Bihar Board - Class 12 Chemistry - Chapter 11: Alcohols, Phenols and Ethers Long Answer Question

Long Question Answer

Question.1. Explain giving reasons :

(i) Boiling point of an alcohol is higher than the corresponding alkane.

(ii) Lower alcohols are soluble in water, while the higher alcohols are not.

(iii) Boiling point of monohydric alcohols increases with increase in carbon atoms.

Answer: (i) The boiling points of alcohols are higher than those of the corresponding alkanes, due to the intermolecular association of a large number of alcohol molecules through hydroge bonding. Thus

Since a large amount of energy is needed to break the extensive hydrogen bonds, which hold a large number of alcohol molecules together, the boiling points of alcohols are significantly higher than the corresponding alkanes, in which no such hydrogen bonding can take place.

(ii) The solubility of an alcohol in water depends upon its capacity to form hydrogen bonds with water molecules. However, the nonpolar hydrocarbon part of the alcohol molecule is unable to form hydrogen bonds with others. Consequently, the lower alcohols are soluble in water, due to the formation of hydrogen bonds between highly polarized hydroxyl groups (-OH), present in alcohols as well as water. However, in higher alcohols the hydrocarbon part becomes larger in size and, therefore, the solubility effect due to hydroxyl groups (-OH) alcohol is outweighed by the counter large ‘repulsive effect’ of the nonpolar hydrocarbon part of the alcohol molecule. Hence, higher alcohols become soluble in water. The large inductive effect (+I effect) of the hydrocarbon part (or alkyl group) in higher alcohols also reduces the polar character of hydroxyl group ( OH) in higher alcohols, which in turn reduces their capacity to form hydrogen bonding with water.

(iii) With the increase in molar mass of alcohol, the intermolecular Van der Waals forces of attraction increases, thereby the boiling point also increases. Thus b.p. Of

C₄H₅OH > C₃H₇OH > C₂H₅OH>CH₃OH

Question. 2. How are the following conversions carried out ?

(i) Propane → Propane-2-ol.

(ii) Benzyl chloride → Benzyl alcohol.

(iii) Ethyl magnesium chloride → Propane-1-ol.

(iv) Methyl magnesium bromide → 2-methy14) propane-2-ol.

Answer . (i)

Question. 3. Given the structure and IUPAC name of the expected product of the following reactions :

(i) Catalytic reduction of butanal.

(ii) Hydroboration of but-1-ene.

(iii) Hydration of propylene in presence of dil. sulphuric acid.

(iv) Reaction of propanone with ethyl magnesium bromide followed by hydrolysis of the product.

Answer ⇒ (i) CH₃CH₂CH₂CH₂OH Butan-1-ol

(ii) CH₂CH₂CH₂CH₂OH Butan-1-ol

Question. 4. Write IUPAC names of the following compounds :

Answer : (i) 2 , 2 , 4-Trimethyl pentane-3-ol.

(ii) 5-Ethyl heptane-2 , 4-diol

(iii) Butane-2 , 3-diol.

(iv) Propane-1 , 2 , 3-triol.

(v) 2-Methyl phenol.

(vi) 4-Methyl phenol.

(vii) 2 , 5-Dimethyl phenol.

(viii) 2 , 6-Dimethyl phenol.

(ix) 1-Methoxy-2-methyl propane.

(x) Ethoxy benzene.

(xi) 1-Phenoxy heptane.

(xii) 2-Ethoxy butane.

Question. 5. Write structures of the compounds whose IUPAC names are as follows :

(i) 2-Methyl butane-2-ol.

(ii) 1-Phenyl propane-2-ol.

(iii) 3 , 5-Dimethyl hexane-1 , 3 , 5 triol.

(iv) 2 , 3-Diethyl phenol.

(v) 1-Ethoxypropane.

(vi) 2-Ethoxy-3-methyl pentane.

(vii) Cyclohexyl ethanol.

(viii) 3-Cyclohexyl pentane-3-ol.

(ix) cyclopent-3-en-1-ol.

(x) 3-Chloro methyl pentane-1-ol.

Question. 6. Draw the structure of all isomeric alcohols of molecular formula C3H12O and give their IUPAC names.

Ans⇒ Eight isomeric alcohols are possible.

(i) CH₃CH₂CH₂CH₂CH₂OH , Pentan -1-ol.

Question . 7. Give two reåctions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Answer :  (i) Reaction with active metals : Phenol reacts with active metals like Na , K etc. and H₂ is evolved.

(ii) Reaction with alkalis : Phenol neutralizes the caustic alkalies such as NaOH or KOH to form salt and water.

Phenol shows a considerable acidic character as compared to that of ethanol. It is because of structural differences in C₆H₅ and C₂H₅ parts of the two molecules.

CH₃-CH₂-O-H

(i) Oxygen connected to sp₃ hybridized C atom which is less electronegative and has electron repelling nature. This makes- O-H bond less polar and release of H becomes more difficult. Alkoxide ions formed is not stabilized by resonance.  

(ii) Oxygen atom connected to sp2 hybridized C atom of benzene ring which is more electronegative than  sp₃ hybridized carbon atom. This makes O-H bound more polar and His released easily. Phenoxide ions formed are stabilized by resonance.

Question. 8. Give IUPAC names of the following ethers :

(i) C₂H₅OCH₃-CH-CH₃

Answer :  (i) 1-Ethoxy-2-methyl propane.

(ii) 2-Chloro-1-methoxy ethane.

(iii) 4-Nitro anisole.

(iv) 1-Methoxy propane.

(v) 1-Ethoxy-3-methyl cyclohexane.

(vi) Ethoxy benzene.

Question. 9. How is chloroform obtained from acetone ?

Answer :  Acetone on warming with moist bleaching powder chloroform is obtained on distillation. The series of reactions are

Ca(OCl) Cl +H₂O Ca (OH)₂+Cl₂

Question. 10. Give two methods of preparation of ethers from alcohol with chemical equations.

Answer:  Following methods of preparation of ethers from alcohol :

(i) From the reaction between alcohol and conc. H₂SO₄.

ROH +ROH R-O-R+H₂O

(ii) From dehydration of alcohol :

R-OH+R-OH R-O-R+H₂O

Question.11 . Give the equations of the reaction for the preparation of phenol from cumene.

Answer: This process has a great industrial importance because it gives the preparation of two very useful compounds i.e. phenol and acetone. The raw materials are benzene and propene and it initially proceeds by Friedel Crafts alkylation of benzene.

Oxygen is bubbled through the above solution to form cumene hydroperoxide which is decomposed with aqueous acid solution to form phenol and acetone as follow

Question.12. Show how will you synthesize

(i) 1-phenylethanol from a suitable alkene.

(ii) cyclohexylmethanol using an alkyl halide by an SN₂ reaction.

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

Answer : (i) Addition of H₂O to ethylbenzene is presence of dil H₂SO₄.

(ii) Hydrolysis of cyclohexylmethyl bromide by aqueous NaOH gives cyclohexylmethanol.

(iii) Hydrolysis of 1-bromopentane by aqueous NaOH gives pentan-1-ol

Question.13. Give two reactions that show the acidic nature of phenol. Compare its acidity with that of ethanol.

Answer : The reactions showing acidic nature of phenol are:

(a) Reaction with sodium: Phenol reacts with active metals like sodium to liberate H, gas.

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

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 less of a proton from ethanol, is not.

Question.14. Explain why ortho nitrophenol is more acidic than ortho methoxyphenol?

Answer : Nitro (NO₂) group is an electron withdrawing group while methoxy (OCH₃) The group is electron releasing in nature. The release of H⁺ ion is therefore easier from o-nitrophenol while it is quite difficult from o-methoxyphenol. Apart from that, o-nitrophenoxide ion is stabilized due to resonance o-nitrophenol is steam volatile while p-nitrophenol is not. This is on account of intramolecular hydrogen bonding in the molecules of o-nitrophenol. As a result, its boiling point is less than that of p-nitrophenol in which the molecules are linked by intermolecular hydrogen bonding.

It is interesting to note that in the substituted phenols, the nature and position of the substituent influences the boiling point of phenol.
For example: o-nitrophenol is steam volatile while p-nitrophenol is not. This is supported by the fact that the boiling point temperature of o-nitrophenol (100C) is less than that of p-nitrophenol, (279C). In o-nitrophenol, there is intramolecular hydrogen bonding in OH and NO₂ groups placed in adjacent positions. However, these are linked by intermolecular hydrogen bonding in the p-isomers. It is quite obvious that extra energy is needed to cleave the hydrogen bonds in the p-isomer. Consequently, its boiling point

o-nitrophenol with lower boiling point is steam volatile while p-nitrophenol is not. This helps in the separation of the two isomers present in the liquid mixture.  On passing steam, o-nitrophenol volatilises and its vapours rise along with steam and after condensation, collected in the receiver p-nitrophenol is left behind in the distillation flask. e-nitrophenol p-nitrophenol.
On the contrary, o-methoxy phenoxide is destabilised since the electron density on the negatively charged oxygen tends to increase due to the electron releasing tendency of the methoxy (OCH₃) group.

In the light of the above discussion, we may conclude that o-nitrophenol is a stronger acid (pKa=7.23) than o-methoxyphenyl (pKa=9.98) 

Question.15. Explain how does the -OH a group attached to a carbon of benzene ring activates it towards electrophilic substitution?

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

As a result of +R effect of the -OH group, the electron density in the benzene ring increases thereby facilitating the attack of 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.

Question.16. Write the mechanism of acid dehydration of ethanol to yield ethene.

Answer : 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:

Question.17. Name the reagents used in the following reactions:

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

(ii) Oxidation of a primary alcohol to aldehyde.

(iii) Bromination Of Phenol To 2 , 4 , 6-tribromophenol

(iv) Benzyl alcohol to benzoic acid.

(v) Dehydration of propane-2-oI to propene.

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

Answer : (i) Acidified potassium dichromate or neutral/ acidic/ alkaline potassium permanganate.

(ii) Pyridinium chlorochromate (PCC) , (C₅H₅NH)+ ClCrO₃- in CH₂Cl₂

or Pyridinium dichromate (PDC) , [(C₅H₅NH)₂]₂+ Cr₂O7₂-in CH₂Cl₂.

(iii) Aqueous bromine, i.e., Br2 / H₂O.

(iv) Acidified or alkaline potassium permanganate.

(v) 85% H₂SO₄ at 440 K.

(vi) Ni/H2 or NaNH₄ or LiAlH₄.

Question.18. Illustrate with examples the limitations of Willamson synthesis for the preparation of certain types of ethers. 

Answer : 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 reactants 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 alkyl halides are primary and the alkoxide may be primary, secondary or tertiary. 

For example- tert-butyl ethyl 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 conditions an alkene, i.e. isobutylene is the main product.

Aryl and vinyl halides cannot be used as substrates because they are less reactive in nucleophilic substitution.

Question.19. Explain the fact that in alkyl aryl ethers, alkoxy group :

(i) activates the benzene ring towards electrophilic substitution.

(ii) directs the incoming substituents towards ortho and para positions in the ring.

Answer :

(i) The alkoxy group (RO-) with lone electron pairs on the oxygen atom activates the ortho and para positions in the ring by +M (or +R) effect as shown below

As the ortho and para positions in the ring become points of high electron density, the electrophiles prefer to attack these positions.

(ii) The alkoxy group directs the incoming group which is an electrophile towards the ortho and para positions in the ring. As a result, a mixture of isomeric products is formed.

Question.20. Show how you will synthesize the following from appropriate alkenes.

Answer : All the alcohols are formed by the hydration of alkenes in the acidic medium. The addition follows Markovnikov's rule. 1-Methylcyclohexene can be used   in the reaction

(ii)  4-Methylpent-3-ene upon hydration in the acidic medium will give the desired alcohol.

(iii) Pent-2-ene gives the desired alcohol upon hydration in the presence of acid.

A

(iv) The cyclic alkene used in this reaction is 2-cyclohexyl but-2-ene.

Question.21. Write the mechanism of acid dehydration of ethanol to yield ethene.

Answer: The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:

Step 1: Protonation of ethanol to form ethyl oxonium ion :

Step 2: Formation of carbocation (rate determining step):

Step 3: Elimination of a proton to form ethene:

The acid consumed in step I is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

Question.22. Your teacher gives you two bottles, one with ethanol and the other with methanol without labeling.

(i) Write a suitable test to distinguish them.

(ii) By using the same test, can you distinguish between propane-1-ol and ethanol? Write the chemical equation.

Answer:

(i) The liquid which gives the iodoform test is ethanol. Methanol does not give an iodoform test. Ethanol reacts with an alkaline solution of Iodine (I2/NaOH) to form yellow precipitate of iodoform.

CH₃CH₂OH +3I₂+4NaOH CHI₃+HCOONa +3H₂O +3NaI

CH₃oh +3I₂+₄NaOH – No reaction

(ii) Yes. Ethanol gives an iodoform test while propane-1-ol does not give an iodoform test.

CH₃-CH₂CH₂OH+I₂+NaOH No reaction

Question.23 Differentiate between Phenol and Alcohol and write Libermann’s reaction related to phenol.

Answer: Differences between Phenol and Alcohol:

Phenol:

1. Physical properties: Characteristic phenolic odor, sparingly soluble in water.
2. It is acidic and dissolves in bases to form salt.
3. On oxidation, hybrid coloured products are formed.
4. Produce characteristic color with Ferric chloride.
5. It does not react with halogen acid.
6. With PCl5, mainly from triaryl phosphate.

Alcohol:

1. Pleasant odor, fairly soluble in water.
2. It is neutral and does not react with bases.
3. It can easily oxidize to Aldehydes and ketones.
4. It does not react with ferric chloride.
5. Forms Alkyl halide.
6. Alkyl chloride is formed.

Libermann’s Reaction: On adding a few drops of concentrated sulphuric acid and little sodium nitrite in phenol first dark blue color is produced on adding water color becomes red and on adding an alkali red color again changes to blue color.

Question.24. What is Williamson’s continuous etherification process? Is it a continuous process? Explain. Give a labeled diagram.

Or,

Describe the laboratory method of preparation of diethyl ether. How ether thus obtained is purified?

Answer:

Laboratory Method for the Preparation of Diethyl Ether (Sulphuric Ether): Diethyl ether is prepared in the laboratory and industry by the Williamson continuous etherification process, i.e., by heating ethanol (in excess) with concentrated sulphuric acid.

Sulphuric acid is regenerated in the reaction hence, it appears as if only a small amount of acid may convert excess alcohol into the ether. So, this method is called Williamson’s continuous etherification process but actually, we cannot get ether continuously.

This is due to the following two reasons:

Water formed in the reaction dilutes the acid and its reactivity decreases.

A part of sulphuric acid is reduced by alcohol into sulfur dioxide.

Method: Ethanol and H₂SO₄ (2:1) are taken in a flask and heated in a sand bath at ‘140C . Ethanol is added at the same rate at which ether is distilled over and is collected in a receiver cooled in ice-cold water.

Purification: Ether contains ethanol, water and sulphuric acid as impurities. It is washed with NaOH to remove sulphuric acid and then agitated with 50% solution of calcium chloride tremove the alcohol. It is then washed with water, dried over anhydrous calcium chloride and redistilled.

Question.25. Write a note on fermentation.

Answer: Fermentation: In this method, molasses or starch are used as raw materials. This is the old method and is used at present also. Fermentation (Latin-fermentare means to boil) proceeds with fast evolution of CO2 producing much foam giving the appearance as if the solution is boiling. “It is a process of decomposition of complex large molecules into simple and small molecules slowly and by the enzyme.” Enzymes are of many kinds and are present in yeast which is a living and complex substance containing several types of bacteria which are called enzymes. It is a good ferment. The enzymes present in yeast are zymase, maltase, invertase, etc.

When yeast is mixed in glucose solution and kept at proper conditions, ethyl alcohol is formed as a result of fermentation.

Favorable conditions for Fermentation:

1. Favorable temperature: It is between 25-35C.
2. Other substances: Some inorganic salts like ammonium sulfate or ammonium nitrate function as food for fermentation.
3. Concentration: Solution should be dilute (Concentration 8-10%).
4. Air: The process occurs in the presence of air.

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