Alcohols, Ethers and Phenols

ALCOHOLS, ETHERS AND PHENOLS E M RAO

ALCOHOLS Physical properties: Solubility: Alcohols are quite soluble in water as they can form hydrogen bond with water. But alcohols having equal to or more than 6 carbons per OH group are insoluble in water as they have predominating hydrophobic (hydrocarbon) chain. Boiling point: Alcohols have higher boiling points than hydrocarbons of similar molecular weight due to hydrogen bonding. For a given molecular formula, more the branching lesser is the boiling point.

0 0

3 2 2 2 3 3 20 0

3 2 3 3 3

CH CH CH CH OH 118 C CH CH(CH )CH OH 108 C

CH CH CH(OH)CH 100 C (CH ) COH 83 C

More the number of OH groups more is the boiling point. 0 03 2 2 2CH CH OH 97 C HOCH CH OH 198 C Methods of preparation: 1. Hydration of alkenes: Water can be added to alkenes in three different ways. Some alkenes give three different isomers in the three methods.

CH3-CH-CH=CH2

CH3dil. H2SO4

CH3

CH3-C-CH2-CH3

OH

Acid catalysed hydration:

CH3-CH-CH=CH2

CH31.Hg(OAc)2/H2O

2. NaBH4

CH3

CH3-CH-CH-CH3

OH

Oxymercuration-demercuration:

CH3-CH-CH=CH2

CH31. B2H6;THF2. H2O2;OH-

CH3-CH-CH2CH2OH

CH3

Hydroboration-oxidation

Acid catalyzed hydration: Mechanism:

CH3-CH=CH2H+

slowCH3-CH-CH3

+ H2O CH3-CH-CH3

OH2+

CH3-CH-CH3

OH-H+

Acids that have weakly nucleophilic conjugate bases are used in this reaction e.g. H2SO4, H3PO4 etc. Being weakly nucleophilic, HSO4- and H2PO4- offer little competition to H2O. In the case of H2SO4, small amount of ROSO3H may be formed. As soon as formed, it undergoes hydrolysis to form ROH because HSO4- is a very good leaving group. Reactivity of alkenes towards acid catalyzed hydration: As formation carbocation is the rate limiting step, the alkene which forms the more stable carbocation will be more reactive.

ALKENE Relative rate of reaction with H3O+

CH2=CH2 1 CH3-CH=CH2 61.6 10 (CH3)2C=CH2 112.5 10


Examples:

H+OH

H+OH

OH

1.

2.

and its enantiomer and its enantiomer

+

Oxymercuration-demercuration:

Mechanism:

(Q)CH3-CH-CH=CH2

CH3 Hg(OAc)2

Hg(OAc)

H2OCH3

CH2-CH-CH-CH2 -Hg(OAc)

OH2+

CH3CH2-CH-CH-CH2 -Hg(OAc)

OH

CH2-CH-CH-CH2-Hg(OAc)OH

NaBH4CH3

CH2-CH-CH-CHOH

1.

2. H2O as a nucleophile attacks the carbon that can form more stable carbocation (as in the case of cyclic haloniumion).

(cyclic mercurinium ion)CH

CH3H3C

+

Hg(OAc)

CH

CH3H3C

+

CH3

Cyclic mercurinium ion is similar to cyclic halonium ion. Like in the case of halogenations, there is no fully fledged carbocation in this reaction too. The net addition of H and OH is syn and anti both (although oxymercuration is anti addition, demercuration step is a mix of both syn and anti). If H2O is replaced with any other polar solvent, the product is obtained accordingly. Examples:

1.Hg(OAc)2/H2O2. NaBH4 OH

1.Hg(OAc)2/MeOH2. NaBH4 OCH3

OH 2. NaBH4

O1.Hg(OAc)2

1.

2.

3.

Hydroboration-oxidation:

Diborane in THF exists as the following complex.

O BF3+

THF can be replaced by diglyme, CH3OCH2CH2OCH2CH2OCH3. Mechanism:


CH2CH3CH

H BH2

CH3CH2-CH2BH2

Both H and BH2 are added from the same side i.e.syn addition

1.

Same reaction repeats until two more alkenes molecules are added. Finally we get tri-alkyl boride

B

CH2CH2CH3

H3CH2CH2C CH2CH2CH3Tri-alkyl boride

Boron attacks less hindered carbon because of the crowding in trialkyl boride. Trialkyl boride undergoes oxidation in the second step.

2. H2O2 + OH- H-O-O- + H2O

3. (CH3CH2CH2)3B + H-O-O- (CH3CH2CH2)3B O-OH

B

CH2CH2CH3

H3CH2CH2C CH2CH2CH3

O OH

4.

(CH3CH2CH2)2B(OCH2CH2CH3)

Step 4 involves migration of alkyl group from boron to oxygen. As is the casewith all migrations, here also it is simultaneous. Steps 3 and 4 repeat two more times until trialkyl borate is formed. Trialkyl borate undergoes hydrolysis to give 3 moles of alcohol and H3BO3.

5. (CH3CH2CH2O)3B + OH- 3 CH3CH2CH2OH + H3BO3 The net addition of H and OH is syn. Examples:

1.B2H6;THF2.H2O2, OH-

OH

1.


OH

2.

+

and its enantiomer

EXERCISE I: 1. OH 1.Hg(OAc)2

2. NaBH4

2.


3. H3O+

4.

1.B2D6;THF2.H2O2, OH-

5.

Et1.Hg(OAc)2 / PhOH2. NaBH4

6.

OCH3OH

H+


2. From the nucleophilic substitution reactions of alkyl halides:

CH3CH2BrNaOH

aq. acetoneCH3CH2OH

(CH3)3CBrH2O

Acetone(CH3)3COH

SN2

SN1

3. From epoxides: Epoxides are three membered cyclic ethers. They are highly reactive due to ring strain. Nucleophiles attack epoxides and thereby convert them into alcohols.

O

1.LiAlH42.H+

CH3CH2OH

O

1.LiAlH42.H+

CH3CHCH3

OH

O 2.H+CH3CH2CH2OH

OCH3CHCH2CH3

OH

1.CH3MgBr/dry ether

2.H+1.CH3MgBr/dry ether

Mechanism:

OCH3CH2OH

H-CH3CH2O-

H+

When epoxides undergo attack directly (without first being protonated) by a nucleophile, the attack occurs at less hindered carbon. But a protonated epoxide undergoes attack by a nucleophile at more substituted carbon. We will see the details about this later in epoxides. Epoxides react with LiAlH4, NaBH4, RLi, RMgX and even with R2CuLi.

O

1.(CH3)2CuLi

2.H2O

OH

2. By reduction of carbonyl compounds (aldehydes, ketones, carboxylic acids

and their derivatives): Alcohols can be prepared by nucleophilic addition reactions of aldehydes and ketones. They can also be prepared by the nucleophilic acyl substitution reactions of carboxylic acids and their derivatives. Let us see the mechanism of both the reactions first and then have a look at different reagents.

O(H)R

R Nu-O-

Nu

R R(H)

OH

Nu

R R(H)H+

rds

Nucleophilic addition reactions of aldehydes/ketones:

When the nucleophile is H- (LiAlH4 or NaBH4) or R- (RLi/RMgX), the product will be an alcohol. ,-unsaturated aldehydes and ketones behave differently toward nucleophile because the nucleophile may attack the C-C double bond in addition to carbonyl carbon. The reactions of ,-unsaturated aldehydes and ketones will be discussed in aldehydes and ketones chapter.

OL

R Nu-O-

Nu

R Lrds

Nucleophilic acyl substitution reactions of acid chlorides, anhydrides and esters:

ONu

RL - Cl / OR / OCOR


The above mechanism excludes carboxylic acids and amides because the mechanism in their case is slightly different due to the presence of acidic hydrogen. The mechanism is discussed later. Different reagents used for reduction of carbonyl compounds 1. LiAlH4 in dry ether followed by acidification.

LiAlH4 reduces aldehydes, ketones, carboxylic acids, amides, anhydrides, esters and acid halides. Alcohols are the products from all the above reductions except from amides. Amides gove amines with LiAlH4. LiAlH4, being a strong hydride donor, can only be used in a non-polar solvents like ethers.

1.LiAlH42.H+

O

H

R

R-CH2-OH

1.LiAlH42.H+

O

R

R

R-CHR-OH

1.LiAlH4R-C-Cl

O

R-C-H

O

The aldehyde further gets reduced and ultimately gives alcohol RCH2OH.

1.LiAlH4O

R-C-H + R'O-O

The aldehyde further gets reduced and the ester ultimately gives two alcohols RCH2OH and R'OH.

R-C-OR'

R-C-H

O

The aldehyde further gets reduced and ultimately gives alcohol RCH2OH.

O

R-C-OHH-

O

R-C-O-H- O

R-C-O-Al+2

H

1.LiAlH42.H+

O

R-C-NH2

R-C-H

NH

R-CH2-NH2

Mechanism:

O

R-C-NH2H-

O

R-C-NH-H- O

R-C-NH-Al+2

Himine

H-NH-

R-C-H

H

H+R-CH2-NH2

1.LiAlH4 R-C-H + R'-C-O-O

Both the carboxylate ion (R'COO-)and the aldehyde (RCHO) further get reduced and the anhydride ultimately gives two alcohols RCH2OH and R'CH2OH.

R-C-O-C-R'

OO O

2. NaBH4 in C2H5OH

NaBH4 reduces aldehydes, ketones and acid halides only. It does not reduce esters, anhydrides, amides and carboxylic acids. And the mechanism is similar to that of LiAlH4. NaBH4, being a very weak hydride donor (due to very high covalent character of B-H bond), can directly be used in polar solvents.

Organometallic compounds: RLi, RMgX and R2CuLi are the commonly used organometallic compounds. Out of these, RLi is the strongest R- donor and R2CuLi the least. Observe the following table.


C-M bond Difference in electronegativity Percent ionic character* C-Li 2.5-1 = 1.5 60 C-Mg 2.5-1.2 = 1.3 52 C-Zn 2.5-1.6 = 0.9 36 C-Cu 2.5-1.9 = 0.6 24

* C MC

E -EPercent ioniccharacter = 100

E

3. RLi in dry ether followed by acidification.

Organo lithium compounds can reduce aldehydes, ketones, esters, anhydrides, acid chlorides into alcohols. They convert carboxylic acids into ketones only. They do not react with amides (at the most acid-base reaction may happen). Mechanism of conversion of acid into ketone is given below.

OLi

R-C-OLi

R

H+O

R-C-OLi + RH

O

R-C-OHR- R-

OH

R-C-OH

R

O

R-C-R

4. RMgX in dry ether followed by acidification.

Grignard reagents convert aldehydes, ketones, acid halides and esters into alcohols. They do not react with carboxylic acids and amides (at the most acid-base reaction may happen in both the cases).

5. R2CuLi/ R2Cd

Gilmans reagent (R2CuLi) and R2Cd can only reduce acid chlorides. They do not react with any other carbonyl compounds including aldehydes and ketones.

or R2Cd

O

R-C-ClR2CuLi

O

R-C-R

6. H2/Ni or Pt

Catalytic hydrogenation converts aldehydes and ketones into alcohols. It converts esters into alcohols only under drastic conditions. H2/Ni does not react with carboxylic acids, amides, acid chlorides and anhydrides. H2 in the presence of Lindlars catalyst (Pd-BaSO4) converts acid chlorides into aldehydes.

O

H

R

R-CH2-OH

O

R

R

R-CHR-OH

H2/Ni

H2/Ni

O

R'O

R

R-CH2-OH + R'OHH2/Ni

O

Cl

R

O

H

RH2

Pd-BaSO4

2000C/10 atm

7.B2H6/THF followed by hydrolysis

Diborane converts aldehydes, ketones and carboxylic acids into alcohols. The mechanism of diborane reduction is similar to that of hydroboration oxidation of alkenes.


O

H

R

R-CH2-OH

O

R

R

R-CHR-OH

1. B2H6/THF

O

HO

R

R-CH2-OH

2. NaOH

1. B2H6/THF

2. NaOH

1. B2H6/THF

2. NaOH

EXERCISE-II 1. Show how do you prepare n-butyl alcohol using C2H5MgBr? 2.

2.H+1.LiAlH4/THF

O

A + B

3.

1.CH3MgBr/THFO

O

A2.H+

4.

1.CH3MgBr/THF

O

O

A2.H+

5.

2.H+1.LiAlH4/THF

O-C-CH3H3CO-C

O O

A + B + C

6.

2.H+1.LiAlH4/THF

O

O C-OEtA + B

Chemical reactions of alcohols 1. Conversion of alcohols into alkyl halides

Alcohols have poor leaving group i.e. OH-. As result, alcohols do not undergo direct nucleophilic substitution reactions under normal conditions. But they can be made to undergo nucleophilic substitution reactions by converting OH- into a better leaving group. Reagents like HCl, PCl3, PCl5 and SOCl2 can convert alcohols into alkyl halides,

ROH + NaCl No reaction

ROHNaCl

H2SO4RCl + H2O

ROH + HCl RCl + H2O

ROH + NaBr No reaction

ROHNaBr

H2SO4RBr + H2O

ROH + HBr RBr + H2O

ROH + NaI No reaction

ROHNaI

H2SO4No reaction

ROH + HI RI + H2O

ROHNaI

H3PO4RI + H2O

Note: Protonation makes OH- into a better leaving group i.e. H2O. As a result H+ is necessary in all the above reactions. H2SO4, being a mild oxidizing agent, oxidizes I- to I2. As a result, ROH cannot be converted into RI with NaI/H2SO4.


I. HX. Observe the following examples. MECHANISM

R-OHH+

R-OH2+ X- RX + H2O SN2 mechanism if the R is primary.

R+ + H2O

R+ + X- RX

SN1 if the R is secondary or tertiary.

Some important points:

1. Reactivity of alcohols toward HX 30 > 20 > 10. 2. Reactivity of different halides HF < HCl < HBr < HI

20 or 10 alcohols react with HCl to give very low yield. In order to increase the yield, a Lewis acid (generally ZnCl2) is added. HCl + ZnCl2 mixture is called as Lucas reagent. ZnCl2 complexes with the alcohol making OH- a better leaving group.

R-OH

ZnCl2

+

-

II. PCl3 / PBr3 / P + I2 Phosphorous trihalides convert alcohols into alkyl halides. Observe the following mechanism.

R-OH +P

Cl ClCl

P Cl

Cl

R-OH

+

Cl-

R-Cl + PCl OH

Cl

Second step is SN2 mechanism when R is primary or secondary. Phosphorous halides are not used for 30 alcohols. P(OH)Cl2 can convert two more moles of alcohols into RCl.

3 ROH + PCl3 3 RCl +P

HO OHOH

P H

OH

O

OH

III. PCl5: PCl5 also converts alcohols into alkyl halides and the mechanism is similar to that of PCl3. But one mole of PCl5 converts only one mole of alcohol into alkyl halide. ROH + PCl5 RCl + POCl3 + HCl IV. SOCl2 (SNi mechanism) SOCl2 converts alcohol into alkyl halide. The mechanism of this reaction depends on presence or absence of a bulky base such as pyridine.


OHH

R

R1 +OS

ClClO

H

R

R1 S

O

Cl + HCl

Substitution Nuceophilic internal,where nucleophile is part of theleaving group. Such a front sideattack leads to retention of configuration.

In the absence of pyridine:

In the presence of pyridine:

OHH

R

R1 +OS

ClClO

H

R

R1 S

O

Cl +NH

+

Cl-A normal SN2 reactionwhich leads to inversion ofconfiguration.

Cl + SO2H

R

R1

R + SO2H

Cl

R1

SOCl2 readily reacts with a solvent like water giving SO2 and HCl. As a result, a solvent like ether is taken in the above reaction. HCl gets evaporated in such a solvent and the internal attack becomes major. A bulky base like pyridine converts HCl into a salt, pyridinium chloride. And as a result, chloride ion is retained in the solution. Yield of this reaction is very high due to the formation of a gas SO2. V. Alkyl tosylates: Another method is converting alcohol into tosylate by treating it with tosyl chloride (TsCl).

H3C S

O

OCl

Toluene sulfonyl chloride (TsCl)

ROH + TsClpyridine ROTs + HCl

Mechanism:

H3C S

O

OCl

ROH

H3C S

O

OORH

+ NH3C S

O

OOR

ROTs

As TsO- is a very good leaving group, ROTs undergoes nucleophilic substitution reactions very easily. Another advantage of ROTs is that it is insoluble in polar media. When the reaction medium is polar, ROTs gets precipitated and the precipitate can be treated with different nucleophiles as illustrated below.


ROTsCl-

RCl+ TsO-

ROTsI-

RI+ TsO-

ROTsCN-

RCN+ TsO-

ROTsN3- RN3+ TsO-

ROTsCH3COO- R-O-C-CH3 + TsO-

O

ROTsCH3O- ROCH3+ TsO-

Illustration:1

OH

HBrPBr3 AB(i)

OH HBr CPBr3D(ii)

OH

HCl EPCl3F

G

(iii)

Write the structures of all the unknowns emphasising on stereochemistry whereever appropriate.

SOCl2

Br

Br

BrBr

ClCl Cl

A B C D E F G

Answers:

+

2. Fischer esterification and ester hydrolysis: Alcohols react with carboxylic acids in acidic medium to form esters. The reaction is reversible and as a result, under suitable conditions, ester can be hydrolysed back to carboxylic acid and alcohol. Although ester cant be formed in basic medium, it can be hydrolysed in basic medium which is called saponification. Mechanism in acidic medium:


R-C-OH + R'H

OH+

R-C-OR' + H2O

O

R-C-OH

OH+

R-C-OH

OH+

R'OH

rdsR-C-OH

OH

R'OH+

R-C-OH

OH

R'O

OH

R'O

+R-C-OR'

OH+

R-C-OR' + H+

O

Esterification:

Rate = kR-C-OH

OH+

= k keq [RCOOH] [H+] [R'OH]

= k' [RCOOH] where k' = k keq [H+] [R'OH]

R-C-OH2

During esterification, alcohol is taken as a solvent to drive the equilibrium in the forward direction. As solvent is generally taken in excess, its concentration hardly changes. The role of acid in esterification is catalyst. Its concentration also hardly changes. As a result, acid catalysed esterification is a pseudo-first order reaction.

R-C-OR' + H2O

OH+

R-C-OH + R'OH

O

R-C-OR'

OH+

R-C-OR'

OH+

rdsR-C-OR'

OH

HOH+

R-C-OR'

OH

OH

R-C-OH-R'

OH

OH

+R-C-OH

OH+

R-C-OH + H+

O

Ester Hydrolysis (AAC2):

Rate = kR-C-OR'

OH+

= k keq [RCOOR'] [H+] [H2O]

= k' [RCOOR'] where k' = k keq [H+] [H2O]

HOH + R'OH

During ester hydrolysis, water is taken as a solvent to drive the equilibrium in the forward

direction. As solvent is generally taken in excess, its concentration hardly changes. The role of acid in ester hydrolysis is catalyst. Its concentration also hardly changes. As a result, acid catalysed ester hydrolysis is a pseudo-first order reaction.

Mechanism in basic medium:

R-C-OH

OR'OHNaOH R-C-O

-

OR'OH

or even R'O-No reaction

Partial positive charge on the carbonyl carbon in carboxylate ion is too small toreact even with alkoxide ion. As a result, one can't prepare ester in basic medium.

But ester can be hydrolysed in basic medium which is called as saponification. Mechanism of ester hydrolysis in basic medium (BAC2):


R-C-OH + R'-

O

R-C-OR'

O

R-C-OR'

O-

OH

OH-

acid base reactionwhich is highly favourable

R-C-O- + R'H

O

first step is rds.

Rate = k [RCOOR'] [NaOH] Intramolecular ester formation:

CH3-CH-C-OH

OHOH+

O

O

O

O

CH3-CH-CH2-C-OH

OH O

-hydroxy ester

-hydroxy ester

CH3-CH=CH-C-OH

O

CH3-CH-CH2-CH2-C-OH

OH O

H+

H+

-hydroxy ester

OO

Intra molecular esterification is difficultbecause of the ring strain. Dehydration is difficult because the carbocationat position is unstable.

Intra molecular esterification is difficultbecause of the ring strain. Dehydration is favorable because of the stability of the alkene.

Intra molecular esterification is favourable.

a lactone

3. Oxidation of alcohols:

R-CH2-OH

Oxidation stateof carbon = -1

[O]R-C-H

O

Oxidation stateof carbon = +1

[O]R-C-OH

O

Oxidation stateof carbon = +2

DIFFERENT REAGENTS USED IN THE OXIDATION OF ALCOHOLS: A. Chromic acid Chromic acid is prepared by dissolving either chromium (VI) oxide or potassium dichromate in aqueous sulfuric acid.

2 43 2 2 4

2 4 22 2 7 2 2 7 2 4

H SOCrO + H O H CrOChromium Chromicacid(VI)oxide

H SO H OK Cr O H Cr O 2H CrOChromicacid

Chromic acid converts primary alcohols into carboxylic acids, secondary alcohols into ketones. It does not react with tertiary alcohols. Thus, the prerequisite for the oxidation of an alcohol is at least one H on the carbon bearing the OH group. A solution of chromic acid

Mechanism of chromic acid oxidation:


OH

H

+ HO-Cr-OH

O

O

fast and

reversibleH

O

O

O-Cr-OH+ H2O

H

O

O

O-Cr-OH

H2O

slow and ratedetermining O + H3O

+ + HCrO3-

Chromium (IV) then participates in further oxidations by a similar mechanism and eventually is transformed to Cr (III) which is green in color. An aldehyde is oxidized further to carboxylic acid through its hydrate form.

O

H2O

+ H3O+ + HCrO3-R-C-H C OHOH

H

R

An aldehyde hydrate

O

O

HO-Cr-OH

COH

H

R

O

O-Cr-OH

O

RCOOH

We will study about aldehyde hydrates in the aldehyde and ketones chapter.

B. KMnO4: KMnO4 is similar to chromic acid in every respect. It oxidises primary alcohols to carboxylic acids, secondary alcohols to ketones and does not react with tertiary alcohol. But when KMnO4 is used in acidic medium at high temperature, the tertiary alcohol may first convert into alkene and the alkene is further oxidized.

O

C CH3OH

CH3

H3CH+/ KMnO4

CH3-C-CH3 + CO2

C. PCC, PDC and CrO3/Pyridine/Cold (Collins reagent):

N

H

+ CrO3Cl-

Pyridinium chloro chromate

N

H

+2

Cr2O7

Pyridinium dichromate These three reagents convert primary alcohols into aldehydes, secondary alcohols into ketones and they do not react with tertiary alcohols. PCC oxidations are generally carried out in aprotic solvents such as CH2Cl2. PCC has no effect on carbon-carbon double bonds

OHPCC

CH2Cl2 H

O

PCC does not oxidize aldehydes further because the PCC is not used in water but rather in an organic solvent, CH2Cl2. Without water, the aldehyde cant be in equilibrium with its hydrate. Recall that only the hydrate of the aldehyde is susceptible to further oxidation not the aldehyde itself.


C. Cu/3500C

Copper at high temperature oxidises primary alcohols into aldehydes, secondary alcohols into ketones. Tertiary alcohols get dehydrated by copper at 3500C.

CH2H3C

H3C

CH3CH2OHCu/3500C

CH3-C-H

O

CH3CHOHCH3Cu/3500C

CH3-C-CH3

O

C CH3CH3

OH

H3CCu/3500C

D. MnO2 MnO2 oxidises only allyl and benzyl alcohols. If they are primary, aldehydes are the final products. If they are secondary, ketones are the final products.

OHMnO2

H

O

CH2OH CHO

MnO2

Exercise-III: 1.

O1.CH3MgBr

2. H2OA (Mixture of two compounds)

AH+

B (Mixture of two compounds)

Describe A and B. 2.

CO

OC2H5C2H5ODiethylcarbonate

1.ex CH3MgBr2. H2O

A + B

If both A and B are alcohols, give their structures. 3.

A (C4H8O2)1.ex CH3MgBr

2. H2OB

If A is an ester and B is the only alcohol produced in the reaction, find out the structure

4. Starting with butane, synthesize the following two compounds . (A) CH3CH2CH2CH2D (B) CH3CHDCH2CH3

5.


OCH3

O

HO

O

1.ex CH3MgBr2. H2O OH

HO

OWhat is the number of moles of CH3MgBr required for the above conversion?

6. Using ethyl bromide as the starting compound, synthesize the following alcohols in not more than three steps. (A) 2-Butanol (B) 1-Propanol (C) 1-Butanol

7.

Benzyl methyl ketone

1.LiAlH42.NH4Cl3.PCl34.KO-t-Bu

A

B1.H2/Ni

2. H+/ 8.

3-Ethyl-3-pentanol1. i-PrMgBr

2. Br

9.

1-Hexen-3-ol1.NaH

2.S

O

O

OMeMeO

10.

Ph O

O1.PhMgCl

2. H+

1.LiAlH42.H+

11.

2-Butanol

1.HBr2. LDA

3.BH3/THF4.H2O2/OH-

12. 1.LiAlH4

2.H+

OO

OMe

OPh

O

1.NaBH42.H+

H2/Pt 13.


1.LiAlH42.H+

1.NaBH42.H+

OO

1.CH3MgBr2.H+

14.

1.LiAlH4

2.H+

1.NaBH42.H+

HN

O

1.H2/Pt

15. 1.ex CH3MgBr2. H2O

O OH

OH

O OH OH

OH

O

What is the number of moles of Grignard reagent used in the above reaction? 16.

CO2EtO

1.LiAlH42.H2O

A + B

If A and B are isomers, what is the relation between them?

POLYHYDROXY COMPOUNDS

Methods of preparation 1. Reaction of Baeyers reagent with alkenes: Alkenes give vicinal diols with cold alkaline KMnO4 (Baeyers reagent). Both hydroxyl groups are added from the same side (syn addition). Examples:


CH3

HH3C

Hcold KMnO4

CH3

CH3

H OH

HO H

CH3

CH3

HO H

H OH+

cold KMnO4 OH

OH

+ OH

OH

enantiomers

enantiomers

cold KMnO4 OH

OH

1.

2.

3.

2. Reaction of alkenes with OsO4/H2O; NaHSO3 Alkenes with OsO4/H2O; NaHSO3 give vicinal diols. Similar to the previous case, here also the hydroxyl groups are added from the same side (syn addition). 3. Reaction of alkenes first with peroxy acetic acid followed by treatment with H+ or OH-. Alkenes react with peroxy carboxylic acids to give epoxides which then react with H+ or OH- to give vicinal diols. Overall addition of hydroxyl groups is anti. Examples

CH3

HH3C

H1. RCOOOH

CH3

CH3

H OH

H OH

1.2. H+

OH

OH

+ OH

OH

enantiomers

OH

OH

2.

3.

1. RCOOOH2. H+

1. RCOOOH2. H+

OH

OH

+

enantiomers Chemical Reactions

1. Pinacol-pinacolone rearrangement When vicinal diols are treated with H+, aldehydes or ketones are obtained.


H3C-C-C-CH3

O OHH

C CH3H3

H+H3C-C-C-CH3

OH

C CH3H3

+H3C-C-C-CH3

O CH3H

CH3

+

+H3C-C-C-CH3

O CH3H

CH3

H3C-C-C-CH3

O CH3

CH3

Pinacol

Pinacolone

Methyl shift

Formation of the first carbocation is the rate determining step. It is surprising, at first

look, that a 30 carbocation is getting rearranged. But see the relative stability of the carbocations:

CH3-O-CH2+ > Ph-CH2

+Ph2CH > (CH3)3C >

+ +

It is often confusing in this reaction as to which group migrates. Whether it depends upon the migratory aptitude or the stability of the carbocation that is obtained after the migration. It is found that in some examples the first factor (migratory aptitude) predominates, in other examples, the second factor (carbocation stability) predominates.

Migratory aptitude

H > Ph > R(30) > R(20) > R(10) > CH3

OCH3 NO2

OCH3> >

Phenyl ring migration is an example of intramolecular electrophilic substitution. Presence of an electron donating group at para position makes benzene ring a better migrating group. Presence of electron withdrawing group does the opposite. But presence of any group at the ortho group makes migration of benzene ring difficult because of steric hindrance.

Stereochemistry of rearrangements: The configuration of the migrating group is retained during migration because the migrating group never becomes completely free. The configuration of the migratory terminus (the carbon atom to which migration takes place) is found to be predominantly inverted in some cases and completely inverted in other cases. The configuration is found to be completely inverted in the case of amino alcohols and cyclic diols. This suggests the formation of ion pair in the rate determining step.

Ph

HO Ph

HH3CPh

H3C HPh

HO Ph

NH2

HH3C NaNO2HCl N2

+

OPh

Examples:


OHCH3

CH3

OH

H+

O

CH3

CH32.

C C CH3

CH3

Br

CH3

OH

H3CAgNO3 C C CH3

CH3

CH3O

H3C

C C CH3

CH3

NH2

CH3

OH

H3CNaNO2 C C CH3

CH3

CH3O

H3CHCl

3.

4.

C C CH3

Ph

OH

Ph

OH

PhH+

C C Ph

Ph

PhO

H3C1.

cold KMnO4 H+

CHO

5.

cold KMnO4 H+

O

6.

O

+

OHHO

H+

H O

7.


Retropinacol rearrangement

O

O

H+

OH

O

+

H2O

OH

O

OH

tautomerism

OH

OH

tautomerism

O

O

OH

Major

O

H+

OH

+

OH

+

OH

1.

2.

2. Reaction of vicinal diols with HIO4 (periodic oxidation) Vicinal diols are oxidized by HIO4. And in the process HIO4 is reduced to HIO3. If AgNO3 is

added to after the reaction, a white precipitate of AgIO3 is obtained.

C

C

OH

OH

HIO4C

C

O

OI O

O

O_

C

C

O

O+ + IO3-

Examples

OH

OH

OH

OH

OH

OH

OHOH

HIO4

HIO4

H

O

O

H

H

O

O

H

1.

2.

=

=

OH

OH

HIO4 No reaction

In chair form, dihedral angle between two equatorial bonds is 600. Dihedral angle between

one equatorial form and one axial form is also 600. As a result, there is no much difference between the reactivity of cis-1,2-cyclohexanol and its trans isomer. Trans isomer is slightly less


reactive because some of it exists in another conformation in which the two hydroxyl groups are anti to each other.

OH

OH

OH

OH

OH

HIO4 H

O

O

H3

(H3C)3C

OH

OH

4.

(H3C)3C OH

(H3C)3C

(H3C)3C

HIO4 no reaction

C(CH3)3

=

=

C

C

O

OH

HIO4C

C

O

O

HO+

5.

C

C

O

OHIO4

C

C

O

O

OH

OH

+6.

C

C

C

O

OH

HIO4

OHC

C

C

O

O

O

O+

+7.

CH2

C

C

OH

HIO4

OH

8 No reaction

Now it is easily understood that an alternate reagent for O3; Zn/H+ can be cold alkaline

KMnO4 followed by HIO4. Cold KMnO4 followed by HIO4 is called as Limaeux reagent.

Exercise-IV

C C CH3

Ph

OH

Ph

OH

PhH+/

1.

OHCH3

CH3

OH

2. H+/

C C CH3

CH3

Br

CH3

OH

H3CAgNO33.

C C CH3

CH3

NH2

CH3

OH

H3CNaNO2HCl

4.

cold KMnO45. H

+/

cold KMnO46. H+/

O

OHH+

7.

O

O

H3O+8.


O

H3O+9.

H+/OH

HO10.

Exercise-V 1. When one mole of each of the following compounds is treated with HIO4, what will the products be, and how many moles of HIO4 will be consumed? (A) CH3CHOHCH2OH (B) CH3CHOHCHO (C) CH2OHCHOHCH2OCH3 (D) CH2OHCH(OCH3)CH2OH (E) cis-1,2-Cyclopentanediol (F) CH2OH(CHOH)3CHO (G) CH2OH(CHOH)3CH2OH 2. Assign the structures to each of the following compounds.

4 3 3

4 2 4

4 2 4

4

4

4 2

4

A+1mol HIO CH COCH +HCHO

B+1mol HIO OHC(CH ) CHOC+1mol HIO HOOC(CH ) CHO

D+1mol HIO 2HOOC-CHOE+3mol HIO 2HCOOH+2HCHO

F+3mol HIO 2HCOOH+HCHO+CO

G+5mol HIO 5HCOOH+HCHO

3.

OH

OH

HIO4 1.LiAlH42.H2O

A B

A is pure while B is a mixture. How many isomers does B contain? 4.

HO

OH

ex.Cu/

HIO 4

PCCCH2 Cl2

H +/

ex.C

rO 3/H

+

E

A+B

F

C

D

1.

Exercise-VI Convert the following.

CH2OHOH

OH

2.


C6H5OH

C6H5OH

OH

+ its enantiomer

C6H5OH

OH

+ its enantiomer

A

B

3. O H3C OH

O

4.

Br

CH3 CHO

5.

HH

O

O

Without using ozone.

6. H

O

O

EXERCISE-VII Write the structures of all the unknowns.

O

O

O

1.AlCl3 A Zn-HgHCl2.H2O

B PCl3 1.AlCl32.H2O

Zn-HgHCl

Se

C D E F+

1.LAH2.H2O

GPPA

H J

I K N

P

L M O

1.LAH2.H2 O

NaBH 4/

C 2H 5OH

C6H6/PPAC6H6/PPA

Zn-HgHCl

C6H6/PPA

NaBH4/C2H5OH

C6H6/PPA

Se/

Se/

NaBH4/C2H5OH

Note: PPA stands for polyphosphoric acid, which is a source of H+.

ETHERS

Physical properties


Ethers and alcohols have comparable solubility in water as both of them can form hydrogen bonding with water. But boiling points of ethers are low when compared to those of alcohols as ethers cant form hydrogen bonds among themselves.

CH3CH2OCH2CH3 Diethyl ether

CH3CH2CH2CH2OH 1-Butanol

Boiling point: 35C 117C Solubility in water: 7.5 g/100 mL 9 g/100 mL

Methods of preparation 1.Williamsons synthesis:

R-L + RO- R-O-R + L-

L - Cl, Br, I, OSO3CH3,

OTs, N2+ etc.

As RO- is a strong base, best yields are obtained when R in R-L is CH3 or 10. With 20 and 30 substrates, RO- gives predominantly elimination products.

Examples:

CH3CH2ICH3ONaCH3OH

CH3CH2OCH3 + NaI1.

CH3-O-S-O-CH3CH3ONaCH3OH

CH3OCH3 +2.O

OCH3-O-S-O-Na

O

O

3. CH3CH2OH + CH2-N_ +

diazomethaneCH3CH2O- + CH3-N NN

+CH3CH2OCH3 + N2

4. HO-CH2-CH2-CH2-CH2BrNaOH

O + NaBr

acid-basereaction

2. Alkoxymercuration-demercuration: This method is a slight variation of oxymercuration-demercuration. In oxymercuration-

demercuration, water is taken as solvent in the first step. In alkoxymercuration-demercuration, alcohol is taken as solvent. Everything else is same. In alkoxymercuration-demercuration, there is no formation of fully fledged carbocation and as a result there are no rearrangements. But the addition is according to Markonikovs rule.

Examples:

1. (CH3)2C=CH21. Hg(OAc)2/CH3OH

2. NaBH4(CH3)2C-CH3

OCH3

2. HO-CH2-CH2-CH2-CH=CH21. Hg(OAc)22. NaBH4

O

CH3

3. Intermolecular dehydration of alcohols: When alcohols are heated at temperatures lower than those required for intra-molecular

dehydration, they give ethers. This method best works for methyl and 10 alcohols. 20 and 30 alcohols give alkenes rather.


CH3CH2OHH+

17O0CCH2=CH2

CH3CH2OHH+

14O0CCH3CH2OCH2CH3

Elimination

Substitution

Chemical reactions: Ethers do not contain acidic hydrogen. As a result, they do not react with Na, NaOH or any

other base. As such they do not undergo nucleophilic substitution reactions because RO- is not a good leaving group. But they do undergo nucleophilic substitution reactions under acidic conditions.

1.Ether cleavage reactions:

R-O-R dil.H+

R-O-RH

+

R-O-RH

+

H2O

ROH + ROH

R+ + ROH 2 ROHSN1

SN2

1. With dil. acids

R-O-R R-O-RH

+

R-O-RH

+

Br-

RBr + ROH

R+ + ROHSN1

SN2

conc. HBr

Br-RBr + ROH

2. With conc. acids

Whether it is SN1 or SN2, depends on the nature of the two alkyl groups. It is

experimentally found that when one of the alkyl groups is 30, benzylic or allylic (or any other carbon chain that can form stable carbocation), the mechanism follows SN1 pathway. Otherwise it is SN2. Examples:

1. CH3CH2OCH3

dil. acid

conc. HBr

CH3CH2OH + CH3OH

CH3CH2OH + CH3Br

ex. HBr CH3CH2Br + CH3Br

2. CH3CHOCH3

dil. acid

conc. HBr

CH3CH(OH)CH3 + CH3OH

CH3CH(OH)CH3 + CH3Br

ex. HBr CH3CH(Br)CH3 + CH3Br

CH3

SN2

SN2


3. (CH3)3C-O-CH3

conc. HI

ex. HI

(CH3)3C-I + CH3OH

(CH3)3C-I + CH3ISN1

4. O

conc. HI

ex. HI

No reaction

No reaction

5. O NO2conc.HI

OH

I

NO2

+

ex. HI

OH

I

NO2

+

ArSN

O ex. HI

conc. HI No reaction

No reaction

6.

7. O CH=CH2dil. H+

O CH-CH3

OH

a hemiacetal

OH

+ CH3CHO

In this example, first water gets added to the double bond as it is highly reactive.The resultant hemiacetal, being unstable, decomposes.

O CH-CH3

OHH

+

8. CH3-OCH2-CH2-OCH3ex. HI

2CH3I + CH3CH2I

In this example, the two carbon chain is first converted into I-CH2-CH2-I which,being unstable, decomposes into ethene. Ethene further reacts with HI.

2.Claisen rearrangement:

When allyl phenyl ethers or allyl vinyl ethers are heated, they undergo rearrangement. Examples:


O

O O OH1.

OO

O

OH O

2.

O

OO

Allyl-vinyl ether

3.

EPOXIDES Epoxides are three membered cyclic ethers.

O OEthylene oxide(or) oxirane(or) epoxy ethane

Propylene oxide(or) 2-methyloxirane(or) 1,2-epoxypropane

Methods of preparation: 1.Reaction of alkenes with peroxy acids: Alkenes, when treated with an organic peroxy acids, yield epoxides.

CH3

HH

H3C RCO3HCH3

HH

H3C

C

O H

OOR

O

CH3H3C

+

RCOOH

CH2Cl2


All the bond fissions and bond formations are simultaneous. As a result, there is no scope for bond rotation in alkene. Consequently, the reaction is stereospecific. Cis alkenes give cis epoxides and trans alkenes give trans epoxides exclusively. 2. Reactions of halohydrins with bulky bases: Halohydrins, when treated with bulky bases, yield epoxides. Bulky bases are used to avoid direct replacement of halogen.

C C

BrHOpyridine O

pyridineO

H3C

H

H

CH3

CH3

CH3

H OH

H Br

(S,R)(S,S)

Chemical reactions: Unlike open chain ethers, epoxides are highly reactive toward nucleophiles. This is because of the ring strain.

ONu- -O-CH2-CH2-Nu

H+HO-CH2-CH2-Nu

In the case of unsymmetrical epoxides, two factors may influence the preferred site of nucleophilic attack. 1. Magnitude of partial positive charge 2. Steric hindrance. Which of the two predominates depends on the conditions. Nucleophile attacks epoxide under two conditions: in acidic medium and in basic medium. In acidic medium (SN1 like), epoxide is first protonated. Nucleophile attacks in the second step. Under these conditions, it is invariably the first factor (magnitude of partial positive charge) which predominates. The nucleophile preferably attacks the carbon which can stabilize the carbocation better. The important point here is that there is no formation of fully fledged carbocation and as a result there are no rearrangements.

O

H+

OH

+CH3OH CH3-CH-CH2OH

OCH3

In basic medium (SN2 like), the nucleophile directly attacks the epoxide and the protonation occurs in the second step. Under these conditions, it is invariably the second factor (steric hindrance) which predominates. The nucleophile preferably attacks the less hindered carbon.

O

H+CH3O- CH3-CH-CH-OCH3 CH3-CH-CH-OCH3

O- OH

Examples:


O

1. CN-

2. H3O+CH3-CH-CH-CN

OH

O

OH-CH3-CH-CH-OH

OH1818

O

H+CH3-CH-CH-OH

OH

18

18

H2O

O

1.AlCl3 CH3-CH-CH-OH

CH3

2.CH3MgBr3.H2O

OCH3-CH-CH-CH3

OH1.CH3MgBr2.H2O

1.

2.

3.

4.

5.

H2O

PHENOLS Compounds in which a hydroxyl group is bonded to an aromatic ring are called phenols. The chemical behavior of phenols is different in some respects from that of the alcohols. A corresponding difference in reactivity was observed in comparing aryl halides, such as bromobenzene, with alkyl halides, such as butyl bromide and tert-butyl chloride. Thus, nucleophilic substitution and elimination reactions were common for alkyl halides, but rare with aryl halides. This distinction carries over when comparing alcohols and phenols, so for all practical purposes substitution and/or elimination of the phenolic hydroxyl group does not occur.

PHYSICAL PROPERTIES

Phenol is not appreciably soluble in water (8.2 g/ 100 mL of water). But it is quite soluble in aqueous NaOH because of the soluble salt it forms with NaOH. At the same time it is not soluble in aqueous NaHCO3.

OH

NaOH

ONa

+ H2O

OH

NO2 NaHCO3 No reaction

OH

NaHCO3 No reaction


OHNO2 NaHCO3

NO2

ONaNO2

NO2

+ H2O + CO2

OHNO2 NaHCO3

NO2

ONaNO2

NO2

+ H2O + CO2

O2N O2N

METHODS OF PREPARATION

1.Dows process:

Cl

NaOH / 3500CHigh pressure

ONa

+ O +

ONa

Minor products

Majorproduct

Dows process occurs through benzyne mechanism. The major product is obtained when OH- attacks the benzyne. The minor products are obtained when phenoxide ion attacks the benzyne.

2. Alkali fusion of benzene sulfonic acid

SO3H

NaOH / 3500CHigh pressure

ONa

Majorproduct

Mechanism of this reaction is similar to that of Dows process i.e, benzyne mechanism. 3. Preparation from benzene diazonium chloride When benzene diazonium chloride is warmed with water, phenol is produced.

NO2 NH2 N2Cl OH

H2Ni

NaNO2HCl

H2Owarming

4. Preparation of phenol from p-nitrochlorobenzene

p-Nitrochlorobenzene is first treated with NaOH. Chlorine is replaced with OH in a nucleophilic aromatic substitution reaction (addition-elimination). Later the unwanted NO2 group can be removed.


Cl ONa ONa OH

NaOH H2Ni

NaNO2

NO2 NO2 NH2 N2Cl

OH

HCl

H3PO2

5. Preparation of phenol from cumene hydroperoxide When cumene hydroperoxide is treated with H3O+, phenol and acetone are obtained.

CH2=CHCH3H3PO4

C CH3H3CO-OH

cumene hydroperoxide

H+

OH

+ CH3COCH3

cumene

O2

Mechanism

Ph-C-O-OHCH3

CH3

H+ Ph-C-O-OH2CH3

CH3

+ CH3

CH3C-OPh+

CH3

CH3C-OPhH2O

+

CH3

CH3C-OPhHO

Hemiacetal

CH3

CH3HO H

+C OPh

+H

CH3COCH3 + PhOH

H2Ords

Migration and removal of water occurs simultaneously in the rate determining step. Hemiacetals are unstable in acidic medium and decompose to aldehydes or ketones (hemiacetals and acetals are discussed in detail in aldehydes and ketones). Migratory aptitude: H > Ph > R(30) > R(20) > R(10) > CH3

OCH3 NO2

OCH3> >

Phenyl ring migration is an example of intramolecular electrophilic substitution (ipso attack). Presence of an electron donating group at para position makes benzene ring a better migrating group. Presence of electron withdrawing group does the opposite. But presence of any group at the ortho group makes migration of benzene ring difficult because of steric hindrance. Examples:

p-CH3-C6H4-CCH3

PhOOH

H+ p-Cresol + acetophenone1.

2. Ph-CH2-OOH H+

Benzaldehyde


OOHH+

O

HOH

HH

O

O

O

as a result of hydride migration

as a result of alkyl migration

as a result of alkenyl migration

3.

6. Fries rearrangement

O-C-RO

OH

C-R

+

OHC-RO

O

AlCl3

Major at lowtemperature

Major at hightemperature

The mechanism of Fries rearrangement is similar to that of Friedel-Crafts acylation. It is often possible to select conditions so that either ortho or para is the major product. High temperature favors ortho isomer and low temperature favors para product. There is no clarity as to whether the reaction is intermolecular or intramolecular (whether there is a formation of fully fledged acylium ion or not). CHEMICAL REACTIONS OF PHENOL

1. Phenol as an acid:

Phenol + Na PhONa + H2

Phenol + NaOH PhONa + H2O


2. Reactions with RX, acid chlorides (RCOCl), and anhydrides (RCOOCOR):

OH

NaOH

O CH2CH2-Cl_ OCH2CH3

OH OCCH3CH3C-Cl

O

O

OH OCCH3CH3C-O-CCH3

O

O

Opyridine

+ CH3COOH

+ HCl

+ NaCl

Pyridine (in the case of anhydride) helps converting phenol into phenoxide ion, which is a better nucleophile. Pyridine is not generally required in the case of acid chloride as it is highly reactive.

3. Reaction of phenol with Br2/H2O

OH

Br2H2O

OH

Br Br

Br

OH

Br2CS2

OH

Br Mechanism

OH OH+

_ Br-Br

OH+

Brtautomerism

OHBr

Reactions proceeds until all ortho and para positions are substituted. 2,4,6-Tribromophenol is not soluble in water and is obtained as a white precipitate. In CS2, only monobromination occurs because of the following reasons: 1. Phenol remains undissociated in CS2. Phenoxide ion is far more reactive than phenol. 2. Heterolytic fission of Br-Br bond is not favorable in non polar media. 3. Introduction of each bromine makes the benzene ring less reactive. Other compounds, which react with bromine water similar to phenol, are anisole and aniline.


NH2 NH2

BrBr

Br

Br2H2O

OCH3 OCH3

BrBr

Br

Br2H2O

4. Reimer-Tiemann Reaction OH

1.CHCl3/ KOH

OHCHO

Major

2.H+

Mechanism

CHCl3 + KOH CCl2 + KCl + H2O Dichlorocarbene is an electrophile as its central atom has only sextet of electrons. But it is a weak electrophile due to back bonding and as a result it attacks only highly activated rings, eg: phenol, pyrrole, furan etc.

OCCl2

OCCl2

tautomerism

O-CHCl2

__ OH

CHOhydrolysis

Salicylaldehyde The reaction is thermodynamically controlled and as a result, more stable product is the major product. Some important points regarding Reimer-Tiemann reaction. 1.CHCl3/KOH gives abnormal products with some aromatic compounds. See the example of pyrrole. Pyrrole, when treated with CHCl3/KOH, gives 3-chrolopyridine as the major product. In the first step dichlorocarbene undergoes cycloaddition to pyrrole and then the ring expands. Although there is a loss of aromaticity in the first step, it is regained ultimately.

N N

H

CHCl3KOH

H

Cl

Cl

N

Cl

H

+N

Cl

NH

CHCl3/ KOHNH

CHO

Minor product

Normal Reimer-Tiemann reactionfollowed by H+

2. Aniline, which is otherwise similar to phenol in its properties, does not undergo Reimer-Tiemann reaction. It gives a different product with CHCl3/KOH. We will discuss this in the chapter amines.

NH2

CHCl3KOH

NC

Isocyanide

Carbylamine reaction

3. Phenol with CCl4/KOH gives salycilic acid.


CHCl3/ KOHfollowed by H+

OH OH

Salicylic acid

COOH

4. Furan reacts with CHCl3/KOH and gives the expected product.

O

CHCl3KOH

O CHOMajor product

5. Tropolone (a highly activated aromatic ring) too undergoes Reimer-Tiemann reaction.

CHCl3/ KOHfollowed by H+

OHO

OHO

OHCTropolone

5.Kolbe-Schmitt Reaction (Carboxylation of phenoxide ion) Carefully observe the following example.

OH

1. NaOH2.CO2

OH

COOH

Major product

3.H3O+

RMgX1.CO22.H3O+

RCOOH

O C OR_ RCO

O_

Carboxylation generalised:

O OCOO

tautomerism

O-COO

_OH

COOHhydrolysis

Salicylic acid

O C O_ _

The Kolbe-Schmitt reaction is equilibrium controlled and as a result the major product is the more stable isomer i.e, salicylic acid. Salicylic acid is used in the preparation of acetyl salicylic acid which is an ingredient of aspirin, an analgesic.

OH

COOH

OCCH3

COOH

O

CH3CCl

Oor

CH3COCCH3O O acetyl salicylic acid

Summary of the reactions of phenol: When reactions of phenol are carefully observed, one thing becomes obvious. That is, in some reactions oxygen is the attacking site and in other reactions it is carbon.


OH

Br2 / H2O

ReimerTeimann Reaction

RX / RCOCl /anhydride

Kolbe's reactio Carbon attacksCarbon attacks

Carbon attacksOxygen attacks

Why phenol or phenoxide ion behaves differently? This is generally explained on the basis of Hard and Soft acid-base concept which is beyond the scope of JEE syllabi. Practical organic chemistry of phenols: Phenols can be identified by the following reagents: 1. Na: Phenols, like any other compounds having acidic hydrogen, liberate H2 with Na. Other compounds which liberate H2 with Na are; terminal alkynes, alcohols, carboxylic acids, sulfonic acids etc. 2. Br2/H2O: Phenols decolorize bromine water and give white precipitate. Other compounds which also decolorize bromine water are alkenes, alkynes, aniline, anisole, and some aldoses (we will study about aldoses in Biomolecules). 3. FeCl3: Phenol/substituted phenols give colored (ranging from orange to red) with FeCl3. In fact all enols (phenol is an enol) give colored compounds with FeCl3. See the following examples.

O O

O

O

Both the above compounds give colored compounds with FeCl3, as their enols are in appreciable amounts. Acetone, acetaldehyde etc do not give colored compounds with FeCl3, as enols are in negligible amounts.

PhOH + FeCl3 Fe(OPh)3a colored compound

Exercise-VIII

1. Show how each of the following ethers is prepared by Williamsons synthesis? If there is any ether which cant be prepared by Williamsons synthesis, propose an alternate method.

(A)

O

(B)

O

(C) OMe

(D) O

(E) OEt

2. Draw the structural formulas of the major products obtained when each of the following reactions.

(A)

O1 mol HI

(B) OMe

1 mol HI

(C) O1 mol HI

(D)

OEt1 mol HI

(E)

OCH3

ex HI

(F)

O

1 mol HI


(G)

O2N Oex HI

(H) ex HI

O

O

(I) O 1 eq HI PCC

CH2Cl2

(J) 1.PhCO3H

2.PhOH,H+

3. Write the major products for the following reaction.

1.

OH

1. NaOH

2. Br

3.

2. +

2 414 12

H /H O HIOA (C H O) B 2mol Benzaldehyde

A is optically active while B is not. Find the structural formulas of A and B.

4. Write all the unknown reagents in the following sequence of reactions.

Br

Br

OH

OH

O

OH

OMe

OH

C CH

OMe

OH H

O O

O

1 2

34

56

7

8

9

10

11

12

Exercise-IX

1.Complete the following equations.


(A) phenol + Br2 0

25 C,CS (B) phenol + conc. H2SO4 025 C

(C) phenol + conc. H2SO4 0100 C

(D) p-cresol + p-toluenesulfonyl chloride -OH

(E) phenol + phthalic anhydride 3AlCl (F) p-cresol + Br2 2H O

(G) phenol + C6H5COCl pyridine (H) phenol + (C6H5CO)2O pyridine

(I) phenol + NaOH (J) product of (I) + CH3OSO2OCH3

(K) product of (I) + CH3I

2. Desribe a simple chemical test for distinguishing following pairs of compounds a) 4-Chlorophenol and 4-chloro-1-phenylbenzene b) 4-methylphenol and 2,4,6-trinitrophenol c) 4-methylphenol and 4-methylbenzoic acid d) ethyl phenyl ether and 4-ethyl phenol e) phenyl vinyl ether and ethyl phenyl ether

3. Describe simple chemical tests that would serve to distinguish between; f) phenol and 0-xylene g) p-ethylphenol, p-methylanisole, and p-methylbenzyl alcohol h) 2,5-dimethylphenol, phenylbenzoate, m-toluic acid i) anisole and 0-toluidine j) acetylsalicylic acid, ethyl acetylsalicylate, ethyl salicylate, and salicylic acid k) m-dinitrobenzene, m-nitroaniline, m-nitrobenzoic acid, and m-nitrophenol

4. Describe simple chemical methods for the separation of the compounds of problem 3, parts a),c),d),and f), recovering each component in essentially pure form.

Exercise-X

Write the major products of the following reactions. 1. OH

CH=CH2

H+

2.

O

Br

CH3ONa

3.

O O O

NaOH

MeOH

4.

OH

OH H+

5.

COOHHO

O

HOOC

H+

6.

O

OHO

1.NaBH42. H+


EXERCISE-I

1. O

2. OH

and its enantiomer

3.

OH

+ 4.

D

OH

5.

EtOPh

6.

O OCH3

EXERCISE-II 1. n-Butyl alcohol can be prepared by treating C2H5MgBr with ethylene oxide (epoxide)

followed by hydrolysis. 2.

OH OH

&

3.

OH

OH

4.

O

OH

5. OHCH2OH

+ C2H5OH + CH3OH

6.

HO CH2OH HO CH2OH

+ + C2H5OH

Answers: Exercise-III


1. OH HO

&

A (mixture of diastereomers)

&

B (mixture of enantiomers) 2. (CH3)3COH, 2C2H5OH 3.

C

O

H O-CHCH3

CH3 4. (A) 1.Br2/hv 2.(CH3)3COK/(CH3)3COH 3.HBr/H2O2 4.Mg/THF 5.D2O

(B) 1.Br2/hv 2.Mg/THF 3.D2O 5. 3 6. (A)1.Mg/THF 2.CH3CHO 3. H+

(B)1.Mg/THF 2.HCHO 3. H+

O1.Mg/THF 2. 3.H+(C)

7. CH2-CH=CH2(A) CH=CH2-CH2(B)

8.

9.

CH2=CH-CH-CH2-CH2-CH3

OCH3

10.

Ph3COH + CH3CH2CH2CH2OH PhCH2OH + CH3CH2CH2CH2OH

11. CH3CH2CH2CH2OH 12.

CH2OH

OH

HO HO

OMe

O

OPh

O

HO

OMe

O

OPh

O 13.

HOOH No reaction HO

OH

14.

HN

No reaction No reaction

15.

3

16.

HO CH2OH HO CH2OH

Diastereomers Exercise-IV

1.

C C CH3

Ph

Ph

Ph

O

2. O

CH3

CH3 3.

C C CH3

CH3

CH3O

H3C

4.

C C CH3

CH3

CH3O

H3C


5.

CHO O

&

6. O

7.

O

O

8. OH

OH

OH 9. OH

10. O

Exercise-V 1.

(A) CH3CHO, HCHO,1 (B) CH3CHO, HCOOH,1 (C) HCHO, OHCCH2OCH3,1 (D) No reaction (E) OHCCH2CH2CH2CH2CHO (F) HCHO,4HCOOH,4 (G) 2HCHO, 3HCOOH, 4

2. A (CH3)2C(OH)CH2OH B OH

OH (Cis or trans) C O

OH

D HOOC(CHOH)2COOH

E HOCH2(CHOH)2CH2OH F HOCH2COCH(OH)CHO G OHC(CHOH)4CH2OH 3. O

O

OH

OH

(A) (B)

A pair of enantiomers and a meso compound

4. O CHO COOH

OHCHO

OH

(A) (B) HCHO (C) (D)

CHO

(E) (F)

Exercise-VI

1. 1.HIO4 2.H2/Ni 2. (A) 1.H+/KMnO4/OH-/cold (B) 1.H+/2.RCO3H/H+.

3. 1.CH3MgBr/THF 2. H+/ KMnO4/OH-/cold 4.CrO3/H+.

4. 1.Mg/THF 2.H2O 3.NBS/CCl4 4.H2O 5.PCC/CH2Cl2 (or MnO2)

5. 1.Br2/hv 2.ROH/KOH/ KMnO4/OH-/cold 4.HIO4

6. 1. Br2/hv 2.NaOH 3. CrO3/H+ 4. CH3MgBr/THF 5. H+ 6. H+/. O3; Zn/H+

Exercise-VII


HO2C

O

(A)

HO2C

(B)

ClOC

(C)

O

(D)

(E)

(F)

(G)

OH HO2C

OH

(H)

O

(I)

(J)

OH

OH

(K)

OH

(L)

(M)

OH

(N)

(O)

(P)

Exercise-VIII

1. (A) Ethyl chloride + Sodium isopropoxide (B) Ethyl chloride + Sodium tertiarybutoxide

(C) ONa

CH3I +

(D) 1.Hg(OAc)2/(CH3)3COH2.NaBH4

(E)


ONa

+ EtBr

2.

(A) (CH3)3CI + (CH3)2C=CH2 + CH3CH2OH (B) I

+ + MeOH

(C) Cyclohexanol + Tertiarybutyl iodide +

Isobutene (D) Cyclohexanol + Ethyl iodide

(E) Phenol + Ethyl iodide (F) HOI

HO+

(G)

IO2N + OH

(H) 2 mol CH3CH2I

(I) ICHO

(J)

OHOPh

3.

1. OH

(Claisen rearrangement)

2.

OPh

H

H

Ph

Ph

Ph

H OH

H OH

(A) (B)

4.

1. CCH3

OHH3C CH CH3

OCH3Pair of enatiomers

2. CCH3

H3CO

H3C CH CH3OH

Configuration is retained 3.

CCH3

OH

C2H5 CH2 CH3

4. CCH3

OH

C2H5 CH2 SH

Configuration is retained 5.

CH(OH)CH3

Configuration is retained

6.

C

OH

C-CH3

+ its enantiomer

5.

1. m-CPBA/CH2Cl2 2. CHCNa followed by H2O 3. Baeyers reagent 4. O3 followed by Zn/H+ 5. HIO4 6. One eq TsCl followed by CH3ONa 7. KMnO4 8. H+/Pinacol-pinacolone rearrangement 9. HBr 10. NBS/CCl4


11. CH3ONa/CH3OH/

Exercise-IX

1.

(A) Para-bromophenol (B) p-Hydroxysulphonic acid (C) o-Hydroxysulphonic acid

(D)

CH3TsO

(E) O COOH

HO

(F) OHBrBr

CH3 (G)

C OPh

O

(H) C OPh

O

+ CH3COOH

(I) Sodium phenoxide (J) Anisole (K) Anisole

2.

a) Na (H2 gas is evolved in the first case) b) NaHCO3 (CO2 gas is evolved in the second case) c) NaHCO3 (CO2 gas is evolved in the second case) d) Na (H2 gas is evolved in the second case) e) Baeyers reagent or Br2/H2O (Both are decolorised by the first compound)

3.

a) Na (H2 gas is evolved in the first case) or Br2/H2O (decolorised by the first compound) b) p-ethyl phenol decolorises Br2/H2O. p-methyl benzyl alcohol evolves H2 with Na. c) 2,5-Dimethyl phenol releases H2 gas with Na. m-Toluic acid evolves CO2 with NaHCO3. d) o-Toluidine gives offensive smell with CHCl3/KOH (Carbyl amine reaction) e)

O-C-CH3

COOH

O

Acetyl salicylic acid

O-C-CH3

COOC2H5

O

Ethyl acetyl salicylate

OH

COOC2H5

Ethyl acetyl salicylate

OH

COOH

Salicylic acid First add NaHCO3 to all beakers. 1st and 4th compounds liberate CO2 gas. Then add Br2/H2O to those two beakers. Only the fourth compound decolorises Br2/H2O. Add Na to the other two beakers. Only the third compound liberates H2.

f) m-Nitroaniline gives offensive smell with CHCl3/KOH. m-Nitrobenzoic acid liberates CO2 gas with NaHCO3. mNitrophenol gives H2 gas with Na.

4.

a) Aq. NaOH


c)

Mixtureaq.NaHCO3

aqueous layer

organic layer

COONa

CH3

H+

COOH

CH3

2,5-Dimethylphenol + Phenyl benzoate

aq.NaOHaqueous layer

organic layer

ONa

Me

OH

Me

Me

H+

Me

Phenyl benzoate d) Aq.HCl f)

Mixtureaq.NaHCO3

aqueous layer

organic layer

COONa

H+

COOH

m-Nitrophenol +m-Nitroaniline +m-Dinitrobenzene

aq.NaOHaqueous layer

organic layer

ONa OH

H+

m-Nitroaniline +m-Dinitrobenzene

NO2 NO2

NO2 NO

aq.HCl

aqueous layer organic layer

NH3+

aq.NaOH

NH2

NO2NO2

Exercise-X 1. O

2. OCH2OCH3

3.

OH3CO

OOH

4. O

5.

OH

O

O

+ CO2

6.

O

O

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