bt alcohols _theory-01_ .pdf

8/19/2019 BT Alcohols _Theory-01_ .pdf

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ALCOHOLS, PHENOLS AND ETHERS OP-CMIII-174

East Delhi Centre : Ph.: 22792226-27, West Delhi Centre : Ph.: 25527517-18, North Delhi Centre : Ph.: 25221424-25, South Delhi Centre : Ph.: 26537392-95 Fax: 26537396

The monohydric alcohols can further be classified according to the type of carbon that

bears the OH group. An alcohol in which OH group is attached to carbon atom, which in turn islinked to only one carbon atom, is called primary (1°) alcohol. Similarly, an alcohol in which OH

group is attached a carbon atom which is linked to two carbon atoms is called secondary (2°)alcohol and an alcohol in which the carbon atom bearing OH group is attached to three carbonatoms is called tertiary (3°) alcohol. For example,

R C OH

H

Primary(1°)

H

R C OH

R

Secondary(2°)

H

R C OH

R

Tertiary(3°)

R

2. IUPAC NOMENCLATURE OF ALCOHOLS

For certain alcohols, common names are used extensively with respect to their IUPAC names. The common names are derived by adding the suffix alcohol to the name of alkyl group. Common names of few alcohols are

Ethyl alcohol

CH3 –CH–CH3

OH

CH3CH2OH

Isopropyl alcohol

CH3 –CH–CH2OH

CH3

Isobutyl alcohol

CH2OH

Benzyl alcohol

CH3COH

tertbutyl alcohol

CH3

CH3

CH3CCH2OH

Neopentyl alcohol

CH3

CH3

The IUPAC rules of naming alcohols are as follows:

(a) Select the parent chain structure as the longest continuous carbon chain that contains the

OH group, then consider the compound to have been derived from this structure byreplacement of hydrogen by various groups. The parent structure is known as ethanol,propanol, butanol etc. depending on the number of carbon atoms. Each name is derived by

replacing the terminal −e of the corresponding alkane by −ol.

(b) Indicate by a number, the position of the OH group in the parent chain, using the lowest

possible number.(c) Indicate by numbers, the positions of other groups attached to the parent chain.

2Phenylethanol

CH3CH2 –CH–CH2OH

CH3

CH2CH2OH

2Methyl1butanol

CH3OHMethanol

2Methyl2butanol

CH3 –CH–CH–CH3

OH

CH3CH2CCH3

3Methyl2butanol

Cl–CH2CH2 –OH

CH3

2ChloroethanolOH

CH3

CH3 –CH–CH=CH2

OH3Buten2ol


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Alcohols containing two hydroxyl groups are called glycols. They have both common nameand IUPAC names. Common names are given in parenthesis.

1, 2ethanediol(Ethylene glycol)

CH3 –CHCH2OH

CH2 –CH2 CH2CH2CH2OH OH OH

1, 2propanediol(Propylene glycol)

OH OH1, 3propanediol

(Trimethylene glycol) cis1, 2cyclopentanediol

H

HH

H

H

HH

OH

H

HO

3. METHODS OF PREPARATION OF ALCOHOLS

Monohydric alcohols can be prepared by the following methods:3.1 OXYMERCURATION−DEMERCURATION

Alkenes can be converted into alcohols by oxymercurationdemercuration reaction. In thisreaction, addition of water takes place according to Markownikoff’s rule.

Mercuricacetate

C = C + Hg(OAc)2 + H2O C C

OH

NaBH4C C

HgOAc

OH HMarkownikoff

addition

AcOH –Hg

–OAc –

For example,

CH3CCH=CH2 CH3C CHCH2

CH3

CH3

Hg(OAc)2

+ H2O

CH3

CH3 OH

HgOCOCH3NaBH4

CH3C CHCH3

CH3

CH3 OH

3.2 HYDROBORATION−OXIDATION

Alkenes react with diborane to form trialkyl boranes, which upon treatment with alkaline

H2O2 give alcohols via antiMarkownikoff’s addition of water.

Alkyl borane

C = C + (BH3)2 H2O2, OH

H BH2

C COHH

C C + B(OH)3

AntiMarkownikoff addition of H2O

THF

For example,

CH3

(i) (BH3)2 (ii) H2O2, OH

Synadditon

OH

CH3

H

H

1Methylcyclopentene 2Methyl1cyclopentanol


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CH3CCH=CH2

CH3

CH3

(i) (BH3)2 (ii) H2O2 , OH

3, 3Dimethyl1butene

CH3CCH2CH2OH

CH3

CH33, 3Dimethyl1butanol

(No rearrangement)

(i) Hydroboration–oxidation of non–terminal alkynes leads to the formation of ketones. For example,

3RCCR BH3 (RCH=CR)3BH2O2/OH

3 R–CH=C–R 3 RCH2COR

OH

THF

(ii) Noncatalytic hydrogenation: Hydroboration–oxidation can be used to reduce doublebonds in alkene without using molecular hydrogen and a transition metal catalyst. For example,

Me2C=CHMeBH3

(Me2CHCHMe)3B 3 Me2CHCH2Me3 CH3COOH

THF

(iii) Selective reduction: Less hindered double bonds can be selectively reduced by

hydroborationoxidation. For example,

(CH3)2CHCH BH +

CH3

CH=CH2

Me

CH2CH2B(CHCHCH3)2

Me

CH3

CH3

H2O2/OH

CH3COOH

CH2CH2OH

Me

CH2CH3

Me

2

3.3 GRIGNARD SYNTHESIS

When a solution of an alkyl halide in dry ethyl ether, (C2H5)2O, is allowed to stand over turnings of metallic magnesium, a vigorous reaction takes place. The solution turns cloudy, beginsto boil and the magnesium metal gradually disappears. The resulting solution is known asGrignard reagent. It is one of the most useful and versatile reagents known to the organicchemists.

RX + Mg ether Dry

RMgX

Alkyl halide Alkyl magnesium halide

The Grignard reagent has the general formula RMgX , and the general name

alkyl magnesium halide. The carbonmagnesium bond is covalent but highly polar, with carbonpulling away electrons from electropositive magnesium but the magnesiumhalogen bond is

essentially ionic, R MgX.

Since magnesium becomes bonded to the same carbon that previously held halogen, the

alkyl group remains intact during the preparation of the reagent. Thus, npropyl chloride yields

npropyl magnesium chloride and isopropyl chloride yields isopropyl magnesium chloride.


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CH3CH2CH2Cl + Mg ether Dry

CH3CH2CH2MgCl

nPropyl chloride nPropyl magnesium chloride

(CH3)2CHCl + Mg ether Dry

(CH3)2CHMgCl

Isopropyl chloride Isopropyl magnesium chloride

The Grignard reagent belongs to a class of compounds called organometallic compounds,in which carbon is bonded to a metal like lithium, potassium, sodium, zinc, mercury, lead, thalliumor to almost any metal known. Each kind of organometallic compound has its own set of propertiesand its particular uses depend on these. But whatever the metal, it is less electronegative than

carbon and the carbonmetal bond is always highly polar. Although the organic group is not afullfledged carbanion but has considerable carbanionic characters. Thus, organometalliccompounds can serve as a source of carbon bearing negative charge.

The Grignard reagent has the formula RMgX and is prepared by the reaction of metallicmagnesium with the appropriate organic halide. This halide can be alkyl (1°, 2°, 3°), allylic, aryl

alkyl (e.g. benzyl), or aryl (phenyl) or substituted phenyl. The halogen may be Cl, Br or I, (Arylmagnesium chlorides must be made in the cyclic ether tetrahydrofuran instead of ethyl ether). Aldehydes and ketones resemble each other closely in most of their reactions. The carbonyl group

is also unsaturated and like the carboncarbon bond, it also undergoes addition. One of the typicalreaction is cis addition of the Grignard reagent.

The electrons of the carbonyl double bond hold together atoms of different

electronegativity, thus, the electrons are not equally shared, the mobile cloud is pulled stronglytowards the more electronegative atom, oxygen. The addition of an unsymmetrical reagenthappens such that the nucleophilic (basic) portion attaches itself to carbon and the electrophilic(acidic) portion attaches itself to oxygen.

The carbonmagnesium bond of the Grignard reagent is a highly polar bond, carbon beingnegative relative to electropositive magnesium. When Grignard reagent is added to carbonylcompounds, the organic group attaches to carbon and magnesium to oxygen.

C=O + R: MgX COMgX

R

H2O COH + Mg(OH)X

R

+ +

H+

Mg2+ + X + H2O

An alcohol

The product is a magnesium salt of the weakly acidic alcohol and is easily converted intothe alcohol by the addition of the stronger acid, water. The Mg(OH)X thus formed is a gelatinousmaterial, which forms coating over carbonyl compound, thus dilute mineral acid (HCl, H 2SO4) is

commonly used instead of water, so that water soluble magnesium salts are formed.

PRODUCTS OF THE GRIGNARD SYNTHESIS:

The type of alcohol that is obtained from a Grignard synthesis depends upon the type of carbonyl compound used. Formaldehyde (HCHO) yields primary alcohols, other aldehydes(RCHO) yield secondary alcohols and ketones (R2CO) yield tertiary alcohols.

The number of hydrogens attached to the carbonyl carbon defines the carbonyl compoundas formaldehyde, higher aldehydes or ketone. The carbonyl carbon is the one that finally bears the


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OH group in the product and the number of hydrogen defines the alcohol as primary, secondary,or tertiary.

For example,

CH3CH2CHCH3 + HC=O

MgBr Formaldehyde

H

CH3CH2CHCH2OMgBr H2O

CH3

CH3CH2CHCH2OH

CH3

secButyl carbinol(1°alcohol)

secbutyl magnesiumbromide

MgBr + CH3C=O CHOMgBr

CH3H2O

CHOH

CH3

Phenyl magnesiumbromide

AcetaldehydePhenylmethylcarbinol

(1Phenylethanol)

(1° alcohol)

H

nC4H9MgBr + CH3C=O Acetone

nC4H9COMgBr H2O

CH3

nC4H9COH

CH3

nButyldimethyl carbinol(3°alcohol)

CH3

nButyl magnesiumbromide

CH3 CH3

C=O + RMgX COMgX

R

H2O COH + Mg++ + X

R

HC=O + RMgX HCOMgX

R

H2OHCOH

R

H HH

(1°alcohol)

Formaldehyde

RC=O + RMgX RCOMgX

R

H2O RCOH

R

H HH

(2°alcohol)

Other aldehydes

RC=O + RMgX RCOMgX

R

H2ORCOH

R

R

(3°alcohol)

Ketone

R R

Reactivity order of the substrates with a Grignard reagent is

Active H compounds > –CHO > C=O > –COCl > –CO2R > –CH2X.

EXCEPTIONAL BEHAVIOUR OF GRIGNARD REAGENT:

Sometimes, Grignard reagent does not react with compounds containing functional groupnormally capable of undergoing reaction. Generally, branching of the carbon chain near thefunctional group prevents the reaction. For example, methyl magnesium bromide or iodide doesnot react with hexamethyl acetone (CH3)3CCOC(CH3)3. It has also been found that if Grignard

reagent contains large alkyl groups, reaction may be prevented. For example, isopropyl methyl


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acetone reacts with methyl magnesium iodide but not with tbutyl magnesium iodide. In other cases, abnormal reaction may take place. For example, when isopropyl magnesium bromide isadded to diisopropyl ketone, the expected tertiary alcohol is not formed, instead the secondary

alcohol, diisopropyl carbinol is obtained resulting from the reduction of the ketone.

(CH3)2CHCOCH(CH3)2 CHMgBr )CH( 23 (CH3)2CHCH(OH)CH(CH3)2 + CH3CH=CH2

, unsaturated carbonyl compounds adds on Grignard reagents at the 1, 2 or 1, 4positions.

1, 2−addition:

R1CH=CHC=OR MgX

H3O+

R1CH=CHCOMgX

R

R2

R1CH=CHCOH

R3

R2

R2

1, 4−addition:

R1CH=CHC=OR

3MgX

H3O+

R1CHCH=COMgX

R

R1CHCH=COH

R

R2 R3

tautomerizesR1CH=CH2CR

2

OR3 R3

[enol]

The reaction of dihalides of the type Br(CH2)nBr with magnesium depends on the value of n. For n = 1, no Grignard reagent is formed. For example,

CH2CH2Mg

CH2=CH2, CH2CH2CH2ether Mg

ether

Br Br Br Br

When n 4, the Grignard reagent can be possibly made with dibromo compound i.e,Br(CH2)nBr.

ORGANOLITHIUM COMPOUNDS:

Alkyl lithium compounds can be prepared by direct displacement, the chlorides give bestyields. For example,

Bu–Cl + 2Lihexane

N2 Bu–Li + LiCl

Other organolithium compounds can be prepared by the halogenmetal exchange with thebutyl lithium compound.

BuLi + RX RLi + BuX


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Organolithium compounds behave like the Grignard reagent but the lithium compounds areusually more reactive and the yield of the product is often better. Because of their sensitivity tooxygen and to water, reaction with lithium alkyls is best carried out in an atmosphere of dry

nitrogen. Alkyl lithium reacts with CO2 to give carboxylate salt.

RLi + CO2 RCO

OLi+

It has been found that sterically hindered talcohols cannot be prepared by the Grignardreaction. On the other hand, many of these alcohols can be prepared by means of lithium alkyls.For example,

(CH3)2CHCOCH(CH3)2OH

LiHC)CH(

2

23 [(CH3)2CH]3COH + LiOH

3.4 HYDROLYSIS OF ALKYL HALIDES

Alkyl halides on hydrolysis gives alcohols either by SN1 or SN2 route depending upon thestructure of alkyl halide and the reaction conditions employed.

RX + OH (or H2O) ROH + X (or HX)

3.5 REDUCTION OF CARBONYL COMPOUNDS

Aldehydes can be reduced to primary alcohols and ketones to secondary alcohols, either by catalytic hydrogenation or by the use of chemical reducing agents like lithium aluminiumhydride (LiAlH4), H2/Ni, B2H6/THF etc.

3.5.1 By LiAlH4

It can reduce all functional groups of column ‘A’ into Column ‘B’. Generally, it is unable to

reduce double bond, which are in conjugation with C=O group the double bond is reduced

when a phenyl group is attached to the carbon.

For example,

C6H5CH=CHCHO 4LiAlH C6H5CH2CH2CH2 –OH

“A” “B”

CHO CH2OH

C=O CHOH

COOH CH2OH

COOR CH2OH + ROH

COCl CH2OH

(RCO)2O RCH2OH

lactone diol

epoxide alcohol


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3.5.2 By NaBH4

It can reduce all the above groups of column ‘A’ into groups of column ‘B’ except, COOH

and COOR. NaBH4 is able to reduce the double bonds, which are in conjugationwith C=O group.

3.5.3 By H2 /Ni

It can reduce all the above groups of column ‘A’ into groups of column ‘B’. It can also

reduce double and triple bond present in carbonyl compounds irrespective of the positionin carbon chain whether it is in conjugation with carbonyl group or not.

CH3CH=CHCHO Ni/H2 CH3CH2CH2CH2 –OH

CH2=CHCH2CHO Ni/H2 CH3CH2CH2CH2 –OH

3.5.4 By B2H6 /THFIt can reduce all above groups of column ‘A’ into groups of column ‘B’ except, COCl.

REDUCTION BY LiAlH4 OR NaBH4

The commonly used reagent for the reduction of aldehydes and ketones are LiAlH4 or

NaBH4. The reaction proceeds by successive transfer of hydride ions (H) from boron or aluminium

to four different carbonyl carbons. As all four of the hydrides are transferred, there are distinct

reducing agent used in every step of reduction. In the first step, the reducing agent is AlH4 (I),

while in the second step, the reducing agent is RCH2OAlH3 (II).

In the same manner, species [RCH2O]2 AlH2 (species III) and [RCH2O]3 AlH

(species IV)

will be obtained. Reduction with NaBH4 is usually done in water or in alcohol which are used as the

solvent. The solvent destroys the intermediate alkoxide ion and yields alcohol. If water or analcohol is not used as the solvent, aqueous acid can be added after the reduction to convert thealkoxide to alcohol.

The mechanism for reduction with lithium aluminium hydride is very similar to that of NaBH4. As LiAlH4 violently reacts with water and other polar protic solvents to give molecular hydrogen, reduction with LiAlH4 are done in aprotic solvents like anhydrous ether. Ethyl acetate isadded to decompose the aluminium complex. But reduction with NaBH4 can be carried out inwater or ethanol solution.

Mechanism of reduction by LiAlH4:

RCHO

H AlH3ether

RCOAlH3

H

H

RCH=O

[RCH2O]2 AlH2

RCH=O

[RCH2O]3 AlHRCH=O

RCH2O AlOCH2R

OCH2R

OCH2R

CH3CO2Et4RCH2OH + Al(OH)3

Aldehydes are reduced to primary alcohols by both LiAlH4 and NaBH4 while these reducingagents reduce ketones to secondary alcohols.

(i) OOH

HLiAlH4


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(ii) Ph

OLiAlD4

H3O PhCDOH.

CH3

(iii) OOH

HNaBH4

(iv)H

O2NaBH4

O

H

OHOH

3.6 REDUCTION OF ESTERS

In chemical reduction of esters, the acid portion of ester is converted into primary alcohol.

RCOOR agentducingRe RCH2OH + ROH

For example,

CH3(CH2)8COOCH32

422

in/5000,C150

OCuCr ,CuO,H

CH3(CH2)8CH2OH + CH3OH

CH3(CH2)10COOC2H5 4LiAlH CH3(CH2)10CH2 –OH + C2H5 –OH

Carboxylic acids and esters produce alcohols by LiAlH4 but not by NaBH4.

OR

OLiAlH4

O OH

OHNaBH4

OR

OOH

ROH

4. GENERAL PHYSICAL PROPERTIES OF THE ALCOHOLS

The properties of the alcohols are largely determined by OH group. However, the alkylgroup (which is inert) also plays important role to explain some of the physical properties likeboiling point and solubility in water.

4.1 BOILING POINT

The lower members like methanol, ethanol, 1propanol have higher boiling points. Theboiling point rises as the molecular weights of the alcohol increases. It is quite evident that withinthe homologous series, the alcohols of normal chain show a rise in the boiling points with theincrease in molecular weights. Like alkanes, the branched chain isomers of alcohol have

lower boiling points. Thus among the four isomeric butyl alcohols, tbutyl has the lowest boilingpoints (80°C).

The order of boiling points of isomeric butyl alcohols is

nButyl alcohol, CH3(CH2)3OH > isobutyl alcohol, (CH3)2CHCH2OH > secbutyl

alcohol, CH3CH(OH)CH2CH3 > tbutyl alcohol, (CH3)3COH


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Compactness of tbutyl alcohol reduces the surface area and hence lowers the boilingpoint. The boiling point of alcohols are much higher than alkanes of comparable molecular weights. The high boiling point of the alcohols is due to hydrogen bonding by which the alcohol

molecules remain associated in the liquid state.

HO HO HO HO

R R R R

The hydrogen of one (OH) group forms a loose bond with the oxygen of OH group of

another molecule i.e. they remain in the molecular association through inter molecular hydrogenbonding, which accounts for their high boiling points. But the hydrogen bonding is not so extensiveas in water molecules. So, water boils at a higher temperature than methyl and ethyl alcohols.

4.2 SOLUBILITY IN WATER

The lower members of alcohols are highly soluble in water but as the size of the alkyl groupincreases, the solubility decreases. This phenomenon is common with other organic compoundshaving atleast one electronegative atom or group like ethers, aldehydes, ketones, acids, amides,sugars etc. and they all are soluble in water. The solubility of alcohols is attributed to its ability toform hydrogen bonds with water.

RO HO

H H

+

But as the molecular weight increases, the solubility decreases. For example, methanol is

infinitely soluble but only 0.6 g of nhexyl alcohol dissolves in 100 ml of water. In general, organiccompounds having atleast one electronegative element become gradually insoluble in water as the

hydrocarbon chain increases. However, branching of the alcohol increases the solubility.Thus, tbutyl alcohol is infinitely soluble but 1butanol is slightly soluble in water. This is again dueto the compactness of the molecule. Better and easy surrounding by water increases thesolubility. Increase in the number of OH groups increases the solubility. For example, glycol(two OH groups) and glycerol (three OH groups) are more soluble in water than methanol andethanol.

5. GENERAL CHEMICAL PROPERTIES OF THE ALCOHOLS

Chemical reactions are mainly based on OH group of alcohol. The oxygen atom of OH

group polarize both the CO bond and the OH bond of any alcohol. Polarization of the OH bondmakes the hydrogen partially positive and explain why alcohols are weak acids. Polarization of theCO bond makes the carbon atoms partially positive. The polarization of CO bond is responsiblefor the weak basic character of alcohols.

On the basis of the above explanation of polarization of CO and OH bonds, we canclassify the reaction of alcohols into two parts, first due to breaking of the CO bond, with removalof OH group and second due to the breaking of OH bond, with removal of ‘H’. Except thesereactions, alcohols show some other reactions like oxidation, reduction, elimination etc. Alkylgroup of alcohols is also responsible for some chemical reactions.


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5.1 REACTION DUE TO O−H BOND CLEAVAGE

5.1.1 REACTION WITH ALKALI METALS:

Active metals (Na, K, Mg, Al etc) when treated with alcohols give hydrogen gas. In thisreaction, order of reactivity of alcohols is CH3OH > 1° > 2° > 3°. This reaction exhibitsacidic character of alcohols.

ROH + Na RONa+ + ½ H2

In OH group of alcohols, oxygen is more electronegative than hydrogen, this results inpolarization of OH bond due to which acidic nature arises in alcohols. Reaction of activemetals with alcohols shows that alcohols are acidic in nature.

RONa+ + HOH NaOH + ROHStronger Stronger Weaker Weaker

base acid base acid

The order of acidity for some compounds is

H2O > ROH > HCCH > NH3 > RH

The order of basicity is

R > 2NH > HCC > OR > OH

The above order is based on the reactions of alcohols with other species.

C2H5OH + Na C2H5O Na

+ ½ H2

HCCNa+ + ROH HCCH + RONa+

5.1.2 ESTERIFICATION:

A direct reaction between a carboxylic acid and alcohol under the catalytic effectof sulphuric acid yields an ester. This is a reversible reaction and is known as the”Fischer esteri f icat ion ”.

RCO

OH

FastRC

OH

+ H+

OH

RC

OH

OH:

..

RC

OH

OH ..Ia

..:

Slow RC O

OH

OH

RCOR

OH2

OH

RC

OH

OH

OH:

R

IMPEHR

..

+

Ib

..

..

: ..

..

H+

RCOR

OH

RCOR

H2O

Fast

..:

RCOR

O OH..


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5.2 REACTIONS DUE TO C−O BOND CLEAVAGEPolarization of the CO bond makes the carbon atom partially positive, so this carbon

would be susceptible to the nucleophilic attack and if it were not for the fact that OH is a strongbase and they are very poor leaving group. Protonation of the alcohol converts a poor leaving

group (OH) into a good one. It also make the carbon atom even more positive (because 2OH is

more electron withdrawing than OH) and therefore, even more susceptible to nucleophilic attack.Now, nucleophilic substitution reactions are possible in alcohols according to the givenmechanism.

5.2.1 REACTION WITH H−X:

For example,

CH3CHCH3Conc. HBr

OHor (NaBr + H2SO4)

Isopropyl alcohol

CH3CHCH3

Br

Isopropyl bromide

CH3CH2CH2CH2CH2OHConc. HCl / ZnCl2

CH3CH2CH2CH2CH2Cl

npentyl alcohol npentyl chloride

CH3CCH3Conc. HCl

OH

CH3

CH3CCH3

Cl

CH3

tert Butyl alcohol tert Butyl chloride

In SN1 reactions of alcohol (when R group is 3°), R may rearrange. Primary alcohols and

methanol apparently react through a mechanism that we recognize as an SN2 type.

With HBr, alcohols produce alkyl bromide.

ROH + H.. ROH2

..

Br

Slow

SN2Br R+ H2O

or for stable carbocation R

, racemic mixture will be obtained for optically active alcohol bySN1 mechanism.

But 3pentanol reacts with HBr to produce 2 and 3bromopentane derivatives.

CH3CHCHCH2CH3

H OH

H

CH3CHCHCH2CH3

H OH2

CH3CHCHCH2CH3

H

CH3CHCH2CH2CH3

I (More +I groupsattached to C

+)

Br CH3CH2CHCH2CH3

Br

(II a)

Br CH3CHCH2CH2CH3

(II b)II (More members of hyperconjugation structure)

Br

In SNi reaction, retention of configuration is observed & mechanism operates through

intimate ionpair formation.


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3C6H5CHCH3PBr 3

OH

3C6H5CHCH3 + H3PO3

Br

5.2.3 REACTION WITH SOCl2:

Alcohols react with thionyl chloride in presence of pyridine to give alkyl chloride withinverted configuration, while in absence of pyridine, an alkyl chloride with retention of configurationis obtained via SNi mechanism.

ROH + SOCl2 RCl + SO2 + HCl

5.2.4 ACID CATALYSED DEHYDRATION OF ALCOHOLS:

Alcohols in presence of dilute acid undergo dehydration forming alkenes. The reactionproceeds by E1 mechanism.

H

CH2OH..

CH2OH2 H

H

CH2

Dehydration of cyclic alcohol is accompanied by expansion in the above reaction. Stability of the ring is given by Baeyer strain theory, according to which the stability order of the rings is6 > 7, 5 > 8, 9 >> 4 > 3 .

5.3 OXIDATION OF 1°, 2°, 3° ALCOHOLS WITH CrO3 OR K2Cr 2O7 IN ACID

Alcohols with atleast one hydrogen atom on 1° and 2° carbinol are oxidised to carbonylcompounds (aldehydes and ketones). PCC (Mixture of pyridine, HCl and CrO3) oxidises 1° alcoholto aldehydes but K2Cr 2O7 or KMnO4 in acid converts 1° alcohol directly to carboxylic acids. Under

mild conditions, 3° alcohols are not oxidised.

Jone’s reagent (chromic acid in aqueous acetone solution). This is a sufficiently mildoxidising agent, so that it oxidises alcohols without oxidising or rearranging double bonds. MnO2can oxidise 1° allylic or 1° benzylic alcohols selectively into aldehydes.

For example,

CH3 –CH=CH–CH2OHacetonein

CrOH 42 CH3 –CH=CH–CH=O

H MnO2

OHO

CH2OHMnO2

CH=O

CHROMIC ACID (JONES REAGENT)

The most convenient reagent for the oxidation of alcohols is 8N chromic acid in sulphuricacid (Jones reagent). Two millimoles of this reagent oxidizes 3 millimoles of monohydric alcoholaccording to the equation:

3R2CHOH + 2H2CrO4 3R2C=O + 2Cr(OH)3 + 2H2OThe function of sulphuric acid is to prevent complex formation of Cr(VI) with its reduced

form Cr (III) to a salt having much less oxidation potential. This ensures that all the Cr( VI) is usedin oxidation, which thus becomes rapid and complete. Acetone is the usual solvent used at ice


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bath temperature. Alcohols having double or triple bonds in the molecule can be selectivelyoxidized to ketones in good yields. For example,

CH3(CH2)3CCCH – CH3 CH3(CH2)3CCC – CH3

OH O

The most likely mechanism for the oxidation of alcohols by Jones reagents has been shownto be

R2CHOH + Cr 6+ R2C=O + Cr

4+ + 2H+

Cr 4+ + Cr 6+ 2Cr 5+

3Cr 5+ + 2R2CHOH 2R2C=O + 2Cr 3+ + 4H+

3R2CHOH + 2Cr 6+ 3R2C=O + 2Cr

3+ + 4H+

Thus, apparently, 3 moles of alcohols react with 2 moles of Cr(VI) to give 3 moles of ketone and 2 moles of Cr(III). It is clear, however, that only one mole of the alcohol is oxidizeddirectly by Cr(VI) and the other two part in the oxidation with Cr (V).

5.4 CLEAVAGE OF 1, 2−GLYCOLS

Lead tetraacetate or periodic acid are commonly used for the cleavage of 1, 2glycols.The former reagent is used in anhydrous solvent, whereas the later in organic solvent. Periodic

acid is more selective and readily cleaves 1, 2glycols at room temperature. But cleavage of an hydroxy ketone or acid by this reagent even at higher temperature is slow. Lead tetraacetate,however, oxidizes hydroxyketone or acid as well as 1, 2glycols more easily. This is explainedon the basis of a fivemembered cyclic intermediate.

C

CI

O

O

O

O

OH

C O + C O

C

CPb(OAc)2

O

O

C O + C O

A 1, 2glycol need not necessarily be cis to undergo cleavage with lead tetraacetate.For instance, trans9, 10decalindiol undergoes cleavage to cyclodecane1, 6dione.

OH

OH

Pb(OAc)4OO

5.5 HALOFORM REACTIONThe general reaction is represented as

RCHCH3 + X2 + NaOH

OH

RCO Na + CHX3 + NaX + H2O

O


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The CH3 carbon is lost as CHX3 and the remaining part exists as acid salt, which can beacidified to liberate free acid.

The structural feature essential in the compound to show haloform reaction is that any of

the following moieties should be present in the molecule attached to some electronwithdrawinggroup or electron donating group by +I only.

CH3CH or XCH2CH or X2CHCH or X3CCH

OH OH OH OH

CH3C or

O

XCH2C or

O

X2CHC or

O

X3CC

O

or any other grouping that can be converted to any of the above moieties.

The mechanism of the reaction can be outlined as:

The reaction has 3 important steps. Step I is the oxidation, caused by mild oxidizing agent

(hypo halite ion). The second step is basepromoted halogenation and the third step is cleavageof C–C bond.

RCHCH3

OH

RCO + CHI3

O

OI

(Oxidisingagent)

RCCH3O

OH

RCCH2O

..

Carbanion I(Resonance stabilized)

I I+

IRCCH2I

O

RCHCH2I

OH

OH

RCCHI

O

..

Carbanion II(more resonance stabilized

due to I effect of iodine)

II I

+

RCCHI2O

OI

OI

RCHCHI2

OH

I2/OH

RCCI3O

OI

RCHCI3

OHOH

RCCI3O

RCOH + CI3

O

..

Protonexchange

OH

Yellow crystalline ppt.

(M.P = 119°C)

I2 + 2OH – I – + OI – + H2O

H2O

Some of the compounds which responds positively to iodoform test are

CH3CH2OH (only primary alcohol)

CH3CHCH3

OH,

CH3CHCH2CH3

OH,

PhCHCH3OH

(secondary alcohols)

CH3CH

O(only aldehyde) ,

CH3CCH3O

,CH3CCH2CH3

O,

PhCCH3O

(ketones)


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CH3CHCH3X

,

CH3CCH3X

X

,

CH3CCH2CCH3O O

,

CH3COEt

O(after heating with OH –)

CH3CCH2CO2H

O

(after heating)

The compounds that respond negatively to iodoform test are

CH3COH

O,

CH3CCl

O,

CH3CNH2O

,CH3COR

O,

CH3COCCH3O O

CH3CCH2COH

O O

,CH3CCH2COEt

O O

Give a simple chemical test that can distinguish between 2–pentanol and 3–pentanol.

6. METHODS FOR DISTINGUISHING PRIMARY, SECONDARY ANDTERTIARY ALCOHOLS

The following methods are used for distinguishing three types of monohydric alcohols.6.1 LUCAS TEST

Alcohols react with concentrated hydrochloric acid in the presence of anhydrous zincchloride to form alkyl chlorides. The alkyl chlorides appear as cloudiness because of itsinsolubility in Lucas reagent.

ROH + HCl 2ZnCl

RCl + H2O Alkyl halide

The three types of alcohols undergo this reaction at different rates. The rates of reactionwith Lucas reagent [conc. HCl + ZnCl2 (anhydrous)] follow the given order:

Tertiary alcohol > Secondary alcohol > Primary alcohol

An unknown alcohol (monohydric) is mixed with conc. HCl and anhydrous ZnCl2 at roomtemperature. The alkyl chloride formed is insoluble in the medium, thus the solutionbecomes cloudy before it separates out as a distinct layer. The following observations aremade,

(a) If cloudiness (white turbidity) appears immediately, the alcohol is tertiary.

(b) If cloudiness appears within 5 minutes, the alcohol is secondary.

(c) If the solution remains clear, i.e., no cloudiness is formed the alcohol is primary.

Remember that the benzyl and allyl alcohol react as rapidly as tertiary alcohol with Lucasreagent because their cations are resonance stabilised and as stable as 3° carbocations.


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Can Lucas test distinguish CH 3CH 2 –OH, PhCH 2 –OH and Me3C–OH ? Explain.

6.2 DICHROMATE TEST (OXIDATION TEST)

This test is based on the fact that three types of monohydric alcohols give differentoxidation products on oxidation.

The unknown alcohol is treated with sodium dichromate in dilute sulphuric acid(orange solution) at room temperature. Then the oxidation products are identified.

(i) A carboxylic acid with same number of carbon atoms as in the alcohol, if formedconfirms the primary alcohol. The colour of the solution changes from orange to green.

(ii) A ketone with same number of carbon atoms as in the alcohol if formed confirms thesecondary alcohol. The colour of the solution also changes from orange to green.

(iii) In case the colour of the solution does not change, i.e. it remains same, then it is

3° alcohol.

6.3 VICTOR MEYER’S TEST

In this test, the following steps are involved,

(i) Alcohol is reacted with conc. HI or red phosphorous and iodine to form correspondingalkyl iodide.

(ii) Alkyl iodide is then treated with silver nitrite when corresponding nitroalkanes is formed.

(iii) Nitro alkane is treated with nitrous acid (NaNO2 + H2SO4) and the solution is madealkaline by addition of excess of caustic soda.

Primary Secondary Tertiary

RCH2OH

HI

RCH2I

AgNO2

RCH2NO2

HNO2

RCNO2

NaOH

NOH

Blood red colour

Nitrolic acid

R2CHOH

HI

R2CHI

AgNO2

R2CHNO2

HNO2

R2CNO2

NaOH

N=O

Blue colour

Nitrolic acid

R3COH

HI

R3CI

AgNO2

R3CNO2

HNO2

No reaction

NaOH

Colourless


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PROFICIENCY TEST− I

The following 10 questions deal with the basic concepts of this section. Answer thefollowing briefly. Go to the next section only if your score is greater than or equal to 8.Do not consult the study material while attempting the questions.

1. If 1° alcohol is oxidised by PCC, which product do you expect?

2. If 3° alcohol is oxidised by K2Cr 2O7 in mild condition, will it be oxidised?

3. Will tbutanol respond to haloform test?

4. R1R2R3COH 2SOCl

(A).

The product (A) will have of configuration.

5. ROH 3OCH

(A)

Will (A) be ether?

6. Dehydration of alcohol always leads to Saytzeff/Hoffman product.

7. CH2CH2CH2Cl

14OH

O

Product mixture

18

In the product mixture, do you expect the product which will have O18 linked with C14 ?

8. Which one is a strong base, CH3O or OH ?

9. Which can replace the other one between OTs , OCH3 ?

10. Alcohols contain bond among them.


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bt alcohols _theory-01_ .pdf

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