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OrgChemI-Review2
廖若川 1
Chapter 7
NBS: N-Bromosuccinimide
OrgChemI-Review2
廖若川 2
Problem 1
Radical Formation by a chlorine radical at room temperature
Ans.
OrgChemI-Review2
廖若川 3
Problem 2:
Analysis
3o radical most stable
allylic bromination
OrgChemI-Review2
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CH3C=CHCH3
CH3 NBS
CH3C=CHCH3
CH2
CH3C=CHCH3
CH2Br
CH3C-CHCH3
CH2
Br
CH3C-CHCH3
CH2
+
CH3C=CHCH3
CH3 NBS
CH3C=CHCH2
CH3
CH3CH-CH=CH2
CH3
CH3C=CHCH2Br
CH3
CH3C-CH=CH2
CH3
Br
CH3C=CHCH3
CH3
2:1
Problem 3:
Analysis:
allylic bromination
Ans.
3o carbon radical is the most stable
Problem 4:
Analysis
allylic bromination
OrgChemI-Review2
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Chapter 8 Nucleophilic substitution
SN reactions
SN2 Mechanism
SN1
OrgChemI-Review2
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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
OrgChemI-Review2
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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
OrgChemI-Review2
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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)
OrgChemI-Review2
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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
OrgChemI-Review2
廖若川 16
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 SN1 SN2 E1 E2
# of steps 2 1 2 1
intermediate species
carbocation carbocation
order of reactivity of substrates
3°>2°>1° 1°>2°>3° 3°>2°>1° 3°>2°>1°
stereochemistry racemization inversion trans trans
acidic or basic conditions
acidic basic acidic basic
Transition state carbocation trigonal bipyramid 三方雙錐
Carbocation
anti- periplanar
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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
OrgChemI-Review2
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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)
OrgChemI-Review2
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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
OrgChemI-Review2
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OH
?
Synthetic Problem (1)
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:
OrgChemI-Review2
廖若川 24
BrCH2CH2CH2CH=CH2
H2C CH2
CH
O
H2CCH3
?
OCH3
BrCH2CH2CH2CH-CH3
OH
BrCH2CH2CH2CH=CH2
Ans.
Synthetic Problem (3)
ANALYSIS
1. Product – starting material = C5H10O – C5H9Br = + OH, -Br
2. Disconnect:
Ans
BrCH2CH2CH2CH=CH2BrCH2CH2CH2CH-CH3
OH OCH3H
+/H2O base
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Synthetic problem (4)
ANALYSIS
1. starting material – Product = C6H10O – C7H12 = - O, + CH2
2. Disconnect:
Ans
OrgChemI-Review2
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Synthetic Problem (5)
ANALYSIS
1. Product – starting material = + OH, + OCH3
2. cis, trans
3. Disconnect:
OrgChemI-Review2
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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).
OrgChemI-Review2
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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
OrgChemI-Review2
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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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OrgChemI-Review2
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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.
OrgChemI-Review2
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Synthetic problem (7)
Analysis
Ans
Fridel Craft Alkylation
Need R-Br
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OCH3
O
O
O
1. AlCl3
2. H2O
+
O
O
O
O
O
O
OO
O
OCH3
O
O
O
AlCl3O-AlCl3
O
O
OCH3
OH
O
O OCH3
OCH3
O
OH
O
+
(b)
anhydride
Friedel Craft Acylation
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
OrgChemI-Review2
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HNO3
H2SO4NO2
Br2
FeBr3
NO2
Br
NH2
NO2
Cl
Ans
Synthetic Problem (9)
Ans
Synthetic Problem (10)
analysis
OrgChemI-Review2
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Cl2
FeCl3
Cl
HNO3
H2SO4
Cl
NO2
Cl
Cl
NH2
Cl
NO2
Cl
NO2
+
Sn, HCl
or
H2, Pd
Ans
Synthetic Problem (11)
Ans
OrgChemI-Review2
廖若川 39
NaNO2
HCl
HNO3
H2SO4
Br
N2
NO2
CN
Br
CuCN
Br2
FeBr3
NO2
Br
NO2
CN
Br
Br
Br
NH2
?
Sn, HCl
or
H2, Pd
CHO
Cl
CHO
Synthetic problem (12)
Synthetic Problem (13)
Analysis
Ans.