OrgChemI-Review2
Chapter 7
NBS: N-Bromosuccinimide
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Problem 1
Radical Formation by a chlorine radical at room temperature
Ans.
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Problem 2:
Analysis
3o radical most stable
allylic bromination
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Problem 3:
Analysis:
allylic bromination
Ans.
3o carbon radical is the most stable
Problem 4:
Analysis
CH3
allylic bromination
CH3
CH3C=CHCH3
CH2
NBS
CH3C=CHCH3
CH3C=CHCH3
CH2
CH3C-CHCH3
CH2
CH2Br
CH3C=CHCH3
2:1
+
CH3C-CHCH3
Br
CH3
CH3
NBS
CH3C=CHCH3
CH3
CH3C=CHCH2Br
CH3C=CHCH2
CH3
CH3CH-CH=CH2
CH3
CH3C-CH=CH2
Br
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Chapter 8 Nucleophilic substitution
SN reactions
SN2 Mechanism
SN1
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Nucleophilicity
(1) Negative anion > neutral nucleophile
Methoxide ion (CH3O-) has nonbonding electrons that are readily available
for bonding.
Methanol (CH3OH) has no negative charge
(2) Small anions are solvated much more strongly than large anions in a
protic solvent
a small anion forms stronger hydrogen bonds and less reactive
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(3) large and small atom size
F- has tightly bound electrons that cannot begin to form a bond until the
atoms are close together. I- has more loosely bound outer electrons that
begin bonding earlier in the reaction.
Polar protic solvents: SN1
Solvation stabilized the SN1 transition state
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Stereochemistry
SN2:
Back-side attack, inversion
SN1:
Carbocation
both
Comparison between SN2 & SN2
An SN2 reaction is favored by a high concentration of a good nucleophile
An SN1 reaction is favored by a low concentration of a nucleophile or by a
poor nucleophile
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Problem: Rank the following sets of compounds with respect to SN2
reaction.
Rank the following sets of compounds with respect to SN1 reaction.
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+
H2O is more polar than ethanol
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Give the major product of the following reactions:
high concentration of a good nucleophile  SN2 (inversion product)
low concentration of a nucleophile or a poor nucleophile  SN1 (racemic)
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Elimination & Alkenes
E2 reaction:
 a one-step mechanism
 Dehydrohalogenation of Alkyl Halides
 Relative reactivities of alkyl halides: 3o > 2o > 1o
 leaving group to be anti to the proton being eliminated.
 Antiperiplanar
 Follow Zaitzev’s rule: most substituted alkene product will form
Zaitsev's Rule
it is the alkene formed when a hydrogen is removed from the -carbon
bonded to the fewest hydrogens.
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Summary of Anti-Zaitzev product Formation:
The Major product of E2 is More substituted alkene unless:
 The base is large
 Alkyl halide is fluoride
 Alkyl halide contains 1 or more C=C bond
E2- Antiperiplanar
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Problem 1: E2 products of (3R, 4R)-3-bromo-3,4-dimethylhexane and (3R,
34)-3-bromo-3,4-dimethylhexane
Problem 2:
E2 products of cis-1-bromo-2-ethylcyclohexane and
trans-1-bromo-2-ethylcyclohexane ?
cis-1-bromo-2-ethylcyclohexane
trans-1-bromo-2-ethylcyclohexane
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2.
E1 mechanism
 E: elimination, 1: unimolecular
 Carbocation is formed, Carbocation rearrangement
 3o > 2o > Io
 follow Zaitzev’s rule
 cis & trans alkenes will form (trans alkene ∵ more stable)
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Feature
# of steps
intermediate
species
order of
reactivity of
substrates
stereochemistry
acidic or basic
conditions
Transition state
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SN1
2
carbocation
SN2
1
E1
2
carbocation
E2
1
3°>2°>1°
1°>2°>3°
3°>2°>1°
3°>2°>1°
racemization inversion
acidic
basic
trans
acidic
trans
basic
carbocation
Carbocation
antiperiplanar
trigonal
bipyramid
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Dehydration of alcohol
 proceed through an E1 mechanism, due to the acid-catalysis necessary
to protonize -OH → leaving group
 Acid: conc. sulfuric acid (H2SO4), or 85% phosphoric acid
Problem
 Dehydration of alcohol, proceed through an E1 mechanism
 Carbocation rearrangement
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10. Alcohol
1. Conversion to halide
relative rate: tertiary > secondary > primary
only for primary and secondary alcohols
2. Conversion to sulfonate ester
3. Conversion of an activated alcohol (an alkyl halide or a sulfonate ester) to
a compound with a new group bonded to the sp3 carbon (SN2 , inversion)
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Transition state
3. Cleavage of ethers
4. ring opening of Epoxides
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5. Reaction of a Grignard reagent with an epoxide
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6. Reactions of Thiols, sulfides, and sulfonium salts
Functional groups transformation
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Synthetic Problem (1)
?
OH
1. Difference between Product & Starting material
Product
- Starting material = - OH
2. Alkane only useful reaction: halogenation
3. Br can be used as leaving groups
4. –Br, + OH
Ans.
Synthetic Problem (2)
1. Difference between product and starting material = 2 C & 1 O
2. Disconnect:
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Ans.
Synthetic Problem (3)
O
?
H2C
BrCH2CH2CH2CH=CH2
H2C
CH
CH3
CH2
ANALYSIS
1. Product – starting material = C5H10O – C5H9Br
= + OH, -Br
2. Disconnect:
O
CH3
OH
BrCH2CH2CH2CH=CH2
BrCH2CH2CH2CH-CH3
Ans
BrCH2CH2CH2CH=CH2
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H+/H2O
OH
O
base
CH3
BrCH2CH2CH2CH-CH3
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Synthetic problem (4)
ANALYSIS
1. starting material – Product = C6H10O – C7H12
= - O, + CH2
2. Disconnect:
Ans
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Synthetic Problem (5)
ANALYSIS
1. Product – starting material = + OH, + OCH3
2. cis, trans
3. Disconnect:
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Chapter 11 Aromaticity
The rules for aromaticity are:
1. Planar (flat molecule)
2. Cyclic (ring compounds)
3. All atoms on cycle are sp2 hybridized
4. (4n+2) electrons total. (n= integral) (Huckle rule)
Examples
not aromatic, sp3 nonplanar (the bottom carbon is sp3 hybridized).
Aromatic. It is planar (each carbon is sp2 hybridized) and has 6 -electrons
(the 4n + 2 rule is satisfied).
not aromatic because it is nonplanar (the top carbon is sp3 hybridized).
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Aromatic. One pair of nonbonding electrons from O participates in the
-electron system. This gives the molecule a total of 6 -electrons.
This molecule is not aromatic. It is not cyclic.
The molecule is not aromatic. It has 4 -electrons and, consequently, does
not satisfy the 4n+2 rule
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12 Benzene Reactions
1. Electrophilic aromatic substitution reactions親電子性芳香取代反應
a. Halogenation
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b. Nitration, sulfonation, and desulfonation
c. Friedel–Crafts acylation and alkylation
2. Reduction of a carbonyl group to a methylene group
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1. Reactions of Substitutents on a Benzene Ring
2. Reactions of amines with nitrous acid
3. Replacement of a diazonium group
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Multi-functional compounds:
1) If the effects of 2 substituents point to reactivity at one particular
carbon
- the electrophile will go
2) The strongest directing group almost always wins
(e.g. OH > alkyl, NR2 > Br)
3) The order of precedence is
a) Strong o,p directors (OR, NR2)
b) Alkyl groups and halogen
c) All meta-directors.
4) Keep a close eye on the steric environment.
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Synthetic problem (6)
Analysis
Ans.
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Synthetic problem (7)
Analysis
Ans
Fridel Craft Alkylation
 Need R-Br
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O
(b)
OCH3
1. AlCl3
+
O
2. H2O
O
O
O
O
O
O
O
O
anhydride
O
O
OCH3
O
O
O AlCl3
O
O
OCH3
OH
OCH3
OO
+
O-AlCl3
O
OH
Friedel Craft Acylation
OCH3
O
Synthetic problem (8)
Analysis
(1) -Br & -NO2 relationship: meta
(2) -Br: ortho-, para- director; -NO2: meta-director
Add NO2 first, direct Br to meta-position
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Ans
Br
HNO3
Br2
H2SO4
FeBr3
NO2
NO2
Synthetic Problem (9)
Ans
Synthetic Problem (10)
analysis
Cl
NO2
NH2
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Ans
Cl
Cl2
FeCl3
Cl
Cl
Cl
NO2
HNO3
+
H2SO4
NO2
Cl
Cl
Sn, HCl
or
H2, Pd
NO2
NH2
Synthetic Problem (11)
Ans
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Synthetic problem (12)
Br
NO2
Br
?
CN
Br
NO2
HNO3
Br2
H2SO4
FeBr3
Br
Sn, HCl
NO2
or
H2, Pd
NH2
Br
Br
CuCN
NaNO2
HCl
CN
N2
Synthetic Problem (13)
CHO
Cl
Analysis
CHO
Ans.
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