CHAPTER 16 - CHEMISTRY OF BENZENE ...
CHAPTER 16 - CHEMISTRY OF BENZENE: ELECTROPHILIC AROMATIC SUBSTITUTION As stated in the previous chapter, benzene and other aromatic rings do not undergo electrophilic addition reactions of the simple alkenes but rather undergo electrophilic substitution reactions to preserve the stable aromatic ring system. The general reaction is: H
E + E
+
+ H+
In each of the cases we will study, the reaction mechanism is the same - what is different in each is the attacking electrophile, E+, and the reagents that produce that electrophile and the subsequent products. The general reaction mechanism for all electrophilic aromatic substitutions (EAS) is: H
+ E+
H
H
E
E
H E
E
+ H+
Resonance stabilized intermedite
In each of the cases we will study, there will be an overall reaction that should show starting material (e.g. benzene), the reagents used for that reaction and the product obtained. There will also be a mechanism for the particular reaction, which is the same in all cases except for the attacking electrophile and how it is generated from the reagents. In writing a mechanism you must show how the E+ is generated, all resonance forms of the intermediate and the product.
1
Below is a summary of the five electrophilic substitution reactions covered in this chapter, showing Reagents Generation of E+ Product 1. Halogenation: A. Bromination and Chlorination X2/FeX3
d d X3Fe----Br----Br
X X + FeX3
X
"Polarized halogen"
X = Cl, Br
B. Iodination I I2/CuCl2
2Cu+ + I +
I2 + 2Cu2+
2. Nitration HNO3/H2SO4
HOHO2 +2H 2SO4
NO2 H3O + 2HSO4 + NO2
3. Sulfonation O S
SO3/H2SO4 O
OH HSO4 + S O O
+H2SO4 O
2
SO3H
4. Friedel-Crafts Alkylation RX, AlCl3
R X
+ AlCl3
R
AlCl3X + R
R = alkyl When R = methyl or a primary halide, the carbocation does not form as a separate entity but rather is a polarized complex with the aluminum tetrahalide anion. Secondary and tertiary carbocations form but remember with any carbocation (except methyl and ethyl) rearrangement is possible and will occur if a more stable cation can be formed. H + CH3CH2CH 2Cl
AlCl3
In the reaction CH3CH 2CH2-
H3C
rearranges to
CH H3C
5. Friedel-Crafts Acylation O
O R C X
AlCl3
O R C X
R C O + AlCl3
AlCl3X +
R = alkyl,aryl, vinyl
R
Now let's look at some reactions:
3
C O
R
O
H
O
C
AlCl3
+ H3C C
CH3
Cl H
NO2
HNO3 H2SO4
H
Br
Br2 FeBr3
Remember that when asked for a mechanism, you must show generation of the attacking electrophile, then the attack on the benzene ring showing all resonance forms of the intermediate and finally the formation of product with loss of H+ . (See beginning of chapter notes). Show complete mechanism for the nitration of benzene:
Directional and Activation/deactivating Properties of Ring Substituents What happens when there is already a substituent on the ring? In which position, relative to that substituent, will the incoming groups go - ortho, meta or para???? 4
First group on the ring determines the reactivity of the ring toward further substitution relative to benzene itself, and also determines the where the second incoming group will go. These are two separate aspects to consider in further substitution on the ring. An activating group makes the ring more reactive than benzene to further substitution. In some cases the activation is so great that polysubstitution occurs and cannot be controlled. A deactivating group makes the ring less reactive than benzene to further substitution. In some cases the ring is deactivated sufficiently to prevent certain reactions from occurring. With one group already on the ring, a second substituent can come on the ring in ortho, meta or para positions. X
There are two ortho positions relative to group X, two meta positions and one para position that are available for subsequent substitution.
ortho
ortho
meta
meta
para
The group, X, already on the ring determines where the incoming group will go......regardless of what the second reaction is. Experimentally we find that some groups give a mixture of ortho and para products almost exclusively with little or no meta product. Some groups give mostly meta product with some ortho and para product in the mixture. (See Table 16.1 on p. 606 of your text). When we examine these cases we find three groups - those that direct ortho, para and activate the ring, those that direct o, p and deactivate the ring (halogens) and those that direct meta and deactivate the ring. Below is a chart summarizing these effects. The vertical double line separates the ortho/para directors from the meta directors. This simply means that when one of the o-p directors is on the ring further substitution will result in products where the new group is ortho or para to that first group. When a 5
meta director is on the ring, further substitution will produce primarily the meta product. Mixtures will occur is almost all cases. Note that with the exceptions of phenyl, vinyl and alkyl, all of the o-p directors have a lone pair of electrons on the atom attached to the ring. In the meta director, with the exception of -CF3, the atoms attached to the ring have a multiple bond to an electronegative element. All of the meta directors are electron-withdrawing groups.
META DIRECTORS
ORTHO,PARA DIRECTORS -NH2 -NHR -NR2 -OH
-OCH3 -OR -NHCOCH3
strong
moderate
Phenyl-CH=CH2 Alkyl-
weak
ACTIVATORS
-F -Cl -Br -I
-CHO -COOH -COOR -COCl -CONH2 -SO3 H -CN
weak
strong
-CF3 -NO2
very strong
DEACTIVATORS
The bold vertical line separates the activating groups from the deactivating groups. Note that of the deactivating groups, only the halogens are o-p directors.
Two effects are working here. One is induction, which is an effect that occurs through the sigma bond system. Electron withdrawing groups will "pull" electron density away from the ring making the electrons of the ring less available for attack by an electrophile. Electron donating groups do the opposite. Induction also plays a role in some of the directional effects. The other effect is resonance, which occurs through the pi system of bonds in the molecule. Resonance plays an important role in the stability of intermediates of the reactions. Sometimes these two effects are opposite to one another. One effect may be stronger and exert greater influence. We will point these out as we discuss the 6
reactions. As an example, look at phenol (benzene with a hydroxyl group on ring). OH is an electron withdrawing group (induction) but OH is a very strong activator and o, p director. Resonance plays a more important role as can be seen by resonance structures of phenol shown below.
OH
OH
OH
OH
Let's examine some general reactions with one group already on the ring. 1. Electrophilic substitution of toluene (methylbenzene). Alkyl groups are electron donators so we can understand why these groups activate the ring. Why are alkyl groups o, p directors? If we examine toluene under attack by some electrophile and examine the intermediate if the electrophile goes in the para position we get:
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CH3
CH3
CH3
CH3
E+ H
H
E
H
H
E
E
resonance forms of intermediate In the circled contributing resonance structure the positive charge is directly stabilized by the electron donating effect of the alkyl group. If you examine attack in the ortho position, you will find a similar phenomenon. DRAW THIS FOR PRACTICE!
Now examine meta attack on toluene: CH3
CH3
CH3 E+
H
H H
CH3
E
H
E E There is no one very stable form of the intermediate. The intermediate in ortho or para attack is more stable, more easily formed and reaction goes in that direction. 2. Electrophilic substitution of Phenol (hydroxy benzene)
8
Examine ortho attack on phenol: OH
OH
OH H
E+
E
OH
OH
H
H
E
E
Why is the circled resonance contributing structure very stable? Draw the resonance forms of the intermediate in meta attack on phenol. Convince yourself that the lone pair of electrons on the oxygen cannot be part of resonance with the ring.
3. Electrophilic substitution of Nitrobenzene If the incoming group goes in the para position
O
O N
NO2
O
E
H
N
O
NO2
E+ H
H
E
H
E
The circled intermediate resonance form is especially UNSTABLE. WHY?
9
H E
Meta attack on nitrobenzene: O
O N
NO2
NO2
NO2
E+
H
H H
E
H E
E
In meta attack there is no very unstable form, nor is there an especially stable form. Both substitutions are difficult, but meta is more likely.
EXCEPTIONS!!!!!!!!!!!!! 1. Hydroxy and amine are very strong activators. In bromination of phenol and aniline, reaction can be accomplished with Br2 in H20 (no iron catalyst needed) and does not stop with monosubstitution. A trisubstituted product is obtained. OH
OH Br2 H2O
Br
Br
Br Solution: If the OH or the NH2 is first acylated to give -OCOCH3 or NHCOCH3, the ring is not so active (resonance forms that draw electrons to side group rather than back into ring can be drawn) Monobromination can be accomplished, usually in the para position because the ortho positions are now stericly hindered. The acyl groups can be hydrolyzed off in a later reaction.
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O HN
NH2
CH3
CH3COCl
NHCOCH3
Br2
H2O
HOAc
HO_ Br
NH2
Br
2. The NH2 group on aniline is a strong base. In a Friedel-Crafts reaction, the NH2 group will react with the Lewis acid, AlCl3 to form a complex. Then the Lewis acid is not available for the F-C reaction. Again acylation of the amine group will help in eliminating this problem NH2
H2N
AlCl3
CH3CH 2Cl AlCl3 Not alkylation product. 3. Friedel-Craft electrophiles are not strong electrophiles. If any deactivating group stronger than halogen is already on the ring, F-C will not occur.
Trisubstituted Benzenes: If two groups are already on the ring, where will a third group go? Same inductive and resonance effects govern the reaction. 1. If directing effects of the two groups reinforce each other, no problem. OH
NO2
11
2. If the two groups oppose each other, the stronger activating group will direct the incoming group. (Mixtures often result). OH
NOT
NOT
CH3
3. Substitution rarely occurs in the position between two groups that are 1,3(meta) to each other because of steric hindrance. Cl NOT CH3
Side Chain Reactions: 1. Oxidation of Alkylbenzene Side Chains The aromatic ring is not oxidized by strong oxidizing agents such and Na2Cr2O7 of KmnO4 which will cleave normal alkenes. But the ring activated alkyl groups to oxidization (Remember alkanes themselves do not react with these reagents). CH3
COOH
KMnO4 H2O, 95˚C
As long as the carbon directly adjacent to the ring has at least one hydrogen on it, that carbon is subject to oxidation. Regardless how long the alkyl group is, the product is always benzoic acid.
12
CH2CH(CH3)2
COOH
KMnO4 H2O, 95˚C
t-Butyl benzene is not subject to oxidation because there is no hydrogen on the carbon adjacent to the ring. CH3 C CH3 CH3
KMnO4 H2O, 95˚C
N.R.
2. Bromination of Alkyl Side Chains: If an alkyl benzene is treated with NBS (N-Bromosuccinimide), in the presence of benzoyl peroxide (a radical stimulator) bromination occurs exclusively on the benzylic C - that is the carbon adjacent to the ring. The intermediate is stabilized by resonance with the ring. CH2CH 2CH3
Br
NBS
CHCH2CH3
ROOR
Benzylic carbon
CH-CH2CH3
CH-CH2CH3
ETC.
3.Reduction of Aromatic compounds: A. Reduction of benzene ring Under normal reduction conditions of H2/Pd in ethanol for example, the benzene ring cannot be reduced. In the reaction below notice that the carbonyl (ketone) and ring are not reduced, only the alkene.
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O CH CH C CH3
O CH2 CH2 C CH3
H2,Pd ethanol
Excessive conditions are necessary to reduce the ring to a cyclohexane. a. H2 with a Platinum catalyst at 2000 psi pressure will reduce the ring. b. Normal pressure but using a more powerful catalyst such as rhodium on carbon will reduce the ring. CH3
CH3
H2,Pt,ethanol 2000 psi, 25˚C
CH3
CH3
Or: H2,Rh/C,ethanol. 1 atm. 25˚C
B. Reduction of Aryl Alkyl Ketones The aromatic ring activates adjacent carbonyl groups toward reduction. O C
CH2CH 3
CH3 H2,Pd ethanol
This gives us a way to overcome the problem in F-C alkylation of rearrangement of alkyl carbocations.
14
l COC
O C
CH 2 CH 3 l AlC 3 CH 3 CH 2 CH AlC 2 Cl l3
CH2CH 3 H2,Pd ethanol
CH3 CH CH3
CH2CH 2CH3
CH2CH 2CH3 +
C. Reduction of Other Side Groups Nitro groups can be reduced with H2, Pd/C in ethanol. Remember this also reduces adjacent ketones and alkenes anywhere in the molecule. O CH3
CH2CH 3
H2, Pd/C ethanol
NO2
NH2 Nitro groups can also be reduced with tin and hydrochloric acid (SnCl2/HCl), which will not reduce the carbonyl or the alkene. This gives us a way to put an amine group on the ring. O
O CH3 1.SnCl , H O+ 2 3
CH3
2. H2O, HONO2
NH2
SYNTHESIS: In all preparations (syntheses) assume that ortho and para products can be separated. Show only the one you need for your synthesis. Prepare p-bromobenzoic acid from benzene 15
Prepare 4-chloro-1-nitro-2-propylbenzene from benzene
Prepare 2-bromo-4-nitrotoluene from benzene
16
Prepare CH2OH
CH2OH from benzene.
Omit Sections 16.8 & 16.9
17
E + E
+
+ H+
In each of the cases we will study, the reaction mechanism is the same - what is different in each is the attacking electrophile, E+, and the reagents that produce that electrophile and the subsequent products. The general reaction mechanism for all electrophilic aromatic substitutions (EAS) is: H
+ E+
H
H
E
E
H E
E
+ H+
Resonance stabilized intermedite
In each of the cases we will study, there will be an overall reaction that should show starting material (e.g. benzene), the reagents used for that reaction and the product obtained. There will also be a mechanism for the particular reaction, which is the same in all cases except for the attacking electrophile and how it is generated from the reagents. In writing a mechanism you must show how the E+ is generated, all resonance forms of the intermediate and the product.
1
Below is a summary of the five electrophilic substitution reactions covered in this chapter, showing Reagents Generation of E+ Product 1. Halogenation: A. Bromination and Chlorination X2/FeX3
d d X3Fe----Br----Br
X X + FeX3
X
"Polarized halogen"
X = Cl, Br
B. Iodination I I2/CuCl2
2Cu+ + I +
I2 + 2Cu2+
2. Nitration HNO3/H2SO4
HOHO2 +2H 2SO4
NO2 H3O + 2HSO4 + NO2
3. Sulfonation O S
SO3/H2SO4 O
OH HSO4 + S O O
+H2SO4 O
2
SO3H
4. Friedel-Crafts Alkylation RX, AlCl3
R X
+ AlCl3
R
AlCl3X + R
R = alkyl When R = methyl or a primary halide, the carbocation does not form as a separate entity but rather is a polarized complex with the aluminum tetrahalide anion. Secondary and tertiary carbocations form but remember with any carbocation (except methyl and ethyl) rearrangement is possible and will occur if a more stable cation can be formed. H + CH3CH2CH 2Cl
AlCl3
In the reaction CH3CH 2CH2-
H3C
rearranges to
CH H3C
5. Friedel-Crafts Acylation O
O R C X
AlCl3
O R C X
R C O + AlCl3
AlCl3X +
R = alkyl,aryl, vinyl
R
Now let's look at some reactions:
3
C O
R
O
H
O
C
AlCl3
+ H3C C
CH3
Cl H
NO2
HNO3 H2SO4
H
Br
Br2 FeBr3
Remember that when asked for a mechanism, you must show generation of the attacking electrophile, then the attack on the benzene ring showing all resonance forms of the intermediate and finally the formation of product with loss of H+ . (See beginning of chapter notes). Show complete mechanism for the nitration of benzene:
Directional and Activation/deactivating Properties of Ring Substituents What happens when there is already a substituent on the ring? In which position, relative to that substituent, will the incoming groups go - ortho, meta or para???? 4
First group on the ring determines the reactivity of the ring toward further substitution relative to benzene itself, and also determines the where the second incoming group will go. These are two separate aspects to consider in further substitution on the ring. An activating group makes the ring more reactive than benzene to further substitution. In some cases the activation is so great that polysubstitution occurs and cannot be controlled. A deactivating group makes the ring less reactive than benzene to further substitution. In some cases the ring is deactivated sufficiently to prevent certain reactions from occurring. With one group already on the ring, a second substituent can come on the ring in ortho, meta or para positions. X
There are two ortho positions relative to group X, two meta positions and one para position that are available for subsequent substitution.
ortho
ortho
meta
meta
para
The group, X, already on the ring determines where the incoming group will go......regardless of what the second reaction is. Experimentally we find that some groups give a mixture of ortho and para products almost exclusively with little or no meta product. Some groups give mostly meta product with some ortho and para product in the mixture. (See Table 16.1 on p. 606 of your text). When we examine these cases we find three groups - those that direct ortho, para and activate the ring, those that direct o, p and deactivate the ring (halogens) and those that direct meta and deactivate the ring. Below is a chart summarizing these effects. The vertical double line separates the ortho/para directors from the meta directors. This simply means that when one of the o-p directors is on the ring further substitution will result in products where the new group is ortho or para to that first group. When a 5
meta director is on the ring, further substitution will produce primarily the meta product. Mixtures will occur is almost all cases. Note that with the exceptions of phenyl, vinyl and alkyl, all of the o-p directors have a lone pair of electrons on the atom attached to the ring. In the meta director, with the exception of -CF3, the atoms attached to the ring have a multiple bond to an electronegative element. All of the meta directors are electron-withdrawing groups.
META DIRECTORS
ORTHO,PARA DIRECTORS -NH2 -NHR -NR2 -OH
-OCH3 -OR -NHCOCH3
strong
moderate
Phenyl-CH=CH2 Alkyl-
weak
ACTIVATORS
-F -Cl -Br -I
-CHO -COOH -COOR -COCl -CONH2 -SO3 H -CN
weak
strong
-CF3 -NO2
very strong
DEACTIVATORS
The bold vertical line separates the activating groups from the deactivating groups. Note that of the deactivating groups, only the halogens are o-p directors.
Two effects are working here. One is induction, which is an effect that occurs through the sigma bond system. Electron withdrawing groups will "pull" electron density away from the ring making the electrons of the ring less available for attack by an electrophile. Electron donating groups do the opposite. Induction also plays a role in some of the directional effects. The other effect is resonance, which occurs through the pi system of bonds in the molecule. Resonance plays an important role in the stability of intermediates of the reactions. Sometimes these two effects are opposite to one another. One effect may be stronger and exert greater influence. We will point these out as we discuss the 6
reactions. As an example, look at phenol (benzene with a hydroxyl group on ring). OH is an electron withdrawing group (induction) but OH is a very strong activator and o, p director. Resonance plays a more important role as can be seen by resonance structures of phenol shown below.
OH
OH
OH
OH
Let's examine some general reactions with one group already on the ring. 1. Electrophilic substitution of toluene (methylbenzene). Alkyl groups are electron donators so we can understand why these groups activate the ring. Why are alkyl groups o, p directors? If we examine toluene under attack by some electrophile and examine the intermediate if the electrophile goes in the para position we get:
7
CH3
CH3
CH3
CH3
E+ H
H
E
H
H
E
E
resonance forms of intermediate In the circled contributing resonance structure the positive charge is directly stabilized by the electron donating effect of the alkyl group. If you examine attack in the ortho position, you will find a similar phenomenon. DRAW THIS FOR PRACTICE!
Now examine meta attack on toluene: CH3
CH3
CH3 E+
H
H H
CH3
E
H
E E There is no one very stable form of the intermediate. The intermediate in ortho or para attack is more stable, more easily formed and reaction goes in that direction. 2. Electrophilic substitution of Phenol (hydroxy benzene)
8
Examine ortho attack on phenol: OH
OH
OH H
E+
E
OH
OH
H
H
E
E
Why is the circled resonance contributing structure very stable? Draw the resonance forms of the intermediate in meta attack on phenol. Convince yourself that the lone pair of electrons on the oxygen cannot be part of resonance with the ring.
3. Electrophilic substitution of Nitrobenzene If the incoming group goes in the para position
O
O N
NO2
O
E
H
N
O
NO2
E+ H
H
E
H
E
The circled intermediate resonance form is especially UNSTABLE. WHY?
9
H E
Meta attack on nitrobenzene: O
O N
NO2
NO2
NO2
E+
H
H H
E
H E
E
In meta attack there is no very unstable form, nor is there an especially stable form. Both substitutions are difficult, but meta is more likely.
EXCEPTIONS!!!!!!!!!!!!! 1. Hydroxy and amine are very strong activators. In bromination of phenol and aniline, reaction can be accomplished with Br2 in H20 (no iron catalyst needed) and does not stop with monosubstitution. A trisubstituted product is obtained. OH
OH Br2 H2O
Br
Br
Br Solution: If the OH or the NH2 is first acylated to give -OCOCH3 or NHCOCH3, the ring is not so active (resonance forms that draw electrons to side group rather than back into ring can be drawn) Monobromination can be accomplished, usually in the para position because the ortho positions are now stericly hindered. The acyl groups can be hydrolyzed off in a later reaction.
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O HN
NH2
CH3
CH3COCl
NHCOCH3
Br2
H2O
HOAc
HO_ Br
NH2
Br
2. The NH2 group on aniline is a strong base. In a Friedel-Crafts reaction, the NH2 group will react with the Lewis acid, AlCl3 to form a complex. Then the Lewis acid is not available for the F-C reaction. Again acylation of the amine group will help in eliminating this problem NH2
H2N
AlCl3
CH3CH 2Cl AlCl3 Not alkylation product. 3. Friedel-Craft electrophiles are not strong electrophiles. If any deactivating group stronger than halogen is already on the ring, F-C will not occur.
Trisubstituted Benzenes: If two groups are already on the ring, where will a third group go? Same inductive and resonance effects govern the reaction. 1. If directing effects of the two groups reinforce each other, no problem. OH
NO2
11
2. If the two groups oppose each other, the stronger activating group will direct the incoming group. (Mixtures often result). OH
NOT
NOT
CH3
3. Substitution rarely occurs in the position between two groups that are 1,3(meta) to each other because of steric hindrance. Cl NOT CH3
Side Chain Reactions: 1. Oxidation of Alkylbenzene Side Chains The aromatic ring is not oxidized by strong oxidizing agents such and Na2Cr2O7 of KmnO4 which will cleave normal alkenes. But the ring activated alkyl groups to oxidization (Remember alkanes themselves do not react with these reagents). CH3
COOH
KMnO4 H2O, 95˚C
As long as the carbon directly adjacent to the ring has at least one hydrogen on it, that carbon is subject to oxidation. Regardless how long the alkyl group is, the product is always benzoic acid.
12
CH2CH(CH3)2
COOH
KMnO4 H2O, 95˚C
t-Butyl benzene is not subject to oxidation because there is no hydrogen on the carbon adjacent to the ring. CH3 C CH3 CH3
KMnO4 H2O, 95˚C
N.R.
2. Bromination of Alkyl Side Chains: If an alkyl benzene is treated with NBS (N-Bromosuccinimide), in the presence of benzoyl peroxide (a radical stimulator) bromination occurs exclusively on the benzylic C - that is the carbon adjacent to the ring. The intermediate is stabilized by resonance with the ring. CH2CH 2CH3
Br
NBS
CHCH2CH3
ROOR
Benzylic carbon
CH-CH2CH3
CH-CH2CH3
ETC.
3.Reduction of Aromatic compounds: A. Reduction of benzene ring Under normal reduction conditions of H2/Pd in ethanol for example, the benzene ring cannot be reduced. In the reaction below notice that the carbonyl (ketone) and ring are not reduced, only the alkene.
13
O CH CH C CH3
O CH2 CH2 C CH3
H2,Pd ethanol
Excessive conditions are necessary to reduce the ring to a cyclohexane. a. H2 with a Platinum catalyst at 2000 psi pressure will reduce the ring. b. Normal pressure but using a more powerful catalyst such as rhodium on carbon will reduce the ring. CH3
CH3
H2,Pt,ethanol 2000 psi, 25˚C
CH3
CH3
Or: H2,Rh/C,ethanol. 1 atm. 25˚C
B. Reduction of Aryl Alkyl Ketones The aromatic ring activates adjacent carbonyl groups toward reduction. O C
CH2CH 3
CH3 H2,Pd ethanol
This gives us a way to overcome the problem in F-C alkylation of rearrangement of alkyl carbocations.
14
l COC
O C
CH 2 CH 3 l AlC 3 CH 3 CH 2 CH AlC 2 Cl l3
CH2CH 3 H2,Pd ethanol
CH3 CH CH3
CH2CH 2CH3
CH2CH 2CH3 +
C. Reduction of Other Side Groups Nitro groups can be reduced with H2, Pd/C in ethanol. Remember this also reduces adjacent ketones and alkenes anywhere in the molecule. O CH3
CH2CH 3
H2, Pd/C ethanol
NO2
NH2 Nitro groups can also be reduced with tin and hydrochloric acid (SnCl2/HCl), which will not reduce the carbonyl or the alkene. This gives us a way to put an amine group on the ring. O
O CH3 1.SnCl , H O+ 2 3
CH3
2. H2O, HONO2
NH2
SYNTHESIS: In all preparations (syntheses) assume that ortho and para products can be separated. Show only the one you need for your synthesis. Prepare p-bromobenzoic acid from benzene 15
Prepare 4-chloro-1-nitro-2-propylbenzene from benzene
Prepare 2-bromo-4-nitrotoluene from benzene
16
Prepare CH2OH
CH2OH from benzene.
Omit Sections 16.8 & 16.9
17
