cm1501 finals section b answers
TRANSCRIPT
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Question 21 (6 marks)Compound (A) undergoes methylation selectively at one of the two phenoxyl groups to give (B) as a major isolated product. Provide an explanation for the above observation.
(a)
(b)
1. When one equivalent of NaOH is used, it will deprotonate the relatively more acidic proton, either (a) or (b), to form phenoxide anions C or D, respectively.
2. In A, the negative charge could be partially delocalized into the electron‐withdrawing NO2 group by resonance, as indicated in C1 ↔ C2.
3. So –OH in (a) is relatively more acidic, and selectively being deprotonated when NaOH is added to form only anion C.
4. Then anion C reacts with MeI via a SN2 reaction to give the product B.
C D
N
O
OH
O
O
N
O
OH
O
O
C1
C2
A B
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Question 22 (8 marks)Provide a synthetic scheme for the synthesis of trans‐stilbene (C) from benzene. You may use any organic reagent with two or less carbon atoms. You may also use any inorganic reagent.
OH
MgBrOHCH3O+
conc H2SO4
catalyst
Mg
BrBr2
FeBr3
1. 2. H3O+O
H H
CH2OH
Dess–Martinoxidation
C
NOTE: There are other acceptable synthetic routes.
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Question 23 (8 marks)Illustrate how you could convert compound (D) to compound (E). You may use any organic and inorganic reagents. D E
conc HNO3conc H2SO4
NO2 NH2
NH2
BrBr
NHCOCH3
NH3+
conc HNO3conc H2SO4
(see NOTE 1)
(see NOTE 2)
(see NOTE 3)
CH3COCl
Br2
Br2FeBr3
NHCOCH3
Br
conc HNO3conc H2SO4
NHCOCH3
Br NO2
H2O/OH
NH2
Br NO2
(see NOTE 4)
SnCl2
H3O+
NOTES: 1. The bulky t‐butyl group strongly discourages ortho‐substitution. Only para‐nitration is preferred.
2. ‒NH2 is a very strong activating group. Dibromination will predominate.
3. ‒NH2 is a basic group. Addition of strong acids protonate it to form ‒NH3
+ that is strongly deactivating, meta‐directing and the reaction is very slow.
4. Converting the ‒NH2 to ‒NHCOCH3reduces its N basicity and thus its activity. Mono‐bromination and nitration are now possible.
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Question 24 (6 marks)Which of the following species should be aromatic by the Hückel’s rule?
B CH2CH3
Boron has an empty orbital in conjugation.Monocyclic, planar, conjugated, 4π: anti‐aromatic.
CH2
The molecule is NOT conjugated.There is a saturated CH2 unit as indicated: non‐aromatic. Oxygen atom is re‐hybridized to sp2.
One lone pair electrons in a p orbital are in conjugation. Monocyclic, planar, conjugated, 6π: aromatic.
ON H
Nitrogen atom is re‐hybridized to sp2.One lone pair electrons in a p orbital are in conjugation. Monocyclic, planar, conjugated, 6π: aromatic.
Look at the molecular periphery.Monocyclic conjugated system in red, planar, 10π(4n+2, n=2): aromatic.
Look at the structure carefully. There are three possible monocyclic, planar, 6π, conjugated systems. These will exhibit aromatic character.
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Question 25 (8 marks)Illustrate how you could convert compound (L) to compound (M). You may use any organic and inorganic reagents. L M
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Question 26 (6 marks)Show all the possible products when compound (N) is treated with HBr. N
1,2-addition1,2-addition
1,4-addition 1,4-addition
BrHBr
BrH
HBr HBr
HBr HBr
H
Br
BrBr H
Br Br
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Question 26 (6 marks)Show all the possible products when compound (N) is treated with HBr. N
O O
HHO
NaBH2
O OH
HHO
OH
OH
O
H+
OH
OH
OH
OH
OHHO
O
OH2OH
H2O
O
O
NOTES:1. Using LiAlH4 may risk some reduction of the COOH group. NaBH4 is milder and
will reduce only the CHO group.2. The key aspect is where protonation occurs for reaction to proceed.
Protonation is as indicated to make the C=O more electrophilic for a weak OH nucleophile to attack. This will form the six‐membered ring in Q as given in the question.
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Question 27 (12 marks)Compound (O) is converted to compound (Q) via compound (P).a. What could be the appropriate structures for reagent (X) and compound (P), respectively. b. Provide a mechanism for the conversion of compound (P) to compound (Q).
O O
HHO
NaBH4
O OH
HHO
OH
OH
O
H+
OH
OH
OH
OH
OHHO
O
OH2OH
H2O
O
O
H B H
H
H
NOTES:1. Using LiAlH4 may risk some reduction of the COOH group. NaBH4 is milder and
will reduce only the CHO group.2. The key aspect is where protonation occurs for the reaction to proceed.
Protonation is as indicated to make the C=O more electrophilic for a weak OH nucleophile to attack. This will form the six‐membered ring in Q as given in the question.
O
P X
Q
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Question 28 (6 marks)One or more products could be obtained in the following reaction. Give all possible products and indicate the major product. Explain why the major product is preferred.
1. It is an example of Hofmann elimination.2. The controlling factor is steric factor due to the
bulky leaving group, (CH3)3N.3. The preferred route is the abstraction of the
sterically less hindered proton, in this case the 2o H.
4. Elimination gives both the trans and cis isomers.
5. Trans isomer is the major product because trans isomer is relatively more stable based on steric factor.