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Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
2
1 . Predict all the products of the following reactions and show stereochemistry whereimportant (2 points each for a total of 20).
OCH3
Br
NO2
Br
O
O
NO2
O O
O
OH
O
Br BrOCH3
NH2
OCH3 OCH3
NH2 H2N
O
H
O
Via
O
HH
O
NaOCH3
2. O33. (CH3)2S
1. Cl2, AlCl32. Zn(Hg), HCl
CH3OH
O
NO2
HNO3
H2SO4
2. H3O+
+Br2
FeBr3
NaNH2
NO2
NH3
CH3O
Cl
1. AcCl, AlCl3
1. Na (0), NH3, t-BuOH
2. HNO3, H2SO4
O
NO2
O2N
1. AlCl32. O33. (CH3)2S
1. I2, HNO3
CH3
O
2. t-BuCl, AlCl3
O
O
O
H
1. CH3Li (2 eq.)
+
H
+
+
I
O
Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
3
2 . Provide a mechanism for the following reaction. Show all intermediates and all resonanceforms for any intermediates that can be described by more than one resonance form (5points). Draw an energy diagram for the reaction (2 points). Will the rate of reactionchange if benzene is replaced with hexadeuterobenzene? why? (1 point).
NO2D
D
D
D
D
D
HNO3
H2SO4+ H2O
hexadeuterobenzene
O OHN
O
O OHS
O
H
O
O OH2N
O
O OHS
O
O
N OO
HNO3H2SO4
NO2 NO2 NO2
+ H2O +
H2SO4
NO2H H H
nitronium ion
O OHS
O
NO2
sigma complex
Ene
rgy
O
Reaction Coordinate
Mechanism:
+ H2O
NO2
NO2H
HSO3
Reaction Energy Diagram:
RateDetermining
Step
The reaction proceeds at the same rate forboth benzene and hexadeuterobenezene. C-H bond cleavage is not occuring in the ratedetermining step.
+ +
Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
4
3 . A by-product of the sulfonation of benzene is diphenylsulfone. Propose a mechanism forthe formation of diphenylsulfone under the reactions conditions below (5 points). You donot need to show all the possible resonance forms the intermediates.
SO3
H2SO4S
O
O
OH + S
O
O
diphenylsulfone
SH
O OS
O
O OHS
O
H
OO O
S
OH
O OS
OH
O
O
OH
OHO S
O
O
O OHS
O
O
S
O
O
OHO OHS
O
H
O
S
O
O
OH2S
O
O
O OHS
O
O
S
O
O H
S
O
O
+
+
O OHS
O
O
+ H2O
Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
5
4 . Devise a synthesis of the following compounds from the indicated starting material and anyother reagents you need (5 points each for a total of 10).
H
H
O
O
H
H
O
O
O
+ From
AcCl
AlCl3
O Na(0), NH3
t-BuOH
O
1. O32. (CH3)2S
Targets
O2N
O
From
HNO3
H2SO4
NO2
1. KMnO4, NaOH, Heat 2. H3O+
NO2
O
HO
1. 2eq CH3Li
2. H3O+Target
Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
6
5 . Provide reagents that will accomplish the following transformations (14 points).
O
Br
OH O
Br
OH
OH O
H
OCH3OCH3
NO2
O
H
O
Br
O2N
1. AcCl, AlCl3
2. Br2, FeBr3
1. Ac2O, DMSO or PCC, CH2Cl2
2. NBS, hv
1. H2, Rh, pressure
2. NaIO4
1. AcCl, AlCl3
Cl
O
2. HNO3, H2SO4
1. AlCl3
2. Na(0), NH3, t-BuOH3. Zn(Hg), HCl
1. NBS, hv
2. 1,3-dithiane, n-BuLi3. I2, H2O
1. Br2, FeBr3
2. HNO3, H2SO4
Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
7
6 . Reaction of acetophenone with Br2 and FeBr3 gives the product indicated below.
Does this reaction occur faster or slower than the treatment of benzene under similarconditions? (1 point) Why? (1 point)
Using both pictures and words explain, in detail, the regiochemistry observed in thebromination of acetophenone (5 points). In order to get full credit, you need to provide adetailed mechanism showing all resonance forms of all intermediates.
O O
Br
Br2, FeBr3
Acetophenone
The reaction will be slower for acetophenone than for benzene because the acyl group is adeactivating (electron withdrawing) group.
FeBr3
Br2
ortho
meta
para
BrH
BrH
BrH
BrH BrH BrH
Br
H
Br
HBr
H
Especially destabilized (+ charge adjacent to EWG)
Especially destabilized (+ charge adjacent to EWG)
O
O O O
O O O
O O O
O
Br
O
Br
O
Br
The sigma complexes leading to ortho and para products are destabilized relative to thesigma complex leading to the meta product. Therefore, the activation energy leading to themeta product will be less than for the ortho and para products and thus the meta productwill predominate.
Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
8
7 . Chloropyridines undergo nucleophilic substitution reactions with sodium methoxide toprovide methoxypyridines. For example, 2-chloropyridine gives 2-methoxypyridine(below). Provide a mechanism for this reaction (2 points).
N Cl
NaOCH3
CH3OHN OCH3
Order the following chloropyridine derivatives from high to low in terms of their reactivitytowards nucleophilic aromatic substitution. The one that reacts the most quickly should belabeled 1, the slowest 3. Explain your answer (5 points).
NN
Cl
N Cl
CH3
Cl
Br
Mechanism:
N Cl
OCH3
N Cl
OCH3
N OCH3
Rate of Reaction:
NN
Cl
N Cl
CH3
Cl12
3
Br
For the pyridinium salt, one of the resonance forms of the intermediate bears no charge.This will be a stable, neutral species and the activation energy is, accordingly, low relativeto the others. In the case of 3-chloropyridine, the negative charge in the itermediate cannotbe localized on the electronegative atom (nitrogen). Therefore, this reaction would beexpected to be slower than for 4-chloropyridine where the negative charge can be localizedon the nitrogen.
N
Cl
N Cl
CH3
N
Cl
OCH3N
Cl
OCH3
N
Cl
OCH3
N Cl
OCH3
CH3
N Cl
OCH3
CH3
N Cl
OCH3
CH3
N
Cl
N
ClCH3O
stable (netural) species
N
Cl
negative chargeon electronegative atom
CH3O
Br
OCH3
N
Cl
OCH3
CH3OCH3O
Chemistry H252 Winter 2000 Exam 3 KEY February 16, 2000
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8 . Propose a mechanism for the following reaction (4 points).
O+H2SO4
+ H2O
O O OHS
O
H
O
O H OH
OH
H
O OHS
O
O O OHS
O
O
OH
O OHS
O
H
O
OH2
H
O OHS
O
O
Product
+ H2O