1 four reactions of alcohols. 2 4 reactions of alcohols 1.substitution with halides to form alkyl...

1

FOUR REACTIONS

OF ALCOHOLS

2

4 REACTIONS OF ALCOHOLS

1. SUBSTITUTION WITH HALIDES TO FORM ALKYL HALIDES

2. ELIMINATION OF WATER (DEHYDRATION TO ALKENES)

3. ACID/BASE REACTIONS WITH STRONG BASES/ACIDS

4. OXIDATION TO ALDEHYDES, KETONES, OR CARBOXYLIC ACIDS.

Alcohols have several polar atoms, several sites of reactivity.

Alcohols can undergo substitution, elimination and acid/base reactions.

An alcohol, like HOH, can act as a Nu:- or base as well as an E+ or acid.

H3C O H..

..

+ +

E+Nu:-

& base

E+

& acid

3

1. SUBSTITUTION WITH HALIDES

1. SUBSTITUTION WITH HALIDES TO FORM ALKYL HALIDES

Hydrohalic acids (HCl, HBr, HI) convert alcohols to alkyl halides

+ XHR OH

alcohol

SN1 SN2or

+ H OHR X

alkyl halide

Even in concentrated form, these acids contain a lot of H2O

H2O is a protic solvent. It stabilizes C+’s and weakens (solvates) Nu:’s

alcohol (substrate)

HI

HBr

HCl

Me

1 unhindered

2

3

SN2

SN2

SN1

SN1

At room temperature, the SN2 reaction is slow.

3° alcohols react rapidly 2° alcohols react moderately 1° and Me°alcohols are unreactive

4


Mechanism:

fast SN1 rxn.

conc. (37%)

C OH

C

C

C

(3º alcohol)t-butyl alcohol

H Cl

..

..

..

.. :

C

C

C

C

H

H

+O..

H2O

C

C

C

C

+

3º C+

Cl -: ....

:

+ H2O

t-butyl chloride

C

C

C

C

Cl.... :

conc. (37%)

C OH

C

C

(2º alcohol)

isopropyl alcohol

H Cl.... :

....

SN1 rxn.moderate

C

C

CH

H

+O.. C

C

C +

2º C+

Cl -: ....

:

..

..: + H2O

chloride

C

C

C Cl

isopropyl

H2O

The hydroxyl (-OH) group is a poor leaving group unless protonated.

5


(1º alcohol)ethyl alcohol

C OHC

conc. (37%)

H Cl.... :

..

..H

H

+OCC ..

needs heat slow SN2 rxn.

CC +

1º C+

not formedtoo unstable

H2O

Cl -: ....

:

+ H2OCC Cl

chlorideethyl

....:

In these mechanisms, even though H-Cl is shown as the acidic species, it is understood that H2O is present (37% HCl contains 63% H2O).

H-Cl is largely present as hydronium ion (H3O+ Cl-) and H3O+ protonates the alcohol in the first step.

Name the mechanism & show the product of the following reaction conc. aq. HBr

CH2 CH CH2 OH

allyl alcohol2-propen-1-ol

+ H2OCH2 CH CH2 Br

allyl bromide3-bromo-1-propene

SN1

6


1° and 2°alcohol reactivity can be improved by reacting with a polar aprotic reagent/solvent, such as phosphorus tribromide (PBr3) or thionyl chloride (SOCl2).

These are fuming, corrosive liquids (nasty) but they do the job well. 1° and 2° alcohols react quickly via SN2 without heating.

(1º alcohol)

+OHR S

O

Cl Cl

Cl-

S

O

ClR O+

H

R Cl + SO2 + HCl

SN2Fast

Br-

R O

H

PBr2+

R Br HOPBr2 +

(1º alcohol)P

Br Br

Br

+OHR

SN2Fast

Note that SOCl2 produces alkyl chlorides & PBr3 produces alkyl bromides.

E+

E+

E+

E+

7


The difference in the rate of reaction of alcohols with conc. HCl is used as a qualitative test (the Lucas Test) to determine the degree of substitution of an alcohol.

Lucas reagent is conc. HCl, saturated with ZnCl2 salt. The Zn+2 ion coordinates (bonds) with the alcohol oxygen even better than H+ and speeds up the rate at which the C+ can form.

About ½ mL of alcohol and 3 mL Lucas reagent are mixed in a test tube. 3°, allyl and benzyl alcohols react instantly (via SN1) to produce an insoluble alkyl halide, which appears initially as a hazy emulsion that separates into two liquid layers on standing.

2° alcohols also react quickly (SN1) in Lucas reagent (R-Cl separates).

1° alcohols do not react (no R-Cl layer forms). 2° and 3° alcohols can be differentiated by repeating the test in conc.

HCl (no ZnCl2 present). 3° alcohols react quickly but the solution remains clear in 2° alcohols.

8

2. DEHYDRATION OF ALCOHOLS TO ALKENES

2. DEHYDRATION TO ALKENES WITH H2SO4 OR H3PO4 AND HEAT

H2SO4 and H3PO4 acids are dehydrating; they react strongly with water and will remove H2O from other chemicals. H2SO4 is sometimes used as a desiccant.

When heated with alcohols, elimination (E1) occurs. The more stable, more substituted alkene (Zaitsev product) is the major product.

H+HSO4-

CC

OHH

C C + H2O

alcohol alkene

dehydrationCC

OHH

C C + H2O

alcohol alkene

dehydration

E1

With only gently heating (50°C) 20% H2SO4 dehydrates 3° alcohols.

2°alcohols require 60% H2SO4 @ 100°C.

1°alcohols need 95% H2SO4 @ 170°C (!), however, much of the product is lost. The 1° alcohol is badly charred.

9

H2O

CH3 +

H

H

3º C+

E1

..O

CH3

H

H

+


Mechanism:

HSO4-

CH3

H

+ H2O

1-methylcyclohexene

moderate

C OH

C

C

(2º alcohol) E1 rxn.

isopropyl alcohol

60%

H+ HSO4-

100ºC

..

..C

C

CH

H

+O C

C

C +

2º C+

H

H2O

E1 HSO4-

C

C

C + H2O + H2SO4

propene

(1º alcohol)ethyl alcohol

95%

170ºCC OHC

H+ HSO4-

slow E1 rxn.

..

..

H

H

H

+OCC ..

Unstable 1 C+ requires high temp.

H

CC +

1º C+H2O

HSO4-

CC + H2O + H2SO4

ethylene

E1

..

..

1-methylcyclohexanol(3º alcohol)

H+ HSO4-

THF50ºC

O H

CH3

Fast E1 rxn.20%

10


Note that even 1° alcohols dehydrate via the E1 mechanism. The high temperature (170°C) allows the formation of the unstable 1°C+.

With weak nucleophiles, no reaction will occur unless a C+ forms. HSO4

- (or H2PO4-) are very weak, non basic Nu:-’s. They have no

propensity to remove H+ from a -carbon via the E2 mechanism. Recall that E1 and SN1 occur in competition. High temperature

favors elimination over substitution. Draw and name the major products of the following reactions.

CH3

OH

H2SO4

CH3

H1-methylcyclopentene

Zaitsev product(major product)

2-methylcyclopentanol

11

C

CH3

H3C

CH3

CH2 CH2 OHH2SO4

neohexyl alcohol

C

CH3

H3C

CH3

CH2 CH2 + C

CH3

H3C

CH3

CH CH3 +


C

CH3

H3C CH CH3

CH3

+C

H3C

H3C

C

CH3

CH3

2,3-dimethyl-2-butene

Zaitsev product(major product)

Recall that in unimolecular reactions, C+ rearrangements can occur.

With that in mind, draw the major (Zaitsev) product from the dehydration of neohexyl alcohol.

H2O +

12


1° and 2° alcohols are charred by the severe conditions of acid dehydration and the E1 mechanism.

There is a way to safely dehydrate these alcohols with good yields even at room temperature. The reagents needed however are unpleasant and corrosive.

Phosphorus oxychloride (POCl3) in pyridine solvent (C5H5N:) will dehydrate 1°, 2° and 3° alcohols to alkenes via an E2 mechanism.

OH....

+ POCl

ClClE+

..N:

O

P

HHH

O

Cl

Cl

+

pKb = 5.3 pyridine is a moderatelystrong base & very bulky

good leavinggroup cyclopentene

E2

Recall that strong, bulky bases like t-butoxide cause E2 even w. 1°substrates.

13

2. DEHYDRATION OF ALCOHOLS TO ALKENES Pyridine, like t-butoxide causes E2 reactions even w. 1°substrates. Pyridine is very bulky because of its large -system of overlapping p-orbitals.

alkyl halide

(substrate)

good Nu-

nonbasic e.g., bromide

Br-

good Nu-

strong base e.g., ethoxide

C2H5O

-

good Nu-

strong bulky base e.g., t-butoxide

(CH3)3CO

-

very poor Nu-

nonbasic e.g., acetic acid

CH3COOH

Me

1°

2

3

SN1, E1

SN2

E2

SN2

SN2

SN1

SN2

SN2

E2

E2

SN2

E2 (SN2)

E2

no reaction

no reaction

SN1, E1

HSO4-

H2ON:

HSO4- and H2O are nonbasic, weak nucleophiles (like CH3COOH).

They will not react unless a C+ forms. This occurs rapidly with 3° alcohols, moderately with 2° alcohols and only at high temp. with 1° alcohols.

14

PROBLEMS

Write equations showing how the following transformations can be carried out. More than one step may be necessary. Mechanism are not required but show all reagents used.

CH2 CH2 OH CH CH3

OH

CH CH2

MarkovnikovproductH2SO4

H2OPOCl3,

N

H2O

E2

CH2

Br

CH3

OH1°

CH2

Markovnikovproduct

H2SO4

H2O

CH2

OH

1°

POCl3,

N E2

KOHSN2

15

3. ACID/BASE REACTIONS OF ALCOHOLS

3. ACID & BASE REACTIONS WITH STRONG BASES & ACIDS

Very strong bases abstract (remove) the weakly acidic H+ (pKa = 16).

The acid/base behavior of alcohols is much the same as that of H2O.

(as acid)

C O H +

alcohol

B:

strong base

..

.. C O-

+ BH+....:

alkoxidepKa = 16 pKb = -2

Very strong acids protonate the weakly basic –OH group (pKb = 16).

C O H +

alcohol

H A

(as base)strong acid

..

.. C O +

H

H+

A-

..

protonated alcohol

pKb = 16 pKa = -2

16

Draw products, show mechanisms and calculate extent of reaction (%).

PROBLEMS

CH3 CH2 O H....

Na+ OH-ethanol

pKa = 16

pKb = -1.74CH3 CH2 O

..

..: Na+ + H2O

sodium ethoxide

pKeq = pKa + pKb -14

pKeq = 16 -1.74 -14 = +.3Extent of reaction < 50%

pKb = -21

CH3 CH2 O H....

Na+ NH2-:

..

CH3 CH2 O....: Na+ + :NH3

CH3 CH2 O H....

Na+HCO3-

pKb = 7.6

pKeq = 16 -21 -14 = -19Extent of reaction = 100%

CH3 CH2 O....: Na+ + H2CO3

pKeq = 16 + 7.6 -14 = 9.4Extent of reaction = 0%

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3. REACTION OF ALCOHOLS WITH A BASE

Active metals such as Na, Li, K, Ca, etc. are very strong bases. They deprotonate alcohols liberating H2 gas

Draw products and show mechanisms for the following.

pKb = ca. -30

+ NaCH3CH2O H

ethanol

Na+ + H2CH3CH2O-

sodium ethoxide

1 /2

pKeq = 16 -30 -14 = -28Extent of reaction = 100%

isopropyl alcohol

+CH

H3C

H3C

HO K

.+CH

H3C

H3C

O- K+ H21 /2

potassium isopropoxide

isobutyl alcohol

CH3CHCH2O

CH3

+ CaH .

.

CH3CHCH2O

CH3

H

H2(CH3CHCH2O- )2 Ca+

CH3

+

calcium diisobutoxide

18

3. REACTION OF ALCOHOLS WITH A BASE

Show the reaction of neopentyl alcohol with aluminum metal.

+ AlCH3CCH2OH

CH3

CH3

...

3

neopentyl alcohol

H2

CH3

(CH3CCH2O- )3

CH3

Al+3 + / 23

aluminum trineopentoxide

The evolution of H2 gas from alcohols in the presence Na metal is a quick qualitative test for alcohols. The reaction is vigorous but not explosive as it is in H2O.

Aldehydes (pKa = 17) and ketones (pKa = 19) release only trace amounts of H2 gas when mixed with Na metal. [The aldehydes and ketones tested (and the test tube) must be free of water, i.e., dry! or a false positive result will be seen.]

19

ACIDITY OF ALCOHOLS

Alcohols typically have a pKa of ca. 16, e.g., CH3CH2OH and CH3OH.

Tert-butyl alcohol is less acidic because its conjugate base (alkoxide) is bulky and not easily solvated by water.

Electron withdrawing groups such as –F make alcohols more acidic. They stabilize the alkoxide (by induction). A weaker conjugate base necessitates a stronger acid.

Note the increased acidity of 2,2,2-trifluoroethanol and especially nonafluoro – t-butyl alcohol.

Unlike resonance effects, the inductive effect is only significant over a short range.

15.741618

15.5125.4

20

PROBLEMS Write equations showing how the following transformations can be carried

out. More than one step may be necessary. Mechanism are not required but show all reagents used.

OH

CH3

O

CH3

CH2CH3

O- Na+

CH3

CH3CH2Cl

SN2

Na or

Na+NH2-

CH2 OH CH2

CH2 BrHBr

fast SN1

Gilman Reagent

Cu2

LI+1

2 H3O+

21

4. OXIDATION OF ALCOHOLS

4. OXIDATION OF ALCOHOLS ( TO CARBONYLS)

The most common oxidants are oxides of nonmetals that are in a high oxidation state, e.g.,HNO3, KMnO4, Na2Cr2O7 in H2SO4, and

NaOCl in HAc. Calculate the oxidation state of:

N in HNO3

Mn in MnO4-

Cr in Na2Cr2O7

Cl in OCl-

C

O: :

C

O: :H

alcohol

carbonylcompound

oxidation

reduction

Alcohols are oxidized to carbonyl compounds: aldehydes, ketones and carboxylic acids.

In turn, these carbonyl compounds can be reduced to alcohols.

(1) + N + (-2)3 = 0 N = +5

Mn + (-2)4 = -1 Mn = +7

(1)2 + Cr2 + (-2)7 = 0 Cr = 12/2 = +6

(-2) + Cl = -1 Cl = +1

22

4. OXIDATION OF ALCOHOLS

Oxidants are usually classified into one of 3 groups (strengths).

1. Mild oxidants include Collins reagent (CrO3 in pyridine), and

PCC (pyridinium chlorochromate, (C5H5NCrO3Cl in

dichloromethane solvent.) These oxidants are dissolved in cold anhydrous solvents. (no water– this weakens the oxidant.)

2. Moderate-to-strong oxidants are cold to warm, aqueous solutions of HNO3, acidic or basic KMnO4, NaOCl in aq. HAc, Jones reagent (CrO3 in aq. H2SO4 + acetone).

3. Severe oxidants are hot, aqueous solutions of HNO3, acidic or basic aq. KMnO4, or Na2Cr2O7 in H2SO4

KMnO4 and Cr+6 reagents are often used for qualitative tests. When reacting, these oxidizing agents are reduced and show distinct color changes. List the color changes that occur.Cr+6

Cr+3

MnO4-

MnO2

Mn+2

orange green orblue

neutral or OH-

H+purple

brownprecipitate

colorless

23

4. OXIDATION OF ALCOHOLS 1 alcohols are easily oxidized. They have 2 -hydrogens. Mild oxidants remove 1 hydrogen yielding an aldehyde. Moderate-to-strong (or severe) oxidants remove both hydrogens

yielding a carboxylic acid.

R C

O

H

aldehyde

R C

O

OHcarboxylic acid

R CH2

OH

1º alcohol

(oxidants in non aq. solvents)

PCC (in CH2Cl2) Collins reagent (CrO3 in pyridine)

mild oxidation

moderate or strong oxidation(oxidants in aq. solvents)

Jones reagent (CrO3 in aq. H2SO4 & acetone)KMnO4, HNO3, etc.

24

4. OXIDATION OF ALCOHOLS 2 alcohols have only 1 -hydrogen. Mild or moderate-to-strong

alcohols remove the only -hydrogen yielding a ketone. Under severe oxidation conditions, the ketone product is further

oxidized. In the study of alkynes, we learned that a ketone is always in equilibrium with a minute amount of an enol, its ‘tautomer’.

In our study of alkenes we learned that hot KMnO4, hot H2CrO4, or

hot conc. HNO3 oxidize enols, cleaving the C-to-C -bond and yielding carboxylic acids or ketones (which will again be cleaved until only carboxylic acids remain).

2º alcohol

(in non aq. solvents or any cold aq. solvent) PCC (in CH2Cl2)

Collins reagent (CrO3 in pyridine)

mild to moderate oxidation

severe oxidation

Jones reagent (CrO3 in aq. H2SO4 & acetone)cold acidic or basic KMnO4, HNO3, etc.

(hot aq. KMnO4 or HNO3)

R CH

OH

R ketone

R C

O

R

R C

O

R

ketone enol

R C

O

R

H

carboxylic acid

2 R C

O

OH

25

4. OXIDATION OF ALCOHOLS 3 alcohols are not oxidized by any of the above oxidants except hot

KMnO4 or hot conc. HNO3 which dehydrate them to alkenes and subsequently cleave the alkenes to carboxylic acids.

Recall that 3 alcohols are easiest to dehydrate to alkenes.

The mechanism of oxidations are either not fully understood or are very complex. The student will need to memorize the oxidation products of alcohols with the various classes of oxidants.

3º alcohol severe oxidation(hot aq. KMnO4 or HNO3)

dehydration

R C

OH

R

R

any anhydrous or cold aq. oxidant no rxn.

R CH R

alkene carboxylic acid

2 R C

O

OH

26

PROBLEMS Draw the products of the following oxidations.

1-hexanol

Jonesreagent

CH2OHCOH

O

hexanoic acid

OH O

2-hexanone

CH2OH

1-hexanol

PCC orCollins reagent CH

O

hexanal

2-hexanol

coldNa2CrO4

in H2SO4

27

PROBLEMS Draw the products of the following oxidations.

OH

H3C cold KMnO4 aqueous

O

H3C

Cleavage occurs on either side of the -carbon

OH

H3C hot concentrated HNO3

C

C

H3CO

OH

OOH

C

CH3C

O

OH

O

OH+

3-methylcyclohexanol

28

IODOFORM REACTION (OXIDATION)

The Iodoform reaction: is a qualitative oxidation test used to identify the structure of a 1° or 2° alcohol. The alcohol is treated with a solution of iodine and a solution of sodium hydroxide.

NaOH + I2 NaOI (sodium hypoiodite)

NaOI is analogous to sodium hypochlorite (NaOCl, the so-called ‘liquid chlorine’ used as laundry bleach,e.g., Javex, and used as swimming pool disinfectant).

NaOI is a mild oxidant. It oxidizes 1° alcohols to aldehydes and 2° alcohols to ketones; but will not oxidize 3° alcohols.

If the 1° or 2° alcohol had a methyl group (-CH3) attached to the -carbon, the three hydrogens of the –CH3 group are substituted by iodine (-CH3 -CI3) and the product cleaves producing iodoform (CHI3), a pale yellow precipitate.

CH3CH2OHNaOI

1° alcohol aldehyde

CH3-C-H

O 3 NaOI

-CH3 -CI3

CI3-C-H

O cleaves

CHI3 + Na-O-C-H

O+

iodoform

Na+ OH-

29

IODOFORM REACTION (OXIDATION) Ethanol is the only 1° alcohol that gives a positive iodoform test

because it is the only 1° that has a –CH3 group attached directly to the -carbon.

By contrast, many 2° alcohols have –CH3 group attached directly to their -carbon and this can be identified by the iodoform test.

Write equations for the iodoform reaction with 2-butanol.

CH3CH2CHCH3

OH

NaOI+

OXIDATIONCH3CH2CCH3

O

+ + H2ONaI

+CH3CH2CCH3

O

3 NaOIIODINATION

NaOH+CH3CH2CCI3

O

CLEAVAGE

+CHI3 Na-O-C-CH2-CH3

O+

3 NaOH+CH3CH2CCI3

O

30

PROBLEMS Using the Lucas test, oxidation with Jones reagent, and the

iodoform test, show how a student could distinguish between 2-pentanol and 3-pentanol.

OH

2-pentanol OH

3-pentanol

Lucas testZnCl2, con. HCl

Jones reagentCrO3 in H2SO4

IodoformNaOI

Cl

3-chloropentaneCl

2-chloropentane

O

2-pentanone O

3-pentanone

O- Na+

O CHI3+no iodoformprecipitate

31

PROBLEMS Write equations showing how the following transformations can be carried


OH OH

CH3

O

CH3MgBr

H3O+

1

2

Cr+6, H+

CH2 CH2 OH CH2OH

C OH

O

1 LiAlH4

2 H3O+

CH CH2

hot KMnO4

POCl3,

N

H2O KMnO4

or

32

SOLUBILITY AND BOILING POINTS OF ALCOHOLS Alcohols are the first organics we have studied that are polar. Alkanes, alkenes, alkynes and arenes are all hydrocarbons with

non polar bonds. Even alkyl halides are only weakly polar. These compounds all exhibit characteristically low boiling points are are insoluble in polar solvent like water.

Describe the trend in solubility of alcohols re: MW and branching.

Explain the unusually high solubility of a polyol such as 1,4-butanediol.Explain why cyclohexanol boils at 160 °C and cyclohexene at 83 °C.

33

PHENOLS

Phenols have the hydroxyl (-OH) functional group of an alcohol directly bonded to an aromatic ring; a fully conjugated ring; a ring with a complete pattern of alternating single and double bonds.

cyclohexyl alcohol phenol

isobuty alcohol -naphthol

Despite their similar appearance, the chemistry of phenols is different than that of alcohols.

OH OH

alcohols phenols

OH

CH3CHCH2OH

CH3

34

PHENOLS

Phenol is manufactured in two steps from benzene. Benzene undergoes Electrophilic Aromatic Substitution (EAS) with

H2SO4 and SO3 (a mixture called ‘oleum’) producing benzenesulfonic acid.

The sulfonic acid group is replaced by the –OH group by reacting with NaOH at high temperature (300°C).

Write equations showing the preparation of phenol from benzene.

benzenesulfonic acid

S

O

O

OH1 NaOH

2 H3O+

300°COH

phenol

The hydroxyl group donates electron density to the aromatic ring by resonance, making phenols reactive toward EAS.

SO3, H2SO4

40 ºC

35

REACTIONS OF PHENOLS

Phenols have the same hydroxyl (-OH) functional group as alcohols but, while alcohols and H2O are essentially neutral (pKa = 16), phenols are weak acids (pKa = 10).

..

..+ Na+ OH-

O H

phenol

....:O-

Na+

+ H2O

pKa = 9.9sodium phenoxide

Phenols react 100% with aqueous NaOH but not with NaHCO3 (pKb = 7.6). This difference is often used differentiate phenols from other stronger organic acids, such as carboxylic acids (pKa = 5).

Carboxylic acids react 100% with both NaOH and NaHCO3.

..

..+

O H

Na+ HCO3-

pKb = 7.6

....:O-

Na+

+ H2CO3

pKeq = 9.9 + 7.6 -14 = 3.5Extent of reaction = 0%

pKeq = 9.9 -1.74 -14 = -5.8Extent of reaction = 100%

pKb = 4.1

36


Phenols are 106 times more acidic than alcohols (pKa 10 vs. pKa 16) because the conjugate base, phenoxide ion (pKb = 4), is 106 times more stable than the alkoxide (pKb = -2).

In the alkoxide, the negative charge resides on only one atom, the oxygen.

In the phenoxide, the negative charge is ‘delocalized’ (spread out) over 4 atoms by resonance throughout the fully conjugated system of the aromatic ring.

O:.... O

..

....

O....

: O....

..

Sodium phenoxide has 4 different resonance structures.

37


Phenols become stronger acids when electron withdrawing groups are attached to the phenyl ring, e.g., -Br, -NO2, -CF3, etc.

Phenols become weaker acids when electron donating groups are attached to the phenyl ring, e.g., -CH3, -OCH3, -NH2, etc.

Draw these structures

HO NH2

......

HO NO2

..

..

HO NO2

O2N

O2N

..

..

38


Recall from our study of electrophilic aromatic substitution (EAS) reactions that electron donor groups make the ring a better Nu:- (electron donor); thus accelerating EAS reactions of aromatics.

These same ring activators (electron donors) will reduce the acidity of phenols by making conjugate base (phenoxide) less stable.

Ring deactivators are electron withdrawing groups. They increase the acidity of phenols by making phenoxides more stable.

NH2

OH

CH3OCH3 F

Cl

Br

I

C OH

O

C CH3

O

C H

O

NH C CH3

O

SO3H NO2

N+R3C N

COCH3

OH

Reactivity Reactivity

o- and p- directing

activators

o- and p- directing

deactivatorsdeactivators

m-directing

39

PROBLEM

Rank the following phenols in order of acidity, where 1 is most acidic.

Then name their conjugate bases.

HO CH3OH

OH

NO2C

O

OH

HO

Cl

m-cresolphenol

benzoic acidp-chlorophenol

m-nitrophenol

1 2345

O CH3O

O

NO2C

O

O

O

Cl

..

...... ..

..

..

..

..

: ::

:: :

m-cresoxide phenoxide benzoate p-chlorophenoxide m-nitrophenoxide

40

PROBLEM Write equations showing how the following transformation can be carried


O CH2CH2CH3

SO3H

H2SO4

SO3

OHNaOH

H3O+

1

2

300° C

NaOH O- Na+

acid/base

CH3CH2CH2I

SN2

sodium phenoxide

an etherphenyl n-propyl ether

benzenesulfonic acid

41

Do the practice problems in your purchased notes.

42

SN1, E1

SN2

E2

SN2

SN2

SN1

SN2

SN2

E2

E2

SN2

E2 (SN2)

E2

no reaction

no reaction

SN1, E1

1 four reactions of alcohols. 2 4 reactions of alcohols 1.substitution with halides to form alkyl...

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dehydration of alcohols

rapidly2 alcohols

benzyl alcohols

alcohol reactivity

lucas test

degree of substitution

alcohol oxygen

alkyl halideseven