NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers PDF – eSaral

# NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers PDF

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Question 1. Classify the following as primary, secondary and tertiary alcohols:

(i)

(ii) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}$

(iii) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}$

(iv)

(v)

(vi)

Solution: (i)

OH is attached with primary carbon. So, it is primary alcohol.

(ii)

$-$ OH attached with primary carbon. So, it is primary alcohol

(iii)

−OH is attached with primary carbon so it is primary alcohol . (iv)

−OH group is attached with secondary carbon. So, it is secondary alcohol.

(v)

-OH group is attached with secondary carbon. So, it is secondary -alcohol

(vi)

−OH group is attached with tertiary carbon. So, it is tertiary -alcohol

Primary alcohols: (i), (ii), (iii)

Secondary alcohols: (iv) and (v)Tertiary alcohols: (vi)

Question 2. Identify allylic alcohols in the above examples.

Solution. In allylic alcohols, the $-0 \mathrm{H}$ group is attached to carbon next to the carbon-carbon double bond.

Therefore, alcohols given in

(ii)

(vi)

Question 3. Name the following compounds according to IUPAC system.

(i)

(ii)

(iii)

(iv)

(v)

Solution: (i)

Parent Carbon $-$ chain $=5$ members(pent)

Many functional group $=-0 \mathrm{H}$ and suffix $=\mathrm{ol}$

Substituent $-\mathrm{CH}_{2} \mathrm{Cl}=$ Chloromethyl and

IUPAC name: 3-Chloromethyl-2-isopropylpentan-1-0l

(ii)

Parent Carbon Chain $=6$ members (hex)

Multi function group $=-0 \mathrm{H}$ and suffix $=\mathrm{ol}$

Substituents $=-\mathrm{CH}_{3}$ (Methyl) IUPAC name: 2,5 -Dimethylhexane-1, 3 -diol

(iii)

Parent Carbon chain ring $\rightarrow 6$ members (cyclohex)

Main functional group $=-\mathrm{OH}$ and suffix $-\mathrm{ol}$

Substituent $=-\mathrm{Br}$ (Prefix $=$ bromo $)$

IUPAC name: 3 -Bromocyclohexanol

(iv)

Main Carbon chain $=6-$ carbon (hex)

Primary Suffix $=$ ene,

main function group $=-0 \mathrm{H}$ (secondary Suffix $=\mathrm{ol}$ )

IUPAC name: Hex $-1-$ en $-3-$ ol

(v)

Main carbon $-$ chain $=4($ but $)$

Main functional group $=-0 \mathrm{H}$ (secondary Suffix $=\mathrm{ol}$ )

Substituent $-\mathrm{Br} \Rightarrow$ prefix $=$ bromo

$-\mathrm{CH}_{3}=$ methyl

Numbering should start from that carbon such that $-0 \mathrm{H}$ group gets the least number.

IUPAC name: $2-$ Bromo $-3-$ methylbut $-2-$ en $-1-$ ol

Question 4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanol?

(i)

(ii)

Solution.

(i)

(ii)

$\mathrm{R}-\mathrm{MgX}$ where

Question 5. Write structures of the products of the following reactions:

(i) $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \stackrel{\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}}{\longrightarrow}$

(ii)

(iii)

Solution.

(i)

(ii) $\mathrm{NaBH}_{4}$ reduces pure carbonyl (ketone) into $2^{\circ}$ Alcohol

(iii) $\mathrm{NaBH}_{4}$ reduces pure carbonyl (aldehyde) into $1^{\circ}$ alcohol.

Question 6. Give structures of the products you would expect when each of the following alcohol reacts with

(a) $\mathrm{HCl}-\mathrm{ZnCl}_{2}$

(b) $\mathrm{HBr}$ and

(c) $\mathrm{SOCl}_{2}$

(i) $\quad$ Butan $-1-o l$

(ii) $2-$ Methylbutan $-2-$ ol

Solution: (a) Rate of reaction with Lucas’ reagent is $3^{\circ}$ Alcohol $>2^{\circ}$ Alcohol $>$ $1^{\circ}$ alcohol.

(i) Primary alcohols do not react appreciably with Luca’s reagent $\left(\mathrm{HCl}-\mathrm{ZnCl}_{2}\right)$ at room temperature. Because reaction with Luca’s reagent is followed by $\mathrm{S}_{\mathrm{N}} 1$ mechanism.

(ii) Tertiary secondary alcohols react with Luca’s reagent because reaction with Luca’s reagent is followed by $\mathrm{S}_{\mathrm{N}} 1$ mechanism gives text in 5 to 10 minutes.

(iii) Tertiary alcohols react immediately with Luca’s reagent because reaction with Luca’s reagent is followed by $\mathrm{S}_{\mathrm{N}} 1$ mechanism

carbocation is $\left(3^{\circ}\right)$ which is stable due to 8 hyperconjugations and +I of the alkyl group. So, it gives turbidity immediately.

(b) (i)

(ii)

(c) (i)

(ii) Question 7. Predict the major product of acid catalysed dehydration of

(i) 1-methylcyclohexanol and (ii) butan-1-ol

Solution.

(i) In the above reaction, $\beta-\mathrm{H}$ will get eliminated and major product formed should be more stable alkene.

(ii)

Question 8. Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

Solution: Resonance structure of the phenoxide ion

Resonance structures of p-nitrophenoxide ion

Resonance structures of o-nitrophenoxide ion

After losing $\mathrm{H}^{+}$ion, the phenoxide ion is formed. In ortho and meta nitro group substituted, phenoxide ions are stabilised due to $(-\mathrm{M} / \mathrm{electro}$ withdrawing) effect of $-\mathrm{NO}_{2}$ group. Hence, the conjugate base (anion) is stable and acidic character increases.

Question 9. Write the equations involved in the following reactions:

(i) Reimer – Tiemann reaction

(ii) Kolbe’s reaction

Solution: (i) Reimer – Tiemann reaction: It is the $\epsilon^{\oplus}$ substitution reaction

Reactant $=$ Phenol

Reagent $=\mathrm{CHCl}_{3}$ or aq. $\mathrm{NaOH}$

(ii) Kolbe’s reaction

Question 10. Write the reactions of Williamson synthesis of $2-$ ethoxy $-3-$ methyl pentane starting from ethanol and $3-$ methyl pentane $-2-$ ol.

Solution: An alkyl halide reacts with an alkoxide ion in Williamson’s synthesis. Also, it is an $\mathrm{S}_{\mathrm{N}} 2$ reaction. In this reaction, alkyl halides should be primary having the least steric hindrance. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from $3-$ methylpentan $-2-$ ol.

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \stackrel{\mathrm{HBr}}{\longrightarrow} \quad \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}$

Ethanol substitution Bromoethane

Question 11. Which of the following is an appropriate set of reactants for the preparation of 1 methoxy-4-nitrobenzene and why?

(i)

(ii)

Solution: Set (ii) is an appropriate set of reactants for the preparation of 1 -methoxy-4nitrobenzene.

In set (i), sodium methoxide $\left(\mathrm{CH}_{3} \mathrm{ONa}\right)$ is a strong nucleophile as well as a strong base. Hence, an elimination reaction predominates over a substitution reaction.

Question 12. Predict the products of the following reactions:

(i) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}+\mathrm{HBr} \rightarrow$

(ii)

(iii)

(iv)

$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OC}_{2} \mathrm{H}_{5} \stackrel{\mathrm{HI}}{\longrightarrow}$

Solution: (i)

(ii) Reaction proceeded by $\mathrm{S}_{\mathrm{N}} 2$ mechanism. So, $\mathrm{Br}^{-}$attacks on the more hindered group.

(iii) $\rightarrow \mathrm{OC}_{2} \mathrm{H}_{5}$ is $+\mathrm{M}$ group. So, it is a ring activating group for the attack of electrophile and $0 / p$ directing and product $p>0$ because hinderance at ortho is more than para.

(iv)

Exercise:

Question 1. Write IUPAC names of the following compounds:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Solution. (i)

Parent carbon chain $=5$ (pent)

Primary suffix $=$ ane

Substituent $=-\mathrm{CH}_{3}$ (methyl)

Main functional group $=-0 H:$ secondary suffix $=o l$

$-$ OH group should get least number

IUPAC name: $2,2,4$, trimethylpentan-3-ol

(ii)

Parent carbon Chain $=7$ (hept)

Primary suffix $=$ ane

Main functional group $=-0 \mathrm{H} ;$ Secondary suffix $=$ ol

$-0 H$ group should get least number

IUPAC name: $5-$ Ethylheptane $-2,4-$ diol

(iii)

Parent carbon chain $=4$

Primary suffix $=$ ane

Main functional group $=-\mathrm{OH}$

Secondary suffix $=$ ol

$-$ OH group should get least number

IUPAC name: butane-2, $3-$ diol

(iv)

Parent carbon chain $=3$

Primary suffix $=$ ane

Main functional group $=-0 H$; Secondary suffix $=$ ol

$-$ OH group should get least number

IUPAC name: Propane-1,2,3-triol

(v)

Main functional group $=$ Phenol

Substituent $=-\mathrm{CH}_{3}$ (methyl)

$-$ OH group should get least number

IUPAC name: 2 -Methylphenol

(vi)

Main functional group $=$ Phenol

Substituent $=-\mathrm{CH}_{3}$ (methyl)

– OH group should get least number

IUPAC name: 4-Methylphenol

(vii)

Main functional group $=$ Phenol

Substituent $=-\mathrm{CH}_{3}$ (methyl)

– OH group should get least number

IUPAC name: 2,5 -Dimethylphenol

(viii)

Main functional group $=$ Phenol

Substituent $=-\mathrm{CH}_{3}$ (methyl)

$-0 H$ group should get least number

IUPAC name: 2,6 -Dimethylphenol

(ix)

(x)

(xi) Ethoxy benzene

$2-$ Ethoxybutane

Question 2. Write structures of the compounds whose IUPAC names are as follows: (i) $\quad$ 2-Methylbutan-2-ol

(ii) 1-Phenylpropan-2-ol

(iii) 3,5 -Dimethylhexane $-1,3,5$-triol’

(iv) $2,3-$ Diethylphenol

(v) 1 – Ethoxypropane

(vi) 2-Ethoxy-3-methylpentane

(vii) Cyclohexylmethanol

(viii) 3-Cyclohexylpentan-3-ol

(ix) Cyclopent-3-en-1-ol

(x) 4-Chloro-3-ethylbutan-1-ol.

Solution. (i) $\quad$ 2-Methylbutan-2-ol

(ii) 1-Phenylpropan-2-ol

(iii) 3,5 -Dimethylhexane $-1,3,5$-triol

(iv) $2,3-$ Diethylphenol

(v) $1-$ Ethoxypropane

Functional group $=$ Ether

(vi) 2-Ethoxy-3-methylpentane

(vii) Cyclohexylmethanol

(viii) 3-Cyclohexylpentan-3-ol

(ix) Cyclopent-3-en-1-ol

(x) $\quad$ 4-Chloro-3-ethylbutan-1-ol.

The above name of the structure is wrong according to IUPAC. Correct name is 3 -methylchloro-pent-1-ol.

Question 3. (i) Draw the structures of all isomeric alcohols of molecular formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$ and give their IUPAC names.

(ii) Classify the isomers of alcohols of molecular formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$ (i) as primary, secondary and tertiary alcohols.

Solution: (i) The structures of all isomeric alcohols of molecular formula, $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$ are shown below:

Structural isomers:- Compounds that have different structural formula but the same molecular formula and different IUPAC name are called structural isomers.

(a)

(b) Pentan-1-ol $\left(1^{\circ}\right.$ alcohol $)$

(c) 2-Methylbutan-1-ol $\left(1^{\circ}\right.$ alcohol $)$

(d) 3-Methylbutan-1-ol $\left(1^{\circ}\right.$ alcohol $)$

(e) 2,2 -Dimethylpropan-1-ol (1o alcohol)

(f) Pentan-2-ol $\left(2^{0}\right.$ alcohol $)$

(g) 3-Methylbutan-2-ol $\left(2^{\circ}\right.$ alcohol $)$

(h) Pentan-3-ol $\left(2^{0}\right.$ alcohol $)$

2-Methylbutan-2-ol $\left(3^{0}\right.$ alcohol $)$

(ii) Primary alcohol: $-\mathrm{OH}$ group attached with $1^{\mathrm{O}}$-carbon

Pentan-1-ol; $\quad$ 2-Methylbutan-1-ol; $\quad$ 3-Methylbutan-1-ol; $\quad 2,2$ Dimethylpropan-1-ol

Secondary alcohol: $-0$ H group attached with $2^{\circ}$-carbon

Pentan-2-ol; 3-Methylbutan-2-ol; Pentan-3-ol

Tertiary alcohol: $-0 H$ group attached with $3^{\circ}$-carbon

2-methylbutan-2-ol

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

Solution. Propanol undergoes intermolecular H-bonding because of the presence of a $-\mathrm{OH}$ group. On the other hand, butane does not form H-bonding. Butane is bonded by Van der waal’s forces which is weaker than H-bond.

Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol’s boiling point is higher than the hydrocarbon butane.

Question 5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Solution: Alcohols form hydrogen bonds with water due to the presence of $-0 H$ group since $\mathrm{H}$ has $+\delta$ and 0 has lone pair. However, hydrocarbons cannot form $\mathrm{H}$-bonds with water.

Therefore, alcohols are more soluble in water than hydrocarbons of comparable molecular masses.

Question 6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

Solution: The addition of borane and followed by oxidation is known as the hydroborationoxidation reaction.

Example: Propan-1-ol is formed by the hydroboration-oxidation reaction of the compound propene. In this reaction, propene reacts with diborane $\left(\mathrm{BH}_{3}\right)_{2}$ to form trialkyl borane as an electrophilic addition product. This electrophilic addition is

followed by anti-Markownikoff rule. This electrophilic addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

Question 7. Give the structures and IUPAC names of monohydric phenols of molecular formula, $\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}$

Solution: The isomers of monohydric phenols having the molecular formula $\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}$ are given below:

Question 8. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Solution. Intramolecular H-bonding is present in o-nitrophenol while in p-nitrophenol, the molecules are strongly associated due to the presence of intermolecular bonding. Hence, due to intramolecular H-bond, o-nitrophenol is steam volatile.

Question 9. Give the equations of reactions for the preparation of phenol from cumene.

Solution. To prepare phenol from cumene, first, the cumene is oxidised to cumene hydroperoxide in the presence of air $\left(\mathrm{O}_{2}\right)$ by a free radical mechanism at a temperature of $368 \mathrm{~K}-408 \mathrm{~K}$.

Then, this cumene hydroperoxide is treated with dilute acid to prepare phenol and acetone.

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

Solution. Chlorobenzene is fused with $\mathrm{NaOH}$ under the conditions of $623 \mathrm{~K}$ temperature and 320 atm pressure to produce sodium phenoxide, which then gives phenol on acidification.

Question 11. Write the mechanism of hydration of ethene to yield ethanol.

Solution. The mechanism of hydration of ethene to yield ethanol involves three steps:

Step 1:

Protonation of ethene to form carbocation by electrophilic attack of $\mathrm{H}_{3} \mathrm{O}^{+}:$

$\mathrm{H}_{2} \mathrm{O}+\mathrm{H}^{+} \stackrel{\text { acid-base reaction }}{\longrightarrow} \mathrm{H}_{3} \mathrm{O}^{+}$(electroplile)

Step 2:

Nucleophilic attack of water on carbocation: lone pair of oxygen (from water, $\mathrm{H}_{2} \mathrm{O}$ ) attacks the carbocation.

Step 3:

Deprotonation to form ethanol:

Question 12. You are given benzene, conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{NaOH}$. Write the equations for the preparation of phenol using these reagents.

Solution: (i) The reaction of Benzene with concentrated sulphuric acid is called sulphonation. It is followed by electrophilic substitution.

Question 13. Show how will you synthesize:

(i) 1-Phenylethanol from a suitable alkene.

(ii) Cyclohexylmethanol using an alkyl halide by an $\mathrm{S}_{\mathrm{N}} 2$ reaction.

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

Solution: (i) By acid-catalyzed hydration of ethylbenzene (styrene), 1-phenylethanol can be synthesized by the electrophilic addition reaction.

(ii) When chloromethylcyclohexane is treated with sodium hydroxide, $\mathrm{S}_{\mathrm{N}} 2$ reaction takes place and cyclohexylmethanol is obtained.

(iii) When 1 -chlopropentane is treated with $\mathrm{NaOH}$, pentan-1-ol is produced by $\mathrm{S}_{\mathrm{N}} 2$ reaction mechanism.

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

Solution. The acidic nature of phenol can be represented by the following two reaction:

(i) Phenol reacts with sodium to give sodium phenoxide, liberating $\mathrm{H}_{2}$ gas.

(ii) Phenol reacts with sodium hydroxide base to give sodium phenoxide salt and water as side products. This reaction exhibits the acidic nature of phenol.

Acidic character depends on the stability of anion. Anion is formed after losing $\mathrm{H}^{\oplus}$. The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized.

Ethanol forms $\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{O}^{-}$anion which is less stable due to $+\mathrm{I}$ effect of $\mathrm{CH}_{3}-\mathrm{CH}_{2}-$. Thus, phenoxide ion is more stable than ethoxide ion.

Question 15. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

Solution. The nitro-group is an electron-withdrawing group (-M group). The presence of nitro group in the ortho position decreases the electron density in the $0-\mathrm{H}$ bond. Thus, it is easier to lose a proton. Also, the o-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid. Because the anion formed is more stable. So acidic (+M) character increases.

On the other hand, methoxy group is an electron-releasing group. As a result, it increases the electron density in the $0-\mathrm{H}$ bond, and hence, the proton cannot be given easily. Because the conjugate base is stable. Hence acidic character decreases.

Due to, this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

Question 16. Explain how does the $-$ OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Solution. The $-$ OH group is an electron-donating group (+M group). Thus, it increases the electron density in the benzene ring at ortho and para as shown in the given resonance structure of phenol.

$-0 H$ group activates the ring for the electrophile to come at ortho and para.

As a result, the benzene ring is activated towards electrophilic substitution.

Question 17. Give equations of the following reactions:

(i) Oxidation of propan-1-ol with alkaline $\mathrm{KMnO}_{4}$ solution.

(ii) Bromine in $\mathrm{CS}_{2}$ with phenol.

(iii) Dilute $\mathrm{HNO}_{3}$ with phenol.

(iv) Treating phenol with chloroform in presence of aqueous $\mathrm{NaOH}$.

Solution: (i) Oxidation of propan-1-ol with alkaline $\mathrm{KMnO}_{4}$ solution gives propanoic acid.

(ii) Bromination of phenol: Reaction of phenol with $\mathrm{Br}_{2}$ in $\mathrm{CS}_{2}$. $\mathrm{Br}^{+}$is electrophilic and phenol show electrophilic substitution react with

electrophilic $\left(\stackrel{\oplus}{\mathrm{N} O}_{2}\right) \cdot-\mathrm{OH}$ group is ring activating group. It increases the

electron density at ortho and para position. Hence, the electrophilic attack occurs at ortho and para. Para is the major product than ortho because there is less hindrance at para.

(iii) Nitration of phenol

(iv)

Question 18. Explain the following with an example.

(i) Kolbe’s reaction.

(ii) Reimer-Tiemann reaction.

(iii) Williamson ether synthesis.

(iv) Unsymmetrical ether.

Solution. (i) Kolbe’s reaction.

When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe’s reaction.

(ii) Reimer-Tiemann reaction.

When phenol is treated with chloroform $\left(\mathrm{CHCl}_{3}\right)$ in the presence of sodium hydroxide, a -CHO group is introduced at the ortho position of the benzene ring.

When phenol is treated with chloroform $\mathrm{CHCl}_{3}$ is presence of potassium hydroxide, a -CHO group is introduced at the otho position of the benzene ring.

(iii) Williamson ether synthesis.

Williamson ether synthesis is a method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.

This reaction involves $S_{N} 2$ attack of the alkoxide ion on the alkyl halide. Better results are obtained when primary alkyl halides are used.

If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

(iv) Unsymmetrical ether:

An unsymmetrical ether is an ether where both alkyl groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon alkyl atoms). For example: ethyl methyl ether $\left(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}\right)$.

Question 19. Write the mechanism of acid dehydration of ethanol to yield ethene.

Solution. 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):

Carbocation is the intermediate formed in the reaction, hence carbocation rearrangement is possible during the reaction.

Step 3:

Elimination of a proton to form ethene:

The acid consumed in step 1 is released in step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

Question 20. How are the following conversions carried out?

(i) Propene $\rightarrow$ Propan-2-ol.

(ii) Benzyl chloride $\rightarrow$ Benzyl alcohol.

(iii) Ethyl magnesium chloride $\rightarrow$ Propan-1-ol.

(iv) Methyl magnesium bromide $\rightarrow 2$-Methylpropan-2-ol.

Solution. (i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained by electrophilic addition reaction:

(ii) If benzyl chloride is treated with $\mathrm{NaOH}$ (followed by acidification) then benzyl alcohol is produced.

(iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol on hydrolysis.

(iv) When methyl magnesium bromide is treated with propanon, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.

Question 21. Name the reagents used in the following reactions:

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

(ii) Oxidation of a primary alcohol to aldehydes.

(iii) Bromination of phenol 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.

Solution. (i) Acidified potassium permanganate

Acidified potassium dichromate $\left(\mathrm{K}_{2} \mathrm{Cr}_{3} \mathrm{O}_{7}\right)$ and acidified potassium permanganate $\left(\mathrm{KMnO}_{4}\right)$

(ii) Pyridinium chlorochromate (PCC)

Mild oxidizing agent like pyridium chloro chromate (PCC), pyridium dichloro chromate (PDC) or $\mathrm{CrO}_{3}$ anhydrous convert primary alcohol into aldehyde.

(iii) Bromine water:

Bromination of phenol is taking place in presence of $\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}$ (bromine water) and so $2,4,6$-tribromophenol formed.

(iv) Acidified potassium permanganate

Acidifies potassium permanganate convert benzylic $0, H$, bond into $-C O O H$ group.

(v) $85 \%$ phosphoric acid

Dehydration of propan-2-ol to propane is followed by elimination (E1) and reagent used is concentration $\mathrm{H}_{2} \mathrm{SO}_{4}, 85 \%$ phosphoric acid.

(vi) $\mathrm{NaBH}_{4}$ or $\mathrm{LiAIH}_{4}$

Butan-2-one to butan-2-ol

By sodium borohydride $\left(\mathrm{NaBH}_{4}\right)$, Lithium aluminium hydride $\left(\mathrm{LiAlH}_{4}\right)$ or by Na/EtOH (Bouveault-Blanc reduction)

Question 22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Solution. Boiling point depends on force of attraction.

Ethanol undergoes intermolecular H-bonding due to the presence of $-\mathrm{OH}$ group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. It is bonded by vanderwall force of attraction (dipole -dipole attraction). Vander wall’s force is weaker than H-bond. Hence, the boiling point of ethanol is higher than that of methoxymethane.

Hydrogen bonding in ethanol. Question 23. Give IUPAC names of the following ethers:

(i)

(ii) $\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$

(iii) $\mathrm{O}_{2} \mathrm{~N}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{OCH}_{3}(\mathrm{p})$

(iv) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}$

(v)

(vi)

Solution: (i)

1-Ethoxy-2-methylpropane

(ii)

2-Chloro-1-methoxyethane

(iii)

4-Nitroanisole

(iv)

1-Methoxynronane

(v)

1-Ethoxy-4,4-dimethylcyclohexane

(vi)

Ethoxybenzene

Question 24. Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:

(i) 1-Propoxypropane

(ii) Ethoxybenzene

(iii) 2-Methoxy-2-methylpropane

(iv) 1 -Methoxyethane

Solution: (i)

(ii)

(iii)

(iv)

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

Solution. The reaction of Williamson synthesis involves $S_{N} 2$ attack of an alkoxide ion on a primary alkyl halide.

If secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides having $(\beta-\mathrm{H})$ which results in an elimination reaction. Alkenes are formed as a product.

Question 26. How is 1 -propoxypropane synthesized from propan-1-ol? Write mechanism of this reaction.

Solution. 1-propoxypropane can be synthesized from propan-1-ol by dehydration.

Propan-1-ol undergoes dehydration in the presence of protic acids (such as $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{4}$ ) to give 1-propoxypropane.

The mechanism of this reaction involves the following three steps:

Step 1: Protonation

$\mathrm{H}^{+}$(acidic) attack on lone pair on oxygen by lewis acid-base reaction.

Step 2: Nucleophilic attack

Step 3: Deprotonation

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

Solution. The formation of ethers by dehydration of alcohol is a bimolecular reaction $\left(\mathrm{S}_{\mathrm{N}} 2\right)$ involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of ethers, alkenes are formed.

Question. 28. Write the equation of the reaction of hydrogen iodide with:

(i) 1 -propoxypropane

(ii) methoxybenzene and

(iii) benzyl ethyl ether.

Solution. (i)

(ii)

(iii)

Question 29. 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.

Solution.

Aryl alkyl ether

In aryl alkyl ethers, due to the $+\mathrm{R}$ effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

(ii) It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 30. Write the mechanism of the reaction of HI with methoxymethane.

Solution. The mechanism of the reaction of HI with methoxymethane is $\mathrm{S}_{\mathrm{N}} 2$ and it involves the following steps:

Step 1: Protonation of methoxymethane:

Step 2: Nucleophilic attack of $\mathrm{I}^{-}$:

Step 3: When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide by $\mathrm{S}_{\mathrm{N}} 2$ reaction

Chemical EquilibriumChemical Equilibrium

$\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}+2 \mathrm{HO} \rightarrow 2 \mathrm{CH}_{3} \mathrm{I}+\mathrm{H}_{2} \mathrm{O}$

Concept insight: The reaction of ether with excess of Hydrogen halide results in the formation of alkyl halide.

Question 31. Write equations of the following reactions:

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

(ii) Nitration of anisole.

(iii) Bromination of anisole in ethanoic acid medium.

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

Solution. (i)

$-\mathrm{OCH}_{3}$ group is electron releasing (+M) group. So, it increases electron density at ortho and para. So, electrophilic $\mathrm{R}^{+}$(Carbonation) attack at ortho and nara. Para will he maior than ortho due fo less hindrance.

(ii)

(iii) In the presence of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{HNO}_{3}$, nitration will take place and electrophile $\stackrel{+}{N} O_{2}$ is formed.

(iv) Acylation

Question 32. Show how would you synthesize the following alcohols from appropriate alkenes?

(i)

(ii)

(iii)

(iv)

Solution: The given alcohols can be synthesized by applying Markovnikov’s rule of acidcatalyzed hydration of appropriate alkenes.

(i)

(ii)

(iii)

Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along with pentan-3-ol.

Thus, the first reaction is preferred over the second one to get pentan-2-ol.

(iv)

Question 33. When 3 -methylbutan-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.

(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

Solution. The mechanism of the given reaction involves the following steps:

Step 1: Protonation (acid base reaction)

Step 2: Formation of $2^{\circ}$ carbocation by the elimination of a water molecule

Step 3: Re-arrangement by the hydride-ion shift.

$3^{\circ}$ carbocation is more stable than $2^{\circ}$ carbocation due to more hyper conjugation in $3^{\circ}(8 \mathrm{HC})$

Step 4: Nucleophilic attack