Chemistry 232 Summer 2016 Quiz 2a

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1 1 Chemistry 232 Summer 2016 Quiz 2a KEY Name: (Please print, surname first) ID: Do all questions. Time available: 20 min. Total available marks: 20. Score for this quiz: Answers written partially or completely in pencil cannot be remarked. 1. Multiple choice. Clearly indicate your choice corresponding to the correct answer in each part below. Be certain to read the entire set of choices before making your selection. 6 marks i) From the list, choose the correct combination of reagents and conditions required to synthesize 4-nitrotoluene (shown) from toluene. a) CH 3 Br, AlCl 3 b) NaNO 3, peroxides c) NH 3, H 2 SO 4 d) Br 2, FeBr 3 e) NaNO 2, H 2 SO 4, 0 C f) HNO 3, H 2 SO 4 The correct IUPAC name for the compound shown here is: a) acetophenone b) carboxyphenyl c) acetaldehyde d) phenol e) benzaldehyde f) benzoic acid Clearly indicate the most stable intermediate formed during the chlorination of anisole (shown at right), from the choices below. In the most stable intermediate, every atom has an octet of electrons.

2 2 iv) From the following five compounds, clearly indicate the one that would react the slowest, and the one that would react the fastest in an electrophilic aromatic substitution reaction. slowest fastest The slowest and fastest reactants are the ones that withdraw electron density by resonance and donate electron density by resonance, respectively. 2. While working for Superior Chemical Aromatic Resources Inc., you are asked to evaluate the two synthetic methods shown below. Clearly indicate the one that will work (0.5 mark) and provide a reason for your choice (1.5 marks). 2 marks total Since both the chlorination and Friedel-Crafts acylation of benzene are possible, the question is really asking you to look at the second reaction and decide if it will work as written. Both reactions will go (Friedel-Crafts on aromatic halides are OK), but it s the orientation that is important. Route A looks good, the acyl group is deactivating, and a meta director. Route B has a problem. The chloro group is a weakly deactivating group, and is an ortho/para director, while we want the meta compound. Route A it is. Sorry, but you can t say I d never work for SCARI. 3. Give the structural formula for the major product of each of the following reactions. The third question is on the next page. 6 marks i) Both methyl groups are oxidized to carboxylic acids. This is the Clemmensen reduction, removing the carbonyl.

3 3 Both groups are o/p directors, and three sites are activated. But the site between the groups is too hindered to be brominated. The site ortho to the ethyl and para to the amine is less hindered than the other site. 4. Use a complete, step-wise, curved arrow mechanism to demonstrate the following result. Friedel- Crafts reaction of 1-chlorobutane with benzene (in the presence of aluminum chloride) results in the formation of two monosubstituted products. You will need to show the generation of the electrophile(s), their reactions with benzene, and of course the two products. 6 marks For the generation of the electrophile, the aluminum trichloride is a strong Lewis acid, forming a complex that reacts much like a primary carbocation, including rearranging. Secondary carbocations are more stable than primary carbocation complexes, and will be formed here by hydride migration. All hydrogens have been shown here, though it is only necessary to draw the one that is migrating. Both the complex and the carbocation will react with benzene in the Friedel-Crafts alkylation. Two electrons from the system bond to the carbocation (forming a carbon-carbon bond) and leave a resonance-stabilized cation behind. The base (AlCl 4 - ) removes the proton shown (this proton must be shown here!) and the two electrons from the C-H bond form a carbon-carbon double bond and reform the -system of the aromatic ring. The byproducts of HCl and AlCl 3 are not shown. END

4 1 Chemistry 232 Summer 2016 Quiz 2b KEY Name: (Please print, surname first) ID: Do all questions. Time available: 20 min. Total available marks: 20. Score for this quiz: Answers written partially or completely in pencil cannot be remarked. 1. Multiple choice. Clearly indicate your choice corresponding to the correct answer in each part below. Be certain to read the entire set of choices before making your selection. 6 marks i) From the list, choose the correct combination of reagents and conditions required to synthesize 4-nitrotoluene (shown) from toluene. a) NH 3, H 2 SO 4 b) Br 2, FeBr 3 c) NaNO 3, peroxides d) HNO 3, H 2 SO 4 e) NaNO 2, H 2 SO 4, 0 C f) CH 3 Br, AlCl 3 The correct IUPAC name for the compound shown here is: a) acetophenone b) carboxyphenyl c) acetaldehyde d) phenol e) benzaldehyde f) benzoic acid Clearly indicate the most stable intermediate formed during the chlorination of anisole (shown at right), from the choices below. In the most stable intermediate, every atom has an octet of electrons.

5 2 iv) From the following five compounds, clearly indicate the one that would react the slowest, and the one that would react the fastest in an electrophilic aromatic substitution reaction. slowest fastest The slowest and fastest reactants are the ones that withdraw electron density by resonance and donate electron density by resonance, respectively. 2. While working for Superior Chemical Aromatic Resources Inc., you are asked to evaluate the two synthetic methods shown below. Clearly indicate the one that will work (0.5 mark) and provide a reason for your choice (1.5 marks). 2 marks total Since both the bromination and Friedel-Crafts acylation of benzene are possible, the question is really asking you to look at the second reaction and decide if it will work as written. Both reactions will go (Friedel-Crafts on aromatic halides are OK), but it s the orientation that is important. Route A looks good, the acyl group is deactivating, and a meta director. Route B has a problem. The bromo group is a weakly deactivating group, and is an ortho/para director, while we want the meta compound. Route A it is. Sorry, but you can t say I d never work for SCARI. 3. Give the structural formula for the major product of each of the following reactions. The third question is on the next page. 6 marks i) Both groups are o/p directors, and three sites are activated. But the site between the groups is too hindered to be brominated. The site ortho to the ethyl and para to the amine is less hindered than the other site. Both methyl groups are oxidized to carboxylic acids.

6 3 This is the Clemmensen reduction, removing the carbonyl. 4. Use a complete, step-wise, curved arrow mechanism to demonstrate the following result. Friedel- Crafts reaction of 1-chlorobutane with benzene (in the presence of aluminum chloride) results in the formation of two monosubstituted products. You will need to show the generation of the electrophile(s), their reactions with benzene, and of course the two products. 6 marks For the generation of the electrophile, the aluminum trichloride is a strong Lewis acid, forming a complex that reacts much like a primary carbocation, including rearranging. Secondary carbocations are more stable than primary carbocation complexes, and will be formed here by hydride migration. All hydrogens have been shown here, though it is only necessary to draw the one that is migrating. Both the complex and the carbocation will react with benzene in the Friedel-Crafts alkylation. Two electrons from the system bond to the carbocation (forming a carbon-carbon bond) and leave a resonance-stabilized cation behind. The base (AlCl 4 - ) removes the proton shown (this proton must be shown here!) and the two electrons from the C-H bond form a carbon-carbon double bond and reform the -system of the aromatic ring. The byproducts of HCl and AlCl 3 are not shown. END

Electrophilic Aromatic Substitution Reactions

Electrophilic Aromatic Substitution Reactions, Course Notes Archive, 1 Electrophilic Aromatic Substitution Reactions An organic reaction in which an electrophile substitutes a hydrogen atom in an aromatic