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Solved Examples on Alcohols, Phenols & Ethers

Question 1:  Explain the following:

  1. When equal volumes of ethanol and  water are mixed, the total volum

  2. ROH’s with three or fewer C’s are H2O soluble; those with five or more C’s are insoluble, and those with four C’s are marginally soluble.

  3. e is less than the sum  of the two individual volumes.

  4. Propanol (MW=60) has a higher boiling point than butane (MW=58)

Solution:

  1. The water solubility of alcohols is attributed to intermolecular H–bonding with H2O . As the molecular weights of the alcohols increase, their solubility in water decreases, because greater carbon content makes the alcohols less hydrophilic. Conversely, their solubility in hydrocarbon solvents increases.

  2. H–bonding between ethanol and water molecules.

  3. Alcohol molecules attract each other by relatively strong H–bonds and somewhat weaker dipole – dipole interactions, resulting in a higher bp. In alkanes only weaker vander Waals attractive forces must be overcome to vaporize the hydrocarbon.

Question 2

  1. What is absolute alcohol?

  2. How is it prepared?

  3. What is denatured alcohol and why is it dangerous to drink?

  4. What is the meaning of the term proof used to describe the alcohol content of wines and spirits?

Solution

  1. Absolute alcohol is pure anhydrous ethanol.

  2. Since ethanol forms a lower boiling azeotropic mixture containing 95.6 : 4.4 (volume %) alcohol : water it cannot be dried by distillation. The water can be removed chemically by treatment with Mg turnings, which react with the water forming Mg(OH)2 and releasing H2 gas. Alternatively, it can be distilled after the addition of some benzene. The benzene forms a ternary azeotrope with the alcohol – water mixture which boils at 65°, lower than the bp of the binary azeotrope. Thus it distills first, removing the water and leaving behind water – free alcohol, which is further distilled. Absolute alcohol purified in this manner should not be drunk because it contains small amounts of toxic, carcinogenic benzene.

  3. Denatured alcohol is alcohol rendered undrinkable by the addition of small quantities of benzene, toxic methanol. .

  4. Proof is defined as twice the percentage by volume of alcohol in an ethanol water mixture. Thus, a wine with alcohol 12% by volume is 24 proof and an on so proof alcoholic liquid has 40% ethanol.

Question 3: Give the product of each reaction

  1. CH3CH2CH = CHCH2OH + MnO2       

  2.  Ph2CHOH+H2/Pd(HClO4)

  3.  m–O2NC6H4CH2COOH+ LiAlH4

  4. Ph2C = CHCH3 + BH3 THF followed by H2O2/OH

  5.  PhCOCH2CH2Br + LiAlD4, followed by H2O

  6. PhCH = CHCH(OH)CH3 + PBr3

Solution:

  1. CH3CH2CH = CHCHO        

  2.  Ph2CH2

  3.  m–H2NC6H4CH2CH2OH

  4. Ph2CH2CH(OH)CH3

  5. PhCD(OH)CH2CH2D

  6. PhCH = CHCHBrCH3

Question 4:   

  1. Write the structure of the product of the general reaction given and name the type of product: RCH2CH2OCH2R + O2  →?

  1. Why should ethers be purified before being  used in a synthesis where they must be removed by distillation?

  2. How are ethers purified?

Solution:

  1. RCH2CHOCH2CH2R.ROOH compounds are hydroperoxides; this product is a hydroperoxide ether.

  2. Hydroperoxides which are often solids, explode when heated or rubbed abrasively. The hydroperoxides are formed by long exposure of ethers to atmospheric O2. They are dissolved in the ether, and are concentrated when the ether is purified by distillation.

  3. Ethers are often purified extraction with FeSO4 solution, which reduces the —COOH group to the nonexplosive, innocuous —OH group.

Question 5:  Place the following groups of compounds in decreasing order of acidity and justify your answers.

  1. Phenol (A), o-nitrophenol (B), m-nitrophenol (C), p-nitrophenol (D);

  2. Phenol (A), o-chlorophenol (E), m-chlorophenol (F), p-chloroophenol (G); and \

  3. (A), o-cresol (H), m-cresol, (I), p-cresol (J).

Solution:

  1. NO2 is electron – withdrawing and acid –strengthening by both induction and resonance. Its resonance effect is effective from only the ortho and para positions to about an equal extent. It predominates over the inductive effect which operates from all positions but at decreasing effectiveness with increasing separation of NO2 and OH. Hence all the nitrophenols are more acidic than phenol with m-nitrophenol 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-bond in o-nitrophenol must be broken and this requires some energy. The decreasing order  is D > B > C > A.

  2. Although Cl is electron – donating by resonance, its electron – withdrawing inductive effect that decreases with increasing separation of Cl and OH predominates, making all the chlorophenols more acidic than phenol. The decreasing order is E > F > G > A.

  3. Me is electron – donating inductive from all positions and hyperconjugatively from the ortho and parra positions. The three isomers are weaker acids than phenol. m-Cresol is the strongest because its acidity is not weakened by hyperconjugation. The decreasing order is A > I > H = J

Question  6: Give a simple test tube reaction that distinguishes between the compounds in each of the following pairs. What would you do, see and conclude?

  1. t-butyl and n-butyl alcohol,

  2. ethyl and n-propyl alcohol,

  3. allyl and n-propyl alcohol,

  4. benzyl methyl ether and benzyl alcohol, and

  5. cyclopentanol and cyclopentyl chloride.

Solution: 

  1. Add Acid Cr2O72– (orange). The 1° n-butyl alcohol is oxidized; its solution changes colour to green Cr(III). The 3° t-butyl alcohol is unchanged. Alternatively, when Lucas reagent (HCl+ZnCl2) is added, the 3° ROH quickly reacts to form the insoluble t-butyl chloride that appears as a second (lower) layer or a cloudiness. The 1° ROH does  not react and remains dissolved  in the reagent.

  1. Add I2 in OH until the I2 colour persists. A pale yellow precipitate of CHI3 appears, indicating that ethyl alcohol is oxidized. n-Propyl alcohol does not have the —CH(OH)CH3 goup and is not oxidized.

  1. Add Br2 in CCl4; as the Br2 adds to the C = C of the colorless allyl alcohol, its orange colour  disappears. The orange colour  persists in the unreactive n-propyl alcohol.

  1. Add  acid Cr2O72–. It oxidizes the alcohol, and the colour changes to green. The ether is unreactive. Alternately, if the two compounds are absolutely dry, add a small piece of Na (caution, use hood and wear goggles!) to each. H2 is released from the alcohol; the ether does not react.

  1. The simplest test is to add conc. H2SO4 to each dry compound. There  will be only one layer as the alcohol dissolves, evolving some heat. The layers will be discernable for the chloride, which is not soluble in H2SO4.

Question 7:  

  1. Upon treatment with sulphuric acid, a mixture of ethyl and n-propyl alcohols yield a mixture of three ethers. What are they?

  2. On the other hand, a mixture of tert-butyl alcohol and ethyl alcohol gives a good yield of a single ether. What ether is this likely to be? How do you account for this good yield?

Solution:       

  1. CH3CH2OH (ethyl alcohol) and CH3CH2CH2OH (n-propyl alcohol) by dehydration make a mixture of three ethers since each alcohol is 1° and follow SN2 reaction Since protonated 3° alcohol (tert-butyl alcohol) can make most stable carbonium ion, hence this leads to the single ether in good quantity.

  2.  

Question 8:  Prepare (a) Pricric acid 2,4,6-trinitrophenol, and (b) 2,4-dichlorophenol from benzene

Solution: Even though the NO2’s are in the o,o,p positions where OH directs electrophilic substitution, phenol cannot be nitrated because the ring is susceptible to oxidative ring cleavage by nitric acid. Instead, we take advantage of a nucleophilic addition – elimination.

PhCl cannot be trinitrated because the Cl and two NO2’s deactive the ring toward further electrophilc substitution. 2,4-Dintrophenol can be nitrated because the two deactivating NO2’s prevent ring oxidation. (b) Phenol cannot be chlorinated because the ring is susceptible to oxidation by Cl2. Again nucleophilic addition – elimination is used.

the displaced Cl is ortho to one other Cl and para to the other and is activated by both. Each of the other Cl’s is meta to at least one deactivating Cl.

Question 9: Place the following groups of compounds in decreasing order of acidity and justify your answers.

  1. Phenol (A), o-nitrophenol (B), m-nitrophenol (C), p-nitrophenol (D);

  2. Phenol (A), o-chlorophenol (E), m-chlorophenol (F), p-chloroophenol (G); and

  3. (A), o0cresol (H), m-cresol, (I), p-cresol (J).

Solution: 

  1. NO2 is electron – withdrawing and acid –strengthening by both induction and resonance. Its resonance effect is effective from only the ortho and para positions to about an equal extent. It predominates over the inductive effect which operates from all positions but at decreasing effectiveness with increasing separation of NO2 and OH. Hence all the nitrophenols are more acidic than phenol with m-nitrophenol 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-bond in o-nitrophenol must be broken and this requires some energy. The decreasing order  is D > B > C > A.

  2. Although Cl is electron – donating by resonance, its electron – withdrawing inductive effect that decreases with increasing separation of Cl and OH predominates, making all the chlorophenols more acidic than phenol. The decreasing order is E > F > G > A.

  3. Me is electron – donating inductive from all positions and hyperconjugatively from the ortho and parra positions. The three isomers are weaker acids than phenol. m-Cresol is the strongest because its acidity is not weakened by hyperconjugation. The decreasing order is A > I > H = J

Question 10:  Write mechanism for the oxidation of a 2° alcohol with Cr(VI) as HCrO4.

Solutions:  A chromate ester is formed in the first step:

The Cr(IV) and Cr(VI) species react to form 2Cr(V), which in turn also oxidizes alcohols giving Cr(III) with its characteristics colour.

Click here to access the Organic Chemistry Revision Notes and IIT JEE Chemistry Syllabus

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