For pyridine.....
(a) Show each potential intermediate that can be generated when an electrophile adds to pyridine. Use a generic E for the electrophile.
(b) Draw all the resonance structures for each potential intermediate
(c) State whether the most stable intermediate occurs for ortho, meta or para substitution. Provide an argument (15 words or less).
(d) You are going to start with benzene, and proceed through molecule A on your way to molecule B. Use the NMR/IR spectra to solve for molecule A and B. Next show the synthetic route to get to molecule B.
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For pyridine.....
(a) Show each potential intermediate that can be generated when
an electrophile adds to...
You are going to start with benzene, and proceed through molecule A on your way to...
You are going to start with benzene, and proceed through
molecule A on your way to molecule B. Use the NMR/IR spectra to
solve for molecule A and B. Next show the synthetic route to get to
molecule B.
Molecule A:C13H180 100 50 way D 4 DOO 5000 2000 1500 1000 sbo NAVENUHBERI- ЗН ЗН ЗН 4H 2H IH 2H HA 8 7 6 5 2 0 PPM Molecule B:C14H220
2. Starting with benzene, proceed through molecule E on your way to molecule F. Use the...
2. Starting with benzene, proceed through molecule E on your way
to molecule F. Use the NMR/IR spectra to solve for molecule E and
F. Show the synthetic route to get to molecule F.
Molecule E:C12H16O
Molecule F: C14H22O
100 롷 50 4000 3000 2000 1500 1000 500 HAVENUMBERI-1 ЗН 4H 2H ЗН 1Н 2H 1Н 10 нар 0 PPM مد =[ [ 400 3000 [0 1500 100 500 أ- HAVEHLIHE EFI ЗН ЗН 2H ЗН 4H ІН 2H 1Н...
You are going to start with benzene, and proceed through molecule A on your way to...
You are going to start with benzene, and proceed through
molecule A on your way to molecule B. Use the NMR/IR spectra to
solve for molecule A and B. Next show the synthetic route to get to
molecule B.
Molecule A:C13H18O
Molecule B:C14H22O
100 50 D 4000 9000 2000 1500 1000 500 HAVENUHBERI- ЗН ЗН ЗН 4H 2H 1Н 2H 8 7 6 5 з 2 1 о 4 PPM 100 ㅋ 50 0 4000 3000 2000 1500 1000 500...
5. When benzamide (the reactant shown) is treated with the acylium cation electrophile shown (generated by...
5. When benzamide (the reactant shown) is treated with the acylium cation electrophile shown (generated by combining concentrated acetyl chloride and aluminum trichloride), the first step of the reaction involves the benzene ring attacking the carbocation to form a sigma complex intermediate. (This curved arrows have been included in the reaction below.) Draw the sigma complex intermediate (including all resonance structures) resulting from meta attack (acyl group adds to meta carbon) and para attack (acyl group adds to para carbon)...
250 ChemActivity 29 Electrophilic Aromatic Substitution no ChemActivity 29 Part A: Electrophilic Aromatic Substitution (What products...
250 ChemActivity 29 Electrophilic Aromatic Substitution no ChemActivity 29 Part A: Electrophilic Aromatic Substitution (What products are formed when a strong electrophile is added to benzene?) Model 1: (review) Electrophilic Addition of HCI Rani o g cyclohexene carbocation intermediate Run 2 U X benzene This product carbocation intermediate DOES NOT Critical Thinking Questions 1 For Rxn I (above) draw curved arrows showing the mechanism of electrophilic addition of HCl. Include an appropriate carbocation intermediate in the box above. Figure 1:...
You are going to start with benzene, and proceed through
molecule A on your way to molecule B. Use the NMR/IR spectra to
solve for molecule A and B. Next show the synthetic route to get to
molecule B.
Molecule A:C13H180 100 50 way D 4 DOO 5000 2000 1500 1000 sbo NAVENUHBERI- ЗН ЗН ЗН 4H 2H IH 2H HA 8 7 6 5 2 0 PPM Molecule B:C14H220
2. Starting with benzene, proceed through molecule E on your way
to molecule F. Use the NMR/IR spectra to solve for molecule E and
F. Show the synthetic route to get to molecule F.
Molecule E:C12H16O
Molecule F: C14H22O
100 롷 50 4000 3000 2000 1500 1000 500 HAVENUMBERI-1 ЗН 4H 2H ЗН 1Н 2H 1Н 10 нар 0 PPM مد =[ [ 400 3000 [0 1500 100 500 أ- HAVEHLIHE EFI ЗН ЗН 2H ЗН 4H ІН 2H 1Н...
You are going to start with benzene, and proceed through
molecule A on your way to molecule B. Use the NMR/IR spectra to
solve for molecule A and B. Next show the synthetic route to get to
molecule B.
Molecule A:C13H18O
Molecule B:C14H22O
100 50 D 4000 9000 2000 1500 1000 500 HAVENUHBERI- ЗН ЗН ЗН 4H 2H 1Н 2H 8 7 6 5 з 2 1 о 4 PPM 100 ㅋ 50 0 4000 3000 2000 1500 1000 500...
5. When benzamide (the reactant shown) is treated with the acylium cation electrophile shown (generated by combining concentrated acetyl chloride and aluminum trichloride), the first step of the reaction involves the benzene ring attacking the carbocation to form a sigma complex intermediate. (This curved arrows have been included in the reaction below.) Draw the sigma complex intermediate (including all resonance structures) resulting from meta attack (acyl group adds to meta carbon) and para attack (acyl group adds to para carbon)...
250 ChemActivity 29 Electrophilic Aromatic Substitution no ChemActivity 29 Part A: Electrophilic Aromatic Substitution (What products are formed when a strong electrophile is added to benzene?) Model 1: (review) Electrophilic Addition of HCI Rani o g cyclohexene carbocation intermediate Run 2 U X benzene This product carbocation intermediate DOES NOT Critical Thinking Questions 1 For Rxn I (above) draw curved arrows showing the mechanism of electrophilic addition of HCl. Include an appropriate carbocation intermediate in the box above. Figure 1:...
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