Chemistry Alcohols

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Alcohols are organic compounds containing a hydroxyl group attached to a saturated carbon atom andhaving the general formula R—OH, where R can be an alkyl or a substituted alkyl group.

Alcohols can be monohydric or polyhydric depending on whether they contain one or more hydroxylgroups. Alcohols can also be classified as primary (1°), secondary (2°), or tertiary (3°) depending onwhether the hydroxyl-bearing carbon atom is attached to one, two, or three carbon atoms.

Preparation of AlcoholsAlcohols are starting point for organic synthesis of a number of compounds and can be obtained fromorganic raw materials in nature like petroleum, natural gas, coal, and biomass. Some important methods forpreparing alcohols are

1. Grignard Synthesis

C = O + R — MgX — C — OMgX

R

– +– ++ – H2O

R

— C — OH + Mg(OH)X

Mg(OH)X is a gelatinous material and is difficult to handle. It is converted to water-soluble magnesiumsalts by mineral acids.

Mg(OH)X Mg2+ + X + H2OH+ –

Grignard synthesis is an example of nucleophilic addition—the characteristic reaction of aldehydes andketones. The class of alcohol obtained in the reaction depends upon the type of carbonyl compoundused. Formaldehyde yields primary alcohols, aldehydes with more than one carbon atom yield secondaryalcohols, and ketones yield tertiary alcohols.

C = O + R — MgX+

H — C — OMgX

R

–

HH2O

R CH2OH1° alcohol

H

H

Formaldehyde

C = O + R — MgX+

R — C — OMgX

R

–

HH2O RR CH OH

2° alcohol

H

RHigher aldehydes

C = O + R — MgX+

R — C — OMgX

R

–

RH2O

R

R

Ketones

R — C — OH

R

R

3° alcohol

ALCOHOLS, PHENOLSAND ETHERS


A convenient synthesis of primary alcohol uses ethylene oxide, and the primary alcohol formed hastwo more carbon atoms than the Grinard reagent.

H2C — CH2 + R — MgX H2C CH2R R CH2CH2OH

OMgX1°alcohol

OEthylene oxide

Organolithium can also be used for the synthesis of alcohols.

+C = O + R — Li

Alcohol

C — O– Li+ H2O C — OH

R

–

Aldehyde/Ketone

The synthesis of alcohols by using organolithiums is a better method than by using Grignard reagents,because organolithium compounds are less prone to unwanted side reactions and are more reactive thanGrignard reagents. Besides, organolithiums can be used to prepare alcohols from highly crowded carbonylcompounds.

2. Hydrolysis of Alkyl Halides

This method involves nucleophilic substitution of a halide ion by a hydroxyl ion.

R – Xaq. NaOH

R — OHor aq. KOH

SNl or SN2

Example:

aq. NaOHCH2Cl CH2OH

3. Hydroxylation of Alkenes

1, 2 – diols can be prepared by using this method.

C = C

Cold alk. KMnO4

or OsO4

RCO3H, H+

(Peroxy acids)

C

OH

C

OH

C

OH

C

OH

Anti-hydroxylation

Syn-hydroxylation


4. Oxymercuration-Demercuration

It involves anti-Markovnikov’s addition of water to double bond without rearrangement.

C = C — C — C —

RR

OH

Alkyl boraneExample:

CH3 — C = CH2

CH3 Hg(OAC)2

CH3 — C — CH2

CH3

Hg(OAC)2

H2O

HgOACNaBH4 — C — C —

R

OH

H

DemercurationOxymercuration

H2O

OH HgOAC

NaBH4 CH3 — C — CH3

CH3

OH

5. Hydroboration-OxidationIt involves anti-Markovnikov’s addition of water to a double bond without rearrangement.

C = C + (BH3)2 — C — C

B

RH2O2

R— C — C — + B(OH)3

R

H

H

OH–

OH

Alkyl borane

Example: CH3 — C = CH2

CH3 (BH3)2 H2O2/OH–

CH3 — CH — CH2 — OH

CH3

Diborane

6. Reduction of Carbonyl CompoundsAldehydes can be reduced to primary alcohols and ketones to secondary alcohols by using chemicalreducing agents.

RCHOH2/Ni

RCH2OH 1° alcohol

C = OR

R

LiAlH4

or NaBH4

R

RCHOH 2° alcohol

Carboxylic acids can be reduced to primary alcohols by using LiAlH4.

4R — COOH + 3LiAlH4 4H2 + 2LiAlO2 + (R CH2O)4AlLi

4RCH2OH1°alcohol

H2O


Acyl chlorides can also be reduced by LiAlH4.

LiAlH4R C Cl

O

R CH2 OH1° alcohol

In the chemical reduction of esters, the acid part is reduced to primary alcohol.

RCOOR R CH2 OH + ROHReducing Agent

Copper chromite, CuO.CuCr2O4, can be used for reducing esters at a high pressure and temperature.Example:

CH3CH2COOCH3

H+ LiAlH4

CH3CH2CH2OH

H2, CuO.CuCr2O4

high T, P CH3CH2CH2OH + CH3OH

In , unsaturated carbonyl compounds, if the double bond is to be protected while reducing thecarboxyl group to an alcohol, special catalysts can be used.

C = C — C —

OLiAlH4

or NaBH4C = C — CH OH (unsaturated alcohol)

For reducing a double bond, hydrogenation can be carried out in the presence of Ni or Pt.

C = C — C —

OH2

Ni/Pt— C — C — CH —

H H OH

(saturated alcohol)

Physical Properties1. Alcohols have a higher boiling point than hydrocarbons or alkyl halides of comparable molecular weights.This is because alcohols are capable of hydrogen bonding to their fellow alcohol molecules.

R — O H — O H — O H — O

RR

RH

The boiling point increases with an increasing number of carbon atoms and branching for a given number ofcarbon atoms reduces the boiling point.


2. Alcohols are highly soluble in water, because of the ability of alcohol molecules to form hydrogen bondswith water.

R — O

H

H — O

H

The solubility decreases with an increase in the number of carbon atoms.

Chemical PropertiesThe chemical properties of an alcohol are characteristic of the hydroxyl group. The reactions of alcoholscan either involve breaking of a C—OH bond with the removal of an —OH group or breaking of aO—H bond with the removal of a proton, H+.

(I) Reactions Involving C—OH Bond Cleavage1. Reaction with phosphorus halides or SOCl2

3ROH + PX3 3R — X + H3PO3(PX3 = PBr3, PI3)

ROH + PX5 R — X + POX3 + HX(PX5 = PCl5)

ROH + SOCl2 R — Cl + SO2 + HCl

2. Reaction with hydrogen halidesR — OH + HX R — X + H2O

The reaction is catalysed by acids. The order of reactivity of alcohols towards HX isallyl, benzyl > 3° > 2° > 1° < CH3

The mechanism for the reaction is SNl:

(a)Fast

R — OH + HX R — OH2 + :X+

+

(b) Slow

R — OH2 R+ + H2O +

(c) FastR+ + :X– R – X

Rearrangement can occur in this mechanism since carbocation intermediates are involved.Primary alcohols react by SN2 mechanism:

:X– + R – OH2 [X-----R-----OH2]+ +

R – X + H2O+

Primary alcohols do not undergo this type of rearrangement, because no carbocation intermediatesare involved.


Reactivity of hydrogen halides: HI > HBr > HCl > HFHydrogen chloride is quite unreactive and requires the presence of ZnCl2 for reaction with primary andsecondary alcohols.Example:

CH3CH2CH2CH2OHdry HBr

orCH3CH2CH2CH2Br

NaBr, H2SO4,

CH3CH2CH2OHHCl + ZnCl2

CH3CH3CH2Cl

CH3 — C — CH3

CH3

OH

conc. HCl

room temp. CH3 — C — CH3

CH3

Cl

3. Dehydration: Acid catalysed

— C — C —

OHH

acid — C = C — + H2O

Reactivity of ROH : 3° > 2° > 1°Dehydration of alcohols in E1 mechanism involving carbocation intermediate has the posibility ofrearrangement.

(II) Reactions involving O—H Bond Cleavage1. Reactions as acids: with active metals

RO — H + M RO–M+ + 12 H2

(M = Na, K, Mg, Al etc.)

Reactivity of ROH CH3OH > 1° > 2° > 3°When alcohols react with active metals, hydrogen gas is released; this is because of the acidicnature of alcohols. Alcohols are weaker acids than water as shown by the reaction where astronger acid (H2O) displaces a weaker acid (alcohol) from its salt.

RO– Na+ + H – OH Na+OH + ROHstrongerbase

strongeracid

weakerbase

weakeracid

–

Relative acidities: H2O > ROH > NH3 > RHRelative basicities: OH– < OR– < NH2

– < R–


Example: CH3CH2OHNa

CH3CH2O- Na+ +12

H2

CH3 — C — OH Al

H

CH3

CH3 — CH — O– 3Al

CH3

Aluminimum isopropoxide

potassium tert-butoxide

K(CH3)3C — OH (CH3)3CO– K+

2. Ester formation

R — OH + R — C

OH+

OH

R — CO

OR+ H2O

R — OH + R — C

O

Cl

R — CO

OR+ HCl

Alcohol Acid

R — OH + S — ClCH3

O

O

CH3

O

O

S — OR + HCl

Tosyl chloride Alkyl tosylate(TSOR)

TS =

O

CH3 S — Cl

OReactivity of alcohols in esterfication: CH3OH > 1° > 2° > 3°

(III) Miscellaneous Reactions1. Oxidation

A primary alcohol containing two –hydrogens can either lose one of them to form an aldehyde orboth of them to form a carboxylic acid.

R — CH2OHPyridinium chlorochromate

KMnO4

R — C = O (An aldehyde)

R — COOH (A carboxylic acid)or K2Cr2O7

C5H5NH+CrO3Cl–

H

(1°)

–R — CH2OH + KMnO41° alcohol Purple

RCOO K + MnO2 + KOHBrown

+


A secondary alcohol can lose its only –hydrogen to form a ketone.

RCHOH (2°)

or CrO3

RK2Cr2O7

R — C = O

R

Prolonged oxidation or oxidation under harsh conditions oxidises secondary alcohols to carboxylicacids involving breaking of carbon-carbon bonds.

CH3CHCH2CH3

alk. KMnO4

OH

CH3CCH2CH3

O

CH3COOH + CH3COOHacid. KMnO4

A tertiary alcohol contains no -hydrogens and is not oxidised. Under harsh conditions(e.g., acidic oxidising agent), a tertiary alcohol dehydrates to an alkene which can further beoxidised to acids having a lesser number of carbon atoms than the parent molecule.

2. Reaction with Reduced Copper

Primary and secondary alcohols are dehydrogenated to form aldehydes and ketones respectively,whereas tertiary alcohols are dehydrated to form olefins.

CH3CH2OHreduced Cu

OH

Primary alcohol300°C CH3CHO + H2

CH3CH(OH)CH3reduced Cu

Secondary alcohol300°C CH3CCH3 + H2

O

CCH3CH3

CH3reduced Cu

300°C C = CH2 + H2OCH3

CH3

Tertiary Alcohol

3. Action of Chlorine or Bromine

Chlorine and bromine, being mild oxidising agents, oxidise primary and secondary alcoholsto aldehydes and ketones respectively, which then undergo halogenation on -carbon.

CH3CH2OHChloral

Cl2 CH3CHO3Cl2 CCl3CHO

CH3CH(OH)CH31, 1, 1 –Tribromo propanone

Br2 CH3COCH33Br2 CBr3COCH3


Tests For AlcoholsPrimary, secondary, and tertiary alcohols can be distinguished by the following methods:

(I) Lucas Test

This test is based on different reactivities of alcohols towards Lucas reagent—a mixture of concentratedhydrochloric acid and anhydrous zinc chloride—at room temperature.

ROH + HClZnCl2

RCl + H2O

Formation of alkyl halides which are insoluble is indicated by the appearance of turbidity in the reactionmixture. Since the reaction involves carbocation intermediate, the order of reactivity of alcohols isbenzyl, allyl > 3° > 2° > 1° > CH3OH.Tertiary alcohols (benzyl and allyl alcohols as well) produce turbidity immediately, secondary alcoholsgive turbidity within 5-10 minutes, and primary alcohols do not give turbidity at all.

(II) Oxidation

Primary alcohols on oxidation give aldehydes which are further oxidised to carboxylic acid having thesame number of carbon atoms as in alcohol.Secondary alcohols are oxidised to ketones which on prolonged oxidation or under drastic conditionsgive carboxylic acids containing lesser carbon atoms than alcohol. Tertiary alcohols cannot be oxidisedbut under drastic conditions dehydrate to alkenes which are oxidised to a mixture of ketones and acidscontaining lesser carbon atoms.

(III) Victor-Meyer Test

This test involves the following steps:(i) Alcohol is treated with concentrated hydriodic acid or red phosphorus and iodine to form the

corresponding alkyl iodide.

(ii) Alkyl iodide is reacted with silver nitrite to form nitroalkane.

(iii) Nitroalkane formed is treated with nitrous acid (HONO).(2NaNO2 + H2SO4 2HONO + Na2SO4)and then with alkali (NaOH or KOH). Different products are obtained for different alcohols.Primary alcohols produce blood red colour, secondary alcohols produce blue colour, and tertiaryalcohols produce no colour.

Some other tests to determine definite linkages in alcohols are:

(iv) Iodoform testThis test is given by methyl ketones which on treatment with iodine and sodium hydroxide (sodiumhypoiodite, NaOI) yields a yellow precipitate of iodoform (CHI3).

R — C — CH3 + 3NaOI

O

R — C — CI3 + 3NaOH

OR — C — CI3 + NaOH

O

RCOO– Na+ + CHI3Iodoformyellow precipitate


Alcohols of the structure

R — C — CH3

H

OH

(R = H or alkyl or aryl group)

also give positive iodoform test.Hypohalites oxidise an alcohol having the above-mentiond structure to methyl ketone which gives apositive iodoform test.

R — C — CH3 + NaOI R — C — CH3 + NaI + H2O

Ogives positiveIodoform test

H

OH

Example:

CH3CHOH, CH3CHCH2CH2CH3, C6H5CHCH3

H OH OH

give positive iodoform test.

(V) Analysis of 1, 2 –Diols. Periodic Acid OxidationCompounds containing two or more = O or –OH groups attached to adjacent carbon atoms undergooxidation with cleavage of carbon-carbon bonds. A C = O group is oxidised to carboxylic group,

— C — OH

H

group is oxidised to –CHO (aldehyde) and if the carbon atom in C—OH group does not

have any hydrogen atom, it is oxidised to a ketonic group. Example:

R

OH

R — C — CH — R + HIO4

O

RCOOH + RCHO

OH

R — C — CH — R

OH

HIO4

R

R — C = O + RCHO

R — CH — CH2 — CH —R

OH

no reaction

OH

HIO4


PHENOLSPhenols are compounds having the general formula ArOH, where Ar is phenyl, substituted phenyl,or some other aryl group like naphthyl group. Phenols have an —OH group attached directly to the aromaticring. Some of the common phenols are

OH

0 – Chlorophenol

ClOH

m – Cresol

CH3

OH

Catechol

OHOH

Resorcinol

OH

OH

HydroquinoneOH

OH

2 – Napthol

OH

m – Hydroxybenzoic acid

COOH

Preparation(I) Dow Process

Chlorobenzene is treated with aqueous NaOH at a temperature of 360°C under a high pressure. Thereaction involves nucleophilic substitution through elimination-addition mechanism involving benzyneintermediate.

Chlorobenzene

+ClNaOH, 360°C

4500 1b/in2O Na– HCl

OH

Sodium phenoxide Phenol

(II) From Cumene (Isopropyl Benzene)Cumene is converted by air oxidation into cumene hydroperoxide which is converted by aqueous acidinto phenol and acetone.

CH3 CH3 CH3

OCH

Cumene

CH3 — C — O — O — H

Cumene hydroperoxide

OH

+ CH3 — C — CH3

Phenol Acetone

O2 H2O, H+

The reaction involves rearrangement of hydroperoxides involving 1, 2–shift to electron-deficient oxygenthrough the following steps:

CH3 — C — O — O — HH+

CH3 CH3

CH3 — C — O — OH2

+ –H2O CH3 — C = O

CH3

H2O

CH3 — C — O

CH3

OH2+

CH3

H

CH3

OHCH3 — C ++

O

CH3 — C — O

OH

–H +


Acid converts the peroxide I into protonated peroxide which loses a molecule of water to form an intermediatein which oxygen bears only six electrons. A 1, 2 –shift of phenyl group from carbon to electron-deficientoxygen yields carbocation II which reacts with water to yield hydroxy compound III, a hemiketal, whichbreaks down to give phenol and acetone.

(III) Hydrolysis of Diazonium Salts

Ar — N2+ + H2O Ar — OH + H+ + N2

N2+ HSO4

Cl

H2O, H+

OH

Cl+ N2

–

Diazonium salts can be obtained as follows:

HNO3

NO2Reduction

NH2

HONO0°C

(NaNO2, HCl)

N2 + Cl–

RCO +

Physical Properties

1. The simplest phenols are colourless liquids or low-melting solids.2. Phenols have a high boiling point because of inter-molecular hydrogen bonding.3. Phenol itself is soluble in water but most other phenols are essentially insoluble in water. However, the

salts of phenols are soluble in water.

ArOH ArO–

phenol (acid)Insoluble in water

phenoxide ion (salt)soluble in water

Chemical Properties

Phenols are highly reactive compounds and show two kinds of reactions due to(I) —OH group(II) electrophilic aromatic substitution on the ring

1. Acidic Nature of Phenols (Salt Formation)

Phenols are fairly acidic compounds. Aqueous hydroxides convert phenols into their salts and aqueousmineral acids convert the salts back into free phenols.

OH–

ArOH ArO–

Phenol Phenoxide ionH+

The acidity of various compounds vis-á-vis water and phenol is as follows:Alcohols < Water < Phenols < Carbonic Acid < Carboxylic AcidBoth phenol and phenoxide ions are resonance stabilised but stabilisation is greater for phenoxide ions.Thus, resonance shifts equilibrium towards ionisation, making phenol more acidic than alcohol, whichdoes not show resonance.


+O — H O — H

+O — H O — H

+

–

O — H O O

–

O O

–

O

–

–

Phenol

OH–

H+

–

–

Phenoxide ion

Electron-withdrawing substituents like —X, or —NO2 tend to disperse the negative charge on phenoxideion, thereby increasing the acidity of phenols.Electron-releasing substituents like —CH3 or —OR tend to intensify the negative charge on phenoxide ion,thereby decreasing the acidity of phenols.Unlike alcohols, phenols are more acidic than water and thus dissolve in aqueous hydroxides.

ArOH + NaOH(aq) ArO–Na+ + H2OStronger acid Weaker acid

However, being less acidic than carbonic acid, phenols are insoluble in aqueous bicarbonates or carbonates.ArOH + NaHCO3(aq) no reaction

2. Ester Formation Fries RearrangementPhenols are converted into their esters by the action of acids, acid chlorides, or anhydrides.

ArO — H

RCOClNaOHAr SO2Clpyridine(RCO)2O

RCO — OAr

ArSO2 — OAr

RCO — OAr

O

OH + C — ClNaOH

C — O

O

Phenyl benzoate

Schotten Baumann Reaction

When esters of phenol are heated with aluminimum chloride, the acyl group migrates from the phenolicoxygen to an ortho or para position of the ring yielding a ketone. This reaction is called the Fries rearrangement.


Example:

O — C — C2H5

O

AlCl3

CS2

C — C2H5

OOH

+

C — C2H5

OH

OPhenyl propionate 0–Hydroxy phenol

ethyl Ketonep–Hydroxyphenylethyl Ketone

The reaction involves self-acylation by the acylium ion, RCO + .

3. Ether Formation

ArO– + R — X ArO – R + X –

(1°)

Phenols are converted into alkyl-aryl ethers by reaction with alkyl halides in an alkaline solution, involvinga nucleophilic attack by a phenoxide ion on the alkyl halide, thereby displacing the halide ion.

CH2Br + HOaq. NaOH

CH2 — O

Aryl-methyl ethers can be prepared using methyl sulfates, (CH3)2SO4.

OH + CH3OSO2OCH3aq.NaOH

OCH3 + CH3OSO3 –Na+

Anisole (Methoxy benzene)

4. Ring Substitution Reactions

A phenolic group powerfully activates an aromatic ring towards electrophilic aromatic substitution andtherefore special precautions are needed to prevent polysubstitution or oxidation.(a)Nitration

OH

Phenol

conc. HNO3

diluate HNO3

OHNO2

NO2

O2N 2, 4, 6 - Trinitrophenol

OHNO2

OH

NO2

+0–Nitrophenol

p–Nitrophenol

20°C


(b) Sulfonation

H2SO4

15 — 20°COH

100°C

OH

OH

SO3H

H2SO4, 100°C

0-Phenolsulfonic acid

SO3H

p-Phenolsulfonic acid

(c) HalogenationTreating phenols with an aqueous solution of bromine results in the displacement of every hydrogen,ortho or para, to the —OH group and may even cause displacement of certain other groups.

Br2, H2OOH

4, 6 –Dibromo –2–methyl phenolCH3 Br

OHCH3

Br

Br2, H2OOH

BrOH

BrSO3H

Br2, 4, 6 –Tribromo phenol

If halogenation is carried out in a solvent of low polarity like chloform or carbon disulfide, the reactioncan be limited to monohalogenation.

OHBr2, CS2

OH

BrO°C

+

OHBr

(Major product)

(d) Friedel-Crafts Alkylation/AcylationFriedel-Crafts alkylation of phenols yields alkyl phenols; however, the yield is poor. This is becauseAlCl3, instead of generating a carbocation, gets involved in forming a complex with electron-richphenol.

OH

+ AlCl3

O AlCl3+

A positively charged oxygen is electron-withdrawing, thereby decreasing the reactivity of ring towardselectrophilic substitution reactions.Friedel-Crafts acylation of phenols also has a low reactivity and behaves the same way.


(e)NitrosationPhenols are one of the few compounds reactive enough to undergo attack by nucleophilic nitrosoniumion, +NO.Nitrous acid converts phenols into nitrosophenols.

NaNO2, H2SO4

OH

7 – 8°C

OH

NO

(f) Coupling with diazonium salts

HO

OH weaklyalkaline+

+

Phenol

N = N

Diazonium salt p-Hydroxyazobenzene

N2

(g)Carbonation: Kolbe reactionTreatment of salt of phenol with carbon dioxide brings about substitution of carboxyl group for hydrogenin ring.

–O Na+–

–

OHNaOH + C

O+

O

125°C, 4–7 atmOH

COONaSodium salicylate(major product)

H+

OHCOOH

Salicylic acid

Some p-hydroxybenzoic acid is also formed but it can be easily separated by fractional distillation. Ifthe reaction is carried out on potassium phenoxide, the salt of p-acid becomes the majorproduct.

Salicylic acid is used to produce a number of important derivatives used as medicines.

OHCOOH

Salicylic acid

CH3OH, H+

C6H5OH, H+

(CH3CO)2OH+

OHCOOCH3

OHCOOC6H5

OCOCH3COOH

Methyl salicylate(Oil of winter green)

(flavouring agent)

Phenyl salicylateSalol

(internal antiseptic)

Acetyl salicylic acid (Aspirin)

(an analgesis and antipyretic)


(h)Reimer-Tiemann reaction+–

OHSalicyladehyde

CHCl3aq. NaOH

CHOO Na

H+ /H2OOH

CHO

+–

OHSalicylic acid

CCl4

aq. NaOHCOOH

O NaH+ /H2O

OHCOOH

The reaction involves electrophilic substitution of dichloro carbene, :CCl2 on the highly reactivephenoxide ring.

:CCl3–CCl3 + OH

HChloroform

– + H2O Cl – + CCl2

Dichlorocarbene

O–

CHO

O: :..O: :

:CCl2..

O

CCl2

HO–

CHCl2

OHOHCHO

H+

Phenoxide ion

The product is largely 0-aldehyde with a small amout of the p-isomer. If both O-position in phenoxide ionare substituted the reaction yields p-isomer.

(i) Reaction with formaldehydePhenol reacts with formaldehyde in the presence of an acid or alkali forming 0–or p-hydroxy methylphenol.

Basic Catalysis

O: :..

HO

CH2O–CH2OH

OH

+ C = OH

+ –H

H2O

H+

Acidic Catalysis

OH:..

H

Protonatedaldehyde

+ C = OHH CH2OH

OHH H2O CH2OH

OH

+

+


Analysis of Phenols

Phenols react with neutral ferric chloride solution and form coloured complexes (red, blue, green, andviolet). This reaction is also shown by enols (e.g., CH2 = CHOH).

Phenols can also be identified by their unique acidic strength. They are soluble in aqueous hydroxides butinsoluble in aqueous carbonates or bicarbonates.

ETHERS

Ethers are compounds of the general formula R — O — R, Ar — O — R or Ar — O — Ar, where Ar isan aromatic group. An ether is symmetrical if the two groups attached to the oxygen atom are the same, andit is unsymmetrical if the groups are different.

CH3CH2CH2CHCH2CH3

OCH3

3-Methoxyhexane

CH3CH2OCH2CH2CH3

1-Ethoxypropane

Preparation

1. Dehydration of alchols

2R — O — HH2SO4

R — O — R + H2O

A water molecule is lost for every pair of alcohol molecules. Dehydration is limited to the preparation ofsymmetrical ethers, because a combination of two different alcohols yields a mixture of three ethers.

Alcohols can also dehydrate to alkenes, but dehydration to ethers is controlled by the choice of reactionconditions.Example:

CH3CH2OH

140°C

180°C

C2H5OC2H5

CH2 = CH2

Diethyl ether

Ethene

Conc.H2SO4

Ether formation by dehydration is an example of nucleophilic substitution: protonated alcohol is thesubstrate and the second molecule of alcohol is the nucleophile.The reaction is SN

1 for 2° and 3° alcohols and SN2 for 1° alcohol.

R — OH2

R — OH + H

–H2O

ROH

R+

[R — O ---R----OH2]

+R — OH2 (Protonated alcohol)

ROHR O — R

H

+H+ + R — O — R

+

R — O — R

H

+R — O — R + H+

+H

+

+

alcohols)SN1(3° & 2°

SN2(1° alcohols)


2. Williamson synthesis

RX

RO Na+

ArO–Na+

R – OR

R – OAr

–

Yield from RX: CH3 > 1° > 2° > 3°

The reaction involves the nucleophic substitution of an alkoxide (or phenoxide) ion for a halide ion. Thismethod can be used for preparing symmetrical as well as asymmetrical ethers.

(CH3)2 CH(OH) Na (CH3)2CHO–Na+ + CH3CH2CH2Br CH3(CH2)2O CH(CH3)2

OH + CH3CH2Br aq. NaOH

O — CH2CH3

Ethoxybenzene

The reaction gives the best yield with 1° alkyl halides. With tertiary alkyl halides, elimination becomes animportant reaction and no ether is obtained.

CH — C — Br + C2H5OH

CH3

CH3

aq.NaOHCH3 — C = CH2

CH3

Physical Properties1. The C — O — C bond angle in ethers is not 180° and the dipole moments of the two C — O bonds do

not cancel each other. Hence, ethers possess a small net dipole moment.

R

net dipole momentRO110°

2. The boiling point of ethers are the same as those of alkanes of comparable molecular weights. Theboiling points of alcohols are much higher than those of ethers, as ethers are incapable of intermolecularhydrogen bonding.

3. The solubility of ethers in water is comparable to that of alcohols, because ethers can form hydrogenbonds with water molecules.

R — O ------ H — O

H

R


Chemical PropertiesEthers are unreactive compounds. The ether linkage is quite stable towards bases, oxidising, and reducingagents.

1. Cleavage by acids

R — O — R + HX R — X + R

R — X

HX

Reactivity of HX: HI > HBr > HCl.Cleavage takes place under vigorous conditions using concentrated acids and high temperature.A dialkyl ether initially yields an alkyl halide and an alcohol; the alcohol may further react to form secondmole of alkyl halide.

CH3 — CH — O — CH — CH3

BrCH3

2CH3CHCH3

48% HBr

CH3

130 – 140°C

The initial reaction between an ether and an acid results in the formation of a protonated ether. Thecleavage then involves a nucleophilic attack by a halide ion on this protonated ether with the displacementof the weakly basic alcohol molecule.

R — O — R+ HXH

+

SN1R — O — R+ X–or SN2

R — X + ROH HX

RXProtonated ether

A primary alkyl group tends to undergo SN2 displacement and a tertiary alkyl group tends to undergo SN1displacement.

CH3 — O — CH2CH2CH3

HI

HICH3I + CH3CH2CH2I

SN2

excess

CH3I + CH3CH2CH2OH

3 – Methoxy propane

CH3 — O — C — CH3

CH3

CH3

HISN1

CH3OH + (CH3)3CI

2 – Methoxy–2–methyl propane


EpoxidesEpoxides are compounds containing the three-membered ring.

— C — C —

O

(Epoxide or oxirane ring)

Epoxides belong to a class of compounds called cyclic ethers. The three-membered ring makes epoxidesan exceedingly important class of compounds.

1, 2 – Epoxybutane

CH2 — CH — CH2 — CH3

OEpoxyethane

CH2 — CH2

O

Preparation

Epoxides are commonly obtained by oxidation of alkenes by peroxy acids.

RCO3HO

C = C

Alkene

PeroxyacidC — C

Epoxide

Silver oxide can also oxidise alkenes to epoxides. An internal SN2 reaction in a chlorohydrin can be used toprepare three membered cyclic ethers.

OH

1, 2 – Epoxypropane

Cl — CH2 — CH — CH3

OH O

Cl — CH2 — CH — CH3–Cl

CH2 — CH — CH3

O

– –

–

Reactions

Epoxides have highly strained three-membered rings that can undergo acid- or base-catalysed ringcleavage.1. Acid-catalysed cleavage

— C — C —

HO

— C — C —

O+

— C — C —

ZZ :

OH

z: = nucleophileAt first, the epoxide is protonated by an acid and the protonated epoxide can then undergo an attack bynucleophilic reagents.

H+— C — C —

O +

— C — C —

OH


+

— C — C —

OH

ROH

H2O — C — C —

OH2 OH

— C — C —

OH

OHR

— C — C —

X OH

–H+

–H+— C — C —

OH OH

— C — C —

OR OH

1, 2, - diol

Alkoxy alcohol

HalohydrinX–

+

2. Base-catalysed cleavage

Z: + — C — C —

O

— C — C —

O–HZ

Z

— C — C —

OH

Z

Under alkaline conditions, an epoxide itself undergoes nucleophilic attack.

O

C2H5O– Na+ + CH2 — CH2 C2H5O CH2 CH2 OH2-Ethoxyethanol

O

NH3 + CH2 — CH2 H2N — CH2 CH2 OH2-Aminoethanol

3. Reaction with Grignard reagent

R — MgX + CH2 — CH2+

CH2 — CH2 — OMgX–

RCH2 CH2 OHPrimary alcohol(Chain lengthened by two carbon atoms)

OR

H +

MgBr + H2C — CH2

O

CH2CH2OMgBr

CH2CH2OH

H2O

2-Phenylethanol

Ethyleneoxide

Phenyl magnesiumbromide


SOLVED OBJECTIVE EXAMPLES

Example 1: Phenols is less acidic than(a) acetic acid (b) p-methoxyphenol(c) p-nitrophenol (d) ethanol

Solution: Phenol is less acidic than carboxylic acids and more acidic than aliphatic alcohols. Moreover,the presence of an electron-withdrawing group (eg. NO2) increases the acidic strength ofphenol but electron-releasing groups decrease the acidic strength. Hence, phenol is lessacidic than (a) and (c).

Example 2: Hydrogen bonding is possible in(a) ethers (b) hydrocarbons(c) water (d) alcohols

Solution: For hydrogen bonding, a compound should have hydrogen attached to an electronegativeatom (F, O, N). The answers are (c) and (d).

Example 3: Which of the following compounds will give a yellow precipitate with iodine and alkali ?(a) 2–Hydroxypropane (b) Acetophenone(c) Methyl acetate (d) Acetamide

Solution: Methyl ketones and alcohols having the structure CH3CH(OH) – give a yellow ppt. withiodine and alkali. The answers are (a) and (b).

Example 4 : Equimolar quantities of ethanol and methanol are heated with conc. H2SO4. The productformed is(a) C2H5OC2H5 (b) CH3OCH3(c) C2H5OCH3 (d) all the three

Solution : 3R1 – OH + 3R2 – OH H2SO4

R1 – O – R2 + R2 – O – R2 + R1 – O – R1 + 3H2O

The answer is (d)

Example 5: The ether — O — CH2 when treated with HI produces

(a) — CH2I (b) CH2OH

(c) — I (d) — OH

Solution: Phenyl group is unreactive towards substitution but an sp3 hybridised carbon undergoessubstitution easily.


— O — CH2 — — OH + — CH2I

H – I

The answers are (a) and (d)

Example 6: The compound that reacts fastest with Lucas reagent (conc. HCl + ZnCl2) at roomtemperature is(a) butan–1–ol (b) butan–2–ol(c) 2–methyl propan–1–ol (d) 2–methyl propan–2–ol

Solution: Tertiary alcohols react fastest with Lucas reagent followed by 2° and 1° alcohols.The answer is (d)

Example 7: In the following compounds,

OH

CH3

(I)

OH

(II)

OH

(III)

NO2

OH

(IV)NO2

the decreasing order of acidic strength is(a) III > IV > I > II (b) I > IV > III > II(c) II > I > III > IV (d) IV > III > I > II

Solution: Electron-withdrawing effect of a group increases the acidity of phenols and electron-releasinggroup has the opposite effect.In (II ), there is a (+I) effect of –CH3 group. [decreases acidity]In (III), there is a (–I) effect of –NO2 group [increase acidity]In (IV), there is (–I) and (–R) effect of –NO2 group. [strongly increases acidity]The answer is (d)

Example 8: The product in the reactionCH3

O ?HBr

is

(a)

CH3

OHBr (b)

CH3

(c)

CH3

BrOH (d) Br

CH3

Solution: Acid-catalysed ring opening of epoxides involves a nucleophilic attack at a moresubstituted carbon atom.


BrCH3

Br–OOH

CH3

The answer is (c)

Example 9: Which of the following compounds is resistant to oxidation by periodic acid (HIO4)?(a) HOCH2CO CH2 OH (b) HO CH2 CH2 CH2 OH(c) HO CH2 CH(OH) CH2 OH (d) HO CH2 CH2 OH

Solution: For oxidation with HIO4, a compound should have carbonyl groups and/or hydroxyl groupson adjacent carbon atoms. In (b), two hydroxyl groups are not adjacent to each other.The answer is (b)

Example 10: Which of the following reagents can be used to separate a mixture of phenol and carboxylicacid?(a) NaOH (b) Na2CO3(c) lime water (d) NaHCO3

Solution: Phenol is less acidic than carbonic acid whereas carboxylic acids are more acidic. Hence,phenols are insoluble in a solution of NaHCO3 but carboxylic acids dissolve in NaHCO3solution.The answer is (d)

Example 11: The compound with the lowest boiling point, that is, the most volatile compound is

(a)OH

NO2

(b)

OH

NO2

(c)

OH

NO2O2N(d)

OH

NO2O2N

NO2

Solution: Phenols capable of forming intermolecular hydrogen bonding have a high boiling point. But(a) has intramolecular rather than intermolecular H–bonding and is the most volatilecompound.

O — H

NO

OThe answer is (a)

Example 12: When phenol is reacted with CHCl3 and NaOH followed by acidification,salicylaldehyde is obtained. Which of the following species are involved in the above reactionas intermediates?


(a) CCl2

H–

O

(b)CHCl2

OH

(c) CHClH

O

OH

(d)CHCl2

O–

Solution: This is Reimer-Tiemann reaction in which (b) and (d) are the intermediates formed.

Example 13: Phenol can be distinguished from aliphatic alcohol with(a) Tollen’s reagent (b) Schiff’s base(c) FeCl3 (d) HCl

Solution: Phenol (all enols) can be oxidised by FeCl3 to form coloured complexes. Alcohols, however,cannot get oxidised by FeCl3.The answer is (c)

Example 14: When sodium phenoxide is heated with CO2 under pressure followed by acidification withHCl, the product obtained is(a) salicylic acid (b) salicyladehyde(c) benzoic acid (d) benzaldehyde

Solution: This is Kolbe-Schmidt reaction and the product formed will be salicylic acid.The answer is (a)

Example 15: An alcohol contains 60% carbon and 13.3% hydrogen and gives positive iodoform test.The alcohol is(a) 2-propanol (b) 1-propanol(c) 2-butanol (d) 1-butanol

Solution: C : H : O = 60 13.3 26.7: :12 1 16 =

5 13.3 1.67: :1.67 1.67 1.67

= 3 : 8 : 1The formula of the compound is C3H8O.Since the alcohol gives positive iodoform test, it is 2-propanol.

CH3CHCH3

OH

The answer is (a)


Example 16: What is the product in the following reaction?

H2C = CH — CH — OH

CH3

H2CrO4

(a) H2C = CH — C = O

CH3

(b) OHC — C = O

CH3

(c) HOOC — C = O

CH3

(d) CH3CHO

Solution: Unlike KMnO4, which oxidises both a double bond and a hydroxyl group, chromic acid(H2CrO4) does not oxidise double bond.The answer is (a)

OBJECTIVE PROBLEMS

Section A

Choose the Correct Answer

1. Which of the following is produced when an aqueous solution of butan-2-ol is refluxed with diluteacidic KMnO4?(a) butanol (b) butanoic acid(c) potassium butanoate (d) butanone

2. Consider the reaction: CH3CH2CH2OH 5PCl alc.KOHA B . The compound ‘B’ is

(a) propane (b) propene(c) propyne (d) propanal

3. Which of the following has the lowest solubility in water?

(a) CH3CH2CH2CH2OH (b)

CH3

CH3—CHCH2OH

(c) HOH2C — CH2OH (d) C6H5CH2CH2OH

4. Chlorine reacts with ethanol to give(a) ethylchloride (b) chloroform(c) chloral (d) acetaldehyde


5. Ethanol is heated with concentrated H2SO4. The product formed is

(a)

O

CH3 — C — O — C2H5 (b) C2H6

(c) C2H4 (d) C2H2

6. Which of the following is also known as ‘picric acid’?(a) phenol (b) carboxylic acids(c) trinitrophenol (d) nitroalkane

7. Which of the following has the highest value of pKa?(a) CH3 — CH2OH (b) Cl — CH2 — CH2OH

(c) F3C — CH2 — OH (d)

CH3

CH3 — CH — CH2OH

8. Which of the following has the highest boiling point?

(a)

OH

CH3 — CH2 — CH — CH3 (b) CH3 — CH2 — CH2 — CH2OH

(c)

CH3

CH3 — CH — CH2OH (d)

CH3

CH3 — C— OH

CH3

9. Wood spirit contains(a) only methanol (b) only ethanol(c) methanol + ethanol (d) a mixture of a number of alcohols

10. The compound which reacts faster with Lucas reagent at room temperature is(a) butan–1–ol (b) butan–2–ol(c) 2–methyl propan–1–ol (d) 2–methyl propan–2–ol

11. Lucas test is used to determine the type of(a) amines (b) alcohols(c) acids (d) phenols

12. When glycerol is heated with oxalic acid, the compound formed is(a) allyl alcohol (b) formic acid(c) tartaric acid (d) formic acid and allyl alochol

13. Salol is prepared from(a) Salicylic acid and phenol (b) Salicylic acid and methanol(c) both (a) and (b) (d) phenol and methylchloride


14. Separation of a higher phenol and an aromatic carboxylic acid can be easily carried out by(a) NaOH (b) Na2CO3(c) lime (d) NaHCO3

15. A compound ‘A’ reacts with sodium metal and also undergoes iodoform reaction. ‘A’ is(a) phenol (b) methanol(c) n-propanol (d) iso-propanol

16. Which of the following pairs gives the same compound on heating with zinc dust?

(a)

OHCH3

and

OH

(b)

OHCHO

and

OH

CHO

(c) CH3CH2CH2OH and CH3CHOHCH3 (d) 1- butanol and 2-butanol17. Which of the following ethers is cleaved even by HCl at room temperature?

(a) C6H5OCH2CH3 (b) CH3CH2OCH2CH3(c) (CH3)3COCH2CH3 (d) (CH3)3COC(CH3)3

18. The ether O—CH2— when treated with HI produces

(a) CH2I (b) CH2OH

(c) I (d) OH

19. The reaction between sodium ethoxide and bromoethane yields(a) methyl ethyl ether (b) dimethyl ether(c) diethyl ether (d) propane

20. Diethyl ether on heating with concentrated HI gives two moles of(a) ethanol (b) iodoform(c) ethyl iodide (d) methy liodide

Section B

Choose the Correct Answer

1. When phenol is treated with bromine water in excess, it gives(a) m-bromophenol (b) o-and p-bromophenol(c) 2, 4–dibromophenol (d) 2, 4, 6–tribromophenol


2. The compound that does not change the orange colour of chromic acid to blue green is(a) 2° alcohol (b) 1° alcohol(c) 3° alcohol (d) none of the above

3. HBr reacts the fastest with(a) 2-methyl propan-2-ol (b) propan-1-ol(c) propan-2-ol (d) 2-methyl propan-1-ol

4. A compound ‘X’ on oxidation gave ‘B’ and then again on oxidation gave an acid. After the firstoxidation, it reacted with ammoniacal silver nitrate to produce a black precipitate. ‘X’ is(a) a primary alcohol (b) a tertiary alcohol(c) acetaldehyde (d) acetone

5. R—CH2—CH2 OH can be converted into RCH2CH2COOH. The correct sequence of reagent is(a) P Br3, KCN, H+ (b) P Br3, KCN, H2(c) KCN, H+ (d) HCN, PBr3, H+

6. Phenol forms benzene on reaction with(a) NaHCO3 (b) NaOH(c) HCl (d) zinc

7. Which of the following statements is correct?(a) phenol is acidic in nature, gives effervescence with NaHCO3(b) phenol is basic in nature due to the presence of OH(c) phenol gives violet colour with neutral FeCl3(d) phenol as well as ethanol can be reduced to corresponding alkanes by heating with zinc dust

8. When phenol is reacted with CHCl3 and NaOH followed by acidification, the compoundsalicylaldehyde is obtained. Which of the following species is involved in the above-mentioned reactionas an intermediate?

(a)

O

CCl2–

H(b)

OHCHCl2

(c)

OHCHCl

OH

(d)

O–

CHCl2

9. Reimer-Tiemann reaction is/are

(a) C6H6 + CH3Cl 3

anhydrousAlCl C6H5CH3 + HCl

(b) CH3COOAg + Br2 CH3Br + AgBr + CO2

(c) CH2 = CH2 + H2 Ni CH3 – CH3

(d) C6H5OH + CHCl3 + 3KOH C6H4OHCHO

+ 3KCl + 3H2O


10. The characteristic dark colour with neutral FeCl3 is given by

(a)OH

(b) CH3CH2 — C = CH — CH3

OH

(c)

OH

CH3

(d) CH3CH2CH2OH

11. The following reaction

OH+ HCl + HCN

2

anhydrousZnCl

CHO

OH

is

(a) Perkin reaction (b) Gattermann reaction(c) Kolbe reaction (d) Gattermann koch reaction

12. Reimer-Tiemann reaction involves(a) carbonium-ion intermediate (b) carbene inter mediate(c) carbanion intermediate (d) free radical intermediate

13. Which of the following has a higher pH?(a) phenol (b) o-cresol(c) p-nitrophenol (d) glycerol

14. A mixture of o-nitrophenol and p-nitrophenol can be separated by(a) sublimation (b) steam distillation(c) fractional crystallisation (d) distillation

15. Which of the following alcohols is a stronger acid?(a) CH3OH (b) CH3 — CH2 — OH

(c) CHOH

CH3

CH3

(d) CH3 — C — OH

CH3

CH3

16. The reaction of anisole with HI leads to the formation of(a) phenol and methanol (b) phenol and methyl iodide(c) iodobenzene and methanol (d) iodobenzene and methyl iodide

17. Methyl-tert-butyl ether on heating with HI gives(a) CH3I + (CH3)3COH (b) CH3OH + (CH3)3CCI(c) CH3I + (CH3)3CI (d) (CH3)3CCI + CH3OH


18. Which of the following compounds is resistant to nucleophilic attack by hydroxyl ions?(a) methyl acetate (b) acetonitrile(c) dimethyl ether (d) acetamide

19. Diethyl ether absorbs oxygen to form(a) a red-coloured sweet-smelling compound (b) acetic acid(c) ether suboxide (d) ether peroxide

20. On boiling with concentrated HBr, phenyl ethyl ether will yield(a) phenol and ethyl bromide (b) bromobenzene and ethanol(c) phenol and ethane (d) bromobenzene

SOLVED SUBJECTIVE EXAMPLES

Example 1: An organic compound (A) [M.F = C7H8O] is insoluble in aqueous sodium bicarbonate butdissolves in aqueous sodium hydroxide and gives characteristic colour with aqueous ferricchloride. When treated with bromine, (A) forms a compound (B) C7H5OBr3. Identify (A)and (B).

Solution: (A) must be a phenolic compound because(i) it is soluble in aq. NaOH but not in aq. NaHCO3(ii) it forms coloured compounds with FeCl3(A) has a phenolic group and it should also have a methyl group at meta position to

–OH group. [Only then (B) is formed on bromination]

Br2

LewisAcid

OHOH

CH3

Br

CH3

Br

Br

(A) (B)

Example 2: A compound (A) with the molecular formula C6H14O does not react with Na, but whenheated with hot HI, it converts into a mixture of two isomeric compounds of the molecularformula C3H7I. Identify A.

Solution: Double bond equivalent (DBE) = nC – – 12 2 2

NH X nn n

where nC = no. of carbon atomsnH = no. of Hydrogen atoms, nX = no. of halogen atomsnN = no. of nitrogen atoms

For compound (A), DBE = 6 – 142 + 1 = 0

(A) is a saturated compound but cannot be an alcohol, because it does not react withNa. Hence, (A) is an ether.


Now, C3H7I has only two possible isomers:

CH3CH2CH2(I)

CH3

(B)

and CH(I)

CH3

(C)

Hot HI cleaves (A) to (B) and (C)

Hence, (A) is CH3CH2CH2 — O — CHCH3

CH3

Example 3: Identify the products, giving the mechanism

(a) CH3CH2CH — CH2

O

NaOCH3 ?CH3OH,

(b) CH3CH2CH — CH2

O

H+?

CH3OH,

Solution: The reactions involved are the cleavage of the expoxide ring.(a) This is a base-catalysed ring opening. The mechanism is SN2 and the nucleophilic attack is on the less-crowded carbon.

CH3CH2CH — CH2

O(Product)

OCH3 CH3CH2CHCH2OCH3 CH3CH2CHCH2OCH3

OH

1-Methoxybutan-2-ol

H+

O–

(b)This is an acid-catalysed ring opening; therefore, mechanism is SN1 and nucleophilicattack is on the more-substituted carbon.

CH3CH2CH — CH2

OCH3CH2CHCH2OH

H+

OCH3 (3, methoxybutanol) (Product)

CH3OH

Example 4: An organic compound (A) has 76.6% C and 6.38% H. Its vapour density is 47. It givescharacteristic colour with FeCl3 solution. (A) when treated with CO2 and NaOH at 140°Cunder pressure gives (B) which an acidification gives (C). (C) reacts with acetyl chloride togive (D) which is a well-known painkiller. Identify A, B, C and D and also explain thereactions involved.


Solution: Molar mass of A = 2 V.D = 2 47 = 94 g/molThe ratio of atoms is

C : H : O = 76.6 6.38 17.02: :12 1 16

= 6.38 : 6.38 : 1.064= 6 : 6 : 1

Hence, the empirical formula of A = C6H6OSince empirical formula mass = 94 = molar mass of ATherefore, the formula of A is C6H6O.Since (A) gives characteristic colour with FeCl3 it should be phenol. The reactions involvedare

OH + CO2

(A)

NaOH140°C, high P

OH

COO–Na+

H +

OH

COOH

(B) (C)Salicylic Acid

+ OH

(C)

O — C — CH3

COOH

(D)Aspirin (pain-killer)

COOH

CH3CCl

O O

Example 5: An optically active alcohol (A), (C6H10O), absorbs two moles of hydrogen per mole of(A) upon catalytic hydrogenation and gives a product (B). Compound (B) is resistant tooxidation by CrO3 and does not show any optical activity. Identify (A) and (B).

Solution: D.B.E of A = 6 – 102 + 1 = 2

(A) must have following characteristics:(i) is an alcohol(ii) contains two bonds(iii) has an asymmetric carbon(iv) is a tertiary alcohol being resistant to oxidation by CrO3

A is


OH

C

CH2CH3

CH3 C CH 3–Methyl pent –1–yn–3–ol

And, B is

OH

C

CH2CH3

CH3 C2H5

Example 6: (A)Al2O3

250°. (B)

(i) HI

(ii) AgOH (C)

Al2O3

150°C (B) (A)

(i) B2H6

(ii) H2O2, OH–

(A) and (C) are isomers. (B) has a formula of C5H10 which can also be obtained from theproduct of reactions of CH3CH2MgBr and acetone. Identify (A), (B), and (C).

Solution: (B) [M.F = C5H10] can be obtained as

O

CH3CH2MgBr + CH3CCH3H+

CH3CCH3

CH2CH3

OMgBr

C2H5 — C — CH3

CH3

OH

CH3CH = C(CH3)2(B)

Since (B) is formed by heating (A) with Al2O3, (A) must be an alcohol. Moreover, (A) and (C)are isomers. Hence

(A) is CH3CH — CHCH3

CH3 OH

(B) is CH3CH = C — CH3

CH3

(C) is CH3 — C — CH2CH3

OH

CH3

; ;

Example 7: Compound (A), C10H12O gives off hydrogen on treatment with sodium metal and decolourisesBr2 in CCl4 to give (B), C10H12OBr2. (A) on treatment with I2 in NaOH gives iodoformand an acid (C) after acidification. Give the structures of (A) to (C) and also of all geometricaland optical isomers of (A).

Solution: D.B.E. of (A) = 10 – 122 + 1 = 5

Since (A) decolouries bromine water and adds only a molecule of bromine, (A) shouldhave a double bond.


D.B.E. of A = 5; it should also have a benzene ring.(A) gives positive iodoform test and should have structure of the typeR — CHCH3

OH

(R = alkyl/aryl group)

Hence (A) =

CH = CHCHCH3

OHBr2

CCl4

CHBrCHBrCH(OH)CH3

(B)

CH = CHCHCH3

OH NaOH

(C)

I2 CHI3 + Iodoform

CH = CHCOOH

(A)

(A) has a chiral carbon and exists in two enantiomeric forms:

CH = CHC6H5

OHH

CH3

CH = CHC6H5

HHO

CH3

(A) has two geometrical isomers:

C = CCH(OH)CH3

C6H5

H HC = C

CH(OH)CH3H

C6H5H

(trans)(cis)

Example 8: Two different Grignard reagents (X) and (Y) produce C6H5CH2C(CH3)2OH on reactionwith (P) and (Q) respectively. Give the structures of (X), (Y), (P), and (Q).

Solution: The compound

CH3

C6H5CH2 — C — CH3

OH

can be synthesised as

C6H5CH2 CCH3

O(P)

+ CH3MgX(X)

CH3 C CH3

O

(Q)

+ C6H5 CH2 MgX

(Y)

C6H5CH2 — C — CH3

OH

CH3


Example 9: A phenolic compound (A), C7H8O2 on mild oxidation gives a highly volatile oil (B). (A)forms (C) on reaction with dimethyl sulphate in alkali. The oxidation of (C) with hot KMnO4gives (D) which then reacts with bromine water to give (E), containing about 72% bromine.Identify (A) to (E) with proper reasoning.

Solution: D.B.E. of [A] = 7 – 4 + 1 = 4D.B.E. = 4 indicates that [A] might contain a benzene ring.[A] should contain a benzene ring as well as a phenolic group.The molecular formula will be C6H5OH – CH2O.Since (A) undergoes mild oxidation, the ring should contain a CH2OH group and since onoxidation a hightly volatile oil is obtained, this group should be at ortho-position.

(highly volatile due to intra-molecular H-bonding)

mil

OH

CH2OH

[A]oxidation

OHCHO

OCH3

CH2OH

[C]

hot

KMnO4

OCH3

COOH

[D]

BromineWater

OCH3

Br

[E]

Br

Br

Etherfication (CH3)2SO4/OH–

Percentage of Br in [E] = 69.56 % (

Example 10: An organic compound (A) on treatment with CHCl3 and KOH gives (B) and (C), both ofwhich, in turn, give the same compound (D) when distilled with Zn dust. The oxidation of(D) yields (E) of formula C7H6O2. Sodium salt of (E) on heating with soda-lime gives (F)which can also be obtained by distilling (A) with zinc dust. Identify (A) to (F).

Solution: The summary of reactions is

[A]CHCl3

KOH[B] + [C]

Zn-dust

[F](i) NaOH

(ii) soda -lime[E] [O]

[D]

Zn-dust

C7H6O2

D.B.E. of E(C7H6O2) = 7 – 62 + 1 = 5

[E] should contain a benzene ring and it undergoes decarboxylation with soda-lime.


[E] can be benzoic acid

COOH

[F] can be benzene (C6H6)[A] is undergoing Reimer-Tiemann reaction; thus, [A] is phenol.The reactions involved are

OH

(A)

CHCl3

KOHOH

CHO(B)

+OHC OH

(C)

COOH(i) NaOH

(ii) Soda-simeCHO

[O]

[F] [E] [D]

Zn-DustZn-Dust

Example 11: Compound (A) gives positive Lucas test in 5 minutes. When 6.0 g of (A) is treated withsodium metal, 1120 ml of hydrogen is evolved at STP. Assuming (A) to contain one atomof oxygen per molecule, write the structural formula of (A).Compound (A) when treated with PBr3 gives compound (B) which when treated withbenzene in the presence of anhydrous AlCl3 gives compound (C). Write the structuralformula of (B) and (C).

Solution: Let the molecular formula of (A) be ROH.[given that (A) contains one oxygen atom/molecule]

(A)ROH + Na RO– Na+ +

6g1120 ml at STP

12 H2

1 mole of (A) gives = 11200 ml at STPSo, the molar mass of (A) = 60 gm (6 gm of A gives 1120 ml of H2 at STP)Weight of alkyl group (R) = molar mass of A – (mass of O + mass of H)in one mole of (A)

= 60 – 16 – 1 = 43Let R = CnH2n + 1 12n + 2n + 1 = 43 n = 3Thus, the molecular formula of A is C3H7OHSince (A) gives positive Lucas test in 5 minutes, it is a secondary alcohol.


(A) = CH3CHCH3

OH

The reactions involved are

(A) (B) (C)

(CH3)2CHOHPBr3

(CH3)2CHBrC6H6

AlCl3CH

CH3

CH3

SUBJECTIVE PROBLEMS (BASIC LEVEL)

1. An ether was cleaved with excess of concentrated HI, giving two isomers of an alkyl iodide thatcontained 74.69% iodine. Identify the ether.

2. Compound (A), C7H8O is insoluble in aqueous NaHCO3 and aqueous NaOH. When treated withbromine water, (A) rapidly forms compound (B), C7H5OBr3. Identify (A) and (B).

3. An optically active alcohol A (C6H10O) absorbs two moles of hydrogen per mole of A upon catalytichydrogenation and gives a product B. The compound B is resistant to oxidation by CrO3 and doesnot show any optical activity. Deduce the structures of A and B.

4. Compound ‘X’ (molecular formula = C5H8O) does not react appreciably with Lucas reagent atroom temperature but gives a precipitate with ammonical silver nitrate. With excess of MeMgBr,0.42 g of ‘X’ gives 224 ml of CH4 at S.T.P. Treatment of ‘X’ with H2 in the presence of Pt catalystfollowed by boiling with excess of HI gives n-pentene. Suggest the structure for ‘X’ and write theequations involved.

5. Write mechanism for the following reaction:

(a)H+

CH2OH

6. Complete the following reactions:

(a) (CH3)2CHOCH3HI(excess)

Two products

(b) CH3CHO[A]dil. H2SO4

Hg2+ (CH3CO)2O [B]

(c) OH?

OH

CHO

(d)CH3MgBr

H3C — CHO ?

OH? H3C — C — CH3

H


(e) moistC2H5I

Ag2O(A)

H2SO4(B) C2H5OH

140°C(C)

(f)

OH

NO2

(C2H5)2SO4

OH–[A] Zn, HCl [B] NaNO2, HCl

5°C[C] [D]

C6H5OH

(g)OH

H+?

7. An organic compound [A] of formula C2H6O on treatment with concentrated H2SO4 givesa neutral compound [B] (C4H10O). The later on treatment with PCl5 gives a product whichon subsequent treatment with KCN yields [C] (C3H5N). Compound (C) on hydrolysis givesan acid, C3H6O2 and on reduction with sodium amalgam gives base C3H9N. Identify[A], [B], and [C].

8. An organic liquid [X] has a sweet smell and boiling point of 78°C. It contains C, H and O and onheating with concentrated H2SO4 forms a gaseous product (Y) of empirical formula CH2. (Y)decolorises bromine water and adds one mole of H2 in the presence of Ni. Identify (X) and (Y).

9. An organic compound (X) containing C, H and O reacts with sodium with the release of hydrogen.Controlled oxidation of (X) gives (Y) which forms a precipitate with sodium bisulphite but does notgive Ag mirror test. A solution of 7.41 g of (X) in 100 g of water freezes at –1.86 °C. Identify (X) and(Y). [Kf for water = 1.86°C/molal].

SUBJECTIVE PROBLEMS (ADVANCED LEVEL)1. An organic compound containing C, H, and O exists in two isomeric forms (A) and (B). An amount

of 0.108 g of one of the isomers gives on combustion 0.308 g of CO2 and 0.072 g of H2O. (A) isinsoluble in NaOH and NaHCO3 whereas (B) is soluble in NaOH. (A) reacts with concentrated HIto give compounds (C) and (D). (C) can be separated from (D) by the ethanolic AgNO3 solution and(D) is soluble in NaOH. (B) reacts readily with bromine water to give compound (E) of molecularformula, C7H5OBr3. Identify (A), (B), (C), (D), and (E).

2. An organic compound (A) containing 60% C and 13.3 % H gave the following results(i) 0.2 gm of the compound displaced 74.66 ml. of air at N.T.P.(ii) On treatment with PCl5, (A) gave another compound (B) which contained 45.2% chlorine(iii) On dehydration, (A) produced a hydrocarbon (C) containing 85.7% C and 14.7% H.(iv) On successive treatment with HI and moist silver oxide, (C) gave a compound (D), isomeric with(A).

Identify (A), (B), (C), and (D).

3. An organic compound (A), (C7H8O) yields two isomeric mono-nitro derivatives (B) and (C) onnitration. Treatment of (A) with acetyl chloride produces (D), which on oxidation with chromylchloride gives (E) whose oxidation with neutral KMnO4 followed by hydrolysis gives (F).Compound (F) on heating with soda-lime gives phenol. (A) on treatment with benzene sulfonylchloride produces (G) which on oxidation with KMnO4 gives (H). Hydrolysis of (H) also gives (F).Identify (A) to (H).


4. An organic compound (A), C7H8O2 on mild oxidation gives a highly volatile oil (B) which forms(C) on reaction with dimethyl sulphate in alkali. Oxidation of (C) with KMnO4 gives (D) which thenreacts with bromine water to give (E) containing about 72% bromine. Identify (A) to (E).

5. A 0.450 g of an aromatic compound (A) on ignition gives 0.905 g of CO2 and 0.185 g ofH2O. 0.350 g of (A) on boiling with HNO3 and adding AgNO3 solution gives 0.574 g of AgCl. Thevapour density of (A) is 87.5. (A) on hydrolysis with Ca(OH)2 yields (B) which on mild reductiongives an optically active compound (C). On heating (C) with I2 and NaOH, iodoform is producedalongwith (D). With HCl, (D) gives a solid which is markedly more soluble in hot water than in cold.Identify (A) to (D) with proper justification.

6. An organic compound (A) containing 78.69% of C, 8.19 % of H, and 13.11 % of O does not reactwith metallic sodium. (A) yields two compounds (B) and (C) on treatment with HI. (B) reacts withsodium to liberate hydrogen and also forms a salt with sodium hydroxide. (B) can be brominated andthe product contains two bromine atoms. Compound (C) containing iodine on treatment with Mg inether followed by the addition of CO2 followed by acidification with dilute acid gives acetic acid.Identify (A), (B), and (C).

7. An organic compound (A) is treated with moderately concentrated NaOH solution and aftersometime produces (B) and (C). The percentage of carbon, hydrogen, and oxygen in compound (B)is 77.8, 7.4, and 14.8 respectively. Oxidation of (B) yields (C) which on acidification with a mineralacid and further treatment with lithium aluminium hydride gives (B). Identify (A), (B), and (C).


ANSWERS

OBJECTIVES PROBLEMS

SECTION A

1. d 2. b 3. d 4. c 5. c6. c 7. d 8. b 9. a 10. d11. b 12. d 13. a 14. d 15. d16. b 17. d 18. a, d 19. c 20. c

SECTION B

1. d 2. c 3. a 4. a 5. a

6. d 7. c 8. a, b 9. d 10. a, b

11. b 12. b 13. d 14. b 15. a16. b 17. c 18. c 19. d 20. a

SUBJECTIVE PROBLEMS (BASIC LEVEL)

1. CH3CH2CH2OCHCH3

CH3

2. (A) = OCH3, (B) = Br OCH3

Br

Br

3.

CH3

(A) = CH C — C — OH

C2H5

,

CH3

(B) = CH3CH2 — C — OH

CH2CH3

4. (X) = HC C — CH2 — CH2 — CH2OHThe reactions involved are

(i) HC CCH2CH2CH2OH AgNo3

NH4OH AgC CCH2CH2CH2OH

(ii) HC C CH2CH2CH2OH 2CH3MgBr MgBrC C CH2CH2CH2OMgBr + 2CH4

(iii) HC CCH2CH2CH2OH H2/Pt CH3CH2CH2CH2CH2OHHI CH3CH2CH2CH= CH2


5. H+CH2OH CH2–OH2

+ –H2OCH2

+ +

HH

–H

6. (a) (CH3)2CHI + CH3I (b) CH CH[A], CH3CH(OCOCH3)2 [B] (c) CO2, 125°C, 3 – 5 atm. or CHCl3 + NaOH

(d)CH3 C

H

OMgBr

CH3 , H2O

(e)C2H5OH, C2H5HSO4, C2H5OC2H5

(f) [B] :[A] :

NO2

OC2H5

NH2

OC2H5

[C] :

N2Cl

OC2H5

+ –

[D] : HO N = N OC2H5

(g)CH3O

7. [A] : C2H5OH [B] : C2H5OC2H5 [C] : C2H5CN8. (X) : C2H5OH, (Y) : CH2 = CH2

9. (X) : CH3CH2CHOH

CH3

(Y) : CH3CH2C

CH3

O


, [E] and [F] :

COOH

[D] :Br

Br

COOH

Br

,Br

CH2COOH

Br

7. [A] : HCOOCH (CH3)2, [B] : CH3CH(OH)CH3

8. , [B] :

CH3

[A] :

COOH

COOH

,Br

COOH

C2H5

[C] :

COOH

9. [X] : CH3CH2COOCH2CH2CH3, [Y] : CH3CH2COOH[Z] : CH3CH2CH2OH

10. , [B] :

CH = CHCH3

[A] :

CH2CH = CH2

Cl

,

CH2CH2CH3

Cl

[C] :

ClOC2H5 OC2H5 OC2H5

11. [A] :CO

COO, [B] : C6H5

COOH

O

, [C] : OO

C6H5

[D] : NN

C6H5

Chemistry Alcohols

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