E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò · 2019-05-29 · Alcohol hol Eh E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò 104 9. Give the equations of reactions of preparation of phenol from cumene

Alcohol, Phenol and Ether

NCERT Textual Exercise (Solved)

100

1. Write the IUPAC names of the following compounds:

(a) � � � �

3

3 3

3 3

CH

|CH CH CH C CH

| | |CH OH CH

(b) � � � �3 2 3

2 5

CH CH CH CH CH CH

| | |OH OH C H

(c) CH – CH –CH – CH3 3

OHOH

(d) HOCH2 – CHOH – CH2OH

(e)

CH3

OH

(f)

CH3

OH

(g)

CH3

CH3

OH

(h)

CH3

CH3

OH

(i) � � � �3 2 3

3

CH O CH CH CH|CH

(j) C6H5 – O – C2H5

(k) C6H5 – O C7 H15

(l) � � � � �3 2 2 3

3

CH CH O CH CH CH

|CH

Sol. (a) 2, 2, 4 – Tri methylpentane – 3 – ol (b) 5-Methylheptan -2, 4-diol (c) Butan-2, 3-diol (d) Propane -1, 2, 3-triol (e) 2-Methylphenol (f) 4-Methylphenol (g) 2, 5-Dimethylphenol (h) 2,6-Diemethylphenol (i) 1-Methoxy-2-methylpropane (j) Ethoxybenzene (k) 1-Phenoxyheptane (l) 2-Ethoxy butane



101

2. Write the structures of the compounds whose IUPAC names are as follows: (a) 2-Methylbutan-2-ol (b) 1-Phenylpropan-2-ol (c) 3, 5-Dimethylhexane-1, 3, 5-triol (d) 2, 3-Diethylphenol (e) 1-Ethoxypropane (f) 2-Ethoxy-3-methylpentane (g) Cyclohexylmethanol (h) 3-Cyclohexylpentan-3-ol (i) Cyclopent-3-en-1-ol (j) 3-Chloromethylpentane-l-ol

Sol. (a)

3

3 2 3

CH

|

CH C CH CH|OH

� � � (b) – CH – CH – CH2 3

OH

(c) 3 3

2 2 2 3

CH CH

| |CH CH C CH C CH| | |OH OH OH

� � � � � (d)

OH

C H2 5

C H2 5

(e) CH3CH2 – O – CH2CH2CH3 (f) CH CH CH CH CH

CH CH O CH

3 2 3

3 2 3

− − −

− −| |

(g)

CH OH2

(h)

(i)

OH

(j) 2 2 2 3

2

HO CH CH CH CH CH|

CH Cl

− − − − −

3. (a) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.

(b) Classify the isomers as primary, secondary and tertiary alcohols.



102

Sol. Eight isomers are possible. These are:

(a) CH CH CH CH CH OH3 2 2 2 2Pentan− −

°1

1ol

( )

(b)

Pentan-2-ol(2º)

OH

CH CH CH CHCH|

3 2 2 3−

(c)

Pentan-3-ol(2º)

OH

CH CH CH CH CH|

3 2 2 3− (d)

2-Methylbutan-1-ol(1º)

CH

CH CH CHCH OH

3

3 2 2

|

(e)


CH

CH CHCH CH OH

3

3 2 2

| (f)


CH

CH C CH CH

OH

3

3 2 3

|

|− −

(g)

CH

CH C CH OH

CHDimethylpropan ol

3

3 2

32 2 1

1

|

|

,( º)

− −

− − −

(h)


CH OH

CH CH CH CH3

3 3

| |− − −

4. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Sol. The molecules of butane are held together by weak van der Waal’s forces of attraction while those of propanol are held together by stronger intermolecular hydrogen bonding.

H O H O H O

CH CH CH C

| | |2 2 3 HH CH CH CH CH CH2 2 3 2 2 3

Therefore, the boiling point of propanol is much higher than that of butane. 5. Alcohols are comparatively more soluble in water than hydrocarbons of

comparable molecular masses. Explain this fact. Sol. Alcohols can form hydrogen bonds with water and break the hydrogen bonds

already existing between water molecules. Therefore, they are soluble in water.



103

On the other hand, hydrocarbons cannot form hydrogen bonds with water

and hence are insoluble in water. 6. What is meant by hydroboration-oxidation reaction? Illustrate with an

example. Sol. The addition of diborane to alkenes to form trialkyl boranes followed by

their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation. For example,

7. Give the structures of IUPAC name of monohydric phenols of molecular

formula C7H8O. Sol. The three isomers are:

OH OH OH

2-methylphenolo-Cresol

3-methylphenolm-Cresol

4-methylphenolp-Cresol

CH3

CH3

CH3

8. While separating a mixture of ortho- and para-nitrophenols by steam distillation, name the isomer which is steam volatile. Give reasons.

Sol. o-Nitrophenol is steam volatile due to chelation (intramolecular H – bonding) and hence can be separated by steam distillation from p-nitrophenol which is not steam volatile because of intermolecular H-bonding.



104

9. Give the equations of reactions of preparation of phenol from cumene. Sol.

10. Write chemical reaction for the preparation of phenol from chlorobenzene.

Sol.

11. The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

(i) What would be the major product of this reaction? (ii) Write a suitable reaction for the preparation of t-butylethyl ether. Sol. (i) The major product of the given reaction is 2-methylprop-1-ene. It is

because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

(ii)

12. You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.

Sol. C H C H SO HBenzene

conc H SOSulphonation

Benzones

6 62 4

6 5 3. ,

uulphonic acid

NaOH fuseK

Sodiumphenoxide

Dil HC H ONa,573 6 5 CCl

NaClPhenol

C H OH 6 5

13 Show how will you synthesise (i) 1-phenylethanol from a suitable alkene. (ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction. (iii) propan-1-ol using a suitable alkyl halide.



105

Sol. (i) Addition of H2O to ethenylbenzene is presence of dil H2SO4.

(ii) Hydrolysis of cyclohexylmethyl bromide by aqueous NaOH gives

cyclohexylmethanol.

CH Br2 CH OH2

+ NaOH + NaBr�

S 2, hydrolysisN

Cyclohexyl-methanol

(iii) Hydrolysis of 1-bromopropane by aqueous NaOH gives propan-1-ol. CH3CH2CH2Br + NaOH

SN Hydrolysisl

CH CH CH OH NaBr2 3 2 2, Propan-1-o

14. Give two reactions that show the acidic nature of phenol. Compare its acidity with that of ethanol.

Sol. The reactions showing acidic nature of phenol are: (a) Reaction with sodium: Phenol reacts with active metals like sodium

to liberate H2 gas.

OH ONa

+ 2Na + H22 2

Phenol Sodiumphenoxide

(b) Reaction with NaOH: Phenol dissolves in NaOH to form sodium phenoxide and water.

OH ONa

+ NaOH + H O22

Phenol is more acidic than ethanol. This is due to the reason that phenoxide ion left after the loss of a proton from phenol is stabilized by resonance, while ethoxide ion left after loss of a proton from ethanol, is not.



106

15. Explain why ortho-nitrophenol is more acidic than ortho-methoxy phenol? Sol. Due to strong –R and – I-effect of the –NO2 group, electron density in the

O – H bond decreases and hence the loss of a proton becomes easy.

Further, after the loss of a proton, the o-nitrophenoxide ion left behind is

stabilized by resonance.

o-Nitrophenoxide ion is stabilized by resonance, thereby making

o-nitrophenol a stronger acid. In contrast, due to +R effect of the OCH3– group, it increases the electron

densityintheO–Hbondtherebymakingthelossofprotondifficult.

–OCH3

Further more, the o-methoxyphenoxide ion left after the loss of a proton is destabilized by resonance.

–OCH3

The two negative charges repel each other thereby destabilizing the o-methoxy phenoxide ion. Thus, o-nitrophenol is more acidic than o-methoxy phenol.



107

16. Explain how does – OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.

Sol. Phenol may be regarded as a resonance hybrid of structures I-V, shown below.

OH

I II III IV V

OH OH OHOH.. .. .. ....

+.. .. + +

As a result of +R effect of the OH-group, the electron density in the benzene ring increases thereby facilitating the attack by an electrophile. In other words, presence of OH-group, activates the benzene ring towards electrophilic substitution reactions. Further, since the electron density is relatively higher at the two o-and one p-position, therefore electrophilic substitution occurs mainly at o-and p-positions.

17. Give equations of the following reactions: (a) Oxidation of propan-1-ol with alkaline KMnO4 solution. (b) Bromine in CS2 with phenol. (c) Dilute nitric acid with phenol. (d) Treating phenol with chloroform in presence of aqueous NaOH. Sol. (a) CH3CH2OH + 2 [O]

(b)

(c)

(d) Cl

2



108

18. Explains the following with an example: (i) Kolbe’s reaction (ii) Reimer - Tiemann reaction (iii) Williamson ether synthesis (iv) Unsymmetrical ether Sol. (i) Kolbe’s reaction: Sodium phenoxide when heated with CO2 at 400K

underapressureof4-7atmospheresfollowedbyacidificationgives2-hydroxybenzoic acid (salicylic acid) as the major product along with a small amount of 4-hydroxy benzoic acid. This reaction is called Kolbe’s reaction.

(ii) Reimer-Tiemann reaction: Treatment of phenol with CHCl3 in

presence of aqueous sodium or potassium hydroxide at 340 K followed by hydrolysis of the resulting product gives 2-hydro-benzaldehyde (salicylaldehyde) as the major product. This reaction is called Reimer-Tiemann reaction.

OH

+ CHCl3 –NaCl,–2H O2

NaOH, 340 K

ONa

CHCl2

ONaOH

CHO

H O2CHO

ONa

CHOH

OH2NaOH

Dil HCl

–NaCl

–NaCl

(iii) Williamson’s ether syntheses: It involves the treatment of an alkyl halide with a suitable sodium alkoxide. The sodium alkoxide needed for the purpose is prepared by the action of sodium on a suitable alcohol.

(iv) Unsymmetrical ether: If the alkyl or aryl groups attached to the oxygen atom are different, ethers are called unsymmetrical ethers. For example, ethylmethylether, methylphenylether, 4-chlorophenyl-4-nitrophenyl ether, etc.



109

19. Write the mechanism of acid-catalysed dehydration of ethanol to ethene. Sol. The mechanism of dehydration of alcohols to form alkenes occur by the

following three steps: (a) Formation of protonated alcohol:

(b) Formation of carbocation :

(c) Elimination of a proton to form ethene:

20. How will the following conversions be carried out? (a) Propene → Propan -2-ol (b) Benzyl chloride → Benzyl alcohol (c) Ethylmagnesium chloride → Propan-1-ol (d) Methylmagnesium bromide → 2-Methylpropan-2-ol Sol. (a)

(b)

(c)



110

(d)

CH3

21. Name the reagents in the following reactions: (a) Oxidation of primary alcohol to a carboxylic acid. (b) Oxidation of a primary alcohol to an aldehyde. (c) Bromination of phenol to 2, 4, 6 – tribromophenol (d) Benzyl alcohol to benzoic acid. (e) Dehydration of propan-2-ol to propene. (f) Butan-2-one to butan-2-ol. Sol. (a) Acidified potassium dichromate or, neutral acidic or alkaline

potassium permanganate. (b) Pyridinium chlorochromate (PCC), C5H5N

+HClCrO3– in CH2Cl2

or Pyridinium dichromate (PDC), (C5H5N+H)2Cr2O7

2– in CH2Cl2

(c) Aqueous bromine, i.e., Br2/H2O. (d) Acidifiedoralkalinepotassiumpermanganate. (e) Conc. H2SO4 at 433 – 433 K or 85% phosphoric acid at 443 K. (f) Ni/H2 or NaBH4 or LiAlH4. 22. Give reasons for the higher boiling point of ethanol in comparison to methoxy

methane. Sol. Ethanol undergoes intermolecular H-bonding due to the presence of a

hydrogen atom attached to the electronegative oxygen atom. As a result, ethanol exists as associated molecules.

Consequently, a large amount of energy is required to break these hydrogen

bonds. Therefore, the boiling point of ethanol is higher than that of methoxy methane which does not form H-bonds.

23. Give the IUPAC names of the following ethers. (a) 3 2 3

3

CH OCH CHCH|CH

(b) CH3 — O — CH2CH2Cl



111

(c) O2N — C6H4 — OCH3 (p) (d) CH3CH2CH2OCH3

(e)

H C3 CH 3

OC H2 5

(f)

OC H2 5

(g) CH3

Sol. (a) 1-methoxy-2-methyl propane (b) 2-Chlorlo-1-methoxyethane (c) 4-Nitroanisole (d) 1-Methoxy propane (e) 4-Ethoxy-1,1-dimethyl benzene (f) Ethoxybenzene (g) 2- Ethoxy butane 24. Write the names of reagents and equations for the preparation of the following

ethers by Williamson synthesis: (a) 1-Propoxypropane (b) Ethoxybenzene (c) 2-Methyl-2-methoxy propane (d) 1-Methoxyethane Sol. (a) CH CH CH O Na CH CH CH Br

Sodium propoxide Bromopropane3 2 2 3 2 2

1

−

−+ −

→∆ CCH CH CH O CH CH CH NaBropoxypropane

3 2 2 2 2 31− − +−Pr

(b)

(c)

CH CH

CH C O Na CH Br CH C OCH NaBr

CBromoethane

3 3

3 3 3 3

| |

| |− − + − → − − +− + ∆

HH CHSodium methyl propoxide methyl methoxypropane

32 2

32 2− − − − − − −

(d) CH CH O Na CH Br CH CH OCHSodiumethoxide

Bromoethane3 2 3 3 2 3

1

− +

−+ − →∆

MMethoxyethaneNaBr+



112

25. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain type of ethers.

Sol. Williamson’s synthesis is a versatile method for the synthesis of both symmetrical and unsymmetrical ethers. However, for the synthesis of unsymmetrical ethers, a proper choice of reactions is necessary. Since Williamson’s synthesis occurs by SN2 mechanism and primary alkyl halides are most reactive in SN2 reaction, therefore, best yields of unsymmetrical ethers are obtained when the alkylhalides are primary and the alkoxide may be primary, secondary or tertiary. For example, tert-butylethyl ether is prepared by treating ethyl bromide with sodium tert-butoxide.

The above ether cannot be prepared by treating sodium ethoxide with

tert-butyl chloride or bromide since under these condition an alkene, i.e., isobutylene is the main product.

26. How is 1-propoxypropane synthesized from propan-1-ol? Sol. (a) Williamson’s synthesis (i) 3 33 2 2

13 3 2 2

13CH CH CH OH PBr CH CH CH Br H

ol BromoprapanePropan− − −+ → + PPO3

(ii) 2 2 23 2 21

3 2 2CH CH CH OH Na CH CH CH O Naol Sodium prooxidePropan− −

− ++ → + HH2

CH3CH2CH2O–Na+ + CH3CH2CH2 – Br

Dry etherHeat

propoxy propaneCH CH CH O CH CH CH → − −

−3 2 2 2 2 3

1 + NaBr

(b) By dehydration of 1-propanol with conc. H2SO4 at 413 K.

CH3CH2CH2OH + H+

+

413

2

KH O H,− + → CH3CH2CH2 — O — CH2CH2CH3



113

27. Preparation of ethers by acid catalysed dehydration of secondary and tertiary alcohols is not a suitable method. Give reason.

Sol. Acid catalysed dehydration of primary alcohols to ethers occurs by SN2 reaction involving nucleophilic attack by the alcohol molecule on the protonated alcohol molecule.

CH3CH2CH2 — O — CH2CH2CH3

Under these conditions, 2º and 3º alcohols, however, give alkenes rather than ethers. The reason being that due to steric hindrance, nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not occur. Instead protonated 2º and 3º lose a molecule of water to form stable 2º and 3º carbocation. These carbocations prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecule to form ethers.

Similarly, 3º alcohols give alkenes rather than ethers.

28. Write the equation for the reaction of hydrogen iodide with (a) 1-Propoxypropane (b) Methoxy benzene, and (c) Benzylethylether Sol. (a) CH3CH2CH2OCH2CH2CH3

HI

Kol Iodopropane

CH CH CH OH CH CH CH I373 3 2 21

3 2 2 → − +− −Propan



114

(b)

(c)

29. Explain the fact that in aryl 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. Sol. In alkyl aryl ethers, the +R-effect of the alkoxy (OR) group increases the

electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.

+ + +

Since the electron density increases more at the two ortho and one para position as compared to meta portions therefore, electrophilic substitution reactions mainly occur at o-and p-positions.

30. Write the mechanism of the reaction of HI with methoxy methane. Sol. With equimolar amounts of HI and methoxy methane, a mixture of

methylacohol and methyl iodide are formed by the following mechanism:

(a)

(b) If however, excess of HI is used, methyl alcohol formed in step (b)

is also converted into methyl iodide by following mechanism:



115

(c)

(d) 31. Write equations of the following reactions: (a) Friedel-crafts reaction –Alkylation in anisole (b) Nitration of anisole. (c) Bromination of anisole in ethanoic acid medium (d) Friedel-Crafts acetylation of anisole.

Sol. (a)

p-methylanisole(Major)

CH3

(b)

(c)



116

(d)

32. Show how will you synthesize the following alcohols from appropriate alkene:

(a)

CH 3

OH (b)

OH

H

(c) OH

(d) OH

Sol. (a)

Addition of H2O to both these alkenes gives the desired alcohol.

(b)

OH

DehydrateH O2

Addition of H2O to 4-methylhept-3-ene in presence of an acid gives the desired alcohol.



117

(c)

Addition of H2O to pent-1-ene gives the desired alcohol.

However, addition of H2O to pent-2-ene gives a mixture of two

alcohols.

(d) OH +DehydrateH O2

Now addition of H2O to both in the presence of an acid gives the desired alcohol.

33. When 3-methyl butant-2-ol is treated with HBr, following reaction takes

place:

Give a mechanism for this reaction.



118

Sol.

Protonation of the given alcohol followed by loss of water gives a 2º carbocation(I), which being unstable rearranges by 1, 2-hydride shift to form the more stable 3º carbocation (II). Nucleophilic attack by Br– ion onthiscarbocation(II)givesthefinalproduct.

Documents

E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò · 2019-05-29 · Alcohol hol Eh E Zdd øãç ½ ø Ù ®Ý ~^Ê½ò 104 9. Give the equations of reactions of preparation of phenol from cumene