ALCOHOL

1. INTRODUCTION 2

2. METHOD OF FORMATION 3

3. PHYSICAL PROPERTIES OF ALCOHOL 7

4. CHEMICAL PROPERTIES OF ALCOHOL 7

5. DISTINCTION BETWEEN PRIMARY, SECONDARY

AND TERTIARYALCOHOLS 12

PHENOL

1. INTRODUCTION 16

2. METHOD OF PREPARATION OF PHENOL 16

3. PHYSICAL PROPERTIES OF PHENOL 17

4. CHEMICAL PROPERTIES OF PHENOL 19

5. DISTINCTION BETWEEN ALCOHOLAND PHENOL 22

ETHERS

METHOD OF PREPARATION OF ETHERS1. 23

2. PHYSICAL PROPERTIES OF ETHERS 24

3. CHEMICAL PROPERTIES OF ETHERS 24

1

CONTENTS

S.NO. TOPIC PAGE NO

Alcohol

1. INTRODUCTION :(i) These are the organic compounds in which –OH group is directly attached with carbon.

(ii) These are hydroxy derivatives of alkanes and mono alkyl derivatives of water.(iii) Their general formula is CnH2n+1OH or CnH2n+2O.

Classification ofAlcoholsMono, Di, Tri or Polyhydric Compounds

Alcohols and phenols may be classified as mono–, di–, tri- or polyhydric compounds depending onwhether they contain one, two, three or many hydroxyl groups respectively in their structures as givenbelow:

(i) Compounds containing C OH bond: In this class ofalcohols, the –OH group is attached to ansp3

sp3 hybridised carbon atom of an alkyl group. They are further classified as follows(a) Primary, secondary and tertiary alcohols: In these three types of alcohols, the –OH group is

attached primary, secondary and tertiary carbon atom, respectively as depicted below:

(b) Allylic alcohols: In these alcohols, the —OH group is attached to a sp3 hybridised carbon next tothe carbon-carbon double bond, that is to an allylic carbon. For example

(c) Benzylic alcohols: In these alcohols, the —OH group is attached to a sp3—hybridised carbonatom next to an aromatic ring. For example

Allylic and benzylic alcohols may be primary, secondary or tertiary.

2

ALCOHOL

Alcohol

(ii) Compounds containing C OH bond: These alcohols contain —OH group bonded to a car-sp2

bon-carbon double bond i.e., to a vinylic carbon or to an aryl carbon. These alcohols are alsoknown as vinylic alcohols.Vinylic alcohol: CH = CH –OH

2

Phenols :

2. METHOD OF FORMATION :From alkenes

(1) By acid catalyzed hydration: Alkenes react with water in the presence of acid as catalyst to form

alcohol. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with

Markonikov’s rule

Mechanism : The mechanism of the reaction involves the following three steps:

Step 1:

H O + H+ H O+

3 3

Step 2:

Step 3:

(2) By hydroboration–oxidation: Diborane (BH ) reacts with alkenes to give trialkyl boranes as3 2

addition product. This is oxidized to alcohol byhydrogen peroxide in the presence of aqueous sodiumhydroxide.

H BH2

CH3 – CH = CH2

CH3 – CH = CH2

(CH3 – CH2 – CH2)3B (CH3 – CH2 – CH2)2BH–

2H2O2, OHH2O

CH3 – CH2– CH2–OH + B(OH)3

Propan–1–ol

Note : In this reaction addition of water at double bond according to Anti Markonikov Rule

3

Alcohol

Mechanism of Hydroborations-deboration

B2H6 THF/H 2O2 /OH e.g., CH3 —CH = CH2 CH3 —CH2 — CH2 —OH

H– transfer+ CH3 – CH = CH2 CH – CH – CH2 CH3 —CH2 – CH2

. . O.

.BH2BH3

BH3

These steps are repeated twice to form (CH3 – CH2 – CH2)2B and then :

R R R. . . .

– H . . . .H – .O. – .O. – H

R —B R —B —O —O —H R —B —O. . —.O. —H

R R H R

– OH

R —B —OR

R

OR|

RO —B —ORWith H2O2 , finally is formed by above mentioned method

OR|

RO —B —OR 3 NaOH

3. Oxymercuration

Na BO + 3ROH3 3

Involves an electrophilic attack on the double bond by the positively charged mercury species. The product

is a mercurinium ion, an organometallic cation containing a three-membered ring.

With mercuric acetate, the product is 3-methyl-2-butanol (Markonikov addition with no rearrangement,

Oxymercuration demercuration reaction).

NaBH4(CH3)2CHCH– CH2

Hg(OCOCH3)2 (CH ) CHCHCH(CH3)3CHCH = CH2

3-Methylbutene-1

3 3 3

HgOCOCH3OH OH

3-Methyl-1butanolnot-isolated

OAc

Hg+

+Hg(OAc)C = C —C —C —

mercurinium ion

Mercuration commonly takes place in a solution containing water and an organic solvent to dissolve the

alkene. Attack on the mercurinium ion by water gives (after deprotonation) an organomercurial alcohol.

+Hg(OAc)

Hg(OAc) Hg(OAc)–H+

—C —C — —C —C — —C —C —

. .H2O:

+ :.O.Horganomercurial alcohol

H —.

O:.

H O: H2

4

Alcohol

Hg(0Ac)|

H| |

4 —C —C —+NaB(0H)4 + 4Hg + 4 -0Ac| |

H0alcohol

|+ NaBH4 + 4 -0H 4 —C —C —

|0H

|

organomercurial alcohol

The second step is demercuration, to form the alcohol. Sodium borohydride (NaBH4, a reducing agent)

replaces the mercuric acetate rearrangement with hydrogen.

4. By Reduction of Carbonyl compounds :

LiAlH /NaC H OHRCH + 2H

|| O

RCR + 2H||O

425 R - CH20H

1º alcoholLiAlH / NaC H OH

425 RCHR|

OH

2º alcohol

Note :(i) We cannot obtain 3º alcohol from this method

(ii) If we use NaH as reductant then the process is called as 'Darzen's process'.

5. By Reduction of Acid & its derivatives

R C OH + 4H LiAlH4 RCH20H|| O

+ 4H LiAlH4 R CH20H + HXR C X||O

LiAlH4

R- C- 0R' + 4H

||0

RCH 0H + R'0H2

RC00C0R + 8H LiAlH4

2 RCH20H + H20

6. From Grignard reagent (RMgx)

(i) Reaction with oxyrene:

- 2

R : MgX H 2 C —CH2 R —CH 2 — CH 2 —0 MgX

0- H30

R —CH2 — CH2 —0H

primary alco.

5

Alcohol

(ii) Reaction with Carbonyl compounds:

i)etherR : Mg — X C 0 R —C —0 —H MgX2ii) H 0 , X-3

2

R : MgX C 0 R —C — 0 Mg X

- -

2 R —C —0 Mg X H —0 —H R —C —0 —H 0 —H MgX 2X

HH

(iii) Reaction with acetaldehyde:

CH3

CH3

H3CCH3 —CH2 — MgBr H0

Et20

0MgBrH

H30

CH3

H3C 0H

butan-2-ol

(iv) Reaction with ketone :

CH3

CH3H3C

H3C 0CH3 —CH2 —CH2 —MgBr Et20

0MgBrH3C

H30

CH3H3C

0H

CH3

2-methylpentan-2-ol

7. By Fermentation

Fermentation is a slow decomposition of complex organic compounds into simpler compounds in the

presence of suitable micro-organisms which are the source of biochemical catalyst known as enzymes.

(C6H1005) n

Starch

CH3.CH2.CH2.CH20H + CH3.C0.CH3

n-Butyl alcohol

6

Alcohol

3. PHYSICAL PROPERTIES OF ALCOHOL :

(1) The lower alcohols are liquids while higher having more than 12 carbon atoms are solids. They are colorless

neutral substances with characteristic sweet alcoholic odour and burning taste.

The lower alcohols are readily soluble in water and the solubility decreases with the increase in molecular

weight.

The solubility of alcohols in water can be explained due to the formation of hydrogen bond between the

highly polarized -0H groups present both in alcohol and water.

(2)

H - 0

H - 0

H - 0

H R H

Hydrogen bonding between alcohol and water molecules

However, in higher alcohols the hydrocarbon character (alkyl chain) of the molecule increases and thus

alcohols tend to resemble hydrocarbon (which are insoluble in water) and hence the solubility in water

decreases. When the ratio of C to 0H is more than 4, alcohols have little solubility in water.

Boiling points of alcohols are much higher than those of the corresponding alkanes. It is again due to the

formation of hydrogen bonding between the hydroxyl groups of the two molecules of an alcohols with the

result several molecules are associated to form a large molecule.

(3)

H - 0

H - 0

H - 0

R RR

Hydrogen bonding in alcohol molecules

Among the isomeric alcohols, b.p. and m.p. show the following trend.

Primary > Secondary > Tertiary

This is because of the fact that in secondary and tertiary alcohols, the alkyl part (hydrogen character)

outweighs the -0H group due to branching.

Lower alcohols form solid addition compounds with anhydrous metallic salts like CaCl2 and MgCl2,

viz., CaCl2.4C2H50H and MgCl2.6C2H50H

By analogy to water of crystallization, these alcohol molecules are referred to as alcohol of crystallization.

For this reason, alcohols cannot be dried over anhydrous calcium chloride.

(4)

4. CHEMICAL PROPERTIES OF ALCOHOL1. Reaction with Na:

2R- O – H + N a –––––– 2R - 0 - Na + H2

The acidic order of alcohols isMe0H > 1º > 2º > 3º

2. Esterification / Reaction with carboxylic : When reaction of alcohol with carboxylic acid inpresence of sulphuric acid to form ester.

conc.H SO2 4 R–C–O–R + H2O

||O

ester

7

Alcohol

Mechanism :-H2S04 ----- H+ + HSO

4

R

. .

RCO H+ R C

|| O

ROH RC||O

ORH + --- O C R. . H2O H|| O

| ||H O

Note :- The above reaction is laboratory method of ester preparation.

3. Reaction with Acid derivatives :-

R-0-H + X C R||O

Conc.H2SO4 + HXR O C R||O

Conc.H2SO4 R C O R||O

R O H R C O C R|| O

||O

4. Reaction with Isocyanicd Acid :-

? ?

R O H

C||O

+ H - N = H - N = H - NH - C O||O

R C O R|OH

amino ester (urethane)Reaction with ethylene oxide :-5.

0- H+ ROH

H2OR - - +

1,2-dialkoxy ethaneReaction with Diazomethane :-6.

R - 0 -H + CH2 N2 R -0 - CH3(ether)

7. Reaction with H2SO4 :-

..

(i) CH3 - CH2 - 0H + H2S04(excess) 140ºC CH3 - CH2 -O - CH -CH2 3..

Mechanism :-

H2S04 + HS04¯

CH3 -CH2 - O : + H+ CH3- CH2

|H

H+

O - H| H

CH3 –CH2 –OHCH3- C H2

(protonated ether)

. .

CH3 CH2 O CH2 CH3. .

8

Alcohol

(ii) CH3 -CH2 - 0H + H2S04

(excess)Mechanism :-

160ºC CH2=

CH2

H+ + HS04-H2S04

CH3 - CH2 - - H + H+ CH3 -CH2- H-H HO CH3 - CH2 = CH2CH2 2

Note : In the above reaction excess of ethanol is present so, intermediate carbocation satisfies itself by

elimination.

8. Action of halogen acids

anhy. ZnCl2C2H50H + HCl C2H5Cl + H20

The reactivity of halogen is in the order : HI > HBr > HCl.SN1 reaction with the Lucas reagent (fast)

CH3 CH3

H —C —0+

CH3

H —C —Cl

-

ZnCl2

H

CH3

CH3

. . ZnCl2 -

ClH — C+H —C —0. . — H

CH3 CH3 CH3

. . -H 0. . —ZnCl2

SN2 reaction with the Lucas reagent (slow)

9. Action of thionyl chloride : When alcohol react with thionyl cloride to form alkyl hylide and reaction iscalled darzen process.

C2H50H + S0Cl2Mechanism

. . Cl

C2H5Cl + HCl + S02

Cl 0 0. . . . . . . .R —0. . —S

-R —0 — S —0. .:R — 0: S = 0 R —0 —S + HCl

+ +Cl

Cl¯

ClClH H Cl H

thionyl chloride chlorosulfite ester

. .0

. .0 ¯ 0

(fast)R+R S = 0 S = 0 R S = 0

Cl Cl Cl

chlorosulfite ester ion pair

9

. .

. .

. .

. .. .

. .

. .

Alcohol

This mechanism resembles the SN1, except that the nucleophile is delivered to the carbocation by the

leaving group, giving retentionof configurationas shown in thefollowingexample. (Under differentconditions,

retention of configuration may not be observed.)

H H0H Cl

S0Cl2C

CH3(CH2)4CH2 CH3

(R) 2-octanol

C

CH3(CH2)4CH2

(R)-2-chlorooctane (84%)

CH3

di neoxa(solvent)

10. Action of phosphorus halides (PX5 and PX3). For example,

(i) C2H50H + PCl5 C2H5Cl + HCl + P0Cl3

Ethyl chloride

Mechanism

3 R - 0H + PCl5 P(0R)3Cl2 + 3HCl

P (0R)3 Cl2 + R0H P(0R)4Cl + HCl

Cl(0R)3P . .

.0. (R0) P = 0 + RCl3

R

(R0)3P = 0 + 3HCl + 3R0H

11. Action of ammonia

Al203 R0H

Al203

R0H

Al203

R0H + NH3 R3N

3º amine

R2NH

2º amine

RNH2

1º amine

The ease of dehydration of alcohols is in the order :

Tertiary > Secondary > Primary

12. Dehydration: Alcohols undergo dehydration (removal of a molecule of water) to form alkenes ontreating with acid e.g., concentrated H S0 or H P0 , or catalysts such as anhydrous zinc chloride or

2 4 3 4

alumina

10

du

Alcohol

Mechanism of dehydration

Step 1:

Step 2:

Step 3:

The acid used in step 1 is released in step 3. To drive the equilibrium to the right, ethene is removed as

it is formed. The relative ease of dehydration, i.e. 3º > 2º > 1º, of alcohols follows the order of stability of

carboniumions.

Oxidation

(a) Primary alcohol initially forms aldehyde on oxidation and on further oxidation forms respective acid.

(b) Secondary alcohol initially forms respective ketone on oxidation which on further oxidationforms acid of less carbon. 0xidation of ketone is slightly difficult than aldehyde due to stability so,we use strong oxidizing agent for oxidation.

(c) Tertiary alcohols are resistant to oxidation in normal conditions but on taking strongest oxidizing agent like chromic acid in dilute nitric acid then they form less carbon ketone.

(d) For oxidation of 1º alcohol, acidic KMn04 is used as oxidant while for 2º alcohol acidicK2Cr207 (more strong than KMn04) is used.

(e) For both (1º and 2º alcohol) we can use common oxidizing agent also, like chromic acid indil ute H 2S04. Colour of this solution is orange and it turns into green colour due to Cr+3.

(f) Reacting species of solution is HCr04¯.

13.

[o]acidic KMnO4

acidic K2Cr2O7

[o]

H2CrO4 dil. HNO3

(ketone of lesser carbon)

11

Alcohol

14. Dehydrogenation / Catalytic Oxidation

Cu / ZnO

300º

1º alcohol aldehyde

Cu / ZnO

300º

secondary alcohol ketone

Cu / ZnO

300º

tert. alcohol alkene

(Note : – This is dehydration process.)

5. DISTINCTION BETWEEN PRIMARY, SECONDARY AND TERTIARY ALCOHOLS :

(1) Lucas test : This test is based on the difference in the three types of alcohols (having 6 or less carbon)

towards Lucas reagent (a mixture of conc. hydrochloric acid and anhydrous zinc chloride)ZnCl2 RCl + H20ROH + HCl

Since alkyl halides are insoluble, their formation is indicated by the appearance of a turbidity in the reaction

mixture. The order of reactivity is tertiary > secondary > primary, the teritary alcohols produce turbidity

immediately, the secondary alcohols give turbidity within 5 - 10 minutes, and the primary alcohols do not

give turbidity at all, at room temperature.

Catalytic dehydrogenation (action of reduced copper at 300ºC). Discussed ealier.

(i) Primary alcohols form aldehydes

(ii) Secondary alcohols form ketones.

(iii) Tertiary alcohols form olefins.

Victor Meyer test : This test is based on the different behaviour of primary, secondary and tertiary

nitroalkanes towards nitrous acid. The 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 the corresponding nitroalkane.

(iii) The nitroalkane is treated with nitrous acid (NaN02 + H2S04) followed by treatment with alkali

(Na0H or K0H). Upon such treatment different alcohols give different colours.

(a) Primary alcohols produce a blood red colour

(b) Secondary alcohols produce a blue colour.

(c) Tertiay alcohols produce no colour.

(2)

(3)

12

Alcohol

Primary

CH3CH20H

P + I2

CH3CH2I

AgN02

Secondary

(CH3)2CH0H

P + I2

Tertiary

(CH3)3C0H

P + I2

(CH3)2CHI

AgN02

(CH3)3Cl

AgN02

CH3CH2N02 (CH3)2CHN02 (CH3)3CN02

HONO HONO H0N0

CH3CN02

| |

N0H

Nitrolic acid Na0H

CH3CN02

| |

N0Na

Sod. salt of nitrolic acid

(Red colour)

(CH3)2.CN02

|

N0

(Pseudonitril)

Na0H

No reaction

(Blue colour)

No reaction

(Colourless)

Ex.1 Give the structure ofHg(0AC)2 / H20; NaBH4.

CH2CH3

the major organic product when 3-ethylpent-2-ene is treated with

CH2CH3

Hg(0AC)2 / H20

(Mark addition)

CH3CH2 —C — CH —CH3Sol. CH3CH2 —C = CH —CH33-Ethylpent-2-ene Hg0AC0H

CH2CH3

CH3CH2 —C — CH2CH3NaBH4

Reduction 0H3-Ethylpentan-3-ol

13

Alcohol

WASH - ABSOLUTEALCOHOL

Wash [10-15% EtOH]

Distillation

Raw spirit [90% EtOH]

Fractional distillation

20-21ºC 78ºC

Mainly CH CHO Rectified spirit [95.5% EtOH] Fuel oil3

Excess benzene Azeotropic distillation65ºC

[Glycerol + CH COCH3

+ Higher alcohol]3

C H + C H OH + H O6 6 2 5 2

(74.1%) (18.5%) (7.4%) (Ternary mixture)

680C

C H + EtOH6 6

(80%) (20%) (Binary mixture)

78.30C

Absolute alcohol[100% EtOH]

14

Alcohol

ALCOHOL

GRGMPHX or PX or PX3 5

or KI + H 3PO 4 or SOCl2 or SO2Cl2(1) RX

H2O (1) Alkene

Re dP/HI (2)

(3)

RHaq. NaOH or aq. KOHor aq. K 2CO3 or moist Ag2O

dil. H2SO4

(2) RX

NH3 1º , 2º , 3º amines

(3) R - O - RH2S(4) R - SH Thiol

dil. H2SO4

Th02

(4) RCOORRCOOH

Na (5)

(6)

RONaHNO2 (5) 1º amine

CH3MgX CHException -Methyl amine givesCH -O-CH or ether

4

R' OR

ald.R'CHO3 3

(7) C AcetalNaHDarzon reduction

dry HCl(6) Aldehyde or ketoneH OR

(1º alc.) (2º alc.)

Na/ EtOH

R-OHR' OR

(7) Acid orKetoneR'COR'

dry HCl

BouveaultBlanc reduction(8) C Ketal

Acid derivative

HCHO orAld.or ketone RMgX H2O

R' OR(8)

R'COZ R'COOR ester (Z = OH,Cl, OCOCH )(9)

(10)

(11)

(12)

(13)

(14)

3

(1º alc) (2ºalc) (3º alc.)

O2H2 SO4 ROSO OH (Alkyl hydrogen sulphate)

2

RONO (Alkyl nitrate)

H2O(9) RMgX

HNO3 2

PhSO2ClCH2CH2

O

RSO Ph (Alkyl benzene sulphonate)2

H C - CH (OR) Acetal3 2

(10)CH MgBr CHCH3 H3O

CH2N2 R - O - CH Ether3

O

CH 2 CH 2

(11) Sugar Fermentation

(15) RO - CH - CH - OH2 2

Alkoxyalkanol

CH2 CO ROCOCH(16)

(17)

Ester3

Dehydration Alkene

Catalytic dehydrogenation (18) Aldehyde or ketone10 or alcohol , Cu or ZnO , 3000 C20

Formation of EtOH by fermention - Exception - 3º alc AlkeneCrystallization

Sucrose(1) Cane sugar Molasses 1ºalc.[O]Aldehyde O]Acid (same no. of C-atom)[(19)

(20)

(21)

2ºalc.[O ]Ketone [[O]Invertase zymaseFermentation

Invert sugarhydrolysis

EtOH3ºalc.[O]

OHCrO 4(22) 1º or 2ºalc. Aldehyde or ketone + Cr+3 (green)DiastaseHOH

(2) Grain Starch Maltose (orange)

O, HCrO Maltase Zymase 4hydrolysis

EtOH

Fermentation

Glucose (23) 3º alc No reaction (No. green colour)(orange)

15

Alcohol

PHENOL1. INTRODUCTION :When 0H group attached at benzine ring is known as phenol.

Nomenclature of Phenols

0H 0H 0H 0H

CH3 H3C CH3

CH3

CH3

4-methyl phenol(p-cresol)

2-methyl phenol(o-cresol)

3-methyl phenol(m-cresol)

2-6 dimethyl phenol

Some dihydric and trihydric phenols are given below:

0H 0H 0H

0H

0H1-2-Benzenediol

(Catechol)1-3-Benzenediol

(Resorchinol) 0H

1-4 Benzenediol(Qunol)

0H 0H 0H

0H

0H 0H H0 0H

0H

1-2-4 Benzenetriol(Hydroxyqionol)

1-2-3 Benzenetriol(pyrogallol

1-3-5 Benzene-triol(Phloroglucinol)

2. METHOD OF PREPARATION OF PHENOL :(a) From haloarenes

16

Alcohol

2. From benzenesulphonic acid

3. From diazonium saltsWhen diazoniumsalts react with water vapour gives phenol.

4. From cumeneWhen Cumene (isopropylbenzene) is oxidised in the presence of air and acid gives phenoland acetone.

3. PHYSICAL PROPERTIES OF PHENOL :

(i) Pure phenols are generally colourless solids or liquids. The light colour usually associated with phenols

is due to its oxidation by air in presence of light.

(ii) Phenols, in general, are insoluble in water; but phenol itself, and polyhydric phenols are fairly soluble in

water which is believed to be due to the formation of hydrogen bonds with water.

(iii) Due to intermolecular hydrogen bonding, phenols usually have relatively high boiling points than the

corresponding hydrocarbons, aryl halides and alcohols. For example, phenol (mol. wt. 94) boils at 182ºC

while toluene (mol. wt 92) boils at 110ºC.

Higher b.p. than alcohols is due to higher polarityof the 0 -H bond and consequently stronger intermolecular

hydrogen bonding in phenols than in alcohols.Appreciable solubilities of the phenol and polyhydric phenols

in water is also due to strong hydrogen bonding between phenols and water molecules.

0Ar 0 Ar H

0

00 H Ar 0 HH HH N

0o-Nitrophenol (IntramolecularH-bonding possible due tocloseness of -N02 and -0H groups)

Hydrogen bonding betweenphenols and water molecules

Intermolecular hydrogenbonding in phenols

17

Alcohol

Phenols containing groups like -N02 or -C00H in the ortho position to the -0H group can also form

intramolecular hydrogen bonds (e.g. o-nitrophenol) which is responsible for their lower boiling points and

less solubility in water than the corresponding meta or para isomer.

Due to possibility of intramolecular hydrogen bonding (also known as chelation) in the ortho isomer,

intermolecular hydrogen bonding is not possible and hence the ortho isomer can neither get associated nor

can form hydrogen bonding with water with the result it has a low b.p. and less solubility in water than the

meta and para isomers which can associate (union of two or more molecules of the same species) as well

as can form hydrogen bonding with water.

0 — H 0 0 H

H00H

N

NN0 0

0 — H 0 H H0

0Hydrogen bonding between p-nitrophenol and waterp-Nitrophenol (2 molecules) (Intramolecular H-bonding is

not possible due to large distance between -N02 and -0H

groups; hence intermolecular H-bonding is possible)

(iv) They possess characteristic colour. They are highly toxic in nature and possess antiseptic properties. They

may produce wounds on skin.

Note :

(i) Phenol exists as a resonance hybrid of the following structures.

..: 0H

..: 0H +

: 0H+

: 0H+

: 0H

:– – :

..

IV VI IIIII

Due to resonance, oxygen atom of the -0H group acquires a positive charge (see structures III

to V) and hence attracts electron pair of the 0 - H bond leading to the release of hydrogen atom as

proton.

..: 0 -H

..: 0:

H++

Phenol Phenoxide ion

Since resonance is not possible in alcohols (due to absence of conjugation of the lone pair of electron of

oxygen with a double bond), the hydrogen atom is more firmly linked to the oxygen atom and hence

alcohols are neutral in nature.

18

Alcohol

(ii) 0nce the phenoxide ion is formed, it stablises itself by resonance, actually phenoxide ion is more stable

than the parent phenol.

.. –: 0 :

.. –: 0 :

..: 0

..: 0

..: 0

: – – :

..

IX XVI VIIIVII

3. Comparison of acidity of phenols and carbonic acid.

Relative acidity of the various common compounds.

> R0H

Alcohols

RC00H

Carboxylic acid

> H2C03 > C6H50H

Phenol

> H0H

WaterCarbonic acid

4. CHEMICAL PROPERTIES OF PHENOL :1. Nitration:

(a) When phenol react with dilute nitric acid at low temperature (298 K), gives a mixture of ortho and para nitrophenols.

(b) When phenol is react with concentrated nitric acid, gives 2,4,6-trinitrophenol.

2. Halogenation:(a) When the reaction is carried out in solvents of low polarity such as CHCl or CS and at low

3 2

temperature, monobromophenols are formed.

19

Alcohol

(b) When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate.

3. Kolbe’s reaction :

Mechanism of Reaction

O

0Na 0 C H 0tautomerisationO

Na

0

H H0 00H 0

0 0Na

H30

Salicyclic acid Sodium salicylate

4. Reimer-Tiemann reaction : 0n treating phenol with chloroform in the presence of sodiumhydroxide, a -CH0 group is introduced at ortho position of benzene ring. This reaction is known asReimer - Tiemann reaction.The intermediate substituted benzal chloride is hydrolysed in the presenceof alkali to produce salicylaldehyde.

The mechanism ofthe Reimer-Tiemann reaction is believed to involve the formation ofdichlormethylene.Na0H+CHCl :CCl +NaCl+H 0

3 2 2

20

Alcohol

Phenols with blocked p-positions give cyclohexadienones containing the dichromethyl group.

0H 0H 0

0Na0H CHCl3

H3C CH3CH3 CH3

In the Reimer-Tiemann reaction, the o-isomer predominates, but if one of the o-positions is occupied,the aldehyde group tends to go the p-positions; e.g., guaiacol forms vanillin

0H0H

00CH3CH3 Na0H

CHCl3

0

Libermann’s Reaction : When phenol is treated with sodiumdissolved in conc. Sulphuric acid, a red colouration

appears, which changes to blue on adding aqueous Na0H. This reaction is called Libermann’s reaction

2NaN02 + H2S04 Na2S04 + 2HN02

Nitrous acid

N -0H0H0 — —H + H -0 - N = 0 H0 — —N = 0- H2 0

p-Nitrosophenol

H - -0H-Na0H +

N — — 0H N —0 0 — 0Na- H2 0- H2 0

Blue

Indophenol (Red)

5. Reaction of phenol with zinc dust : When phenol is heated with zinc dust gives benzene.

6. Oxidation:0xidation of phenol with chromic acid produces a conjugated diketone known asbenzoquinone. In the presence of air, phenols are slowly oxidised to dark coloured mixtures containing quinones.

21

Alcohol

5. DISTINCTION BETWEEN ALCOHOLAND PHENOL :(i)

(ii)

Phenols turns blue litmus red but alcohols do not.

Phenols neutralize base, while alcohols do not.

0H0Na

+ Na0H + H2 0

No reactionR -0H + Na0H

(iii) Phenols give violet colour with FeCl3, while alcohols do not.

00H

Fe + 3HCl3 + FeCl3

3

VioletNo reactionR - 0H + FeCl3

Identify the major products in the following reactions

0H

Ex.1

N02

CH3

Bromine(i) (ii)HN03 A

WaterBH2S04

0

CH3

BrCH3

0 0H

Sol. A= B =02N

Br CH3

N02

22

Alcohol

ETHERS1. METHOD OF PREPARATION OF ETHERS :

1. Williamson’s Synthesis: Heating ofalkyl halide with sodium or potassium alkoxidegives ether. This is a good method for preparation of simple as well as mixed ethers.

R-X + Na-O- R R-O- R + NaX

This method is not applicable to tert. alkyl halides because the alkoxide ions being both powerful nucleo-philes and bases could being dehydrogenation of the tertiary alkyl halides to form alkenes.

R -ONa R -O Na

R -O -Na

Ar -O -Na

R -O -R

R -O -ArAryl Ether

R -X

aq. NaOH

-OH CH3 -CH2 -Br -O -CH2 -CH3

The reactivity of primary (1°) alkyl halide is in the order CH > CH -CH > CH -CH -CH and3 3 2 3 2 2

the tendency of the alkyl halide to undergo elimination is 3° > 2° > 1°. Hence for better yield thealkylhalide should be primary of the alkoxide should be secondary or tertiary.

C2H5Br NaO -C - C2H5 -O -C - NaBr

2. By heating excess of alcohols with conc. H2SO4, e.g.,

conc.H2SO4

140ºC

C2H5 - O - C2H5 + H2O

Diethyl etherC2H5 - OH + HO - C2H5

Ethanol (2 molecules)

Recall that 2º and 3º alcohols under the above conditions give alkenes as the main product. Moreover,

this method is limited only for the preparation of simple ethers.

By heating alkyl halide with dry silver oxide (only for simple ethers)3.

C2H5I + Ag2O + IC2H5 C2H5.O.C2H5 + 2 AgI

Remember that reaction of alkyl halides with moist silver oxide (Ag2O + H2O = AgOH) gives alcohols.

By the use of diazomethane to form methyl ethers.

BF3

4.

n-C7H15OH + CH2N2 n-C7H15OCH3 + N2Methyl n-heptyl ether

BF3 C6H5OCH3Anisole

C H OH + CH N6 5 2 2

23

Alcohol

2. PHYSICAL PROPERTIES OF ETHERS :

1. Their boiling points are much lower than the isomeric alcohols. This is because of absence of -OH group in

ethers and hence they are incapable of forming intermolecular hydrogen bonds.

2. Since the two C - O bonds in ethers are not linear (180º) but they are at an angle of about 110º, i.e., the

molecule is bent, the dipole moment of the two C - O bonds do not cancel each other. Consequently, ethers

are slightly polar and have a small net dipole moment (e.g. 1.18 D for diethyl ether).

Note that the bond angle in ether is somewhat greater than water (105º). It is due to the repulsion between

bulky alkyl groups.

3. CHEMICAL PROPERTIES OF ETHERS :

Ethers are much less reactive than compounds containing other functional group. They do not react with

active metals like Na, strong base like NaOH, reducing or oxidising agents.

1. Formation of peroxides : On standing in contact with air, ethers are converted into unstable peroxides (R2O

O) which are highly explosive even in low concentrations. Hence ether is always purified before distillation.

Purification (removal of peroxide) can be done by washing ether with a solution of ferrous salt (which reduces

peroxide to alcohols) or by distillation with conc. H2SO4 (which oxidises peroxides).

The presence of peroxides in ether is indicated by formation of red colour when ether is shaken with an

aqueous solution of ferrous ammonium sulphate and potassium thiocyanate. The peroxide oxidises

Fe2+ to Fe3+ which reacts with thiocyanate ion to give red colour of ferric thiocyanate.CNS-

Peroxide + Fe2+ Fe3+ Fe(CNS)3

Red

However, the formation of peroxide is prevented by adding a little Cu2O to it.

Basic nature : Owing to the presence of unshared electron pairs on oxygen, ethers are basic. Hence they

dissolve in strong acids (e.g., HCl, conc. H2SO4) at low temperature to form oxonium salts.

2.

[(C2H5)2 OH]+ HSO4-]

Diethyloxonium hydrogen sulphate

(C2H5)2O + H2SO4

Diethyl ether

On account of this property, ether is removed from ethyl bromide by shaking with conc. H2SO4.

The oxonium salts are stable only at low temperature and in a strongly acidic medium. On dilution,

they decompose to give back the original ether and acid.

Ethers also form coordination complexes with Lewis acids like BF3, AlCl3 RMgX, etc.R

R2O..

R2O: + BF3MgR2O BF3 (b) 2R2O + RMgX(a)

R2O X

It is for this reason that ethers are used as solvent for Grignard reagents.

Action of dilute H2SO4 (hydrolysis).

dil. H2 SO4 , heat

3.

C2H5-O-C2H5 2 C2H5-OHPressure

4. Action of concentrated H2SO4.

C2H5-O-C2H5 + H2SO4 (conc.) heat C2H5OH + C2H5HSO4

24

Alcohol

5. Action of conc. HI or HBr.

(i)C2H5 - O-C2H5 + HI (cold)

(ii) C6H5-O-C2H5 + HI

C2H5OH + C2H5I

C6H5OH + C2H5I

Mechanism of reaction : S 2 and S 1 mechanisms for the cleavage of ethers with HI. S 2 cleavageN N N

occurs at a faster rate with HI than with HCl?

H

+Step 1: R O R HI

R

I-O

R'base2acid 2 acid 1base1

H

+O

Step 2 for SN 2 I slowRI HOR (R is 1°) R R'

Step 3 for S 1 R+ + I- RI (R be 3°)N

(b) The transfer of H+ to ROR in step 1 is greater with HI, which is a stronger acid, than with HCl.Furthermore, in step 2, I-, being a better nucleophile than Cl -, reacts at a faster rate.

25

Alcohol

SOLVED EXAMPLES Ex.1 Place the following groups of compounds indecreasing order of acidity and justify your answers.

(a) Phenol (A), o-nitrophenol (B), m-nitrophenol (C), p-nitrophenol (D);(b) Phenol (A), o-chlorophenol (E), m-chlorophenol (F), p-chloroophenol (G); and(c) (A), ocresol (H), m-cresol, (I), p-cresol (J).(a) NO is electron -withdrawing and acid -strengthening by both induction and resonance. Its reso-Sol.

2

nance effect is effective from only the ortho and para positions to about an equalextent. It predominatesover the inductive effect which operates fromall positions but at decreasing effectiveness with increasing separation of NO and OH. Hence all the nitrophenols are more acidic than phenol with m-nitrophenol

2

being the weakest of the three. Since the inductive effective from the closer o positions is the strongest,one might expect o-nitrophenol to be stronger than p-nitrophenol. However, the intramolecular H-bondin o-nitrophenol must be broken and this requires some energy. The decreasing order is DBCA.(b) Although Cl is electron - donating by resonance, its electron - withdrawing inductive effect thatdecreases with increasing separation of Cl and OH predominates, making all the chlorophenols moreacidic than phenol. The decreasing order is E F G A.(c) Me is electron - donating inductive from all positions and hyperconjugatively from the ortho and parapositions. The three isomers are weaker acids than phenol. m-Cresol is the strongest because its acidityis not weakened by hyperconjugation. The decreasing order isA I J H.

Hydrolysis ofcompound (A) ofmolecular formula C9H10Cl Br yields (B) ofmolecular formula C9H10O.

(B) gives the haloform reaction. Strong oxidation of (B) yields a dibasic acid which forms only onemononitro derivative. What isA?

Ex.2

Cl Br

C -CH3 COONaO = C -CH3

H2O I2Sol. + CHI 3NaOH

CH3

(A)

CH3CH3

(B)

C9H10O

O = C -CH3 COOHCOOHNO2

HNO3[O]H2SO4

COOH(One mononitroderivative)

CH3(B)

COOH

Ex.3 When Bromobenzene is monochlorinated two isomeric compounds (A) and (B) are obtained.Monobromination of (A) gives several isomeric products of molecular formula C6H3Cl Br2, while

monobromination of (B) yields only two isomers (C) and (D). Compound (C) is identical withone of thecompounds obtained from the bromination of (A). Give the structures of (A), (B), (C) and (D) and alsostructures offour isomeric monobrominated products of (A). Support your answer with reasoning.

26

Alcohol

BrBr BrCl

Cl2/FeCl3-HCl

Cl2/FeCl3-HClSol.

Bromobenzene-Bromochlorobenzene

(A)Cl

p-Bromochlorobenzene

(B)

Br BrBrCl ClCl

Br2 +Br

(B)2-Bromochlorobenzene

(A)Br

2-Chloro-1,4-dibromobenzene

(ii)

2-Chloro-1,3-dibromobenzene

(i)

Br BrCl Br Cl

+Br

4-Chloro-1,3-dibromobenzene(C6H3ClBr)

(iii)

3-Chloro-1,2-dibromobenzene

(iv)

BrBrBrBr

Br2 +Br

Cl4-Chloro-1,3-dibromobenzene

(C)

Cl4-Chloro-1,2-dibromobenzene

(D)

Cl4-Bromochlorobenzene

(B)

(C) is identical with (iii).

Ex.4 An organic liquid with sweet smell and b.pt 78°C contains C, H and O. On heating with conc. H SO2 4

gives a gaseous product (B) of empirical formula (CH )n. Compound (B) decolourises bromine water2

and alk. KMnO . (B) also reacts with one mole of H in presence of Ni. What are (A) and (B)?4 2

Sol. Compound (empirical formula CH ) decolourizes Br water, reacts with alk. KMnO and adds one2 2 4

mole of H and so (B) is alkene.2

Compound (B) is obtained from (A) by the action of H SO and so (A) is alcohol.2 4

Ahas b.pt. 78°C. So (A) is ethanol

C2H5OH C2 H4Conc.H2SO4

Reactions: (i)Ethanol Ethylene

C2H4 Br2(B) water decolourises (C H Br )(ii)

2 4 2

C2H4KMnO4 decolourises (HCOOK) Pot. formate(B)

(iii) C H + H C H2 4 2 2 6

27

Alcohol

Ex.5 Two isomeric compounds (A) and (B), have same formula C11H13OCl. Both are unsaturated, andyieldthe same compound (C) on catalytic hydrogenation and produce 4-Chloro-3-ethoxybenzoic acid onvigorous oxidation.(A) exists in geometrical isomers, (D) and (E), but not (B). Give structures of (A) to(E) withproper reasoning.

As oxidation of (A) and (B) gives 4-Chloro-3-ethoxybenzoicacid, the structures of (A) and (B) can be writtenas

Sol.

COOHR

OC2H5OC2H5vigorousoxidation Cl

Cl4-Chloro-3-ethoxybenzoic acid

(C)(A) and (B)

Since the molecular formula ofAand B is C11H13OCl, therefore R must be C3H5.As (A) and (B)areunsaturated compounds, both give the same compound (C) on catalytic hydrogenation. Therefore Rmust be -CH = CH -CH3 or -CH2 -CH = CH2. Thus structures of (A) and (B) can be

CH = CH -CH3 CH2 -CH = CH2 CH2 -CH2 -CH3

OC2H5 OC2H5 OC2H5and H2/Ni

Cl

(A)Cl(B)

Cl(C)

Only (A) can exist as geometrical isomers and not B. Thus structures of (D) and (E) are

H HH

CH3

CH3

H

C = C C = C

OC2H5 OC2H5Cl

(D)Cl(E)

Ex.6 An alkyl halide, X, of formula C6H13Cl on treatment with potassium tert. butoxide gives twoisomericalkenes Y and Z (C6H12). Both alkenes on hydrogenation give 2,3-Dimethyl butane. Predictstructures of X,Yand Z.

H3C -CH -CH -CH3

- + H2/Pt.C6H13Cl

(X)

[(CH3)3CO] K C6H12(Y and Z)

Sol.CH3 CH3

2,3-Dimethylbutane

Pot. tert. butoxide-KCl, -(CH3)3COH

Product, 2, 3-dimethyl butane suggest that (Y) and (Z) must be and X must beCH2

CH3

CH3

CH3

CH3

CH3

H3C

CH3

H3C

H3C

H3C

H3C

C = C CH -C CH -C

Cl(Y)

C6H12

(Z)C6H12

(X)(C6H13Cl)

28

Alcohol

This series of reactions are

CH3

CH3

H3C

H3C

CH2

CH3

CH3

CH3

H3C

H3C

H CC = C 3

H3CCH -C C -C

Cl +[(CH3)3CO]-K+

2,3-Dimethylbut-1-ene

(Z)2,3-Dimethylbut-2-ene

(Y)2-Chloro-2,3-dimethylbutane

(X)

H2/NiH /Ni

2

CH2

CH3

H3C

H3CCH -CH

2,3-Dimethylbutane

Ex.7 Amixture of ethyl alcohol and n-propyl alcohol on treatment with H2SO4 gives a mixture of three ethers.On the other hand, a mixture of tert. butyl alcohol and ethyl alcohol gives the good yield of a single ether.Identify the ether and account for its good yield.

H2SO4

-H OH3C -CH2 - CH2 -OH

n-Propyl alcohol

H3C - CH2 - O - CH2 - CH3

Diethyl ether

H3C -CH2 -OH

Ethanol

Sol. (a) +2

+ H3C - CH2 - CH2 - O - CH2 - CH2 - CH3 H3C - CH2 - CH2 - O - CH2 - CH3+Ethyl n-propyl etherDi-n-propyl ether

CH3CH3H2SO4

-H OH3C - C - O - CH2 - CH3(b) H3C -CH2 -OH

Ethyl alcohol

+ H3C - C - O - H2

CH3CH3tert. butyl alcohol

The mixture of ethyl alcohol and tert. butyl alcohol on treatment with H2SO4 gives tert. butyl ethyletherin good yield, because tert. butyl alcohol gives stable tert. butyl carbocation which rapidly undergoesnucleophilic attack byethyl alcohol to give tert. butyl ethyl ether.

CH3 CH3 CH3..C2H5 -O.. -HH+

-H O

+H3C -C +H3C -C -OH H3C -C -O - C2H5

2

CH3tert. butyl carbocation

CH3 CH3 H

CH3..

H3C - C -O.. -CH2 - CH3-H+

CH3

Ethyl tert. butylether

On the other hand, in first case, both ethyl alcohol and n-propyl alcohol yield almost equally stablecarbocations, each one of which can be attacked either by ethyl or n-propyl alcohol giving a mixtureof three ethers.

29

Alcohol

Ex.8 0.535gm ethanol and acetaldehyde mixture when heated with Fehling solution gave 1.2 gm of a redprecipitate. What is the percentage of acetaldehyde in the mixture? [Atomic weight. of Cu = 63.8]. C H OH + CH CHO = 0.535Sol.

2 5 3

Let ‘a’gm C H OH and ‘b’gm CH CHO be present in mixture2 5

a + b = 0.5353

Now mixture reacts with Fehling solution to give a red precipitate, which suggests a characteristicreaction for aldehyde, isCH CHO + 2CuO ® CH COOH + Cu O

3

Fehling solution3 2

red ppt.Q 143.6 gm Cu O is given by 44gm CH CHO

2 3

441.21.2 gm Cu O is given by = 0.368 gm CH CHO

2

So, b = 0.368 gm143.6 3

0.368 % CH CHO = 100 = 68.73%

3 0.535

Ex.9 How will you prepare (a) carbitol (water soluble organic solvent) and (b) 18-crown-6 (a crown-ether)from ethylene oxide.

+HSol. (a) C H -OH + CH -CH C H O -CH CH -OH2 5 2 2

O

2 5 2 2

CH2 -CH2

O

C2H5 -O - CH2 -CH2 -O -CH2 -CH2 -OH

Carbitol

HCl(b) 3 CH -CH HO - CH -CH - O - CH - CH - O - CH -CH OH2 2

O

2 2 2 2

(A) Triethyl glycol

HCl

2 2

ClCH2CH2OCH2CH2 - OCH2 - CH2Cl

(B)

OOO

NowA+ B aq. KOH

OO

O

18-Crown-6

Ex.10 Predict the product(s) of each reaction below. Be sure to specify the stereochemistry where appropri-ate. (Ph, phenyl; *, isotope label)

(a) Ph-O-CH -Ph excess HI2

(b) (H C) CH-O-C(CH )3 2 3 3

(c)1equiv HI

30

Alcohol

O

1. PhO 2. H(d)

Ph

OH (catalyst )

PhOH(e)Ph

O

(f)

H CH2CH3

H(catalyst )H H H2O

O

(g) H CH2CH3

1. OHH H 2. H

*-O -C H 2 -CH CH2 (h)

(i) CH3 - S -CH3 (CH3CH2 )3 O BF4

H

O

CH3

H

O

CH3

HCl(j)

1. PhLi(k) 2. H

Sol. (a) Ph -OH + I-CH -Ph S 1 cleavage at the benzyl carbon of the protonated ether2 N

(b) (H3C)2 CH -OH Br -C(CH3 )3 SN1 cleavage at the more substituted carbon of the proto-

nated ether

OH

(c) S 2 cleavage at the less hindered carbon ofthe protonated ether CH3IH N

(d) S 2 attack at the less hindered carbon of the oxiraneN

Ph -CH -CH 2 -O -Ph

OH

Ph -CH -CH2 -OH L(e)

OPh

31

Alcohol

HCH2CH3

(f) HO -CH2 - Like S 1 attack by water on the more substituted carbon of theproto-

NOH

nated oxirane

CH3

(g) HO -CH 2 - S 2 attack at the less hindered carbon of the oxirane, with retention ofN

H CH2CH3

configuration at the other carbon

OH

CH2

(h) Claisen rearrangement

CH3+

(i) H3C S BF4 (Ch3CH 2 )2 O S 2 attack by the sulfide on the trialkyloxonium ionN

CH2CH3

H

OH

CH3(j) Like S 1 attack by Cl on the more substituted carbon of the protonated oxiraneN

-

Cl

(some Z isomer may also be formed)

PhH

(k) S 2 attack by “Ph-” at the less substituted carbon of the oxiraneOH N

CH3

Ex.11 3,3-Dimethylbutane-2-ol loses a molecule of water in the presence of concentrated sulphuric acid togive tetramethylethylene as a major product, suggest a suitable mechanism.

CH3CH3+HSol. CH3 - C CH -CH3CH3 -C -CH -CH3 -H O2

H3C +OH2CH3 OH

CH3 CH3 CH3+ 1,2-Methyl shiftCH3 -C -CH -CH3

CH3

2º CarbocationLess stable

C = C-H+

CH3 CH3

Tetramethylethylene

0.037 g of an alcohol, ROH was added to CH3MgI and the gas evolved measured 11.2 cm3 at S.T.P.

What is the molecular mass of ROH ? On dehydration ROH gives an alkene which on ozonolysis gives

acetone as one of the products. ROH on oxidation easily gives an acid containing the same number of

carbon atoms. Give structrue of ROH and the acid with proper reasoning.

Ex.12

Sol. CH4 + ROMgIROH + CH3 MgIAlcohol

11.2 cm3 methane = 0.037 g alcohol

32

Alcohol

0.037 22,40022,400 cm3 methane = = 74 g alcohol

11.2

Molar mass of alcohol (CnH2n+1.OH) is 74 g mol-1

12n + 2n + 1 + 17 = 74 or 14n + 18 = 74

56

14n = 56 or n = = 414

Molecular formula of alcohol is C4H9OH. Since the alcohol can be easily oxidised to acid, it must be1º

alcohol. Ondehydration alcohol gives alkene which on ozonolysis gives acetone as one of the proudcts.Thus alcohol must consist of the fragment (CH3)2CH- and hence the alcohol is isobutyl alcohol. The

concerned reactions are as follows :CH3

CH3 -CH -

CH2OHIsobutyl alcohol

CH3

CH3 -C = CH2 + H2OIsobutylene

Conc.H2SO4

OCH3

CH3 -C = CH2 + O3

CH3

CHCCl4

Low temp.C CH H2O/Zn

2

3

O OOzonideCH3

C = O + HCHO + H2O2CH3

AcetoneCH3 CH3CH3

[O]

-H O[O]CH3 -CH -CH2OH

2-Methylpropan-1-ol

CH3 -CH -COOH

2-Methylpropanoicacid

CH3 -CH -CHO

2-Methylpropana12

Ex.13 Amixture of two aromatic compoundsAand B was separated by dissolving in chloroform followed byextraction with aqueous KOH solution. The organic layer containing compound A, when heated withalcoholic solution of KOH produced a compound C(C7H5N) associated with an unpleasant odour. Thealkaline aqueous layer on the other hand, when heated with chloroform and then acidified gave a mixtureof two isomeric compounds D and E ofmolecular formula C7H6O2. Identify the compoundA, B, C, D,

E and write their structures.

KOH(alc.)

Organic layer(A)

C7H5N (C)mixture of

CHCl /H+

3Sol. A and B

C H O 7 6 2

(D)CHCl3/H+Aqueous layer

(B)C7H6O2

(E)

Since the compound C having the molecular formula C7H5N has an unpleasant smell, it must be an isocyanide corresponding to the structure C6H5 -N C (C).Consequently, (A) must be an amine corresponding to the structure C6H5NH2 (A) (Aniline)Since (B) gives positive Reimer-Teimann reaction, it must be a phenol corresponding to the

structure C6H5OH.

CHCl3/H+

Thus, OH + OHC OHOH

(E)(B) CHO (D)

33

Alcohol

Q.1

Q.2

Draw the structures of all isomeric alcohols ofmolecular formula C H O and give their IUPAC names.5 12

While separating a mixture ofortho and para nitrophenols bysteam distillation, name the isomer which

will be steamvolatile. Give reason.

Show how will you synthesis:

(i) 1-phenylethanol from a suitable alkene.

(ii) cyclohexylmethanol using analkyl halide by an SN reaction.

Q.3

2

(iii) pentan-1-ol using a suitable alkyl halide?

Q.4

Q.5

Q.6

Give two reactions that show the acidic nature ofphenol. Compare acidity ofphenol with that ofethanol.

Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?

Explain how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic

substitution?

Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.

Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method.

Give reason.

Write the equation of the reaction of hydrogen iodide with:

(i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether.

Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards

electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in

benzene ring.

Write the mechanism of the reaction ofHI with methoxymethane.

Describe the mechanism of alcohols reacting bothas nucleophiles and as electrophiles in their reactions.

Write one chemical reaction each to illustrate the following :

(i) Reimer -Tiemann reaction

(ii)Williamson's synthesis

How would you account for the following :

(i) Phenols are much more acidic than alcohols.

(ii) The boiling points of ethers are much lower than those of the alcohols of comparable molar masses ?

Write the reaction and the conditions involved in the conversions of

Q.7

Q.8

Q.9

Q.10

Q.11

Q.12

Q.13

Q.14

Q.15

Q.16

(i) Propene to 1-propanol (ii) Phenol to Salicylic acidQ.17 How are the following conversions carried out ? (Write the reactions and conditions in each case)

i) Ethanol to 2-propanol

(ii) Phenol toAcetophenone.

Q.18 How are the following conversions carried out ?

(i) Phenol to Toluene

(ii) Ethanol to 1, 1-dichloroethane.

34

EXERCISE-I

Alcohol

Q.19 How are the following conversions carried out ? (Write reactions with conditions).

(i) 1-propanol to 1-chloro-2-propanol (ii) Phenol to salicylic acid

Write reactions and conditions for the following conversions

(i) 2-Propanone into 2-methyl-2-propanol (ii) n-Propyl alcohol into hexane.

Explain hydration of alkene ?

Write the method of formation of alcohol fromacid derivatives ?

How willyou distinguish between the following pairs by chemical reactions ?

Q.20

Q.21

Q.22

Q.23

(i) 1-Propanol and 2-Propanol (ii) Ethanoland phenol

Q.24

Q.25

Give reasons for the fact that methanol is miscible with water while iodomethane is not.

How willyou carry out following conversions

(i) EthylAlcoholto ethylene glycol (ii) 2-Propanol to 1-bromopropane

Q.26 Explain the mechanism of following reactions

2C2H2OH 24 C2H5OC2H5 H2Oconc.H SO

413K

Give equations ofthe following reactions:

(i) Oxidation of propan-1-ol with alkaline KMnO solution.

Q.27

4

(ii) Bromine in CS with phenol.2

(iii) Dilute HNO with phenol.3

(iv) Treating phenol with chloroform in presence of aqueous NaOH.

Explain the following with an example.

(i) Kolbe’s reaction.

(ii) Reimer-Tiemann reaction.

(iii) Williamson ether synthesis.

(iv) Unsymmetrical ether.

Write the mechanism of Reimer-Tiemann reaction ?

Exmplain estrification reaction with mechanism ?

Q.28

Q.29

Q.30

35

Alcohol

Q.1 Arrange the following as mentioned.(1) C6H5CH2CH2Br,

(I)C6H5CH(Br)CH3 and

(II)C6H5CH = CH-Br (Reactivity withAgNO3)

(III)

Cl

ClCl

NO2

(Reactivity with NaOt)(2)NO2 NO2 NO2

(I)H2O, (I)

(II)CH3O- and(III)

(III) CH3COO-

(IV)OH-

(II)(3) (Nucleophillicity rate

SN2 reaction)

CH3COO-, (I)

OH

C6H5O-

(II)C6H5SO3

-

(III)(4) (Leaving group ability)

(5) , C2H5OH, HCOOH, CH3COOH (Acidic strength)

(I) (II) (III) (IV)

CH2-OH CH2-OH

(6) C6H5CH2OH, (C6H5)2CHOH, (C6H5)3COH,

NO2

(IV)

Cl

(I) (II) (III) (V)(Reactivity with HBr decreasing order)

(7) Phenol,(I)

o-chlorophenol(II)

m-chlorophenol,(III)

p-chlorophenol(IV)

(Decreasing acidic character)(8) CH3CH2CH2CH2-OH,

(I)CH3CH(OH)CH3.

(II)(CH3)2C(OH)C2H5

(III)(decreasing order ofreactivity for esterification)

[Classifies SN1, SN2, E1 or E2][Classify as SN1, SN2, E1, E2]

[Classify as SN1, SN2, E1 or E2]

(9)(10)

CH3CH2CH2Cl + I-(CH3)3CBr + CN-(alc)

CH3CH(Br)CH3 + OH-(11)

(12) t-Butyl chloride, sec-butyl chloride and CCl4 with alc silver nitrate.

36

EXERCISE-II

Alcohol

Q.2 Identity the productsA, B, C, D etc.

Me

dil. KMnO4 HIO4 OH

CA B(1)

Me

CHCl2

C6H6 C H2 O/H+(A)(2) CH4 BCH3Br D

AlCl3(anhyd)

OH

G(3) HA + HB (Elb’s oxidation)

H(4) X (with intermediate)

OHHI (excess)

Heat(5) (CH3)2CHOCH3

OH

CH3(CH2)4COOH

ZnCl2

Zn-HgA B

(6) HClOH

NO2

(C2H5)2SO4

OH-Zn/HCl NaNO2 /HCl C6H5OH

5ºCCA B D

(7)

OH

PBr5 alcohol Br2B C(8) A CH3CH2CH2Br KOH

alcohol followed by NaNH2 dil. H2SO4

Hg2+

NH2OHED F

Q.3 Explain the following giving reasons (equations wherever necessary)(1)

(2)

Sodium metal can be used for drying diethyl ether and benzene but not ethanol.

The boiling point of the three alcohols lie in the order : n-Butyl alcohol > sec-Butyl alcohol > terButyl alcohol.

No esterification takes place between ethyl alcohol and excess of sulphuric acid at 170ºC.

Sodium chloride solution in water is added to decrease the solubility of organic compounds in

water.Acid catalysed dehydration of t-butanol is faster than that of n-butanol.

Alcohols cannot be used as solvent with Grignard reagents or with LiAlH4.

Phenols are more acidic than alcohols.

(3)

(4)

(5)

(6)

(7)

37

Alcohol

Alkylation of PhO- with an active alkyl halide such as CH2 = CHCH2Cl gives phenylallyl ether(8)and also some o-allyl phenol.

Although phenoxide ion has more number of resonating structures than benzoate ion, benzoic acid is a stronger acid than phenol.

Although n-alcohol and acetic acid both have the same molecular mass (60), the former boils at a lower temperature (97º) than the latter (118ºC).

Drygaseous hydrohalic acids and not their aqueous solutions are used to prepare alkyl halides from alkenes.

Isobutane gas dissolves in 63% H2SO4 to form a deliquescent white solid. If the H2SO4 solution is diluted with water and heated, the organic compound obtained is a liquid boiling at 83ºC.

tert-Butyl iodide undergoes following reactions. (CH3)3Cl + H2O (CH3)3COH + HI(CH3)3Cl + OH- (CH3)2C = CH2 + H2O + I-

CH3CH2I undergoes loss of HI with strong base faster than CD3CH2I for loss of DI

The carbocation is F3C -C+ is destabilized while the carbocation Me3C+ is stabilized.

p-Dihalobenzenes have higher melting points and lower solubilities than that of their o- andm-isomers.

Iodoform gives precipitate with silver nitrate on heating, while chloroform does not.

Ethyl iodide becomes violet on standing inpresence of light.

When 2-chloro-3-methylbutane is treated with alcoholic potash, 2-methyl-2-butene is themain product.

(9)

(10)

(11)

(12)

(13)

(14)

(15)

(16)

(17)

(18)

(19)

Q.4 What happens when (Give equations also)(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)

(13)

Ethylidene bromide is hydrolysed with aq. potassium hydroxide.

Chloral is treated with aqueous sodiumhydroxide.

Chloroform is boiled with aqueous potassium hydroxide.

Phenol is treated with carbon tetrachloride in presence of excess of alkali.

Propene is heated with chlorine at about 400ºC under a pressure of 70 atmosphere.

Silver acetate is treated with bromine.

Carbon tetrachloride is treated with hydrogen fluoride in presence ofantimony fluoride.

2-Butanol is treated with acidified potassium dichromate.

Diethyl ether is heated with sulphuric acid under pressure.

Excess of ethanol is heated with concentrated sulphuric acid at 140ºC.

Diethyl ether is heated with concentrated sulphuric acid.

Diethyl ether is treated with concentrated sulphuric acid at 0ºC.

Phenol is treated with excess of bromine water.

Tert. butyl iodide forms tert. butyl alcohol with water but with OH- yields Isobutylene. Explain.

Why SN reaction proceeds faster in methanol than in C6H6 ?

Whydoes aqueous sodium bicarbonate solution dissolve carboxylic acids but not phenols though they are also acidic ?

What is the function of ZnCl2 in Lucas reagent employed in the reaction of alcohols ?

Q.5

Q.6

Q.7

1

Q.8

38

Alcohol

Q.9 Dehydration of alcohols to alkenes is always carried out in the presence of conc. H2SO4 and not withconc. HCl or HNO3.

Give structures of (W), (X), (Y) and (Z) in the following reactions :

CrO3

Q.10

H2C = CH - CH2OH (W) ;(CH Cl )2 2

+- H3OCH3OH,H+ KMnO4.OH(W) (X) C H O (Y) C H O (Z) C H O .5 10 2 5 12 4 3 6 3

(cold, dilute) H2O

Q.11 An organic compound (A), C9H12O was subjected to a series of tests in the laboratory. It was found that this compound :(i)

(ii)

(iii)

(iv)

(v)

Rotates the plane polarized light.

Evolves hydrogen gas with sodium.

Reacts with I2 and NaOH to produce a pale yellow solid compound.

Does not react with Br2/CCl4.

Reacts with hot KMnO4 to form compound (B), C7H6O2 which can also be synthesized by the reaction of benzene and carbonyl chloride followed by hydrolysis.

Loses optical activity as a result of formation of compound (C) on being heated with HI and P.

Reacts with Lucas reagent in about 5 minutes.

(vi)

(vii)

Give structures ofAto C with proper reasoning and draw Fischer projection for (A). Give reactionsforthe steps wherever possible.

Compound (A), C4H8O, on catalytic hydrogenation gave (B) C4H10O. Compound (A) decolorizedBr2 in CCl4 and cold aqueous KMnO4 solution. On refluxing, (B) with HBr, compound (C) is formed.(C), on treatment with Mg in dry ether followed by hydrolysis, gave a hydrocarbon gas weighing 1.5 gper 1.12 litres at S.T.P. Give a structure for (A).

Explain briefly the formation of products giving the structure of the intermediates.

Q.12

Q.13

(a)

NaNH2

NH3(b) OCH3 OCH3

Br H2NO

H3C C-OC2H5

CH

O

OC-OC2H5(i) OH-

(ii) H+NaOEt Br(c) H2C H C-CH-C-OH3

C-OC2H5CH2-C-OH

OO

39

Alcohol

Q.14 (a) An organic compound C8H18 on monochlorination gives a single monochloride. Write thestructureof the hydrocarbon.Write the structures of possible major monosubstituted products formed when Br+ attacks the following molecules.

(b)

,OCH3,COO

HO3S NHCOCH3

Q.15 Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic A. Theorganometallic reacts with ethanal to give an alcohol B after mild acidification. Prolonged treatmentof alcohol B with an equivalent amount of HBr gives 1-Bromo-1-Methylcyclopentane (C). Writethe structures ofA, B and explain how C is obtained from B.

Write the products in the following reactionsQ.16

Peroxide(i) H3C-CH = CH2 + BrCCl3

H3C-CH=C(CH3)2 + HClPeroxide

(ii)

Q.17 Write down the structures forA, B, C, D and F.

NaNH2

Mel

Na/NH3(l)PhC CH A B

H + .-.+ KNH2

Ph(i) CC = C

Ph Br

(ii) DMe I + Cu + Heat

MeCH3

+ H3C- C- CH2-Br + Anhy. AlCl3 Heat

H

(iii) E

Q.18 Dehydrobromination of compounds (A) and (B) yield the same alkene (C).Alkene (C) can regenerate(A) and (B) by addition of HBr in the presence and absence of peroxide respectively. Hydrolysis of (A)and (B) give isomeric products (D) and (E) respectively, 1, 1-Diphenylethane is obtained on the reactionof (C) with benzene in the presence of H+. Give structures of (A) to (E) with reasons.

CompoundA, C8H8O can be oxidised with KMnO4 to B having molecular formula C8H6O4. CompoundB is dicarboxylic acid, but does not form anhydride on heating. B When treated with bromine inpresence of iron, gives only one monoderivative C, C8H5BrO4. Give the structural formula of A, Band C.

How are the following conversions carried out?

(i) PropenePropan-2-ol.

(ii) Benzyl chlorideBenzyl alcohol.(iii) Ethyl magnesium chloridePropan-1-ol.

(iv) Methyl magnesium bromide2-Methylpropan-2-ol.

Q.19

Q.20

40

Alcohol

Q.21 Write the name the reagents and complete equation in following conversion reactions ?

(i) Oxidation of a primary alcohol to carboxylic acid.

(ii) Oxidation of a primary alcohol to aldehyde.

(iii) Bromination ofphenol to 2,4,6-tribromophenol.

(iv) Benzyl alcohol to benzoic acid.

(v) Dehydration of propan-2-ol to propene.

(vi) Butan-2-one to butan-2-ol.

Write the names of reagents and equations for the preparation of the following ethers byWilliamson’s

synthesis:

Q.22

(i) 1-Propoxypropane

(iii) 2-Methoxy-2-methylpropane

(ii) Ethoxybenzene

(iv) 1-Methoxyethane

Q.23 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of

ethers.

Write equations ofthe following reactions:Q.24

(i) Friedel-Crafts reaction -alkylation of anisole.

(iii) Bromination of anisole in ethanoic acid medium.

(ii) Nitration of anisole.

(iv) Friedel-Craft’s acetylation of anisole.

Q.25 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

Give a mechanismfor this reaction.

(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocationby

a hydride ion shift from 3rd carbon atom.

Give equations only to distinguish between primary, secondary and tertiary alcohols byLucas Test.

(i) Name the reagents and write the chemical equations for the preparation of the following compounds

byWillamson's synthesis

(a) Ethoxybenzene

(b) 2-Methyl-2-methoxypropane

(ii) Whydo phenols not give protonation reaction readily ?

The following is not an appropriate reaction for the preparation of t-butylethyl ether.

Q.26

Q.27

Q.28

(i) What would be the major product of this reaction ?

(ii) Write a suitable reaction for the preparation of tert-butylethyl ether.

Give the structures and IUPAC name of the products expected from the following reactions :

(i) Catalytic reduction of butanal.

(ii) Hydration of propene in the presence of dilute sulphuric acid.

(iii) Reaction ofpropanone with methylmagnesium bromide followed byhydrolysis.

Write the chemical properties of ehter ?

Q.29

Q.30

41

Alcohol

Q.1 Hydrogen bonding is possible in-

(A) Ethers (B) Hydrocarbons (C) Alkanes (D) Alcohols

Q.2 The increasing order of boiling points of 1º, 2º, 3º alcohol is -

(A) 1º > 2º > 3º (B) 3º > 2º > 1º (C) 2º > 1º > 3º (D) None

Q.3 The solubility of lower alcohols in water is due to -(A) Formation of hydrogen bond between alcohol and water molecules

(B) Hydrophobic nature of alcohol

(C) Increases in boiling points

(D) None of these

Conversion of CH3OH to CH3COOH can suitably be carried out with the reagent (under high

pressure condition) :

Q.4

(A) CO2/H2SO4 (B) CO/BF3 (C) CO2/BF3 (D) CO/H2SO4

Q.5 Which of the following reactions of an alcohol does not involve O - H bond breaking :

(A) Reaction with alkali metals(C) Reaction with sulphonyl chloride

(B) Reaction with an acyl chloride(D) Reaction with conc. sulphuric acid.

Q.6 Alkyl chloride is formed when alcohol is treated with HCl in presence of anhydrous ZnCl2. The

order of reactivity with respect to alcohol is :(A) 3º > 2º > 1º (B) 1º > 2º > 3º (C) 2º > 1º > 3º (D) 1º > 3º > 2º

Q.7 When ethyl alcohol reacts with acetic acid, the products formed are-

(A) Sodium ethoxide + hydrogen(C) Ethyl acetate + soap

(B) Ethyl acetate + water(D) Ethyl alcohol + water

Q.8 Methyl alcohol reacts with phosphorus trichloride to form-

(A) Methane (B) Methyl chloride (C) Acetyl chloride (D) Dimethyl ether

Q.9 The -OH group of methyl alcohol cannot be replaced by chlorine by the action of-

(A) Chlorine(C) Phosphorus trichloride

(B) Hydrogen chloride(D) Phosphorus pentachloride

Q.10 Reaction of alcohol does not show cleavage of R-O linkage-

(A) ROH + PCl5 (B) ROH + SOCl2 (C) ROH + HCl (D) ROH + Na

Q.11 Replacement of -OH group in alcohol by -Cl cannot be carried out with-

(A) PCl5 (B) SO2Cl2 (C) PCl3 (D) SOCl2

Q.12 Which alcohol does not give a ketone on oxidation-

(A) Isopropyl alcohol(C) Ethylmethylcarbinol

(B) Allyl alcohol(D) Methylphenylcarbinol

Q.13 A compound X with molecular formula C3H8O can be oxidised to a compound Y with the

molecular formula C3H6O2, X is most likely to be-

(A) Primary alcohol(C) Aldehyde

(B) Secondary alcohol(D) Ketone

42

EXERCISE-III

Alcohol

Q.14 Tertiary alcohols are resistant to oxidation because :(A) They do not have hydrogen atom

(B) Of large +I effect of alkyl groups

(C) Of greater steric hindrance

(D) All the above

Ether in contact with air for a long time form peroxides. The presence of peroxide in ether can be tested by adding Fe2+ ion in it and then adding -

Q.15

(A) KCNS (B) SnCl2 (C) HgCl2 (D) KI

Q.16 Match list I with list II and then select the correct answer from the codes given below the lists -List I

(A) (CH3)3C -OH(B) CH3 -O -C (CH3)3(C) C2H5 -O -C2H5(D) CH3 -O -CH (CH3)2Codes :

List II(a) Sulphuric ether(b) Chain isomer of 1-butanol(c) Mixed ether(d) Has one secondary carbon

Abbbc

Bdccb

Caada

Dcdad

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

Q.17 Ether reacts with chlorine in the dark to form-(A) CH2Cl -CH2 -O -CH2 -CH3(B) CH2Cl -CH2 -O -CH2 -CH2Cl(C) CH3 -CH (Cl) -O -CH(Cl)-CH3(D) CCl3 -CH2 -O -CH2 -CCl3Ether does not form oxonium salt on reaction with-

Q.18(A) Cold conc. H2SO4(C) Conc. HI

(B) Cold conc. HCl(D) None of the above

Q.19 The ordinary alkyl ethers are cleaved by -(A) Ethanol (B) Ethyl halide (C) BF3 (D) Hydrogen iodide

Q.20 The decomposition of ethers by HI or HBr is called -(A) Zerewitinoff's reaction(C) Williamson's method

Ether is not formed in this reaction

(B) Ziesel's method(D) Hell-Volhard-Zelinsky reaction

Q.21Conc.H2SO4

140º(A) 2C2H5OH

(B) (CH3)3 C-Cl + C2H5ONa (C) C2H5Cl + (CH3)3 C-ONa

(D) Oxygen of ether can be replaced by chlorine when treated with PCl5

Unsymmetrical ethers are best prepared by -(A) Willamson's continuous etherification process(B) Reacting Grignard reagent with alkyl halide(C) Treating sodium alkoxides with alkyl bromides(D) Heating an alkanol with conc. H2SO4

Which of the following is used as an additive by fire departments under the name Rapid-Water -

Q.22

Q.23(A) Ethylene glycol(C) Epichlorohydrin

(B) Polyethylene oxide(D) Epoxy oxide

43

Alcohol

Q.24 In the Williamson's synthesis for diethyl ether, which species works as a nucleophile -(A) Halide ion (B) Ethoxide ion (C) Ethyde ion (D) Hydride ion

Q.25 The structure of the compounds formed by the reaction of diethyl ether with oxygen of air is -(A) CH3CH2 - O - O - CH2CH3

CH3|

(B) CH3CH2 - O -CH O - O - H

(C) CH3CH2 - O - O - CH2 - O - CH3

(D) CH2 (OCH3) - CH2 - O - C2H5

Ether bottles should not be kept open in air because -Q.26(A) Ether is an anaesthetic(C) Ether is costly

(B) Ether forms an explosive peroxide(D) Ether gets oxidised to ethanol

Q.27 Isopropyl alcohol vapour is passed over alumina heated at about at about 240ºC. The productformed is -(A) diisopropyl ether(B) propene(C) a mixture of diisopropyl ether and propene(D) 3-hexene

Q.28 Which of the followingmethoxide -

(A) CH3CH2CH2Cl(C) PhCH2Cl

is not expected to give ether on react ion with sodium

(B) CH2 = CHCH2Cl(D) CH2 = CHCl

Q.29 Consider the following reaction,

1.(CH3COO)2Hg,CH3OHCH3CH = CH2

2.NaBH4

The product formed in the reaction is -

(A) (B)

(C) (D)

Q.30 When phenol is reacted with CHCl3 & NaOH followed by acidification, salicyladehyde is

obtained.

Which of the following species are involved in the above mentioned reaction as intermediates -O

H CHCl2(A) (B) (C) (D) Both (A) and (B)

Q.31 The presence of -OH on adjacent carbon atoms can be detected by the reaction of the compoundwith -

(A) Conc. H2SO4 (B) Conc. HNO3 (C) HIO4 (D) Acidic KMnO4

44

Alcohol

Q.32 In phenols -(A) - OH group is attached in side chain(B) - OH group is directly attached to benzene nucleus(C) Both (A) & (B) (D) None

The compound containing hydrogen bond is-Q.33

(A) Toluene (B) Phenol (C) Chlorobenzene (D) Nitrobenzene

Q.34 Phenol on treatment with ammonia gives -(A) Benzene (B) Benzoic acid (C)Aniline (D) None

Q.35 Salicylic acid, aspirin, nylon, plastics and picric acid have a common raw material, namely -(A) Methane (B) Formic acid (C) Phenol (D) Alcohol

Q.36 Which of the following will not be soluble in sodium carbonate solution -

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

Q.37 Under suitable conditions C6H5CH2OH (A) C6H5OH (B), and C6H5COOH (C) can act as acids.

The increasing order of their acidic strength is -

(A) A < B < C (B) A < C < B (C) B < A < C (D) C < B < A

Q.38 Kolbe’s reaction consists in obtaining -(A) Anisol from phenol(B) Salicylaldehyde from phenol and CHI3

(C) Salicylic acid from sodium phenate and CO2

(D) Salicylic acid from phenol and CO2

The most suitable method of separation of a 1 : 1 mixture of o- and p- nitrophenol is -Q.39(A) Sublimation (B) Chromatography (C) Crystallisation (D) Distillation

Q.40 p-Nitrophenol is stronger acid than phenol because nitro group is -(A) Electron withdrawing(C) Basic

(B) Electron donating(D)Acidic

Q.41 Which derivative of phenol gives effervescence with NaHCO3 -(A) o-Cresol(C) 2,4,6- Trinitrophenol

(B) Catechol(D) 2,4,6- Tribromophenol

Q.42 The product obtained by the reaction of phenol with benzene diazonium chloride is -(A) Phenyl hydroxylamine(C) Phenyl hydrazine

(B) Para amino azobenzene(D) Para hydroxy azobenzene

Q.43 Phenol and benzoic acid can be distinguished by -(A) Aqueous NaHCO3

(C) Aqueous NaOH

Phenol is converted into salicylaldehyde by-

(A) Etard reaction

(B) Aqueous NaNO3

(D) Conc. H2SO4

Q.44

(B) Kolbe reaction

(C) Reimer- Tiemann reaction (D) Cannizzaro reaction

45

Alcohol

Q.45 Alcohols which can give white turbidity immediately on reaction with Lucas reagent are

CH3|

H3C C OH|

H

(A) (B) Ph3C-OH (C) H3C-OH (D)OH

Q.46 Correct method to prepare Me3C-O-Me in good yield is / are

(i) Hg ( OAc ) 2MeOH(ii)SBH

(A) Me3C-Cl + MeONa (B) Me2C = CH2

H

MeOH

(C) Me3C - ONa + MeCl

Which is not cleaved by HIO4?

(D) Me2C = CH2

Q.47(A) glycerol (B) glycol (C) 1,3-propandiol (D) 1-methoxy-2-propanol

Ph

.concHI product(s) of this reaction is / are?Q.48

O

Ph Ph I

OH(A) (B) (C) (D)

I I

Q.49 In which of the following reactions 3°alcohol will be obtained as a product.

O||

MgBr (excess) + H C Cl (A)H

O||

(B) PhMgBr (excess) + CH3 C ClH

O O|| ||

(C) CH3MgBr (excess) + CH3 C O C CH3H

O||

(D) CH3MgBr (excess) + Cl C O EtH

Which of the following reagents or process are suitable to distinguish MeOH & EtOH ?

OH

Q.50

+& HCOOH

(A) NaOI (B)

(C) anhydrous ZnCl2 + conc. HCl (D) Victor Mayor's process

46

Alcohol

Matrix Match Type QuestionQ.51 Match the column:

Column-I and column-II contains fourentrieseach. Entryof column-I are to be uniquelymatched with only one entry of column-II.

Column I(Compound)

H3C-CH2-CH2-CH2-CH2-OH

H3C CH CH3|OH

Column II(Boiling point in °C)

(A)

(B)

(P)

(Q)

290°C

138°C

(C) H2C -CH -CH 2 (R) 102°C| |

OH OH|

OH

CH3|

H3C C CH 2 CH3|

OH

(D) (S) 82.4°C

Q.52 Column I

(Compound)

Column II

(A) (P) Gives orange colour with Brady's reagent (2,4-DNP)O

(B) (Q) Gives haloform reactionCl

O

(C) (R) Gives positive Beilstein's test

Cl

(D) (S) Can give aldol reaction with dil. NaOH

(T) Gives white ppt. with alcoholic

AgNO3

47

Alcohol

ANSWER KEY

EXERCISE - IQ.1 1.

4.7.10.

II > I > III III > I > IIII > III > IV > I E1

Me

2.5.8.11.

II > III > IIII > IV > I > II I > II > IIIE1

O

3.6.9.12.

III > II > IV > IIII > II > I > V > IV SN-2(t- > S- > CCl4)

Me MeOH

MeQ.2 1. A = B = C =

OH O OOOH

CHOCH32. A = Br2/hv B = C = 2 Cl2/hv D =

OHOH

OH3. + 4.

OOHOH

OH OH

OH

5. 6. A= B =CH - CH -CH + CH OH3 3 3 OH OH

C - (CH2)4 - CH3 CH2 - (CH2)4 - CH3

O

NO2 N2Cl

NH2

7. A = B = C =

OC2H5 OC H OC H2 5 2 5

D = C2H5O N = N OH

8. A = CH3CH2CH2 - OH B = CH3 - CH = CH2 C = CH3 - CH - CH2

Br Br

CH3

3D = CH3 - C CH E = CH3 - C - CH2 F = CH

C = N - OH

OOH

OCH3 Br Br

Q.11 A= B = m-Cresol C = Methyl iodide D = Phenol E =CH3

Br

OH Br

Q.12 A = B = C =

48

Alcohol

Q.14 a = 2, 2, 3, 3-tetramethylbutane

Br

-NH - C - CH3-C - O - -Br SO3H -b =

O O

OHMe

Br

MgBrCH - CH3Q.15 A= B = C =

CH3

Q.16 CH3 - CH2 - C - CH3CH3 - CH - CH2 - CCl3

Br Cl

Ph

H

H

Me

Me

Q.17 A = Ph -C C Me B = C = Ph - C C -PhC = C

D = Me - E = Me - -C - Me- -Me

Me

Ph - CH - CH3Q.18 A= Ph - CH2 - CH2 - Br B =

Br

D = Ph - CH2 - CH2 - OHC =

E =

Ph - CH = CH2

Ph - CH - CH3

OH

CHO COOH COOH

-Br

Q.19

Me COOH COOH

EXERCISE - IIIQ.1Q.8Q.15Q.22Q.29Q.36Q.43Q.50

DBACBCAA,B

Q.2Q.9Q.16Q.23Q.30Q.37Q.44Q.51

AABBDAC

Q.3Q.10Q.17Q.24Q.31Q.38Q.45

ADCBCCA,B

Q.4Q.11Q.18Q.25Q.32Q.39Q.46

BBCBBDB,C,DQ.52

Q.5Q.12Q.19Q.26Q.33Q.40Q.47

DCDBBAC,D

Q.6Q.13Q.20Q.27Q.34Q.41Q.48

AABCCCA,B,C

Q.7Q.14Q.21Q.28Q.35Q.42

BABDCD

Q.49 B,C,D(A) Q; (B) S; (C) P; (D) R (A) P,Q,S (B) Q,R,T (C) P,S (D) R,T

49