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A B A B+

7 Free Radicals 自由基

7.1 Introduction

甚麼是自由基?

自由基包含一個未成對電子 unpaired electrons 的原子或原子團

A自由基 B自由基

H2 H-H

O2 O=O

H2O H-O-H

Anion

HOΘ

H-OΘ

Free radical

HO․ H-O․

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7.2 Alkanes reactions

Alkanes are quite inert. C-H, C-C bond energies are high

(C-H = 96-99 kcal/mol, C-C = 83-85kcal/mol.)

alkanes undergo

(1) free radical halogenation

These reactions require high temperatures or light

(2) combustion燃燒 when activated.

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7.3 Carbon Radicals

- carbon atom that shares three of its valence electrons with

other atoms in bonds while the fourth remains unshared and

unpaired

- It is the result of homolytic cleavage.(= homolytic fission均勻

分裂)

The radical occupies a p orbital; the carbon is sp2 hybridized.

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7.4 Free radical halogenation of methane

Cl. Are very high-energy reactive radical.

Attack inert high energy C-H bonds

Free Radicals are formed by homolytic cleavage. (electrons

break away from covalent bond and go back to original

atom)

Use Excess Alkane to Get Monochlorination

Using excess alkane minimizes dichlorination, trichloriation

7.5 Free Radical reaction Steps

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7.5.1 Initiation引發

Creating a radical - Starts the reaction

A compound that can easily form radicals is used to

generate a reactive radical

Halogens split equally across the bond to yield radicals in

the presence of light (UV light or Heating)

Bonds in peroxides過氧化合物 undergo homolytic cleavage

on heat or impact to yield radicals

Peroxide過氧化物 R-O-O-R:

H2O2, hydrogen peroxide, tert-butyl peroxide, benzoyl peroxide

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7.5.2 Propagation增殖 Reaction

Reaction of one radical with non-radical to form another

radical + products

Propagation Keeps the reaction going

7.5.3 Termination終止 Reaction

When reagents are depleted, or during the reaction, 2 radicals

combine to give a non-radical Kills the reaction

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7.6 Stability of Radicals

Alkyl groups can donate electron density through their s

bonds.

The more alkyl groups attached to radical, the greater the

inductive donation of electron density

Radicals have an unpaired electron in an orbital (electron

deficient)

Electron donation stabilizes the radical and makes it stable

Radical Stability:

Benzyl ~ allyl > 3o > 2

o > I

o

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7.7 Factors that determine product distribution

The rate-determining step of the overall reaction is

hydrogen abstraction

2o radical more stable than 1

o radical

The more stable the radical, the more easily it is formed.

(It is easier to remove a hydrogen atom from a 2o carbon

than from a 1o carbon)

The stable alkyl radical is formed faster

∴ 2-chlorobutane is formed faster

Alkyl Radical Formation Rates at room temperature

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Determining the relative amounts of products obtained

Probability:

The number of hydrogens that can be abstracted that will lead

to the formation of the particular product

Reactivity:

The relative rate at which particular hydrogen is abstracted

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Radical monochlorination of 2,2,5-trimethylhexane results in

the formation of five monochlorination products.

⊕ Because radical halogenation of an alkane can yield several

different monosubstitution products and products that contain

more then one halogen, it is not a good method to use for

synthesis.

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7.7.3 The Reactivity–Selectivity Principle

Alkyl Radical Formation by chlorine radical at room

temperature

bromine radical is less reactive

∴ more selective than a chlorine radical

Radical bromination of butane leads to a 98% yield of

2-bromobutane

When a bromine radical is the abstracting agent, the reactivity

factor is more important than the probability factor.

71% yield of 2-chlorobutane when butane is chlorinated.

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Bromination of 2,2,5-trimethylhexane results in an 82% yield in

which bromine replaces the tertiary hydrogen.

Chlorination of the same alkane results in a 14% yield of the

tertiary alkyl halide.

Reactions with the bromine radical are higher in energy than

reactions with the chlorine radical.

Relative rates of formation when a bromine radical is used are

different from the relative rates of formation when a chlorine

radical is used.

The more reactive a species is, the less selective it will be

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7.7.4 Alkanes undergo chlorination and bromination but not

fluorination or iodination.

The standard heat of combustion for the sum of the two

propagating steps shows that:

fluorination of alkanes is too reactive, not practical for the

synthesis.

iodination of alkanes will not undergo

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7.8 More examples of Halogenenation of Alkanes

Isobutane

1o C-H: 9 probabilities

3o: radical stability

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7.9 Addition of HX in the Presence of Peroxide

7.9.1 HBr-peroxide

In the reaction of 1-butene and HBr, follow Markovnikov’s rule

In the presence of the peroxide, a free radical reaction takes

place. (Anti-Markovnikov)

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7.9.2 Peroxide Effect

Markovnikov product

Why don't the other hydrogen halides behave in the same way

as HBr/peroxide (anti-Markovnikov rule)?

Hydrogen fluoride

The hydrogen-fluorine bond is so strong that fluorine

radicals aren't formed in the initiation step.

Hydrogen chloride

With hydrogen chloride, the second half of the propagation

stage is very slow.

This is due to the relatively high hydrogen-chlorine bond

strength.

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Hydrogen iodide

the first step of the propagation is endothermic and this slows

the reaction down. Not enough energy is released when the

weak carbon-iodine bond is formed.

In the case of hydrogen bromide, both steps of the

propagation stage are exothermic.

A peroxide has no effect on the addition of hydrogen chloride

or hydrogen iodide to an alkene.

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CH3CH2C CH CH3CH2CH CHBr+ HBr

peroxide

7.9.3 Free radical Hydrohalogenation of Alkynes

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7.11 Radical Substitution of Allylic or Benzylic Hydrogens

The more stable radicals form faster

Halogenation of a benzylic or allylic carbon will result in the

preferential substitution of that halogen.

A halogen adds to the carbon bonded to the benzene ring

resulting in a benzylic substituted product.

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7.10 Bromination with NBS

N-Bromosuccinimde NBS: a mild brominating reagent

The bromine radical is obtained as a result of homolytic

cleavage of the N-Br bond.

allylic radical is stabilized by resonance

∵ the resonance hybrid of the allyl radical is symmetrical,

∴ only one substitution product is formed.

N

O

O

Brh

N•

O

O

•Br+

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CH3CH2CH CH2CH3CHCH CH2

BrCH3CH=CHCH2Br++ NBS

Advantage of NBS Bromination:

the low concentration of Br2 and HBr present cannot be added

to the double bond

The allylic radical has two resonance contributors

The two resonance contributors are not the same, so two

substitution products are formed.

2 resonance form

∴ 2 Products: 3-bropmo-1-butene & 1-bromo-2-butene

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Br2

NBS

Br

Br

Br

Cl

Cl+ Cl2

+h + HBr

+ HBr

+peroxide

7.11 Summary of free radical halogenations with

alkases/alkenes

Reactions of Cyclic Compounds

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7.12 Stereochemistry of Radicals

If a radical substitution reaction creates a chirality center in the

product, both R and s enantiomers will be formed.

The product will be a racemic mixture.

Radical Intermediate

planar structure intermediate

Br can react either with the top or bottom of the molecule

forming both R and S enantiomers

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HBr/H2O2

.

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7.13 Radicals In the Ozone Layer

UV light initiate free radical reactions

UV cause cancer & cataract

Ozone layer:24km above the earth

The earth’s surface has been protected from too much UV

by a layer of Ozone

Chlorofluorocarbons (CFCs) are used as coolants

CFCs → Cl․ radical ozone

Ozone hole

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The Ozone Depleting Reaction

Chlorofluorocarbons are the Ozone Depleting Reagents

Because They Form Chlorine Radicals

Two approaches to solve the problems

1. replacing compounds do not have C-Cl bond

CH2FCF3 (hydrofluorocarbons) now is used in refrigerator

CH3CHF2: used as aerosal propellant

2. compounds that are more quickly to degrade

CH3CCl2F: used in foam insulation

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7.14 Anti-oxidation & Health

7.14.1 生物體中自由基的產生

(1) 體內正常新陳代謝所產生:

a. 白血球也是利用氧自由基去殺死外來的細菌

b. 細胞有氧呼吸中粒腺體電子傳遞鏈因氧氣反應不完全 O2

-.

(2) 外界不正當的影響:

輻射線、吸菸、病毒、毒藥、缺氧或過多高濃度的氧

7.14.2 自由基對人體的危害

自由基攻擊附近的分子造成細胞的死亡

細胞膜及脂蛋白中的多元不飽和脂肪酸 脂質過氧化(Lipid

peroxidation) 細胞膜功能嚴重受損

攻擊蛋白質使蛋白質斷裂或凝集等 影響離子通道及細胞功能

破壞 DNA 致基因突變或致癌等，細胞也可能因此而死亡。

自由基有關的疾病，包括：

在身體發生發炎反應時，白血球也可以釋放出自由基來殺死細菌

腎臟病、動脈硬化、缺血性心臟病、器官移植、糖尿病、發炎性

疾病、癌症、神經系統疾病、高血壓、白內障、藥物毒性、等等

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HOCl

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Fe2+ + H2O2 Fe3+ + OH + OH

H2O2 + Fe3+ OOH + H + Fe2+

OH + Fe2+ Fe3+ + OH

Fe3+ OOH Fe2+

OH

+

H2O2+ OOH H2O+

+ H + O2

O2 H2O2

+

O2 + OH OH+

+血紅素 Fe2+

O2 血紅素 Fe3+ O2

7.14.3 生物體中自由基/過氧化物

超氧陰離子自由基 (super oxide anion) O2-.

hydroxyl radical OH.

過氧化氫 ( Hydrogen peroxide) H2O2

單線態氧 (singlet oxygen) 1O2

氫過氧化物 (alkyl peroxide) ROOH

Fenton’s reaction

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O CH3

CH3 CH3 CH3CH3

R1

HO

R2

CH3

7.14.4 Lipid Peroxidation脂質過氧化

7.14.5 抗氧化維他命/小分子

(1) Vitamin E (Tocopherol)

O R

CH3

CH3

HO

H3C

CH3

O R

CH3

CH3

O

H3C

CH3

OH

[O]

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OO

HO OH

OH

OHO

O

O O

OH

OH

Ascorbic acid Semidehydroascorbateradical

OO

O O

OH

OH+ e

Dehydroascorbate

(2) Vitamin C

(3) Hydroquinone/Semiquinone

The semiquinone that results is stabilized by resonance and is

therefore unreactive in comparison with other radicals.

Ubiquinone (Coenzyme Q)

n= 10 (Q-10)

Oxidative phosphorylation

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OHOO

OH

OH

HO

CH2OH

HO

OH

OH

O O

Arbutin熊果素

Shiseido懷捷皙嫩白露

(4) Antioxidants in Food食品中常用之抗氧化劑

Propyl gallate沒食子酸丙酯

BHT BHA

2-tert-Butyl-hydroxytoluene 2-tert-Butyl-hydroxy-anisole

CH3

OH

OCH3

OH

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Exercise

1. Please write down the major product of the following reactions:

2. Please write down the major product of the following reactions:

3. Please write down the major product of the following reactions:

Give the major product of the reaction of 1-methylcyclohexene with the following reagents:

a. NBS//CH2Cl2

b. Br2/ CH2Cl2

c. HBr

d. HBr/peroxide

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8 Alkyl Halides (I)鹵烷類

8.1 Introduction

Alkyl halides is a carbon-halogen bond

Halogens: fluorine (F), chlorine(Cl), bromine(Br) & iodine (I).

Uses

Major use as solvents 溶劑: CCl4, CH2Cl2, CH3Cl

anesthetics麻醉劑:

CHCl3 chloroform哥羅芳, Halothane (CF3CHClBr)

dry cleaning乾洗: Trichloroethylenes CHCl2CH2Cl

CFCs (chloroflurocarbons): CF3CFCl2 propellant for inhalants

Refrigerator coolants冰箱冷凍劑: Freons ex. CCl3F, CCl2F2

Ozone depletants. Radicals destroy the O3 layer

Fire retardants 滅火劑: halons CF3Br

Insecticides殺虫藥: DDT (Dichlorodiphenyl-trichloroethane)

ClCl

Cl

Cl

Cl

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I I

CH3

Br

8.2 Nomenclature

(1) Find the longest Chain. Assign lowest # to substituents

(2) Halogens have the same priority as alkyl groups and nitro

groups.

(3) If more than 1 group is competing for the lowest number on a

chain, assign lowest number according to alphabetic priority

CH2Cl2 CHCl3 (CH3)3CCl

Common Methylene chloride

Chloroform

Cyclopropyl chloride

tert-Butyl chloride

IUPAC Dichloromethane Trichloro- methane

2-chloro- propane

2-chloro-2- methylpropane

(CH3)2CHCHBrCH2CH3

3-Bromo-2-methylpentane

CH3CHFCH2CH3 2-Flurorobutane (lowest no.)

3,3-diiodo-4-methylhexane

(lowest sum of the numbers)

1-bromo-2-methylcyclopentane

BrBr

1,2-Dibromopentane

Cl

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8.3 Reactivity反應性:

∵ C-X bonds容易斷裂 → Alkyl halides are reactive

(1) Atomic numbers & atomic weight

Group VII: F < Cl < Br < I

(2) Bond length

C-F < C-Cl < C-Br < C-I

∵ atom number ↑, proton↑ → electrons ↑ → bond length↑

(3) Covalent bond strength:

C- -halogen covalent bonds

C-F >> C-C bond > C-H bond.

∴ alkyl fluorides and fluorocarbons are chemically stable

polyfluoroalkane (Teflon): coating in frying pans

C-Cl: slightly weaker than C-C bond,

Bond strength: C-F > C-Cl > C-Br > C-I

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Bond Energies (kcal/mole at 25 deg C)

C-H 98.7 C-C 82.6 C=C 145.8 C-O 85.5

C-F 116 C-Cl 81 C-Br 68 C-I 51

(4) Electronegativities陰電性 of Halogens >> carbon

∵ larger atomic no. ∴ more protons in the nucleus → attract

more electrons → more electronegativity

electronegativities: I < Br < Cl < F

∴ carbon is electrophilic親電子性 and the halogen is

nucleophilic親核性

(5) Stability of the corresponding halide anions

Cl- < Br

- < I

- ∴ Good leaving groups

(6) Relative reactivity

R-F < R-Cl , R-Br < R-I

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CH2 CH2 Hal

R

8.4 Nucleophiles and Electrophiles

Electrophile 親電子基: (Electro: 電子; phile: love, 親)

An electron deficient 電子缺乏 atom原子, ion 離子 or molecule

分子 that has an affinity 親和性 for an electron pair, and will bond

to a base or nucleophile.

Nucleophile 親核基: (Nucleo: 核; phile: love, 親)

An atom, ion or molecule that has an electron pair that may be

donated in forming a covalent bond to an electrophile

8.5 Reactions of alpha & beta carbon of alkyl halides

Functional group-bearing carbon as and the carbon atom

adjacent to it as

replacement or substitution of the halogen on the -carbon

Elimination reaction脫去反應；消去反應

the halogen group is eliminated along with a hydrogen.

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8.5 Nucleophilic Substitutions 親核性取代反應

nucleophilic substitution because the atom or group

replacing the leaving group is a nucleophile

can take place in one step (SN2) or two steps (SN1)

SN2

The nucleophile attacks the partially positively charged carbon.

As the nucleophile approaches the carbon and forms a new

bond

SN1

The carbon-halogen bond breaks, forming a carbocation.

Next step, the nucleophile attacks the carbocation to form a

substitution product.

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8.6 SN2 Reactions

SN2 Reactions Can Be Used to Make a Variety of Compounds

8.6.1 SN2 mechanism

S: Substitution取代, N : Nucleophilic親核

2: bimolecular

HO- attack back side of C-Br

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back-side背面 bonding by the nucleophile is an inversion 反轉 of

configuration about the -carbon

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8.6.2 Rate of Reaction

Speed at which the reaction occurs

i.e. rate: reactants反應物 → product產物

R-Br + NaOH R-OH + NaBr

(dissolving NaOH & alkyl bromide in dimethyl sulfoxide &

heating)

rate influenced by the concentration of both reactants

(bimolecular)

↑ R-Br & ↑ NaOH → ↑ rate

reaction rate = k[R-Br][NaOH] (bimolecular)

k: Reaction rate constant

∴ rate-determining step of te reaction involves both reactants

→ second-order reaction (first-order in each reactant)

Energetics

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8.6.3 影響 SN2反應機轉要素

8.6.3.1. The Alkyl Moiety

Reaction rate:

1o alkyl bromides react faster > 2

o alkyl bromides

3o alkyl bromides are unreactive or undergo elimination

reactions.

1o > 2

o > 3

o

Reactivity: 3,000,000 100,000 2,500 <1

The nucleophile must approach 接近 the electrophilic

alpha-carbon atom from the side opposite the halogen.

"steric hindrance" effect立體障礙:

the rear-side 背後 approach接近 of the nucleophile to the

-carbon will be hindered阻礙 by neighboring鄰近 alkyl

substituents on the- & -carbons.

H

HH

Br

H

H3C HBr

CH3

H3C HBr

CH3

H3C CH3

Br> > >

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8.6.3.2. Leaving Group ability

Leaving group ability Reactivity

The weaker the base, Stability of anion ; the better it is as a

leaving group

Relative acidities of HX

HI > HBr > HCl > HF

Relative basicities of halide ions

Relative leaving abilities of halide ions

Iodides are the weakest base and the most stable base.

Fluorides are the strongest base and the least stable base.

The larger the halogen, the faster the reaction for the alkyl

halide.

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The weaker the base, the better it is as a leaving group.

Weak bases are stable bases; they readily bear the electrons

and do not share their electron well.(強鹼容易提供電子對；反之

弱鹼則不易提供電子對)

Large atoms are more polarizable than small atoms

The high polarizability of a large iodide atom causes it to

react

Alkyl iodide is the most reactive alkyl halide

alkyl fluoride is the least reactive.

The lower the pKa, the stronger the acid, the weaker the

conjugate base

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An SN2 reaction proceeds in the direction that allows the

stronger base to displace the weaker base.

8.6.3.3. Nucleophilicity親核基性

Basicity鹼性:

Strength of a base shares its lone pair with a proton

Basicity is measured by the acid dissociation constant (Ka)

Nucleophilicity:

Strength of a compound (a nucleophile) is able to attack an

electron-deficient atom

measured by a rate constant (k)

Nucleophile’s strength

(i) negatively charged species (anions) are more nucleophilic

(and basic) than are equivalent neutral species.

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(ii) nucleophilicity (& basicity) ↑ from left to right of the

periodic table C- > N

- > O

- > F

-

less electronegativity = it holds the e- around it less firmly

Base Strength and Nucleophile Strength

When comparing molecules with attacking atoms of

approximately the same size (atoms are in the same row), the

strongest base is the best nucleophile

Nucleophilicity is affected by steric effects.

tert-Butoxide is a stronger base than ethoxide ion since

tert-butanol is a weaker acid than ethanol.

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(iii) ↑ size, nucleophilicity↑

F- < Cl

- < Br

- < I

-;

HO- < HS

-

The larger the nucleophile the better the nucleophilicity-

ability to donate electrons and form s bonds.

Larger anions are more stable (i.e., less basic) than smaller

anions

I- is larger and more polarizable than a F

-

∴ better nucleophile than the fluoride ion.

The relatively loosely held (polarizable) electrons of the I- can

overlap from farther away with the orbital of carbon undergoing

nucleophillic attack.

The tightly bound electrons of the fluoride ion cannot start to

overlap until the atoms are closer together.

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Problem: List the following species in order of decreasing

nucleophilicity in an aqueous solution

Negative nucleophiles are stronger than neutral nucleophiles.

Compare the pKas for the ranking of the other nucleophiles

A carboxylic acid is a stronger base than a phenol, which is a

stronger acid than water

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8.6.4. Solvent Effects

F– is the best nucleophile in an aprotic solvent 不具氫基溶劑

because it is the strongest base.

I– is the best nucleophile in a protic solvent具氫基溶劑

because it more polarizable and it is poorly solvated.

In aprotic solvents, sizes and nucleophilicities of the halide

ions parallel one another. As the size increases on a halogen,

the basicity decreases, the nucleophilicity decreases in an

aprotic solvent, and the nucleophilicity increases in a protic

solvent

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Polar, protic solvents 極性具氫基溶劑

are hydrogen bond donors (a hydrogen is bonded to oxygen

or nitrogen).

ex. water and alcohols 醇 solvate anions by hydrogen

bonding interactions.

Nucleophile/Water (protic solvents) Relationship

The partially positive charged hydrogens on the protic solvent

point toward the negative charged species.

These solvated species are more stable and less reactive than

the unsolvated "naked" anions.

Cl

H3C

O H

H3CO

H

CH3

OH

CH3O

H

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easier to break the ion-dipole interactions between I- & solvent

than between the more basic F- & the solvent.

The iodide ion is a better nucleophile in a protic solvent.

Polar, aprotic solvents 極性不具氫基溶劑

are not hydrogen bond donors (no hydrogen bonded to an

oxygen of nitrogen)

dimethyl sulfoxide DMSO, dimethylformamide DMF

(CH3)2NCHO

do not solvate anions

good solvation of the cations

∴ anions are freer to participate in SN reactoin

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Summary

(1) SN2 reaction involve a neutral alkyl halide & charged

nucleophile:

↑ polarity of a solvent → strong stabilizing the (-)-charge

nucleophile → ↓ rate of reaction

(2) SN2 reaction involve a neutral alkyl halide & neutral

nucleophile,

Charge on transition state > charge on neutral nucleophile,

↑ polarity of a solvent → ↑ rate of reaction

(3) SN2 reaction will be favored by high conc. of negative

charged nucleophile in an aprotic polar solvent

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8.6.5 Stereochemistry of SN2 Reactions

SN2 : Nucleophiles Back-side attack ∴ inversion of

configuration

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trans-1-t-butyl-4-chlorocyclohexane

Br

I

Cis-1-bromo-4-t-butylcyclohexane

NaI

acetone

trans-1-t-butyl-4-iodo-cyclohexane

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8.6.6 Intermolecular分子間 Versus Intramolecular分子內

Reactions

:

Carrying out an internal SN2 reaction

low concentration of reactant favors an intramolecular 分子內

reaction

The intramolecular reaction is also favored when a five- or

six-membered ring is formed

Use sodium metal to generate the oxygen anion

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Use a non-nucleophilic base to generate the oxygen anion.

The Intramolecular Reaction is Favored When a Five- or

Six-Membered Ring Can Be Formed

Three- and four-membered rings are less easily formed

Three-membered ring compounds are formed more easily than

four-membered ring compounds

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8.7 SN1 Reaction

8.7.1 Mechanism of the SN1 Reaction

The leaving group departs before the nucleophile approaches

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8.7.2 Experimental Evidence for an SN1 Reaction

kinetic

Reaction rate = k[ alkyl halide ]

rate-determining step is the alkyl halide → alkyl carbocation

∴ unimolecular and follow a first-order rate equation SN1

two-step mechanism

the rate-determining step: ionization of the alkyl halide

carbocation碳陽離子 is formed as a high-energy intermediate 反

應中間化合物, and this species bonds immediately to nearby

nucleophiles.

If the nucleophile is a neutral molecule, the initial product is an

"onium" cation陽離子

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8.7.3 SN1 Reactivity影響 SN1反應機轉要素

8.7.3.1. The Alkyl Moiety

reaction rate: 3o > 2

o > 1

o

reverse reactivity order in SN1 reactions and SN2 reactions

SN1 reactivity is governed by intermediate carbocation 碳陽離子

stability

3o cations are most stable and 1

o cations are least stable

SN1 mechanism:

R-X [ R+ ] R-Nu

carbon-halogen bond breaking

formation carbocation - reactive intermediate

reaction of carbocation & nucleophile

∴ stabilized carbocation → stabilized transition state →

↓ activation energy → ↑ rate of reaction

The stability安定性 of carbocations :

CH3+ < CH3CH2

+ < (CH3)2CH

+ < CH2=CH-CH2

+ < C6H5CH2

+ <

(CH3)3C+

3 o

alkyl halides will be more reactive than 2 o

3o > 2

o > 1

o > CH3

(opposite to the reactivity order observed for the SN2

mechanism)

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carbocation can be stabilized by sharing electrons from adjacent

C-H bond

hyperconjugation

8.7.3.2 Allylic and benzylic halides are reactive by SN1 & SN2

mechanism.

CH2 CH2 CH2

CH2 CH2

CH2

Benzylic carbocation

Allylic carbocation

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their carbocations have the same stability as 2 o

carbocations.

The SN1 Reaction of Allylic Halides Can Form Two Products

8.7.3.3. Leaving Group Influence on SN1 Reactivity

The alkyl iodide is the most reactive and the alkyl fluoride is the

least reactive.

Electrophiles with less basic (more stable) leaving groups will

react faster by SN1 because the stability of the intermediate

produced by the rate-determining step in the SN1 reaction is a

function of both the carbocation stability and the leaving group

stability

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8.7.3.4. Nucleophilicity親核基性

∵ nucleophiles只參予 in the fast second step,

their relative molar concentrations rather than their

nucleophilicities should be the primary product-determining

factor決定要素.

Solvolysis溶劑分解.

carbocations react with the solvent (nucleophiles) →

substitution product.

(CH3)3C-Br + CH3OH (solvent) (CH3)3C-O-CH3 +

HBr

8.7.3.5. Solvent effect

polarity of solvents 溶劑極性

water > formic acid > dimethyl sulfoxide > acetonitrile > ethanol

> acetone >> methylene chloride & ether

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The higher the dielectric constant, the more polar the solvent.

The more polar the solvent, the faster an SN1 reaction goes.

∵ lowers the activation energy for SN1 reactions.

The more polar the solvent, the faster an SN1 reaction goes.

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Polar solvents stabilize charged transition states of SN1

reactions which resemble a carbocation/leaving group anion

intermediate more than they stabilize neutral reactant

If the reactants are neutral, the charge on the reactants will be

less than the charge on the transition state.

Increasing the Polarity of the Solvent Increases the Rate

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8.7.4. Stereospecificity

carbocation has a trigonal 三面體(flat) configuration (sp2

hybridized)

can bond to a nucleophile equally well from either face 兩面.

If the intermediate from a chiral alkyl halide survives long,

the products are expected to be racemic (a 50:50 mixture of

enantiomers).

SN1 reaction 2 products are formed.

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SN1 Product Stereochemistry

SN1 reaction, the leaving group leaves before the nucleophile

attacks a planar carbocation intermediate.

Two products are formed

SN2 Product Stereochemistry

incoming nucleophile attacks the chirality center on the side

opposite to where the leaving group is bonded (back side attack)

inversion of configuration

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8.8 Vinyl and aryl halides do not undergo SN2 nor SN1

∵ nucleophile is repelled by the electron cloud

Vinyl and aryl halides do not undergo SN1 because

1. vinylic & aryl cations are more unstable than 1o carbcation

2. sp hybridization cannot be +ve

sp2 bonds are difficult to break

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8.9 Carbocation Rearrangement

A. 1,2 hydride shift

In an SN1 reaction of 2-bromo-2-methylbutane, the secondary

carbocation undergoes a 1,2-hydride shift to form a tertiary

carbocation. The tertiary carbocation is then attacked by water.

In an SN2 reaction, no rearrangement occurs.

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1,2-Methyl shift

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8.10 Comparison between SN1 & SN2

SN1 SN2

3o Alkyl halides > 2o > Io

Stabili

Io and 2o Alkyl Halides

Low steric for backside attack

R-X R+ + X-

rate determining step

Rate depends on RX

R-X + Y RY + X-

rate depends on RX &Y

Leaving group stability affects the

rate the most

I- is more stable than Cl-,

∴ I is a better leaving group for

SN1(& SN2)

Nucleophilicity of the incoming

group affects the rate the most

↑ polarity of nucleophile,

↑reaction rate;

higher atomic no. more polarizable;

ex. I- > Cl-

The higher the negative charge

density,

the better the nucleophile

Racemic Products Inverted Product

carbocation rearrangement No carbocation rearrangement

Polar protic solvents stabilize cation

∵ Polar protic solvents assist in

ionizing the bond that is breaking in

the rate determining step

Polar aprotic solvents favor SN2

∵ transition state is polar,

∴ polar solvents ↑the rate.

aprotic is best,

∵ ionization of solvent must be

prevented

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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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8.11 Biological Nucleophilic reaction

Biological systems use SAM as a methylating agent to transfer a

methyl group to a nucleophile

轉換 norepinephrine正腎上腺素 into epinephrine腎上腺素

The cell membrane phospholipid component

phosphatidylethanolamine is converted by three methylation

reactions to phosphatidycholine.

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Exercise

1. Name the following alkyl halides according to IUPAC rules:

2. Draw structures corresponding to the following IUPAC names:

(a) 2,3-Dichloro-4-methylhexane (b) 4-Bromo-4-ethyl-2-methylhexane

(c) 3-Iodo-2,2,4,4-tetramethylpentane

3. Give the major product of the following reactions:

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4. What nucleophiles could be used to react with butyl bromide to prepare the following

compounds?

a. CH3CH2CH2CH2OH B. CH3CH2CH2CH2SCH2CH3 C.

CH3CH2CH2CH2OCH3

D. CH3CH2CH2CH2NHCH3 E. CH3CH2CH2CH2CN F. CH3CH2CH2CH2SH

5. Which one in each of the following pairs will react faster in an SN2 reaction with OH-?

a. CH3Br or CH3I b. CH3CH2I in ethanol or in dimethyl sulfoxide

c. (CH3) 3CCl or CH3Cl d. H2C=CHBr or H2C=CHCH2Br

6. What products would you expect from the reaction of 1-bromopropane with each of the

following?

a. NaNH2 b. KOC(CH3) 3 c. NaI d. NaCN d. LiOH e. NaSH

7. Rank the following sets of compounds with respect to SN2 reaction.

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

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9. Which reaction in each of the following pairs will take place more rapidly?

10. Which of the following alkyl halides form a substituted product from an SN1 reaction that

is different from the substituted product formed from an SN2 reaction?

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11. Draw the products obtained from the solvolysis of each of the following compounds in

ethanol:

12

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9. Alkenes (II)

9.1 Introduction

Alkyl Halides Undergo Substitution and Elimination Reactions

In an elimination reaction, a halogen is removed from one

carbon and a hydrogen is removed from an adjacent carbon.

A double bond is formed between the two carbons from which

the atoms were removed.

Dehydrohalogenation脫鹵化氫反應 of Alkyl Halides

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9.2 Elimination E2 mechanism

Mechanism for an E2 Reaction

The E2 reaction is a concerted, one-step reaction.

Rate = k[alkyl halide][base]

E: elimination; 2: bimolecular

removing H from -carbon, ∴ also called -elimination,

removing adjacent leaving gr. ∴ also called 1,2-elimination

Reaction rate: 3o > 2

o > I

o

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9.2.1 Alkyl halides

The weaker the base, the better it is as a leaving group

9.2.2 The Regioselectivity of the E2 Reaction

The major product of an E2 reaction is the most stable alkene

The greater the number of substituents, the more stable is the

alkene

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9.2.3 Zaitsev's Rule柴瑟夫法則

9.2.3.1 Definition

The more substituted alkene product is obtained when a proton

is removed from the -carbon that is bonded to the fewest

hydrogens

the major product of a -elimination is the more stable (the more

highly substituted) alkene

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9.2.3.2 Anti-Zaitsev product (I)

Conjugated alkene products are preferred over the more

substituted alkene product

9.2.3.3 Anti-Zaitzev product (II)

A sterically hindered alkyl halide + a sterically hindered base

bulky base in an E2 reaction is sterically bulky

→ remove the most accessible hydrogen.

∴ remove one of the more exposed terminal hydrogens, which

leads to the less substituted alkene.

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size of the base ↑ → percent of the less substituted alkene↑

However:

If the alkyl halide is not sterically hindered and the base is only

moderately hindered, the major product will be the more

substituted alkene.

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9.2.3.4 Anti-Zaitzev product (III)

Fluoride Ion is a Poor Leaving Group

The reaction with fluoride leaving group yields the less stable

terminal alkene product predominantly.

With fluoride as the leaving group, the transition state resembles

the carbanion obtained by removing hydrogen from a carbon

adjacent to the carbon bearing the leaving group more than it

(the transition state) resembles the product alkene.

Since a primary (terminal) carbanion is more stable than a

secondary (internal) carbanion, this reaction yields mostly

terminal olefin.

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9.2.3.5 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

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9.2.4 Stereochemistry of the E2 Reaction

The bonds to the eliminated groups (H and X) must be in the

same plane

Anti elimination is favored in an E2 reaction.

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H X

H

X

H

X

plane

H

X

1. Anti requires the molecule to be in a staggered

conformation.

2. Back-side attack achieves the best overlap of interacting

orbitals

3. It avoids repulsion of the electron-rich base with the

electron-rich leaving group.

Antiperiplanar 反側同平面

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The alkene with the bulkiest groups on opposite sides of the

double bond will be formed in greater yield,

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The conformation of cyclohexane with the substituent equatorial

is the most stable.

The less stable conformation, with the chloro substituent in the

axial position, readily undergoes an E2 reaction.

A hydrogen is antiperiplanar to the chlorine.

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The more stable conformation has a hydrogen antiperiplanar to

the chlorine substituent whereas the less stable conformation

does not

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.

The less stable conformation has a hydrogen antiperiplanar to

the chlorine substituent whereas the more stable conformation

does not.

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CH3 O:-

H

H

H

H

Cl

CH3 OH :Cl-

1-Isopropyl-cyclohexene

2

1

6 + +E2

More stable chair (no H is anti and

coplanar to Cl)

Less stable chair(H on carbon 6 is

anti and coplanar to Cl)

2

2

11

6 6Cl

H

H

H

H

Cl

HH

HH

H

Cl

H

H

HCH3 O:

-

CH3 OH :Cl-

3-Isopropyl-cyclohexene

21

6E2

+ +

in the more stable chair of the trans isomer, there is no H anti

and coplanar with X, but there is one in the less stable chair

in the more stable chair of the trans isomer, there is no H anti

and coplanar with X, but there is one in the less stable chair

it is only the less stable chair conformation of this isomer that

can undergo an E2 reaction

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meso-1,2-Dibromo-

1,2-diphenylethane

(E)-1-Bromo-1,2-

diphenylethylene

C C

Br H

C6 H5C6 H5

CH3 O- Na+

CH3 OHC6 H5 CH-CHC6 H5

Br Br

E2 of Enantiomers

E2 reaction of the meso-dibromide gives only the E-alkene

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9.3 Elimination E1 mechanism

E: elimination, 1: unimolar

Carbocation is formed

3o > 2

o > I

o

follow Zaitzev’s rule ∴ regioselective

not regiospecific, cis & trans alkenes will form

major product: trans alkene ∵ more stable

The Mechanism for an E1 Reaction

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9.3.1 Follow Zaitzev’s rule

The major product is generally the more substituted alkene.

The More Stable Alkene is the Major Product

9.3.2 Alkyl halides

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9.3.3 Carbocation rearrangement

E1 reaction forms a carbocation intermediate

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9.3.4 Dehydration脱水 of alcohol

proceed through an E1 mechanism, due to the acid-catalysis

necessary to protonize -OH → leaving group

Can forms cis or trans alkene,

Major product- trans alkene, ∵ more stable

Acid: conc. sulfuric acid (H2SO4), or 85% phosphoric acid

Reaction rate: 3o > 2

o > I

o

∵ most stable Carbocation forms

Follows Zaitzev's Rule:

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C C

H

OH

CH3

H

CH3CH3 C C

HH3C

CH3H3CC C

H3C

CH2

H3C

3-Methyl-2-butanol2-methyl-2-butenemajor product

3-methyl-1-buteneminor product

the most-substituted double bond will form.

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9.3.5 E1 From Cyclic Compounds

When a substituted cyclohexane undergoes an E1 reaction, the

two groups that are eliminated do not have to be diaxial.

The carbocation may undergo rearrangement.

An E1 reaction involves both syn and anti elimination

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9.4 Competition Between Substitution and Elimination

OH- ion can act as a nucleophile and hit the back side of the

alpha carbon, or it can remove a beta hydrogen.

Both reactions occur for the same reason:

The electron-withdrawing halogen, which causes the carbon to

which it is bonded to have a partial positive charge.

Both use high concentrations of a good nucleophile or strong

base.

9.4.1 Primary alkyl halides under SN2 and E2 conditions.

9.4.1.1 with good nucleophile

when reacted with a good nucleophile such as the methoxide

ion.

The SN2 product is favored in the case of a primary alkyl halide

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9.4.1.2 Sterically hindered primary alkyl halides in SN2 and E2

reactions

The nucleophile will have a difficult time getting to the back side

of the -carbon.

Elimination is the favored reaction.

9.4.1.3 Sterically hindered base

A bulky base encourages elimination over substitution

The nucleophile is sterically hindered

∴ it has difficulty getting to the back side of the alpha carbon.

∵ bulky nucleophile has difficulty getting to the back side of the

alpha carbon, the elimination product will predominate.

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9.4.2 Secondary alkyl halide under SN2/E2 conditions.

(1) Strong base E2

Weak base SN2

• SN2/E2 reactions are favored by a high concentration of

nucleophile/strong base

• SN1/E1 reactions are favored by a poor nucleophile/weak base

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(2) higher temperature E2

9.4.3 Tertiary alkyl halides under SN2/E2 conditions.

A tertiary alkyl halide is the most reactive under E2 conditions

and the least reactive under SN2 conditions

Only the elimination product is formed when a 3o alkyl halide

reacts with a nucleophile under SN2/E2 conditions.

weak base encourages substitution over elimination

Tertiary (SN1/E1): Substitution is Favored

Tertiary (SN2/E2): Only Elimination

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9.5 Substitution and Elimination Reactions in Synthesis

9.5.1 Williamson Ether Synthesis

In synthesizing an ether,

the less hindered group should be provided by the alkyl halide

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9.5.2 To synthesize an alkene

use the most hindered alkyl halide to maximize the elimination

product and minimize the substitution product

2-Bromopropane is more hindered than 1-bromopropane

Reaction conditions to favor alkenes or alcohols from tertiary

alkyl halides

To maximize the alkene product from 2-bromo-2-methylbutane, a

strong base is used.

Both elimination and substitution products are formed with a low

concentration of hydroxide ion.

If the reactant is a tertiary alkyl halide, use SN2/E2 conditions

because it gives only elimination

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9.5.3 Consecutive連續的 E2 Elimination Reactions

3,5-Dichloro-2,6-dimethylheptane can undergo two consecutive

dehydrohalogenation reactions to yield

1,6-dimethyl-2,4-heptadiene

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9.6 How to design a synthesis?

Retrosynthesis

Alkane only reaction: halogeantion + functional gr.

E2, high conc. t-BuO-, elimination , substitution

NBS allylic bromination

?

X

- HXX

- HX

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9.7 Summary

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Alkyl halide SN2 vs. E2 SN1 vs. E1

1o Mainly substitution

Unless there is steric

hindrance

No SN1/E1 REACTIONS

2o Both substitution &

elimination,

Stronger bulkier base,

temp↑ →

↑elimination

Both substitution &

elimination,

temp↑

→ ↑% of elimination

3o Only elimination

Both substitution &

elimination,

temp↑

→ ↑% of elimination

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Exercises

1. Give the major elimination product obtained from an E2 reaction of each of the following

alkyl halides with hydroxide ion (HO-):

2. Which alkyl halide would you expect to be more reactive in an E2 reaction:

3. For each of the following reactions, draw the major elimination product; if the product can

exist as stereoisomers, indicate which stereoisomer is obtained in greater yield.

a. ( R )-2-bromohexane + high concentration of CH3O -

b. ( R )-3-bromo-3-methylhexane + CH3OH

c. trans -1-chloro-2-methylcyclohexane + high concentration of CH3O -

d. trans -1-chloro-3-methylcyclohexane + high concentration of CH3O -

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4. Give the major elimination product obtained from an E1 reaction of each of the following alkyl

halides

5. What alkene will be formed in an E2 reaction of each of the following compounds?

a. (1S, 2S)-1-bromo-1,2-diphenylpropane

b. (1S,2R)-1-bromo-1,2-diphenylpropan

4. How do you make the product with the starting material?

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7. How do you make the product with the starting material?

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10 Alcohols, Ethers, Epoxides, Amines, and Thiols

10.1 Alcohol 醇

Introduction

Alcohols are compounds in which an H of an alkane has been

replaced by an OH (hydroxy group).

10.1.1 Nomenclature命名

A. Common name:

changing the "ane" ending on the alkane to "yl" and adding

alcohol.

B. IUPAC

(1) parent chain is numbered in the direction that gives the

functional group, OH, the lowest possible number

suffix: ol

On longer chains, the location of the -OH determines chain

numbering.

For example: (CH3)2CHCH2CH(OH)CH3 is 4-methylpentan-2-ol

CH3CH2OH

Ethyl

alcohol

n-butyl

alcohol

isopropyl

alcohol

sec-butyl

alcohol

tert-butyl alcohol

Ethanol Butanol 2-propanol 2-butanol 2-methyl-2-propanol

(2) If there is a functional group and a substituent, the

functional group gets the lowest number.

(CH3) 2C=CHCH(OH)CH3 is 4-methyl-3-penten-2-ol

OHOHOH

OH

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OH

H3C

H3C CH3

H3C CH2Cl

OH

HO OH

OH

(3) A number is not needed to designate the position of a

functional group in a cyclic compound since it is assumed to be

at the 1-position.

4-t-butylcyclohexanol

(4) If more than one substituent is present, they are named in

alphabetical order.

2-(chloromethyl)-6-methyl-cyclohexanol

(5) If the substituents are the same, use the prefixes di, tri, tetra,

etc.

Glycerol, 1,2,3-propanetriol

(6) With certain types of complex structures, use the prefix

hydroxy for –OH gr.

HOCH3CH3CH3CH3CH3CH3COOH 6-Hydroxyhexanoic acid

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Nomenclature priority

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10.1.2 PHYSICAL PROPERTIES

10.1.2.1 Alcohols and Hydrogen Bond

Alcohols have a polar C-O single bond and a very polar O-H

bond

C-O bonds are polar

∵ Oxygen is more electronegative than Carbon

Alcohols have strong hydrogen bonding

(F, O, N, X make polar bonds with H that make them mildly

acidic).

Hydrogen Bonding 氫鍵

Intermolecular 分子間氫鍵

→ ↑the boiling point沸點

→ Solubility 溶解度 in water ↑

the higher the electronegativity difference, the stronger the

hydrogen bonding

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Low-molecular-weight alcohol (~C5 alcohol) dissolves in water

∵ R-O-H hydrogen bonding with water

R-OH: R > C5, is insoluble in water

Hydrophilic親水性 (water-seeking) < hydrophobic疏水性

R group (hydrocarbon) too large

>1 OH, ex glycerol (1,2,3-propantriol) & sugar are soluble in

water infinitely

10.1.2.2 Acidity of Alcohol

Alcohols can act as either Brønsted acids or Brønsted bases

- amphoterism兩性化作用.

The lone pair of the oxygen on the hydroxy group can act as a

proton acceptor (this group is now a good leaving group)

The proton of the hydroxy group can be removed.

RO-H + H2O RO- + H2O

+

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The ease of deprotonation depends on the inductive effect

exerted by the R group.

Group 4 Group 5 Group 6

C-H N-H O-H

Example Alkane Ammonium alcohol

pKa 48 33 15

Acidity↑ electronegativity↑

Compound 2,2,2-Trifluoroethanol Ethanol t-Butanol

pKa 12.4 15.9 18

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10.1.3 Acid Base Reactions of Alcohols

Alcohols are weaker acids than water

Alcohols do not react with mild bases

such as Na2CO3 or NaHCO3

pKa: ROH ~16, pKa HCO3- ~10.6

Formation of Alkoxides

Alcohols react with active metals to produce alkoxides with

release of hydrogen gas

R-O-H + Metal (ex. Na) R-O- M

+ + H2↑

Alkoxides

Sodium methoxide (NaOCH3, NaOMe)

Sodium ethoxide (NaOC2H5, NaOEt)

Potassium t-butoxide (KOBut, t-BuOK)

Basicity:

CH3O- < CH3CH2O

- < (CH3) 3CO

-

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10.1.4 Reactions of ALCOHOLS

10.1.4.1 Reaction of Alcohol with Hydrogen Halides

Substitution Reactions of Alcohols

∵ a weakly basic halide ion cannot displace the strongly basic

HO- group

∴ No Substitution Reactions

SN1 or SN2 of the alcohols

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OH of alchohol is a poor leaving group (pKa ~15).

∴ should be protonated first.

∵ H2O is stable, ∴ good leaving group

Primary alcohols undergo SN2 reactions with hydrogen halides:

1o alcohols

cannot undergo SN1 reactions because primary carbocations

are unstable.

1o alcohols have to undergo SN2 reactions

need catalyst: ZnCl2

ZnCl2 increases the rate of the reaction by creating a better

leaving group than water

2o and 3

o alcohols undergo SN1 reactions

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2o alcohol requires heating

3o alcohols + HX react at room temp.

∵ 3o carbocations are easier to form than 2

o carbocations

∴ 3o alcohol is faster than the reaction with a 2

o alcohol.

Carbocation rearrangement

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Nucleophilic Substitution Reactions of Alcohols

SN1: if an alcohol is reacted with HCl, HBr, or HI.

1. protonation (+ H+) of -OH group;

2. leaves as water;

3. halide ion reacts with the carbocation.

involving carbocation formation

rate: 3o > 2

o > I

o alcohol > methanol

Reactivity order of halogen acids: HI > HBr > HCl

Reactivity of alcohols: 3o > 2

o >1

o

3o- SN1

1o – SN2

2o – mainly SN2

Why is It Important to Be Able to Convert Alcohols to Alkyl

Halides?

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10.1.4.2. Reaction with Thionyl chloride

Pyridine is the solvent and acts as a base.

3. Reaction with Phosphorus Tribromide or Phosphorus

Trichloride

Comparison between SOCl2 and PBr3

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10.1.4.4. Sulfonate Ester

Sulfonate ester activate an alcohol for SN reaction with a

nucleophile

p-toluenesulfonyl chloride, tosyl chloride, TsCl :

p-CH3C6H4SO2Cl

p-toluenesulfonic acid, tosyl acid: p-CH3C6H4SO3H

The electrons holding the proton of sulfonic acid are delocalized

over three oxygen atoms when they are no longer bonded to H

∴ sulfonate is good leaving gr.

Common Sulfonyl Chlorides

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Sulfonate Ester Reacts with Nucleophiles

Once the alcohol has been activated by being converted into a

sulfonate ester, the appropriate nucleophile is added, generally

under conditions that favor SN2 reactions.

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Prepare (R)-2-chlorobutane from:

(a) (R)-butanol, (b) (S)-2-butanol

More examples

trans-2-methylcyclohexanol

(R)-2-butanol

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10.1.4.5 Dehydration of alcohol (Elimination)

To prevent the alkene formed in the dehydration reaction from

adding water and re-forming the alcohol, the alkene can be

removed by distillation as it is formed.

10.1.4.5.1 E1 mechanism

The acid first protonates the oxygen of the alcohol

poor leaving group (OH) → into a good leaving group (H2O)

A base removes a H+ from a carbon adjacent to carbonium,

→ alkene product and regenerating the acid catalyst.

The rate of dehydration reflects the ease of carbocation

formation

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Primary Alcohols Undergo Dehydration by an E2 Pathway

10.1.4.5.2 Dehydration of 2o and 3

o alcohols

Zaitsev's Rule: the most-substituted double bond will form.

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10.1.4.5.3 carbocation rearrangement

Ring Expansion

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10.1.4.5.4 A Milder Way to Dehydrate an Alcohol

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H3CH

H

OH

H3C H

OH2CrO3

H3COH

O +Cr

3+

ethanol

orange

greenaceticacid

Ethanal(acetaldehyde)

RR'

H

OH

R R'

OCrO3

2o alcohol Ketone

RR'

R"

OH CrO3

3o alcohol

Noreaction

10.1.4.6 Oxidation 氧化 of Alcohols

10.1.4.6.1 Alcohols oxidize to form aldehydes 醛類, ketones酮類,

or carboxylic acids酸

1o alcohols + KMnO4 or H2CrO4 → carboxylic acids

2o alcohols oxidize → ketones.

No more acidic H

Reaction stop

3o alcohols will not oxidize under these conditions

∵ can not break C-C

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10.1.4.6.2 Potassium permanganate 過錳酸鉀

KMnO4 + RCH2OH MnO2 + RCOOH

Purple brown

10.1.4.6.3 Oxidation of alcohol with chromic acid

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10.1.4.6.4 Oxidation with Pyridinium chlorochromate (PCC)

1o alcohols + PCC (Pyridinium chlorochromate)→ aldehydes

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10.1.4.7 Biosynthesis & Catabolism of fatty acid

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10.1.5 Important alcohols

Methanol (Methyl alcohol, CH3OH, MeOH) 甲醇

also called wood alcohol,

∵ produced from distillation of wood

假酒

toxic to human 對人有毒, leading to blindness 盲、甚至死亡

Ethanol (ethyl alcohol, CH3CH2OH, EtOH) 乙醇

results from fermentation 發酵 of sugars

Isopropyl alcohol (iPrOH, (CH3)2CHOH ) 異丙醇

Rubbing alchool

Solvents for cosmetics化妝品, lotions乳液, perfumes香水

Ethylene glycol (1,2-ethanediol)

Low freezing point (-12℃), high boiling point (199℃)

Anti-freeze抗凝劑, coolant冷凍劑

Menthol薄荷腦

From peppermint oil薄荷油

Cholesterol膽固醇

CH3

HO

H3C CH3

HO

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10.2 Ethers醚

Ethers are functional groups that have two R groups attached to

an oxygen. R-O-R’

10.2.1 Introduction

Acyclic ethers

CnH2n+2O

low-boiling point

relatively unreactive,

Good solvents. Slightly polar

Low-molecular-weight ether can hydrogen-bond to water

∴ slightly soluble in water

similar to alkanes in many respects ex. Boiling point, but can

be distinguished by their solubility in concentrated H2SO4.

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OCH3CH2CH2CH2CH2CH2CH2CH2

10.2.2 NOMENCLATURE

A. common name:

places the two names as branches before the name "ether"

B. IUPAC

simple ethers (< 5C), the longer chain is given precedence;

shorter side is given the ending "-oxy" and placed before the

root

methoxypropane (CH3OCH2CH2CH3)

cyclohexyl octyl ether

(∵ octane has more C than cyclohexane)

Alkyl Name Alkoxy Name

CH3- Methyl CH3O- Methoxy

CH3CH2- Ethyl CH3CH2O- Ethoxy

(CH3) 2CH- Isopropyl (CH3) 2CHO- Isopropoxy

(CH3) 3C- tert-Butyl (CH3)3CO- tert-Butoxy

C6H5- Phenyl C6H5O- Phenoxy

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R O H R O Na R O R

R BrNa

Alcohol Alkoxide alkyl halide Ether

+ NaBr

OHNa O

O

Br

OHNa

O

Br

Cyclopentanol

Ethanol

A

B

Preparation of Ether (Williamson ether synthesis)

The Williamson ether synthesis is an SN2 reaction between an

alkoxide and an alkyl halide.

Route A is better ∵ bromocyclopentane is a 2o halide &

elimination side reaction may occur

OH

O

X

HOX

AB

A

B

?

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Synthesis of t-butyl ethyl ether (2-ethoxy-2methylpropane)

Route B is better

Route A → eliminating product

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10.2.4 Ether Reaction:

Ethers can undergo nucleophilic substitution reactions with

hydrogen bromide and hydrogen iodide.

The oxygen is protonated (a good leaving group)

SN1 Mechanism of an Ether

(i) The oxygen is protonated.

(ii) The alcohol departs to leave a carbocation.

(iii) The nucleophile attacks the carbocation.

SN2 Mechanism of an Ether

(i) oxygen is protonated.

(ii) The nucleophile attacks the less sterically hindered carbon.

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O

O

O

Tetrahydrofuran Dioxane

10.2.5 CYCLIC ETHERS

oxygen incorporated into the ring.

good solvents,

eg. tetrahydofuran (five-membered cyclic ether, THF)

most common cyclic ethers are the three-membered ring

(epoxide)

10.2.5.1 NOMENCLATURE

IUPAC: three-membered ethylene oxides are called oxiranes,

with branches numbered by the usual priority rules.

The historical name for three-membered ethers is epoxide, and

the ending "-oxide" is attached to the parent alkene.

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10.2.5.2 Preparation of Epoxide

10.2.5.2.1. An alkene and a peracid result in an epoxide.

10.2.5.2.2. by the Williamson ether synthesis.

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10.2.5.3 Ring Opening of Epoxides

(1) Acid-catalyzed

the oxygen is protonated. epoxide is back-side attacked by the

halide ion.

Protonated epoxides are so reactive that they can ring open by

poor nucleophiles such as water and alcohol.

Back-side attack occurs by the water or alcohol.

The ring opens. The extra proton is removed by base.

Reaction of an unsymmetrical epoxide with an alcohol.

If the protonated epoxide is asymmetrical, two products

The major product is the one resulting from attack of the

nucleophile on the more substituted carbon.

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The more substituted carbon is preferentially attacked because

after protonation, the C-O bond begins to break.

If methanol attacks carbon 2, then that carbon is a developing

secondary carbocation.

If methanol attacks carbon 1, then that carbon is a developing

primary carbocation

SN1 is favored by bulkier epoxides.

Cleavage → trans fashion.

∵ back-side attack on the oxonium ion

HIOCH3OH

H

O

H

OHH3CO OHHO

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(2) Basic conditions

the site of nucleophilic attack differs in acid

The C-O bond does not break until the carbon is attacked by the

nucleophile.

The nucleophile is more likely to attack the less substituted

carbon because it is less sterically hindered.

The alkoxide picks up a proton from the solvent

Epoxides can react with a variety of nucleophiles

negative charged ion attacks the less substituted carbon of the

epoxide.

Because the epoxide is symmetrical, the nucleophile can attack

either carbon of the three-membered epoxide ring.

Acidic condition vs. Basic condition

As the C-O bond starts to break, a partial positive charge

develops on the carbon that is losing its share of the oxygen's

electrons. The bond starts to break before the Nu attacks.

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∵ the strain in the three-membered ring

epoxides are reactive to open without being protonated

nucleophile attacks an unprotonated epoxide → pure SN2

In acidic conditions, the Nu attacks the most substituted carbon.

In basic conditions, the Nu attacks the less substituted carbon.

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HO

OH

NH

NN

N

O

NH

DNAHO

HO

OH

O

OHO

OH

O

O

O

OMe

O

OH

NH

NN

N

O

NH

DNA

O

O

O

OMe

O

O

O

O

OMe

O

O

Aflatoxin

Polyaromatic hydrocarbon as carcinogen 致癌物

Formation of arene oxide addition products as a result of

nucleophilic attack DNA leads to cancer-causing products.

Afflatoxin (花生久置發霉所生的毒素)

經 cytochrome P-450作用後成致癌物

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Epoxy Glues

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10.3 Thiols

Sulfur analogs of alcohols

stronger acids (pKa = 10) than alcohols

not good at hydrogen-binding

In protic solvent, thiolate ions are better nucleophiles than

alkoxide ions (S- > O

-)

Sulfur analogs of ethers: sulfides or thioethers

Sulfur is an excellent nucleophile

∵ its electron cloud is polarized

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10.4 Organometallic Compounds

a compound that contains a carbon-metal bond.

Since most metals are less electonegative than carbon, the

carbon bonded to the metal is nucleophilic.

Organometallic reagents are commonly used as electrophiles.

Organolithium and organomagnesium compounds react as if the

alkyl groups are carbanions, paired with a metal cation.

Preparation of Orgnolithium Compounds

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O O

R Li

R

Li

HO

R

R OH R BrLi

R Li R CH2CH2 OH

1. O

2. H3O+

O

CH3-Li+

O-

CH3

OH

CH3H3O+

Attack goes to the least-substituted carbon.

trans additions

2-Carbon homologation- adding 2 carbons at once to a

compound

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2. H3O+

O 1. C3H7MgBr C3H7 OH

Cyclopenteneoxide

2-Propylcyclopentanol

Gringard reagents (Organomagnesium compounds)

are prepared by adding an alkyl halide to Mg shavings

Alkyl halides, vinyl halides, and aryl halides can all be used to

form organolithium and organomagnesium compounds

cannot be prepared from compounds containing acidic groups

(OH, NH2, NHR, SH, CO2H)

Product alcohol contains 2 more C than the Grignard reagents

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Gilman reagents

The reaction of methyllithium with copper iodide yields a

dimethyl lithium cuprate (Gilman reagents)

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CH3

CH3

OH

+ HCl

D.

CH3

CH3CH2 H

OH

PBr3

pyridine

Y

-CN

Z

B.

Exercise

1. Give the major product of each of the following reactions:

2. Give the structure of W, X, Y, Z (NB. Pay attention to the stereochemistry).

CH3CH2CHCH3 + HBr

OH

A.

CH3C

CH3

CHCH3

CH3

OH

+ HBrB.

CH3

OH

+ HClC.

CH3

CH3CH2 H

OH

W

-CN

XTsCl

pyridine

A.

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HO OH

O

CH3

CH3

O

CH3

CH3 CH3O-

CH3OH

O

CH3

CH3

H3C

H

O

CH3

CH3

H3C

H CH3O-

CH3OH

H+

CH3OH

C.

D.

.B.

H+

CH3OH

A

3. Five the major product formed when each of the following alcohols is heated in the

presence of H2SO4:

A. B. C.

D. E.

4. Draw the chemical structure of the product from the following reactions:

E.

CH3C

CH3

CHCH3

OH

CH3

CH3CH2CH2CH CCH3

OH

CH3

CH3

OH

CH3CH2CH2CH2CH2OH1. methanesulfonyl chloride

2. NaCN

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F.

G.

H.

I.

J.

K.

L.

M.

N.

CH3CH2CH2CH2CH2OH1. p-toluenenesulfonyl chloride

2. Na metal, OH

CH3CH2CH2CH2CH2OHH2SO4

CH3CH2CH2CH2CH2OHSO2Cl

pyridine

CH2MgBr

1. Ethylene oxide

2. H3O+

CH3CHCH2CH2CH2OCH3

CH3

HI

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O.

P.

R.

5. Predict the product(s) of the following transformations:

6. Show how you could prepare the following substances from cyclohexanol:

7. Show how you could prepare the following substances from 1-propanol:

8. How would you prepare the following ethers?

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Cl CH2CH2CNC.

9.What are the reagents a-d in the following reactions?

10. Using the given starting material, any inorganic reagents, and any carbon-containing

compounds with no more than 2 carbon atoms, indicate how the following syntheses could

be carried out:

11. How the following compounds could be prepared using ethylene oxide as one of the

reactants:

a. CH3CH2CH2CH2OH B. CH3CH2CH2CH2Br C. CH3CH2CH2CH3

A. OH

B. OCH3

CH3CH2C CH CH3CH2C CCH2CH2OHD.

CH3CHCH2CH2OH

CH3

CH3CHCH2CH2CH2CH2OH

CH3E.

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11 Aromatic芳香 Hydrocarbons

11.1 What is the Structure of Benzene?

Cyclohexene

H = -28.6 kcal/mol (experimental)

Bezene “cyclohexatriene”

H = -85.8 kcal/mol (calculated) “cyclohexatriene”

H = -49.8 kcal/mol (experimental)

Benzene is stabilized by electron delocalization

The experimental value of the heat of hydrogenation was smaller

than the calculated value.

Benzene is 36 kcal/mol more stable than the hypotheical

cyclohexatriene.

The heat of hydrogenation for benzene is lower than that of the

calculated cyclohexatriene.

The extra stability a compound gains as a result of having

delocalized electrons is called resonance energy.

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Resonance Energy

A measure of the extra stability a compound gains from having

delocalized electrons

11.2 Benzene C6H6

Bond length was 1.44Å on all C-C bonds

Cyclohexene gives off 28.6 kcal/mol when reduced.

Benzene gives off 49.8 kcal/mol

No significant reaction with Br2, KMnO4 without a catalyst

Resonance in Benzene

Electrons are delocalized over the whole system

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11.3 Aromaticity芳香環性

Aromatic Compounds芳香環化合物

The rules for aromaticity are:

1. Planar (flat molecule)

2. Cyclic (ring compounds)

3. All atoms on cycle are sp2 hybridized

4. (4n+2) (n= integral no. 正整數) (Huckle rule)

11.4 Examples of Compounds

11.4.1 Annulenes環烯:

Monocyclic hydrocarbons with alternating single and double

bonds

A prefix in brackets denotes the number of carbons in the ring.

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11.4.2 cyclooctatetraene and Cyclobutadiene

not aromatic because it is nonplanar

Cyclobutadiene

Cyclobutadiene is not aromatic because it has an even number

of electron pairs (4 electrons)

11.4.2 Cyclopropene

Neither cyclopropene nor the cyclopropenyl anion is aromatic

Huckle rule: (4n+2, n = 0) total 2

The cyclopropene cation is aromatic

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11.4.3 cyclopentadienyl anion

Resonance contributors of the cyclopentadienyl anion.

Huckle rule: (4n+2, n = 1) total 6 electrons

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11.4.4 Cycloheptatriene

sp3 carbon interrupted cloud not aromatic

.

Huckle rule: (4n+2, n = 1) total 6

Cycloheptatrienyl cation is aromatic and relatively stable

Cycloheptatrienyl bromide acts like an ionic compound; it is

soluble in water.

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11.5 Heterocyclic aromatic compounds雜環

Heterocyclic雜環 aromatic compounds (heteroaromatics) follow

the rules for aromaticity.

Huckel's rule can be fulfilled by lone pairs as well as pi bonds.

11.5.1 Pyridine

Huckle rule: (4n+2, n = 1) total 6 electrons

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11.5.2 Pyrrole

Pyrrole orients its nitrogen lone pair it becomes part of the

system.

Huckle rule: (4n+2, n = 1) total 6

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11.5.3 Resonance of Furan

Furan has two lone pairs on its oxygen atom.

One of these becomes part of the ring system and the other is

oriented perpendicular to the system in an sp2 hybrid orbital.

Huckle rule: (4n+2, n = 1) total 6 electrons

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11.5.4 Polycyclic heteroaromatics and heteroaromatics with

more than one heteroatom

Aromatic compounds are also involved in many other

biochemical functions.

11.6 Polycyclic aromatic compounds (PAH)

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Summary of 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 = 0, 1, 2, 3,… ) (Huckle rule)

Conditions for resonance

Movement of electrons through p orbitals over other atoms =

delocalization of electrons

Alternating double bonds (multiple bonds)

All atoms should have a p orbital that is not involved in

hybidization sp2 or sp hybridized C, N, O, S or carbocations.

Lone pairs, anions also participate in resonance.

Position of the nucleus does not change. Only the position

of the electron changes.

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12 Reactions of Benzene and Substituted Benzenes

12.1 Nomenclature of benzenoid compounds

"benzene" is the root name

(1) attaching "benzene" after the name of the substituent

(2) common names

(3) Benzene rings with alkyl substituents are named as

alkyl-substituted benzenes or phenyl-substituted alkanes.

mono- substituted benzenes have trivial names

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If benzene is not the principal structure, it is called "phenyl".

When a benzene ring is a substituent, it is called a phenyl group.

A benzene ring with a methylene group is called a benzyl group

(4) 2 substituents on benzene

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(5) Polysubstituted benzene

i. parent name is chosen

ii. numbering as small as possible

iii. substituents in alphabetical order, not numerical order

2,4,6-trinitrotoluene (三硝基甲苯, TNT): 有爆炸性的軍用化學品

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12.2 Reactions of Benzene

Electrophilic Aromatic Substitutions

the substitution product is the most stable.

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mechanism

.

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12.2.1. Halogenation

12.2.1.1 Benzene chlorination

requires Cl2 and FeCl3 (catalyst)

Dichlorobenzene

insecticide, mothball (kill moths that eat woolen garments)

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DDT (dichlorodiphenyltrichloroethane)

Insecticide, degradation slowly,

dehydrogenation to form (ClC6H4)2C=CCl2,

[1,1-dichloro-2,2-di(p-chlorophenyl)ethene (DDE) →

environmental hormone → ↓ male population

banned in the USA in 1970

AlCl3

DDE

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Dioxins 戴奧辛

major sources :

from the incineration (burning) of chlorine-containing

plastics, eg. PVC [polyvinyl chloride or poly(chloroethene)].

Chlorinated pesticides and herbicides used in farming also

produce dioxin.

Because dioxin cannot be removed from the organisms, the

dioxin will concentrate up the food chain

PCBs多氯化聯苯基

contain 1-8 Cl atoms & polychlorinated biphenyls (PCBs

傳熱力強，但不傳電。

不易燃燒。

性質穩定，不會有化學變化。

被廣泛使用，用作電介質，置於電容器及變壓器等電子儀器內，或作為

熱流交換液以調節儀器操作的溫度

神經系統: 反應遲鈍、手腳麻痺震顫、記憶力衰退、智力發展受阻

干擾荷爾蒙分泌，降低成人的生殖能力, 致癌, 特別是肝癌

(Cl)nn(Cl)

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12.2.1.2 Bromination

A lewis acid such as ferric bromide is required as a catalyst

Donating a lone pair to a Lewis acid weakens the Br—Br

The electrophile adds to the benzene ring.

A base in the reaction mixture removes the proton from the

carbon that attacked the electrophile.

The benzene ring is regenerated.

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12.2.1.3 Iodination

Nitric acid is required to generate the electrophile for the

iodination of benzene.

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12.2.2 Nitration

requires nitric acid and sulfuric acid

Usage

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12.2.3 Sulfonation

requires sulfur trioxide and sulfuric acid (Fuming sulfuric acid)

Fuming sulfuric acid is used to generate the electrophile for the

sulfonation of benzene.

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12.2.4. Friedel-Crafts Alkylations

require an alkyl halide and Lewis acid (AlCl3, FeBr3, …) resulting

in an alkylbenzene.

alkyl cation ion is the required electrophile in the electrophilic

aromatic substitution reaction

R-X: X = Cl, Br, I, But not :

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The carbocation is formed from the reaction of an alkyl halide

with aluminum chloride.

The electrophile attaches to the ring. The base removes the

proton from the carbocation intermediate. The benzene ring is

regenerated.

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Other Lewis acid catalyzed reaction:

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Carbocation rearrangement

Friedel-Crafts Alkylation Side Product

(1)

A true primary carbocation is never formed

(2)

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12.2.5 Friedel-Crafts Acylations

require an acid halide (acyl halide) or anhydride

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An acyl chloride or an anhydride

can be used for Friedel-Crafts acylation.

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Friedel–Crafts Acylation vs Friedel–Crafts alkylation

It is not possible to obtain a good yield of an alkylbenzene

containing a straight-chain group via Friedel–Crafts alkylation

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Choices of reduction

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12.3 IMPORTANT AROMATIC COMPOUNDS

Styrene

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12.4 Substituted Benzene

The relative positions of the two substituents are indicated by

numbers or by prefixes (o-, m-, p-)

two substituents are listed in alphabetical order:

If one of the substituents can be incorporated into a name, that

name is used and the incorporated substituent is given the

1-position

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Naming Polysubstituted Benzenes

The incorporated substituent is given the 1-position;

the ring is numbered in the direction that yields the lowest

possible number

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12.5 Reaction of Alkyl Substituents

12.5.1 Free Radical halogenation

Bromaination at benzylic position

12.5.2 Nucleophilic Substitution

The halogen in the benzylic position can be replaced by a

nucleophile by (SN1 or SN2)

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12.5.3. Elimination

A halo-substituted alkyl group can undergo elimination

12.5.4. Hydrogenation of Unsaturated Substituents

A.

B.

C.

D.

Hydrogenation of Benzene requires High temperature &

pressure

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E. Aniline is produced by the reduction of nitrobenzene by

catalytic hydrogenation or by tin (and sometimes iron) in

hydrochloric acid.

ONLY One nitro group is selectively reduced.

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12.5.5 Oxidation of the Alkyl Substituent

KMnO4 will oxidize any alkyl benzene with an alpha proton to

benzoic acid.

An alkyl group bonded to a benzene ring can be oxidized to a

carboxylic acid group by acidic solutions potassium

permanganate or sodium dichromate.

The benzene ring itself is not readily oxidized.

Benzylic hydrogens are removed during the oxidation of the

benzylic carbon.

∵ lack of benzylic hydrogens.

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If a milder oxidizing reagent (MnO2) is used:

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12.6 Aniline and diazonium salt

A primary amine can be converted into a diazonium salt 重氮鹽

by treatment with nitrous acid 亞硝酸.

Nitrous acid is unstable, so it is formed using an aqueous

solution of sodium nitrite and HCl or HBr.

Nucleophiles such as -CN, Cl

-, and Br- will replace the diazonium

group if the appropriate cuprous salt is added to the solution

containing the diazonium salt.

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Nitrosonium Ion亞硝基離子 Formation

The first step in the formation of nitrous acid from sodium nitrite

and HCl is the loss of water.

Loss of water from protonated nitrous acid generates the

nitrosonium ion, which is the reactive species in the reaction of

amines with nitrous acid.

The nitrogen atom of the amine shares its lone pair electrons

with the nitrosonium ion.

Loss of a proton from nitrogen forms a nitrosamine.

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12.6.1 Synthesis of para-chloroethylbenzene

(1) Ethylbenzene as Starting Material

(2)

12.6.2 Iodine displacement of Diazonium salt

12.6.3 Fluorination of Benzene

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12.6.4 Hydroxy group replaces the diazonium group

(1) An acidic aqueous solution of a diazonium salt that warms up

forms a phenol.

(2) Copper(I) oxide and copper(II) nitrate can be added to get a

higher yield of a phenol.

12.6.5 A Diazonium Group Can Be Replaced by a Hydrogen

Reduction by H3PO2

A hydrogen will replace a diazonium group if the diazonium ion

is treated with hypophophourous acid.

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Exercise

1. Which of the following compounds are aromatic?

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.2. Provide the necessary reagents next to the arrows.

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3. Draw the structure of each of the following compounds:

(a) m-chlorotoluene (b) p-bromophenol (c) o-nitroaniline (d) m-chlorobenzonitrile

(e) 2-bromo-4-iodophenol (f) m-dichlorobenzene (g) 2,5-dinitrobenzaldehyde (h) o-xylene

(i) m-ethylphenol (j) p-nitrobenzenesulfonic acid (k) (E)-2-phenyl-2-pentene

(l) o-bromoaniline (m). 2,4-dichlorotoluene (n) (E)-2-phenyl-2-pentene

4. Draw the product of each of the following reactions:

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13. Reactions of Substituted Benzenes

Relative rate of Electrophilic Substitution Reactions

electron-donating substituents reactivity of the benzene

ring and stabilize the carbocation intermediate.

electron-withdrawing substituents reactivity of benzene

ring and destabilize the carbocation intermediate

10-4

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13.1 Inductive Electron Withdrawal and Donation

(i) inductive electron donation

Alkyl gr. Donate electrons through a bond

↑ the rates of electrophilic substitution reactions when

compared to benzene.

(ii) inductive electron withdrawal

-NH3+ is more electropositive than H

Withdrawal of electrons through a bond → ↓ the rates of

electrophilic substitution reactions when compared to benzene.

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13.2 Resonance Electron Donation and Withdrawal

(i) Resonance Electron Donation

Substituents with a lone pair of electrons (OH, OR, NH2, & Cl)

→ donate electrons to the benzene ring by resonance

→ ↑ the rate of a electrophilic substitution reaction when

compared to benzene.

(ii) Resonance Electron Withdrawal

electronegative atom directly bound to a phenyl ring withdraws

electrons from the ring by resonance and decrease the rate of an

electrophilic substitution reaction when compared to benzene.

Substituents such as CO, CN, and NO2 withdraw electrons by

resonance

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13.3 Activating Substituents

13.3.1 Strongly Activating Substituents

strongly activating substituents Donate electrons into the ring

by resonance

→ benzene ring more reactive toward electrophilic substitution

13.3.2 Moderately Activating Groups

donate electrons into the ring by resonance and withdraw

electrons from the ring by induction

substituents donate electrons by resonance to the ring and away

from the ring.

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13.3.3 Weakly Activating Groups

Alkyl, aryl, and alkenyl groups attached to the benzene ring

weakly activate the ring toward electrophilic substitution

reactions.

Donate into the rings by resonance & withdrew electrons from

the rings by resonance

electron-donating > electron-withdrawing

13.4 Deactivating Groups

13.4.1 Weakly Deactivating Groups (Halide)

donate electrons into the ring by resonance and withdraw

electrons from the ring by induction.

electron donation into the ring by resonance < withdrawal of

electrons by induction.

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13.4.2 Moderately Deactivating Groups

have carbonyl groups directly attached to the ring.

withdraw electrons from the ring by resonance and induction.

13.4.3 Strongly Deactivating Groups

withdraw electrons from the ring by both induction & resonance.

4o ammonium groups (-NR3

+) have no resonance effect and thus

only inductively withdraw electrons from the ring.

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13.5 The Effect of Substituents on Orientation

All activating substituents and the weakly deactivating halogens

are ortho–para directors

All substituents that are more deactivating than halogens are

meta directors

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(1) All activating substituents direct an incoming electrophiles

to the ortho & para position

(2) Weakly deactivating substituents (halogen) direct an

incoming electrophiles to the ortho & para position

(3) All moderately deactivating substituents direct an incoming

electrophiles to the meta position

(4) All strongly deactivating substituents direct an incoming

electrophiles to the meta position

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13.6 Ortho/para directors of electrophilic substitution

13.6.1 Toluene

The structures of the carbocation intermediates formed from the

reaction of an electrophile with toluene.

∵ substituent is attached directly to the (+) charged carbon,

∴ stabilize by inductive electron donation

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Effects of steric hindrance of the substituted group

↑sterically hindered ortho position

↑ para position

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13.6.2 Anisole

ortho and para directors

The methoxy and hydroxy substituents are so strongly

activating that bromination is done without the Lewis acid

catalyst.

if a catalyst is used, substitution occurs at all ortho and para

positions

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13.6.3 Phenol

Acidity of phenol

Electron withdrawal decreases reactivity toward electrophilic

substitution and increases acidity

Electron-withdrawing groups stabilize a base and therefore

increase the strength of its conjugate acid

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The more electronic deficient a substituent on phenol, the

stronger the acid:

Stable through resonance of lone pair into nitro gr.

Electron donation increases reactivity toward electrophilic

substitution and decreases acidity

Electron-donating groups destabilize a base and thus decrease

the strength of its conjugate acid

unstable

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13.6.4 Halobenzene

Weakly deactivating substituents (halogens) are ortho–para

directors

13.6.5 Benzaldehyde

CHO gr withdraw electrons, meta director

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13.6.6 Anilinium cation

Substituents with a positive charge on the atom attached to the

benzene ring withdraw electrons from the ring

→ directing the incoming electrophile to the meta position.

The lone pair on the amino group forms a complex with the

Lewis acid catalyst, which converts the substituent to a meta

director.

Aniline and N-substituted anilines cannot undergo Friedel-Crafts

reactions.

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Benzenesulfonic acid & Nitrobenzene

If there is a meta director on the ring (moderate or strong

deactivators), the ring will be too unreactive to undergo either

Friedel-Crafts acylation or alkylation.

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Summary of electro withdrawal groups

Deactivating substituents are those that can withdraw electrons

and destabilize the carbocation in the ring.

Electrophilic substitutions with these type of substituent tend to

occur meta position

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Summary of electron donating groups

Activating substituents are those that can provide a lone pair of

electrons to stabilize the carbocation in the ring.

Electrophilic substitutions with these type of substituent tend to

occur ortho and para

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13.7 Designing a disubstituted benzene

(I) Less Steps is the best way

Two different routes can be used for the synthesis of

2-phenylethanol.

(a)

(b)

The route (b) requires excess benzene and involves a radical

reaction that can produce unwanted side products.

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(2) The order次序 in which the substituents are to be placed on

the ring must be considered

(i)

a.

b.

(ii) nitro-substituted aniline

aniline is first converted to an amide.

(Nitrobenzene: no Friedel-Crafts acylation)

The acetyl group is a protecting group which prevents oxidation

of the amine group.

After nitration, the protecting group can be removed by

hydrolysis.

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(iii)

Since the tertiary amino group is a strong activator, nitration can

be carried out in milder conditions than nitric acid and sulfuric

acid.

(iv) para-chlorobenzoic acid

The methyl group is oxidized after it directs the chloro

substituent to the para position

(v) meta-chlorobenzoic acid

The methyl group is oxidized before chlorination because a meta

director is needed to obtain the desired product.

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(vi) synthesis of m-nitroacetophenone

Friedel-Crafts acylation is carried out first because nitrobenzene

is too deactivated to undergo a Friedel-Crafts reaction.

(vii)

Friedel-Crafts acylation must be carried out before sulfonation

and the sulfonic acid group must be put on the ring after the

carbonyl group is reduced to a methylene group.

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13.8 Synthesis of Tri-substituted Benzenes

(1) Both substituents direct to the same position

(2) Both direct the incoming substituents to different positions

Three positions are activated, but steric hindrance makes the

position between the substituents less accessible.

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(3) If the two substituents direct to different positions, the strong

activating substituent will have the greater effect.

(4) If the two substituents have similar activating properties, a

mixture of products will be obtained.

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OCH3

O

O

O

1. AlCl3

2. H2O

+

CH3 1. NBS/

2. Mg

3. ethylene oxide

4. H+

N + Br2

Exercises

1. Draw the structure of the major product(s) of mononitration of the following substances:

(a) bromobenzene (b) benzonitrile (c) benzoic acid (d) nitrobenzene (e) phenol (f)

benzaldehyde

2. Draw the structure of the major product(s) of electrophilic monochlorination of the following

substances:

(a) m-nitrophenol, (b) o-methylphenol, (c) p-chloronitrobenzene

3. Draw the structure of the major product(s) of sulfonation of the following substances:

(a) o-chlorotoluene (b) m-bromophenol (c) p-nitrotoluene

4. Give the product of the reaction of excess benzene with each of the following reagents:

a. isobutyl chloride + AlCl3 b. neopentyl chloride + AlCl3

c. propene + HF d. dichloromethane + AlCl3

5. give the products of the following reactions:

a.

b.

c.

d.

e.

OCCH3

O

H2SO4

+ HNO3

Br

CH3

1. Mg

2. D2O

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CF3

FeCl3+ Cl2

CCH3

Br

O

NO2

COOH

Cl

Cl

O CH3

Cl

CH3

NO2

OCCH3CH3OC

OO

f.

7. Show how the following compounds can be prepared from benzene:

a. m -chlorobenzenesulfonic acid b. m -chloroethylbenzene

c. m –bromobenzonitrile d. 1-phenylpentane

e. m -bromobenzoic acid f. m -hydroxybenzoic acid

g. p –cresol h. benzyl alcohol

i. benzylamine j. N,N,N -trimethylanilinium iodide

k. 2-methyl-4-nitrophenol l. p -benzylchlorobenzene

m. benzyl methyl ether n. p -nitroaniline

o. m –bromoiodobenzene p. p -nitro- N -methylaniline

i. 1-bromo-3-nitrobenzene

8. For each of the following compounds, indicate the ring carbon that would be nitrated if the

compound was treated with HNO3/H2SO4:

a. b. c.

d. e. F

G. g.

NHCCH3

O

HO COOH