBS reacts with trace amounts of HBr to produce a low enough concentration of bromine to facilitate the allylic bromination reaction.
Allylic Bromination Mechanism
Step 1: Initiation
Once the pre-initiation step involving NBS produces small quantities of Br2, the bromine molecules are homolytically cleaved by light to produce bromine radicals.
Step 2: Propagation
One bromine radical produced by homolytic cleavage in the initiation step removes an allylic hydrogen of the alkene molecule. A radical intermediate is generated, which is stabilized by resonance. The stability provided by delocalization of the radical in the alkene intermediate is the reason that substitution at the allylic position is favored over competing reactions such as addition at the double bond.
The intermediate radical then reacts with a Br2 molecule to generate the allylic bromide product and regenerate the bromine radical, which continues the radical chain mechanism. If the alkene reactant is asymmetric, two distinct product isomers are formed.
Step 3: Termination
The radical chain mechanism of allylic bromination can be terminated by any of the possible steps shown below.
Radical Allylic Chlorination
Like bromination, chlorination at the allylic position of an alkene is achieved when low concentrations of Cl2 are present. The reaction is run at high temperatures to achieve the desired results.
Industrial Uses
Allylic chlorination has important practical applications in industry. Since chlorine is inexpensive, allylic chlorinations of alkenes have been used in the industrial production of valuable products. For example, 3-chloropropene, which is necessary for the synthesis of products such as epoxy resin, is acquired through radical allylic chlorination (shown below).
Problems (Answers are attached as a file)
1. Cyclooctene undergoes radical allylic bromination. Write out the complete mechanism including reactants, intermediates and products.
2. Predict the two products of the allylic chlorination reaction of 1-heptene.
3. What conditions are required for allylic halogenation to occur? Why does this reaction outcompete other possible reactions such as addition when these conditions are met?
4. Predict the product of the allylic bromination reaction of 2-benzylheptane. (Hint: How are benzylic hydrogens similar to allylic hydrogens?)
5. The reactant 5-isopropyl-1-hexene generates the products 3-bromo-5-isopropyl-1-hexene and 1-bromo-5-isopropyl-2-hexene. What reagents were used in this reaction? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Free_Radical_Reactions_of_Alkenes/Radical_Allylic_Halogenation.txt |
This page looks at the production of alcohols by the direct hydration of alkenes - adding water directly to the carbon-carbon double bond.
Manufacturing ethanol
Ethanol is manufactured by reacting ethene with steam. The reaction is reversible.
Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion. A flow scheme for the reaction looks like this:
Manufacturing other alcohols
If you start from an unsymmetrical alkene like propene, you have to be careful to think about which way around the water adds across the carbon-carbon double bond. Markovnikov's Rule says that when you add a molecule HX across a carbon-carbon double bond, the hydrogen joins to the carbon atom which already has the more hydrogen atoms attached to it.
Thinking of water as H-OH, the hydrogen will add to the carbon with the more hydrogens already attached. That means that in the propene case, you will get propan-2-ol rather than propan-1-ol.
The conditions used during manufacture vary from alcohol to alcohol.
Contributors
Jim Clark (Chemguide.co.uk)
Hydration of Alkenes
Alkenes can be converted to alcohols by the net addition of water across the double bond. There are multiple ways that are commonly used to do this transformation.
Hydration of Alkenes
The net addition of water to alkenes is known as hydration.
The result involves breaking the pi bond in the alkene and an OH bond in water and the formation of a C-H bond and a C-OH bond. The reaction is typically exothermic by 10 - 15 kcal/mol,1 but has an entropy change of -35 - -40 cal/mol K. Consequently, the net free energy change for the process tends to close to 0, and the equilibrium constant for the direct addition is close to 1. Nonetheless, there are multiple approaches that allow this transformation to be carried out to completion.
Acid-Catalyzed Hydration
The direct addtion of water to an alkene is too slow to be of any significance. However, the addition can be catayzed by Lewis or Bronsted acids.
The mechanism of hydration involves electrophlic addition of the proton (or acid) to the double bond to form a carbocation intermediate. Addition of water in the second step results in formation of an oxonium ion, which, upon deprotonation, gives the alcohol.
The proton in the oxonium intermediate can be deprotonated by any base present, including the conjugate base of the acid used as a catalyst, or even by another alkene molecule, which would generate another carbocation intermediate and propagate the chain mechanism.
Using Le Châtelier's Principle to get the product
The extent of reaction of the acid catalyzed hydration of alkenes is determined by the equilibrium constant, which, as noted above, is near unity. Consequently, the reaction can be carried out by adding excess water to increase the yield of products, an application of Le Châtelier's Principle. Alternatively, the reverse reaction, acid-catalyzed dehydration of the alcohol to form the alkene can promoted by removing water from the reaction by using a Dean-Stark trap.
Regioselectivity of addition
Protonation of the alkene in the first step of the reaction can occur at either carbon. However, the more stable carbocation is preferably formed. The order of stability of alkyl cations is
3o > 2o > 1o
Therefore, protonation will occur at the less substituted carbon, to create the more substituted carbocation, where the water adds. The addition of a proton at the less substituted carbon and the -OH to the more substituted carbon is known as Markovnikov's Rule.
The carbocation formed in the reaction is prone to rearrangement, if possible. Therefore, hydration of an olefiin next to a branched aliphatic center will result in the alcohol forming in a position that was not part of the original double bond.
Oxymercuration
Transition metals can also be used as the acids for hydration reactions. In some cases, coordination of the alkene to a metal leaves it susceptible to reaction with a nucleophile such as water. The classic case of nucleophilic donation to a coordinated alkene occurs with mercury (II) salts such as mercuric chloride, \(HgCl_2\), or mercuric acetate, \(Hg(OAc)_2\). The reaction, or rather the sequence of reactions, is called oxymercuration - demercuration or oxymercuration - reduction.
We will break the two different reactions in this sequence apart and focus only on the first one: oxymercuration. This reaction qualifies as an electrophilic addition because, as in the previous cases, it begins with donation of a π-bonding pair to an electrophile. In this case, we will consider the electrophile to be aqueous \(Hg^{2+}\) ion.
That electrophilic addition (from the alkene's perspective) results in the formation of an alkene complex. In reality, the mercury ion is also coordinated by several water molecules, but we will ignore them for simplicity.
The complex formed by addition of mercury is not a localized carbocation, as is formed by protonation, but is better considered a bridged or cyclic structure, which results from addition of d electrons to the empty orbital in the cation.
This situation is something like formation of a cyclic bromonium ion formed in the bromination of alkenes. As with bromination, the cyclic intermediate can be opened by attack of a nucleophile, in this case, water.
The final part of the reaction sequence is displacement of mercury from the hydroxyalkylmercury complex, effected through addition of sodium borohydride. The details of the reaction are usually dismissed in textbooks because they have little to do with electrophilic addition, the topic we are focusing on. However, the result is that the mercury is replaced by a hydrogen atom. The metal is converted to silvery, liquid, elemental mercury.
Regioselectivy of hydration by oxymercuration
As shown above, oxymercuration leads to Markovnikov addition of water to the double bond. The reason for this can be seen by considering the electronic structure of the cyclic cation. The electronic structure of the cyclic intermediate can be deduced by using resonance structures as shown below.
By considering these resonance structures, it can be seen that the postive charge is distributed over both of the carbons of the olefin. However, because the resonance structure on the left has the positive charge on a secondary carbon, it is energetically more favorable than the structure in the middle, where the charge is a primary cation. Consequently, although the charge is distributed over both carbons, the more substituted carbon has more positive charge density in the overall resonance hybrid. Therefore, it is more electrophilic, and is more susceptibleable to nucleophlic attack.
An important advantage of oxymercuration over simple acid catalysis is that the cyclic structure is not prone to rearrangement, and can therefore is amenable to hydration of alkenes with branched substituents. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Hydration_of_Alkenes/Formation_of_alcohols_from_alkenes.txt |
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH3 or BHR2) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene bouble bond. Furthermore, the borane acts as a lewisAnti-Markovnikov acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Anti-MarkovnikovHydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
Introduction
Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An Additional feature of this reaction is that it occurs without rearrangement.
The Borane Complex
First off it is very imporatnt to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B2H6 (diborane). Additionally, the dimer B2H6 ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a lewis acid-base complex, which allows boron to have an electron octet.
$2BH_3 \rightarrow B_2H_6$
The Mechanism
Step #1
• Part #1: Hydroboration of the alkene. In this first step the addittion of the borane to the alkene is initiated and prceeds as a concerted reaction because bond breaking and bond formation occurs at the same time. This part consists of the vacant 2p orbital of the boron electrophile pairing with the electron pair of the ? bondof the nucleophile.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
• Part #2: The Anti Markovnikov addition of Boron. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition).
Oxidation of the Trialkylborane by Hydrogen Peroxide
Step #2
• Part #1: the first part of this mechanism deals with the donation of a pair of electrons from the hydrogen peroxide ion. the hydrogen peroxide is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from hydroboration.
EpoxidationEpoxidation
• Part 2: In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the removal of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
• Part 3: This is the final part of the Oxidation process. In this part the trialkylborate reacts with aqueous NaOH to give the alcohol and sodium borate.
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well.
Problems
What are the products of these following reactions?
#1.
#2.
#3.
Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown.
#4.
#5. (E)-3-methyl-2-pentene
If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the alkenes.
#1.
#2.
#3.
#4.
#5.
Contributors
• Gilbert Torres (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Hydroboration-Oxidation_of_Alkenes.txt |
Oxacyclopropane rings, also called epoxide rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with peroxycarboxylic acid.
Oxacyclopropane Synthesis by Peroxycarboxylic Acid
Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH.
Mechanism
Peroxycarboxylic acids are generally unstable. An exception is meta-chloroperoxybenzoic acid, shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions.
Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as chloroform, ether, acetone, or dioxane is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal diol, a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions.
Example
Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the epoxide ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal diol, a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is anti since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one S and one R stereocenter. But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of S, S and R, R enantiomers.
Problems
1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid)
a) in acetone solvent.
b) in an aqueous medium with acid or base catalyst present.
2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent.
3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent.
4. Predict the reaction of propene with MCPBA.
a) in acetone solvent
b) after aqueous work-up.
5. Predict the reaction of cis-2-butene in chloroform solvent.
Answers
1. a) Cis-2-methyl-3-propyloxacyclopropane
b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
2. Trans-3-ethyl-2-methyloxacyclopropane.
3. Trans-3,4-diethyloxacyclopropane.
4. a) 1-ethyl-oxacyclopropane
b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol
5. Cis-2,3-dimethyloxacyclopropane
Contributors
• Kristen Perano | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Oxacyclopropane_Synthesis.txt |
The carbon-carbon double bonds in alkenes such as ethene react with potassium manganate(VII) solution (potassium permanganate solution).
Oxidation of alkenes with cold dilute potassium manganate(VII) solution
Alkenes react with potassium manganate(VII) solution in the cold. The colorchange depends on whether the potassium manganate(VII) is used under acidic or alkaline conditions.
• If the potassium manganate(VII) solution is acidified with dilute sulfuric acid, the purple solution becomes colorless.
• If the potassium manganate(VII) solution is made slightly alkaline (often by adding sodium carbonate solution), the purple solution first becomes dark green and then produces a dark brown precipitate.
Chemistry of the Reaction
We'll look at the reaction with ethene. Other alkenes react in just the same way. Manganate(VII) ions are a strong oxidizing agent, and in the first instance oxidize ethene to ethane-1,2-diol (old name: ethylene glycol). Looking at the equation purely from the point of view of the organic reaction:
This type of equation is quite commonly used in organic chemistry. Oxygen written in square brackets is taken to mean "oxygen from an oxidizing agent". The reason for this is that a more normal equation tends to obscure the organic change in a mass of other detail - as you will find below!
The full equation depends on the conditions.
• Under acidic conditions, the manganate(VII) ions are reduced to manganese(II) ions.
• Under alkaline conditions, the manganate(VII) ions are first reduced to green manganate(VI) ions . . .
. . . and then further to dark brown solid manganese(IV) oxide (manganese dioxide).
This last reaction is also the one you would get if the reaction was done under neutral conditions. You will notice that there are neither hydrogen ions nor hydroxide ions on the left-hand side of the equation.
You might possibly remember that further up the page it says that potassium manganate(VII) is often made slightly alkaline by adding sodium carbonate solution. Where are the hydroxide ions in this? Carbonate ions react with water to some extent to produce hydrogencarbonate ions and hydroxide ions.
\[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^- \]
It is the presence of these hydroxide ions that gives sodium carbonate solution its pH in the 10 - 11 region.
Using the reaction to test for carbon-carbon double bonds
If an organic compound reacts with dilute alkaline potassium manganate(VII) solution in the cold to give a green solution followed by a dark brown precipitate, then it may contain a carbon-carbon double bond. But equally it could be any one of a large number of other compounds all of which can be oxidized by manganate(VII) ions under alkaline conditions.
The situation with acidified potassium manganate(VII) solution is even worse because it has a tendency to break carbon-carbon bonds. It reacts destructively with a large number of organic compounds and is rarely used in organic chemistry.
You could use alkaline potassium manganate(VII) solution if, for example, all you had to do was to find out whether a hydrocarbon was an alkane or an alkene - in other words, if there was nothing else present which could be oxidized. It isn't a useful test. Bromine water is far more clear cut.
Oxidation of alkenes with hot concentrated acidified potassium manganate(VII) solution
The diols, such as ethane-1,2-diol, which are the products of the reaction with cold dilute potassium manganate(VII), are themselves quite easily oxidized by manganate(VII) ions. That means that the reaction will not stop at this point unless the potassium manganate(VII) solution is very dilute, very cold, and preferably not under acidic conditions. If you are using hot concentrated acidified potassium manganate(VII) solution, what you finally end up with depends on the arrangement of groups around the carbon-carbon double bond.
The formula below represents a general alkene. In organic chemistry, the symbol R is used to represent hydrocarbon groups or hydrogen in a formula when you don't want to talk about specific compounds. If you use the symbol more than once in a formula (as here), the various groups are written as R1, R2, etc.
In this particular case, the double bond is surrounded by four such groups, and these can be any combination of same or different - so they could be 2 hydrogens, a methyl and an ethyl, or 1 hydrogen and 3 methyls, or 1 hydrogen and 1 methyl and 1 ethyl and 1 propyl, or any other combination you can think of. In other words, this formula represents every possible simple alkene:
The first stage of the extended oxidation
The acidified potassium manganate(VII) solution oxidizes the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds.
The products are known as carbonyl compounds because they contain the carbonyl group, C=O. Carbonyl compounds can also react with potassium manganate(VII), but how they react depends on what is attached to the carbon-oxygen double bond. So we need to work through all the possible combinations.
If both attached R groups in the products are alkyl groups
Carbonyl compounds which have two hydrocarbon groups attached to the carbonyl group are called ketones. Ketones aren't that easy to oxidize, and so there is no further action. (But see note in red below.) If the groups attached either side of the original carbon-carbon double bond were the same, then you would end up with a single ketone. If they were different, then you would end up with a mixture of two. For example:
In this case, you would end up with two identical molecules called propanone. On the other hand, if one of the methyl groups in the original molecule was replaced by an ethyl group, you would get a mixture of two different ketones - propanone and butanone.
What would you get if there was a methyl and an ethyl group on both sides of the original carbon-carbon double bond? Again, you would get a single ketone formed - in this case, butanone. If you aren't sure about this, draw the structures and see.
This last section is a gross over-simplification. In practice, ketones are oxidized by potassium manganate(VII) solution under these conditions. The reaction is untidy and results in breaking carbon-carbon bonds either side of the carbonyl group. Potassium manganate(VII) is such a devastating oxidizing agent that it is rarely used in organic chemistry.
If a product has one hydrocarbon group and one hydrogen
For example, suppose the first stage of the reaction was:
In this case, the first product molecule has a methyl group and a hydrogen attached to the carbonyl group. This is a different sort of compound known as an aldehyde. Aldehydes are readily oxidized to give carboxylic acids, containing the -COOH group. So this time, the reaction will go on a further step to give ethanoic acid, CH3COOH.
The acid structure has been turned around slightly to make it look more like the way we normally draw acids, but the net effect is that an oxygen has been slotted in between the carbon and hydrogen. The overall effect of the potassium manganate(VII) on this kind of alkene is therefore:
Obviously, if there was a hydrogen atom attached to both carbons at the ends of the carbon-carbon double bond, you would get two carboxylic acid molecules formed - which might be the same or different, depending on whether the alkyl groups were the same or different.
Play around with this until you are happy about it. Draw a number of alkenes, all of which have a hydrogen attached at both ends of the carbon-carbon double bond. Vary the alkyl groups - sometimes the same on each end of the double bond, sometimes different. oxidize them to form the acids, and see what you get.
If a product has two hydrogens, but no hydrocarbon group
You might have expected that this would produce methanoic acid, as in the equation:
But it doesn't! That's because methanoic acid is also easily oxidized by potassium manganate(VII) solution. In fact, it oxidizes it all the way to carbon dioxide and water. So the equation in a case like this might be, for example:
The exact nature of the other product (in this example, propanone) will vary depending on what was attached to the right-hand carbon in the carbon-carbon double bond. If there were two hydrogens at both ends of the double bond (in other words, if you had ethene), then all you would get would be carbon dioxide and water.
Example
Working back from the results helps you to work out the structure of the alkene. For example, the alkene C4H8 has three structural isomers:
Work out which of these would give each of the following results if they were treated with hot concentrated potassium manganate(VII) solution.
• Isomer A gives a ketone (propanone) and carbon dioxide.
• Isomer B gives a carboxylic acid (propanoic acid) and carbon dioxide.
• Isomer C gives a carboxylic acid (ethanoic acid).
Solution
Acids are produced when there is a hydrogen atom attached to at least one of the carbons in the carbon-carbon double bond. Since in C there is only one product, the alkene must be symmetrical around the double bond. That's but-2-ene. If you have got two hydrogens at one end of the bond, this will produce carbon dioxide. A is 2-methylpropene, because the other molecule is a ketone. B must be but-1-ene because it produces carbon dioxide and an acid.
Summary
Think about both ends of the carbon-carbon double bond separately, and then combine the results afterwards.
• If there are two alkyl groups at one end of the bond, that part of the molecule will give a ketone.
• If there is one alkyl group and one hydrogen at one end of the bond, that part of the molecule will give a carboxylic acid.
• If there are two hydrogens at one end of the bond, that part of the molecule will give carbon dioxide and water.
• If there are two hydrogens at one end of the bond, that part of the molecule will give carbon dioxide and water. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Oxidation_of_Alkenes_with_Potassium_Manganate.txt |
Ozonolysis is a type of weak oxidative cleavage where we cleave alkenes (double bonds) into either ketones, aldehydes or carboxylic acid using ozone.
Topic: Weak Oxidative Cleavage
The History of Ozonolysis
Ozonolysis, or “oxidative cleavage” originated in the 1800’s with its inventor, Christian Friedrich Schönbein. The reaction also is attributed to Carl Dietrich Harries, therefore you may hear this reaction termed “Harries ozonolysis”. You may be thinking, “Umm... Why do I need to know this?”. The answer is, well you don’t! (But in case you were wondering, now you know & can tell your friends).
Ozone Structure
It is in simplest terms using ozone, or O3 (the structure is shown above with correct formal charges) to cleave carbon-carbon double bonds (C=C) to produce various carbonyls.
Pathways
1. Aldehydes (CHO), and ketones (CH3COCH3) can be formed through reductive workup. This refers to what is shown under the arrow below. Me2S is also DMS and stands for dimethyl sulfide.
Ozonolysis Retrosynthesis
Now I’m sure you’re wondering, which structure was our starting material? For that answer scroll down just a bit, but first try to draw it out!
2. Carboxylic acid (COOH, CO2H) can also be formed as well except it is through an oxidative workup step instead. H2O2 (hydrogen peroxide) is the reagent to look out for because as seen below it turns:
- Ends of alkenes with 1 –H ≠ Aldehydes but Carboxylic Acids instead
Ozonolysis with Oxidative Workup
Answer from above:
Di-substituted Alkene
How Does Ozonolysis Work?
Now we will discuss the specifics of the mechanism of ozonolysis.
Reagents
Commonly seen is the use of O3, which is ozone (structure shown above), in the presence of a reducing agent such as dimethyl sulfide (DMS, Me2S, S(CH3)2) or zinc & acetic acid (Zn/HOAc). Therefore a full reaction may look like this:
Ozonolysis Reaction
Now as we said, instead of DMS under the arrow you may also see zinc and acetic acid. Beware of the other ways “acetic acid” can be written in addition to HOAc. They are: [AcOH, CH3COOH, CH3CO2H]
This type of mechanism is referred to as ozonolysis with reductive workup.
In case you need to know:
After the reduction takes place our Zn or DMS will attach to the remaining third oxygen from our O3that is not seen in our final product. Because of this, along with our carbonyl products we will also find DMSO (dimethyl sulfoxide) or ZnO (zinc oxide) produced as well.
Ozonolysis Mechanism
DMSO is simply the blue structure you see at the top right of the image. A couple other intermediates you may see are the molozonide and the ozonide which we will discuss below.
Reaction Mechanism and Intermediates
When the reaction takes place, the first intermediate you will see is called your molozonide. It is a cyclic structure containing 3 oxygen atoms connected in a row, also referred to as a 1,2,3-trioxolane.
Molozonide Intermediate
This will be your unstable intermediate which will further undergo a rearrangement to form your ozonide. An ozonide is more stable than the molozonide, however it still exhibits a fair amount of instability.
Ozonide Intermediate
These are not isolated either and will further be reduced or oxidized depending on if you follow the “reductive workup” pathway or the “oxidative workup” pathway.
*Rudolf Criegee proposed the mechanism therefore if you hear words such as the ‘Criegee zwitterion’ or ‘Criegee intermediate’ that is why! It is referring to an intermediate that appears between the molozonide and the ozonide.
Criegee Intermediate
Products
The products of ozonolysis will vary depending on two things:
1) The R groups that are attached to the alkene:
1. Ends of alkenes with –R groups on both sides = Ketones
2. Ends of alkenes with 1 –H = Aldehydes
3. Ends of alkenes with 2 –Hs (yielding single carbon fragments) = Formaldehyde
2) Instead of a “reductive workup” with either zinc (Zn) and acetic acid (HOAc), we use an “oxidative workup” with hydrogen peroxide (H2O2).
Ozonolysis with Oxidative Workup of Cyclic Alkene
What this will do is alter our products so that any aldehydes (CHO) that were formed in our cleavage step will be oxidized into carboxylic acids (COOH).
Exercise \(1\)
Can you guess what the product will be here?
Hint
Cut across the double bond and add in the missing hydrogen on the red C. Then use the rule above to predict the final product. Scroll down for the answer :)
Special Ozonolysis Cases
# 1 - Fragments
- You may sometimes hear the products of this reaction referred to as “fragments”. All this is referring to is the different molecules that were formed when we broke apart our original C-C double bond.
- For example if 2 aldehydes (CHO) were your products, you could say that 2 aldehyde fragments were formed while performing an oxidative cleavage.
- This word is more commonly used when multiple alkenes are cleaved.
#2 - Pyridine
- Pyridine is commonly seen as a buffer when any kind of acid is generated in the reaction.
#3 - Alcohols
- If NaBH4 (sodium borohydride) is used as the reagent during the reductive workup we will yield alcohols instead of aldehydes & ketones as seen when Zn, or DMS is used.
- This is explained much better in the video below where we focus strictly on reduction. To watch the full explanation just click here and scroll to 7:30!
Reduction with Sodium Borohydride
# 4 - Dichloromethane
- Dichloromethane (CH2Cl2) in particular may be seen as a solvent to help with cleavage of the ozonide intermediate to yield our product.
# 5 - Triphenylphosphine
- Triphenylphosphine (PPH3, PH3P) is one of the other reagents used, just as we learned with Zn and DMS to achieve our product in a reductive workup mechanism.
Answer to above example with H2O2:
Dicarboxylic Acid
“I’m confused about....”
a. Which step of the mechanism is sometimes known as reductive ozonolysis?
- The last step where we go from our ozonide to the product is known as the reductive workup. It is commonly seen more than the oxidative one. Sometimes seen to accomplish this along with DMS & Zn is PPH3(Triphenylphosphine).
b. What is meant by the word ‘quench’?
- Generally speaking this means to stop the reaction by getting rid of any unreacted reagents, usually with an aqueous solution.
c. What are some of the reaction conditions? (Mainly for lab)
- Temperature wise it is suggested to run this reaction at around -78 ̊C.
d. Are they any indicators I should know about that are used in this reaction?
- Yes, blue usually means you are done reacting with the alkene. Also, violet can be seen along with Sudan Red III which is specifically used for special cases where multiple alkenes can react at different rates.
Summary
Remember, if you see O3 + alkene these are conditions for the ozonolysis mechanism. Your product will be some form of a carbonyl, either an aldehyde, ketone, or carboxylic acid if using H2O2.
Contributors
• Johnny Betancourt, Clutchprep. Source page can be accessed here.
Ozonolysis of Alkenes and Alkynes
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis.
Introduction
The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the Carbon-Carbon double bond and is replaced by a Carbon-Oxygen double bond instead.
Reaction Mechanism
Step 1:
The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the Carbon-Carbon double bond, which then form the molozonide intermediate. Due to the unstable molozonide molecule, it continues further with the reaction and breaks apart to form a carbonyl and a carbonyl oxide molecule.
Step 2:
The carbonyl and the carbonyl oxide rearranges itself and reforms to create the stable ozonide intermediate. A reductive workup could then be performed to convert convert the ozonide molecule into the desired carbonyl products. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Ozonolysis.txt |
This page looks at the polymerisation of alkenes to produce polymers like poly(ethene) (usually known as polythene, and sometimes as polyethylene), poly(propene) (old name: polypropylene), PVC and PTFE. It also looks briefly at how the structure of the polymers affects their properties and uses.
Poly(ethene) (polythene or polyethylene)
Low density poly(ethene): LDPE
Manufacture
In common with everything else on this page, this is an example of addition polymerisation. An addition reaction is one in which two or more molecules join together to give a single product. During the polymerisation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene.
The number of molecules joining up is very variable, but is in the region of 2000 to 20000. Conditions:
Temperature: about 200°C
Pressure: about 2000 atmospheres
Initiator: a small amount of oxygen as an impurity
Properties and uses
Low density poly(ethene) has quite a lot of branching along the hydrocarbon chains, and this prevents the chains from lying tidily close to each other. Those regions of the poly(ethene) where the chains lie close to each other and are regularly packed are said to be crystalline. Where the chains are a random jumble, it is said to be amorphous. Low density poly(ethene) has a significant proportion of amorphous regions.
One chain is held to its neighbours in the structure by van der Waals dispersion forces. Those attractions will be greater if the chains are close to each other. The amorphous regions where the chains are inefficiently packed lower the effectiveness of the van der Waals attractions and so lower the melting point and strength of the polymer. They also lower the density of the polymer (hence: "low density poly(ethene)").
Low density poly(ethene) is used for familiar things like plastic carrier bags and other similar low strength and flexible sheet materials.
High density poly(ethene): HDPE
Manufacture
This is made under quite different conditions from low density poly(ethene). Conditions:
Temperature: about 60°C
Pressure: low - a few atmospheres
Catalyst: Ziegler-Natta catalysts or other metal compounds
Ziegler-Natta catalysts are mixtures of titanium compounds like titanium(III) chloride, TiCl3, or titanium(IV) chloride, TiCl4, and compounds of aluminium like aluminium triethyl, Al(C2H5)3. There are all sorts of other catalysts constantly being developed.
These catalysts work by totally different mechanisms from the high pressure process used to make low density poly(ethene). The chains grow in a much more controlled - much less random - way.
Properties and uses
High density poly(ethene) has very little branching along the hydrocarbon chains - the crystallinity is 95% or better. This better packing means that van der Waals attractions between the chains are greater and so the plastic is stronger and has a higher melting point. Its density is also higher because of the better packing and smaller amount of wasted space in the structure.
High density poly(ethene) is used to make things like plastic milk bottles and similar containers, washing up bowls, plastic pipes and so on. Look for the letters HDPE near the recycling symbol.
Poly(propene) (polypropylene): PP
Poly(propene) is manufactured using Ziegler-Natta and other modern catalysts. There are three variants on the structure of poly(propene) which you may need to know about, but we'll start from the beginning with a general structure which fits all of them.
The general structure
If your syllabus simply mentions the structure of poly(propene) with no more detail, this is adequate.
The trick is to think about the shape of the propene in the right way:
Now line lots of them up in a row and join them together. Notice that the double bonds are all replaced by single bonds in the process.
In a simple equation form, this is normally written as:
The three variations on this structure
You have got to remember that the diagrams above are 2-dimensional. Real poly(propene) chains are 3-dimensional. There are three different sorts of poly(propene) depending in how the CH3 groups are arranged in space. These are called isotactic, atactic and syndiotactic poly(propene). The commonly used version is isotactic poly(propene).
Isotactic poly(propene)
A bit of the isotactic poly(propene) chain looks like this:
This very regular arrangement of the CH3 groups makes it possible for the chains to pack close together and so maximise the amount of van der Waals bonding between them. That means that isotactic poly(propene) is quite strong either as a solid object or when it is drawn into fibres. This is the common form of poly(propene) which is used to make plastic crates and ropes amongst many other things. Look for the letters PP near the recycling symbol.
Atactic poly(propene)
In atactic poly(propene) the CH3 groups are orientated randomly along the chain.
This lack of regularity makes it impossible for the chains to lie closely together and so the van der Waals attractions between them are weaker. Atactic poly(propene) is much softer with a lower melting point. It is formed as a waste product during the manufacture of isotactic poly(propene) and its uses are limited. It is used, for example, in road paint, in making roofing materials like "roofing felt", and in some sealants and adhesives.
Syndiotactic poly(propene)
Syndiotactic poly(propene) is a relatively new material and is another regularly arranged version of poly(propene). In this case, every alternate CH3 group is orientated in the same way.
This regularity means that the chains can pack closely, and van der Waals attractions will be fairly strong. However, the attractions aren't as strong as in isotactic poly(propene). This makes syndiotactic poly(propene) softer and gives it a lower melting point.
Because syndiotactic poly(propene) is relatively new, at the time of writing uses were still being developed. It has uses in packaging - for example, in plastic film for shrink wrapping food. There are also medical uses - for example, in medical tubing and for medical bags and pouches. There are a wide range of other potential uses - either on its own, or in mixtures with isotactic poly(propene).
Poly(chloroethene) (polyvinyl chloride): PVC
Poly(chloroethene) is commonly known by the initials of its old name, PVC.
Structure
Poly(chloroethene) is made by polymerising chloroethene, CH2=CHCl. Working out its structure is no different from working out the structure of poly(propene) (see above). As long as you draw the chloroethene molecule in the right way, the structure is pretty obvious.
The equation is usually written:
It doesn't matter which carbon you attach the chlorine to in the original molecule. Just be consistent on both sides of the equation. The polymerisation process produces mainly atactic polymer molecules - with the chlorines orientated randomly along the chain. The structure is no different from atactic poly(propene) - just replace the CH3 groups by chlorine atoms.
Because of the way the chlorine atoms stick out from the chain at random, and because ot their large size, it is difficult for the chains to lie close together. Poly(chloroethene) is mainly amorphous with only small areas of crystallinity.
Properties and uses
You normally expect amorphous polymers to be more flexible than crystalline ones because the forces of attraction between the chains tend to be weaker. However, pure poly(chloroethene) tends to be rather hard and rigid.
This is because of the presence of additional dipole-dipole interactions due to the polarity of the carbon-chlorine bonds. Chlorine is more electronegative than carbon, and so attracts the electrons in the bond towards itself. That makes the chlorine atoms slightly negative and the carbons slightly positive. These permanent dipoles add to the attractions due to the temporary dipoles which produce the dispersion forces.
Plasticisers are added to the poly(chloroethene) to reduce the effectiveness of these attractions and make the plastic more flexible. The more plasticiser you add, the more flexible it becomes. Poly(chloroethene) is used to make a wide range of things including guttering, plastic windows, electrical cable insulation, sheet materials for flooring and other uses, footwear, clothing, and so on and so on.
Poly(tetrafluoroethene): PTFE
You may have come across this under the brand names of Teflon or Fluon.
Structure
Structurally, PTFE is just like poly(ethene) except that each hydrogen in the structure is replaced by a fluorine atom.
The PTFE chains tend to pack well and PTFE is fairly crystalline. Because of the fluorine atoms, the chains also contain more electrons (for an equal length) than a corresponding poly(ethene) chain. Taken together (the good packing and the extra electrons) that means that the van der Waals dispersion forces will be stronger than in even high density poly(ethene).
Properties and uses
PTFE has a relatively high melting point (due to the strength of the attractions between the chains) and is very resistant to chemical attack. The carbon chain is so wrapped up in fluorine atoms that nothing can get at it to react with it. This makes it useful in the chemical and food industries to coat vessels and make them resistant to almost everything which might otherwise corrode them.
Equally important is that PTFE has remarkable non-stick properties - which is the basis for its most familiar uses in non-stick kitchen and garden utensils. For the same reason, it can also be used in things like low-friction bearings.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Polymerization_of_Alkenes.txt |
In each case, we will look at ethene as typical of all of the alkenes. There are no complications as far as the basic facts are concerned as the alkenes get bigger.
Ethene and fluorine
Ethene reacts explosively with fluorine to give carbon and hydrogen fluoride gas. This isn't a useful reaction.
\[\ce{CH2=CH2 + 2F2 -> 2C + 4HF}\]
Ethene and Chlorine or Bromine or Iodine
In each case you get an addition reaction. For example, bromine adds to give 1,2-dibromoethane.
The reaction with bromine happens at room temperature. If you have a gaseous alkene like ethene, you can bubble it through either pure liquid bromine or a solution of bromine in an organic solvent like tetrachloromethane. The reddish-brown bromine is decolourized as it reacts with the alkene. A liquid alkene (like cyclohexene) can be shaken with liquid bromine or its solution in tetrachloromethane. Chlorine reacts faster than bromine, but the chemistry is similar. Iodine reacts much, much more slowly, but again the chemistry is similar. You are much more likely to meet the bromine case than either of these.
Alkenes decolorize bromine water.
If you shake an alkene with bromine water (or bubble a gaseous alkene through bromine water), the solution becomes colorless. This is complicated by the fact that the major product is not 1,2-dibromoethane. The water also gets involved in the reaction, and most of the product is 2-bromoethanol.
However, there will still be some 1,2-dibromoethane formed, so at this sort of level you can probably get away with quoting the simpler equation:
Addition Reactions Initiated by Electrophilic Halogen
• Addition Reactions Initiated by Electrophilic Halogen
• Addition Reactions involving other Cyclic Onium Intermediates
• Brønsted Acid Additions
• Hydrogenation of Alkenes
• Oxidations
• Epoxidation
• Hydroxylation
• Oxidative Cleavage of Double Bonds
As illustrated in the drawing below, the pi-bond fixes the carbon-carbon double bond in a planar configuration, and does not permit free rotation about the double bond itself. We see then that addition reactions to this function might occur in three different ways, depending on the relative orientation of the atoms or groups that add to the carbons of the double bond: (i) they may bond from the same side, (ii) they may bond from opposite sides, or (iii) they may bond randomly from both sides. The first two possibilities are examples of stereoselectivity, the first being termed syn-addition, and the second anti-addition. Since initial electrophilic attack on the double bond may occur equally well from either side, it is in the second step (or stage) of the reaction (bonding of the nucleophile) that stereoselectivity may be imposed.
If the two-step mechanism described above is correct, and if the carbocation intermediate is sufficiently long-lived to freely-rotate about the sigma-bond component of the original double bond, we would expect to find random or non-stereoselective addition in the products. On the other hand, if the intermediate is short-lived and factors such as steric hindrance or neighboring group interactions favor one side in the second step, then stereoselectivity in product formation is likely. The following table summarizes the results obtained from many studies, the formula HX refers to all the strong Brønsted acids. The interesting differences in stereoselectivity noted here provide further insight into the mechanisms of these addition reactions.
Reagent H–X X2 HO–X RS–Cl Hg(OAc)2 BH3
Stereoselectivity mixed anti anti anti anti syn
Stereoselectivity in Addition Reactions to Double Bonds
The halogens chlorine and bromine add rapidly to a wide variety of alkenes without inducing the kinds of structural rearrangements noted for strong acids (first example below). The stereoselectivity of these additions is strongly anti, as shown in many of the following examples.
An important principle should be restated at this time. The alkenes shown here are all achiral, but the addition products have chiral centers, and in many cases may exist as enantiomeric stereoisomers. In the absence of chiral catalysts or reagents, reactions of this kind will always give racemic mixtures if the products are enantiomeric. On the other hand, if two chiral centers are formed in the addition the reaction will be diastereomer selective. This is clearly shown by the addition of bromine to the isomeric 2-butenes. Anti-addition to cis-2-butene gives the racemic product, whereas anti-addition to the trans-isomer gives the meso-diastereomer.
We can account both for the high stereoselectivity and the lack of rearrangement in these reactions by proposing a stabilizing interaction between the developing carbocation center and the electron rich halogen atom on the adjacent carbon. This interaction, which is depicted for bromine in the following equation, delocalizes the positive charge on the intermediate and blocks halide ion attack from the syn-location.
The stabilization provided by this halogen-carbocation bonding makes rearrangement unlikely, and in a few cases three-membered cyclic halonium cations have been isolated and identified as true intermediates. A resonance description of such a bromonium ion intermediate is shown below. The positive charge is delocalized over all the atoms of the ring, but should be concentrated at the more substituted carbon (carbocation stability), and this is the site to which the nucleophile will bond.
Stereoelectronic Effect
The stereoselectivity described here is in large part due to a stereoelectronic effect.
Because they proceed by way of polar ion-pair intermediates, chlorine and bromine addition reactions are faster in polar solvents than in non-polar solvents, such as hexane or carbon tetrachloride. However, in order to prevent solvent nucleophiles from competing with the halide anion, these non-polar solvents are often selected for these reactions. In water or alcohol solution the nucleophilic solvent may open the bromonium ion intermediate to give an α-halo-alcohol or ether, together with the expected vic-dihalide. Such reactions are sensitive to pH and other factors, so when these products are desired it is necessary to modify the addition reagent. Aqueous chlorine exists as the following equilibrium, Keq ≈ 10-4. By adding AgOH, the concentration of HOCl can be greatly increased, and the chlorohydrin addition product obtained from alkenes.
$\ce{Cl_2 + H_2O <<=> HOCl + HCl}$
The more widely used HOBr reagent, hypobromous acid, is commonly made by hydrolysis of N-bromoacetamide, as shown below. Both HOCl and HOBr additions occur in an anti fashion, and with the regioselectivity predicted by this mechanism (OH bonds to the more substituted carbon of the alkene).
$CH_3CONHBr + H_2O \rightarrow HOBr + CH_3CONH_2$
Vicinal halohydrins provide an alternative route for the epoxidation of alkenes over that of reaction with peracids. As illustrated in the following diagram, a base induced intramolecular substitution reaction forms a three-membered cyclic ether called an epoxide. Both the halohydrin formation and halide displacement reactions are stereospecific, so stereoisomerism in the alkene will be reflected in the epoxide product (i.e. trans-2-butene forms a trans-disubstituted epoxide). A general procedure for forming these useful compounds will be discussed in the next section. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Reactions_of_Alkenes_with_Halogens.txt |
Sulfenyl chloride additions are initiated by the attack of an electrophilic sulfur species on the pi-electrons of the double bond. The resulting cationic intermediate may be stabilized by the non-bonding valence shell electrons on the sulfur in exactly the same way the halogens exerted their influence. Indeed, a cyclic sulfonium ion intermediate analogous to the bromonium ion is believed to best represent this intermediate (see drawing below).
Figure 1: cyclic sulfonium ion intermediate
Two advantages of the oxymercuration method of adding water to a double bond are its high anti-stereoselectivity and the lack of rearrangement in sensitive cases. These characteristics are attributed to a mercurinium ion intermediate, analogous to the bromonium ion discussed above. In this case it must be d-orbital electrons that are involved in bonding to carbon. A drawing of this intermediate is shown below.
Figure 2: mercurinium ion intermediate
Hydroboration Stereoselectivity
The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. As noted, this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that the geometry of this addition must be syn. Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene in the following equation. Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with retention of configuration so the overall addition of water is also syn.
The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored blue in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl.
Hydrogenation of Alkenes
Addition of hydrogen to a carbon-carbon double bond is called hydrogenation. The overall effect of such an addition is the reductive removal of the double bond functional group. Regioselectivity is not an issue, since the same group (a hydrogen atom) is bonded to each of the double bond carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H2), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst, as shown in the following diagram.
Catalysts are substances that changes the rate (velocity) of a chemical reaction without being consumed or appearing as part of the product. Catalysts act by lowering the activation energy of reactions, but they do not change the relative potential energy of the reactants and products. Finely divided metals, such as platinum, palladium and nickel, are among the most widely used hydrogenation catalysts. Catalytic hydrogenation takes place in at least two stages, as depicted in the diagram. First, the alkene must be adsorbed on the surface of the catalyst along with some of the hydrogen. Next, two hydrogens shift from the metal surface to the carbons of the double bond, and the resulting saturated hydrocarbon, which is more weakly adsorbed, leaves the catalyst surface. The exact nature and timing of the last events is not well understood.
As shown in the energy diagram, the hydrogenation of alkenes is exothermic, and heat is released corresponding to the ΔE (colored green) in the diagram. This heat of reaction can be used to evaluate the thermodynamic stability of alkenes having different numbers of alkyl substituents on the double bond. For example, the following table lists the heats of hydrogenation for three C5H10 alkenes which give the same alkane product (2-methylbutane). Since a large heat of reaction indicates a high energy reactant, these heats are inversely proportional to the stabilities of the alkene isomers. To a rough approximation, we see that each alkyl substituent on a double bond stabilizes this functional group by a bit more than 1 kcal/mole.
Alkene Isomer (CH3)2CHCH=CH2
3-methyl-1-butene
CH2=C(CH3)CH2CH3
2-methyl-1-butene
(CH3)2C=CHCH3
2-methyl-2-butene
Heat of Reaction
( ΔHº )
–30.3 kcal/mole –28.5 kcal/mole –26.9 kcal/mole
From the mechanism shown here we would expect the addition of hydrogen to occur with syn-stereoselectivity. This is often true, but the hydrogenation catalysts may also cause isomerization of the double bond prior to hydrogen addition, in which case stereoselectivity may be uncertain.
The formation of transition metal complexes with alkenes has been convincingly demonstrated by the isolation of stable platinum complexes such as Zeise's salt, K[PtCl3(C2H4)].H2O, and ethylenebis(triphenylphosphine)platinum, [(C6H5)3P]2Pt(H2C=CH2). In the latter, platinum is three-coordinate and zero-valent, whereas Zeise's salt is a derivative of platinum(II). A model of Zeise's salt and a discussion of the unusual bonding in such complexes may be viewed by clicking here. Similar complexes have been reported for nickel and palladium, metals which also function as catalysts for alkene hydrogenation.
A non-catalytic procedure for the syn-addition of hydrogen makes use of the unstable compound diimide, N2H2. This reagent must be freshly generated in the reaction system, usually by oxidation of hydrazine, and the strongly exothermic reaction is favored by the elimination of nitrogen gas (a very stable compound). Diimide may exist as cis-trans isomers; only the cis-isomer serves as a reducing agent. Examples of alkene reductions by both procedures are shown below. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Addition_Reactions_involving_other_Cyclic_Onium_Intermediates.txt |
• Epoxidation
• Hydroxylation
• Oxidative Cleavage of Double Bonds
Oxidations
Some oxidation reactions of alkenes give cyclic ethers in which both carbons of a double bond become bonded to the same oxygen atom. These products are called epoxides or oxiranes. An important method for preparing epoxides is by reaction with peracids, RCO3H. The oxygen-oxygen bond of such peroxide derivatives is not only weak (ca. 35 kcal/mole), but in this case is polarized so that the acyloxy group is negative and the hydroxyl group is positive (recall that the acidity of water is about ten powers of ten weaker than that of a carboxylic acid). If we assume electrophilic character for the OH moiety, the following equation may be written.
It is unlikely that a dipolar intermediate, as shown above, is actually formed. The epoxidation reaction is believed to occur in a single step with a transition state incorporating all of the bonding events shown in the equation. Consequently, epoxidations by peracids always have syn-stereoselectivity, and seldom give structural rearrangement. You may see the transition state by clicking the Change Equation button. Presumably the electron shifts indicated by the blue arrows induce a charge separation that is immediately neutralized by the green arrow electron shifts.
The previous few reactions have been classified as reductions or oxidations, depending on the change in oxidation state of the functional carbons. It is important to remember that whenever an atom or group is reduced, some other atom or group is oxidized, and a balanced equation must balance the electron gain in the reduced species with the electron loss in the oxidized moiety, as well as numbers and kinds of atoms. Starting from an alkene (drawn in the box), the following diagram shows a hydrogenation reaction on the left (the catalyst is not shown) and an epoxidation reaction on the right. Examine these reactions, and for each identify which atoms are reduced and which are oxidized.
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups.
Hydroxylation
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown below.
Oxidative Cleavage of Double Bonds
Ozonolysis
In determining the structural formula of an alkene, it is often necessary to find the location of the double bond within a given carbon framework. One way of accomplishing this would be to selectively break the double bond and mark the carbon atoms that originally formed that bond. For example, there are three isomeric alkenes that all give 2-methylbutane on catalytic hydrogenation. These are 2-methyl-2-butene (compound A), 3-methyl-1-butene (compound B) and 2-methyl-1-butene (compound C), shown in the following diagram. If the double bond is cleaved and the fragments marked at the cleavage sites, the location of the double bond is clearly determined for each case. A reaction that accomplishes this useful transformation is known. It is called ozonolysis.
Ozone, O3, is an allotrope of oxygen that adds rapidly to carbon-carbon double bonds. Since the overall change in ozonolysis is more complex than a simple addition reaction, its mechanism has been extensively studied. Reactive intermediates called ozonides have been isolated from the interaction of ozone with alkenes, and these unstable compounds may be converted to stable products by either a reductive workup (Zn dust in water or alcohol) or an oxidative workup (hydrogen peroxide). The results of an oxidative workup may be seen by clicking the "Show Reaction" button a second time. Continued clicking of this button repeats the cycle. The chief difference in these conditions is that reductive workup gives an aldehyde product when hydrogen is present on a double bond carbon atom, whereas oxidative workup gives a carboxylic acid or carbon dioxide in such cases. The following equations illustrate ozonide formation, a process that is believed to involve initial syn-addition of ozone, followed by rearrangement of the extremely unstable molozonide addition product. They also show the decomposition of the final ozonide to carbonyl products by either a reductive or oxidative workup.
From this analysis and the examples given here, you should be able to deduce structural formulas for the alkenes that give the following ozonolysis products:
Glycol Cleavage
The vicinal glycols prepared by alkene hydroxylation (reaction with osmium tetroxide or permanganate) are cleaved to aldehydes and ketones in high yield by the action of lead tetraacetate (Pb(OAc)4) or periodic acid (HIO4). This oxidative cleavage of a carbon-carbon single bond provides a two-step, high-yield alternative to ozonolysis, that is often preferred for small scale work involving precious compounds. A general equation for these oxidations is shown below. As a rule, cis-glycols react more rapidly than trans-glycols, and there is evidence for the intermediacy of heterocyclic intermediates (as shown), although their formation is not necessary for reaction to occur.
Vicinal Syn Dihydroxylation
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
Introduction
The reaction with $OsO_4$ is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an anti dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures.
$OsO_4$ is formed slowly when osmium powder reacts with gasoues $O_2$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
$Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4$
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO4 was potassium permanganate, $KMnO_4$. Although syn diols will result from the reaction of KMnO4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation.
Mechanism
• Electrophilic attack on the alkene
• Pi bond of the alkene acts as the nucleophile and reacts with osmium (VIII) tetroxide (OsO4)
• 2 electrons from the double bond flows toward the osmium metal
• In the process, 3 electron pairs move simultaneously
• Cyclic ester with Os (VI) is produced
• Reduction
• H2S reduces the cyclic ester
• NaHSO4 with H2O may be used
• Forms the syn-1,2-diol (glycol)
Example: Dihydroxylation of 1-ethyl-1-cycloheptene
Chemical Highlight
Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate.
Problems
Questions:
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
Answers:
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol.
• Shivam Nand | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Oxidations/Epoxidation.txt |
Alkenes are usually prepared from either alcohols or haloalkanes (alkyl halides), although there are several methods for creating alkenes.
Synthesis of Alkenes
Sodium or potassium salt of a dicarboxylic acid on electrolysis gives an alkene. An alkene is generated when an aqueous solution of sodium or potassium salt of a dibasic acid (with adjacent carboyxlic groups) is electrolyzed.
Example: Electrolysis of Sodium Succinate
The electrolysis of sodium succinate gives ethene.
Oxidation at the anode
Reduction at Cathode
$\ce{2H2O + 2e^{-} -> 2OH^{-} + H2(g)} \nonumber$
Alkenes by Dehydration of Alcohols
Alkenes are usually prepared from either alcohols or haloalkanes (alkyl halides).
Dehydration of Alcohols
Alkenes are obtained by the dehydration of alcohols. The dehydration of alcohols can be affected by two common methods.
1. By passing the vapors of an alcohol over heated alumina.
2. By heating an alcohol with concentrated mineral acid, such as concentrated \(H_2SO_4\) or concentrated \(H_3PO_4\). Anhydrous zinc chloride can also be used as a dehydrating agent.
By passing the vapors of an alcohol over alumina (\(Al_2O_3\)) at 623 K (350°C).
The order of the ease of dehydration of alcohols is: tertiary > secondary > primary. Secondary and tertiary alcohols are best dehydrated by dilute sulfuric acid.
By heating an alcohol with concentrated sulfuric acid at 453 K (180°C).
Other dehydrating agents like phosphoric acid and anhydrous zinc chloride may also be used.
Example
Cyclohexanol on dehydration gives cyclohexene.
cyclohexanol cyclohexene
The loss of water from an alcohol to give an alkene does not occur in just one step; a series of steps are involved in the mechanism of dehydration of alcohols. In the dehydration reaction given above, the following steps are involved.
1. First, the acid protonates (adding a proton or H+) the alcohol on the most electronegative atom, namely oxygen. This process is usually reversible.
2. In the second step, the protonated alcohol loses water to give a positively charged species known as a carbonium ion or carbocation.
3. Finally the carbonium ion loses a proton to give alkene.
The mechanism of dehydration of ethyl alcohol is described below.
Alkenes from Aldehydes and Ketones - Wittig Reaction
The Wittig reaction or Wittig olefination is a chemical reaction of an aldehyde or ketone with a triphenyl phosphonium ylide (often called a Wittig reagent) to give an alkene and triphenylphosphine oxide.
The Wittig reaction was discovered in 1954 by Georg Wittig, for which he was awarded the Nobel Prize in Chemistry in 1979. It is widely used in organic synthesis for the preparation of alkenes. It should not be confused with the Wittig rearrangement.
Wittig reactions are most commonly used to couple aldehydes and ketones to singly substituted phosphine ylides. With unstabilised ylides this results in almost exclusively the Z-alkene product. In order to obtain the E-alkene, stabilised ylides are used or unstabilised ylides using the Schlosser modification of the Wittig reaction can be performed.
Reaction mechanism
Classical mechanism
The steric bulk of the ylide 1 influences the stereochemical outcome of nucleophilic addition to give a predominance of the betaine 3 (cf. Bürgi–Dunitz angle). Note that for betaine 3 both R1 and R2 as well as PPh3+ and O− are positioned anti to one another.
Carbon-carbon bond rotation gives the betaine 4, which then forms the oxaphosphetane 5. Elimination gives the desired Z-alkene 7 and triphenylphosphine oxide 6. With simple Wittig reagents, the first step occurs easily with both aldehydes and ketones, and the decomposition of the betaine (to form 5) is the rate-determining step. However, with stabilised ylides (where R1 stabilises the negative charge) the first step is the slowest step, so the overall rate of alkene formation decreases and a bigger proportion of the alkene product is the E-isomer. This also explains why stabilised reagents fail to react well with sterically hindered ketones.
Mechanism
Mechanistic studies have focused on unstabilized ylides, because the intermediates can be followed by NMR spectroscopy. The existence and interconversion of the betaine (3a and 3b) is subject of ongoing research. Phosphonium ylides 1 react with carbonyl compounds 2 via a π²s/π²a [2+2] cycloaddition to directly form the oxaphosphetanes 4a and 4b. The stereochemistry of the product 5 is due to the addition of the ylide 1 to the carbonyl 2 and to the equilibration of the intermediates. Maryanoff and Reitz identified the issue about equilibration of Wittig intermediates and termed the process "stereochemical drift". For many years, the stereochemistry of the Wittig reaction, in terms of carbon-carbon bond formation, had been assumed to correspond directly with the Z/E stereochemistry of the alkene products. However, certain reactants do not follow this simple pattern. Lithium salts can also exert a profound effect on the stereochemical outcome.
Mechanisms differ for aliphatic and aromatic aldehydes and for aromatic and aliphatic phosphonium ylides. Evidence suggests that the Wittig reaction of unbranched aldehydes under lithium-salt-free conditions do not equilibrate and are therefore under kinetic reaction control. Vedejs has put forth a theory to explain the stereoselectivity of stabilized and unstabilized Wittig reactions.
Wittig reagents
Preparation of phosphorus ylides
Wittig reagents are usually prepared from a phosphonium salt, which is in turn prepared by the quaternization of triphenylphosphine with an alkyl halide. The alkylphosphonium salt is deprotonated with a strong base such as n-butyllithium:
[Ph3P+CH2R]X + C4H9Li → Ph3P=CHR + LiX + C4H10
One of the simplest ylides is methylenetriphenylphosphorane (Ph3P=CH2). It is also a precursor to more elaborate Wittig reagents. Alkylation of Ph3P=CH2 with a primary alkyl halide R−CH2−X, produces substituted phosphonium salts:
Ph3P=CH2 + RCH2X → Ph3P+CH2CH2R X
These salts can be deprotonated in the usual way to give Ph3P=CH−CH2R.
Structure of the ylide
Ball-and-stick model of Ph3P=CH2, as found in the crystal structure
The Wittig reagent may be described in the phosphorane form (the more familiar representation) or the ylide form:
The ylide form is a significant contributor, and the carbon is nucleophilic.
Reactivity
Simple phosphoranes are reactive. Most hydrolyze and oxidize readily. They are therefore prepared using air-free techniques. Phosphoranes are more air-stable when they contain an electron withdrawing group. Some examples are Ph3P=CHCO2R and Ph3P=CHPh. These ylides are sufficiently stable to be sold commercially
From the phosphonium salts, these reagent are formed more readily, requiring only NaOH, and they are usually more air-stable. These are less reactive than simple ylides, and so they usually fail to react with ketones, necessitating the use of the Horner–Wadsworth–Emmons reaction as an alternative. They usually give rise to an E-alkene product when they react, rather than the more usual Z-alkene.
Scope and limitations
The Wittig reaction is a popular method for the synthesis of alkene from ketones and aldehydes. The Wittig reagent can generally tolerate carbonyl compounds containing several kinds of functional groups such as OH, OR, aromatic nitro and even ester groups. There can be a problem with sterically hindered ketones, where the reaction may be slow and give poor yields, particularly with stabilized ylides, and in such cases the Horner–Wadsworth–Emmons (HWE) reaction (using phosphonate esters) is preferred. Another reported limitation is the often labile nature of aldehydes which can oxidize, polymerize or decompose. In a so-called Tandem Oxidation-Wittig Process the aldehyde is formed in situ by oxidation of the corresponding alcohol.
As mentioned above, the Wittig reagent itself is usually derived from a primary alkyl halide. Quaternization of triphenylphosphine with most secondary halides is inefficient. For this reason, Wittig reagents are rarely used to prepare tetrasubstituted alkenes. However the Wittig reagent can tolerate many other variants. It may contain alkenes and aromatic rings, and it is compatible with ethers and even ester groups. Even C=O and nitrile groups can be present if conjugated with the ylide- these are the stabilised ylides mentioned above. Bis-ylides (containing two P=C bonds) have also been made and used successfully.
One limitation relates to the stereochemistry of the product. With simple ylides, the product is usually mainly the Z-isomer, although a lesser amount of the E-isomer is often formed also – this is particularly true when ketones are used. If the reaction is performed in DMF in the presence of LiI or NaI, the product is almost exclusively the Z-isomer. If the E-isomer is the desired product, the Schlosser modification may be used. With stabilised ylides the product is mainly the E-isomer, and this same isomer is also usual with the HWE reaction.
Schlosser modification
The major limitation of the traditional Wittig reaction is that the reaction proceeds mainly via the erythro betaine intermediate, which leads to the Z-alkene. The erythro betaine can be converted to the threo betaine using phenyllithium at low temperature. This modification affords the E-alkene.
Allylic alcohols can be prepared by reaction of the betaine ylid with a second aldehyde. For example:
Examples
Because of its reliability and wide applicability, the Wittig reaction has become a standard tool for synthetic organic chemists.
The most popular use of the Wittig reaction is for the introduction of a methylene group using methylenetriphenylphosphorane (Ph3P=CH2). Using this reagent even a sterically hindered ketone such as camphor can be converted to its methylene derivative. In this case, the Wittig reagent is prepared in situ by deprotonation of methyltriphenylphosphonium bromide with potassium tert-butoxide. In another example, the phosphorane is produced using sodium amide as a base, and this reagent converts the aldehyde shown into alkene I in 62% yield. The reaction is performed in cold THF, and the sensitive nitro, azo and phenoxide groups are tolerated. The product can be used to incorporate a photostabiliser into a polymer, to protect the polymer from damage by UV radiation.
Another example of its use is in the synthesis of leukotriene A methyl ester. The first step uses a stabilised ylide, where the carbonyl group is conjugated with the ylide preventing self condensation, although unexpectedly this gives mainly the cis product. The second Wittig reaction uses a non-stabilised Wittig reagent, and as expected this gives mainly the cis product. Note that the epoxide and ester functional groups survive intact.
Methoxymethylenetriphenylphosphine is a Wittig reagent for the homologation of aldehydes. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkanes_from_Reduction_of_Carboxylates_-_Kolbe%27s_Electrolytic_method.txt |
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond.
Introduction
The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
The required range of reaction temperature decreases with increasing substitution of the hydroxy-containing carbon:
• 1° alcohols: 170° - 180°C
• 2° alcohols: 100°– 140 °C
• 3° alcohols: 25°– 80°C
If the reaction is not sufficiently heated, the alcohols do not dehydrate to form alkenes, but react with one another to form ethers (e.g., the Williamson Ether Synthesis).
Alcohols are amphoteric; they can act both as acid or base. The lone pair of electrons on oxygen atom makes the –OH group weakly basic. Oxygen can donate two electrons to an electron-deficient proton. Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion +OH2 (The pKa value of a tertiary protonated alcohol can go as low as -3.8). This basic characteristic of alcohol is essential for its dehydration reaction with an acid to form alkenes.
Mechanism for the Dehydration of Alcohol into Alkene
Different types of alcohols may dehydrate through a slightly different mechanism pathway. However, the general idea behind each dehydration reaction is that the –OH group in the alcohol donates two electrons to H+ from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid (the nucleophile) then attacks the hydrogen adjacent to the carbocation and form a double bond.
Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). The relative reactivity of alcohols in dehydration reaction is ranked as the following
Methanol < primary < secondary < tertiary
Primary alcohol dehydrates through the E2 mechanism
Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. Then the nucleophile HSO4 back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism
Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon, forming a double bond. Notice in the mechanism below that the aleke formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Therefore, the trans diastereomer of the 2-butene product is most abundant.
Dehydration reaction of secondary alcohol: The dehydration mechanism for a tertiary alcohol is analogous to that shown above for a secondary alcohol.
When more than one alkene product are possible, the favored product is usually the thermodynamically most stable alkene. More-substituted alkenes are favored over less-substituted ones; and trans-substituted alkenes are preferred compared to cis-substituted ones.
1. Since the C=C bond is not free to rotate, cis-substituted alkenes are less stable than trans-subsituted alkenes because of steric hindrance (spatial interfererence) between two bulky substituents on the same side of the double bond (as seen in the cis product in the above figure). Trans-substituted alkenes reduce this effect of spatial interference by separating the two bulky substituents on each side of the double bond (for further explanation on the rigidity of C=C bond, see Structure and Bonding in Ethene- The pi Bond).
2. Heats of hydrogenation of differently-substituted alkene isomers are lowest for more-substituted alkenes, suggesting that they are more stable than less-substituted alkenes and thus are the major products in an elimination reaction. This is partly because in more --substituted alkenes, the p orbitals of the pi bond are stabilized by neighboring alkyl substituents, a phenomenon similar to hyperconjugation.
Hydride and Alkyl Shifts
Since the dehydration reaction of alcohol has a carbocation intermediate, hydride or alkyl shifts can occur which relocates the carbocation to a more stable position. The dehydrated products therefore are a mixture of alkenes, with and without carbocation rearrangement. Tertiary cation is more stable than secondary cation, which in turn is more stable than primary cation due to a phenomenon known as hyperconjugation, where the interaction between the filled orbitals of neighboring carbons and the singly occupied p orbital in the carbocation stabilizes the positive charge in carbocation.
• In hydride shifts, a secondary or tertiary hydrogen from a carbon next to the original carbocation takes both of its electrons to the cation site, swapping place with the carbocation and renders it a more stable secondary or tertiary cation.
Similarly, when there is no hydride available for hydride shifting, an alkyl group can take its bonding electrons and swap place with an adjacent cation, a process known as alkyl shift.
Practice Problems
Test your understanding by predicting what product(s) will be formed in each of the following reactions:
1.
2.
Solutions
1. Did you notice the reaction temperature? It is only 25°, which is much lower than the required temperature of 170°C for dehydration of primary alcohol. This reaction will not produce any alkene but will form ether.
2. . Notice that the reactant is a secondary -OH group, which will form a relatively unstable secondary carbocation in the intermediate. Thus hydride shift from an adjacent hydrogen will occur to make the carbocation tertiary, which is much more stable. The products are a mixture of alkenes that are formed with or without carbocation rearrangement (A number of products are formed faster than hydride shift can occur).
• Thuy Hoang | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkenes_from_Dehydration_of_Alcohols.txt |
Alkenes can be obtained from haloalkanes (alkyl halides). These haloalkanes are usually bromo and iodo and less commonly, chloro derivatives. Haloalkanes on heating with alcoholic \(KOH\) loses one molecule of hydrogen halide to give alkene.
Example \(2\)
If two alkenes may be formed due to dehydrohalogenation of a haloalkane, the one which is most substituted is the main product.
Example \(3\)
For example, dehydrohalogenation of 2-bromobutane gives,
The order of reactivity of haloalkanes in dehydrohalogenation is, Tertiary > Secondary > Primary. Note: Reactions in which a small molecule like \(H_2O\) or \(HX\) is lost are known as elimination reactions.
Alkenes from Vicinal Dihaloalkanes
Vicinal dihaloalkanes are those dihalogen derivatives of alkanes in which two halogen atoms are on the adjacent carbon atoms. Alkenes can be obtained from vicinal dihaloalkanes by dehalogenation. When such a dihaloalkane is heated with zinc in methanol, an alkene is formed.
Example \(4\)
1,2-dibromoethane gives 1,2-dibromoethane ethene
Alkenes from Hydrogenation of Alkynes
Alkenes can be easily obtained by hydrogenation of alkynes. An alkyne on controlled hydrogenation with hydrogen in the presence of Ni or Pd at 200°C give a corresponding alkene.
\[\ce{C_{n}H_{2n-2} + H_2 ->[Ni, 200 C] C_{n}H_{2n}}\]
Example
Ethyne gives ethene on hydrogenation.
Prepartion of Alkenes
This page looks at ways of preparing alkenes in the lab by the dehydration of alcohols.
Dehydration of alcohols using aluminium oxide as catalyst
Example 1: Dehydration of Ethanol to Product Ethene
This is a simple way of making gaseous alkenes like ethene. If ethanol vapor is passed over heated aluminum oxide powder, the ethanol is essentially cracked to give ethene and water vapor.
\[ CH_3CH_2OH \overset{Al_2O_3}{\longrightarrow} CH_2=CH_2 + H2O\]
To make a few test tubes of ethene, you can use this apparatus:
It wouldn't be too difficult to imagine scaling this up by boiling some ethanol in a flask and passing the vapor over aluminum oxide heated in a long tube.
Dehydration of alcohols using an acid catalyst
The acid catalysts normally used are either concentrated sulfuric acid or concentrated phosphoric(V) acid, H3PO4. Concentrated sulphuric acid produces messy results. Not only is it an acid, but it is also a strong oxidizing agent. It oxidizes some of the alcohol to carbon dioxide and at the same time is reduced itself to sulfur dioxide. Both of these gases have to be removed from the alkene. It also reacts with the alcohol to produce a mass of carbon. There are other side reactions as well.
Example 1: Dehydration of Ethanol to produce Ethene
Ethanol is heated with an excess of concentrated sulfuric acid at a temperature of 170°C. The gases produced are passed through sodium hydroxide solution to remove the carbon dioxide and sulfur dioxide produced from side reactions. The ethene is collected over water.
\[ CH_3CH_2OH \overset{conc. H_2SO_4}{\longrightarrow} CH_2=CH_2 + H2O\]
The concentrated sulfuric acid is a catalyst. Write it over the arrow rather than in the equation.
Example 2: Dehydration of Cyclohexanol to produce Cyclohexe
This is a preparation commonly used at this level to illustrate the formation and purification of a liquid product. The fact that the carbon atoms happen to be joined in a ring makes no difference whatever to the chemistry of the reaction. Cyclohexanol is heated with concentrated phosphoric(V) acid and the liquid cyclohexene distils off and can be collected and purified. Phosphoric(V) acid tends to be used in place of sulphuric acid because it is safer and produces a less messy reaction.
The dehydration of more complicated alcohols
You have to be wary with more complicated alcohols in case there is the possibility of more than one alkene being formed. Butan-2-ol is a good example of this, with no less than three different alkenes being formed when it is dehydrated. Butan-2-ol is an example to illustrate the problems. It is important that you understand it so that you can work out what will happen in similar cases. When you dehydrate an alcohol, you remove the -OH group, and a hydrogen atom from the next carbon atom in the chain. With molecules like butan-2-ol, there are two possibilities when that happens.
That leads to these products:
The products are but-1-ene, CH2=CHCH2CH3, and but-2-ene, CH3CH=CHCH3. In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below. Which isomer gets formed is just a matter of chance.
The overall result
Dehydration of butan-2-ol leads to a mixture containing:
• but-1-ene
• cis-but-2-ene (also known as (Z)-but-2-ene)
• trans-but-2-ene (also known as (E)-but-2-ene)
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkenes_from_Dehydrohalogenation_of_Haloalkanes.txt |
Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). The halogen atoms significantly alters the physical properties of the molecules including electronegativity, bond length, bond strength, and molecular size.
Properties of Alkyl Halides
The haloalkanes, also known as alkyl halides, are a group of chemical compounds comprised of an alkane with one or more hydrogens replaced by a halogen atom (fluorine, chlorine, bromine, or iodine).
There is a fairly large distinction between the structural and physical properties of haloalkanes and the structural and physical properties of alkanes. As mentioned above, the structural differences are due to the replacement of one or more hydrogens with a halogen atom. The differences in physical properties are a result of factors such as electronegativity, bond length, bond strength, and molecular size.
Halogens and the Character of the Carbon-Halogen Bond
With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge.
The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
Haloalkanes Have Higher Boiling Points than Alkanes
When comparing alkanes and haloalkanes, we will see that haloalkanes have higher boiling points than alkanes containing the same number of carbons. London dispersion forces are the first of two types of forces that contribute to this physical property. You might recall from general chemistry that London dispersion forces increase with molecular surface area. In comparing haloalkanes with alkanes, haloalkanes exhibit an increase in surface area due to the substitution of a halogen for hydrogen. The incease in surface area leads to an increase in London dispersion forces, which then results in a higher boiling point.
Dipole-dipole interaction is the second type of force that contributes to a higher boiling point. As you may recall, this type of interaction is a coulombic attraction between the partial positive and partial negative charges that exist between carbon-halogen bonds on separate haloalkane molecules. Similar to London dispersion forces, dipole-dipole interactions establish a higher boiling point for haloalkanes in comparison to alkanes with the same number of carbons.
The table below illustrates how boiling points are affected by some of these properties. Notice that the boiling point increases when hydrogen is replaced by a halogen, a consequence of the increase in molecular size, as well as an increase in both London dispersion forces and dipole-dipole attractions. The boiling point also increases as a result of increasing the size of the halogen, as well as increasing the size of the carbon chain.
Contributors
• Rachael Curtis (UC Davis)
Haloalkanes
Halogen containing organic compounds are relatively rare in terrestrial plants and animals. The thyroid hormones T3 and T4 are exceptions; as is fluoroacetate, the toxic agent in the South African shrub Dichapetalum cymosum, known as "gifblaar". However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large amounts of halogen. Some examples are shown below. The ocean is the largest known source for atmospheric methyl bromide and methyl iodide. Furthermore, the ocean is also estimated to supply 10-20% of atmospheric methyl chloride, with other significant contributions coming from biomass burning, salt marshes and wood-rotting fungi. Many subsequent chemical and biological processes produce poly-halogenated methanes.
Synthetic organic halogen compounds are readily available by direct halogenation of hydrocarbons and by addition reactions to alkenes and alkynes. Many of these have proven useful as intermediates in traditional synthetic processes. Some halogen compounds, shown in the box. have been used as pesticides, but their persistence in the environment, once applied, has led to restrictions, including banning, of their use in developed countries. Because DDT is a cheap and effective mosquito control agent, underdeveloped countries in Africa and Latin America have experienced a dramatic increase in malaria deaths following its removal, and arguments are made for returning it to limited use. 2,4,5-T and 2,4-D are common herbicides that are sold by most garden stores. Other organic halogen compounds that have been implicated in environmental damage include the polychloro- and polybromo-biphenyls (PCBs and PBBs), used as heat transfer fluids and fire retardants; and freons (e.g. CCl2F2 and other chlorofluorocarbons) used as refrigeration gases and fire extinguishing agents. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Haloalkanes/Alkyl_Halide_Occurrence.txt |
Halogen containing organic compounds are relatively rare in terrestrial plants and animals. The thyroid hormones T3 and T4 are exceptions; as is fluoroacetate, the toxic agent in the South African shrub Dichapetalum cymosum, known as "gifblaar". However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large amounts of halogen. Some examples are shown below.
The ocean is the largest known source for atmospheric methyl bromide and methyl iodide. Furthermore, the ocean is also estimated to supply 10-20% of atmospheric methyl chloride, with other significant contributions coming from biomass burning, salt marshes and wood-rotting fungi. Many subsequent chemical and biological processes produce poly-halogenated methanes.
Synthetic organic halogen compounds are readily available by direct halogenation of hydrocarbons and by addition reactions to alkenes and alkynes. Many of these have proven useful as intermediates in traditional synthetic processes. Some halogen compounds, shown in the box. have been used as pesticides, but their persistence in the environment, once applied, has led to restrictions, including banning, of their use in developed countries. Because DDT is a cheap and effective mosquito control agent, underdeveloped countries in Africa and Latin America have experienced a dramatic increase in malaria deaths following its removal, and arguments are made for returning it to limited use. 2,4,5-T and 2,4-D are common herbicides that are sold by most garden stores. Other organic halogen compounds that have been implicated in environmental damage include the polychloro- and polybromo-biphenyls (PCBs and PBBs), used as heat transfer fluids and fire retardants; and freons (e.g. CCl2F2 and other chlorofluorocarbons) used as refrigeration gases and fire extinguishing agents.
Alkyl halides provide nice examples for learning about two very important organic reaction mechanism types: nucleophilic substitution and beta-elimination. In learning about these mechanisms in the context of alkyl halide reactivity, we will also learn some very fundamental ideas about three main players in many organic reactions: nucleophiles, electrophiles, and leaving groups. We'll start with an overview of the substitution and elimination reactions which alkyl halides undergo.
Introduction to Alkyl Halides
Propose a substitution mechanism for the following reactions. Pay special attention to stereochemistry if indicated. Look at the conditions given to determine if the substitution is unimolecular or bimolecular (SN1 or SN2).
Solution
Propose an elimination mechanism for the following reactions. Pay special attention to stereochemistry if indicated. Look at the conditions given to determine if the elimination is unimolecular or bimolecular (E1 or E2).
Solution
Show the major elimination product and draw the mechanism for each of the following reactions.
Solution
Alkyl halide practice problems
a. Because we see a tertiary alkyl halide and a weak base, we know that this is an E1 elimination. The most substituted alkene is the major product.
b. Because we see a secondary alkyl bromide and a strong base, we know that this is an E2 elimination. The more substituted alkene is the major product.
Elimination Solutions
The above elimination mechanism is unimolecular (E1) because there is a weak base in a high concentration. Remember: a weak base favors E1 and high concentration of the base will also favor E1.
The above elimination mechanism is bimolecular (E2) because there is a strong base present. Remember: a strong base favors E2.
Substitution Solutions
The above substitution mechanism is bimolecular (SN2) because there is a strong nucleophile given as well as an aprotic solvent. Remember: a strong nucleophile favors SN2 and an aprotic solvent will also favor SN2.
The above substitution mechanism is unimolecular (SN1) because there is not a strong nucleophile present. Remember: a weak nucleophile favors SN1 and because the electrophile leaves first, a carbocation is formed. Therefore, the nucleophile, CH3CH2CH2OH can attack both the front and the back, resulting in a racemic mixture. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Alkyl_halide_practice_problems/Elimination_Solution.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/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Competition_between_substitution_and_elimination.txt |
In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, i.e. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the SN1 vs. SN2 character of a nucleophilic substitution reaction.
Consider two hypothetical SN2 reactions: one in which the electrophile is a methyl carbon and another in which it is tertiary carbon.
Because the three substituents on the methyl carbon electrophile are tiny hydrogens, the nucleophile has a relatively clear path for backside attack. However, backside attack on the tertiary carbon is blocked by the bulkier methyl groups. Once again, steric hindrance - this time caused by bulky groups attached to the electrophile rather than to the nucleophile - hinders the progress of an associative nucleophilic (SN2) displacement.
The factors discussed in the above paragraph, however, do not prevent a sterically-hindered carbon from being a good electrophile - they only make it less likely to be attacked in a concerted SN2 reaction. Nucleophilic substitution reactions in which the electrophilic carbon is sterically hindered are more likely to occur by a two-step, dissociative (SN1) mechanism. This makes perfect sense from a geometric point of view: the limitations imposed by sterics are significant mainly in an SN2 displacement, when the electrophile being attacked is a sp3-hybridized tetrahedral carbon with its relatively ‘tight’ angles of 109.4o. Remember that in an SN1 mechanism, the nucleophile attacks an sp2-hybridized carbocation intermediate, which has trigonal planar geometry with ‘open’ 120 angles.
With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step.
Elimination by the E1 mechanism
Just as there were two mechanisms for nucleophilic substitution, there are two elimination mechanisms.
The E1 mechanism is nearly identical to the SN1 mechanism, differing only in the course of reaction taken by the carbocation intermediate. As shown by the following equations, a carbocation bearing beta-hydrogens may function either as a Lewis acid (electrophile), as it does in the SN1 reaction, or a Brønsted acid, as in the E1 reaction.
Thus, hydrolysis of tert-butyl chloride in a mixed solvent of water and acetonitrile gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration. The alcohol is the product of an SN1 reaction and the alkene is the product of the E1 reaction. The characteristics of these two reaction mechanisms are similar, as expected. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific.
(CH3)3CCl + H2O ——> [ (CH3)3C(+) ] + Cl(–) + H2O ——> (CH3)3COH + (CH3)2C=CH2 + HCl + H2O
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution product.
2. The cation may transfer a beta-proton to a base, giving an alkene product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
Since the SN1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Electrophiles.txt |
An overview of the E2 mechanism
At the beginning of this module, we saw the following reaction between tert-butyl bromide and cyanide. Clearly, this is not a substitution reaction.
(3) (CH3)3C-Br + CN(–) ——> (CH3)2C=CH2 + Br(–) + HCN
We know that t-butyl bromide is not expected to react by an SN2 mechanism. Furthermore, the ethanol solvent is not sufficiently polar to facilitate an SN1 reaction. The other reactant, cyanide anion, is a good nucleophile; and it is also a decent base, being about ten times weaker than bicarbonate. Consequently, a base-induced elimination seems to be the only plausible reaction remaining for this combination of reactants.
Below is a mechanistic diagram of an elimination reaction by the E2 pathway:
.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanim of this reaction is single-step, and is referred to as the E2 mechanism.
In the methanol solvent used here, methanethiolate has greater nucleophilicity than methoxide by a factor of 100. Methoxide, on the other hand is roughly 106 times more basic than methanethiolate. As a result, we see a clear-cut difference in the reaction products, which reflects nucleophilicity (bonding to an electrophilic carbon) versus basicity (bonding to a proton). Kinetic studies of these reactions show that they are both second order (first order in R–Br and first order in Nu:(–)), suggesting a bimolecular mechanism for each. The substitution reaction is clearly SN2. The corresponding designation for the elimination reaction is E2. An energy diagram for the single-step bimolecular E2 mechanism is shown on the right. We should be aware that the E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the R–groups on the beta-carbon enhance the acidity of that hydrogen, then substantial breaking of C–H may occur before the other bonds begin to be affected. Similarly, groups that favor ionization of the halogen may generate a transition state with substantial positive charge on the alpha-carbon and only a small degree of C–H breaking. For most simple alkyl halides, however, it is proper to envision a balanced transition state, in which there has been an equal and synchronous change in all the bonds. Such a model helps to explain an important regioselectivity displayed by these elimination reactions.
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1º-beta-hydrogens compared with one 3º-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called the Zaitsev Rule.
The main factor contributing to Zaitsev Rule behavior is the stability of the alkene. We noted earlier that carbon-carbon double bonds are stabilized (thermodynamically) by alkyl substituents, and that this stabilization could be evaluated by appropriate heat of hydrogenation measurements. Since the E2 transition state has significant carbon-carbon double bond character, alkene stability differences will be reflected in the transition states of elimination reactions, and therefore in the activation energy of the rate-determining steps. From this consideration we anticipate that if two or more alkenes may be generated by an E2 elimination, the more stable alkene will be formed more rapidly and will therefore be the predominant product. This is illustrated for 2-bromobutane by the energy diagram on the right. The propensity of E2 eliminations to give the more stable alkene product also influences the distribution of product stereoisomers. In the elimination of 2-bromobutane, for example, we find that trans-2-butene is produced in a 6:1 ratio with its cis-isomer.
The Zaitsev Rule is a good predictor for simple elimination reactions of alkyl chlorides, bromides and iodides as long as relatively small strong bases are used. Thus hydroxide, methoxide and ethoxide bases give comparable results. Bulky bases such as tert-butoxide tend to give higher yields of the less substituted double bond isomers, a characteristic that has been attributed to steric hindrance. In the case of 2-bromo-2,3-dimethylbutane, described above, tert-butoxide gave a 4:1 ratio of 2,3-dimethyl-1-butene to 2,3-dimethyl-2-butene ( essentially the opposite result to that obtained with hydroxide or methoxide). This point will be discussed further once we know more about the the structure of the E2 transition state.
The importance of maintaining a planar configuration of the trigonal double-bond carbon components must never be overlooked. For optimum pi-bonding to occur, the p-orbitals on these carbons must be parallel, and the resulting doubly-bonded planar configuration is more stable than a twisted alternative by over 60 kcal/mole. This structural constraint is responsible for the existence of alkene stereoisomers when substitution patterns permit. It also prohibits certain elimination reactions of bicyclic alkyl halides, that might be favorable in simpler cases. For example, the bicyclooctyl 3º-chloride shown below appears to be similar to tert-butyl chloride, but it does not undergo elimination, even when treated with a strong base (e.g. KOH or KOC4H9). There are six equivalent beta-hydrogens that might be attacked by base (two of these are colored blue as a reference), so an E2 reaction seems plausible. The problem with this elimination is that the resulting double bond would be constrained in a severely twisted (non-planar) configuration by the bridged structure of the carbon skeleton. The carbon atoms of this twisted double-bond are colored red and blue respectively, and a Newman projection looking down the twisted bond is drawn on the right. Because a pi-bond cannot be formed, the hypothetical alkene does not exist. Structural prohibitions such as this are often encountered in small bridged ring systems, and are referred to as Bredt's Rule.
Bredt's Rule should not be applied blindly to all bridged ring systems. If large rings are present their conformational flexibility may permit good overlap of the p-orbitals of a double bond at a bridgehead. This is similar to recognizing that trans-cycloalkenes cannot be prepared if the ring is small (3 to 7-membered), but can be isolated for larger ring systems. The anti-tumor agent taxol has such a bridgehead double bond (colored red), as shown in the following illustration. The bicyclo[3.3.1]octane ring system is the smallest in which bridgehead double bonds have been observed. The drawing to the right of taxol shows this system. The bridgehead double bond (red) has a cis-orientation in the six-membered ring (colored blue), but a trans-orientation in the larger eight-membered ring.
Stereochemistry of the E2 Reaction
E2 elimination reactions of certain isomeric cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram.
Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state.
The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers an anti orientation of the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference for equatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates.
A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.
An orbital drawing of the anti-transition state is shown on the right. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the SN2 mechanism. In this drawing the α and β carbon atoms are undergoing a rehybridization from sp3 to sp2 and the developing π-bond is drawn as dashed light blue lines. The symbol R represents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints syn-elimination has been observed. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Elimination_by_the_E2_mechanism.txt |
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. In this module, we are focusing on substitution reactions in which the leaving group is a halogen ion, although many reactions are known, both in the laboratory and in biochemical processes, in which the leaving group is something other than a halogen.
In order to understand the nature of the leaving group, it is important to first discuss factors that help determine whether a species will be a strong base or weak base. If you remember from general chemistry, a Lewis base is defined as a species that donates a pair of electrons to form a covalent bond. The factors that will determine whether a species wants to share its electrons or not include electronegativity, size, and resonance.
As Electronegativity Increases, Basicity Decreases: In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, electronegativity increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons.
As Size Increases, Basicity Decreases:In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons.
Resonance Decreases Basicity:The third factor to consider in determining whether or not a species will be a strong or weak base is resonance. As you may remember from general chemistry, the formation of a resonance stabilized structure results in a species that is less willing to share its electrons. Since strong bases, by definition, want to share their electrons, resonance stabilized structures are weak bases.
Now that we understand how electronegativity, size, and resonance affect basicity, we can combine these concepts with the fact that weak bases make the best leaving groups. Think about why this might be true. In order for a leaving group to leave, it must be able to accept electrons. A strong bases wants to donate electrons; therefore, the leaving group must be a weak base. We will now revisit electronegativity, size, and resonance, moving our focus to the leaving group, as well providing actual examples.
As Electronegativity Increases, The Ability of the Leaving Group to Leave Increases
As mentioned previously, if we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basisity decreases, and the ability of the leaving group to leave increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them. The following diagram illustrates this concept, showing -CH3 to be the worst leaving group and F- to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that SN2 reactions of fluoroalkanes are rarely observed.
As Size Increases, The Ability of the Leaving Group to Leave Increases:Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Leaving_groups.txt |
What is a nucleophile?
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates.
More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
Of course, carbons can also be nucleophiles - otherwise how could new carbon-carbon bonds be formed in the synthesis of large organic molecules like DNA or fatty acids? Enolate ions (section 7.5) are the most common carbon nucleophiles in biochemical reactions, while the cyanide ion (CN-) is just one example of a carbon nucleophile commonly used in the laboratory. Reactions with carbon nucleophiles will be dealt with in chapters 13 and 14, however - in this chapter and the next, we will concentrate on non-carbon nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Now, lets discuss some of the major factors that affect nucleophilicity.
Protonation states and nucleophilicity
The protonation state of a nucleophilic atom has a very large effect on its nucleophilicity. This is an idea that makes intuitive sense: a hydroxide ion is much more nucleophilic (and basic) than a water molecule, because the negatively charged oxygen on the hydroxide ion carries greater electron density than the oxygen atom of a neutral water molecule. In practical terms, this means that a hydroxide nucleophile will react in an SN2 reaction with methyl bromide much faster ( about 10,000 times faster) than a water nucleophile.
Periodic trends and solvent effects in nucleophilicity
There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity:
The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond.
This horizontal trends also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions.
Recall that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity).
The vertical periodic trend for nucleophilicity is somewhat more complicated that that for basicity: depending on the solvent that the reaction is taking place in, the nucleophilicity trend can go in either direction. Let's take the simple example of the SN2 reaction below:
. . .where Nu- is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile in that case where both are iodide). If this reaction is occurring in a protic solvent (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile.
Relative nucleophilicity in a protic solvent
This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the least basic. What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile?
As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction running in a protic solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile:
In order for the nucleophile to attack the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the cage.
The picture changes if we switch to a polar aprotic solvent, such as acetone, in which there is a molecular dipole but no hydrogens bound to oxygen or nitrogen. Now, fluoride is the best nucleophile, and iodide the weakest.
Relative nucleophilicity in a polar aprotic solvent
The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile.
A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile.
Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO).
In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a cysteine amino acid, for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS-) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do.
Resonance effects on nucleophilicity
Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance.
The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group.
Steric effects on nucleophilicity
Steric hindrance is an important consideration when evaluating nucleophility. For example, tert-butanol is less potent as a nucleophile than methanol. This is because the comparatively bulky methyl groups on the tertiary alcohol effectively block the route of attack by the nucleophilic oxygen, slowing the reaction down considerably (imagine trying to walk through a narrow doorway while carrying three large suitcases!).
It is not surprising that it is more common to observe serines acting as nucleophiles in enzymatic reactions compared to threonines - the former is a primary alcohol, while the latter is a secondary alcohol.
Example
Which is the better nucleophile - a cysteine side chain or a methionine side chain? Explain.
Example
In each of the following pairs of molecules/ions, which is the better nucleophile in a reaction with CH3Br in acetone solvent? Explain your choice.
1. phenolate ion (deprotonated phenol) or benzoate ion (deprotonated benzoic acid)
2. water and hydronium ion
3. trimethylamine and triethylamine
4. chloride anion and iodide anion
5. CH3NH- and CH3CH2NH | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Nucleophiles.txt |
The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing on the right. Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength. The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions, which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pKa = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI.
Alkyl haides undergo both substitution and elimination reactions. In describing these two reaction pathways, 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/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Reactions_of_alkyl_halides_-_an_overview.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
Substitution reactions of alkyl halides: two mechanisms
Solutions | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Substitution_reactions_of_alkyl_halides%3A_two_mechanisms/Substituion_reaction_PP.txt |
Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). We will only look at compounds containing one halogen atom. For example:
alkyl halides fall into different classes depending on how the halogen atom is positioned on the chain of carbon atoms. There are some chemical differences between the various types.
Primary alkyl halides
In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group.Some examples of primary alkyl halides include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) halogenoalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different. Examples:
Tertiary alkyl halides
In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Physical properties of alkyl halides
Boiling Points
The chart shows the boiling points of some simple alkyl halides.
Notice that three of these have boiling points below room temperature (taken as being about 20°C). These will be gases at room temperature. All the others you are likely to come across are liquids. Remember:
• the only methyl halide which is a liquid is iodomethane;
• chloroethane is a gas.
The patterns in boiling point reflect the patterns in intermolecular attractions.
van der Waals dispersion forces
These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up. This is why the boiling points increase as the number of carbon atoms in the chains increases. Look at the chart for a particular type of halide (a chloride, for example). Dispersion forces get stronger as you go from 1 to 2 to 3 carbons in the chain. It takes more energy to overcome them, and so the boiling points rise.
The increase in boiling point as you go from a chloride to a bromide to an iodide (for a given number of carbon atoms) is also because of the increase in number of electrons leading to larger dispersion forces. There are lots more electrons in, for example, iodomethane than there are in chloromethane - count them!
van der Waals dipole-dipole attractions
The carbon-halogen bonds (apart from the carbon-iodine bond) are polar, because the electron pair is pulled closer to the halogen atom than the carbon. This is because (apart from iodine) the halogens are more electronegative than carbon. The electronegativity values are:
C 2.5 F 4.0
Cl 3.0
Br 2.8
I 2.5
This means that in addition to the dispersion forces there will be forces due to the attractions between the permanent dipoles (except in the iodide case). The size of those dipole-dipole attractions will fall as the bonds get less polar (as you go from chloride to bromide to iodide, for example). Nevertheless, the boiling points rise! This shows that the effect of the permanent dipole-dipole attractions is much less important than that of the temporary dipoles which cause the dispersion forces. The large increase in number of electrons by the time you get to the iodide completely outweighs the loss of any permanent dipoles in the molecules.
Example 1: Boiling Points of Some Isomers
The examples show that the boiling points fall as the isomers go from a primary to a secondary to a tertiary halogenoalkane. This is a simple result of the fall in the effectiveness of the dispersion forces.
The temporary dipoles are greatest for the longest molecule. The attractions are also stronger if the molecules can lie closely together. The tertiary halogenoalkane is very short and fat, and won't have much close contact with its neighbours.
Solubility
Solubility in water
The alkyl halides are at best only slightly soluble in water. For a halogenoalkane to dissolve in water you have to break attractions between the halogenoalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Both of these cost energy.
Energy is released when new attractions are set up between the halogenoalkane and the water molecules. These will only be dispersion forces and dipole-dipole interactions. These aren't as strong as the original hydrogen bonds in the water, and so not as much energy is released as was used to separate the water molecules. The energetics of the change are sufficiently "unprofitable" that very little dissolves.
Note
This energetic picture of solvation is not the whole story and entropic factors must also be considered to understand solvation properly.
Solubility in organic solvents
alkyl halides tend to dissolve in organic solvents because the new intermolecular attractions have much the same strength as the ones being broken in the separate halogenoalkane and solvent.
Chemical Reactivity
The pattern in strengths of the four carbon-halogen bonds are:
Notice that bond strength falls as you go from C-F to C-I, and notice how much stronger the carbon-fluorine bond is than the rest. To react with the alkyl halides, the carbon-halogen bond has got to be broken. Because that gets easier as you go from fluoride to chloride to bromide to iodide, the compounds get more reactive in that order. Iodoalkanes are the most reactive and fluoroalkanes are the least. In fact, fluoroalkanes are so unreactive that we shall pretty well ignore them completely from now on in this section!
The influence of bond polarity
Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples:
The electronegativities of carbon and iodine are equal and so there will be no separation of charge on the bond.
One of the important set of reactions of alkyl halides involves replacing the halogen by something else - substitution reactions. These reactions involve either:
• the carbon-halogen bond breaking to give positive and negative ions. The ion with the positively charged carbon atom then reacts with something either fully or slightly negatively charged.
• something either fully or negatively charged attracted to the slightly positive carbon atom and pushing off the halogen atom.
You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. If is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Physical_Properties_of_Alkyl_Halides.txt |
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.
Reactivity of Alkyl Halides
The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing below.
Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength. The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions, which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pKa = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI.
The characteristics noted above lead us to anticipate certain types of reactions that are likely to occur with alkyl halides. 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
Ionic 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.
Alkyl Halide Reactions
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. Sodium, for example, reduces elemental chlorine to chloride anion (sodium is oxidized to its cation), as do the other metals under varying conditions. In a similar fashion these same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to halide anion, and the carbon bonds to the metal (the carbon has carbanionic character). Halide reactivity increases in the order: Cl < Br < I. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination). The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them. The other metals mentioned above react in a similar manner, but the two shown here are the most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.
R3C-X + 2Li → R3C-Li + LiX An Alkyl Lithium Reagent
R3C-X + Mg → R3C-MgX A Grignard Reagent
The metals referred to here are insoluble in most organic solvents, hence these reactions are clearly heterogeneous, i.e. take place on the metal surface. The conditions necessary to achieve a successful reaction are critical.
• First, the metal must be clean and finely divided so as to provide the largest possible surface area for reaction.
• Second, a suitable solvent must be used. For alkyl lithium formation pentane, hexane or ethyl ether may be used; but ethyl ether or THF are essential for Grignard reagent formation.
• Third, since these organometallic compounds are very reactive, contaminants such as water, alcohols and oxygen must be avoided.
These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents unique and useful reactants in synthesis.
Reactions of organolithium and Grignard reagents reflect the nucleophilic (and basic) character of the functional carbon in these compounds. Many examples of such reactions will be encountered in future discussions, and five simple examples are shown below. The first and third equations demonstrate the strongly basic nature of these compounds, which bond rapidly to the weakly acidic protons of water and methyl alcohol (colored blue). The nucleophilic carbon of these reagents also bonds readily with electrophiles such as iodine (second equation) and carbon dioxide (fifth equation). The polarity of the carbon-oxygen double bonds of CO2 makes the carbon atom electrophilic, shown by the formula in the shaded box, so the nucleophilic carbon of the Grignard reagent bonds to this site. As noted above, solutions of these reagents must also be protected from oxygen, since peroxides are formed (equation 4).
Another important reaction exhibited by these organometallic reagents is metal exchange. In the first example below, methyl lithium reacts with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Other alkyl lithiums give similar Gilman reagents. A useful application of these reagents is their ability to couple with alkyl, vinyl and aryl iodides, as shown in the second equation. Later we shall find that Gilman reagents also display useful carbon-carbon bond forming reactions with conjugated enones and with acyl chlorides.
2 CH3Li + CuI → (CH3)2CuLi + LiI Formation of a Gilman Reagent
(C3H7)2CuLi + C6H5I → C6H5-C3H7 + LiI + C3H7Cu A Coupling Reaction
The formation of organometallic reagents from alkyl halides is more tolerant of structural variation than were the nucleophilic substitutions described earlier. Changes in carbon hybridization have little effect on the reaction, and 1º, 2º and 3º-alkyl halides all react in the same manner. One restriction, of course, is the necessary absence of incompatible functional groups elsewhere in the reactant molecule. For example, 5-bromo-1-pentanol fails to give a Grignard reagent (or a lithium reagent) because the hydroxyl group protonates this reactive function as soon as it is formed.
BrCH2CH2CH2CH2CH2OH + Mg → [ BrMgCH2CH2CH2CH2CH2OH ] → HCH2CH2CH2CH2CH2OMgBr
Exchange metalation is particularly useful when it can be directed to specific sites in a molecule. One such case is the directed ortho metalation of aromatic rings bearing a suitable directing group. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Reactions_of_Alkyl_Halides_with_Reducing_Metals.txt |
If two halogen atoms are present in a given compound, reactions with reducing metals may take different paths depending on how close the carbon-halogen bonds are to each other. If they are separated by four or more carbons, as in the first example below, a bis-organometallic compound may be formed. However, if the halogens are bonded to adjacent (vicinal) carbons, an elimination takes place with formation of a double bond. Since vicinal-dihalides are usually made by adding a halogen to a double bond, this reaction is mainly useful for relating structures to each other. The last example, in which two halogens are bonded to the same carbon, referred to as geminal (twinned), gives an unusual reagent which may either react as a carbon nucleophile or, by elimination, as a carbene. Such reagents are often termed carbenoid.
The solution structure of the Simmons-Smith reagent is less well understood than that of the Grignard reagent, but the formula given here is as useful as any that have been proposed. Other alpha-halogenated organometallic reagents, such as ClCH2Li, BrCH2Li, Cl2CHLi and Cl3CLi, have been prepared, but they are substantially less stable and must be maintained at very low temperature (ca. -100 º C) to avoid loss of LiX. The stability and usefulness of the Simmons-Smith reagent may be attributed in part to the higher covalency of the carbon-zinc bond together with solvation and internal coordination of the zinc. Hydrolysis (reaction with water) gives methyl iodide, confirming the basicity of the carbon; and reaction with alkenes gives cyclopropane derivatives, demonstrating the carbene-like nature of the reagent. The latter transformation is illustrated by the equation below.
Elimination reactions of the stereoisomeric 1,2-dibromo-1,2-diphenylethanes provide a nice summary of the principles discussed above. The following illustration shows first the meso-diastereomer and below it one enantiomer of the racemic-diastereomer. In each case two conformers are drawn within parentheses, and the anti-relationship of selected vicinal groups in each is colored green. The reaction proceeding to the left is a dehydrohalogenation induced by treatment with KOH in alcohol. Since this is a stereospecific elimination, each diastereomer gives a different stereoisomeric product. The reaction to the right is a dehalogenation (the reverse of halogen addition to an alkene), caused by treatment with iodide anion. Zinc dust effects the same reaction, but with a lower degree of stereospecificity. The mechanism of the iodide anion reaction is shown by red arrows in the top example. A similar mechanism explains the comparable elimination of the racemic isomer. In both reactions an anti-transition state is observed.
The two stereoisomers of 1-bromo-1,2-diphenylethene (shown on the left of the diagram) undergo a second dehydrobromination reaction on more vigorous treatment with base, as shown in the following equation. This elimination generates the same alkyne (carbon-carbon triple bond) from each of the bromo-alkenes. Interestingly, the (Z)-isomer (lower structure) eliminates more rapidly than the (E)-isomer (upper structure), again showing a preference for anti-orientation of eliminating groups.
C6H5CH=CBrC6H5 + KOH → C6H5C≡CC6H5 + KBr + H2O
Preparation of Alkynes by Dehydrohalogenation
The last reaction shown above suggests that alkynes might be prepared from alkenes by a two stage procedure, consisting first of chlorine or bromine addition to the double bond, and secondly a base induced double dehydrohalogenation. For example, reaction of 1-butene with bromine would give 1,2-dibromobutane, and on treatment with base this vicinal dibromide would be expected to yield 1-bromo-1-butene followed by a second elimination to 1-butyne.
CH3CH2CH=CH2 + Br2 → CH3CH2CHBr–CH2Br + base → CH3CH2CH=CHBr + base → CH3CH2C≡CH
In practice this strategy works, but it requires care in the selection of the base and solvent. If KOH in alcohol is used, the first elimination is much faster than the second, so the bromoalkene may be isolated if desired. Under more extreme conditions the second elimination takes place, but isomerization of the triple bond also occurs, with the more stable isomer (2-butyne) being formed along with 1-butyne, even becoming the chief product. To facilitate the second elimination and avoid isomerization the very strong base sodium amide, NaNH2, may be used. Since ammonia is a much weaker acid than water (by a factor of 1018), its conjugate base is proportionally stronger than hydroxide anion (the conjugate base of water), and the elimination of HBr from the bromoalkene may be conducted at relatively low temperature. Also, the acidity of the sp-hybridized C-H bond of the terminal alkyne traps the initially formed 1-butyne in the form of its sodium salt.
CH3CH2C≡CH + NaNH2 → CH3CH2C≡C:(–) Na(+) + NH3
An additional complication of this procedure is that the 1-bromo-1-butene product of the first elimination (see previous equations) is accompanied by its 2-bromo-1-butene isomer, CH3CH2CBr=CH2, and elimination of HBr from this bromoalkene not only gives 1-butyne (base attack at C-1) but also 1,2-butadiene, CH3CH=C=CH2, by base attack at C-3. Dienes of this kind, in which the central carbon is sp-hybridized, are called allenes and are said to have cumulated double bonds. They are usually less stable than their alkyne isomers. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Reactions_of_Dihalides.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.
Substitution and Elimination Reactions of Alkyl Halides
Just as there were two mechanisms for nucleophilic substitution, there are two elimination mechanisms. The E1 mechanism is nearly identical to the SN1 mechanism, differing only in the course of reaction taken by the carbocation intermediate. As shown by the following equations, a carbocation bearing beta-hydrogens may function either as a Lewis acid (electrophile), as it does in the SN1 reaction, or a Brønsted acid, as in the E1 reaction.
Thus, hydrolysis of tert-butyl chloride in a mixed solvent of water and acetonitrile gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration. The alcohol is the product of an SN1 reaction and the alkene is the product of the E1 reaction. The characteristics of these two reaction mechanisms are similar, as expected. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific.
(CH3)3CCl + H2O → [ (CH3)3C(+) ] + Cl(–) + H2O → (CH3)3COH + (CH3)2C=CH2 + HCl + H2O
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution product.
2. The cation may transfer a beta-proton to a base, giving an alkene product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
Since the SN1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place.
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. 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). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/E1_Elimination_Reactions.txt |
We have not yet considered the factors that influence elimination reactions, such as example 3 in the group presented at the beginning of this section.
(3) (CH3)3C-Br + CN(–) → (CH3)2C=CH2 + Br(–) + HCN
We know that t-butyl bromide is not expected to react by an SN2 mechanism. Furthermore, the ethanol solvent is not sufficiently polar to facilitate an SN1 reaction. The other reactant, cyanide anion, is a good nucleophile; and it is also a decent base, being about ten times weaker than bicarbonate. Consequently, a base-induced elimination seems to be the only plausible reaction remaining for this combination of reactants. To get a clearer picture of the interplay of these factors consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles.
In the methanol solvent used here, methanethiolate has greater nucleophilicity than methoxide by a factor of 100. Methoxide, on the other hand is roughly 106 times more basic than methanethiolate. As a result, we see a clear-cut difference in the reaction products, which reflects nucleophilicity (bonding to an electrophilic carbon) versus basicity (bonding to a proton). Kinetic studies of these reactions show that they are both second order (first order in R–Br and first order in Nu:(–)), suggesting a bimolecular mechanism for each. The substitution reaction is clearly SN2. The corresponding designation for the elimination reaction is E2. An energy diagram for the single-step bimolecular E2 mechanism is shown below.
We should be aware that the E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the R–groups on the beta-carbon enhance the acidity of that hydrogen, then substantial breaking of C–H may occur before the other bonds begin to be affected. Similarly, groups that favor ionization of the halogen may generate a transition state with substantial positive charge on the alpha-carbon and only a small degree of C–H breaking. For most simple alkyl halides, however, it is proper to envision a balanced transition state, in which there has been an equal and synchronous change in all the bonds. Such a model helps to explain an important regioselectivity displayed by these elimination reactions.
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1º-beta-hydrogens compared with one 3º-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called the Zaitsev Rule.
The main factor contributing to Zaitsev Rule behavior is the stability of the alkene. We noted earlier that carbon-carbon double bonds are stabilized (thermodynamically) by alkyl substituents, and that this stabilization could be evaluated by appropriate heat of hydrogenation measurements. Since the E2 transition state has significant carbon-carbon double bond character, alkene stability differences will be reflected in the transition states of elimination reactions, and therefore in the activation energy of the rate-determining steps. From this consideration we anticipate that if two or more alkenes may be generated by an E2 elimination, the more stable alkene will be formed more rapidly and will therefore be the predominant product. This is illustrated for 2-bromobutane by the energy diagram below. The propensity of E2 eliminations to give the more stable alkene product also influences the distribution of product stereoisomers. In the elimination of 2-bromobutane, for example, we find that trans-2-butene is produced in a 6:1 ratio with its cis-isomer.
The Zaitsev Rule is a good predictor for simple elimination reactions of alkyl chlorides, bromides and iodides as long as relatively small strong bases are used. Thus hydroxide, methoxide and ethoxide bases give comparable results. Bulky bases such as tert-butoxide tend to give higher yields of the less substituted double bond isomers, a characteristic that has been attributed to steric hindrance. In the case of 2-bromo-2,3-dimethylbutane, described above, tert-butoxide gave a 4:1 ratio of 2,3-dimethyl-1-butene to 2,3-dimethyl-2-butene ( essentially the opposite result to that obtained with hydroxide or methoxide). This point will be discussed further once we know more about the the structure of the E2 transition state.
E2 Elimination Reactions
The importance of maintaining a planar configuration of the trigonal double-bond carbon components must never be overlooked. For optimum pi-bonding to occur, the p-orbitals on these carbons must be parallel, and the resulting doubly-bonded planar configuration is more stable than a twisted alternative by over 60 kcal/mole. This structural constraint is responsible for the existence of alkene stereoisomers when substitution patterns permit. It also prohibits certain elimination reactions of bicyclic alkyl halides, that might be favorable in simpler cases. For example, the bicyclooctyl 3º-chloride shown below appears to be similar to tert-butyl chloride, but it does not undergo elimination, even when treated with a strong base (e.g. KOH or KOC4H9). There are six equivalent beta-hydrogens that might be attacked by base (two of these are colored blue as a reference), so an E2 reaction seems plausible. The problem with this elimination is that the resulting double bond would be constrained in a severely twisted (non-planar) configuration by the bridged structure of the carbon skeleton. The carbon atoms of this twisted double-bond are colored red and blue respectively, and a Newman projection looking down the twisted bond is drawn on the right. Because a pi-bond cannot be formed, the hypothetical alkene does not exist. Structural prohibitions such as this are often encountered in small bridged ring systems, and are referred to as Bredt's Rule.
Bredt's Rule should not be applied blindly to all bridged ring systems. If large rings are present their conformational flexibility may permit good overlap of the p-orbitals of a double bond at a bridgehead. This is similar to recognizing that trans-cycloalkenes cannot be prepared if the ring is small (3 to 7-membered), but can be isolated for larger ring systems. The anti-tumor agent taxol has such a bridgehead double bond (colored red), as shown in the following illustration. The bicyclo[3.3.1]octane ring system is the smallest in which bridgehead double bonds have been observed. The drawing to the right of taxol shows this system. The bridgehead double bond (red) has a cis-orientation in the six-membered ring (colored blue), but a trans-orientation in the larger eight-membered ring. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/E2_Elimination_Reactions/Planar_Conguratio.txt |
E2 elimination reactions of certain isomeric cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram.
Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state.
The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers an anti orientation of the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference for equatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates.
A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.
An orbital drawing of the anti-transition state is shown above. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the SN2 mechanism. In this drawing the α and β carbon atoms are undergoing a rehybridization from sp3 to sp2 and the developing π-bond is drawn as dashed light blue lines. The symbol R represents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints syn-elimination has been observed.
It is also worth noting that anti-transition states were preferred in several addition reactions to alkenes, so there is an intriguing symmetry to these inverse structural transformations.
Having arrived at a useful and plausible model of the E2 transition state, we can understand why a bulky base might shift the regioselectivity of the reaction away from the most highly substituted double bond isomer. Steric hindrance to base attack at a highly substituted beta-hydrogen site would result in preferred attack at a less substituted site. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/E2_Elimination_Reactions/Stereochemistry_o.txt |
SN1 Mechanism
Reaction 7, is clearly different from the other cases we have examined. It not only shows first order kinetics, but the chiral 3º-alkyl bromide reactant undergoes substitution by the modest nucleophile water with extensive racemization. In all of these features this reaction fails to meet the characteristics of the SN2 mechanism. A similar example is found in the hydrolysis of tert-butyl chloride, shown below. Note that the initial substitution product in this reaction is actually a hydronium ion, which rapidly transfers a proton to the chloride anion. This second acid-base proton transfer is often omitted in writing the overall equation, as in the case of reaction 7.
(CH3)3C-Cl + H2O → (CH3)3C-OH2(+) + Cl(–) → (CH3)3C-OH + HCl
Although the hydrolysis of tert-butyl chloride, as shown above, might be interpreted as an SN2 reaction in which the high and constant concentration of solvent water does not show up in the rate equation, there is good evidence this is not the case. First, the equivalent hydrolysis of ethyl bromide is over a thousand times slower, whereas authentic SN2 reactions clearly show a large rate increase for 1º-alkyl halides. Second, a modest increase of hydroxide anion concentration has no effect on the rate of hydrolysis of tert-butyl chloride, despite the much greater nucleophilicity of hydroxide anion compared with water.
The first order kinetics of these reactions suggests a two-step mechanism in which the rate-determining step consists of the ionization of the alkyl halide, as shown in the diagram below. In this mechanism, a carbocation is formed as a high-energy intermediate, and this species bonds immediately to nearby nucleophiles. If the nucleophile is a neutral molecule, the initial product is an "onium" cation, as drawn above for t-butyl chloride, and presumed in the energy diagram. In evaluating this mechanism, we may infer several outcomes from its function.
1. The only reactant that is undergoing change in the first (rate-determining) step is the alkyl halide, so we expect such reactions would be unimolecular and follow a first-order rate equation. Hence the name SN1 is applied to this mechanism.
2. Since nucleophiles only participate in the fast second step, their relative molar concentrations rather than their nucleophilicities should be the primary product-determining factor. If a nucleophilic solvent such as water is used, its high concentration will assure that alcohols are the major product. Recombination of the halide anion with the carbocation intermediate simply reforms the starting compound. Note that SN1 reactions in which the nucleophile is also the solvent are commonly called solvolysis reactions. The hydrolysis of t-butyl chloride is an example.
3. The Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate. The stability of carbocations was discussed earlier, and a qualitative relationship is given below.
Carbocation
Stability
CH3(+) < CH3CH2(+) < (CH3)2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) (CH3)3C(+)
Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an SN1 mechanism. This is opposite to the reactivity order observed for the SN2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism.
Fourth, in order to facilitate the charge separation of an ionization reaction, as required by the first step, a good ionizing solvent will be needed. Two solvent characteristics will be particularly important in this respect. The first is the ability of solvent molecules to orient themselves between ions so as to attenuate the electrostatic force one ion exerts on the other. This characteristic is related to the dielectric constant, ε, of the solvent. Solvents having high dielectric constants, such as water (ε=81), formic acid (ε=58), dimethyl sulfoxide (ε=45) & acetonitrile (ε=39) are generally considered better ionizing solvents than are some common organic solvents such as ethanol (ε=25), acetone (ε=21), methylene chloride (ε=9) & ether (ε=4). The second factor is solvation, which refers to the solvent's ability to stabilize ions by encasing them in a sheath of weakly bonded solvent molecules. Anions are solvated by hydrogen-bonding solvents, as noted earlier. Cations are often best solvated by nucleophilic sites on a solvent molecule (e.g. oxygen & nitrogen atoms), but in the case of carbocations these nucleophiles may form strong covalent bonds to carbon, thus converting the intermediate to a substitution product. This is what happens in the hydrolysis reactions described above.
Fifth, the stereospecificity of these reactions may vary. The positively-charged carbon atom of a carbocation has a trigonal (flat) configuration (it prefers to be sp2 hybridized), and can bond to a nucleophile equally well from either face. If the intermediate from a chiral alkyl halide survives long enough to encounter a random environment, the products are expected to be racemic (a 50:50 mixture of enantiomers). On the other hand, if the departing halide anion temporarily blocks the front side, or if a nucleophile is oriented selectively at one or the other face, then the substitution might occur with predominant inversion or even retention of configuration.
SN1 Substitution Reactions
Molecularity
If a chemical reaction proceeds by more than one step or stage, its overall velocity or rate is limited by the slowest step, the rate-determining step. This "bottleneck concept" has analogies in everyday life. For example, if a crowd is leaving a theater through a single exit door, the time it takes to empty the building is a function of the number of people who can move through the door per second. Once a group gathers at the door, the speed at which other people leave their seats and move along the aisles has no influence on the overall exit rate. When we describe the mechanism of a chemical reaction, it is important to identify the rate-determining step and to determine its "molecularity". The molecularity of a reaction is defined as the number of molecules or ions that participate in the rate determining step. A mechanism in which two reacting species combine in the transition state of the rate-determining step is called bimolecular. If a single species makes up the transition state, the reaction would be called unimolecular. The relatively improbable case of three independent species coming together in the transition state would be called termolecular.
Kinetics
One way of investigating the molecularity of a given reaction is to measure changes in the rate at which products are formed or reactants are lost, as reactant concentrations are varied in a systematic fashion. This sort of study is called kinetics, and the goal is to write an equation that correlates the observed results. Such an equation is termed a kinetic expression, and for a general reaction of the type:
$A + B \rightarrow C + D$
it takes the form:
$\text{Reaction Rate} = k[A]^n[B]^m$
where the rate constant k is a proportionality constant that reflects the nature of the reaction, [A] is the concentration of reactant A, [B] is the concentration of reactant B, and n & m are exponential numbers used to fit the rate equation to the experimental data. Chemists refer to the sum n + m as the kinetic order of a reaction. In a simple bimolecular reaction n & m would both be 1, and the reaction would be termed second order, supporting a mechanism in which a molecule of reactant A and one of B are incorporated in the transition state of the rate-determining step. A bimolecular reaction in which two molecules of reactant A (and no B) are present in the transition state would be expected to give a kinetic equation in which n=2 and m=0 (also second order). The kinetic expressions found for the reactions shown at the beginning of this section are written in blue in the following equations. Each different reaction has its own distinct rate constant, k#. All the reactions save 7 display second order kinetics, reaction 7 is first order.
It should be recognized and remembered that the molecularity of a reaction is a theoretical term referring to a specific mechanism. On the other hand, the kinetic order of a reaction is an experimentally derived number. In ideal situations these two should be the same, and in most of the above reactions this is so. Reaction 7 above is clearly different from the other cases reported here. It not only shows first order kinetics (only the alkyl halide concentration influences the rate), but the chiral 3º-alkyl bromide reactant undergoes substitution by the modest nucleophile water with extensive racemization. Note that the acetonitrile cosolvent does not function as a nucleophile. It serves only to provide a homogeneous solution, since the alkyl halide is relatively insoluble in pure water.
One of the challenges faced by early workers in this field was to explain these and other differences in a rational manner. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/SN1_Substitution_Reactions/Molecularity_an.txt |
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.
Steric Hindranc
The two models displayed below start as methyl bromide, on the left, and ethyl bromide, on the right. These may be replaced by isopropyl, tert-butyl, neopentyl, and benzyl bromide models by pressing the appropriate buttons. (note that when first activated, this display may require clicking twice on the selected button.) In each picture the nucleophile is designated by a large violet sphere, located 3.75 Angstroms from the alpha-carbon atom (colored a dark gray), and located exactly opposite to the bromine (colored red-brown). This represents a point on the trajectory the nucleophile must follow if it is to bond to the back-side of the carbon atom, displacing bromide anion from the front face. With the exception of methyl and benzyl, the other alkyl groups present a steric hindrance to the back-side approach of the nucleophile, which increases with substitution alpha and beta to the bromine. The hydrogen (and carbon) atoms that hinder the nucleophile's approach are colored a light red. The magnitude of this steric hindrance may be seen by moving the models about in the usual way, and is clearly greatest for tert-butyl and neopentyl, the two compounds that fail to give substitution reactions.
SN2 Substitution Reactions and Alkyl Moiet
Some of the most important information concerning nucleophilic substitution and elimination reactions of alkyl halides has come from studies in which the structure of the alkyl group has been varied. If we examine a series of alkyl bromide substitution reactions with the strong nucleophile thiocyanide (SCN) in ethanol solvent, we find large decreases in the rates of reaction as alkyl substitution of the alpha-carbon increases. Methyl bromide reacts 20 to 30 times faster than simple 1º-alkyl bromides, which in turn react about 20 times faster than simple 2º-alkyl bromides, and 3º-alkyl bromides are essentially unreactive or undergo elimination reactions. Furthermore, β-alkyl substitution also decreases the rate of substitution, as witnessed by the failure of neopentyl bromide, (CH3)3CCH2-Br (a 1º-bromide), to react.
Alkyl halides in which the alpha-carbon is a chiral center provide additional information about these nucleophilic substitution reactions. Returning to the examples presented at the beginning of this section, we find that reactions 2, 5 & 6 demonstrate an inversion of configuration when the cyanide nucleophile replaces the bromine. Other investigations have shown this to be generally true for reactions carried out in non-polar organic solvents, the reaction of (S)-2-iodobutane with sodium azide in ethanol being just one example ( in the following equation the alpha-carbon is maroon and the azide nucleophile is blue). Inversion of configuration during nucleophilic substitution has also been confirmed for chiral 1º-halides of the type RCDH-X, where the chirality is due to isotopic substitution.
(S)-CH3CHICH2CH3 + NaN3 ——> (R)-CH3CHN3CH2CH3 + NaI
We can now piece together a plausible picture of how nucleophilic substitution reactions of 1º and 2º-alkyl halides take place. The nucleophile must approach the electrophilic alpha-carbon atom from the side opposite the halogen. As a covalent bond begins to form between the nucleophile and the carbon, the carbon halogen bond weakens and stretches, the halogen atom eventually leaving as an anion. The diagram below shows this process for an anionic nucleophile.
We call this description the SN2 mechanism, where S stands for Substitution, N stands for Nucleophilic and 2 stands for bimolecular (defined below). In the SN2 transition state the alpha-carbon is hybridized sp2 with the partial bonds to the nucleophile and the halogen having largely p-character. Both the nucleophile and the halogen bear a partial negative charge, the full charge being transferred to the halogen in the products. The consequence of rear-side bonding by the nucleophile is an inversion of configuration about the alpha-carbon. Neutral nucleophiles react by a similar mechanism, but the charge distribution in the transition state is very different.
This mechanistic model explains many aspects of the reaction. First, it accounts for the fact that different nucleophilic reagents react at very different rates, even with the same alkyl halide. Since the transition state has a partial bond from the alpha-carbon to the nucleophile, variations in these bond strengths will clearly affect the activation energy, ΔE, of the reaction and therefore its rate. Second, the rear-side approach of the nucleophile to the alpha-carbon will be subject to hindrance by neighboring alkyl substituents, both on the alpha and the beta-carbons. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/SN1_Substitution_Reactions/Nucleophilicity.txt |
This page looks at the reaction between halogenoalkanes (haloalkanes or alkyl halides) and ammonia. The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas. We'll talk about the reaction using 1-bromoethane as a typical primary halogenoalkane. There is no difference in the details of the reaction if you chose a secondary or tertiary halogenoalkane instead. The equations would just look more complicated than they already are! You get a series of amines formed together with their salts. The reactions happen sequentially.
Making a Primary Amine
The reaction happens in two stages. In the first stage, a salt is formed - in this case, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group.
$CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_3^+ + Br^-$
There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
$CH_3CH_2NH_3^+ + NH_3 \rightleftharpoons CH_3CH_2NH_2 + NH_4^+Br^-$
The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine - ethylamine. The more ammonia there is in the mixture, the more the forward reaction is favored.
Making a Secondary Amine
The reaction doesn't stop at a primary amine. The ethylamine also reacts with bromoethane - in the same two stages as before. In the first stage, you get a salt formed - this time, diethylammonium bromide. Think of this as ammonium bromide with two hydrogens replaced by ethyl groups.
There is again the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
The ammonia removes a hydrogen ion from the diethylammonium ion to leave a secondary amine - diethylamine. A secondary amine is one which has two alkyl groups attached to the nitrogen.
Making a Tertiary Amine
And still it doesn't stop! The diethylamine also reacts with bromoethane - in the same two stages as before. In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
The ammonia removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine. A tertiary amine is one which has three alkyl groups attached to the nitrogen.
Making a Quaternary Ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
What do you actually get if you react bromoethane with ammonia?
Whatever you do, you get a mixture of all of the products (including both amines and their salts) shown on this page.
To get mainly the quaternary ammonium salt, you can use a large excess of bromoethane. If you look at the reactions going on, each one needs additional bromoethane. If you provide enough, then the chances are that the reaction will go to completion, given enough time.
On the other hand, if you use a very large excess of ammonia, the chances are always greatest that a bromoethane molecule will hit an ammonia molecule rather than one of the amines being formed. That will help to prevent the formation of secondary (etc) amines. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Reaction_of_Alkyl_Halides_with_Ammonia.txt |
This page looks at how silver nitrate solution can be used as part of a test for halogenoalkanes (haloalkanes or alkyl halides), and also as a means of measuring their relative reactivities.
Testing for halogenoalkanes
Silver nitrate solution can be used to find out which halogen is present in a suspected halogenoalkane. The most effective way is to do a substitution reaction which turns the halogen into a halide ion, and then to test for that ion with silver nitrate solution.
The halogenoalkane is warmed with some sodium hydroxide solution in a mixture of ethanol and water. Everything will dissolve in this mixture and so you can get a good reaction. The halogen atom is displaced as a halide ion:
$R-X + OH^- \rightarrow R-OH + X^-$
with $X$ is any haligen atom.
There is no need to make this reaction go to completion. The silver nitrate test is sensitive enough to detect fairly small concentrations of halide ions. The mixture is acidified by adding dilute nitric acid. This prevents unreacted hydroxide ions reacting with the silver ions. Then silver nitrate solution is added. Various precipitates may be formed from the reaction between the silver and halide ions:
Table 1:
ion present observation
Cl- white precipitate
Br- very pale cream precipitate
I- very pale yellow precipitate
Confirming the precipitates
It is actually quite difficult to distinguish between these colors, especially if there isn't much precipitate. You can sort out which precipitate you have by adding ammonia solution.
Table 2:
original precipitate observation
AgCl precipitate dissolves to give a colorless solution
AgBr precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colorless solution
AgI precipitate is insoluble in ammonia solution of any concentration
Comparing halogenoalkane reactivities
In this case, various halogenoalkanes are treated with a solution of silver nitrate in a mixture of ethanol and water. Nothing else is added. After varying lengths of time precipitates appear as halide ions (produced from reactions of the halogenoalkanes) react with the silver ions present. As long as you are doing everything under controlled conditions (same amounts of everything, same temperature and so on), the time taken gives a good guide to the reactivity of the halogenoalkanes - the quicker the precipitate appears, the more reactive the halogenoalkane.
The halide ion is formed in one of two ways, depending on the type of halogenoalkane you have present - primary, secondary or tertiary. For a primary halogenoalkane, the main reaction is one between the halogenoalkane and water in the solvent.
$\ce{R-X + H_2O \rightarrow R-OH + H^+ + X^-}] A tertiary halogenoalkane ionizes to a very small extent of its own accord. \[\ce{R-X \rightleftharpoons R^{+} + X^-}$
Secondary halogenoalkanes do a bit of both of these.
Comparing the reaction rates as you change the halogen
You would have to keep the type of halogenoalkane (primary, secondary or tertiary) constant, but vary the halogen. You might, for example, compare the times taken to produce a precipitate from this series of primary halogenoalkanes:
Obviously, the time taken for a precipitate of silver halide to appear will depend on how much of everything you use and the temperature at which the reaction is carried out. But the pattern of results is always the same. For example:
• A primary iodo compound produces a precipitate quite quickly.
• A primary bromo compound takes longer to give a precipitate.
• A primary chloro compound probably won't give any precipitate until well after you have lost interest in the whole thing!
The order of reactivity reflects the strengths of the carbon-halogen bonds. The carbon-iodine bond is the weakest and the carbon-chlorine the strongest of the three bonds. In order for a halide ion to be produced, the carbon-halogen bond has to be broken. The weaker the bond, the easier that is.
If you have looked at the mechanisms for these reactions, you will know that a lone pair on a water molecule attacks the slightly positive carbon atom attached to the halogen. It is slightly positive because most of the halogens are more electronegative than carbon, and so pull electrons away from the carbon.
It is tempting to think that the reaction will be faster if the electronegativity difference is greater. The slight positive charge on the carbon will be larger if it is attached to a chlorine atom than to an iodine atom.
That means that there will be more attraction between a lone pair on the water and a carbon atom attached to a chlorine atom than if it was attached to an iodine atom. The electronegativity difference between carbon and iodine is negligible.
However, the fastest reaction is with an iodoalkane. In these reactions, bond strength is the main factor deciding the relative rates of reaction.
Comparing the reaction rates of primary, secondary and tertiary halogenoalkanes
You would need to keep the halogen atom constant. It is common to use bromides because they have moderate reaction rates. You could, for example, compare the reactivity of these compounds:
Again, the actual times taken will vary with reaction conditions, but the pattern will always be the same. For example:
• The tertiary halide produces a precipitate almost instantly.
• The secondary halide gives a slight precipitate after a few seconds. The precipitate thickens up with time.
• The primary halide may take considerably longer to produce a precipitate.
It is more difficult to explain the reason for this, because it needs a fairly intimate knowledge of the mechanisms involved in the reactions. It reflects the change in the way that the halide ion is produced as you go from primary to secondary to tertiary halogenoalkanes. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Reaction_of_Alkyl_Halides_with_Silver_Nitrate.txt |
Halogenoalkanes (haloalkanes or alkyl halides) reaction with cyanide ions from sodium or potassium cyanide solution to generate nitriles.
Replacing a halogen by -CN
If a halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol, the halogen is replaced by a -CN group and a nitrile is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture.
The solvent is important. If water is present you tend to get substitution by -OH instead of -CN. For example, using 1-bromopropane as a typical primary halogenoalkane:
$\ce{CH_3CH_2CH_2Br + CN^{-} \rightarrow CH_3CH_2CH_2CN + Br^{-}}$
You could write the full equation rather than the ionic one, but it slightly obscures what's going on:
$\ce{CH_3CH_2CH_2Br + KCN \rightarrow CH_3CH_2CH_2CN + KBr}$
The bromine (or other halogen) in the halogenoalkane is simply replaced by a -CN group - hence a substitution reaction to generate a nitrile. In this example, butanenitrile is formed. Secondary and tertiary halogenoalkanes behave similarly, although the mechanism will vary depending on which sort of halogenoalkane you are using.
Why these reactions matter
This reaction with cyanide ions is a useful way of lengthening carbon chains. For example, in the equations above, you start with a 3-carbon chain and end up with a 4-carbon chain. There are not very many simple ways of making new carbon-carbon bonds. It is fairly easy to change the -CN group at the end of the new chain into other groups.
The Reaction of Alkyl Halides with Hydroxide Ions
This page looks at the reactions between halogenoalkanes (haloalkanes or alkyl halides) and hydroxide ions from sodium or potassium hydroxide solution. It covers both substitution and elimination reactions.
Substitution vs. elimination Reactions
There are two different sorts of reaction that you can get depending on the conditions used and the type of halogenoalkane. Primary, secondary and tertiary halogenoalkanes behave differently in this respect.
Substitution reactions
In a substitution reaction, the halogen atom is replaced by an -OH group to give an alcohol. For example:
Or, as an ionic equation:
In the example, 2-bromopropane is converted into propan-2-ol.
The halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture.
The solvent is usually a 50/50 mixture of ethanol and water, because everything will dissolve in that. The halogenoalkane is insoluble in water. If you used water alone as the solvent, the halogenoalkane and the sodium hydroxide solution wouldn't mix and the reaction could only happen where the two layers met.
Elimination reactions
Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide.
The 2-bromopropane has reacted to give an alkene - propene.
Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons.
The halogenoalkane is heated under reflux with a concentrated solution of sodium or potassium hydroxide in ethanol. Propene is formed and, because this is a gas, it passes through the condenser and can be collected.
What decides whether you get substitution or elimination?
The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors.
The type of halogenoalkane
This is the most important factor .
type of halogenoalkane substitution or elimination?
primary mainly substitution
secondary both substitution and elimination
tertiary mainly elimination
For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution. However, you can influence things to some extent by changing the conditions.
The solvent
The proportion of water to ethanol in the solvent matters.
• Water encourages substitution.
• Ethanol encourages elimination.
The temperature
Higher temperatures encourage elimination.
Concentration of the sodium or potassium hydroxide solution
Higher concentrations favor elimination.
In summary
For a given halogenoalkane, to favor elimination rather than substitution, use:
• higher temperatures
• a concentrated solution of sodium or potassium hydroxide
• pure ethanol as the solvent
To favor substitution rather than elimination, use:
• lower temperatures
• more dilute solutions of sodium or potassium hydroxide
• more water in the solvent mixture | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/The_Reaction_of_Alkyl_Halides_with_Cyanide_Ions.txt |
This page looks at ways of making halogenoalkanes in the lab starting from alcohols. Halogenoalkanes can be made from the reaction between alkenes and hydrogen halides, but they are more commonly made by replacing the -OH group in an alcohol by a halogen atom. That's the method we'll concentrate on in this page.
Making halogenoalkanes from alcohols using hydrogen halides
The general reaction looks like this:
Making chloroalkanes
It is possible to make tertiary chloroalkanes successfully from the corresponding alcohol and concentrated hydrochloric acid, but to make primary or secondary ones you really need to use a different method - the reaction rates are too slow. A tertiary chloroalkane can be made by shaking the corresponding alcohol with concentrated hydrochloric acid at room temperature.
Making bromoalkanes
Rather than using hydrobromic acid, you usually treat the alcohol with a mixture of sodium or potassium bromide and concentrated sulfuric acid. This produces hydrogen bromide which reacts with the alcohol. The mixture is warmed to distil off the bromoalkane. You will find practical details of a reaction of this kind further down the page.
Making iodoalkanes
In this case the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H3PO4, and the iodoalkane is distilled off. The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide which reacts with the alcohol.
Phosphoric(V) acid is used instead of concentrated sulfuric acid because sulfuric acid oxidises iodide ions to iodine and produces hardly any hydrogen iodide. A similar thing happens to some extent with bromide ions in the preparation of bromoalkanes, but not enough to get in the way of the main reaction.
Making halogenoalkanes from alcohols using phosphorus halides
Making chloroalkanes
Chloroalkanes can be made by reacting an alcohol with liquid phosphorus(III) chloride, PCl3.
They can also be made by adding solid phosphorus(V) chloride, PCl5 to an alcohol. This reaction is violent at room temperature, producing clouds of hydrogen chloride gas. It isn't a good choice as a way of making halogenoalkanes, although it is used as a test for -OH groups in organic chemistry.
There are also side reactions involving the POCl3 reacting with the alcohol.
Making bromoalkanes and iodoalkanes
These are both made in the same general way. Instead of using phosphorus(III) bromide or iodide, the alcohol is heated under reflux with a mixture of red phosphorus and either bromine or iodine. The phosphorus first reacts with the bromine or iodine to give the phosphorus(III) halide.
These then react with the alcohol to give the corresponding halogenoalkane which can be distilled off.
Making bromoethane in the lab
Concentrated sulfuric acid is added slowly with lots of shaking and cooling to some ethanol in a flask, and then solid potassium bromide is added. The flask is then connected to a condenser so that the bromoethane formed can be distilled off.
Bromoethane has a low boiling point but is denser than water and almost insoluble in it. To prevent it from evaporating, it is often collected under water in a flask surrounded by ice. Sometimes it is simply collected in a tube surrounded by ice without any water. The reaction flask is heated gently until no more droplets of bromoethane collect.
Impurities in the bromoethane include:
• hydrogen bromide (although most of that will dissolve in the water if you are collecting the bromoethane under water);
• bromine - from the oxidation of bromide ions by the concentrated sulfuric acid;
• sulfur dioxide - formed when concentrated sulfuric acid oxidises the bromide ions;
• unreacted ethanol;
• ethoxyethane (diethyl ether) - formed by a side reaction between the ethanol and the concentrated sulfuric acid.
The purification sequence
Stage 1: If you have collected the bromoethane under water, transfer the contents of the collection flask to a separating funnel. Otherwise, pour the impure bromoethane into the separating funnel, add some water and shake it.
• Pour off and keep the bromoethane layer.
The water you discard will contain almost all of the hydrogen bromide, and quite a lot of any bromine, sulfur dioxide and ethanol present as impurities.
Stage 2: To get rid of any remaining acidic impurities (including the bromine and sulfur dioxide), return the bromoethane to the separating funnel and shake it with either sodium carbonate or sodium hydrogencarbonate solution.
• This reacts with any acids present liberating carbon dioxide and forming soluble salts.
• Separate and retain the lower bromoethane layer as before.
Stage 3: Now wash the bromoethane with water in a separating funnel to remove any remaining inorganic impurities (excess sodium carbonate solution, etc). This time, transfer the lower bromoethane layer to a dry test tube.
Stage 4: Add some anhydrous calcium chloride to the tube, shake well and leave to stand. The anhydrous calcium chloride is a drying agent and removes any remaining water. It also absorbs ethanol, and so any remaining ethanol may be removed as well (depending on how much calcium chloride you use).
Stage 5:Transfer the dry bromoethane to a distillation flask and fractionally distil it, collecting what distils over at between 35 and 40°C.
In principle, this should remove any remaining organic impurities. In practice, though, any ethoxyethane (which is perhaps the most likely impurity left at this stage) has a boiling point very, very close to that of bromoethane. It is unlikely that you will be able to separate the two. If there is any ethanol left which hadn't been absorbed by the calcium chloride, that would certainly be removed because its boiling point is much higher.
Contributors
Jim Clark (Chemguide.co.uk)
Uses of Alkyl Halides
This page looks at some of the uses of halogenoalkanes (haloalkanes or alkyl halides).
CFCs and their replacements
CFCs are chlorofluorocarbons - compounds containing carbon with chlorine and fluorine atoms attached. Two common CFCs are:
CFC-11 CCl3F
CFC-12 CCl2F2
Uses of CFCs
CFCs are non-flammable and not very toxic. They therefore had a large number of uses. They were used as refrigerants, propellants for aerosols, for generating foamed plastics like expanded polystyrene or polyurethane foam, and as solvents for dry cleaning and for general degreasing purposes. Unfortunately, CFCs are largely responsible for destroying the ozone layer. In the high atmosphere, the carbon-chlorine bonds break to give chlorine free radicals. It is these radicals which destroy ozone. CFCs are now being replaced by less environmentally harmful compounds. CFCs can also cause global warming. One molecule of CFC-11, for example, has a global warming potential about 5000 times greater than a molecule of carbon dioxide. On the other hand, there is far more carbon dioxide in the atmosphere than CFCs, so global warming isn't the major problem associated with them.
Replacements for CFCs
These are still mainly halogenoalkanes, although simple alkanes such as butane can be used for some applications (for example, as aerosol propellants).
Hydrochlorofluorocarbons (HCFCs): These are carbon compounds which contain hydrogen as well as halogen atoms. For example:
HCFC-22 CHClF2
The formula can be worked out from the number in the name in exactly the same way as for CFCs. These have a shorter life in the atmosphere than CFCs, and much of them is destroyed in the low atmosphere and so doesn't reach the ozone layer. HCFC-22 has only about one-twentieth of the effect on the ozone layer as a typical CFC.
Hydrofluorocarbons (HFCs): These are compounds containing only hydrogen and fluorine attached to carbon. For example:
HFC-134a CH2F-CF3
Because these HCFCs don't contain any chlorine, they have zero effect on the ozone layer. HFC-134a is now widely used in refrigerants, for blowing foamed plastics and as a propellant in aerosols.
Hydrocarbons
Again, these have no effect on the ozone layer, but they do have a down-side. They are highly flammable and are involved in environmental problems such as the formation of photochemical smog.
Other uses of organohalogen compounds
Strictly speaking, the compounds we are talking about here are halogenoalkenes, not halogenoalkanes:
• Chloroethene, CH2=CHCl, is used to make poly(chloroethene) - usually known as PVC.
• Tetrafluoroethene, CF2=CF2, is used to make poly(tetrafluoroethene) - PTFE.
Lab uses of halogenoalkanes
If you refer to the halogenoalkanes menu (see below), you will find that they react with lots of things leading to a wide range of different organic products. Halogenoalkanes are therefore useful in the lab as intermediates in making other organic chemicals. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Synthesis_of_Alkyl_Halides/Synthesis_of_Alkyl_Halides_from_Alcohols.txt |
Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule. If a molecule contains both a double and a triple bond, the carbon chain is numbered so that the first multiple bond gets a lower number. If both bonds can be assigned the same number, the double bond takes precedence. The molecule is then named "n-en-n-yne", with the double bond root name preceding the triple bond root name (e.g. 2-hepten-4-yne).
Introduction
Here are the molecular formulas and names of the first ten carbon straight chain alkynes.
Name Molecular Formula
Ethyne \(\ce{C2H2}\)
Propyne \(\ce{C3H4}\)
1-Butyne \(\ce{C4H6}\)
1-Pentyne \(\ce{C5H8}\)
1-Hexyne \(\ce{C6H10}\)
1-Heptyne \(\ce{C7H12}\)
1-Octyne \(\ce{C8H14}\)
1-Nonyne \(\ce{C9H16}\)
1-Decyne \(\ce{C10H18}\)
The more commonly used name for ethyne is acetylene, which used industrially.
Naming Alkynes
Like previously mentioned, the IUPAC rules are used for the naming of alkynes.
Rule 1
Find the longest carbon chain that includes both carbons of the triple bond.
Rule 2
Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes. For example:
4-chloro-6-diiodo-7-methyl-2-nonyne
Rule 3
After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order. For example:
2,2,10-triiodo-5-methyl-3-decyne
If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond.
5- methyl-7-octyn-3-ol
When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne. For example:
4-methyl-1,5-octadiyne
Rule 4
Substituents containing a triple bond are called alkynyl. For example:
4-bromo-1-chloro-1-ethynylcyclohexane
Here is a table with a few of the alkynyl substituents:
Molecule
Ethynyl -C?CH
2- Propynyl -CH2C?CH
2-Butynyl -CH3C?CH2CH3
Rule 5
A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example:
6-ethyl-3-methylnon-4-en-1-yne
Reference
1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
Problems
Name or draw out the following molecules:
1. 4,4-dimethyl-2-pentyne
2. 4-Penten-1-yne
3. 1-ethyl-3-dimethylnonyne
4.
Acidity of Terminal Alkynes
Hybridization due to triple bonds allows the uniqueness of alkyne structure. This triple bond contributes to the nonpolar bonding strength, linear, and the acidity of alkynes. Physical Properties include nonpolar due to slight solubility in polar solvents and insoluble in water. This solubility in water and polar solvents is a characteristic feature to alkenes as well. Alkynes dissolve in organic solvents.
Properties of Alkynes
Alkanes are undoubtedly the weakest Brønsted acids commonly encountered in organic chemistry. It is difficult to measure such weak acids, but estimates put the pKa of ethane at about 48. Hybridizing the carbon so as to increase the s-character of the C-H increases the acidity, with the greatest change occurring for the sp-C-H groups found in terminal alkynes. Thus, the pKa of ethene is estimated at 44, and the pKa of ethyne (acetylene) is found to be 25, making it 1023 times stronger an acid than ethane. This increase in acidity permits the isolation of insoluble silver and copper salts of such compounds.
$\ce{RC \bond{#} C-H + Ag(NH_3)_2^+ ->[\text{in } NH_4OH] RC\bond{#} C-Ag _{(insoluble)} + NH_3 + NH_4^+}$
Despite the dramatic increase in acidity of terminal alkynes relative to other hydrocarbons, they are still very weak acids, especially when compared with water, which is roughly a billion times more acidic. If we wish to prepare nucleophilic salts of terminal alkynes for use in synthesis, it will therefore be necessary to use a much stronger base than hydroxide (or ethoxide) anion. Such a base is sodium amide (NaNH2), discussed above, and its reactions with terminal alkynes may be conducted in liquid ammonia or ether as solvents. The products of this acid-base reaction are ammonia and a sodium acetylide salt. Because the acetylide anion is a powerful nucleophile it may displace halide ions from 1º-alkyl halides to give a more highly substituted alkyne as a product (SN2 reaction). This synthesis application is described in the following equations. The first two equations show how acetylene can be converted to propyne; the last two equations present a synthesis of 2-pentyne from propyne.
H-C≡C-H + NaNH2 (in ammonia or ether) → H-C≡C-Na (sodium acetylide) + NH3
H-C≡C-Na + CH3-I → H-C≡C-CH3 + NaI
CH3-C≡C-H + NaNH2 (in ammonia or ether) → CH3-C≡C-Na (sodium propynylide) + NH3
CH3-C≡C-Na + C2H5-BrCH3-C≡C-C2H5 + NaBr
Because RC≡C:(–) Na(+) is a very strong base (roughly a billion times stronger than NaOH), its use as a nucleophile in SN2 reactions is limited to 1º-alkyl halides; 2º and 3º-alkyl halides undergo elimination by an E2 mechanism.
The enhanced acidity of terminal alkynes relative to alkanes also leads to metal exchange reactions when these compounds are treated with organolithium or Grignard reagents. This exchange, shown below in equation 1, can be interpreted as an acid-base reaction which, as expected, proceeds in the direction of the weaker acid and the weaker base. This factor clearly limits the usefulness of Grignard or lithium reagents when a terminal triple bond is present, as in equation 2.
1) RC≡C-H + C2H5MgBr (in ether) → RC≡C-MgBr + C2H6
2) HC≡C-CH2CH2Br + Mg (in ether) → [ HC≡C-CH2CH2MgBr] → BrMgC≡C-CH2CH2H
The acidity of terminal alkynes also plays a role in product determination when vicinal (or geminal) dihalides undergo base induced bis-elimination reactions. The following example illustrates eliminations of this kind starting from 1,2-dibromopentane, prepared from 1-pentene by addition of bromine. The initial elimination presumably forms 1-bromo-1-pentene, since base attack at the more acidic and less hindered 1 º-carbon should be favored. The second elimination then produces 1-pentyne. If the very strong base sodium amide is used, the terminal alkyne is trapped as its sodium salt, from which it may be released by mild acid treatment. However, if the weaker base KOH is used for the elimination, the terminal alkyne salt is not formed, or is formed reversibly, and the initially generated 1-pentyne rearranges to the more stable 2-pentyne via an allene intermediate.
In the case of non-terminal alkynes, sodium and potassium amide, and related strong bases from 1 º-amines, are able to abstract protons from carbon atoms adjacent to the triple bond. The resulting allenic carbanions undergo rapid proton transfer equilibria, leading to the relatively stable terminal alkyne conjugate base. This isomerization may be used to prepare longer chain 1-alkynes, as shown in the following conversion of 3-heptyne to 1-heptyne. The R and R' substituents on the allenic intermediate range from propyl to hydrogen, as the proton transfers proceed. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Naming_the_Alkynes.txt |
The characteristic of the triple bond helps to explain the properties and bonding in the alkynes.
Importance of Triple Bonds
Hybridization due to triple bonds allows the uniqueness of alkyne structure. This triple bond contributes to the nonpolar bonding strength, linear, and the acidity of alkynes. Physical Properties include nonpolar due to slight solubility in polar solvents and insoluble in water. This solubility in water and polar solvents is a characteristic feature to alkenes as well. Alkynes dissolve in organic solvents.
Boiling Points
Compared to alkanes and alkenes, alkynes have a slightly higher boiling point. Ethane has a boiling point of -88.6 ?C, while Ethene is -103.7 ?C and Ethyne has a higher boiling point of -84.0 ?C.
Acidity
The acidity of terminal alkynes compared to alkenes and alkanes are stronger. Compared with Ethane which has a pKa of 62 (least acidic) and Ethene of a pKa of 45, Ethyne has a pKa of 26. With alkynes having the sp hybridization, this makes it the most acidic hydrocarbon. Therefore terminal alkynes must be deprotonated by stronger bases. The importance of the s orbital being attracted to the nucleus contributes to the electronegativity
Bonding and Hybridization
Bond Name Location Overlap
Bond 1 s (? bond) bond Formed between 2 sp orbitals of carbon and hydrogen atoms End-on overlap
Bond 2 S (? bond) bond Formed between the 2 sp orbital of 2 unsaturated Carbon atoms. End-on overlap
Bond 3 p-bonds (? bonds) Formed between the 2 p-orbitals among the carbon atoms Side-on overlap
Orbital Name
Location
Orbital 1 sp hybrid orbitals Formed in the linear structure model of carbon atom
Orbital 2 p-orbitals Formed on each carbon
Alkynes are High In Energy
Alkynes are involved in a high release of energy because of repulsion of electrons. The content of energy involved in the alkyne molecule contributes to this high amount of energy. The pi-bonds however, do not encompass a great amount of energy even though the concentration is small within the molecule. The combustion of Ethyne is a major contributor from CO2, water, and the ethyne molecule
??H = -311 kcal/mol
To help understand the relative stabilities of alkyne isomers, heats of hydrogenation must be used. Hydrogenation of the least energy, results in the release of the internal alkyne. With the result of the production of butane, the stability of internal versus terminal alkynes has significant relative stability due to hyperconjugation.
Problems
1. What is the carbon-carbon, carbon-hydrogen bond length for alkyne? Is it shorter or longer than alkane and alkene?
2. Which is the most acidic and most stable, alkane, alkene, or alkyne? And depends on what?
3. How many pi bonds and sigma bonds are involved in the structure of ethyne?
4. Why is the carbon-hydrogen bond so short?
5. What is the alkyne triple bond characterizes by? How is this contribute to the weakness of the pi bonds?
6. How is heat of hydrogenation effects the stability of the alkyne?
Contributors
• Bao Kha Nguyen, Garrett M. Chin
Spectroscopy of the Alkynes
Unlike alkenyl hydrogens, alkynyl hydrogens give rise to shielded hydrogens, or relatively high field chemical shifts for H-NMR when subjected to an external magnetic field. This can be explained by the cylindrical \(pi\) cloud around the carbon-carbon triple bond. Long range coupling is also observed in the alkynes. Infrared spectroscopy is a useful complement to NMR data, and displays characteristic peaks for terminal and internal alkynes.
NMR Absorptions of Alkyne Hydrogens
As discussed before, a carbon-carbon triple bond is the functional characteristic of the alkynes, and protons, or hydrogens, bound to these sp-hybridized carbon atoms resonate at ? = 1.7-3.1 ppm. For example, in the NMR spectrum of 3,3-dimethyl-1-butyne, the terminal hydrogen of the alkyne appears at ? = 2.06 ppm.
3,3-dimethyl-1-butyne.
The H-NMR spectrum of 3,3-dimethyl-1-butyne shows a high field signal due to the alkynyl hydrogen on the terminal alkyne. This high field position suggests a relatively shielded hydrogen, which can be explained by the cylindrical electron cloud around the axis of the molecule. When the two ? bonds are subjected to an external magnetic field, these ? electrons will enter into a cylindrical motion that results in this strong shielding effect with high field chemical shifts, something that is absent in the alkenes. This electron cloud can be seen in the figure below.
Another important aspect to keep in mind for alkyne NMR is the splitting of the peaks, or spin spin coupling. In 1-pentyne, for example, the terminal hydrogen is split by the hydrogens across the triple bond, even though it is separated from them by three carbons. As indicated in the figure below, the three peaks for the terminal hydrogen can be explained by this long range coupling, where the -CH2 group adjacent the sp carbon splits the terminal hydrogen.
Alkynes and Infrared Spectroscopy
Infrared Spectroscopy can be helpful in identifying terminal and internal alkynes.
Terminal Alkyne: Internal Alkyne:
3-chloro-1-propyne 4,4-dichloro-2-pentyne
Internal alkynes, with its sp carbons attached to other carbons, will show weak bands for its triple bond at the 2100-2260 cm-1 region. However, this stretch is relatively weak, and is sometimes not present at all if the internal alkyne is symmetrical. In these cases, the IR spectrum loses its value as a helpful tool. Terminal alkynes, where the sp carbon is attached to a hydrogen, will show bands on the IR spectrum for both its alkynyl hydrogen and its triple bond. The C-H stretch on the terminal alkyne tends to appear as a strong, narrow band in the 3260-3330 cm-1 region while the triple bond shows a weak peak at 2100-2260 cm-1. Additional C-H bends will appear between 610-700 cm-1. The figure below shows these different characteristic stretches for internal and terminal alkynes.
Problems
Based on the NMR and IR provided, what is the structure of C6H10?
Answer: The degree of unsaturation is 2, indicating that the structure contains an alkyne, a triple bond. Because we don't see a peak at the 2100-2260 cm-1 range on the IR spectrum as expected, and a C-H stretch for an internal alkyne at 3260-3330 cm-1 is also absent, we can assume that a symmetrical internal alkyne is present. The NMR spectrum is more useful for this problem, and indicates that there are two equivalent -CH3 groups and two equivalent -CH2 groups. With this information, we get a molecule that looks like this:
hexyne
Therefore, using both the IR and NMR spectrum, we get hexyne.
Contributors
• Rebecca Shragge | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Properties_of_Alkynes/Properties_and_Bonding_in_the_Alkynes.txt |
The reactions of alkynes closely parallel the reactions of alkenes. Addition reactions are typical. In these reactions, the \(\ce{C\bond{#}C}\) triple bond is converted into a \(\ce{C\bond{=}C}\) double bond. The alkenes formed from these reactions can undergo a tautomerization (see the hydration reactions below) or react further to produce alkane derivatives.
Reactivity of Alkynes
A carbon-carbon triple bond may be located at any unbranched site within a carbon chain or at the end of a chain, in which case it is called terminal. Because of its linear configuration (the bond angle of a sp-hybridized carbon is 180º), a ten-membered carbon ring is the smallest that can accommodate this function without excessive strain. Since the most common chemical transformation of a carbon-carbon double bond is an addition reaction, we might expect the same to be true for carbon-carbon triple bonds. Indeed, most of the alkene addition reactions also take place with alkynes, and with similar regio- and stereoselectivity.
When the addition reactions of electrophilic reagents, such as strong Brønsted acids and halogens, to alkynes are studied we find a curious paradox. The reactions are even more exothermic than the additions to alkenes, and yet the rate of addition to alkynes is slower by a factor of 100 to 1000 than addition to equivalently substituted alkenes. The reaction of one equivalent of bromine with 1-penten-4-yne, for example, gave 4,5-dibromo-1-pentyne as the chief product.
HC≡C-CH2-CH=CH2 + Br2 → HC≡C-CH2-CHBrCH2Br
Although these electrophilic additions to alkynes are sluggish, they do take place and generally display Markovnikov Rule regioselectivity and anti-stereoselectivity. One problem, of course, is that the products of these additions are themselves substituted alkenes and can therefore undergo further addition. Because of their high electronegativity, halogen substituents on a double bond act to reduce its nucleophilicity, and thereby decrease the rate of electrophilic addition reactions. Consequently, there is a delicate balance as to whether the product of an initial addition to an alkyne will suffer further addition to a saturated product. Although the initial alkene products can often be isolated and identified, they are commonly present in mixtures of products and may not be obtained in high yield. The following reactions illustrate many of these features. In the last example, 1,2-diodoethene does not suffer further addition inasmuch as vicinal-diiodoalkanes are relatively unstable.
As a rule, electrophilic addition reactions to alkenes and alkynes proceed by initial formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond. Such complexes are formed reversibly and may then reorganize to a reactive intermediate in a slower, rate-determining step. Reactions with alkynes are more sensitive to solvent changes and catalytic influences than are equivalent alkenes. For examples and a discussion of mechanisms click here.
Why are the reactions of alkynes with electrophilic reagents more sluggish than the corresponding reactions of alkenes? After all, addition reactions to alkynes are generally more exothermic than additions to alkenes, and there would seem to be a higher π-electron density about the triple bond ( two π-bonds versus one ). Two factors are significant in explaining this apparent paradox. First, although there are more π-electrons associated with the triple bond, the sp-hybridized carbons exert a strong attraction for these π-electrons, which are consequently bound more tightly to the functional group than are the π-electrons of a double bond. This is seen in the ionization potentials of ethylene and acetylene.
Acetylene HC≡CH + Energy → [HC≡CH •(+) + e(–) ΔH = +264 kcal/mole
Ethylene H2C=CH2 + Energy → [H2C=CH2] •(+) + e(–) ΔH = +244 kcal/mole
Ethane H3C–CH3 + Energy → [H3C–CH3] •(+) + e(–) ΔH = +296 kcal/mole
As defined by the preceding equations, an ionization potential is the minimum energy required to remove an electron from a molecule of a compound. Since pi-electrons are less tightly held than sigma-electrons, we expect the ionization potentials of ethylene and acetylene to be lower than that of ethane, as is the case. Gas-phase proton affinities show the same order, with ethylene being more basic than acetylene, and ethane being less basic than either. Since the initial interaction between an electrophile and an alkene or alkyne is the formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond, the relatively slower reactions of alkynes becomes understandable.
A second factor is presumed to be the stability of the carbocation intermediate generated by sigma-bonding of a proton or other electrophile to one of the triple bond carbon atoms. This intermediate has its positive charge localized on an unsaturated carbon, and such vinyl cations are less stable than their saturated analogs. Indeed, we can modify our earlier ordering of carbocation stability to include these vinyl cations in the manner shown below. It is possible that vinyl cations stabilized by conjugation with an aryl substituent are intermediates in HX addition to alkynes of the type Ar-C≡C-R, but such intermediates are not formed in all alkyne addition reactions.
Carbocation
Stability
CH3(+) RCH=CH(+) < RCH2(+) RCH=CR(+) < R2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) R3C(+)
Methyl 1°-Vinyl 2°-Vinyl 1°-Allyl 1°-Benzyl
Application of the Hammond postulate indicates that the activation energy for the generation of a vinyl cation intermediate would be higher than that for a lower energy intermediate. This is illustrated for alkenes versus alkynes by the following energy diagrams.
Despite these differences, electrophilic additions to alkynes have emerged as exceptionally useful synthetic transforms. For example, addition of HCl, acetic acid and hydrocyanic acid to acetylene give respectively the useful monomers vinyl chloride, vinyl acetate and acrylonitrile, as shown in the following equations. Note that in these and many other similar reactions transition metals, such as copper and mercury salts, are effective catalysts.
HC≡CH + HCl + HgCl2 (on carbon) → H2C=CHCl vinyl chloride
HC≡CCH2Cl + HCl + HgCl2 → H2C=CClCH2Cl 2,3-dichloropropene
HC≡CH + CH3CO2H + HgSO4 → H2C=CHOCOCH3 vinyl acetate
HC≡CH + HCN + Cu2Cl2 → H2C=CHCN acryonitrile
Complexes formed by alkenes and alkynes with transition metals are different from the simple pi-complexes noted above. Here a synergic process involving donation of electrons from a filled π-orbital of the organic ligand into an empty d-orbital of the metal, together with back-donation of electrons from another d-orbital of the metal into the empty π*-antibonding orbital of the ligand. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Addition_by_Electrophilic_Reagents.txt |
Most Hydrogen halide reactions with alkynes occur in a Markovnikov-manner in which the halide attaches to the most substituted carbon since it is the most positively polarized. A more substituted carbon has more bonds attached to 1) carbons or 2) electron-donating groups such as Fluorine and other halides. However, there are two specific reactions among alkynes where anti-Markovnikov reactions take place: the radical addition of HBr and Hydroboration Oxidation reactions. For alkynes, an anti-Markovnikov addition takes place on a terminal alkyne, an alkyne on the end of a chain.
HBr Addition With Radical Yields 1-bromoalkene
The Br of the Hydrogen Bromide (H-Br) attaches to the less substituted 1-carbon of the terminal alkyne shown below in an anti-Markovnikov manner while the Hydrogen proton attaches to the second carbon. As mentioned above, the first carbon is the less substituted carbon since it has fewer bonds attached to carbons and other substituents. The H-Br reagent must also be reacted with heat or some other radicial initiator such as a peroxide in order for this reaction to proceed in this manner. This presence of the radical or heat leads to the anti-Markovnikov addition since it produces the most stable reaction. For more on Anti-Markovnikov additions:Radical Additions--Anti-Markovnikov Product Formation
The product of a terminal alkyne that is reacted with a peroxide (or light) and H-Br is a 1-bromoalkene.
Regioselectivity: The Bromine can attach in a syn or anti manner which means the resulting alkene can be both cis and trans. Syn addition is when both Hydrogens attach to the same face or side of the double bond (i.e. cis) while the anti addition is when they attach on opposite sides of the bond (trans).
Stereospecific Hydroboration Oxidation of Alkynes
Step 1: Hydroboration of terminal alkynes reacts in an anti-Markovnikov fashion in which the Boron attacks the less substituted carbon which is the least hindered. It is a stereospecific reaction where syn addition is observed as the hydroboration occurs on the same side of the alkyne and results in cis stereochemistry. However, a bulky borane reagent needs to be used to stop at the alkenyl-borane stage. Otherwise, a second hydroboration will occur. In this example, diisoamyl borane or HBsia2, a fairly large and sterically-hindered borane reagent, is used. Both of the alkyne's pi bonds will undergo hydroboration if BH3 (borane) is used by itself.
Step 2: Oxidation is the next step that occurs. The resulting alkylborane is oxidized to a vinyl alcohol due to the reaction with a hydroxide in a basic solution such as aqueous sodium hydroxide. A vinyl alcohol is an alcohol that has both an alkene and an -OH group. After the vinyl alcohol is formed, tautomerization takes place. Tautomerism is the interconversion of isomeric compounds due to the migration of a proton. The alcohol, which in this case is a terminal enol, spontaneously and rapidly rearranges due to tautomerism to become an aldehyde since the aldehyde form is much more stable than the enol.
For additional information on hydroboration oxidation: Hydroboration Oxidation
Problems
1. What is the product of this reaction?
2. What process causes the conversion of a vinyl alcohol to an aldeyde and what are some of its distinct features?
3. True or False: Only cis products are observed in radical H-Br additions to terminal alkynes.
4. Explain why a bulky borane reagent is necessary for hydroboration oxidation reactions.
5. Draw the product.
Answers
1. Don't be confused by the borane reagent! Just remember, anytime there is a bulky borane reagent reacting with a terminal alkyne, the hydroboration oxidation reaction will occur and be proceeded by tautomerism which will produce an aldeyde as shown below.
2. The interconversion of an enol or vinyl alcohol to an aldehyde is called tautomerism and it is very distinct since it is a spontaneous reaction that proceeds very quickly.
3. False. Both cis and trans products are produced as both syn and anti additions are observed.
4. If a small borane reagent is utilized, both pi bonds will be used and a second hydroboration will occur. This will break the double bond of the alkene and the aldehyde product will not be formed.
5.
• Ali Alvandi | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Anti-Markovnikov_Additions_to_Triple_Bonds.txt |
The catalytic addition of hydrogen to 2-butyne not only serves as an example of such an addition reaction, but also provides heat of reaction data that reflect the relative thermodynamic stabilities of these hydrocarbons, as shown in the diagram below.
From the heats of hydrogenation, shown in blue in units of kcal/mole, it would appear that alkynes are thermodynamically less stable than alkenes to a greater degree than alkenes are less stable than alkanes. The standard bond energies for carbon-carbon bonds confirm this conclusion. Thus, a double bond is stronger than a single bond, but not twice as strong. The difference ( 63 kcal/mole ) may be regarded as the strength of the π-bond component. Similarly, a triple bond is stronger than a double bond, but not 50% stronger. Here the difference ( 54 kcal/mole ) may be taken as the strength of the second π-bond. The 9 kcal/mole weakening of this second π-bond is reflected in the heat of hydrogenation numbers ( 36.7 - 28.3 = 8.4 ).
Since alkynes are thermodynamically less stable than alkenes, we might expect addition reactions of the former to be more exothermic and relatively faster than equivalent reactions of the latter. In the case of catalytic hydrogenation, the usual Pt and Pd hydrogenation catalysts are so effective in promoting addition of hydrogen to both double and triple carbon-carbon bonds that the alkene intermediate formed by hydrogen addition to an alkyne cannot be isolated. A less efficient catalyst, Lindlar's catalyst, prepared by deactivating (or poisoning) a conventional palladium catalyst by treating it with lead acetate and quinoline, permits alkynes to be converted to alkenes without further reduction to an alkane. The addition of hydrogen is stereoselectively syn (e.g. 2-butyne gives cis-2-butene). A complementary stereoselective reduction in the anti mode may be accomplished by a solution of sodium in liquid ammonia. This reaction will be discussed later in this section.
R-C≡C-R + H2 & Lindlar catalyst → cis R-CH=CH-R
R-C≡C-R + 2 Na in NH3 (liq) → trans R-CH=CH-R + 2 NaNH2
Alkenes and alkynes show a curious difference in behavior toward catalytic hydrogenation. Independent studies of hydrogenation rates for each class indicate that alkenes react more rapidly than alkynes. However, careful hydrogenation of an alkyne proceeds exclusively to the alkene until the former is consumed, at which point the product alkene is very rapidly hydrogenated to an alkane. This behavior is nicely explained by differences in the stages of the hydrogenation reaction. Before hydrogen can add to a multiple bond the alkene or alkyne must be adsorbed on the catalyst surface. In this respect, the formation of stable platinum (and palladium) complexes with alkenes has been described earlier. Since alkynes adsorb more strongly to such catalytic surfaces than do alkenes, they preferentially occupy reactive sites on the catalyst. Subsequent transfer of hydrogen to the adsorbed alkyne proceeds slowly, relative to the corresponding hydrogen transfer to an adsorbed alkene molecule. Consequently, reduction of triple bonds occurs selectively at a moderate rate, followed by rapid addition of hydrogen to the alkene product. The Lindlar catalyst permits adsorption and reduction of alkynes, but does not adsorb alkenes sufficiently to allow their reduction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Catalytic_Hydrogenation.txt |
Addition of strong acids to alkynes is quite similar to the addition of strong acids to alkenes. The regiochemistry follows Markovnikov’s Rule, but the stereochemistry is often different. Addition to alkenes is usually not stereospecific, whereas alkynes usually undergo anti addition.
Introduction
Strong acids usually protonate alkenes on their less substituted carbon thus developing a positive charge on the more substituted carbon of the reacting double bond. This intermediate carbocation may rearrange or simply pick up a nucleophile to form the expected addition product. The situation with alkynes is different. Similar protonation of the $\pi$ bond of an alkyne would produce a very unstable vinyl cation so this does not happen in most cases. A vinyl cation can only have one alkyl group adjacent to the vacant p-orbital to stabilize its positive charge so the stability of the vinyl cation is similar to the mono-substituted primary alkyl cation (See below). As a result, a strong acid can form a $\pi$ complex with the alkyne $\pi$ bond but proton transfer to the alkyne is extremely difficult. The strong acid, however, does polarize the $\pi$ bond developing positive charge at both carbons of the $\pi$ bond. The more substituted carbon holds most of the positive charge and becomes the site of attack by any nucleophile present in solution. The nucleophile bonds to the more substituted carbon allowing the acid to protonate the less substituted carbon of the $\pi$ bond.
In nucleophilic addition reactions, a nucleophile attacks an electron poor species. Now we will take a look at electrophilic addition reactions, particularly of alkynes. The reaction mechanisms, as you will notice, are quite similar to the electrophilic addition reactions of alkenes. The triple bonds of alkynes, because of its high electron density, are easily attacked by electrophiles, but less reactive than alkenes due to the compact C-C electron cloud.As with electrophilic addition to unsymmetrical alkenes, the Markovnikov rule is followed, adding the electrophile to the less substituted carbon. Here we will go through the following reactions listed below to learn the mechanisms behind these electrophilic additions of alkynes: (1) HX Addition to Alkenes, (2) Halogenation of Alkynes and (3) Hydration of Alkynes.
Figure 1
The addition of an electrophile to either an alkene or an alkyne will undergo the same steps listed below.
1. Start with a reactant (either an alkene or an alkyne), which has $\pi$ electrons. An electron pair moves from the $\pi$ bond to the electrophilic proton to form a new covalent bond. A carbocation intermediate forms on the most stable carbon. As shown in the diagram above, the intermediate consists of hydrogen covalently bonded to the carbon and a positive charge on the other carbon.
2. The addition of the halide ion to the positive charged intermediate forms the second covalent bond giving you the haloalkane product (as seen on the left of the above diagram) and a haloalkene product (as seen on the right of the above diagram). The halide ion acts as a nucleophile, which is attracted to the positive charged carbon because it has electrons to donate or share. So when the nucleophilic halide (represented as X-) attacks, it aims for the positively charged carbon resulting in that second covalent bond as seen above.
Make Note: The way to determine where the addition of the proton and the halide takes place is based on Markovnikov’s Rule, which states that the proton adds onto the carbon with the most hydrogens and the halogen prefers the most substituted carbon.
Now let's take a look at our first reaction, the addition of Hydrogen Halide.
Reaction 1: Addition of Hydrogen Halide to an Alkyne
Summary: Reactivity order of hydrogen halides: HI > HB r> HCl > HF.
Follows Markovnikov’s rule:
• Hydrogen adds to the carbon with the greatest number of hydrogens, the halogen adds to the carbon with fewest hydrogens.
• Protination occurs on the more stable carbocation. With the addition of HX, haloalkenes form.
• With the addition of excess HX, you get anti addition forming a geminal dihaloalkane.
Addition of a HX to an Internal Alkyne
As described in Figure 1, the $\pi$ electrons will attack the hydrogen of the HBr and because this is a symmetric molecule it does not matter which carbon it adds to, but in an asymmetric molecule the hydrogen will covalently bond to the carbon with the most hydrogens. Once the hydrogen is covalently bonded to one of the carbons, you will get a carbocation intermediate (not shown, but will look the same as depicted in Figure 1) on the other carbon. Again, this is a symmetric molecule and if it were asymmetric, which carbon would have the positive charge?
The final step is the addition of the Bromine, which is a good nucleophile because it has electrons to donate or share. Bromine, therefore attacks the carbocation intermediate placing it on the highly substituted carbon. As a result, you get 2-bromobutene from your 2-butyne reactant, as shown below.
Now, what if you have excess HBr?
Addition due to excess HX present ? yields a geminal dihaloalkane
Here, the electrophilic addition proceeds with the same steps used to achieve the product in Addition of a HX to an Internal Alkyne. The $\pi$ electrons attacked the hydrogen, adding it to the carbon on the left (shown in blue). Why was hydrogen added to the carbon on left and the one on the right bonded to the Bromine?
Now, you will have your carbocation intermediate, which is followed by the attack of the Bromine to the carbon on the right resulting in a haloalkane product.
Addition of HX to Terminal Alkyne
• Here is an addition of HBr to an asymmetric molecule.
• First, try to make sense of how the reactant went to product and then take a look at the mechanism.
The $\pi$ electrons are attacking the hydrogen, depicted by the electron pushing arrows and the Bromine gains a negative charge. The carbocation intermediate forms a positive charge on the left carbon after the hydrogen was added to the carbon with the most hydrogen substituents.
The Bromine, which has a negative charge, attacks the positively charged carbocation forming the final product with the nucleophile on the more substituted carbon.
Reaction: Halogenation of Alkynes
Summary:
• Stereoslectivity: anti addition
• Reaction proceeds via cyclic halonium ion
Addition of Br2
• The addition of Br2 to an alkyne is analogous to adding Br2 to an alkene.
• Once Br2 approaches the nucleophilic alkyne, it becomes polarized.
• The $\pi$ electrons, from the triple bond, can now attack the polarized bromine forming a C-Br bond and displacing the bromine ion.
• Now, you will get an intermediate electrophilic carbocation, which will immediately react with the bromine ion giving you the dibromo product.
First, you see the polarized Br2 being attacked by the $\pi$ electrons. Once you form the C-Br bond, the other bromine is released as a bromine ion. The intermediate here is a bromonium ion, which is electrophilic and reacts with the bromine ion giving you the dibromo product.
Reaction: Hydration of Alkynes
Summary: With the addition of water, alkynes can be hydrated to form enols that spontaneously tautomerize to ketones. Reaction is catalyzed by mercury ions. Follows Markovnikov’s Rule: Terminal alkynes give methyl ketones
• The first step is an acid/base reaction where the $\pi$ electrons of the triple bond acts as a Lewis base and attacks the proton therefore protinating the carbon with the most hydrogen substituents.
• The second step is the attack of the nucleophilic water molecule on the electrophilic carbocation, which creates an oxonium ion.
• Next you deprotonate by a base, generating an alcohol called an enol, which then tautomerizes into a ketone.
• Tautomerism is a simultaneous proton and double bond shift, which goes from the enol form to the keto isomer form as shown above in Figure 7.
Now let's look at some Hydration Reactions.
Hydration of Terminal Alkyne produces methyl ketones
Just as described in Figure 7 the $\pi$ electrons will attack a proton, forming a carbocation, which then gets attacked by the nucleophilic water molecules. After deprotination, we generate an enol, which then tautomerizes into the ketone form shown.
Hydration of Alkyne
As you can see here, the $\pi$ electrons of the triple bond are attacking the proton, which forms a covalent bond on the carbon with the most hydrogen substituents. Once the hydrogen is bound you have a carbocation, which gets attacked by the water molecule. Now you have a positive charge on the oxygen which results in a base coming in and deprotinating the molecule. Once deprotonated, you have an enol, which then gets tautomerized.
Tautomerism is shown here when the proton gets attacked by the double bond $\pi$ electrons forming a covalent bond between the carbon and the hydrogen on the less substituted carbon. Electrons from the Oxygen end up moving to the carbon, forming a double bond with carbon and giving itself a positive charge, which then gets attacked by the base. The base deprotonates the oxygen resulting in the more stable final product at equilibrium, which is a ketone.
Practice Problems
1) Reaction: Addition of Hydrogen Halide to an Alkyne
1a) What product would result from the reaction of 1-Pentyne with Hydrogen Bromide?
1b) Take your answer from 1a and add an excess of Hydrogen Bromide. What would be the product?
1c) What is the mechanism for this reaction?
2) Reaction: Addition of X2 to Alkynes
2a) What product would result from the reaction of 1-Pentyne with Br2?
2b) Take your answer from 2a and add an excess of Br2. What would be the product?
2c) What is the mechanism for this reaction?
3) Reaction: Hydration of Alkynes
3a) If 1-Pentyne were to react with mercury(I) sulfate, water, and sulfuric acid, what would be the product?
4) What is the product when 3-methylbutyne reacts with HCl? Include in your answer:
a) the reaction mechanism
b) a clear explanation for why the hydrogen and chlorine bind to where they do
5) a) Draw the enol form of the Ketone below.
b) Draw the Keto form from the enol below. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Electrophilic_Addition_Reactions_of_Alkynes.txt |
As with alkenes, the addition of water to alkynes requires a strong acid, usually sulfuric acid, and is facilitated by mercuric sulfate. However, unlike the additions to double bonds which give alcohol products, addition of water to alkynes gives ketone products ( except for acetylene which yields acetaldehyde ).
keto-enol Tautomerization
The explanation for this deviation lies in enol-keto tautomerization, illustrated by the following equation. The initial product from the addition of water to an alkyne is an enol (a compound having a hydroxyl substituent attached to a double-bond), and this immediately rearranges to the more stable keto tautomer.
Tautomers are defined as rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers ( acetone, for example, is 99.999% keto tautomer ). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. The three examples shown below illustrate these reactions for different substitutions of the triple-bond. The tautomerization step is indicated by a red arrow. For terminal alkynes the addition of water follows the Markovnikov rule, as in the second example below, and the final product ia a methyl ketone ( except for acetylene, shown in the first example ). For internal alkynes ( the triple-bond is within a longer chain ) the addition of water is not regioselective. If the triple-bond is not symmetrically located ( i.e. if R & R' in the third equation are not the same ) two isomeric ketones will be formed.
HC≡CH + H2O + HgSO4 & H2SO4 → [ H2C=CHOH ] H3C-CH=O
RC≡CH + H2O + HgSO4 & H2SO4 → [ RC(OH)=CH2 ] RC(=O)CH3
RC≡CR' + H2O + HgSO4 & H2SO4 → [ RHC=C(OH)R' + RC(OH)=CHR' ] RCH2-C(=O)R' + RC(=O)-CH2R'
Two factors have an important influence on the enol-keto tautomerizations described here. The first is the potential energy difference between the tautomeric isomers. This factor determines the position of the equilibrium state. The second factor is the activation energy for the interconversion of one tautomer to the other. This factor determines the rate of rearrangement. Since the potential energy or stability of a compound is in large part a function of its covalent bond energies, we can estimate the relative energy of keto and enol tautomers by considering the bonds that are changed in the rearrangement. From the following diagram, we see that only three significant changes occur, and the standard bond energies for those changes are given to the right of the equation. The keto tautomer has a 17.5 kcal/mole advantage in bond energy, so its predominance at equilibrium is expected.
The rapidity with which enol-keto tautomerization occurs suggests that the activation energy for this process is low. We have noted that the rearrangement is acid & base catalyzed, and very careful experiments have shown that interconversion of tautomers is much slower if such catalysts are absent. A striking example of the influence of activation energy on such transformations may be seen in the following hypothetical rearrangement. Here we have substituted a methyl group (colored maroon) for the proton of a conventional tautomerism, and the methyl shifts from oxygen to carbon just as the proton does in going from an enol to a ketone.
H2C=CH-O-CH3 X> CH3-CH2-CH=O
The potential energy change for this rearrangement is even more advantageous than for enol-keto tautomerism, being estimated at over 25 kcal/mole from bond energy changes. Despite this thermodynamic driving force, the enol ether described above is completely stable to base treatment, and undergoes rapid acid-catalyzed hydrolysis with loss of methanol, rather than rearrangement. The controlling difference in this case must be a prohibitively high activation energy for the described rearrangement, combined with lower energy alternative reaction paths. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Hydration_of_Alkynes_and_Tautomerism.txt |
Hydroboration Reactions
Diborane reacts readily with alkynes, but the formation of substituted alkene products leaves open the possibility of a second addition reaction. A clever technique for avoiding this event takes advantage of the fact that alkynes do not generally suffer from steric hindrance near the triple-bond (the configuration of this functional group is linear). Consequently, large or bulky electrophilic reagents add easily to the triple-bond, but the resulting alkene is necessarily more crowded or sterically hindered and resists further additions. The bulky hydroboration reagent needed for this strategy is prepared by reaction of diborane with 2-methyl-2-butene, a highly branched alkene. Because of the alkyl branching, only two alkenes add to a BH3 moiety (steric hindrance again), leaving one B-H covalent bond available for reaction with an alkyne, as shown below. The resulting dialkyl borane is called disiamylborane, a contraction of di-secondary-isoamylborane (amyl is an old name for pentyl).
2 (CH3)2C=CHCH3 + BH3 in ether ——> [ (CH3)2CH-CH(CH3) ]2B-H disiamylborane
An important application of disiamylborane is its addition reaction to terminal alkynes. As with alkenes, the B-H reagent group adds in an apparently anti-Markovnikov manner, due to the fact that the boron is the electrophile, not the hydrogen. Further addition to the resulting boron-substituted alkene does not occur, and the usual oxidative removal of boron by alkaline hydrogen peroxide gives an enol which rapidly rearranges to the aldehyde tautomer. Thus, by the proper choice of reagents, terminal alkynes may be converted either to methyl ketones (mercuric ion catalyzed hydration) or aldehydes (hydroboration followed by oxidation).
RC≡CH + (C5H11)2B-H ——> [ RCH=CH-B(C5H11)2 ] + H2O2 & NaOH ——> [ RCH=CH-OH ] ——> RCH2-CH=O
Hydroboration of internal alkynes is not a particularly useful procedure because a mixture of products will often be obtained, unless the triple-bond is symmetrically substituted. Mercuric ion catalyzed hydration gives similar results.
Oxidations
Reactions of alkynes with oxidizing agents such as potassium permanganate and ozone usually result in cleavage of the triple-bond to give carboxylic acid products. A general equation for this kind of transformation follows. The symbol [O] is often used in a general way to denote an oxidation.
RC≡CR' + [O] ——> RCO2H + R'CO2H
Nucleophilic Addition Reactions and Reduction
The sp-hybrid carbon atoms of the triple-bond render alkynes more electrophilic than similarly substituted alkenes. As a result, alkynes sometimes undergo addition reactions initiated by bonding to a nucleophile. This mode of reaction, illustrated below, is generally not displayed by alkenes, unless the double-bond is activated by electronegative substituents, e.g. F2C=CF2, or by conjugation with an electron withdrawing group.
HC≡CH + KOC2H5 in C2H5OH at 150 ºC → H2C=CH-OC2H5
HC≡CH + HCN + NaCN (catalytic) → H2C=CH-CN
The smallest and most reactive nucleophilic species is probably an electron. Electron addition to a functional group is by definition a reduction, and we noted earlier that alkynes are reduced by solutions of sodium in liquid ammonia to trans-alkenes. To understand how this reduction occurs we first need to identify two distinct reactions of sodium with liquid ammonia (boiling point -78 ºC). In the first, sodium dissolves in the pure liquid to give a deep blue solution consisting of very mobile and loosely bound electrons together with solvated sodium cations (first equation below). For practical purposes, we can consider such solutions to be a source of "free electrons" which may be used as powerful reducing agents. In the second case, ferric salts catalyze the reaction of sodium with ammonia, liberating hydrogen and forming the colorless salt sodium amide (second equation). This is analogous to the reaction of sodium with water to give sodium hydroxide, but since ammonia is 1018 times weaker an acid than water, the reaction is less violent. The usefulness of this reaction is that sodium amide, NaNH2, is an exceedingly strong base (18 powers of ten stronger than sodium hydroxide), which may be used to convert very weak acids into their conjugate bases.
Na + NH3 (liquid, –78 ºC ) → Na(+) + e(–) (a blue solution)
Na + NH3 (liquid, –78 ºC ) + Fe → H2 + NaNH2 (a colorless solution)
Returning to the reducing capability of the blue electron solutions, we can write a plausible mechanism for the reduction of alkynes to trans-alkenes, as shown below. Isolated carbon double-bonds are not reduced by sodium in liquid ammonia, confirming the electronegativity difference between sp and sp2 hybridized carbons. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Hydroboration_Reactions_and_Oxidations.txt |
Conjugate base anions of terminal alkynes (acetylide anions) are nucleophiles, and can do both nucleophilic substitution and nucleophilic addition reactions.
Acidity of Terminal Alkynes: Formation of Acetylide Anions
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion. The origin of the enhanced acidity can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the homolytic C-H BDE is larger.
Compound Conjugate Base Hybridization "s Character" pKa C-H BDE (kJ/mol)
CH3CH3 CH3CH2- sp3 25% 50 410
CH2CH2 CH2CH- sp2 33% 44 473
HCCH HCC- sp 50% 25 523
Consequently, acetylide anions can be readily formed by deprotonation using a sufficiently strong base. Amide anion (NH2-), in the form of NaNH2 is commonly used for the formation of acetylide anions.
Nucleophilic Substitution Reactions of Acetylides
Acetylide anions are strong bases and strong nucleophiles. Therefore, they are able to displace halides and other leaving groups in substitution reactions. The product is a substituted alkyne.
Because the ion is a very strong base, the substitution reaction is most efficient with methyl or primary halides without substitution near the reaction center,
Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.
Nucleophilic Addition of Acetylides to Carbonyls
Acetylide anions will add to aldehydes and ketones to form alkoxides, which, upon protonation, give propargyl alcohols.
With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers.
The triple bond in the propargyl alcohol can be modified by using the reactivity of the alkyne. For example, Markovnikov and anti-Markovnikov hydration of the triple bond leads to formation of the hydroxy-substituted ketone and aldehyde, respectively, after enol-keto tautomerization.
Problems
1. The pKa of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by amide, as shown above.
Answers
1. Assuming the pKa of pent-1-yne is about 25, then the difference in pKas is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 1010
Ozonolysis of Alkenes and Alkynes
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis.
Introduction
The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the Carbon-Carbon double bond and is replaced by a Carbon-Oxygen double bond instead.
Reaction Mechanism
Step 1:
The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the Carbon-Carbon double bond, which then form the molozonide intermediate. Due to the unstable molozonide molecule, it continues further with the reaction and breaks apart to form a carbonyl and a carbonyl oxide molecule.
Step 2:
The carbonyl and the carbonyl oxide rearranges itself and reforms to create the stable ozonide intermediate. A reductive workup could then be performed to convert convert the ozonide molecule into the desired carbonyl products.
Reducing Alkynes-The Reactivity of the Two Bonds
Reactions between alkynes and catalysts are a common source of alkene formation. Because alkynes differ from alkenes on account of their two procurable π bonds, alkynes are more susceptible to additions. Aside from turning them into alkenes, these catalysts affect the arrangement of substituents on the newly formed alkene molecule. Depending on which catalyst is used, the catalysts cause anti- or syn-addition of hydrogens. Alkynes can readily undergo additions because of their availability of two π bonds.
Hydrogenation of an Alkyne
Alkynes can be fully hydrogenated into alkanes with the help of a platinum catalyst. However, the use of two other catalysts can be used to hydrogenate alkynes to alkanes. These catalysts are: Palladium dispersed on carbon (Pd/C) and finely dispersed nickel (Raney-Ni).
Hydrogenation of an Alkyne to a Cis-Alkene
Because hydrogenation is an interruptible process involving a series of steps, hydrogenation can be stopped, using modified catalysts (e.g., Lindlar’s Catalyst) at the transitional alkene stage. Lindar’s catalyst has three components: Palladium-Calcium Carbonate, lead acetate and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis-alkene.
Hydrogenation of an Alkyne to a Trans-Alkene
Alkynes can be reduced to trans-alkenes with the use of sodium dissolved in an ammonia solvent. An Na radical donates an electron to one of the P bonds in a carbon-carbon triple bond. This forms an anion, which can be protonated by a hydrogen in an ammonia solvent. This prompts another Na radical to donate an electron to the second P orbital. Soon after this anion is also protonated by a hydrogen from the ammonia solvent, resulting in a trans-alkene.
Contributors
• Ravjot Takhar (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Nucleophilic_Reactivity_of_Deprotonated_Alkynes.txt |
Alkynes can be a useful functional group to synthesize due to some of their antibacterial, antiparasitic, and antifungal properties. One simple method for alkyne synthesis is by double elimination from a dihaloalkane.
Introduction
One case in which elimination can occur is when a haloalkane is put in contact with a nucleophile. The table below is used to determine which situations will result in elimination and the formation of a $\pi$ bond.
Elimination: Haloalkane-Nucleophile Reaction. Empty Box means no elimination or $\pi$ bond forms
Type of Haloalkane Weak Base, Poor Nucleophile Weak Base, Good Nucleophile Strong, Unhindered Base Strong, Hindered Base
Primary
Unhindered E2
Branched E2 E2
Secondary E1 E2 E2
Tertiary E1 E1 E2 E2
To synthesize alkynes from dihaloalkanes we use dehydrohalogenation. The majority of these reactions take place using alkoxide bases (other strong bases can also be used) with high temperatures. This combination results in the majority of the product being from the E2 mechanism.
E2 Mechanism
Recall that the E2 mechanism is a concerted reaction (occurs in 1 step). However, in this 1 step there are 3 different changes in the molecule. This is the reaction between 2-Bromo-2-methylpropane and sodium hyrdroxide.
Figure: Step 1: the base (blue) will deprotonate the haloalkane; Step 2: the leaving group (red) will depart from the molecule; Step 3: the deprotonated carbon will rehybridize from $sp^3$ to $sp^2$ via the formation of a $pi$ bond.
This is a brief review of the E2 reaction. For further information on why the reaction proceeds as it does visit the E2 reaction page. Now, if we apply this concept using 2 halides on vicinal or geminal carbons, the E2 reaction will take place twice resulting in the formation of 2 Pi bonds and thus an Alkyne.
Dihaloalkane Elimination
This is a general picture of the reaction taking place without any of the mechanisms shown.
or
* With a terminal haloalkane the equation above is modified in that 3 equivalents of base will be used instead of 2.
Lets look at the mechanism of a reaction between 2,3-Dibromopentane with sodium amide in liquid ammonia.
• Liquid ammonia is not part of the reaction, but is used as a solvent
• Notice the intermediate of the alkyne synthesis. It is stereospecifically in its anti form. Because the second proton and halogen are pulled off the molecule this is unimportant to the synthesis of alkynes. For more information on this see the page on preparation of alkenes from haloalkanes.
Preparation of Alkynes from Alkeneshalogenation of an alkene
Lastly, we will briefly look at how to prepare alkynes from alkenes. This is a simple process using first halogenation of the alkene bond to form the dihaloalkane, and next, using the double elimination process to protonate the alkane and from the 2 Pi bonds.
This first process is gone over in much greater detail in the page on halogenation of an alkene. In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is the reactant needed to produce the alkyene using double elimination, as covered previously on this page.
In The Lab
Due to the strong base and high temperatures needed for this reaction to take place, the triple bond may change positions. An example of this is when reactants that should form a terminal alkyne, form a 2-Alkyne instead. The use of NaNH2 in liquid NH3 is used in order to prevent this from happening due to its lower reacting temperature. Even so, most chemists will prefer to use nucleophilic substitution instead of elimination when trying to form a terminal alkyne.
Questions
1. Why would we need 3 bases for every terminal dihaloalkane instead of 2 in order to form an alkyne?
1. What are the major products of the following reactions:
1. 1,2-Dibromopentane with sodium amide in liquid ammonia
2. 1-Pentene first with Br2 and chloromethane, followed by sodium ethoxide (Na+ -O-CH2CH3)
2. What would be good starting molecules for the synthesis of the following molecules:
1. Use a 6 carbon diene to synthesize a 6 carbon molecule with 2 terminal alkynes.
Answers
Answer 1: Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want.
Answer 2:
a.) 1-Pentyne
b.) 1-Pentyne
Answer 3:
Answer 4: Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Synthesis_of_Alkynes/Preparation_of_Alkynes_by_Double_Elimination.txt |
Primary amides
Primary amides are named by changing the name of the acid by dropping the -oic acid or -ic acid endings and adding -amide. The carbonyl carbon is given the #1 location number. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
methanamide or formamide (left), ethanamide or acetamide (center) , benzamide (right)
Exercise
Draw a structure for the following compound: 3-chlorobenzamide.
Exercise
Try to name the following compound:
pentamide
Secondary Amides
Secondary amides are named by using an upper case N to designate that the alkyl group is on the nitrogen atom. Alkyl groups attached to the nitrogen are named as substituents. The letter N is used to indicate they are attached to the nitrogen.
N-methylpropanamide
Exercise
Try to draw a structure for the following compound: N,N-dimethylformamide.
Exercise
Try to name the following compound:
Solution
N-phenylethanamide or N-phenylacetamide
Tertiary Amides
Tertiary amides are named in the same way as secondary amides, but with two N's
Exercise
Try to draw a structure for N,N-dimethylformamide.
Amides Background
Amides are neutral compounds -- in contrast to their seemingly close relatives, the amines, which are basic. The amide linkage is planar -- even though we normally show the C-N connected by a single bond, which should provide free rotation.
Properties of Amides
This page explains what amides are and looks at their simple physical properties such as solubility and melting points.
Amides are derived from carboxylic acids. A carboxylic acid contains the -COOH group, and in an amide the -OH part of that group is replaced by an -NH2 group. So, amides contain the -CONH2 group. The most commonly discussed amide is ethanamide, CH3CONH2 (old name: acetamide).
The three simplest amides are:
HCONH2 methanamide
CH3CONH2 ethanamide
CH3CH2CONH2 propanamide
Notice that in each case, the name is derived from the acid by replacing the "oic acid" ending by "amide". If the chain were branched, the carbon in the -CONH2 group counts as the number 1 carbon atom. For example:
Physical properties
Melting points
Methanamide is a liquid at room temperature (melting point: 3°C), but the other amides are solid. For example, ethanamide forms colorless deliquescent crystals with a melting point of 82°C. A deliquescent substance absorbs water from the atmosphere and then dissolves in it. Ethanamide crystals nearly always look wet.
The melting points of the amides are high for the size of the molecules because they can form hydrogen bonds. The hydrogen atoms in the -NH2 group are sufficiently positive to form a hydrogen bond with a lone pair on the oxygen atom of another molecule.
As you can see, there is the potential for lots of hydrogen bonds to be formed. Each molecule has two slightly positive hydrogen atoms and two lone pairs on the oxygen atom. These hydrogen bonds need a reasonable amount of energy to break, and so the melting points of the amides are quite high.
Solubility in water
The small amides are soluble in water because they have the ability to hydrogen bond with the water molecules. It needs energy to break the hydrogen bonds between amide molecules and between water molecules before they can mix - but enough energy is released again when the new hydrogen bonds are set up to allow this to happen.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Nomenclature_of_Amides.txt |
The important classes of organic compounds known as alcohols, phenols, ethers, amines and halides consist of alkyl and/or aryl groups bonded to hydroxyl, alkoxyl, amino and halo substituents respectively. If these same functional groups are attached to an acyl group (RCO–) their properties are substantially changed, and they are designated as carboxylic acid derivatives. Carboxylic acids have a hydroxyl group bonded to an acyl group, and their functional derivatives are prepared by replacement of the hydroxyl group with substituents, such as halo, alkoxyl, amino and acyloxy.
The following table lists some representative derivatives and their boiling points. An aldehyde and ketone of equivalent molecular weight are also listed for comparison. Boiling points are given for 760 torr (atmospheric pressure), and those listed as a range are estimated from values obtained at lower pressures. The relatively high boiling point of carboxylic acids is due to extensive hydrogen bonded dimerization. Similar hydrogen bonding occurs between molecules of 1º and 2º-amides (amides having at least one N–H bond), and the first three compounds in the table serve as hydrogen bonding examples.
The last nine entries in the above table cannot function as hydrogen bond donors, so hydrogen bonded dimers and aggregates are not possible. The relatively high boiling points of equivalent 3º-amides and nitriles are probably due to the high polarity of these functions. Indeed, if hydrogen bonding is not present, the boiling points of comparable sized compounds correlate reasonably well with their dipole moments.
Structure of Amides
Amides are neutral compounds -- in contrast to their seemingly close relatives, the amines, which are basic. The amide linkage is planar -- even though we normally show the C-N connected by a single bond, which should provide free rotation. Figure 1 below shows this common drawing of an amide.
Figure 1. An amide; usual representation. The amide shown here, and in Figure 2, is the primary amide from ethanoic acid (acetic acid); the amide is called ethanamide (acetamide).
To help understand these properties, we need to look at a more complex -- but better -- representation of the amide structure. This is shown in Figure 2:
Figure 2. Resonance structures for an amide. Remember that the molecule does not actually switch between these structures. Instead, the actual structure is somewhere in between the structures shown. It can be thought of as some average of these structures.
Why is this resonance system better? A qualitative argument is that the O, which is very electronegative, draws electrons toward it. In this case, it draws electrons from the lone pair of the N. Note that in the right hand form, the electrons of the N lone pair have moved in to the double bond (giving the N a + charge), and electrons of the C=O double bond have moved out to the O (giving it a - charge).
The resonance system shown in Figure 2 is based on measurements of the properties of amides. That is, detailed study of amides shows that the properties are better explained by Figure 2 than by Figure 1. As examples:
• The bond length measured for amides is about half way between that typical for C-N single bonds and C=N double bonds. This is easily explained by the resonance system shown in Figure 2, which suggests that the actual bond between C and N is about a 1 1/2 bond.
• A double bonded structure, or a structure with a substantial contribution of double bonding, would be expected to be planar, without free rotation about the C-N bond. This fits with observation.
• The left hand structure in Figure 2 might look like it would accept an H+ on the N, thus acting as a base. However, the right hand structure has no lone pair, and even has a positive charge on the N. These features argue against the N being basic. A resonance system with a substantial contribution of the right hand structure would not be expected to be basic. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Properties_of_Amides/Physical_Properties_of_Carboxylic_Acid_Derivatives.txt |
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.
Reactivity of Amides
1o Amides can be converted to nitriles by reaction by dehydration with thionyl chloride.
Example 1:
Mechanism
1) Nucleophilic attach on thionyl chloride
2) Leaving group removal
3) Deprotonation
4) Leaving group removal
Contributors
Prof. Steven Farmer (Sonoma State University)
Amide to nitrile reduction mechanism
Using cyanuric chloride
Notes
• Conditions are relatively mild: dimethylformamide at room temperature.
• Compatible with most sensitive functionalities.
• Only one third molar equivalent of cyanuric chloride is needed to go to completion.
• Cyanuric chloride should be recrystallized before use.
• The excess cyanuric chloride and the cyanuric acid by-product can be removed by 5% aq.sodium bicarbonate washings.
Conversion of Amides into Amines with LiAlH4
Introduction
Amides can be converted to 1°, 2° or 3° amines using LiAlH4.
General Reaction
Example 1: Amide Reductions
Alkyl groups attached to the nitrogen do not affect the reaction.
Mechanism
1) Nucleophilic attach by the hydride
2) Leaving group removal
3) Nucleophilic attach by the hydride
Contributors
Prof. Steven Farmer (Sonoma State University)
General Mechanism of Amide Reactions
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)
Other Reactions of Amides
This page explains the reason for the lack of basic character in amides, and describes their dehydration to give nitriles, reaction with bromine and sodium hydroxide solution to form primary amines with one less carbon atom (the Hofmann degradation), and their reduction using LiAlH4.
The lack of base character in amides
Unusually for compounds containing the -NH2 group, amides are neutral. This section explains why -NH2 groups are usually basic and why amides are different.
The usual basic character of the -NH2 group
Simple compounds containing an -NH2 group such as ammonia, NH3, or a primary amine like methylamine, CH3NH2, are weak bases. A primary amine is a compound where the -NH2 group is attached to a hydrocarbon group. The active lone pair of electrons on the nitrogen atom in ammonia can combine with a hydrogen ion (a proton) from some other source - in other words it acts as a base.
With a compound like methylamine, all that has happened is that one of the hydrogen atoms attached to the nitrogen has been replaced by a methyl group. It doesn't make a huge amount of difference to the lone pair and so ammonia and methylamine behave similarly.
For example, if these compounds are dissolved in water, the nitrogen lone pair takes a hydrogen ion from a water molecule - and equilibria like these are set up:
$NH_3 (aq) + H_2O (l) \rightleftharpoons NH_4^+(aq) + OH^- (aq)$
$CH_3NH_2 (aq) + H_2O (l) \rightleftharpoons CH_3NH_3^+(aq) + OH^- (aq)$
Notice that the reactions are reversible. In both cases the positions of equilibrium lie well to the left. These compounds are weak bases because they don't hang on to the incoming hydrogen ion very well. Both ammonia and the amines are alkaline in solution because of the presence of the hydroxide ions, and both of them turn red litmus blue.
Why doesn't something similar happen with amides?
Amides are neutral to litmus and have virtually no basic character at all - despite having the -NH2 group. Their tendency to attract hydrogen ions is so slight that it can be ignored for most purposes. We need to look at the bonding in the -CONH2 group.
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.
A $pi$ bond is made by sideways overlap between p orbitals on the carbon and the oxygen. In an amide, the lone pair on the nitrogen atom ends up almost parallel to these p orbitals, and overlaps with them as they form the pi bond.
The result of this is that the nitrogen lone pair becomes delocalized - in other words it is no longer found located on the nitrogen atom, but the electrons from it are spread out over the whole of that part of the molecule. This has two effects which prevent the lone pair accepting hydrogen ions and acting as a base:
• Because the lone pair is no longer located on a single atom as an intensely negative region of space, it isn't anything like as attractive for a nearby hydrogen ion.
• Delocalization makes molecules more stable. For the nitrogen to reclaim its lone pair and join to a hydrogen ion, the delocalization would have to be broken, and that will cost energy.
The dehydration of amides
Amides are dehydrated by heating a solid mixture of the amide and phosphorus(V) oxide, P4O10. Water is removed from the amide group to leave a nitrile group, -CN. The liquid nitrile is collected by simple distillation. For example, with ethanamide, you will get ethanenitrile.
The Hofmann Degradation
The Hofmann degradation is a reaction between an amide and a mixture of bromine and sodium hydroxide solution. Heat is needed. The net effect of the reaction is a loss of the -CO- part of the amide group. You get a primary amine with one less carbon atom than the original amide had.
The general case would be (as a flow scheme):
If you started with ethanamide, you would get methylamine. The full equation for the reaction is:
$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
The Hofmann degradation is used as a way of cutting a single carbon atom out of a chain.
The reduction of amides
Amides can be reduced to primary amines by reaction with lithium tetrahydridoaluminate, LiAlH4, in dry ether (ethoxyethane) at room temperature. The initial reaction is followed by treatment with dilute acid, such as dilute sulphuric or hydrochloric acid.
For example, if you reduce ethanamide, you will get ethylamine.
$CH_3CONH_2 + 4[H] \rightarrow CH_3CH_2NH_2 + H2O$
An overall equation with hydrogen in square brackets is fine for this level. You might notice that this is a slightly different reduction from the one that happens when LiAlH4 reduces the carbon-oxygen double bond in an aldehyde or ketone. In those cases, the oxygen remains in the final molecule, and you get an -OH group formed. Here, the oxygen is removed entirely. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Reactivity_of_Amides/1_Amides_can_be_converted_to_Nitriles_with_Thionyl_Chloride.txt |
This page looks at the structures, formation, hydrolysis and uses of the polyamides, nylon and Kevlar.
What are polyamides?
Polyamides are polymers where the repeating units are held together by amide links. An amide group has the formula - CONH2. An amide link has this structure:
In an amide itself, of course, the bond on the right is attached to a hydrogen atom.
Nylon
In nylon, the repeating units contain chains of carbon atoms. (That is different from Kevlar, where the repeating units contain benzene rings - see below.) There are various different types of nylon depending on the nature of those chains.
Nylon-6,6
Nylon-6,6 is made from two monomers each of which contain 6 carbon atoms - hence its name. One of the monomers is a 6 carbon acid with a -COOH group at each end - hexanedioic acid. The other monomer is a 6 carbon chain with an amino group, -NH2, at each end. This is 1,6-diaminohexane (also known as hexane-1,6-diamine).
When these two compounds polymerise, the amine and acid groups combine, each time with the loss of a molecule of water. This is known as condensation polymeriz ation. Condensation polymerization is the formation of a polymer involving the loss of a small molecule. In this case, the molecule is water, but in other cases different small molecules might be lost.
The diagram shows the loss of water between two of the monomers:
This keeps on happening, and so you get a chain which looks like this:
Nylon-6
Iit is possible to get a polyamide from a single monomer. Nylon-6 is made from a monomer called caprolactam.
Notice that this already contains an amide link. When this molecule polymerizes, the ring ope ns, and the molecules join up in a continuous chain.
Kevlar
Kevlar is similar in structure to nylon-6,6 except that instead of the amide links joining chains of carbon atoms together, they join benzene rings. The two monomers are benzene-1,4-dicarboxylic acid and 1,4-diaminobenzene.
If you line these up and remove water between the -COOH and -NH2 groups in the same way as we did with nylon-6,6, you get the structure of Kevlar:
Making nylon-6,6
Nylon-6,6 is made by polymerising hexanedioic acid and 1,6-diaminohexane exactly as shown further up the page. Because the acid is acidic and the amine is basic, they first react together to form a salt. That is then converted into nylon-6,6 by heating it under pressure at 350°C. The two monomers can both be made from cyclohexane.
• Oxidation of the cyclohexane opens the ring of carbon atoms and produces a -COOH group at each end. That gives you the hexanedioic acid. Some of that can then be converted into the 1,6-diaminohexane.
• The acid is treated with ammonia to produce the ammonium salt.
• The ammonium salt is heated to 350°C in the presence of hydrogen and a nickel catalyst. This both dehydrates the salt and reduces it to the 1,6-diaminohexane.
Making nylon-6,6 in the lab
In the lab, it is easy to make nylon-6,6 at room temperature using an acyl chloride (acid chloride) rather than an acid. The 1,6-diaminohexane is used just as before, but hexanedioyl dichloride is used instead of hexanedioic acid.
If you compare the next diagram with the diagram further up the page for the formation of nylon-6,6, you will see that the only difference is that molecules of HCl are lost rather than molecules of water.
In the lab, this reaction is the basis for the nylon rope demonstration. You make a solution of the hexanedioyl dichloride in an organic solvent, and a solution of 1,6-diaminohexane in water. You carefully float one solution on top of the other in a small beaker, taking care to get as little mixing as possible. Nylon-6,6 forms at the boundary between the two solutions. If you pick up the boundary layer with a pair of tweezers, you can pull out an amazingly long tube of nylon from the beaker.
Hydrolysis of polyamides
Simple amides are easily hydrolysed by reaction with dilute acids or alkalis. Polyamides are fairly readily attacked by strong acids, but are much more resistant to alkaline hydrolysis. Hydrolysis is faster at higher temperatures. Hydrolysis by water alone is so slow as to be completely unimportant. Kevlar is rather more resistant to hydrolysis than nylon is.
If you spill something like dilute sulphuric acid on a fabric made from nylon, the amide linkages are broken. The long chains break and you can eventually end up with the original monomers - hexanedioic acid and 1,6-diaminohexane. Because you produce small molecules rather than the original polymer, the fibers are destroyed, and you end up with a hole!
Uses of polyamides
• Nylon: Apart from obvious uses in textiles for clothing and carpets, a lot of nylon is used to make tire cords - the inner structure of a vehicle tire underneath the rubber. The fibers are also used in ropes, and nylon can be cast into solid shapes for cogs and bearings in machines, for example.
• Kevlar: Kevlar is a very strong material - about five times as strong as steel, weight for weight. It is used in bulletproof vests, in composites for boat construction, in lightweight mountaineering ropes, and for lightweight skis and racquets - amongst many other things.
Contributors
Jim Clark (Chemguide.co.uk)
The Hydrolysis of Amides
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/Supplemental_Modules_(Organic_Chemistry)/Amides/Reactivity_of_Amides/Polyamides.txt |
An acid anhydride is what you get if you remove a molecule of water from two carboxylic acid -COOH groups. For example, if you took two ethanoic acid molecules and removed a molecule of water between them you would get the acid anhydride, ethanoic anhydride (old name: acetic anhydride).
For equation purposes, ethanoic anhydride is often written as (CH3CO)2O. The reactions of acid anhydrides are rather like those of acyl chlorides except that during their reactions, a molecule of carboxylic acid is produced rather than the HCl formed when an acyl chloride reacts.
Figure $1$: Acid Anhydrides react with ammonia, 1o amines and 2o amines to form amides.
If ethanoic anhydride is added to concentrated ammonia solution, ethanamide is formed together with ammonium ethanoate. Again, the reaction happens in two stages. In the first stage, ethanamide is formed together with ethanoic acid.
$\ce{(CH_3CO)_2O + NH_3 \rightarrow CH_3CONH_2 + CH_3COOH}$
Then the ethanoic acid produced reacts with excess ammonia to give ammonium ethanoate.
$\ce{ CH_3COOH + NH_3 \rightarrow CH_3COONH_4}$
When combined together to give one overall equation:
$\ce{ (CH_3CO)_2O + 2NH_3 \rightarrow CH_3CONH_2 + CH_3COONH_4}$
You need to follow this through really carefully, because the two products of the reaction overall can look confusingly similar.
Making Amides from Acyl Chlorides
Acyl chlorides (also known as acid chlorides) have the general formula RCOCl. The chlorine atom is very easily replaced by other things. For example, it is easily replaced by an -NH2 group to make an amide.
Figure 1: Acid chlorides react with ammonia, 1o amines and 2o amines to form amides
To make ethanamide from ethanoyl chloride, you normally add the ethanoyl chloride to a concentrated solution of ammonia in water. There is a very violent reaction producing lots of white smoke - a mixture of solid ammonium chloride and ethanamide. Some of the mixture remains dissolved in water as a colorless solution. The reaction can be thought of as happening in two stages.
In the first stage, the ammonia reacts with the ethanoyl chloride to give ethanamide and hydrogen chloride gas.
$CH_3COCl + NH_3 \rightarrow CH_3CONH_2 + HCl$
Then the hydrogen chloride produced reacts with excess ammonia to give ammonium chloride.
$NH_3 + HCl \rightarrow NH_4Cl$
. . . and you can combine all this together to give one overall equation:
$CH_3COCl + 2NH_3 \rightarrow CH_3CONH_2 + NH_4Cl$.
Contributors
Jim Clark (Chemguide.co.uk)
Making Amides from Carboxylic Acids
The carboxylic acid is first converted into an ammonium salt which then produces an amide on heating. The ammonium salt is formed by adding solid ammonium carbonate to an excess of the acid. For example, ammonium ethanoate is made by adding ammonium carbonate to an excess of ethanoic acid.
$2CH_3COOH + (NH_4)_2CO_3 \rightarrow 2CH_3COONH_4 + H_2O + CO_2$
When the reaction is complete, the mixture is heated and the ammonium salt dehydrates producing ethanamide.
$CH_3COONH_4 \rightarrow CH_3CONH_2 + H_2O$
The excess of ethanoic acid is there to prevent dissociation of the ammonium salt before it dehydrates. Ammonium salts tend to split into ammonia and the parent acid on heating, recombining on cooling. If dissociation happened in this case, the ammonia would escape from the reaction mixture and be lost. You could not get any recombination. The dissociation is reversible:
$CH_3COONH_5 (s) \rightleftharpoons CH_3COOH (l) + NH_3 (g)$
The presence of the excess ethanoic acid helps to prevent this from happening by moving the position of equilibrium to the left.
Some Practical Details
• The ammonium carbonate is added slowly to concentrated ethanoic acid and the reaction is left until all production of carbon dioxide stops.
• It is then heated under reflux for half an hour for the dehydration to take place.
• The mixture is distilled at about 170°C to remove excess ethanoic acid and water - leaving almost pure ethanamide in the flask.
Further purification stages are beyond the scope of this Modules
Activating agent
Carboxylic acid can be converted to amides by using DCC as an activating agent
Direct conversion of a carboxylic acid to an amide by reaction with an amine.
Making Amides from Nitriles
The hydrolysis of nitriles is a satisfactory method for preparation of unsubstituted amides and is particularly convenient when hydrolysis is induced under mildly basic conditions by hydrogen peroxide:
For the preparation of amides of the type $\ce{R_3CNHCOR}$, which have a tertiary alkyl group bonded to nitrogen, the Ritter reaction of an alcohol or alkene with a nitrile or hydrogen cyanide is highly advantageous. This reaction involves formation of a carbocation by action of strong sulfuric acid on an alkene or an alcohol (Equation 1), combination of the carbocation with the unshared electrons on nitrogen of $\ce{RCN}$ (Equation 2), and then addition of water (Equation 3). We use here the preparation of an $\ce{N}$-tert-butylalkanamide as an example; $\ce{RC \equiv N}$ can be an alkyl cyanide such as ethanenitrile or hydrogen cyanide itself:
$\tag{1}$
$\tag{2}$
$\tag{3}$
This reaction also is useful for the preparation of primary amines by hydrolysis of the amide. It is one of the relatively few practical methods for synthesizing amines with a tertiary alkyl group on the nitrogen:
Contributors
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Synthesis_of_Amides/Making_Amides_from_Acid_Anhydrides.txt |
In the IUPAC system of nomenclature, functional groups are normally designated in one of two ways. The presence of the function may be indicated by a characteristic suffix and a location number. This is common for the carbon-carbon double and triple bonds which have the respective suffixes ene and yne. Halogens, on the other hand, do not have a suffix and are named as substituents, for example: (CH3)2C=CHCHClCH3 is 4-chloro-2-methyl-2-pentene. If you are uncertain about the IUPAC rules for nomenclature you should review them now.
Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group. The nomenclature of amines is complicated by the fact that several different nomenclature systems exist, and there is no clear preference for one over the others. Furthermore, the terms primary (1º), secondary (2º) & tertiary (3º) are used to classify amines in a completely different manner than they were used for alcohols or alkyl halides. When applied to amines these terms refer to the number of alkyl (or aryl) substituents bonded to the nitrogen atom, whereas in other cases they refer to the nature of an alkyl group. The four compounds shown in the top row of the following diagram are all C4H11N isomers. The first two are classified as 1º-amines, since only one alkyl group is bonded to the nitrogen; however, the alkyl group is primary in the first example and tertiary in the second. The third and fourth compounds in the row are 2º and 3º-amines respectively. A nitrogen bonded to four alkyl groups will necessarily be positively charged, and is called a 4º-ammonium cation. For example, (CH3)4N(+) Br(–) is tetramethylammonium bromide.
• The IUPAC names are listed first and colored blue. This system names amine functions as substituents on the largest alkyl group. The simple -NH substituent found in 1º-amines is called an amino group. For 2º and 3º-amines a compound prefix (e.g. dimethylamino in the fourth example) includes the names of all but the root alkyl group.
• The Chemical Abstract Service has adopted a nomenclature system in which the suffix -amine is attached to the root alkyl name. For 1º-amines such as butanamine (first example) this is analogous to IUPAC alcohol nomenclature (-ol suffix). The additional nitrogen substituents in 2º and 3º-amines are designated by the prefix N- before the group name. These CA names are colored magenta in the diagram.
• Finally, a common system for simple amines names each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine. These are the names given in the last row (colored black).
Many aromatic and heterocyclic amines are known by unique common names, the origins of which are often unknown to the chemists that use them frequently. Since these names are not based on a rational system, it is necessary to memorize them. There is a systematic nomenclature of heterocyclic compounds, but it will not be discussed here.
Nomenclature of Amines
Nature abounds with nitrogen compounds, many of which occur in plants and are referred to as alkaloids. Structural formulas for some representative alkaloids and other nitrogen containing natural products are displayed below, and we can recognize many of the basic structural features listed above in their formulas. Thus, Serotonin and Thiamine are 1º-amines, Coniine is a 2º-amine, Atropine, Morphine and Quinine are 3º-amines, and Muscarine is a 4º-ammonium salt.
The reader should be able to recognize indole, imidazole, piperidine, pyridine, pyrimidine & pyrrolidine moieties among these structures. These will be identified by pressing the "Show Structures" button under the diagram.
Nitrogen atoms that are part of aromatic rings , such as pyridine, pyrrole & imidazole, have planar configurations (sp2 hybridization), and are not stereogenic centers. Nitrogen atoms bonded to carbonyl groups, as in caffeine, also tend to be planar. In contrast, atropine, coniine, morphine, nicotine and quinine have stereogenic pyramidal nitrogen atoms in their structural formulas (think of the non-bonding electron pair as a fourth substituent on a sp3 hybridized nitrogen). In quinine this nitrogen is restricted to one configuration by the bridged ring system. The other stereogenic nitrogens are free to assume two pyramidal configurations, but these are in rapid equilibrium so that distinct stereoisomers reflecting these sites cannot be easily isolated.
It should be noted that structural factors may serve to permit the resolution of pyramidal chiral amines. Two examples of such 3º-amines, compared with similar non-resolvable analogs, are shown in the following diagram. The two nitrogen atoms in Trögers base are the only stereogenic centers in the molecule. Because of the molecule's bridged structure, the nitrogens have the same configuration and cannot undergo inversion. The chloro aziridine can invert, but requires a higher activation energy to do so, compared with larger heterocyclic amines. It has in fact been resolved, and pure enantiomers isolated. An increase in angle strain in the sp2-hybridized planar transition state is responsible for the greater stability of the pyramidal configuration. The rough estimate of angle strain is made using a C-N-C angle of 60º as an arbitrary value for the three-membered heterocycle.
Of course, quaternary ammonium salts, such as that in muscarine, have a tetrahedral configuration that is incapable of inversion. With four different substituents, such a nitrogen would be a stable stereogenic center.
A Structure Formula Relationship
Recall that the molecular formula of a hydrocarbon (CnHm) provides information about the number of rings and/or double bonds that must be present in its structural formula. In the formula shown below a triple bond is counted as two double bonds.
Rings + Double Bonds in a CnHm Hydrocarbon= (2n + 2 - m)/2
Compound
Molecular
Formula
Revised
Formula
Calculated
Rings + C=Z
Coniine C8H17N C9H18 1
Nicotine C10H14N2 C12H16 5
Morphine C17H19NO3 C18H20 9
This molecular formula analysis may be extended beyond hydrocarbons by a few simple corrections. These are illustrated by the examples in the table above, taken from the previous list of naturally occurring amines.
• The presence of oxygen does not alter the relationship.
• All halogens present in the molecular formula must be replaced by hydrogen.
• Each nitrogen in the formula must be replaced by a CH moiety. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Nomenclature_of_Amines/Natural_Nitrogen_Compounds.txt |
Boiling Point and Water Solubility
It is instructive to compare the boiling points and water solubility of amines with those of corresponding alcohols and ethers. The dominant factor here is hydrogen bonding, and the first table below documents the powerful intermolecular attraction that results from -O-H---O- hydrogen bonding in alcohols (light blue columns). Corresponding -N-H---N- hydrogen bonding is weaker, as the lower boiling points of similarly sized amines (light green columns) demonstrate. Alkanes provide reference compounds in which hydrogen bonding is not possible, and the increase in boiling point for equivalent 1º-amines is roughly half the increase observed for equivalent alcohols.
Compound CH3CH3 CH3OH CH3NH2 CH3CH2CH3 CH3CH2OH CH3CH2NH2
Mol.Wt. 30 32 31 44 46 45
Boiling
Point ºC
-88.6º 65º -6.0º -42º 78.5º 16.6º
The second table illustrates differences associated with isomeric 1º, 2º & 3º-amines, as well as the influence of chain branching. Since 1º-amines have two hydrogens available for hydrogen bonding, we expect them to have higher boiling points than isomeric 2º-amines, which in turn should boil higher than isomeric 3º-amines (no hydrogen bonding). Indeed, 3º-amines have boiling points similar to equivalent sized ethers; and in all but the smallest compounds, corresponding ethers, 3º-amines and alkanes have similar boiling points. In the examples shown here, it is further demonstrated that chain branching reduces boiling points by 10 to 15 ºC.
Compound CH3(CH2)2CH3 CH3(CH2)2OH CH3(CH2)2NH2 CH3CH2NHCH3 (CH3)3CH (CH3)2CHOH (CH3)2CHNH2 (CH3)3N
Mol.Wt. 58 60 59 59 58 60 59 59
Boiling
Point ºC
-0.5º 97º 48º 37º -12º 82º 34º
The water solubility of 1º and 2º-amines is similar to that of comparable alcohols. As expected, the water solubility of 3º-amines and ethers is also similar. These comparisons, however, are valid only for pure compounds in neutral water. The basicity of amines (next section) allows them to be dissolved in dilute mineral acid solutions, and this property facilitates their separation from neutral compounds such as alcohols and hydrocarbons by partitioning between the phases of non-miscible solvents.
Basicity of Amines
A review of basic acid-base concepts should be helpful to the following discussion. Like ammonia, most amines are Brønsted and Lewis bases, but their base strength can be changed enormously by substituents. It is common to compare basicity's quantitatively by using the pKa's of their conjugate acids rather than their pKb's. Since pKa + pKb = 14, the higher the pKa the stronger the base, in contrast to the usual inverse relationship of pKa with acidity. Most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their water solutions are basic (have a pH of 11 to 12, depending on concentration). The first four compounds in the following table, including ammonia, fall into that category.
The last five compounds (colored cells) are significantly weaker bases as a consequence of three factors. The first of these is the hybridization of the nitrogen. In pyridine the nitrogen is sp2 hybridized, and in nitriles (last entry) an sp hybrid nitrogen is part of the triple bond. In each of these compounds (shaded red) the non-bonding electron pair is localized on the nitrogen atom, but increasing s-character brings it closer to the nitrogen nucleus, reducing its tendency to bond to a proton.
Compound
NH3 CH3C≡N
pKa 11.0 10.7 10.7 9.3 5.2 4.6 1.0 0.0 -1.0 -10.0
Secondly, aniline and p-nitroaniline (first two green shaded structures) are weaker bases due to delocalization of the nitrogen non-bonding electron pair into the aromatic ring (and the nitro substituent). This is the same delocalization that results in activation of a benzene ring toward electrophilic substitution. The following resonance equations, which are similar to those used to explain the enhanced acidity of ortho and para-nitrophenols illustrate electron pair delocalization in p-nitroaniline. Indeed, aniline is a weaker base than cyclohexyl amine by roughly a million fold, the same factor by which phenol is a stronger acid than cyclohexanol. This electron pair delocalization is accompanied by a degree of rehybridization of the amino nitrogen atom, but the electron pair delocalization is probably the major factor in the reduced basicity of these compounds. A similar electron pair delocalization is responsible for the very low basicity (and nucleophilic reactivity) of amide nitrogen atoms (last green shaded structure). This feature was instrumental in moderating the influence of amine substituents on aromatic ring substitution, and will be discussed further in the section devoted to carboxylic acid derivatives.
Conjugated amine groups influence the basicity of an existing amine. Although 4-dimethylaminopyridine (DMAP) might appear to be a base similar in strength to pyridine or N,N-dimethylaniline, it is actually more than ten thousand times stronger, thanks to charge delocalization in its conjugate acid. The structure in the gray box shows the locations over which positive charge (colored red) is delocalized in the conjugate acid. This compound is often used as a catalyst for acyl transfer reactions.
Finally, the very low basicity of pyrrole (shaded blue) reflects the exceptional delocalization of the nitrogen electron pair associated with its incorporation in an aromatic ring. Indole (pKa = -2) and imidazole (pKa = 7.0), see above, also have similar heterocyclic aromatic rings. Imidazole is over a million times more basic than pyrrole because the sp2 nitrogen that is part of one double bond is structurally similar to pyridine, and has a comparable basicity.
Although resonance delocalization generally reduces the basicity of amines, a dramatic example of the reverse effect is found in the compound guanidine (pKa = 13.6). Here, as shown below, resonance stabilization of the base is small, due to charge separation, while the conjugate acid is stabilized strongly by charge delocalization. Consequently, aqueous solutions of guanidine are nearly as basic as are solutions of sodium hydroxide.
The relationship of amine basicity to the acidity of the corresponding conjugate acids may be summarized in a fashion analogous to that noted earlier for acids:
Strong bases have weak conjugate acids, and weak bases have strong conjugate acids.
Acidity of Amines
We normally think of amines as bases, but it must be remembered that 1º and 2º-amines are also very weak acids (ammonia has a pKa = 34). In this respect it should be noted that pKa is being used as a measure of the acidity of the amine itself rather than its conjugate acid, as in the previous section. For ammonia this is expressed by the following hypothetical equation:
NH3 + H2O ____> NH2(–) + H2O-H(+)
The same factors that decreased the basicity of amines increase their acidity. This is illustrated by the following examples, which are shown in order of increasing acidity. It should be noted that the first four examples have the same order and degree of increased acidity as they exhibited decreased basicity in the previous table. The first compound is a typical 2º-amine, and the three next to it are characterized by varying degrees of nitrogen electron pair delocalization. The last two compounds (shaded blue) show the influence of adjacent sulfonyl and carbonyl groups on N-H acidity. From previous discussion it should be clear that the basicity of these nitrogens is correspondingly reduced.
Compound C6H5SO2NH2
pKa 33 27 19 15 10 9.6
The acids shown here may be converted to their conjugate bases by reaction with bases derived from weaker acids (stronger bases). Three examples of such reactions are shown below, with the acidic hydrogen colored red in each case. For complete conversion to the conjugate base, as shown, a reagent base roughly a million times stronger is required.
C6H5SO2NH2 + KOH C6H5SO2NH(–) K(+) + H2O a sulfonamide base
(CH3)3COH + NaH (CH3)3CO(–) Na(+) + H2 an alkoxide base
(C2H5)2NH + C4H9Li (C2H5)2N(–) Li(+) + C4H10 an amide base
Important Reagent Bases
The significance of all these acid-base relationships to practical organic chemistry lies in the need for organic bases of varying strength, as reagents tailored to the requirements of specific reactions. The common base sodium hydroxide is not soluble in many organic solvents, and is therefore not widely used as a reagent in organic reactions. Most base reagents are alkoxide salts, amines or amide salts. Since alcohols are much stronger acids than amines, their conjugate bases are weaker than amide bases, and fill the gap in base strength between amines and amide salts. In the following table, pKa again refers to the conjugate acid of the base drawn above it.
Base Name Pyridine Triethyl
Amine
Hünig's Base Barton's
Base
Potassium
t-Butoxide
Sodium HMDS LDA
Formula (C2H5)3N (CH3)3CO(–) K(+) [(CH3)3Si]2N(–) Na(+) [(CH3)2CH]2N(–) Li(+)
pKa 5.3 10.7 11.4 14 19 26 35.7
Pyridine is commonly used as an acid scavenger in reactions that produce mineral acid co-products. Its basicity and nucleophilicity may be modified by steric hindrance, as in the case of 2,6-dimethylpyridine (pKa=6.7), or resonance stabilization, as in the case of 4-dimethylaminopyridine (pKa=9.7). Hünig's base is relatively non-nucleophilic (due to steric hindrance), and like DBU is often used as the base in E2 elimination reactions conducted in non-polar solvents. Barton's base is a strong, poorly-nucleophilic, neutral base that serves in cases where electrophilic substitution of DBU or other amine bases is a problem. The alkoxides are stronger bases that are often used in the corresponding alcohol as solvent, or for greater reactivity in DMSO. Finally, the two amide bases see widespread use in generating enolate bases from carbonyl compounds and other weak carbon acids.
Nonionic Superbases
An interesting group of neutral, highly basic compounds of nitrogen and phosphorus have been prepared, and are referred to as superbases. To see examples of these compounds Click Here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Properties_of_Amines/Advanced_Properties_of_Amines.txt |
This page explains what amines are, and what the difference is between primary, secondary and tertiary amines. It looks in some detail at their simple physical properties such as solubility and boiling points.
The easiest way to think of amines is as near relatives of ammonia, NH3. In amines, the hydrogen atoms in the ammonia have been replaced one at a time by hydrocarbon groups. On this page, we are only looking at cases where the hydrocarbon groups are simple alkyl groups. Amines fall into different classes depending on how many of the hydrogen atoms are replaced.
Primary amines
In primary amines, only one of the hydrogen atoms in the ammonia molecule has been replaced. That means that the formula of the primary amine will be RNH2 where "R" is an alkyl group.
Examples include:
Naming amines can be quite confusing because there are so many variations on the names. For example, the simplest amine, CH3NH2, can be called methylamine, methanamine or aminomethane.
The commonest name at this level is methylamine and, similarly, the second compound drawn above is usually called ethylamine. Where there might be confusion about where the -NH2 group is attached to a chain, the simplest way of naming the compound is to use the "amino" form.
For example:
Secondary amines
In a secondary amine, two of the hydrogens in an ammonia molecule have been replaced by hydrocarbon groups. At this level, you are only likely to come across simple ones where both of the hydrocarbon groups are alkyl groups and both are the same.
For example:
There are other variants on the names, but this is the commonest and simplest way of naming these small secondary amines.
Tertiary amines
In a tertiary amine, all of the hydrogens in an ammonia molecule have been replaced by hydrocarbon groups. Again, you are only likely to come across simple ones where all three of the hydrocarbon groups are alkyl groups and all three are the same. The naming is similar to secondary amines. For example:
Physical properties of amines
Boiling points
The table shows the boiling points of some simple amines.
type formula boiling point (°C)
primary CH3NH2 -6.3
primary CH3CH2NH2 16.6
primary CH3CH2CH2NH2 48.6
secondary (CH3)2NH 7.4
tertiary (CH3)3N 3.5
We will need to look at this with some care to sort out the patterns and reasons. Concentrate first on the primary amines.
Primary amines
It is useful to compare the boiling point of methylamine, CH3NH2, with that of ethane, CH3CH3. Both molecules contain the same number of electrons and have, as near as makes no difference, the same shape. However, the boiling point of methylamine is -6.3°C, whereas ethane's boiling point is much lower at -88.6°C.
The reason for the higher boiling points of the primary amines is that they can form hydrogen bonds with each other as well as van der Waals dispersion forces and dipole-dipole interactions. Hydrogen bonds can form between the lone pair on the very electronegative nitrogen atom and the slightly positive hydrogen atom in another molecule.
The hydrogen bonding is not as efficient as it is in, say, water, because there is a shortage of lone pairs. Some slightly positive hydrogen atoms will not be able to find a lone pair to hydrogen bond with. There are twice as many suitable hydrogens are there are lone pairs. The boiling points of the primary amines increase as you increase chain length because of the greater amount of van der Waals dispersion forces between the bigger molecules.
Secondary amines
For a fair comparison you would have to compare the boiling point of dimethylamine with that of ethylamine. They are isomers of each other - each contains exactly the same number of the same atoms.
The boiling point of the secondary amine is a little lower than the corresponding primary amine with the same number of carbon atoms. Secondary amines still form hydrogen bonds, but having the nitrogen atom in the middle of the chain rather than at the end makes the permanent dipole on the molecule slightly less. The lower boiling point is due to the lower dipole-dipole attractions in the dimethylamine compared with ethylamine.
Tertiary amines
This time to make a fair comparison you would have to compare trimethylamine with its isomer 1-aminopropane. If you look back at the table further up the page, you will see that the trimethylamine has a much lower boiling point (3.5°C) than 1-aminopropane (48.6°C). In a tertiary amine there aren't any hydrogen atoms attached directly to the nitrogen. That means that hydrogen bonding between tertiary amine molecules is impossible. That's why the boiling point is much lower.
Solubility in water
The small amines of all types are very soluble in water. In fact, the ones that would normally be found as gases at room temperature are normally sold as solutions in water - in much the same way that ammonia is usually supplied as ammonia solution.
All of the amines can form hydrogen bonds with water - even the tertiary ones. Although the tertiary amines don't have a hydrogen atom attached to the nitrogen and so can't form hydrogen bonds with themselves, they can form hydrogen bonds with water molecules just using the lone pair on the nitrogen.
Solubility falls off as the hydrocarbon chains get longer - noticeably so after about 6 carbons. The hydrocarbon chains have to force their way between water molecules, breaking hydrogen bonds between water molecules. However, they don't replace them by anything as strong, and so the process of forming a solution becomes less and less energetically feasible as chain length grows.
Smell
The very small amines like methylamine and ethylamine smell very similar to ammonia - although if you compared them side by side, the amine smells are slightly more complex. As the amines get bigger, they tend to smell more "fishy", or they smell of decay. If you are familiar with the smell of hawthorn blossom (and similarly smelling things like cotoneaster blossom), this is the smell of trimethylamine - a sweet and rather sickly smell like the early stages of decaying flesh.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Properties_of_Amines/Basic_Properties_of_Amines.txt |
In comparing the chemistry of the amines with alcohols and ethers, we discover many classes of related compounds in which nitrogen assumes higher oxidation states, in contrast to limited oxidation states of oxygen.
In this context, keep in mind that the oxidation state of elemental oxygen (O2) and nitrogen (N2) is defined as zero. The most prevalent state of covalently bonded oxygen is -2. This is the case for water, alcohols, ethers and carbonyl compounds. The only common higher oxidation state (-1) is found in the peroxides, R–O–O–R, where R=hydrogen, alkyl, aryl or acyl. Because of the low covalent bond energy of the peroxide bond (ca.35 kcal/mole), these compounds are widely used as free radical initiators, and are sometimes dangerously explosive in their reactivity (e.g. triacetone triperoxide used by terrorist bombers). Nitrogen compounds, on the other hand, encompass oxidation states of nitrogen ranging from -3, as in ammonia and amines, to +5, as in nitric acid. The following table lists some of the known organic compounds of nitrogen, having different oxidation states of that element. Some of these classes of compounds have been described; others will be discussed later.
Oxidation State
_3
_2
_1
0
+1
+3
Formulas
(names)
R3N
(amines)
R4N(+)
(ammonium)
C=N–R
(imines)
C≡N
(nitriles)
R2NNR2
(hydrazines)
C=NNR2
(hydrazones)
RN=NR
(azo cpd.)
R2NOH
(hydroxyl amine)
R3NO
(amine oxide)
N2
(nitrogen)
R–N2(+)
(diazonium)
R–N=O (nitroso)
R-NO2
(nitro)
RO–N=O
(nitrite ester)
Amine Oxides and the Cope Elimination
Amine oxides are prepared by oxidizing 3º-amines or pyridines with hydrogen peroxide or peracids (e.g. ZOOH, where Z=H or acyl).
$R_3N: + ZOOH \rightarrow R_3N^{(+)}–O^{(–)} + ZOH$
Amine oxides are relatively weak bases, pKa ca. 4.5, compared with the parent amine. The coordinate covalent N–O function is polar, with the oxygen being a powerful hydrogen bond acceptor. If one of the alkyl substituents consists of a long chain, such as C12H25, the resulting amine oxide is an amphoteric surfactant and finds use in shampoos and other mild cleaning agents.
An elimination reaction, complementary to the Hofmann elimination, occurs when 3º-amine oxides are heated at temperatures of 150 to 200 ºC. This reaction is known as the Cope Elimination. It is commonly carried out by dropwise addition of an amine oxide solution to a heated tube packed with small glass beads. A stream of nitrogen gas flowing through the column carries the volatile alkene products to a chilled receiver. The nitrogen-containing product is a hydroxyl amine. Unlike the Hofmann elimination, this reaction takes place by a concerted cyclic reorganization, as shown in the following diagram. For such a mechanism, the beta-hydrogen and amine oxide moieties necessarily have a syn-relationship.
Cope elimination of diastereomeric amine oxides, such as those shown in examples #2 & 3 above, provide proof of the syn-relationship of the beta-hydrogen and amine oxide groups. These examples also demonstrate a strong regioselectivity favoring the more stable double bond.
Pyrolytic syn-Eliminations
Amine oxides are not the only functions that undergo a unimolecular syn-elimination on heating. To see examples of other cases Click Here
Nitroxide Radicals
2º-Amines lacking α-hydrogens are oxidized by peroxides (ZOOH) to nitroxide radicals of surprising stability. In the example shown at the top of the following diagram it should be noted that resonance delocalization of the unpaired electron contributes to a polar N–O bond. The R=H compound, known by the acronym TEMPO, is a relatively stable red solid. Many other nitroxides have been prepared, three of which are drawn at the lower right. If one or more hydrogens are present on an adjacent carbon, the nitroxide decomposes to mixtures including amine oxides and nitrones, as shown at the lower left. Nitroxides are oxidized to unstable oxammonium cations by halogens.
The spin of the nitroxyl unpaired electron may be studied by a technique called electron paramagnetic resonance (epr or esr). Experiments of this kind have demonstrated that the epr spectra are sensitive to substituents on the radical as well as its immediate environment. This has led to a spin labeling strategy for investigating the conformational structures of macromolecules like proteins. Thus, site-directed spin labeling (SDSL) has emerged as a valuable technique for mapping elements of secondary structure, at the level of the backbone fold, in a wide range of proteins, including those not amenable to structural characterization using classical structural techniques, such as nuclear magnetic resonance and X-ray crystallography.
Phosphorous Analogs of Amines
Phosphorus is beneath nitrogen in the periodic table. To see examples of organophosphorus compounds and their chemistry Click Here | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Properties_of_Amines/Oxidation_States_of_Nitrogen.txt |
The reactivity of amines is similar to ammonia: amines are basic, nucleophilic, and react with alkyl halides, acid chlorides, and carbonyl compounds. Additionally, aromatic amines are highly reactive in electrophilic aromatic substitution.
• Amines as Bases
This page looks at the reactions of amines as bases. Their basic properties include the reactions with dilute acids, water and copper(II) ions.
• Amines as Nucleophiles
A nucleophile is something which is attracted to, and then attacks, a positive or slightly positive part of another molecule or ion. All amines contain an active lone pair of electrons on the very electronegative nitrogen atom. It is these electrons which are attracted to positive parts of other molecules or ions.
• Amine Reactions
Ammonia and many amines are not only bases in the Brønsted sense, they are also nucleophiles that bond to and form products with a variety of electrophiles.
• Reactions of Aryl Diazonium Salts
Aryl diazonium salts are important intermediates. They are prepared in cold (0 º to 10 ºC) aqueous solution, and generally react with nucleophiles with loss of nitrogen.
• Reaction of Amines with Nitrous Acid
Nitrous acid reacts with aliphatic amines in a fashion that provides a useful test for distinguishing primary, secondary and tertiary amines.
• Substitution and Elimination Reactions of Amines
• The Reaction of Amines with Nitrous Acid
Reactivity of Amines
Electrophilic Substitution at Nitrogen
Ammonia and many amines are not only bases in the Brønsted sense, they are also nucleophiles that bond to and form products with a variety of electrophiles. A general equation for such electrophilic substitution of nitrogen is:
2 R2ÑH + E(+) R2NHE(+) R2ÑE + H(+) (bonded to a base)
A list of some electrophiles that are known to react with amines is shown here. In each case the electrophilic atom or site is colored red.
Electrophile RCH2–X RCH2–OSO2R R2C=O R(C=O)X RSO2–Cl HO–N=O
Name Alkyl Halide Alkyl Sulfonate Aldehyde or Ketone Acid Halide or Anhydride Sulfonyl Chloride Nitrous Acid
Alkylation
It is instructive to examine these nitrogen substitution reactions, using the common alkyl halide class of electrophiles. Thus, reaction of a primary alkyl bromide with a large excess of ammonia yields the corresponding 1º-amine, presumably by an SN2 mechanism. The hydrogen bromide produced in the reaction combines with some of the excess ammonia, giving ammonium bromide as a by-product. Water does not normally react with 1º-alkyl halides to give alcohols, so the enhanced nucleophilicity of nitrogen relative to oxygen is clearly demonstrated.
$2 RCH_2Br + \underbrace{NH_3}_{\text{large excess}} \rightarrow RCH_2NH_2 + NH_4^{(+)} Br^{(–)}$
It follows that simple amines should also be more nucleophilic than their alcohol or ether equivalents. If, for example, we wish to carry out an SN2 reaction of an alcohol with an alkyl halide to produce an ether (the Williamson synthesis), it is necessary to convert the weakly nucleophilic alcohol to its more nucleophilic conjugate base for the reaction to occur. In contrast, amines react with alkyl halides directly to give N-alkylated products. Since this reaction produces HBr as a co-product, hydrobromide salts of the alkylated amine or unreacted starting amine (in equilibrium) will also be formed.
2 RNH2 + C2H5Br RNHC2H5 + RNH3(+) Br(–) RNH2C2H5(+) Br(–) + RNH2
Unfortunately, the direct alkylation of 1º or 2º-amines to give a more substituted product does not proceed cleanly. If a 1:1 ratio of amine to alkyl halide is used, only 50% of the amine will react because the remaining amine will be tied up as an ammonium halide salt (remember that one equivalent of the strong acid HX is produced). If a 2:1 ratio of amine to alkylating agent is used, as in the above equation, the HX issue is solved, but another problem arises. Both the starting amine and the product amine are nucleophiles. Consequently, once the reaction has started, the product amine competes with the starting material in the later stages of alkylation, and some higher alkylated products are also formed. Even 3º-amines may be alkylated to form quaternary (4º) ammonium salts. When tetraalkyl ammonium salts are desired, as shown in the following example, Hünig's base may be used to scavenge the HI produced in the three SN2 reactions. Steric hindrance prevents this 3º-amine (Hünig's base) from being methylated.
C6H5NH2 + 3 CH3I + Hünig's base C6H5N(CH3)3(+) I(–) + HI salt of Hünig's base
The Hinsberg Test: Reaction with benzenesulfonyl chloride
Another electrophilic reagent, benzenesulfonyl chloride, reacts with amines in a fashion that provides a useful test for distinguishing primary, secondary and tertiary amines (the Hinsberg test). As shown in the following equations, 1º and 2º-amines react to give sulfonamide derivatives with loss of HCl, whereas 3º-amines do not give any isolable products other than the starting amine. In the latter case a quaternary "onium" salt may be formed as an intermediate, but this rapidly breaks down in water to liberate the original 3º-amine (lower right equation).
The Hinsberg test is conducted in aqueous base (NaOH or KOH), and the benzenesulfonyl chloride reagent is present as an insoluble oil. Because of the heterogeneous nature of this system, the rate at which the sulfonyl chloride reagent is hydrolyzed to its sulfonate salt in the absence of amines is relatively slow. The amine dissolves in the reagent phase, and immediately reacts (if it is 1º or 2º), with the resulting HCl being neutralized by the base. The sulfonamide derivative from 2º-amines is usually an insoluble solid. However, the sulfonamide derivative from 1º-amines is acidic and dissolves in the aqueous base. Acidification of this solution then precipitates the sulfonamide of the 1º-amine. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Reactivity_of_Amines/Amine_Reactions.txt |
We are going to have to use two different definitions of the term "base" in this page. A base is
• a substance which combines with hydrogen ions. This is the Brønsted–Lowry theory.
• an electron pair donor. This is the Lewis theory.
The easiest way of looking at the basic properties of amines is to think of an amine as a modified ammonia molecule. In an amine, one or more of the hydrogen atoms in ammonia has been replaced by a hydrocarbon group. Replacing the hydrogens still leaves the lone pair on the nitrogen unchanged - and it is the lone pair on the nitrogen that gives ammonia its basic properties. Amines will therefore behave much the same as ammonia in all cases where the lone pair is involved.
The reactions of amines with acids
These are most easily considered using the Brønsted–Lowry theory of acids and bases - the base is a hydrogen ion acceptor. We'll do a straight comparison between amines and the familiar ammonia reactions. Ammonia reacts with acids to produce ammonium ions. The ammonia molecule picks up a hydrogen ion from the acid and attaches it to the lone pair on the nitrogen.
If the reaction is in solution in water (using a dilute acid), the ammonia takes a hydrogen ion (a proton) from a hydroxonium ion. (Remember that hydrogen ions present in solutions of acids in water are carried on water molecules as hydroxonium ions, H3O+.)
$\ce{ NH_3 (aq) + H_3O^{+} (aq) -> NH_4^{+} + H_2O (l)} \nonumber$
If the acid was hydrochloric acid, for example, you would end up with a solution containing ammonium chloride - the chloride ions, of course, coming from the hydrochloric acid. You could also write this last equation as:
$\ce{NH_3 (aq) + H^{+} -> NH_4^{+} (aq)} \nonumber$
. . . but if you do it this way, you must include the state symbols. If you write H+ on its own, it implies an unattached hydrogen ion - a proton. Such things don't exist on their own in solution in water. If the reaction is happening in the gas state, the ammonia accepts a proton directly from the hydrogen chloride:
$\ce{NH_3 (aq) + HCl (g) -> NH_4^+ (s) + Cl^- (s)} \nonumber$
This time you produce clouds of white solid ammonium chloride.
The nitrogen lone pair behaves exactly the same. The fact that one (or more) of the hydrogens in the ammonia has been replaced by a hydrocarbon group makes no difference.
Example 1: Ethylamine
If the reaction is done in solution, the amine takes a hydrogen ion from a hydroxonium ion and forms an ethylammonium ion.
$\ce{CH3CH2NH2( aq ) + H3O^{+}(aq) <=> CH3CH2NH3^{+}(aq) + H2O( l )} \nonumber$
or:
$\ce{CH3CH2NH2(aq) + H^{+}(aq) -> CH3CH2NH3^{+}(aq)} \nonumber$
The solution would contain ethylammonium chloride or sulfate or whatever. Alternatively, the amine will react with hydrogen chloride in the gas state to produce the same sort of white smoke as ammonia did - but this time of ethylammonium chloride.
$\ce{ CH_3CH_2NH_2(g) + HCl(g) \rightarrow CH_3CH_2NH_3^{+} (s) + Cl^{-} (s)} \nonumber$
These examples have involved a primary amine, but it makes no real difference if a secondary or tertiary amine were used.; the equations would just look more complicated. The product ions from diethylamine and triethylamine would be diethylammonium ions and triethylammonium ions respectively.
The Reactions of Amines with Water
Again, it is easiest to use the Brønsted–Lowry theory and, again, it is useful to do a straight comparison with ammonia. Ammonia is a weak base and takes a hydrogen ion from a water molecule to produce ammonium ions and hydroxide ions. However, the ammonia is only a weak base, and doesn't hang on to the hydrogen ion very successfully. The reaction is reversible, with the great majority of the ammonia at any one time present as free ammonia rather than ammonium ions.
$\ce{ NH_3(aq) + H_2O(l) <=> NH_4^{+} (aq) + OH^{-} (aq)} \nonumber$
The presence of the hydroxide ions from this reaction makes the solution alkaline. The amine still contains the nitrogen lone pair, and does exactly the same thing. For example, with ethylamine, you get ethylammonium ions and hydroxide ions produced.
$\ce{ CH3CH2NH2 (aq) + H2O (l) <=> CH3CH2NH3^{+} (aq) + OH^{-} (aq)} \nonumber$
There is, however, a difference in the position of equilibrium. Amines are usually stronger bases than ammonia (there are exceptions to this, though - particularly if the amine group is attached directly to a benzene ring).
The reactions of amines with copper(II) ions
Just like ammonia, amines react with copper(II) ions in two separate stages. In the first step, we can go on using the Brønsted–Lowry theory (that a base is a hydrogen ion acceptor). The second stage of the reaction can only be explained in terms of the Lewis theory (that a base is an electron pair donor).
Example 2: Reaction between Ammonia and Copper (II)
Copper(II) sulphate solution, for example, contains the blue hexaaquacopper(II) complex ion -$\ce{[Cu(H_2O)_6]^{2+}}$.
In the first stage of the reaction, the ammonia acts as a Brønsted–Lowry base. With a small amount of ammonia solution, hydrogen ions are pulled off two water molecules in the hexaaqua ion.
This produces a neutral complex - one carrying no charge. If you remove two positively charged hydrogen ions from a 2+ ion, then obviously there isn't going to be any charge left on the ion. Because of the lack of charge, the neutral complex isn't soluble in water, and so you get a pale blue precipitate.
$\ce{[Cu(H_2O)_6]^{2+} + 2NH_3 <=> [Cu(H2O)4(OH)2] + 2NH4^{+}} \nonumber$
This precipitate is often written as Cu(OH)2 and called copper(II) hydroxide. The reaction is reversible because ammonia is only a weak base. That precipitate dissolves if you add an excess of ammonia solution, giving a deep blue solution.
The ammonia replaces four of the water molecules around the copper to give tetraamminediaquacopper(II) ions. The ammonia uses its lone pair to form a co-ordinate covalent bond (dative covalent bond) with the copper. It is acting as an electron pair donor - a Lewis base.
$\ce{[Cu(H_2O)_6]^{2+} + 4NH_3 <=> [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O} \nonumber$
The color changes are:
The corresponding reaction with amines
The small primary amines behave in exactly the same way as ammonia. There will, however, be slight differences in the shades of blue that you get during the reactions.
Example 3: Reaction between Methylamine and Copper (II)
With a small amount of methylamine solution you will get a pale blue precipitate of the same neutral complex as with ammonia. All that is happening is that the methylamine is pulling hydrogen ions off the attached water molecules.
$\ce{[Cu(H2O)6]^{2+} + 2CH3NH2 <=> [Cu(H2O)4(OH)2] + 2CH3NH3^{+}}\nonumber$
With more methylamine solution the precipitate redissolves to give a deep blue solution - just as in the ammonia case. The amine replaces four of the water molecules around the copper.
$\ce{[Cu(H2O)6]^{2+} + 4CH3NH2 <=> [Cu(CH3NH2)4(H2O)2]^{2+} + 4 H2O} \nonumber$
As the amines get bigger and more bulky, the formula of the final product may change - simply because it is impossible to fit four large amine molecules and two water molecules around the copper atom. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Reactivity_of_Amines/Amines_as_Bases.txt |
A nucleophile is something which is attracted to, and then attacks, a positive or slightly positive part of another molecule or ion. All amines contain an active lone pair of electrons on the very electronegative nitrogen atom. It is these electrons which are attracted to positive parts of other molecules or ions.
The reactions of primary amines with halogenoalkanes
You get a complicated series of reactions on heating to give a mixture of products including secondary and tertiary amines and their salts, and quaternary ammonium salts.
Making secondary amines and their salts
In the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide
In the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine - a secondary amine.
Making tertiary amines and their salts
But it doesn't stop here! The diethylamine also reacts with bromoethane - in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.
In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.
The ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine.
Making a quaternary ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
The reactions of amines with acyl chlorides
We'll take the reaction between methylamine and ethanoyl chloride as typical. If you add concentrated methylamine solution to ethanoyl chloride, there is a violent reaction in the cold. N-methylethanamide and methylammonium chloride are formed - partly as a white solid mixture, and partly in solution. The overall equation is:
$CH_3COCl + 2CH_3NH_2 \rightarrow CH_3CONHCH_3 + CH_3NH_3Cl$
The reactions of amines with acid anhydrides
These reactions are chemically similar to those between amines and acyl chlorides, but they are much slower, needing heat. Taking the reaction between methylamine and ethanoic anhydride as typical. The product is N-methylethanamide (as with ethanoyl chloride), but this time the other product is methylammonium ethanoate rather than methylammonium chloride.
$(CH_3CO)_2O + 2CH_3NH_2 \rightarrow CH_3CONHCH_3 + CH_3COO^- + ^+N_3CH_3$
Reaction of Amines with Nitrous Acid
Nitrous acid ($HNO_2$ or $HONO$) reacts with aliphatic amines in a fashion that provides a useful test for distinguishing primary, secondary and tertiary amines.
• 1°-Amines + HONO (cold acidic solution) $\rightarrow$ Nitrogen Gas Evolution from a Clear Solution
• 2°-Amines + HONO (cold acidic solution) $\rightarrow$ An Insoluble Oil (N-Nitrosamine)
• 3°-Amines + HONO (cold acidic solution) $\rightarrow$ A Clear Solution (Ammonium Salt Formation)
Nitrous acid is a Brønsted acid of moderate strength (pKa = 3.3). Because it is unstable, it is prepared immediately before use in the following manner:
Under the acidic conditions of this reaction, all amines undergo reversible salt formation:
This happens with 3º-amines, and the salts are usually soluble in water. The reactions of nitrous acid with 1°- and 2°- aliphatic amines may be explained by considering their behavior with the nitrosonium cation, NO(+), an electrophilic species present in acidic nitrous acid solutions.
Secondary Amines
The distinct behavior of 1º, 2º & 3º-aliphatic amines is an instructive challenge to our understanding of their chemistry, but is of little importance as a synthetic tool. The SN1 product mixtures from 1º-amines are difficult to control, and rearrangement is common when branched primary alkyl groups are involved. The N-nitrosamines formed from 2º-amines are carcinogenic, and are not generally useful as intermediates for subsequent reactions.
Aryl Amines
Nitrous acid reactions of 1º-aryl amines generate relatively stable diazonium species that serve as intermediates for a variety of aromatic substitution reactions. Diazonium cations may be described by resonance contributors, as in the bracketed formulas shown below. The left-hand contributor is dominant because it has greater bonding. Loss of nitrogen is slower than in aliphatic 1º-amines because the C-N bond is stronger, and aryl carbocations are comparatively unstable.
Aqueous solutions of these diazonium ions have sufficient stability at 0º to 10 ºC that they may be used as intermediates in a variety of nucleophilic substitution reactions. For example, if water is the only nucleophile available for reaction, phenols are formed in good yield.
2º-Aryl Amines:
2º-Aryl amines give N-nitrosamine derivatives on reaction with nitrous acid, and thus behave identically to their aliphatic counterparts.
3º-Aryl Amines:
Depending on ring substitution, 3º-Aryl amines may undergo aromatic ring nitrosation at sites ortho or para to the amine substituent. The nitrosonium cation is not sufficiently electrophilic to react with benzene itself, or even toluene, but highly activated aromatic rings such as amines and phenols are capable of substitution. Of course, the rate of reaction of NO(+) directly at nitrogen is greater than that of ring substitution, as shown in the previous example. Once nitrosated, the activating character of the amine nitrogen is greatly diminished; and N-nitrosoaniline derivatives, or indeed any amide derivatives, do not undergo ring nitrosation. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Reactivity_of_Amines/Amines_as_Nucleophiles.txt |
Aryl diazonium salts are important intermediates. They are prepared in cold (0 º to 10 ºC) aqueous solution, and generally react with nucleophiles with loss of nitrogen. Some of the more commonly used substitution reactions are shown in the following diagram. Since the leaving group (N2) is thermodynamically very stable, these reactions are energetically favored. Those substitution reactions that are catalyzed by cuprous salts are known as Sandmeyer reactions. Fluoride substitution occurs on treatment with BF4(–), a reaction known as the Schiemann reaction. Stable diazonium tetrafluoroborate salts may be isolated, and on heating these lose nitrogen to give an arylfluoride product. The top reaction with hypophosphorus acid, H3PO2, is noteworthy because it achieves the reductive removal of an amino (or nitro) group. Unlike the nucleophilic substitution reactions, this reduction probably proceeds by a radical mechanism.
These aryl diazonium substitution reactions significantly expand the tactics available for the synthesis of polysubstituted benzene derivatives. Consider the following options:
1. The usual precursor to an aryl amine is the corresponding nitro compound. A nitro substituent deactivates an aromatic ring and directs electrophilic substitution to meta locations.
2. Reduction of a nitro group to an amine may be achieved in several ways. The resulting amine substituent strongly activates an aromatic ring and directs electrophilic substitution to ortho & para locations.
3. The activating character of an amine substituent may be attenuated by formation of an amide derivative (reversible), or even changed to deactivating and meta-directing by formation of a quaternary-ammonium salt (irreversible).
4. Conversion of an aryl amine to a diazonium ion intermediate allows it to be replaced by a variety of different groups (including hydrogen), which may in turn be used in subsequent reactions.
The following examples illustrate some combined applications of these options to specific cases. You should try to conceive a plausible reaction sequence for each. Once you have done so, you may check suggested answers by clicking on the question mark for each.
Bonding to Nitrogen
A resonance description of diazonium ions shows that the positive charge is delocalized over the two nitrogen atoms. It is not possible for nucleophiles to bond to the inner nitrogen, but bonding (or coupling) of negative nucleophiles to the terminal nitrogen gives neutral azo compounds. As shown in the following equation, this coupling to the terminal nitrogen should be relatively fast and reversible. The azo products may exist as E / Z stereoisomers. In practice it is found that the E-isomer predominates at equilibrium.
Unless these azo products are trapped or stabilized in some manner, reversal to the diazonium ion and slow nucleophilic substitution at carbon (with irreversible nitrogen loss) will be the ultimate course of reaction, as described in the previous section. For example, if phenyldiazonium bisufate is added rapidly to a cold solution of sodium hydroxide a relatively stable solution of sodium phenyldiazoate (the conjugate base of the initially formed diazoic acid) is obtained. Lowering the pH of this solution regenerates phenyldiazoic acid (pKa ca. 7), which disassociates back to the diazonium ion and eventually undergoes substitution, generating phenol.
C6H5N2(+) HSO4(–) + NaOH (cold solution) C6H5N2–OH + NaOH (cold) C6H5N2–O(–) Na(+)
phenyldiazonium bisulfate phenyldiazoic acid sodium phenyldiazoate
Aryl diazonium salts may be reduced to the corresponding hydrazines by mild reducing agents such as sodium bisulfite, stannous chloride or zinc dust. The bisulfite reduction may proceed by an initial sulfur-nitrogen coupling, as shown in the following equation.
Ar-N2(+) X(–)
NaHSO3
Ar-N=N-SO3H
NaHSO3
Ar-NH-NH-SO3H
H2O
Ar-NH-NH2 + H2SO4
The most important application of diazo coupling reactions is electrophilic aromatic substitution of activated benzene derivatives by diazonium electrophiles. The products of such reactions are highly colored aromatic azo compounds that find use as synthetic dyestuffs, commonly referred to as azo dyes. Azobenzene (Y=Z=H) is light orange; however, the color of other azo compounds may range from red to deep blue depending on the nature of the aromatic rings and the substituents they carry. Azo compounds may exist as cis/trans isomer pairs, but most of the well-characterized and stable compounds are trans.
Some examples of azo coupling reactions are shown below. A few simple rules are helpful in predicting the course of such reactions:
1. At acid pH (< 6) an amino group is a stronger activating substituent than a hydroxyl group (i.e. a phenol). At alkaline pH (> 7.5) phenolic functions are stronger activators, due to increased phenoxide base concentration.
2. Coupling to an activated benzene ring occurs preferentially para to the activating group if that location is free. Otherwise ortho-coupling will occur.
3. Naphthalene normally undergoes electrophilic substitution at an alpha-location more rapidly than at beta-sites; however, ortho-coupling is preferred. See the diagram for examples of α / β notation in naphthalenes.
You should try to conceive a plausible product structure for each of the following couplings. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Reactivity_of_Amines/Reactions_of_Aryl_Diazonium_Salts.txt |
Amine functions seldom serve as leaving groups in nucleophilic substitution or base-catalyzed elimination reactions. Indeed, they are even less effective in this role than are hydroxyl and alkoxyl groups. In the case of alcohols and ethers, a useful technique for enhancing the reactivity of the oxygen function was to modify the leaving group (OH(–) or OR(–)) to improve its stability as an anion (or equivalent). This stability is conveniently estimated from the strength of the corresponding conjugate acids.
As noted earlier, 1º and 2º-amines are much weaker acids than alcohols, so it is not surprising that it is difficult to force the nitrogen function to assume the role of a nucleophilic leaving group. For example, heating an amine with HBr or HI does not normally convert it to the corresponding alkyl halide, as in the case of alcohols and ethers. In this context we note that the acidity of the putative ammonium leaving group is at least ten powers of ten less than that of an analogous oxonium species. The loss of nitrogen from diazonium intermediates is a notable exception in this comparison, due to the extreme stability of this leaving group (the conjugate acid of N2 would be an extraordinarily strong acid).
One group of amine derivatives that have proven useful in SN2 and E2 reactions is that composed of the tetraalkyl (4º-) ammonium salts. Most applications involving this class of compounds are eliminations, but a few examples of SN2 substitution have been reported.
C6H5–N(CH3)3(+) Br(–) + R-S(–) Na(+)
acetone & heat
R-S-CH3 + C6H5–N(CH3)2 + NaBr
(CH3)4N(+) OH(–)
heat
CH3–OH + (CH3)3N
Hofmann Elimination
Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below shows a typical Hofmann elimination. Obviously, for an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted earlier in examining elimination reactions of alkyl halides.
In example #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The chief product from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical advantage of those beta-hydrogens, but also the greater stability of the double bond.
Example #3 illustrates two important features of the Hofmann elimination:
1. Simple amines are easily converted to the necessary 4º-ammonium salts by exhaustive alkylation, usually with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction). Exhaustive methylation is shown again in example #4.
2. When a given alkyl group has two different sets of beta-hydrogens available to the elimination process (colored orange & magenta here), the major product is often the alkene isomer having the less substituted double bond.
The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the Hofmann Rule, and contrasts strikingly with the Zaitsev Rule formulated for dehydrohalogenations and dehydrations. In cases where other activating groups, such as phenyl or carbonyl, are present, the Hofmann Rule may not apply. Thus, if 2-amino-1-phenylpropane is treated in the manner of example #3, the product consists largely of 1-phenylpropene (E & Z-isomers).
To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state, first described for dehydrohalogenation. The energy diagram shown earlier for a single-step bimolecular E2 mechanism is repeated below.
The E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the bond to the leaving group (X) is substantially broken relative to the other bond changes, the transition state approaches that for an E1 reaction (initial ionization followed by a fast second step). At the other extreme, if the acidity of the beta-hydrogens is enhanced, then substantial breaking of C–H may occur before the other bonds begin to be affected. For most simple alkyl halides it was proper to envision a balanced transition state, in which there was a synchronous change in all the bonds. Such a model was consistent with the Zaitsev Rule.
When the leaving group X carries a positive charge, as do the 4º-ammonium compounds discussed here, the inductive influence of this charge will increase the acidity of both the alpha and the beta-hydrogens. Furthermore, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations available to substituted beta-carbons. It seems that a combination of these factors acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of the leaving group and beta-hydrogen, noted for dehydrohalogenation, is found for many Hofmann eliminations; but syn-elimination is also common, possibly because the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group.
Three additional examples of the Hofmann elimination are shown in the following diagram. Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation.
Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to sever the nitrogen from the remaining molecular framework.
Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path.
The Reaction of Amines with Nitrous Acid
This page looks at the reactions of nitrous acid with aliphatic amines (those where the amine group isn't attached directly to a benzene ring). Nitrous acid is properly called nitric(III) acid, but that name isn't commonly used.
Testing for the various types of amines
The reaction between amines and nitrous acid was used in the past as a very neat way of distinguishing between primary, secondary and tertiary amines. However, the product with a secondary amine is a powerful carcinogen, and so this reaction is no longer carried out at this level.
Nitrous acid, HNO2, (sometimes written as HONO to show its structure) is unstable and is always prepared in situ. It is usually made by reacting a solution containing sodium or potassium nitrite (sodium or potassium nitrate(III)) with hydrochloric acid. Nitrous acid is a weak acid and so you get the reaction:
$H^+ (aq) + NO_2^- (aq) \rightleftharpoons HNO_2 (aq)$
Because nitrous acid is a weak acid, the position of equilibrium lies well the right. In each of the following reactions, the amine would be acidified with hydrochloric acid and a solution of sodium or potassium nitrite added. The acid and the nitrite form nitrous acid which then reacts with the amine.
Primary amines and nitrous acid
The main observation is a burst of colorless, odorless gas. Nitrogen is given off. Unfortunately, there is no single clear-cut equation that you can quote for this. You get lots of different organic products. For example, amongst the products you get an alcohol where the -NH2 group has been replaced by OH. If you want a single equation, you could quote (taking 1-aminopropane as an example):
$CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CH_2OH + H_2O + N_2$
The propan-1-ol will be only one product among many - including propan-2-ol, propene, 1-chloropropane, 2-chloropropane and others. The nitrogen, however, is given off in quantities exactly as suggested by the equation. By measuring the amount of nitrogen produced, you could use this reaction to work out the amount of amine present in the solution.
Secondary amines and nitrous acid
This time there isn't any gas produced. Instead, you get a yellow oil called a nitrosamine. These compounds are powerful carcinogens - avoid them! For example:
Tertiary amines and nitrous acid
Again, a quite different result. This time, nothing visually interesting happens - you are left with a colorless solution. All that has happened is that the amine has formed an ion by reacting with the acid present. With trimethylamine, for example, you would get a trimethylammonium ion, (CH3)3NH+.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Reactivity_of_Amines/Substitution_and_Elimination_Reactions_of_Amines.txt |
• Azide Reduction
• Gabriel Synthesis
The Gabriel synthesis is a great way to make primary amines. This alkylation procedure doesn’t produce ammonium salts like the SN2 reaction would. Potassium phthalimide is treated with base, then a primary alkyl halide, and then either hydrazine, acid, or base.
• Preparation of Amines
• Preparation of Amines1
This page looks at the preparation of amines from halogenoalkanes (also known as haloalkanes or alkyl halides) and from nitriles.
• The Leuckart Reaction
The Leuckart Reaction is a useful procedure for the reductive alkylation of ammonia, 1º-, & 2º-amines, in which formic acid or a derivative thereof serves as the reducing agent.
Synthesis of Amines
The Gabriel synthesis is a great way to make primary amines. This alkylation procedure doesn’t produce ammonium salts like the SN2 reaction would. Potassium phthalimide is treated with base, then a primary alkyl halide, and then either hydrazine, acid, or base.
Overview
The goal of Gabriel synthesis is to create a primary amine (RNH2). Doing a direct SN2 reaction would result in a quaternary amine salt, which can be pretty useless (unless you use Hofmann elimination). Before we get into the mechanism, let’s look at the reaction pathway.
Gabriel synthesis summary
Mechanism
We start off with phthalimide, treat it with base like KOH or NaOH, add an alkyl halide, and then liberate the resulting amine. Let’s break down each step of the mechanism:
Nitrogen deprotonation
In the very first step, the nitrogen of the phthalimide is deprotonated to create potassium phthalimide. Heads up: some professors might just start off with potassium phthalimide and skip the deprotonation step entirely.
Nitrogen Alkylation
Now that the nitrogen has a negative charge, it can act as a nucleophile and perform an SN2 reaction on an alkyl halide. Once that happens, we’re got a primary amine just waiting to be liberated.
Hydrazine nucleophilic acyl substitution
At this point, there are a couple of variations. Usually, hydrazine (NH2NH2) is used, but sometimes acidic hydrolysis or basic hydrolysis is used. We’ll take a look at those mechanisms later. In any case, the nucleophile kicks off the nitrogen through a nucleophilic acyl substitution mechanism.
Intramolecular N-N proton transfer
The protonated hydrazine is then deprotonated by the nitrogen that was just kicked off of one carbonyl. We still need to get the amine off of the second carbonyl, and the NH2 of the hydrazine repeats the process to kick it off.
Second NAS
The unreacted NH2 of the hydrazine then attacks the other carbonyl through another NAS mechanism to finally kick off the amine. Now there is a negative charge and a positive charge, and those need to be taken care of.
Primary Amine Formation
The negative nitrogen deprotonates the positive nitrogen, and drumroll please… we’ve got our primary amine (as well as the cyclic product phthalhydrazide)! In case you want to see the whole mechanism at once, here it is:
Hydrazine full mechanism
Alternative Mechanisms with Acid or Base
Hydrazine is a very dangerous, reactive molecule. If you’ve seen or read The Martian, you might have an idea of how much care needs to be taken when working with it. Chemists try to minimize the risk of lab accidents, so sometimes they’ll find ways to work without dangerous chemicals like hydrazine. Luckily, the amine liberation portion of this reaction can be done in acidic or basic conditions.
Acidic Hydrolysis
The acidic hydrolysis version of the amine liberation works just like ester or amide hydrolysis, and the phthalimide becomes phthalic acid (o-dicarboxybenzene). Let’s take a look at the mechanism in acidic conditions. The mechanism shows H3O+, which can be made by adding any acid (like HCl) in aqueous solution.
Acidic hydrolysis mechanism
Basic Hydrolysis
The mechanism in base is pretty similar to saponification (aka base-catalyzed ester hydrolysis), but with nitrogen instead of an oxygen attached to the R-group.
Basic hydrolysis mechanism
For the record, look at how few steps the hydrazine and basic hydrolysis mechanisms have compared to the acidic version. If you have the choice on an exam, I’d recommend mechanisms with the fewest steps. This reaction can be used to make tons of different primary amines, including benzylamine, n-butylamine, and many more!
Contributors
• Johnny Betancourt, Clutchprep. Source page can be accessed here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Synthesis_of_Amines/Gabriel_Synthesis.txt |
Preparation of Primary Amines
Although direct alkylation of ammonia by alkyl halides leads to 1º-amines, alternative procedures are preferred in many cases. These methods require two steps, but they provide pure product, usually in good yield. The general strategy is to first form a carbon-nitrogen bond by reacting a nitrogen nucleophile with a carbon electrophile. The following table lists several general examples of this strategy in the rough order of decreasing nucleophilicity of the nitrogen reagent. In the second step, extraneous nitrogen substituents that may have facilitated this bonding are removed to give the amine product.
Nitrogen
Reactant
Carbon
Reactant
1st Reaction
Type
Initial Product
2nd Reaction
Conditions
2nd Reaction
Type
Final Product
N3(–) RCH2-X or
R2CH-X
SN2 RCH2-N3 or
R2CH-N3
LiAlH4 or
4 H2 & Pd
Hydrogenolysis RCH2-NH2 or
R2CH-NH2
C6H5SO2NH(–) RCH2-X or
R2CH-X
SN2 RCH2-NHSO2C6H5 or
R2CH-NHSO2C6H5
Na in NH3 (liq) Hydrogenolysis RCH2-NH2 or
R2CH-NH2
CN(–) RCH2-X or
R2CH-X
SN2 RCH2-CN or
R2CH-CN
LiAlH4 Reduction RCH2-CH2NH2 or
R2CH-CH2NH2
NH3 RCH=O or
R2C=O
Addition /
Elimination
RCH=NH or
R2C=NH
H2 & Ni
or NaBH3CN
Reduction RCH2-NH2 or
R2CH-NH2
NH3 RCOX Addition /
Elimination
RCO-NH2 LiAlH4 Reduction RCH2-NH2
NH2CONH2
(urea)
R3C(+) SN1 R3C-NHCONH2 NaOH soln. Hydrolysis R3C-NH2
A specific example of each general class is provided in the diagram below. In the first two, an anionic nitrogen species undergoes an SN2 reaction with a modestly electrophilic alkyl halide reactant. For example #2 an acidic phthalimide derivative of ammonia has been substituted for the sulfonamide analog listed in the table. The principle is the same for the two cases, as will be noted later. Example #3 is similar in nature, but extends the carbon system by a methylene group (CH2). In all three of these methods 3º-alkyl halides cannot be used because the major reaction path is an E2 elimination.
The methods illustrated by examples #4 and #5 proceed by attack of ammonia, or equivalent nitrogen nucleophiles, at the electrophilic carbon of a carbonyl group. A full discussion of carbonyl chemistry is presented later, but for present purposes it is sufficient to recognize that the C=O double bond is polarized so that the carbon atom is electrophilic. Nucleophile addition to aldehydes and ketones is often catalyzed by acids. Acid halides and anhydrides are even more electrophilic, and do not normally require catalysts to react with nucleophiles. The reaction of ammonia with aldehydes or ketones occurs by a reversible addition-elimination pathway to give imines (compounds having a C=N function). These intermediates are not usually isolated, but are reduced as they are formed (i.e. in situ). Acid chlorides react with ammonia to give amides, also by an addition-elimination path, and these are reduced to amines by LiAlH4.
The 6th example is a specialized procedure for bonding an amino group to a 3º-alkyl group (none of the previous methods accomplishes this). Since a carbocation is the electrophilic species, rather poorly nucleophilic nitrogen reactants can be used. Urea, the diamide of carbonic acid, fits this requirement nicely. The resulting 3º-alkyl-substituted urea is then hydrolyzed to give the amine.
One important method of preparing 1º-amines, especially aryl amines, uses a reverse strategy. Here a strongly electrophilic nitrogen species (NO2(+)) bonds to a nucleophilic carbon compound. This nitration reaction gives a nitro group that can be reduced to a 1º-amine by any of several reduction procedures.
The Hofmann rearrangement of 1º-amides provides an additional synthesis of 1º-amines. To learn about this useful procedure Click Here.
Preparation of Secondary and Tertiary Amines
Of the six methods described above, three are suitable for the preparation of 2º and/or 3º-amines. These are
1. Alkylation of the sulfonamide derivative of a 1º-amine. Gives 2º-amines.
2. Reduction of alkyl imines and dialkyl iminium salts. Gives 2º & 3º-amines.
3. Reduction of amide derivatives of 1º & 2º-amines. Gives 2º & 3º-amines.
Examples showing the application of these methods to the preparation of specific amines are shown in the following diagram. The sulfonamide procedure used in the first example is similar in concept to the phthalimide example #2 presented in the previous diagram. In both cases the acidity of the nitrogen reactant (ammonia or amine) is greatly enhanced by conversion to an imide or sulfonamide derivative. The nucleophilic conjugate base of this acidic nitrogen species is then prepared by treatment with sodium or potassium hydroxide, and this undergoes an SN2 reaction with a 1º or 2º-alkyl halide. Finally, the activating group is removed by hydrolysis (phthalimide) or reductive cleavage (sulfonamide) to give the desired amine. The phthalimide method is only useful for preparing 1º-amines, whereas the sulfonamide procedure may be used to make either 1º or 2º-amines.
Examples #2 & #3 make use of the carbonyl reductive amination reaction (method #4 in the preceding table. This versatile procedure may be used to prepare all classes of amines (1º, 2º & 3º), as shown here and above. A weak acid catalyst is necessary for imine formation, which takes place by amine addition to the carbonyl group, giving a 1-aminoalcohol intermediate, followed by loss of water. The final reduction of the C=N double bond may be carried out catalytically (Pt & Pd catalysts may be used instead of Ni) or chemically (by NaBH3CN). The imine or enamine intermediates are normally not isolated, but are immediately reduced to the amine product.
Another general method for preparing all classes of amines makes use of amide intermediates, easily made from ammonia or amines by reaction with carboxylic acid chlorides or anhydrides. These stable compounds may be isolated, identified and stored prior to the final reduction. Examples #4 & #5 illustrate applications of this method. As with the previous method, 1º-amines give 2º-amine products, and 2º-amines give 3º-amine products. The last example (#6) shows how 4º-ammonium salts may be prepared by repeated (exhaustive) alkylation of amines. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Synthesis_of_Amines/Preparation_of_Amines.txt |
This page looks at the preparation of amines from halogenoalkanes (also known as haloalkanes or alkyl halides) and from nitriles.
Making amines from halogenoalkanes
The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas. We'll talk about the reaction using 1-bromoethane as a typical halogenoalkane. You get a mixture of amines formed together with their salts. The reactions happen one after another.
Making a primary amine
The reaction happens in two stages. In the first stage, a salt is formed - in this case, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group.
$CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_3^+Br^-$
There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
$CH_3CH_2NH_3^+Br^- + NH_3 \rightleftharpoons CH_3CH_2NH_2 + NH_4^+ Br^-$
The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine - ethylamine. The more ammonia there is in the mixture, the more the forward reaction is favored.
Making a secondary amine
The reaction doesn't stop at a primary amine. The ethylamine also reacts with bromoethane - in the same two stages as before. In the first stage, you get a salt formed - this time, diethylammonium bromide. Think of this as ammonium bromide with two hydrogens replaced by ethyl groups.
There is again the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
The ammonia removes a hydrogen ion from the diethylammonium ion to leave a secondary amine - diethylamine. A secondary amine is one which has two alkyl groups attached to the nitrogen.
Making a tertiary amine
The reaction does not stop! The diethylamine also reacts with bromoethane - in the same two stages as before. In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
The ammonia removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine. A tertiary amine is one which has three alkyl groups attached to the nitrogen.
Making a quaternary ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
Reacting bromoethane with ammonia
Whatever you do, you get a mixture of all of the products (including the various amines and their salts) shown on this page. To get mainly the quaternary ammonium salt, you can use a large excess of bromoethane. If you look at the reactions going on, each one needs additional bromoethane. If you provide enough, then the chances are that the reaction will go to completion, given enough time.
On the other hand, if you use a very large excess of ammonia, the chances are always greatest that a bromoethane molecule will hit an ammonia molecule rather than one of the amines being formed. That will help to prevent the formation of secondary (etc) amines - although it won't stop it entirely.
Making primary amines from nitriles
Nitriles are compounds containing the -CN group, and can be reduced in various ways. Two possible methods are described here.
1. Reducing nitriles using LiAlH4. One possible reducing agent is lithium tetrahydridoaluminate(III) - often just called lithium tetrahydridoaluminate or lithium aluminum hydride. The nitrile reacts with the lithium tetrahydridoaluminate in solution in ethoxyethane (diethyl ether, or just "ether") followed by treatment of the product of that reaction with a dilute acid. Overall, the carbon-nitrogen triple bond is reduced to give a primary amine. For example, with ethanenitrile you get ethylamine: $CH_3CN + 4[H] \rightarrow CH_3CH_2NH_2$
2. The reduction of nitriles using hydrogen and a metal catalyst. The carbon-nitrogen triple bond in a nitrile can also be reduced by reaction with hydrogen gas in the presence of a variety of metal catalysts. Commonly quoted catalysts are palladium, platinum or nickel. The reaction will take place at a raised temperature and pressure. It is impossible to give exact details because it will vary from catalyst to catalyst. For example, ethanenitrile can be reduced to ethylamine by reaction with hydrogen in the presence of a palladium catalyst.
The Leuckart Reaction
A useful procedure for the reductive alkylation of ammonia, 1º-, & 2º-amines, in which formic acid or a derivative thereof serves as the reducing agent, is known as the Leuckart Reaction. Some examples of this reaction are shown below.
The manner in which a hydride moiety is transferred from formate to an iminium intermediate is a matter for speculation, but may be summarized roughly as below:
.
Both aldehydes and ketones may be used as the carbonyl reactant. By using ammonia as a reactant, this procedure may be used to prepare 1º-amines; however, care must be taken to avoid further alkylation to 2º & 3º-amines. Polyalkylation is sometimes desired, as in example #3 where dimethylation is accomplished with formaldehyde. This is sometimes referred to as the Eschweiler-Clarke procedure, and it has proven to be a useful method for converting 1º-amines to precursors for Hofmann or Cope elimination reactions. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Synthesis_of_Amines/Preparation_of_Amines1.txt |
The acid anhydride functional group results when two carboxylic acids combine and lose water (anhydride = without water). Symmetrical acid anhydrides are named like carboxylic acids except the ending -acid is replaced with -anhydride. This is true for both the IUPAC and Common nomenclature.
Unsymmetrical acid anhydrides
Unsymmetrical acid anhydrides are named by first naming each component carboxylic acid alphabetically arranged (without the word acid) followed by spaces and then the word anhydride.
propanoic anhydride
ethanoic propanoic anhydride
Try to name the following compound
Try to draw a structure for the following compound�
• 1,2-benzenedicarboxylic anhydride J
Common names that you should know
acetic anhydride (Try to name this anhydride by the proper name. J )
succinic anhydride (Try to name this anhydride by the proper name. J )
Contributors
Prof. Steven Farmer (Sonoma State University)
Properties of Anhydrides
This page explains what acid anhydrides are and looks at their simple physical properties such as boiling points. It introduces their chemical reactivity in a general way. A carboxylic acid such as ethanoic acid has the structure:
If you took two ethanoic acid molecules and removed a molecule of water between them you would get the acid anhydride, ethanoic anhydride (old name: acetic anhydride).
You can actually make ethanoic anhydride by dehydrating ethanoic acid, but it is normally made in a more efficient, round-about way.
Naming acid anhydrides
This is really easy. You just take the name of the parent acid, and replace the word "acid" by "anhydride". "Anhydride" simply means "without water". So . . . ethanoic acid forms ethanoic anhydride; propanoic acid forms propanoic anhydride, and so on.
Physical properties of acid anhydrides
We will take ethanoic anhydride as typical.
Appearance
Ethanoic anhydride is a colorless liquid, smelling strongly of vinegar (ethanoic acid). The smell is because ethanoic anhydride reacts with water vapor in the air (and moisture in your nose) to produce ethanoic acid again. This reaction with water is given in detail on another page. (Find it from the acid anhydrides menu - link at the bottom of this page.)
Solubility in water
Ethanoic anhydride can't be said to dissolve in water because it reacts with it to give ethanoic acid. There is no such thing as an aqueous solution of ethanoic anhydride.
Boiling point
Ethanoic anhydride boils at 140°C. This is because it is a fairly big polar molecule and so has both van der Waals dispersion forces and dipole-dipole attractions. It does not, however, form hydrogen bonds. That means that its boiling point isn't as high as a carboxylic acid of similar size. For example, pentanoic acid (the most similarly sized acid) boils at 186°C.
Reactivity of acid anhydrides
Comparing acid anhydrides with acyl chlorides (acid chlorides)
You have almost certainly come across acid anhydrides for the first time just after looking at acyl chlorides, or you may be studying them at the same time as acyl chlorides. It is much, much easier to think of acid anhydrides as if they were a sort of modified acyl chloride than to try to learn about them from scratch. That is the line I intend to take throughout all this section. Compare the structure of an acid anhydride with that of an acyl chloride - looking carefully at the way it is color in the diagram.
In the reactions of ethanoic anhydride, the red group at the bottom always stays intact. It is behaving in many ways as if it was a single atom - just like the chlorine atom in the acyl chloride. The usual reaction of an acyl chloride is replacement of the chlorine by something else.
Taking ethanoyl chloride as typical, the initial reaction is of this kind:
Hydrogen chloride gas is given off, although that might go on to react with other components of the mixture. With an acid anhydride, the reaction is slower, but the only essential difference is that instead of hydrogen chloride being produced as the other product, you get ethanoic acid instead.
Just like the hydrogen chloride, this might afterwards go on to react with other things present. The reactions (of both acyl chlorides and acid anhydrides) involve things like water, alcohols and phenols, or ammonia and amines. All of these particular cases contain a very electronegative element with an active lone pair of electrons - either oxygen or nitrogen.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Anhydrides/Nomenclature_of_Anhydrides.txt |
Acid anhydrides are a source of reactive acyl groups, and their reactions and uses resemble those of acyl halides. Acid anhydrides tend to be less electrophilic than acyl chlorides, and only one acyl group is transferred per molecule of acid anhydride, which leads to a lower atom efficiency. The low cost, however, of acetic anhydride makes it a common choice for acetylation reactions.
• Acid Anhydrides react with alcohols to form esters
Acid Anhydrides react with alcohols to form esters
• Acid Anhydrides React with Amines to Form Amides
Acid Anhydrides react with amines to form amides
• Acid Anhydrides react with water to form carboxylic acids
Acid Anhydrides react with water to form carboxylic acids
• General Mechanism of Anhydride Reactions
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.
• Reactions of Acid Anhydrides with Nitrogen Compounds
This page looks at the reactions of acid anhydrides with ammonia and with primary amines. These reactions are considered together because their chemistry is so similar. There is also a great similarity between acid anhydrides and acyl chlorides (acid chlorides) as far as these reactions are concerned.
• Reactions of Acid Anhydrides with Oxygen Compounds
This page looks at the reactions of acid anhydrides with water, alcohols and phenols (including the manufacture of aspirin). These reactions are all considered together because their chemistry is so similar. There is also a great similarity between acid anhydrides and acyl chlorides (acid chlorides) as far as these reactions are concerned.
Reactivity of Anhydrides
Acid Anhydrides react with amines to form amides
Example 1:
Mechanism
1) Nucleophilic Attack by the Amine
2) Deprotonation by the amine
3) Leaving group removal
Contributors
Prof. Steven Farmer (Sonoma State University)
Acid Anhydrides react with alcohols to form esters
Acid Anhydrides react with alcohols to form esters
General Reaction
Reactions of anhydrides use Pyridine as a solvent
Example 1:
Mechanism
1) Nucleophilic Attack by the Alcohol
2) Deprotonation by pyridine
3) Leaving group removal
4) Protonation of the carboxylate
Contributors
Prof. Steven Farmer (Sonoma State University)
Acid Anhydrides react with water to form carboxylic acids
Acid Anhydrides react with water to form carboxylic acids
Example 1:
Mechanism
1) Nucleophilic Attack by the Alcohol
2) Deprotonation by pyridine
3) Leaving group removal
4) Protonation of the carboxylate
General Mechanism of Anhydride Reactions
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) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Anhydrides/Reactivity_of_Anhydrides/Acid_Anhydrides_React_with_Amines_to_Form_Amides.txt |
This page looks at the reactions of acid anhydrides with ammonia and with primary amines. These reactions are considered together because their chemistry is so similar. There is also a great similarity between acid anhydrides and acyl chlorides (acid chlorides) as far as these reactions are concerned.
Comparing the structures of ammonia and primary amines
Each substance contains an -NH2 group. In ammonia, this is attached to a hydrogen atom. In a primary amine, it is attached to an alkyl group (shown by "R" in the diagram below) or a benzene ring.
Comparing the reactions of acyl chlorides and acid anhydrides with these compounds
Because the formula is much easier, it helps to start with the acyl chlorides.
The reactions with acyl chlorides
We'll take ethanoyl chloride as typical of the acyl chlorides. Taking a general case of a reaction between ethanoyl chloride and a compound XNH2 (where X is hydrogen, or an alkyl group, or a benzene ring). The reaction happens in two stages:
First:
In each case, the reaction initially produces hydrogen chloride gas - the hydrogen coming from the -NH2 group, and the chlorine from the ethanoyl chloride. Everything left over just gets joined together. But ammonia and amines are basic, and react with the hydrogen chloride to produce a salt. So the second stage of the reaction is:
\[ XNH_2 + HCl \longrightarrow XNH_3^+ + Cl^-\]
The same reactions with acid anhydrides
Again, the reaction happens in two stages. In the first:
If you compare this with the acyl chloride equation, you can see that the only difference is that ethanoic acid is produced as the second product of the reaction rather than hydrogen chloride. Then the ethanoic acid reacts with excess ammonia or amine to give a salt - this time an ethanoate.
\[ CH_3COOH + XNH_2 \longrightarrow CH_3COO^- + {^+}NH_3X \]
This looks more difficult than the acyl chloride case because of the way the salt is written. You get an ethanoate ion and a positive ion with this structure:
This is easier to understand with real compounds - as you will see below. In summary, these reactions are just the same as the corresponding acyl chloride reactions except:
• Initially, ethanoic acid is formed as the second product rather than hydrogen chloride gas.
• The second stage of the reaction involves the formation of an ethanoate rather than a chloride.
• The reactions are slower. Acid anhydrides aren't so violently reactive as acyl chlorides, and the reactions normally need heating.
The individual reactions
The reaction with ammonia
In this case, the "X" in the equations above is a hydrogen atom. So in the first instance you get ethanoic acid and an organic compound called an amide. Amides contain the group -CONH2. In the reaction between ethanoic anhydride and ammonia, the amide formed is called ethanamide.
This is more usually (and more easily!) written as:
The ethanoic acid produced reacts with excess ammonia to give ammonium ethanoate.
. . . and you can combine all this together to give one overall equation:
You need to follow this through really carefully, because the two products of the reaction overall can look confusingly similar. The corresponding reaction with an acyl chloride is:
\[ CH_3COCl + 2NH_3 \longrightarrow CH_3CONH_2 + NH_4^+\,Cl^-\]
The reaction with primary amines
The reaction with methylamine
We'll take methylamine as typical of primary amines where the -NH2 is attached to an alkyl group. The initial equation would be:
The first product this time is called an N-substituted amide.
If you compare the structure with the amide produced in the reaction with ammonia, the only difference is that one of the hydrogens on the nitrogen has been substituted for a methyl group. This particular compound is N-methylethanamide. The "N" simply shows that the substitution is on the nitrogen atom, and not elsewhere in the molecule. The equation would normally be written:
You can think of primary amines as just being modified ammonia. If ammonia is basic and forms a salt with the ethanoic acid, excess methylamine will do exactly the same thing.
The salt is called methylammonium ethanoate. It is just like ammonium ethanoate, except that one of the hydrogens has been replaced by a methyl group. You would usually combine these equations into one overall equation for the reaction:
\[ (CH_3CO)_2O + 2CH_3NH_2 \longrightarrow CH_3CONHCH_3 + CH_3COO^- ^{+}NH_3CH_3\]
The corresponding reaction with an acyl chloride is:
\[ CH_3OCl + 2CH_3NH_2 \longrightarrow CH_3CONHCH_3 + CH_3NH_3^+ \, Cl^- \]
The reaction with phenylamine (aniline)
Phenylamine is the simplest primary amine where the -NH2 group is attached directly to a benzene ring. Its old name is aniline. In phenylamine, there isn't anything else attached to the ring as well. You can write the formula of phenylamine as C6H5NH2. There is no essential difference between this reaction and the reaction with methylamine, but I just want to look at the structure of the N-substituted amide formed.
The overall equation for the reaction is:
\[ (CH_3CO)_2O + 2C_6H_5NH_2 \longrightarrow CH_3CONHC_6H_5 + CH_3COO^- \; ^+NH_3C_6H_5 \]
The products are N-phenylethanamide and phenylammonium ethanoate. This reaction can sometimes look (even more!) confusing if the phenylamine is drawn showing the benzene ring, and especially if the reaction is looked at from the point of view of the phenylamine. For example, the product molecule might be drawn looking like this:
If you stop and think about it, this is obviously the same molecule as in the equation above, but it stresses the phenylamine part of it much more. Looking at it this way, notice that one of the hydrogens of the -NH2 group has been replaced by an acyl group - an alkyl group attached to a carbon-oxygen double bond. You can say that the phenylamine has been acylated or has undergone acylation. Because of the nature of this particular acyl group, it is also described as ethanoylation. The hydrogen is being replaced by an ethanoyl group, CH3CO-.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Anhydrides/Reactivity_of_Anhydrides/Reactions_of_Acid_Anhydrides_with_Nitrogen_Compounds.txt |
This page looks at the reactions of acid anhydrides with water, alcohols and phenols (including the manufacture of aspirin). These reactions are all considered together because their chemistry is so similar. There is also a great similarity between acid anhydrides and acyl chlorides (acid chlorides) as far as these reactions are concerned.
Comparing the structures of water, ethanol and phenol
Each substance contains an -OH group. In water, this is attached to a hydrogen atom. In an alcohol, it is attached to an alkyl group - shown in the diagrams below as "R". In phenols, it is attached to a benzene ring. Phenol (the simplest member of the family of phenols) is C6H5OH.
The reactions with acyl chlorides
Because the formula is much easier, it helps to start with the acyl chlorides. We'll take ethanoyl chloride as typical of the acyl chlorides. Taking a general case of a reaction between ethanoyl chloride and a compound X-O-H (where X is hydrogen, or an alkyl group or a benzene ring):
So . . . in each case, hydrogen chloride gas is produced - the hydrogen coming from the -OH group, and the chlorine from the ethanoyl chloride. Everything left over just gets joined together.
The same reactions with acid anhydrides
If you compare this with the acyl chloride equation, you can see that the only difference is that ethanoic acid is produced as the second product of the reaction rather than hydrogen chloride. These reactions are just the same as the corresponding acyl chloride reactions except:
• Ethanoic acid is formed as the second product rather than hydrogen chloride gas.
• The reactions are slower. Acid anhydrides are not so violently reactive as acyl chlorides.
The reaction with water
Modifying the general equation we've just looked at, you will see that you just get two molecules of ethanoic acid produced.
This is more usually (and more easily!) written as:
\[(CH_3CO)_2O + H_2O \longrightarrow 2CH_3COOH \]
The reaction happens slowly at room temperature (faster on gentle warming) without a great deal exciting to observe - unlike in the acyl chloride case where hydrogen chloride fumes are produced. You mix two colorless liquids and get another colorless liquid!
The equivalent acyl chloride reaction is:
\[ CH_3COCl + H_2O \longrightarrow CH_3COOH + HCl\]
The reaction with alcohols
We'll start by taking the general case of any alcohol reacting with ethanoic anhydride. The equation would be:
or, more simply:
The product this time (apart from the ethanoic acid always produced) is an ester. For example, with ethanol you would get the ester ethyl ethanoate:
This reaction also needs gentle heating for it to happen at a reasonable rate, and again there is not anything visually dramatic. The equivalent acyl chloride reaction is:
\[ CH_3COCl + CH_3CH_2OH \longrightarrow CH_3COOCH_2CH_3 + HCl\]
The reaction with phenols
Phenols have an -OH group attached directly to a benzene ring. In the substance normally called "phenol", there isn't anything else attached to the ring as well. We'll look at that first. The reaction between phenol and ethanoic anhydride isn't particularly important, but you would get an ester just as you do with an alcohol.
Or, more simply:
\[ (CH_3CO)_2O + C_6H_5OH \longrightarrow CH_3COOC_6H_5 + CH_3COOH \]
Especially if you write the equation in this second way, it is obvious that you have just produced another ester - in this case, called phenyl ethanoate. The equivalent acyl chloride reaction is:
\[ CH_3COCl + C_6H_5OH \longrightarrow CH_3COOC_6H_5 + HCl\]
But beware! You may come across the structure of the ester drawn in a variety of other ways which make it look much more as if it was a derivative of phenol (which of course it is!).
Example
Looking at it this way, notice that the hydrogen of the phenol -OH group has been replaced by an acyl group - an alkyl group attached to a carbon-oxygen double bond. You can say that the phenol has been acylated or has undergone acylation. Because of the nature of this particular acyl group, it is also described as ethanoylation. The hydrogen is being replaced by an ethanoyl group, CH3CO-.
Using a similar reaction to make aspirin
The reaction with phenol itself isn't very important, but you can make aspirin by a very similar reaction. The molecule below is 2-hydroxybenzoic acid (also known as 2-hydroxybenzenecarboxylic acid). The old name for this is salicylic acid.
You might find it written in either of these two ways. They are the same structure with the molecule just flipped over in space.
You might also find it with the -OH group at the top and the -COOH group next door and either to the left or right of it. Life can get very confusing! When this reacts with ethanoic anhydride, it is ethanoylated (or acylated, if you want to use the more general term) to give:
You might find all sorts of other variants on drawing this as well. This molecule is aspirin.
Although this reaction can also be done with ethanoyl chloride, aspirin is manufactured by reacting 2-hydroxybenzoic acid with ethanoic anhydride at 90°C. The reasons for using ethanoic anhydride rather than ethanoyl chloride include:
• Ethanoic anhydride is cheaper than ethanoyl chloride.
• Ethanoic anhydride is safer to use than ethanoyl chloride. It is less corrosive and not so readily hydrolysed (its reaction with water is slower).
• Ethanoic anhydride doesn't produce dangerous (corrosive and poisonous) fumes of hydrogen chloride.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Anhydrides/Reactivity_of_Anhydrides/Reactions_of_Acid_Anhydrides_with_Oxygen_Compounds.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 | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Nomenclature_of_Arenes.txt |
Aromaticity is a property of conjugated cycloalkenes in which the stabilization of the molecule is enhanced due to the ability of the electrons in the $\pi$ orbitals to delocalize. This acts as a framework to create a planar molecule.
Properties of Arenes
Aromaticity is a property of conjugated cycloalkenes in which the stabilization of the molecule is enhanced due to the ability of the electrons in the $\pi$ orbitals to delocalize. This act as a framework to create a planar molecule.
Introduction
Why do we care if a compound is aromatic or not? Because we encounter aromatics every single day of our lives. Without aromatic compounds, we would not only be lacking many material necessities, our bodies would also not be able to function. Aromatic compounds are essential in industry; about 35 million tons of aromatic compounds are produced in the world every year to produce important chemicals and polymers, such as polyester and nylon. Aromatic compounds are also vital to the biochemistry of all living things. Three of the twenty amino acids used to form proteins ("the building blocks of life") are aromatic compounds and all five of the nucleotides that make up DNA and RNA sequences are all aromatic compounds. Needless to say, aromatic compounds are vital to us in many aspects.
The three general requirements for a compound to be aromatic are:
1. The compound must be cyclic
2. Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar.
3. The compound must follow Hückel's Rule (the ring has to contain 4n+2 p-orbital electrons).
Among the many distinctive features of benzene, its aromaticity is the major contributor to why it is so unreactive. This section will try to clarify the theory of aromaticity and why aromaticity gives unique qualities that make these conjugated alkenes inert to compounds such as Br2 and even hydrochloric acid. It will also go into detail about the unusually large resonance energy due to the six conjugated carbons of benzene.
The delocalization of the p-orbital carbons on the sp2 hybridized carbons is what gives the aromatic qualities of benzene.
Basic Structue of Benzene
Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. You might ask yourselves how it's possible to have all of the bonds to be the same length if the ring is conjugated with both single (1.47 Å) and double (1.34 Å), but it is important to note that there are no distinct single or double bonds within the benzene. Rather, the delocalization of the ring makes each count as one and a half bonds between the carbons which makes sense because experimentally we find that the actual bond length is somewhere in between a single and double bond. Finally, there are a total of six p-orbital electrons that form the stabilizing electron clouds above and below the aromatic ring.
Evidence of Aromaticity: Heats of Hydrogenation
One of the ways to test the relative amounts of resonance energy in a molecule is to compare the heats of hydrogenation between similar compounds. For instance, if we compare cyclohexene, 1,3-cyclohexadiene, and benzene, we would expect that their heats of hydrogenation will increase since the number of double bonds increases respectively. However, experimental evidence suggests that the actual heat of hydrogenation for benzene is actually 49.3 kcal/mole, making it even more stable than the 1,3-cyclohexadiene even though it has two double bonds, compared to benzene's three double bonds. This characteristic can be attributed to the aromaticity of benzene which delocalizes the electrons of the six pi orbitals.
Problems
1. What is the hybridization of each carbon and the overall shape of benzene?
2. What is the resonance energy of benzene?
3. Place the following compounds in order of heats of hydrogenation from smallest to greatest : Benzene, 1,3-Cyclohexadiene, and Cyclohexene.
Answers
1. All six carbons are sp2 hybridized and the aromaticity of the benzene creates a planar molecule.
2. 29.6 kcal/mol
3. Cyclohexene < Benzene < 1,3-Cyclohexadiene.
• Ramie Hosein
Aromaticity
The adjective "aromatic" is used by organic chemists in a rather different way than it is normally applied. It has its origin in the observation that certain natural substances, such as cinnamon bark, wintergreen leaves, vanilla beans and anise seeds, contained fragrant compounds having common but unexpected properties. Cinnamon bark, for example, yielded a pleasant smelling compound, formula C9H8O, named cinnamaldehyde.
• Aromatic Ions and Antiaromaticity
• Aromatic Systems and Factors Required for Aromaticity
• Benzene and Other Aromatic Compounds
• Fused Benzene Ring Compounds
Because of the low hydrogen to carbon ratio in this and other aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
Aromaticity
Aromatic Ions
Carbanions and carbocations may also show aromatic stabilization. Some examples are:
The three-membered ring cation has 2 π-electrons and is surprisingly stable, considering its ring strain. Cyclopentadiene is as acidic as ethanol, reflecting the stability of its 6 π-electron conjugate base. Salts of cycloheptatrienyl cation (tropylium ion) are stable in water solution, again reflecting the stability of this 6 π-electron cation.
Antiaromaticity
Conjugated ring systems having 4n π-electrons (e.g. 4, 8, 12 etc. electrons) not only fail to show any aromatic properties, but appear to be less stable and more reactive than expected. As noted above, 1,3,5,7-cyclooctatetraene is non-planar and adopts a tub-shaped conformation. The compound is readily prepared, and undergoes addition reactions typical of alkenes. Catalytic hydrogenation of this tetraene produces cyclooctane. Planar bridged annulenes having 4n π-electrons have proven to be relatively unstable. Examples of 8 and 12-π-electron systems are shown below, together with a similar 10 π-electron aromatic compound.
The simple C8H6 hydrocarbon pentalene does not exist as a stable compound, and its hexaphenyl derivative is air sensitive. The 12-π-electron analog heptalene has been prepared, but is also extremely reactive (more so than cyclooctatetraene). On the other hand, azulene is a stable 10-π-electron hydrocarbon that incorporates structural features of both pentalene and heptalene. Azulene is a stable blue crystalline solid that undergoes a number of typical aromatic substitution reactions. The unexpected instability of 4n π-electron annulenes has been termed "antiaromaticity". Other examples may be cited. Thus, all attempts to isolate 1,3-cyclobutadiene have yielded its dimer, or products from reactions with other compounds introduced into the reaction system. Similarly, cyclopentadienyl cation (4 π-electrons) and cycloheptatrienyl anion (8 π-electrons) show very high reactivity when forced to form.
Cyclooctatetraene is a fascinating compound. To see more of its chemistry Click Here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/Aromaticity/Aromatic_Ions_and_Antiaromaticity.txt |
Many unsaturated cyclic compounds have exceptional properties that we now consider characteristic of "aromatic" systems. The following cases are illustrative:
Compound Structural
Formula
Reaction
with Br2
Thermodynamic
Stabilization
1,3-Cyclopentadiene Addition ( 0 ºC ) Slight
1,3,5-Cycloheptatriene Addition ( 0 ºC ) Slight
1,3,5,7-Cyclooctatetraene Addition ( 0 ºC ) Slight
Benzene Substitution Large
Pyridine Substitution Large
Furan Substitution ( 0 ºC ) Moderate
Pyrrole Substitution Moderate
The first three compounds (cyclic polyenes) have properties associated with alkenes in general. Each reacts readily with bromine to give addition products, as do most alkenes. The thermodynamic change on introducing double bonds into the carbon atom ring is also typical of alkenes (a destabilization of ca. 26 kcal/mol for each double bond). Conjugation offsets this increase in energy by a small amount (4-6 kcal/mol).
The remaining four compounds exhibit very different properties, and are considered aromatic. Benzene and pyridine are relatively unreactive with bromine, requiring heat and/or catalysts to force reaction, the result of which is substitution rather than addition. Furan and pyrrole react more rapidly with bromine, but they also give substitution products. This tendency to favor substitution rather than addition suggests that the parent unsaturated ring system has exceptional stability. Thermodynamic measurements support this conclusion. The enhanced stability, often referred to as aromatic stabilization, ranges (in the above cases) from a low of 16 kcal/mol for furan to 36 kcal/mol for benzene.
Factors Required for Aromaticity
• A planar (or near planar) cycle of sp2 hybridized atoms, the p-orbitals of which are oriented parallel to each other. These overlapping p-orbitals generate an array of π-molecular orbitals.
• These π-orbitals are occupied by 4n+2 electrons (where n is an integer or zero). This requirement is known as The Hückel Rule. All the aromatic compounds discussed above have 6 π-electrons (n=1).
1,3-Cyclopentadiene and 1,3,5-cycloheptatriene both fail to meet the first requirement, since one carbon atom of each ring is sp3 hybridized and has no p-orbital. Cyclooctatetraene fails both requirements, although it has a ring of sp2 hybridized atoms. This molecule is not planar ( a geometry that would have 135º bond angles ). Angle strain is relieved by adopting a tub-shaped conformation; consequently, the p-orbitals can only overlap as isolated pairs, not over the entire ring. Furthermore, cyclooctatetraene has 8 π-electrons, a number not consistent with the Hückel Rule.
Benzene is the archetypical aromatic compound. It is planar, bond angles=120º, all carbon atoms in the ring are sp2 hybridized, and the pi-orbitals are occupied by 6 electrons. The aromatic heterocycle pyridine is similar to benzene, and is often used as a weak base for scavenging protons. Furan and pyrrole have heterocyclic five-membered rings, in which the heteroatom has at least one pair of non-bonding valence shell electrons. By hybridizing this heteroatom to a sp2 state, a p-orbital occupied by a pair of electrons and oriented parallel to the carbon p-orbitals is created. The resulting planar ring meets the first requirement for aromaticity, and the π-system is occupied by 6 electrons, 4 from the two double bonds and 2 from the heteroatom, thus satisfying the Hückel Rule.
Four illustrative examples of aromatic compounds are shown above. The sp2 hybridized ring atoms are connected by brown bonds, the π-electron pairs and bonds that constitute the aromatic ring are colored blue. Electron pairs that are not part of the aromatic π-electron system are black. The first example is azulene, a blue-colored 10 π-electron aromatic hydrocarbon isomeric with naphthalene. The second and third compounds are heterocycles having aromatic properties. Pyridine has a benzene-like six-membered ring incorporating one nitrogen atom. The non-bonding electron pair on the nitrogen is not part of the aromatic π-electron sextet, and may bond to a proton or other electrophile without disrupting the aromatic system. In the case of thiophene, a sulfur analog of furan, one of the sulfur electron pairs (colored blue) participates in the aromatic ring π-electron conjugation. The last compound is imidazole, a heterocycle having two nitrogen atoms. Note that only one of the nitrogen non-bonding electron pairs is used for the aromatic π-electron sextet. The other electron pair (colored black) behaves similarly to the electron pair in pyridine.
Annulenes
Monocyclic compounds made up of alternating conjugated double bonds are called annulenes. Benzene and 1,3,5,7-cyclooctatetraene are examples of annulenes; they are named [6]annulene and [8]annulene respectively, according to a general nomenclature system in which the number of pi-electrons in an annulene is designated by a number in brackets. Some annulenes are aromatic (e.g. benzene), but many are not due to non-planarity or a failure to satisfy the Hückel Rule. Compounds classified as [10]annulenes (a Hückel Rule system) serve to illustrate these factors.
As shown in the following diagram, 1,3,5,7,9-cyclodecapentaene fails to adopt a planar conformation, either in the all cis-configuration or in its 1,5-trans-isomeric form. The transannular hydrogen crowding that destabilizes the latter may be eliminated by replacing the interior hydrogens with a bond or a short bridge (colored magenta in the diagram). As expected, the resulting 10 π-electron annulene derivatives exhibit aromatic stability and reactivity as well as characteristic ring current anisotropy in the nmr. Naphthalene and azulene are [10]annulene analogs stabilized by a transannular bond. Although the CH2 bridged structure to the right of naphthalene in the diagram is not exactly planar, the conjugated 10 π-electron ring is sufficiently close to planarity to achieve aromatic stabilization. The bridged [14]annulene compound on the far right, also has aromatic properties.
A modified [10]annulene, aromatic by nmr criteria, was prepared recently by chemists at California Institute of Technology. Remarkably, this hydrocarbon is chemically unstable, in contrast to most other aromatic hydrocarbons.
Barrelene
Formulation of the Hückel rule prompted organic chemists to consider the possible aromaticity of many unusual unsaturated hydrocarbons. One such compound was the 6 π-electron bicyclic structure, now known as barrelene. Although the π-bonds in barrelene are not coplanar, it was believed that transannular overlap might still lead to aromatic stabilization.
A synthesis of barrelene (bicyclo[2.2.2]-2,5,7-octatriene) was accomplished nearly fifty years ago by H. Zimmerman (Wisconsin), using a double Hofmann elimination. As shown in the following diagram, the chemical behavior of this triene confirmed it was not aromatic in the accepted sense of this term. Bromine addition took place rapidly with transannular bond formation, in the same fashion as with norbornadiene (bicyclo[2.2.1]-2,5-heptadiene). Pyrolysis of barrelene gave the expected cycloreversion products benzene and acetylene.
The heat of hydrogenation of barrelene reflects its thermodynamic stability. The value for cyclohexene is -28 kcal/mol, significantly less than one third of the barrelene number. Furthermore, the first double bond of barrelene is reduced with the release of 36.7 kcal/mol heat, indicating destabilization rather than stabilization. The electronic spectrum of barrelene shows a π-electron interaction similar to that in related homoconjugated dienes. (λmax ≅220-230 nm).
An explanation for the lack of aromatic behavior in the case of barrelene may be found by comparing the orbital symmetry of the six component p-orbitals with those of benzene. Benzene is an annulene in which all six p-orbitals may be oriented with congruent overlapping phases. The cylindrical array of p-orbitals in barrelene cannot be so arranged, as shown in the diagram on the right. There will always be one region (a nodal plane) in which the transannular overlap is incongruent. By clicking on this diagram, a Jmol model of barrelene will be displayed in a separate window. This model may be moved about for viewing. The p-orbitals of the double bonds may also be displayed. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/Aromaticity/Aromatic_Systems_and_Factors_Required_for_Aromaticity.txt |
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequaterepresentation of the true molecule.
The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/Aromaticity/Benzene_and_Other_Aromatic_Compounds.txt |
In 1931, German chemist and physicist Erich Hückel proposed a rule to determine if a planar ring molecule would have aromatic properties. This rule states that if a cyclic, planar molecule has $4n+2\; \pi$ electrons, it is aromatic. This rule would come to be known as Hückel's Rule.
Four Criteria for Aromaticity
When deciding if a compound is aromatic, go through the following checklist. If the compound does not meet all the following criteria, it is likely not aromatic.
1. The molecule is cyclic (a ring of atoms)
2. The molecule is planar (all atoms in the molecule lie in the same plane)
3. The molecule is fully conjugated (p orbitals at every atom in the ring)
4. The molecule has $4n+2\; \pi$ electrons (n=0 or any positive integer)
Why 4n+2 π Electrons?
According to Hückel's Molecular Orbital Theory, a compound is particularly stable if all of its bonding molecular orbitals are filled with paired electrons. This is true of aromatic compounds, meaning they are quite stable. With aromatic compounds, 2 electrons fill the lowest energy molecular orbital, and 4 electrons fill each subsequent energy level (the number of subsequent energy levels is denoted by $n$), leaving all bonding orbitals filled and no anti-bonding orbitals occupied. This gives a total of $4n+2 \; \pi$ electrons. You can see how this works with the molecular orbital diagram for the aromatic compound, benzene, below.
Figure 1: Molecular Orbitals levels of Benzene
Benzene has 6 $\pi$ electrons. Its first 2 $\pi$ electrons fill the lowest energy orbital, and it has 4 $\pi$ electrons remaining. These 4 fill in the orbitals of the succeeding energy level. Notice how all of its bonding orbitals are filled, but none of the anti-bonding orbitals have any electrons.
Example $1$: Benzene
Confirm that benzene is aromatic.
Solution
To apply the 4n+2 rule, first count the number of π electrons in the molecule. Then, set this number equal to 4n+2 and solve for n. If is 0 or any positive integer (1, 2, 3,...), the rule has been met. For example, benzene has six π electrons:
4n + 2 = 6
4n = 4
n = 1
For benzene, we find that n=1, which is a positive integer, so the rule is met.
How Can You Tell Which Electrons are π Electrons?
Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually π electrons. Once this is figured out, the rule is quite straightforward. π electrons lie in p orbitals. Sp2 hybridized atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp2 hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the π electrons. A simple way to know if an atom is sp2 hybridized is to see if it has 3 attached atoms and no lone pairs of electrons. This video provides a very nice tutorial on how to determine an atom's hybridization. In a cyclic hydrocarbon compound with alternating single and double bonds, each carbon is attached to 1 hydrogen and 2 other carbons. Therefore, each carbon is sp2 hybridized and has a p orbital. Let's look at our previous example, benzene:
Each double bond (π bond) always contributes 2 π electrons. Benzene has 3 double bonds, so it has 6 π electrons.
Aromatic Ions
Hückel's Rule also applies to ions. As long as a compound has 4n+2 π electrons, it does not matter if the molecule is neutral or has a charge. For example, cyclopentadienyl anion is an aromatic ion. How do we know that it is fully conjugated? That is, how do we know that each atom in this molecule has 1 p orbital? Let's look at the following figure. Carbons 2-5 are sp2 hybridized because they have 3 attached atoms and have no lone electron pairs. What about carbon 1? Another simple rule to determine if an atom is sp2 hybridized is if an atom has 1 or more lone pairs and is attached to an sp2 hybridized atom, then that atom is sp2 hybridized also. This video explains the rule very clearly. Therefore, carbon 1 has a p orbital. Cyclopentadienyl anion has 6 π electrons and fulfills the 4n+2 rule.
Heterocyclic Aromatic Compounds
So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain 1 or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are sp2 hybridized. But is the oxygen atom sp2 hybridized? The oxygen has at least 1 lone electron pair and is attached to an sp2 hybridized atom, so it is sp2 hybridized as well. Notice how oxygen has 2 lone pairs of electrons. How many of those electrons are π electrons? An sp2 hybridized atom only has 1 p orbital, which can only hold 2 electrons, so we know that 1 electron pair is in the p orbital, while the other pair is in an sp2 orbital. So, only 1 of oxygen's 2 lone electron pairs are π electrons. Furan has 6 π electrons and fulfills the 4n+2 rule.
Problems
Using the criteria for aromaticity, determine if the following molecules are aromatic:
Answers
1. Aromatic - only 1 of S's lone pairs counts as π electrons, so there are 6 π electrons, n=1
2. Not aromatic - not fully conjugated, top C is sp3 hybridized
3. Not aromatic - top C is sp2 hybridized, but there are 4 π electrons, n=1/2
4. Aromatic - N is using its 1 p orbital for the electrons in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1
5. Aromatic - there are 6 π electrons, n=1
6. Not aromatic - all atoms are sp2 hybridized, but only 1 of S's lone pairs counts as π electrons, so there 8 π electrons, n=1.5
7. Not aromatic - there are 4 π electrons, n=1/2
8. Aromatic - only 1 of N's lone pairs counts as π electrons, so there are 6 π electrons, n=1
9. Not aromatic - not fully conjugated, top C is sp3 hybridized
10. Aromatic - O is using its 1 p orbital for the elections in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1 | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/Huckel%27s_Rule.txt |
What is an "aromatic" compound? It is common to start by saying that aromatic compounds are compounds related to benzene. However, as you go on in organic chemistry you will find a variety of compounds called aromatic, even though they are not so obviously benzene derivatives. Defining aromatic in terms of benzene is a useful start in an introductory course. As we will see here, it is not easy to give a more complete definition that is satisfying in an introductory course.
Definition
First, why is aromaticity an issue? The special feature of benzene is that it is more stable than we might expect. If we write the structure of "1,3,5-cyclohexatriene", it looks like one of the Kekule structures for benzene. But benzene, a real chemical, does not have the properties we would expect for "1,3,5-cyclohexatriene". Why? Benzene has two resonance structures, and that the actual structure is the resonance hybrid. In general, resonance structures delocalize electrons, and thus stabilize a structure.
The resonance stabilization in benzene is considerably more than we might expect for simply having some double bonds near each other. Clearly, there is something special about benzene, which results in an unusual degree of resonance stabilization -- and results in the property called aromaticity. We can discuss this more, but doing so goes beyond course material. The purpose is to help those who are going on in organic chemistry get a sense of what this topic holds.
The resonance system in benzene involves six electrons. They occupy a series of p orbitals, one on each C atom. The p orbitals on neighboring C atoms overlap sideways, a type of bonding known as π (pi) bonding. Effectively, the six overlapping p orbitals form one large orbital -- a loop around the entire molecule. This "loop orbital" provides an unusual degree of stabilization, and is part of the secret to aromaticity. That is, one key feature of aromatic compounds is that there is a set of electrons in a loop orbital resulting from overlapping p orbitals around a ring. (sometimes referred to as a "closed loop of six electrons", which he also calls the "aromatic sextet").
Let's look at some examples of compounds that are aromatic, but are less obviously "like benzene". As we discuss these examples, some patterns will emerge. Unfortunately, reasons for not discussing this too much in an introductory class will also emerge; understanding aromaticity even at an elementary level beyond "like benzene" requires understanding orbitals.
Example 1
One simple example is pyridine, C5H5N, shown below:
This example is simple enough, because it is very similar to benzene in structure.
Some aromatic compounds with a heteroatom (an atom other than C or H) in the ring have a five-membered ring. These include pyrrole, C4H5N, and furan, C4H4O:
What do these compounds share with benzene and pyridine that makes them aromatic? They appear to have only two double bonds, thus only four p orbital electrons perpendicular to the ring. But they also have one lone pair that is in a p orbital perpendicular to the ring. Thus they actually have six electrons that are in p orbitals perpendicular to the ring.
At this point, then, a pattern may be emerging: it seems that a loop of six electrons in overlapping p orbitals perpendicular to the molecular ring is a condition for aromaticity. Let's test this prediction by looking at some other structures.
Five membered rings of C. Look at cyclopentadiene, shown below. It has 2 double bonds, hence has 4 electrons in p orbitals perpendicular to the ring. But the other C is "saturated"; it is a -CH2- and is sp3-hybridized. No loop, not even 6 p electrons. Therefore not aromatic; this is correct.
Acidity of Aromatic Compounds
But now look at the anion derived from this chemical -- the ion that would result if this chemical behaved as an acid, and gave off an H+; this ion is also shown below. Note that it now has six electrons in p orbitals perpendicular to the ring -- very similar to pyrrole, above. If pyrrole is aromatic, then maybe this ion should be? Yes, it is. A manifestation of this is that the parent compound, cyclopentadiene, is a rather "strong" acid -- by the standards of H attached to hydrocarbons. The acidity of cyclopentadiene is due to the stabilization of the resulting anion. Ka for cyclopentadiene is about 10-16. That certainly isn't strong compared to compounds commonly discussed as acids, but it is 1029 times stronger than for the non-cyclic form of this molecule.
Now we have seen a variety of chemical species, both neutral molecules and ions, that are aromatic. They share the common feature that they all have 6 electrons in a continuous loop of overlapping p orbitals. Turns out that this is still not a sufficiently broad description of what makes a chemical aromatic.
An interesting example is 1,3,5,7-cyclooctatetraene (often simply called cyclooctatetraene), C8H8, shown at the left. At first glance, it would seem to be similar to benzene, except with a larger ring. But it is definitely not aromatic. Its chemical behavior is what you would expect for an alkene, and its shape is not planar, but is "tub-shaped", as shown to the right.
Now, the non-aromatic character of cyclooctatetraene alone is not too hard to explain. After all, if it were aromatic and planar, it would have bond angles of 135 deg. That is considerably more than the simple sp2 bond angle of 120 deg. So we might suggest that the bond angle strain of the planar form more than compensates for any gain due to aromaticity. However, there is more to the story. Turns out that it is fairly easy for cyclooctatetraene to gain 2 electrons, forming the dianion, C8H82-. This ion is planar, and is aromatic. Two drawings of this ion are below, followed by a 3D model.
cyclooctatetraene dianion cyclooctatetraene dianion, shown with the circle that denotes an aromatic ring
cyclooctatetraene dianion, C8H82-, 3D cyclooctatetraene, C8H8, 3D (neutral; repeated from above, for comparison of the 3D shapes of molecule and ion)
So what do we learn from the story of cyclooctatetraene and its dianion? Prior to considering this, all our examples of aromaticity had six electrons in the π electron loop. Cyclooctatetraene has eight, and is not aromatic; its dianion has ten, and is aromatic. So clearly, six is not the only allowed number.
Study of many molecules and ions, as well as theoretical work that is well beyond our course, have indicated that a species will be aromatic if there are 4n+2 electrons in a planar π electron loop -- where n is any integer, starting with 0. (This is known as Hückel's rule.) 0? That would mean that a π electron loop with two electrons is aromatic. In fact, the cation derived from cyclopropene, shown below, is unusually stable, and is considered aromatic.
The cyclopropenyl (or "cyclopropenium") cation, C3H3+. It has 4n+2 electrons in the π electron loop, where n=0; thus it satisfies Hückel's rule and is aromatic.
As another example, with a larger n, consider a particular isomer of 18-annulene -- the isomer with every third double bond cis. The following set of figures show three representations of its structure.
If 18-annulene were aromatic (n=4), we might expect it to be planar. In fact, some programs will calculate a planar structure for it. Actual measurement shows that is very slightly distorted from planar, due to the repulsion of the hydrogens that are inside the ring. The following figure shows 3D representations of both "forms" of 18-annulene; they have been rotated so you view them "edge on". The top structure is the planar form that one might naively expect; the other is the slightly distorted structure, which closely corresponds to what is actually observed.
The annulene name is used generically for cyclic molecules with alternating single and double bonds. The numeric prefix indicates the ring size. Benzene might be considered as 6-annulene, and cyclooctatetraene as 8-annulene. Note that all annulenes have the general formula CxHx (where x must be an even number), and that the term does not in itself imply aromatic character.
Summary
Aromatic compounds are more stable than we might expect when we see a structure showing single and double bonds. We start our study of aromaticity with the classic case of benzene. But as we continue, we find examples of aromatic compounds that contain heteroatoms, charges, and rings of different sizes. As we explore these, we find that the key features of aromatic chemicals are dictacted by Hückel's rule:
Aromaticity requires a planar loop of electrons in overlapping p orbitals. The number of electrons in the loop must be 4n+2, where n is an integer >= 0.
Contributors
>Robert Bruner (http://bbruner.org) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/What_does_aromatic_really_mean.txt |
Benzene rings may be joined together (fused) to give larger polycyclic aromatic compounds. A few examples are drawn below, together with the approved numbering scheme for substituted derivatives. The peripheral carbon atoms (numbered in all but the last three examples) are all bonded to hydrogen atoms. Unlike benzene, all the C-C bond lengths in these fused ring aromatics are not the same, and there is some localization of the pi-electrons.
The six benzene rings in coronene are fused in a planar ring; whereas the six rings in hexahelicene are not joined in a larger ring, but assume a helical turn, due to the crowding together of the terminal ring atoms. This helical configuration renders the hexahelicene molecule chiral, and it has been resolved into stable enantiomers.
As these extended aromatic compounds become larger, the ratio of hydrogen to carbon decreases. For example, the symmetrical hexacyclic compound coronene has a H/C ratio =1/2, compared with 1 for benzene. If we were to imagine fused ring systems of this kind to be further extended in space, the H/C ratio would approach zero, and the resulting compound would be a form of carbon. Such a carbon allotrope exists and is called graphite. Another well-characterized carbon allotrope is diamond. The structures for these two forms of carbon are very different, and are displayed below. Diamond is an extended array of sp3 hybridized carbon atoms; whereas, graphite consists of overlapping sheets of sp2 hybridized carbon atoms arranged in a hexagonal pattern. You may examine models of partial diamond and graphite structures by clicking on the appropriate structure below.
Diamond
Graphite
A comparison of the coronene and corannulene models discloses an interesting difference in their shapes. Coronene is absolutely flat and, aside from the peripheral hydrogens, resembles a layer of graphite. Its very high melting point reflects this resemblance. Corannulene, on the other hand, is slightly curved, resulting in a bowl-like shape.
If we extend the structure of corannulene by adding similar cycles of five benzene rings, the curvature of the resulting molecule should increase, and eventually close into a sphere of carbon atoms. The archetypical compound of this kind (C60) has been named buckminsterfullerene because of its resemblance to the geodesic structures created by Buckminster Fuller. It is a member of a family of similar carbon structures that are called fullerenes. These materials represent a third class of carbon allotropes. Alternating views of the C60 fullerene structure are shown on the right, together with a soccer ball-like representation of the 12 five and 20 six-membered rings composing its surface. Precise measurement by Atomic Force Microscopy (AFM) has shown that the C-C bond lengths of the six-membered rings are not all equal, and depend on whether the ring is fused to a five or six-membered beighbor.
Interest in the fullerenes has led to the discovery of a related group of carbon structures referred to as nanotubes. As shown in the following illustration, nanotubes may be viewed as rolled up segments of graphite. The chief structural components are six-membered rings, but changes in tube diameter, branching into side tubes and the capping of tube ends is accomplished by fusion with five and seven-membered rings. Many interesting applications of these unusual structures have been proposed. A model of a nanotube will be displayed by clicking on the diagram | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Fused_Benzene_Ring_Compounds.txt |
A substituent on a benzene ring can effect the placement of additional substituents on that ring during Electrophilic Aromatic Substitution. How do we know where an additonal substituent will most likely be placed? The answer to this is through inductive and resonance effects. Inductive effects are directly correlated with electronegativity. Substituents can either be meta directing or ortho-para directing.
Introduction
The three general positions of a disubstituted benzene ring are ortho, meta and para.
Figure 1: The Effect of an Electron Donating Groups on a Benzene Ring
The first scenario for adding an electrophile to a monosubstituted benzene ring is when the substituent is an electron donating group. Electron donating groups are alkyl groups, phenyl groups or substituents that have a lone pair of electrons on the atom directly bonded to the ring. Electron donating groups are donating by induction (Activating and Deactivating Benzene Rings) and resonance. Examples of electron donating groups: -CH3, -OCH3, -OH, -NH2
Electron donating groups cause the second subtituent to add on to the para or ortho position on the benzene ring. The reason for this can be explained by the different carbocation resonance structures of the ortho, meta and para positions.
Some electron donating groups have an extra resonance form in which there is a double bond between the atom and the carbon on the benzene. This is a very stable resonance form. This is due to directing effects of substituents in conjugation with the benzene ring.
When the electrophile is added to the ortho position, three different resonance forms are possible. Carbocation forms 1 and 2 are secondary carbocations, but position 3 forms a tertiary carbocation and the positive charge is on the carbon directly attached to the electron donating group, which is the most stable. This carbocation is also stablized by the electrons from the electron donating group. More stable intermediates (the carbocation) have lower transition state energies and therefore a faster reaction rate, forming more of this product. This is the reason that the ortho position is one of the major products.
If the electrophile is added to the monosubstituted benzene ring in the para position one of the three resonance forms of the carbocations will be a tertiary carbocation which is highly stable because of the +I effect if the three -CH3. This carbocation intermediate is the same as the one formed from ortho substitution.
For the meta substituted carbocation resonance structures, there are three possible resonance froms that are secondary carbocations. These forms are not as stable as the tertiary carbocation form in the ortho and para substituted carbocations.
Therefore, the two major products of the reaction of a monosubstituted benzene ring with an electron donating group and additional electrophile are the ortho and para positions. It's important to note that the para product is slightly more common than the ortho product due to steric hindrance.
H-NMR spectroscopy can be used to determine whether or not a compound has a second substituent at the ortho or para position. At the ortho position there are four distinct signals, but for the para position there are only two signals because the molecule is symmetrical. Electron donating groups on a benzene ring are said to be activating, because they increase the rate of the second substitution so that it is higher than that of standard benzene.
Electron donating groups are said to be ortho/para directing and they are activators.
The Effect of an Electron Withdrawing Group on a Benzene Ring
The other circumstance is when you have add an additional electrophile to a monosubstituted benzene ring with an electron withdrawing group on it. Electron withdrawing groups have an atom with a slight positive or full positive charge directly attached to a benzene ring. Examples of electron withdrawing groups: -CF3, -COOH, -CN. Electron withdrawing groups only have one major product, the second substituent adds in the meta position. Again, this can be explained by the resonance forms of the carbocation intermediates.
When the second electrophile is added on to the benzene ring in the ortho position, the same three resonance forms of the carbocation are produced. Again, one form is a tertiary carbocation with the positive charge on the carbon directly attached to the electron withdrawing group. Unlike in the case with an electron donating group, this resonance form is much less stable. This is due to the electron withdrawing group pulling away electrons from the carbon, creating an even stronger positive charge. This situation holds true for the para substituted tertiary carbocation resonace form as well.
For the meta position, all the carbocations formed are secondary. Although these are not entirely stable, they are more favored than the resonance forms of the ortho and para positions.
The major product of a monosubstituted benzene ring with an electron withdrawing group and an additional electrophile is a product with meta substitution. In contrast to electron donating groups, electron withdrawing groups are deactivating. This means that the rate of the second substitution is lower than that of standard benzene.
Table 1: Common Substituents
Orthe- and Para- Directing Meta Directing
Strong Activating Moderately Activating Weakly Activating Weakly Deactivating Moderately Deactivating Strongly Deactivating
-NH2 -NHR -OH -OCH3 -NHCOR -OCOR -CH3 -phenyl -F -Cl -Br -I -COH -COCH3 -COOCH3 -SO3H -NO2 -CF3 -CCl3
Halogens: A Special Case
Halogens are very electronegative. This means that inductively they are electron withdrawing. However, because of their ability to donate a lone pair of electrons in resonance forms, they are activators and ortho/para directing. Resonance forms win out in directing. Because they are electron withdrawing, halogens are very weak activators.
Electron withdrawing groups are meta directors and they are deactivators.
Contributors
• Lauren Rice and Samantha Spragg (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Inductive_Effects_of_Alkyl_Groups.txt |
Introduction
Many unsaturated cyclic compounds have exceptional properties that we now consider characteristic of "aromatic" systems. The following cases are illustrative:
Compound
Structural
Formula
Reaction
with Br2
Thermodynamic
Stabilization
1,3-Cyclopentadiene
Addition ( 0 ºC )
Slight
1,3,5-Cycloheptatriene
Addition ( 0 ºC )
Slight
1,3,5,7-Cyclooctatetraene
Addition ( 0 ºC )
Slight
Benzene
Substitution
Large
Pyridine
Substitution
Large
Furan
Substitution ( 0 ºC )
Moderate
Pyrrole
Substitution
Moderate
The first three compounds (cyclic polyenes) have properties associated with alkenes in general. Each reacts readily with bromine to give addition products, as do most alkenes. The thermodynamic change on introducing double bonds into the carbon atom ring is also typical of alkenes (a destabilization of ca. 26 kcal/mol for each double bond). Conjugation offsets this increase in energy by a small amount (4-6 kcal/mol).
The remaining four compounds exhibit very different properties, and are considered aromatic. Benzene and pyridine are relatively unreactive with bromine, requiring heat and/or catalysts to force reaction, the result of which is substitution rather than addition. Furan and pyrrole react more rapidly with bromine, but they also give substitution products. This tendency to favor substitution rather than addition suggests that the parent unsaturated ring system has exceptional stability. Thermodynamic measurements support this conclusion. The enhanced stability, often referred to as aromatic stabilization, ranges (in the above cases) from a low of 16 kcal/mol for furan to 36 kcal/mol for benzene.
Factors Required for Aromaticity
A planar (or near planar) cycle of sp2 hybridized atoms, the p-orbitals of which are oriented parallel to each other. These overlapping p-orbitals generate an array of π-molecular orbitals.
These π-orbitals are occupied by 4n+2 electrons (where n is an integer or zero). This requirement is known as The Hückel Rule. All the aromatic compounds discussed above have 6 π-electrons (n=1).
1,3-Cyclopentadiene and 1,3,5-cycloheptatriene both fail to meet the first requirement, since one carbon atom of each ring is sp3 hybridized and has no p-orbital. Cyclooctatetraene fails both requirements, although it has a ring of sp2 hybridized atoms. This molecule is not planar ( a geometry that would have 135º bond angles ). Angle strain is relieved by adopting a tub-shaped conformation; consequently, the p-orbitals can only overlap as isolated pairs, not over the entire ring. Furthermore, cyclooctatetraene has 8 π-electrons, a number not consistent with the Hückel Rule.
Benzene is the archetypical aromatic compound. It is planar, bond angles=120º, all carbon atoms in the ring are sp2 hybridized, and the pi-orbitals are occupied by 6 electrons. The aromatic heterocycle pyridine is similar to benzene, and is often used as a weak base for scavenging protons. Furan and pyrrole have heterocyclic five-membered rings, in which the heteroatom has at least one pair of non-bonding valence shell electrons. By hybridizing this heteroatom to a sp2 state, a p-orbital occupied by a pair of electrons and oriented parallel to the carbon p-orbitals is created. The resulting planar ring meets the first requirement for aromaticity, and the π-system is occupied by 6 electrons, 4 from the two double bonds and 2 from the heteroatom, thus satisfying the Hückel Rule.
Four illustrative examples of aromatic compounds are shown above. The sp2 hybridized ring atoms are connected by brown bonds, the π-electron pairs and bonds that constitute the aromatic ring are colored blue. Electron pairs that are not part of the aromatic π-electron system are black. The first example is azulene, a blue-colored 10 π-electron aromatic hydrocarbon isomeric with naphthalene. The second and third compounds are heterocycles having aromatic properties. Pyridine has a benzene-like six-membered ring incorporating one nitrogen atom. The non-bonding electron pair on the nitrogen is not part of the aromatic π-electron sextet, and may bond to a proton or other electrophile without disrupting the aromatic system. In the case of thiophene, a sulfur analog of furan, one of the sulfur electron pairs (colored blue) participates in the aromatic ring π-electron conjugation. The last compound is imidazole, a heterocycle having two nitrogen atoms. Note that only one of the nitrogen non-bonding electron pairs is used for the aromatic π-electron sextet. The other electron pair (colored black) behaves similarly to the electron pair in pyridine.
Annulenes
Monocyclic compounds made up of alternating conjugated double bonds are called annulenes. Benzene and 1,3,5,7-cyclooctatetraene are examples of annulenes; they are named [6]annulene and [8]annulene respectively, according to a general nomenclature system in which the number of pi-electrons in an annulene is designated by a number in brackets. Some annulenes are aromatic (e.g. benzene), but many are not due to non-planarity or a failure to satisfy the Hückel Rule. Compounds classified as [10]annulenes (a Hückel Rule system) serve to illustrate these factors.
As shown in the following diagram, 1,3,5,7,9-cyclodecapentaene fails to adopt a planar conformation, either in the all cis-configuration or in its 1,5-trans-isomeric form. The transannular hydrogen crowding that destabilizes the latter may be eliminated by replacing the interior hydrogens with a bond or a short bridge (colored magenta in the diagram). As expected, the resulting 10 π-electron annulene derivatives exhibit aromatic stability and reactivity as well as characteristic ring current anisotropy in the nmr. Naphthalene and azulene are [10]annulene analogs stabilized by a transannular bond. Although the CH2 bridged structure to the right of naphthalene in the diagram is not exactly planar, the conjugated 10 π-electron ring is sufficiently close to planarity to achieve aromatic stabilization. The bridged [14]annulene compound on the far right, also has aromatic properties.
Barrelene
Formulation of the Hückel rule prompted organic chemists to consider the possible aromaticity of many unusual unsaturated hydrocarbons. One such compound was the 6 π-electron bicyclic structure, now known as barrelene. Although the π-bonds in barrelene are not coplanar, it was believed that transannular overlap might still lead to aromatic stabilization.
A synthesis of barrelene (bicyclo[2.2.2]-2,5,7-octatriene) was accomplished nearly fifty years ago by H. Zimmerman (Wisconsin), using a double Hofmann elimination. As shown in the following diagram, the chemical behavior of this triene confirmed it was not aromatic in the accepted sense of this term. Bromine addition took place rapidly with transannular bond formation, in the same fashion as with norbornadiene (bicyclo[2.2.1]-2,5-heptadiene). Pyrolysis of barrelene gave the expected cycloreversion products benzene and acetylene.
The heat of hydrogenation of barrelene reflects its thermodynamic stability. The value for cyclohexene is -28 kcal/mol, significantly less than one third of the barrelene number. Furthermore, the first double bond of barrelene is reduced with the release of 36.7 kcal/mol heat, indicating destabilization rather than stabilization.
The electronic spectrum of barrelene shows a π-electron interaction similar to that in related homoconjugated dienes. (λmax ≅220-230 nm).
An explanation for the lack of aromatic behavior in the case of barrelene may be found by comparing the orbital symmetry of the six component p-orbitals with those of benzene. Benzene is an annulene in which all six p-orbitals may be oriented with congruent overlapping phases. The cylindrical array of p-orbitals in barrelene cannot be so arranged, as shown in the diagram on the right. There will always be one region (a nodal plane) in which the transannular overlap is incongruent. By clicking on this diagram, a Jmol model of barrelene will be displayed in a separate window. This model may be moved about for viewing. The p-orbitals of the double bnds may also be displayed.
Aromatic Ions
Carbanions and carbocations may also show aromatic stabilization. Some examples are:
The three-membered ring cation has 2 π-electrons and is surprisingly stable, considering its ring strain. Cyclopentadiene is as acidic as ethanol, reflecting the stability of its 6 π-electron conjugate base. Salts of cycloheptatrienyl cation (tropylium ion) are stable in water solution, again reflecting the stability of this 6 π-electron cation.
Antiaromaticity
Conjugated ring systems having 4n π-electrons (e.g. 4, 8, 12 etc. electrons) not only fail to show any aromatic properties, but appear to be less stable and more reactive than expected. As noted above, 1,3,5,7-cyclooctatetraene is non-planar and adopts a tub-shaped conformation. The compound is readily prepared, and undergoes addition reactions typical of alkenes. Catalytic hydrogenation of this tetraene produces cyclooctane. Planar bridged annulenes having 4n π-electrons have proven to be relatively unstable. Examples of 8 and 12-π-electron systems are shown below, together with a similar 10 π-electron aromatic compound.
The simple C8H6 hydrocarbon pentalene does not exist as a stable compound, and its hexaphenyl derivative is air sensitive. The 12-π-electron analog heptalene has been prepared, but is also extremely reactive (more so than cyclooctatetraene). On the other hand, azulene is a stable 10-π-electron hydrocarbon that incorporates structural features of both pentalene and heptalene. Azulene is a stable blue crystalline solid that undergoes a number of typical aromatic substitution reactions. The unexpected instability of 4n π-electron annulenes has been termed "antiaromaticity". Other examples may be cited. Thus, all attempts to isolate 1,3-cyclobutadiene have yielded its dimer, or products from reactions with other compounds introduced into the reaction system. Similarly, cyclopentadienyl cation (4 π-electrons) and cycloheptatrienyl anion (8 π-electrons) show very high reactivity when forced to form.
Questions
1) Of the following compounds (A through J), which would be considered aromatic by application of the Hückel Rule?
A, D, E & G
Click here for the answers
2) Of the following compounds (A through J), which would be considered aromatic by application of the Hückel Rule?
Click here for the answers
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Other_Aromatic_Systems.txt |
Polycyclic aromatics are compounds containing two or more fused aromatic rings. These fused benzene rings share two carbon atoms between them. These are also called polycyclic benoid or polycyclic aromatic hydrocarbons (PAHs). Many PAHs are carcinogenic and highly toxic.
Naming
The naming system for PAHs is quite complex. A series of fused benzene rings that occur in a linear manner are called acenes.
Anthracene Tetracene
Another form of fusion of benzene rings is angular fusion, or "annulation", which is, as its name implies, an angled fusion of the benzene rings.
Figure 1: Phenanthrene
Naphthalene
Figure 2: Naphthalene
Naphthalene is the simplest polycyclic aromatic hydrocarbon since it is only a bicyclic molecule made up of two aromatic benzenes. The pi electrons in this molecule are even more delocalized than those in the simpler benzene molecule. Naphthalene is also planar and has 4n + 2 pi electrons (10) giving it the stabilizing and resonating aromatic properties shared with benzene.
Most Fused Benzenoid Hydrocarbons are aromatic
Aromaticity is a property that the majority of fused benzenoid hydrocarbons share. The extent to which each polycyclic molecule is aromatic varies. As shown above, the structural isomers of anthracene and phenanthrene are very similar molecules made up of three benzene rings. Anthracene is fused linearly, whereas phenanthrene is fused at an angle. This difference in fusions causes the phenanthrene to have five resonance structures which is one more than anthracene. This extra resonance makes the phenanthrene around 6 kcal per mol more stable.
Acidity of Substituted Benzoic Acids
Arenes are aromatic hydrocarbons. The term "aromatic" originally referred to the pleasant smells given off by arenes, but now implies a particular type of delocalized bonding (see below). The arenes you are likely to encounter at this level are based on benzene rings. The simplest of these arenes is benzene itself, C6H6. The next simplest arene is methylbenzene (common name: toluene), which has one of the hydrogen atoms attached to the ring replaced by a methyl group - C6H5CH3.
The structure of Benzene
The structure of benzene is covered in detail on two pages in the organic bonding section of this site. It is important to understand the structure of benzene thoroughly to understand benzene and methylbenzene chemistry. Unless you have read these pages recently, you should spend some time on them now before you go any further on this page.
This diagram shows one of the molecular orbitals containing two of the delocalized electrons, which may be found anywhere within the two "doughnuts". The other molecular orbitals are almost never drawn.
• Benzene, C6H6, is a planar molecule containing a ring of six carbon atoms, each with a hydrogen atom attached.
• The six carbon atoms form a perfectly regular hexagon. All of the carbon-carbon bonds have exactly the same lengths - somewhere between single and double bonds.
• There are delocalized electrons above and below the plane of the ring.
• The presence of the delocalized electrons makes benzene particularly stable.
• Benzene resists addition reactions because those reactions would involve breaking the delocalization and losing that stability.
• Benzene is represented by this symbol, where the circle represents the delocalized electrons, and each corner of the hexagon has a carbon atom with a hydrogen attached.
The structure of methylbenzene (toluene)
Methylbenzene has a methyl group attached to the benzene ring replacing one of the hydrogen atoms.
Attached groups are often drawn at the top of the ring, but you may occasionally find them drawn in other places with the ring rotated.
Physical properties
Boiling points
In benzene, the only attractions between the neighbouing molecules are the van der Waals dispersion forces. There is no permanent dipole on the molecule. Benzene boils at 80°C, which is higher than other hydrocarbons of similar molecular size (pentane and hexane, for example). The higher boiling point is presumably due to the ease with which temporary dipoles can be set up involving the delocalized electrons.
Methylbenzene boils at 111°C. Methylbenzene is a larger molecule, thus, the van der Waals dispersion forces will be increased. Methylbenzene also has a small permanent dipole; thus, there will be dipole-dipole attractions as well as dispersion forces. The dipole is due to the CH3 group's tendency to "push" electrons away from itself. This also affects the reactivity of methylbenzene (see below).
Melting points
You might have expected that methylbenzene's melting point would be higher than benzene's as well, but it isn't - it is much lower! Benzene melts at 5.5°C; methylbenzene at -95°C.
Molecules must pack efficiently in the solid if they are to optimize their intermolecular forces. Benzene is a tidy, symmetrical molecule and packs very efficiently. The methyl group that protrudes from the methylbenzene structure tends to disrupt the closeness of the packing. If the molecules are not as closely packed, the intermolecular forces don't work as well, causing the melting point to decrease.
Solubility in water
The arenes are insoluble in water. Benzene is quite large compared with a water molecule. For benzene to dissolve, it would have to break a significant number of the existing hydrogen bonds between the water molecules. In addition, the quite strong van der Waals dispersion forces between the benzene molecules would require breaking; both of these processes require energy. The only new forces between the benzene and the water would be van der Waals dispersion forces. These forces are not as strong as hydrogen bonds (or the original dispersion forces in the benzene), therefore, only a limited amount of energy is released when they form. It simply isn't energetically profitable for benzene to dissolve in water. It would, of course, be even worse for larger arene molecules.
Reactivity
Benzene is resistant to addition reactions. Adding something new to the ring would require that some of the delocalized electrons form bonds with the substituent being added, resulting in a major loss of stability because the delocalization is broken. Instead, benzene primarily undergoes substitution reactions - replacing one or more of the hydrogen atoms with a new substituent, preserving the delocalized electrons as they were.
The reactivity of a compound like methylbenzene must be considered in two distinct parts:
For example, if you explore other pages in this section, you will find that alkyl groups attached to a benzene ring are oxidized by alkaline potassium manganate(VII) solution. This oxidation does not occur in the absence of the benzene ring. The tendency of the CH3 group to "push" electrons away from itself also has an effect on the ring, making methylbenzene react more quickly than benzene itself. You will find this explored in other pages in this section as well.
Properties of Arenes
Electron-withdrawing groups
The conjugate base of benzoic acid is destabilized by electron-donating groups. This makes the acid less acidic
Electron-withdrawing groups deactivate the benzene ring to electrophilic attack and make benzoic acids more acidic.
Electron-donating groups
The conjugate base of benzoic acid is stabilized by electron-withdrawing groups. This makes the acid more acidic
Electron-withdrawing groups activate the benzene ring to electrophilic attack and make benzoic acids less acidic. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Polycyclic_Aromatics.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
The Phenyl Group
The terms phenyl and phenol, along with benzene and benzyl, can confuse beginning organic chemistry students. Figure 1 shows what the four terms mean.
Figure 1: The top two structures in Figure 1 are molecules. The -ene suffix of benzene might indicate that it is similar to an alkene. The -ol suffix of phenol indicates that it has an -OH group.
The lower two structures in Figure 1 show groups. The line extending off without anything connected is the line that shows this is a group, which should be attached to something. For example, one might have phenyl chloride (C6H5Cl, also called chlorobenzene) or one might have benzyl chloride (C6H5CH2Cl). (The structures of these two compounds are shown below in Figure 2.) The phenyl group is based simply on benzene, with one H removed. The benzyl group is based on methylbenzene (toluene), with one H removed from the methyl group.
Figure 2 shows the molecules benzyl chloride and phenyl chloride; these are based on the groups discussed above.
If you have the misfortune to come across the word benzol, be forewarned that this is an old German word for benzene. Similarly, toluol is a German word for toluene. Benzin is a German word for gasoline; it is related to the uncommon term benzine, used for some types of gasoline.
Contributors
>Robert Bruner (http://bbruner.org) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Properties_of_Benzene.txt |
Benzene is an aromatic compound that is greatly stabilized by its resonance forms. Stable compounds are much harder to react with, therefore a strong electrophile will be needed to attack the $\pi$ electrons in the benzene ring. Electrophiles used in alkene reactions will typically not be strong enough on their own, therefore a Lewis acid catalyst is required to help generate the electrophile.
Reactivity of Arenes
The remarkable stability of the unsaturated hydrocarbon benzene has been discussed in Aromaticity. The chemical reactivity of benzene contrasts with that of the alkenes in that substitution reactions occur in preference to addition reactions, as illustrated in the following diagram (some comparable reactions of cyclohexene are shown in the green box).
Benzene
The conditions commonly used for the aromatic substitution reactions discussed here are repeated in the table below. The electrophilic reactivity of these different reagents varies. We find, for example, that nitration of nitrobenzene occurs smoothly at 95 ºC, giving meta-dinitrobenzene, whereas bromination of nitrobenzene (ferric catalyst) requires a temperature of 140 ºC. Also, as noted earlier, toluene undergoes nitration about 25 times faster than benzene, but chlorination of toluene is over 500 times faster than that of benzene. From this we may conclude that the nitration reagent is more reactive and less selective than the halogenation reagents.
Both sulfonation and nitration yield water as a by-product. This does not significantly affect the nitration reaction (note the presence of sulfuric acid as a dehydrating agent), but sulfonation is reversible and is driven to completion by addition of sulfur trioxide, which converts the water to sulfuric acid. The reversibility of the sulfonation reaction is occasionally useful for removing this functional group.
Halogenation: C6H6 + Cl2 & heat
FeCl3 catalyst
——> C6H5Cl
Chlorobenzene
+ HCl
Nitration: C6H6 + HNO3 & heat
H2SO4 catalyst
——> C6H5NO2
Nitrobenzene
+ H2O
Sulfonation: C6H6 + H2SO4 + SO3
& heat
——> C6H5SO3H
Benzenesulfonic acid
+ H2O
Alkylation:
Friedel-Crafts
C6H6 + R-Cl & heat
AlCl3 catalyst
——> C6H5-R
An Arene
+ HCl
Acylation:
Friedel-Crafts
C6H6 + RCOCl & heat
AlCl3 catalyst
——> C6H5COR
An Aryl Ketone
+ HCl
The Friedel-Crafts acylation reagent is normally composed of an acyl halide or anhydride mixed with a Lewis acid catalyst such as AlCl3. This produces an acylium cation, R-C≡O(+), or a related species. Such electrophiles are not exceptionally reactive, so the acylation reaction is generally restricted to aromatic systems that are at least as reactive as chlorobenzene. Carbon disulfide is often used as a solvent, since it is unreactive and is easily removed from the product. If the substrate is a very reactive benzene derivative, such as anisole, carboxylic esters or acids may be the source of the acylating electrophile. Some examples of Friedel-Crafts acylation reactions are shown in the following diagram. The first demonstrates that unusual acylating agents may be used as reactants. The second makes use of an anhydride acylating reagent, and the third illustrates the ease with which anisole reacts, as noted earlier. The H4P2O7 reagent used here is an anhydride of phosphoric acid called pyrophosphoric acid. Finally, the fourth example illustrates several important points. Since the nitro group is a powerful deactivating substituent, Friedel-Crafts acylation of nitrobenzene does not take place under any conditions. However, the presence of a second strongly-activating substituent group permits acylation; the site of reaction is that favored by both substituents.
A common characteristic of the halogenation, nitration, sulfonation and acylation reactions is that they introduce a deactivating substituent on the benzene ring. As a result, we do not normally have to worry about disubstitution products being formed. Friedel-Crafts alkylation, on the other hand, introduces an activating substituent (an alkyl group), so more than one substitution may take place. If benzene is to be alkylated, as in the following synthesis of tert-butylbenzene, the mono-alkylated product is favored by using a large excess of this reactant. When the molar ratio of benzene to alkyl halide falls below 1:1, para-ditert-butylbenzene becomes the major product.
C6H6 (large excess) + (CH3)3C-Cl + AlCl3 ——> C6H5-C(CH3)3 + HCl
The carbocation electrophiles required for alkylation may be generated from alkyl halides (as above), alkenes + strong acid or alcohols + strong acid. Since 1º-carbocations are prone to rearrangement, it is usually not possible to introduce 1º-alkyl substituents larger than ethyl by Friedel-Crafts alkylation. For example, reaction of excess benzene with 1-chloropropane and aluminum chloride gives a good yield of isopropylbenzene (cumene).
C6H6 (large excess) + CH3CH2CH2-Cl + AlCl3 ——> C6H5-CH(CH3)2 + HCl
Additional examples of Friedel-Crafts alkylation reactions are shown in the following diagram.
The first and third examples show how alkenes and alcohols may be the source of the electrophilic carbocation reactant. The triphenylmethyl cation generated in the third case is relatively unreactive, due to extensive resonance charge delocalization, and only substitutes highly activated aromatic rings. The second example shows an interesting case in which a polychlororeactant is used as the alkylating agent. A four fold excess of carbon tetrachloride is used to avoid tri-alkylation of this reagent, a process that is retarded by steric hindrance. The fourth example illustrates the poor orientational selectivity often found in alkylation reactions of activated benzene rings. The bulky tert-butyl group ends up attached to the reactive meta-xylene ring at the least hindered site. This may not be the site of initial bonding, since polyalkylbenzenes rearrange under Friedel-Crafts conditions (para-dipropylbenzene rearranges to meta-dipropylbenzene on heating with AlCl3).
A practical concern in the use of electrophilic aromatic substitution reactions in synthesis is the separation of isomer mixtures. This is particularly true for cases of ortho-para substitution, which often produce significant amounts of the minor isomer. As a rule, para-isomers predominate except for some reactions of toluene and related alkyl benzenes. Separation of these mixtures is aided by the fact that para-isomers have significantly higher melting points than their ortho counterparts; consequently, fractional crystallization is often an effective isolation technique. Since meta-substitution favors a single product, separation of trace isomers is normally not a problem. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Benzene/Characteristics_of_Specific_Substitution_Reactions_of_Benzenes.txt |
Many other substitution reactions of benzene have been observed, the five most useful are listed below (chlorination and bromination are the most common halogenation reactions). Since the reagents and conditions employed in these reactions are electrophilic, these reactions are commonly referred to as Electrophilic Aromatic Substitution. The catalysts and co-reagents serve to generate the strong electrophilic species needed to effect the initial step of the substitution. The specific electrophile believed to function in each type of reaction is listed in the right hand column.
Reaction Type Typical Equation Electrophile E(+)
Halogenation: C6H6 + Cl2 & heat
FeCl3 catalyst
——> C6H5Cl + HCl
Chlorobenzene
Cl(+) or Br(+)
Nitration: C6H6 + HNO3 & heat
H2SO4 catalyst
——> C6H5NO2 + H2O
Nitrobenzene
NO2(+)
Sulfonation: C6H6 + H2SO4 + SO3
& heat
——> C6H5SO3H + H2O
Benzenesulfonic acid
SO3H(+)
Alkylation:
Friedel-Crafts
C6H6 + R-Cl & heat
AlCl3 catalyst
——> C6H5-R + HCl
An Arene
R(+)
Acylation:
Friedel-Crafts
C6H6 + RCOCl & heat
AlCl3 catalyst
——> C6H5COR + HCl
An Aryl Ketone
RCO(+)
A Mechanism for Electrophilic Substitution Reactions of Benzene
A two-step mechanism has been proposed for these electrophilic substitution reactions. In the first, slow or rate-determining, step the electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate. In the second, fast step, a proton is removed from this intermediate, yielding a substituted benzene ring. The following four-part illustration shows this mechanism for the bromination reaction. Also, an animated diagram may be viewed.
Preliminary step: Formation of the strongly electrophilic bromine cation
Step 1: The electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate
Step 2: A proton is removed from this intermediate, yielding a substituted benzene ring
This mechanism for electrophilic aromatic substitution should be considered in context with other mechanisms involving carbocation intermediates. These include SN1 and E1 reactions of alkyl halides, and Brønsted acid addition reactions of alkenes.
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution or addition product.
2. The cation may transfer a proton to a base, giving a double bond product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
SN1 and E1 reactions are respective examples of the first two modes of reaction. The second step of alkene addition reactions proceeds by the first mode, and any of these three reactions may exhibit molecular rearrangement if an initial unstable carbocation is formed. The carbocation intermediate in electrophilic aromatic substitution (the benzenonium ion) is stabilized by charge delocalization (resonance) so it is not subject to rearrangement. In principle it could react by either mode 1 or 2, but the energetic advantage of reforming an aromatic ring leads to exclusive reaction by mode 2 (ie. proton loss).
Electrophilic Substitution of Disubstituted Benzene Rings
When a benzene ring has two substituent groups, each exerts an influence on subsequent substitution reactions. The activation or deactivation of the ring can be predicted more or less by the sum of the individual effects of these substituents. The site at which a new substituent is introduced depends on the orientation of the existing groups and their individual directing effects. We can identify two general behavior categories, as shown in the following table. Thus, the groups may be oriented in such a manner that their directing influences act in concert, reinforcing the outcome; or are opposed (antagonistic) to each other. Note that the orientations in each category change depending on whether the groups have similar or opposite individual directing effects.
Orientational Interaction of Substituents
Antagonistic or Non-Cooperative
Reinforcing or Cooperative
D = Electron Donating Group (ortho/para-directing)
W = Electron Withdrawing Group (meta-directing)
The products from substitution reactions of compounds having a reinforcing orientation of substituents are easier to predict than those having antagonistic substituents. For example, the six equations shown below are all examples of reinforcing or cooperative directing effects operating in the expected manner. Symmetry, as in the first two cases, makes it easy to predict the site at which substitution is likely to occur. Note that if two different sites are favored, substitution will usually occur at the one that is least hindered by ortho groups.
The first three examples have two similar directing groups in a meta-relationship to each other. In examples 4 through 6, oppositely directing groups have an ortho or para-relationship. The major products of electrophilic substitution, as shown, are the sum of the individual group effects. The strongly activating hydroxyl (–OH) and amino (–NH2) substituents favor dihalogenation in examples 5 and six.
Substitution reactions of compounds having an antagonistic orientation of substituents require a more careful analysis. If the substituents are identical, as in example 1 below, the symmetry of the molecule will again simplify the decision. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization, it will normally exert the product determining influence, examples 2, 4 & 5, even though it may be overall deactivating (case 2). Case 3 reflects a combination of steric hindrance and the superior innate stabilizing ability of methyl groups relative to other alkyl substituents. Example 6 is interesting in that it demonstrates the conversion of an activating ortho/para-directing group into a deactivating meta-directing "onium" cation [–NH(CH3)2(+) ] in a strong acid environment. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Benzene/Electrophilic_Aromatic_Substitution.txt |
An early method of preparing phenol (the Dow process) involved the reaction of chlorobenzene with a concentrated sodium hydroxide solution at temperatures above 350 ºC. The chief products are phenol and diphenyl ether (see below). This apparent nucleophilic substitution reaction is surprising, since aryl halides are generally incapable of reacting by either an SN1 or SN2 pathway.
C6H5–Cl + NaOH solution
350 ºC
C6H5–OH + C6H5–O–C6H5 + NaCl
The presence of electron-withdrawing groups (such as nitro) ortho and para to the chlorine substantially enhance the rate of substitution, as shown in the set of equations presented below.
To explain this, a third mechanism for nucleophilic substitution has been proposed. This two-step mechanism is characterized by initial addition of the nucleophile (hydroxide ion or water) to the aromatic ring, followed by loss of a halide anion from the negatively charged intermediate as illustrated below. The sites over which the negative charge is delocalized are colored blue, and the ability of nitro, and other electron withdrawing, groups to stabilize adjacent negative charge accounts for their rate enhancing influence at the ortho and para locations.
Three additional examples of aryl halide nucleophilic substitution are presented below. Only the 2- and 4-chloropyridine isomers undergo rapid substitution, the 3-chloro isomer is relatively unreactive. Nitrogen nucleophiles will also react, as evidenced by the use of Sanger's reagent for the derivatization of amino acids. The resulting N-2,4-dinitrophenyl derivatives are bright yellow crystalline compounds that facilitated analysis of peptides and proteins, a subject for which Frederick Sanger received one of his two Nobel Prizes in chemistry.
Additional Examples
Such addition-elimination processes generally occur at sp2 or sp hybridized carbon atoms, in contrast to SN1 and SN2 reactions. When applied to aromatic halides, as in the present discussion, this mechanism is called SNAr. Some distinguishing features of the three common nucleophilic substitution mechanisms are summarized in the following table.
Mechanism
Number of Steps
Bond Formation Timing
Carbon Hybridization
SN1
Two
After Bond Breaking
Usually sp3
SN2
One
Simultaneous with
Bond Breaking
Usually sp3
SNAr
Two
Prior to Bond Breaking
Usually sp2
Nucleophilic Elimination Reactions
There is good evidence that the synthesis of phenol from chlorobenzene does not proceed by the addition-elimination mechanism (SNAr) as previously described. For example, treatment of para-chlorotoluene with sodium hydroxide solution at temperatures above 350 ºC gave an equimolar mixture of meta- and para-cresols (hydroxytoluenes). Chloro and bromobenzene reacted with the very strong base sodium amide (NaNH2 at low temperature (-33 ºC in liquid ammonia) to give good yields of aniline (aminobenzene). However, ortho-chloroanisole gave exclusively meta-methoxyaniline under the same conditions. These reactions are described by the following equations.
The mechanism for which is as follows:
The explanation for this curious repositioning of the substituent group lies in a different two-step mechanism we can refer to as an elimination-addition process. The intermediate in this mechanism is an unstable benzyne species, as displayed in the above illustration by clicking the "Show Mechanism" button. In contrast to the parallel overlap of p-orbitals in a stable alkyne triple bond, the p-orbitals of a benzyne are tilted ca.120º apart, so the reactivity of this incipient triple bond to addition reactions is greatly enhanced. In the absence of steric hindrance (top example) equal amounts of meta- and para-cresols are obtained. The steric bulk of the methoxy group and the ability of its ether oxygen to stabilize an adjacent anion result in a substantial bias in the addition of amide anion or ammonia.
Nucleophilic Addition Reactions
Although it does so less readily than simple alkenes or dienes, benzene adds hydrogen at high pressure in the presence of Pt, Pd or Ni catalysts. The product is cyclohexane and the heat of reaction provides evidence of benzene's thermodynamic stability. Substituted benzene rings may also be reduced in this fashion, and hydroxy-substituted compounds, such as phenol, catechol and resorcinol, give carbonyl products resulting from the fast ketonization of intermediate enols. Nickel catalysts are often used for this purpose, as noted in the following equations.
Benzene is more susceptible to radical addition reactions than to electrophilic addition. We have already noted that benzene does not react with chlorine or bromine in the absence of a catalyst and heat. In strong sunlight or with radical initiators benzene adds these halogens to give hexahalocyclohexanes. It is worth noting that these same conditions effect radical substitution of cyclohexane, the key factors in this change of behavior are the pi-bonds array in benzene, which permit addition, and the weaker C-H bonds in cyclohexane. The addition of chlorine is shown below on the left; two of the seven meso-stereoisomers are displayed to the right.
The Birch Reduction
Another way of adding hydrogen to the benzene ring is by treatment with the electron rich solution of alkali metals, usually lithium or sodium, in liquid ammonia. See examples of this reaction, which is called the Birch Reduction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Benzene/Nucleophilic_Reactions_of_Benzene_Derivatives.txt |
Compounds in which two or more benzene rings are fused together were described in an earlier section, and they present interesting insights into aromaticity and reactivity. The smallest such hydrocarbon is naphthalene. Naphthalene is stabilized by resonance. Three canonical resonance contributors may be drawn, and are displayed in the following diagram.
The two structures on the left have one discrete benzene ring each, but may also be viewed as 10-pi-electron annulenes having a bridging single bond. The structure on the right has two benzene rings which share a common double bond. From heats of hydrogenation or combustion, the resonance energy of naphthalene is calculated to be 61 kcal/mole, 11 kcal/mole less than that of two benzene rings (2 * 36). As expected from an average of the three resonance contributors, the carbon-carbon bonds in naphthalene show variation in length, suggesting some localization of the double bonds. The C1–C2 bond is 1.36 Å long, whereas the C2–C3 bond length is 1.42 Å. This contrasts with the structure of benzene, in which all the C–C bonds have a common length, 1.39 Å.
Naphthalene is more reactive than benzene, both in substitution and addition reactions, and these reactions tend to proceed in a manner that maintains one intact benzene ring. The following diagram shows three oxidation and reduction reactions that illustrate this feature. In the last example, catalytic hydrogenation of one ring takes place under milder conditions than those required for complete saturation (the decalin product exists as cis/trans isomers). Electrophilic substitution reactions take place more rapidly at C1, although the C2 product is more stable and predominates at equilibrium. Examples of these reactions will be displayed by clicking on the diagram. The kinetically favored C1 orientation reflects a preference for generating a cationic intermediate that maintains one intact benzene ring. By clicking on the diagram a second time, the two naphthenonium intermediates created by attack at C1 and C2 will be displayed.
The structure and chemistry of more highly fused benzene ring compounds, such as anthracene and phenanthrene show many of the same characteristics described above.
Reactions of Substituent Groups
Oxidation of Alkyl Side-Chains
The benzylic hydrogens of alkyl substituents on a benzene ring are activated toward free radical attack, as noted earlier. Furthermore, SN1, SN2 and E1 reactions of benzylic halides, show enhanced reactivity, due to the adjacent aromatic ring. The possibility that these observations reflect a general benzylic activation is supported by the susceptibility of alkyl side-chains to oxidative degradation, as shown in the following examples (the oxidized side chain is colored). Such oxidations are normally effected by hot acidic pemanganate solutions, but for large scale industrial operations catalyzed air-oxidations are preferred. Interestingly, if the benzylic position is completely substituted this oxidative degradation does not occur (second equation, the substituted benzylic carbon is colored blue).
C6H5CH2CH2CH2CH3 + KMnO4 + H3O(+) & heat C6H5CO2H + CO2
p-(CH3)3C–C6H4CH3 + KMnO4 + H3O(+) & heat p-(CH3)3C–C6H4CO2H
These equations are not balanced. The permanganate oxidant is reduced, usually to Mn(IV) or Mn(II). Two other examples of this reaction are given below, and illustrate its usefulness in preparing substituted benzoic acids.
Reduction of Nitro Groups and Aryl Ketones
Electrophilic nitration and Friedel-Crafts acylation reactions introduce deactivating, meta-directing substituents on an aromatic ring. The attached atoms are in a high oxidation state, and their reduction converts these electron withdrawing functions into electron donating amino and alkyl groups. Reduction is easily achieved either by catalytic hydrogenation (H2 + catalyst), or with reducing metals in acid. Examples of these reductions are shown here, equation 6 demonstrating the simultaneous reduction of both functions. Note that the butylbenzene product in equation 4 cannot be generated by direct Friedel-Crafts alkylation due to carbocation rearrangement. The zinc used in ketone reductions, such as 5, is usually activated by alloying with mercury (a process known as amalgamation).
Several alternative methods for reducing nitro groups to amines are known. These include zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most cases.
Conversion of Halogens to Organometallic Reagents
The reaction of alkyl and aryl halides with reactive metals (usually Li & Mg) to give nucleophilic reagents has been noted. This provides a powerful tool for the conversion of chloro, bromo or iodo substituents into a variety of other groups. Many reactions of these aryl lithium and Grignard reagents will be discussed in later sections, and the following equations provide typical examples of carboxylation, protonation and Gilman coupling. Metal halogen exchange reactions take place at low temperature, and may be used to introduce iodine at designated locations. An example of this method will be displayed below by clicking on the diagram. In this example care must be taken to maintain a low temperature, because elimination to an aryne intermediate takes place on warming.
Hydrolysis of Sulfonic Acids
The potential reversibility of the aromatic sulfonation reaction was noted earlier. The following equation illustrates how this characteristic of the sulfonic acids may be used to prepare the 3-bromo derivative of ortho-xylene. Direct bromination would give the 4-bromo derivative.
Modifying the Influence of Strong Activating Groups
The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Bromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily. Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
By acetylating the heteroatom substituent on phenol and aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH2
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
Deactivation of Phenol to Phenyl Acetate and Aniline to Acetanilide | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Benzene/Reactions_of_Fused_Benzene_Rings.txt |
When substituted benzene compounds undergo electrophilic substitution reactions, two related features must be considered.
Relative reactivity
The first is the relative reactivity of the compound compared with benzene itself. Experiments have shown that substituents on a benzene ring can influence reactivity in a profound manner. For example, a hydroxy or methoxy substituent increases the rate of electrophilic substitution about ten thousand fold, as illustrated by the case of anisole in the virtual demonstration (above). In contrast, a nitro substituent decreases the ring's reactivity by roughly a million. This activation or deactivation of the benzene ring toward electrophilic substitution may be correlated with the electron donating or electron withdrawing influence of the substituents, as measured by molecular dipole moments. In the following diagram we see that electron donating substituents (blue dipoles) activate the benzene ring toward electrophilic attack, and electron withdrawing substituents (red dipoles) deactivate the ring (make it less reactive to electrophilic attack).
The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:
Inductive and Conjugation Effects
The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring (left-hand diagram below).
The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance), as in the middle diagram. Finally, polar double and triple bonds conjugated with the benzene ring may withdraw electrons, as in the right-hand diagram. Note that in the resonance examples all the contributors are not shown. In both cases the charge distribution in the benzene ring is greatest at sites ortho and para to the substituent.
In the case of the nitrogen and oxygen activating groups displayed in the top row of the previous diagram, electron donation by resonance dominates the inductive effect and these compounds show exceptional reactivity in electrophilic substitution reactions. Although halogen atoms have non-bonding valence electron pairs that participate in p-π conjugation, their strong inductive effect predominates, and compounds such as chlorobenzene are less reactive than benzene. The three examples on the left of the bottom row (in the same diagram) are examples of electron withdrawal by conjugation to polar double or triple bonds, and in these cases the inductive effect further enhances the deactivation of the benzene ring. Alkyl substituents such as methyl increase the nucleophilicity of aromatic rings in the same fashion as they act on double bonds.
Isomerization
The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs. Since a mono-substituted benzene ring has two equivalent ortho-sites, two equivalent meta-sites and a unique para-site, three possible constitutional isomers may be formed in such a substitution. If reaction occurs equally well at all available sites, the expected statistical mixture of isomeric products would be 40% ortho, 40% meta and 20% para. Again we find that the nature of the substituent influences this product ratio in a dramatic fashion. Bromination of methoxybenzene (anisole) is very fast and gives mainly the para-bromo isomer, accompanied by 10% of the ortho-isomer and only a trace of the meta-isomer. Bromination of nitrobenzene requires strong heating and produces the meta-bromo isomer as the chief product.
Some additional examples of product isomer distribution in other electrophilic substitutions are given in the table below. It is important to note here that the reaction conditions for these substitution reactions are not the same, and must be adjusted to fit the reactivity of the reactant C6H5-Y. The high reactivity of anisole, for example, requires that the first two reactions be conducted under very mild conditions (low temperature and little or no catalyst). The nitrobenzene reactant in the third example is very unreactive, so rather harsh reaction conditions must be used to accomplish that reaction.
Y in C6H5–Y
Reaction
% Ortho-Product
% Meta-Product
% Para-Product
–O–CH3 Nitration 30–40 0–2 60–70
–O–CH3 F-C Acylation 5–10 0–5 90–95
–NO2 Nitration 5–8 90–95 0–5
–CH3 Nitration 55–65 1–5 35–45
–CH3 Sulfonation 30–35 5–10 60–65
–CH3 F-C Acylation 10–15 2–8 85–90
–Br Nitration 35–45 0–4 55–65
–Br Chlorination 40–45 5–10 50–60
These observations, and many others like them, have led chemists to formulate an empirical classification of the various substituent groups commonly encountered in aromatic substitution reactions. Thus, substituents that activate the benzene ring toward electrophilic attack generally direct substitution to the ortho and para locations. With some exceptions, such as the halogens, deactivating substituents direct substitution to the meta location. The following table summarizes this classification.
Orientation and Reactivity Effects of Ring Substituents
ctivating Substituents
ortho & para-Orientation
Deactivating Substituents
meta-Orientation
Deactivating Substituents
ortho & para-Orientation
–O(–)
–OH
–OR
–OC6H5
–OCOCH3
–NH2
–NR2
–NHCOCH3
–R
–C6H5
–NO2
–NR3(+)
–PR3(+)
–SR2(+)
–SO3H
–SO2R
–CO2H
–CO2R
–CONH2
–CHO
–COR
–CN
–F
–Cl
–Br
–I
–CH2Cl
–CH=CHNO2
The information summarized in the above table is very useful for rationalizing and predicting the course of aromatic substitution reactions, but in practice most chemists find it desirable to understand the underlying physical principles that contribute to this empirical classification. We have already analyzed the activating or deactivating properties of substituents in terms of inductive and resonance effects, and these same factors may be used to rationalize their influence on substitution orientation.
The first thing to recognize is that the proportions of ortho, meta and para substitution in a given case reflect the relative rates of substitution at each of these sites. If we use the nitration of benzene as a reference, we can assign the rate of reaction at one of the carbons to be 1.0. Since there are six equivalent carbons in benzene, the total rate would be 6.0. If we examine the nitration of toluene, tert-butylbenzene, chlorobenzene and ethyl benzoate in the same manner, we can assign relative rates to the ortho, meta and para sites in each of these compounds. These relative rates are shown (colored red) in the following illustration, and the total rate given below each structure reflects the 2 to 1 ratio of ortho and meta sites to the para position. The overall relative rates of reaction, referenced to benzene as 1.0, are calculated by dividing by six. Clearly, the alkyl substituents activate the benzene ring in the nitration reaction, and the chlorine and ester substituents deactivate the ring.
From rate data of this kind, it is a simple matter to calculate the proportions of the three substitution isomers. Toluene gives 58.5% ortho-nitrotoluene, 37% para-nitrotoluene and only 4.5% of the meta isomer. The increased bulk of the tert-butyl group hinders attack at the ortho-sites, the overall product mixture being 16% ortho, 8% meta and 75% para-nitro product. Although chlorobenzene is much less reactive than benzene, the rate of ortho and para-substitution greatly exceeds that of meta-substitution, giving a product mixture of 30% ortho and 70% para-nitrochlorobenzene. Finally, the benzoic ester gave predominantly the meta-nitro product (73%) accompanied by the ortho (22%) and para (5%) isomers, as shown by the relative rates. Equivalent rate and product studies for other substitution reactions lead to similar conclusions. For example, electrophilic chlorination of toluene occurs hundreds of times faster than chlorination of benzene, but the relative rates are such that the products are 60% ortho-chlorotoluene, 39% para and 1% meta-isomers, a ratio similar to that observed for nitration.
The manner in which specific substituents influence the orientation of electrophilic substitution of a benzene ring is shown in the following interactive diagram. As noted on the opening illustration, the product-determining step in the substitution mechanism is the first step, which is also the slow or rate determining step. It is not surprising, therefore, that there is a rough correlation between the rate-enhancing effect of a substituent and its site directing influence. The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites. This can be seen in seven representative substituents beneath the following diagram.
Y–
CH3
Cl or Br
NO2
RC=O
SO3H
OH
NH2
CH3 substitution
In the case of alkyl substituents, charge stabilization is greatest when the alkyl group is bonded to one of the positively charged carbons of the benzenonium intermediate. This happens only for ortho and para electrophilic attack, so such substituents favor formation of those products. Interestingly, primary alkyl substituents, especially methyl, provide greater stabilization of an adjacent charge than do more substituted groups (note the greater reactivity of toluene compared with tert-butylbenzene).
SO3 substitution
Nitro (NO2), sulfonic acid (SO3H) and carbonyl (C=O) substituents have a full or partial positive charge on the atom bonded to the aromatic ring. Structures in which like-charges are close to each other are destabilized by charge repulsion, so these substituents inhibit ortho and para substitution more than meta substitution. Consequently, meta-products predominate when electrophilic substitution is forced to occur.
NH2 substitution
Halogen (X), OR and NR2 substituents all exert a destabilizing inductive effect on an adjacent positive charge, due to the high electronegativity of the substituent atoms. By itself, this would favor meta-substitution; however, these substituent atoms all have non-bonding valence electron pairs which serve to stabilize an adjacent positive charge by pi-bonding, with resulting delocalization of charge. Consequently, all these substituents direct substitution to ortho and para sites. The balance between inductive electron withdrawal and p-π conjugation is such that the nitrogen and oxygen substituents have an overall stabilizing influence on the benzenonium intermediate and increase the rate of substitution markedly; whereas halogen substituents have an overall destabilizing influence. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Benzene/Substitution_Reactions_of_Benzene_Derivatives.txt |
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