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

廖若川 4

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

廖若川 5

Chapter 8 Nucleophilic substitution

SN reactions

SN2 Mechanism

SN1

OrgChemI-Review2

廖若川 6

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

廖若川 7

(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

廖若川 8

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

OrgChemI-Review2

廖若川 9

Problem: Rank the following sets of compounds with respect to SN2

reaction.

Rank the following sets of compounds with respect to SN1 reaction.

OrgChemI-Review2

廖若川 10

H2O is more polar than ethanol

+

OrgChemI-Review2

廖若川 11

OrgChemI-Review2

廖若川 12

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

廖若川 13

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.

OrgChemI-Review2

廖若川 14

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

OrgChemI-Review2

廖若川 15

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)

OrgChemI-Review2

廖若川 17

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

OrgChemI-Review2

廖若川 18

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

廖若川 19

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

廖若川 20

Transition state

3. Cleavage of ethers

4. ring opening of Epoxides

OrgChemI-Review2

廖若川 21

5. Reaction of a Grignard reagent with an epoxide

OrgChemI-Review2

廖若川 22

6. Reactions of Thiols, sulfides, and sulfonium salts

Functional groups transformation

OrgChemI-Review2

廖若川 23

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

OrgChemI-Review2

廖若川 25

Synthetic problem (4)

ANALYSIS

1. starting material – Product = C6H10O – C7H12 = - O, + CH2

2. Disconnect:

Ans

OrgChemI-Review2

廖若川 26

Synthetic Problem (5)

ANALYSIS

1. Product – starting material = + OH, + OCH3

2. cis, trans

3. Disconnect:

OrgChemI-Review2

廖若川 27

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

廖若川 28

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

廖若川 29

12 Benzene Reactions

1. Electrophilic aromatic substitution reactions親電子性芳香取代反應

a. Halogenation

OrgChemI-Review2

廖若川 30

b. Nitration, sulfonation, and desulfonation

c. Friedel–Crafts acylation and alkylation

2. Reduction of a carbonyl group to a methylene group

OrgChemI-Review2

廖若川 31

1. Reactions of Substitutents on a Benzene Ring

2. Reactions of amines with nitrous acid

3. Replacement of a diazonium group

OrgChemI-Review2

廖若川 32

OrgChemI-Review2

廖若川 33

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.

OrgChemI-Review2

廖若川 34

Synthetic problem (6)

Analysis

Ans.

OrgChemI-Review2

廖若川 35

Synthetic problem (7)

Analysis

Ans

Fridel Craft Alkylation

Need R-Br

OrgChemI-Review2

廖若川 36

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

廖若川 37

HNO3

H2SO4NO2

Br2

FeBr3

NO2

Br

NH2

NO2

Cl

Ans

Synthetic Problem (9)

Ans

Synthetic Problem (10)

analysis

OrgChemI-Review2

廖若川 38

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.