NCERT Solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers

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NCERT Solutions for Class 12 Alcohols, Phenols and Ethers Chapter 7

Question1. Write IUPAC names of the following compounds:

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

(ix) 

(x)

(xi)

(xii) 

Solution :
(i) 2, 2, 4-Trimethylpentan-3-ol

(ii) 5-Ethylheptane-2, 4-diol

(iii) Butane-2, 3-diol

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

(v) 2-Methylphenol

(vi) 4-Methylphenol

(vii) 2, 5-Dimethylphenol

(viii) 2, 6-Dimethylphenol

(ix) 1-Methoxy-2-methylpropane

(x) Ethoxybenzene

(xi) 1-Phenoxyheptane

(xii) 2-Ethoxybutane

Question2. Write structures of the compounds whose IUPAC names are as follows:

(i) 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) 3-Chloromethylpentan-1-ol.

Solution :
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Question3. (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.

(ii) Classify the isomers of alcohols in 3 (i) as primary, secondary and tertiary alcohols.

Solution :
(i) The structures of all isomeric alcohols of molecular formula, C5H12O are shown below:

(a)

Pentan-1-ol (1°)

(b)

2-Methylbutan-1-ol (1°)

(c)

3-Methylbutan-1-ol (1°)

(d)

2, 2-Dimethylpropan-1-ol (1°)

(e)

Pentan-2-ol (2°)

(f)

3-Methylbutan-2-ol (2°)

(g)

Pentan-3-ol (2°)

(h)

2-Methylbutan-2-ol (3°)

(ii) Primary alcohol: Pentan-1-ol; 2-Methylbutan-1-ol;

3-Methylbutan-1-ol; 2, 2−Dimethylpropan-1-ol

Secondary alcohol: Pentan-2-ol; 3-Methylbutan-2-ol;

Pentan-3-ol

Tertiary alcohol: 2-methylbutan-2-ol

Question4. Explain why propanol has a higher boiling point than that of the hydrocarbon, butane? \section*

{Solution} Although butane and propanol have comparable molecular masses (58 and 60 respectively), propanol has a higher boiling point due to the presence of a polar -OH group in its structure. The key factors contributing to this higher boiling point are: \begin{itemize} \item \textbf{Hydrogen Bonding in Propanol:} Propanol contains an -OH group, which leads to strong intermolecular hydrogen bonding between its molecules. This hydrogen bonding significantly increases the boiling point. \item \textbf{Weaker Intermolecular Forces in Butane:} Butane, on the other hand, only exhibits weak van der Waals’ forces (dispersion forces) between its molecules, which are much weaker compared to hydrogen bonds. \end{itemize} As a result, propanol has a higher boiling point (\(391 \, \text{K}\)) compared to butane (\(309 \, \text{K}\)). \[ \text{Propanol:} \, \text{C}_3\text{H}_7\text{OH} \quad \text{forms hydrogen bonds:} \quad \delta^+ \text{H} – \delta^- \text{O} \ldots \delta^+ \text{H} – \delta^- \text{O} \] \[ \text{Butane:} \, \text{C}_4\text{H}_{10} \quad \text{has only van der Waals’ forces.} \]

Question5. Explain why alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. \section*

{Solution} Alcohols exhibit higher solubility in water compared to hydrocarbons with similar molecular masses due to the presence of the -OH group in alcohols. The reasons are: \begin{itemize} \item \textbf{Hydrogen Bonding:} The -OH group in alcohols allows them to form hydrogen bonds with water molecules. These strong hydrogen bonds enhance the solubility of alcohols in water. \item \textbf{Lack of Hydrogen Bonding in Hydrocarbons:} Hydrocarbons, lacking the -OH group, cannot form hydrogen bonds with water. They only interact with water through weaker van der Waals forces, which do not significantly increase solubility. \end{itemize} Thus, the ability of alcohols to form hydrogen bonds with water makes them more soluble than hydrocarbons of similar molecular mass. \[ \text{Alcohol:} \quad \text{R-OH} \quad \text{can form hydrogen bonds with water:} \quad \delta^+ \text{H} – \delta^- \text{O} \ldots \delta^+ \text{H} – \delta^- \text{O} \] \[ \text{Hydrocarbon:} \quad \text{R-H} \quad \text{cannot form hydrogen bonds with water.}

Question6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example. : Alcohols, Phenols and Ethers Chapter 7

Solution :
The addition of borane followed by oxidation is known as the hydroboration-oxidation reaction. For example, propan-1-ol is produced by the hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane (BH3)2 to form trialkyl borane as an addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

Question7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.     : Alcohols, Phenols and Ethers Chapter 7

Solution :

Question8. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give a reason. 

    : Alcohols, Phenols and Ethers Chapter 7

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

Question9. Give the equations of reactions for the preparation of phenol from cumene.

Solution :
To prepare phenol, cumene is first oxidized in the presence of air of cumene hydro-peroxide.

Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by-products.

Question10. Write chemical reaction for the preparation of phenol from chlorobenzene. 

   : Alcohols, Phenols and Ethers Chapter 7

Solution :
Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.

Question11. Write the mechanism of hydration of ethene to yield ethanol. 

  : Alcohols, Phenols and Ethers Chapter 7

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

Step 1:

Protonation of ethene to form carbocation by electrophilic attack of H3O+:

Step 2:

Nucleophilic attack of water on carbocation:

Step 3:

Deprotonation to form ethanol:

Question12. You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents. 

    : Alcohols, Phenols and Ethers Chapter 7

Solution :

Question13. 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?

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

(ii) When chloromethylcyclohexane is treated with sodium hydroxide, cyclohexylmethanol is obtained.

(iii) When 1-chloropentane is treated with NaOH, pentan-1-ol is produced.

Question14. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol. 

    : Alcohols, Phenols and Ethers Chapter 7

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

(i) Phenol reacts with sodium to give sodium phenoxide, liberating H2.

(ii) Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products.

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilised whereas ethoxide ion does not.

Question15. Explain why ortho-nitrophenol is more acidic than ortho-methoxyphenol. \section*

{Solution} The acidity of phenolic compounds can be influenced by the substituents present on the benzene ring. Here’s why ortho-nitrophenol is more acidic than ortho-methoxyphenol: \begin{itemize} \item \textbf{Ortho-Nitrophenol:} In ortho-nitrophenol, the nitro group (\(-NO_2\)) is an electron-withdrawing group. It pulls electron density away from the hydroxyl group (-OH), thereby reducing the electron density of the O-H bond. This makes it easier for the phenolic hydrogen to dissociate as a proton (\(H^+\)). \[ \text{Ortho-Nitrophenol:} \quad \text{C}_6\text{H}_4(\text{NO}_2)\text{OH} \] \[ \text{Electron-Withdrawing Nitro Group} \quad \text{decreases electron density} \quad \text{of O-H bond}. \] \item \textbf{Ortho-Methoxyphenol:} In ortho-methoxyphenol, the methoxy group (\(-OCH_3\)) is an electron-releasing group. It increases the electron density of the O-H bond, making it more difficult for the hydrogen to dissociate as a proton. \[ \text{Ortho-Methoxyphenol:} \quad \text{C}_6\text{H}_4(\text{OCH}_3)\text{OH} \] \[ \text{Electron-Releasing Methoxy Group} \quad \text{increases electron density} \quad \text{of O-H bond}. \] \item \textbf{Resonance Stabilization:} After deprotonation, the ortho-nitrophenoxide ion is resonance stabilized due to the presence of the nitro group, which stabilizes the negative charge on the oxygen. This resonance stabilization further enhances the acidity of ortho-nitrophenol. \[ \text{Ortho-Nitrophenoxide Ion:} \quad \text{C}_6\text{H}_4(\text{NO}_2) \text{O}^- \] \item \textbf{Conclusion:} Due to the electron-withdrawing effect of the nitro group and the resulting resonance stabilization of the phenoxide ion, ortho-nitrophenol is a stronger acid compared to ortho-methoxyphenol, where the electron-releasing methoxy group decreases acidity by making proton removal more difficult. \end{itemize}

Question16. Explain how does the −OH group attached to a carbon of benzene ring activate it towards electrophilic substitution? 

    : Alcohols, Phenols and Ethers Chapter 7

Solution :
The −OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.

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

Question17. Give equations of the following reactions:

(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.

(ii) Bromine in CS2 with phenol.

(iii) Dilute HNO3 with phenol.

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

Solution :
(i)

(ii)

(iii)

(iv)

Question18. Explain the following with an example: \begin{enumerate} \item Kolbe’s reaction \item Reimer-Tiemann reaction \item Williamson ether synthesis \item Unsymmetrical ether \end{enumerate} \section*

{Solution} \subsection*{(i) Kolbe’s Reaction} Kolbe’s reaction is a decarboxylative dimerization of two carboxylic acids (or carboxylate ions) to form a carbon-carbon bond of an alkane. This reaction involves the generation of a radical intermediate, which then couples to form the product. \[ 2 \text{R-COO}^- \xrightarrow{2 \text{e}^-} \text{R-R} + 2 \text{CO}_2 \] **Example:** Decarboxylation of sodium acetate: \[ 2 \text{CH}_3\text{COONa} \xrightarrow{\text{electrolysis}} \text{CH}_3\text{CH}_3 + 2 \text{CO}_2 \] \subsection*{(ii) Reimer-Tiemann Reaction} The Reimer-Tiemann reaction involves the ortho-formylation of phenols. Phenol is converted to salicylaldehyde using chloroform (\(\text{CHCl}_3\)) and a strong base such as sodium hydroxide (\(\text{NaOH}\)). \[ \text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3 \text{NaOH} \rightarrow \text{OHC}-\text{C}_6\text{H}_4\text{OH} + 3 \text{NaCl} + 2 \text{H}_2\text{O} \] **Example:** Phenol reacts with chloroform and sodium hydroxide to yield salicylaldehyde: \[ \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{CHCl}_3, \text{NaOH}} \text{OHC}-\text{C}_6\text{H}_4\text{OH} \] \subsection*{(iii) Williamson Ether Synthesis} Williamson ether synthesis is a method for preparing ethers from an alkoxide and an alkyl halide. The alkoxide acts as a nucleophile and attacks the alkyl halide, resulting in the formation of an ether. \[ \text{R-O}^- \text{Na}^+ + \text{R’-X} \rightarrow \text{R-O-R’} + \text{NaX} \] **Example:** Preparation of ethyl ether from ethanol and ethyl iodide: \[ \text{C}_2\text{H}_5\text{ONa} + \text{C}_2\text{H}_5\text{I} \rightarrow \text{C}_2\text{H}_5\text{-O-C}_2\text{H}_5 + \text{NaI} \] \subsection*{(iv) Unsymmetrical Ether} An unsymmetrical ether has two different alkyl groups attached to the oxygen atom. Its general formula is \(\text{R-O-R’}\). **Example:** Ethyl methyl ether: \[ \text{CH}_3\text{-O-C}_2\text{H}_5 \]

Question19. Write the mechanism of acid-catalysed 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):

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.

Question20. How are the following conversions carried out?

(i) Propene → Propan-2-ol

(ii) Benzyl chloride → Benzyl alcohol

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

(iv) Methyl magnesium bromide → 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.

(ii) If benzyl chloride is treated with 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 propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.

Question21. 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 propan-2-ol to propene.

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

Solution :
(i) Acidified potassium permanganate

(ii) Pyridinium chlorochromate (PCC)

(iii) Bromine water

(iv) Acidified potassium permanganate

(v) 85% phosphoric acid

(vi) NaBH4 or LiAlH4

Ques 22: Give reason for the higher boiling point of ethanol in comparison to methoxymethane. \section*

{Solution} Ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) has a higher boiling point than methoxymethane (\(\text{CH}_3\text{OCH}_3\)) due to the presence of intermolecular hydrogen bonding in ethanol. Here is the explanation: \begin{itemize} \item \textbf{Ethanol:} The presence of the \(-\text{OH}\) group in ethanol allows for the formation of hydrogen bonds between ethanol molecules. These hydrogen bonds are strong intermolecular forces that cause ethanol molecules to be more tightly associated with each other. To break these hydrogen bonds during boiling, additional energy is required, resulting in a higher boiling point. \item \textbf{Methoxymethane:} In contrast, methoxymethane does not have an \(-\text{OH}\) group and therefore cannot form hydrogen bonds. The only intermolecular forces present are weaker van der Waals forces. As a result, less energy is required to overcome these forces, leading to a lower boiling point for methoxymethane. \end{itemize} \[ \text{Ethanol:} \quad \text{CH}_3\text{CH}_2\text{OH} \quad \text{(hydrogen bonding present)} \] \[ \text{Methoxymethane:} \quad \text{CH}_3\text{OCH}_3 \quad \text{(hydrogen bonding absent)} \] \textbf{Conclusion:} The higher boiling point of ethanol compared to methoxymethane is due to the presence of strong hydrogen bonding in ethanol, which requires more energy to overcome. 

Question23. Give IUPAC names of the following ethers:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution :
(i) 1-Ethoxy-2-methylpropane

(ii) 2-Chloro-1-methoxyethane

(iii) 4-Nitroanisole

(iv) 1-Methoxypropane

(v) 1-Ethoxy-4, 4-dimethylcyclohexane

(vi) Ethoxybenzene

Question24. 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)

Question25. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.

Solution :
The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide.

But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.

Question26. How is 1-propoxypropane synthesized from propan-1-ol? Write the mechanism of this reaction. \section*

{Solution} 1-propoxypropane can be synthesized from propan-1-ol by the action of protic acids such as sulfuric acid (\(\text{H}_2\text{SO}_4\)) or phosphoric acid (\(\text{H}_3\text{PO}_4\)). Two molecules of propan-1-ol condense to form one molecule of 1-propoxypropane with the loss of a water molecule. \textbf{Reaction:} \[ 2 \, \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O} \] \textbf{Mechanism:} \begin{enumerate} \item \textbf{Protonation of Alcohol:} The hydroxyl group (\(-\text{OH}\)) of propan-1-ol is protonated by the acid catalyst. This increases the electrophilicity of the alcohol group. \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}_2^+ \] \item \textbf{Nucleophilic Attack:} The oxygen atom of the second propan-1-ol molecule, which acts as a nucleophile, attacks the protonated alcohol. This results in the formation of a protonated ether (1-propoxypropane) and a molecule of water. \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}_2^+ + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O} + \text{H}^+ \] \item \textbf{Deprotonation:} The protonated ether loses a proton (deprotonation), resulting in the formation of 1-propoxypropane. \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_2\text{CH}_2\text{CH}_3^+ \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_2\text{CH}_2\text{CH}_3 + \text{H}^+ \] \end{enumerate}

Question27. 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 (SN2) 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.

Question28. Write the equation of the reaction of hydrogen iodide with:

(i) 1-propoxypropane 

(ii) Methoxybenzene and 

(iii) Benzyl ethyl ether

Solution :
(i)

(ii)

(iii)

Question29. 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 :
(i)

In aryl alkyl ethers, due to the +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.

Question30. Write the mechanism of the reaction of HI with methoxymethane.

Solution :
The mechanism of the reaction of HI with methoxymethane involves the following steps:

Step1: Protonation of methoxymethane:

Step2: Nucleophilic attack of I−:

Step3:

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

Question31. 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)

(ii)

(iii)

(iv)

Question32. Show how would you synthesise the following alcohols from appropriate alkenes?

(i) 

(ii) 

(iii) 

(iv) 

Solution :
The given alcohols can be synthesized by applying Markovnikov’s rule of acid-catalyzed 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)

Question33. 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 the 3rd carbon atom.) \section*

{Solution} The mechanism for the reaction between 3-methylbutan-2-ol and HBr is as follows: \begin{enumerate} \item \textbf{Protonation of Alcohol:} The hydroxyl group (\(-\text{OH}\)) of 3-methylbutan-2-ol is protonated by HBr. This step increases the electrophilicity of the alcohol group, making it a better leaving group. \[ \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3\text{)CH}_2\text{CH}_3 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{C(OH}_2^+\text{)(CH}_3\text{)CH}_2\text{CH}_3 + \text{Br}^- \] \textbf{Structure:} \[ \chemfig{CH_3-CH_2-C(OH_2^+)(CH_3)-CH_2-CH_3} \] \item \textbf{Formation of Secondary Carbocation:} The protonated alcohol (\(\text{CH}_3\text{CH}_2\text{C(OH}_2^+\text{)(CH}_3\text{)CH}_2\text{CH}_3\)) loses a water molecule to form a secondary carbocation. \[ \text{CH}_3\text{CH}_2\text{C(OH}_2^+\text{)(CH}_3\text{)CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{C}^+(CH_3)\text{CH}_2\text{CH}_3 + \text{H}_2\text{O} \] \textbf{Structure:} \[ \chemfig{CH_3-CH_2-C^+(CH_3)-CH_2-CH_3} \] \item \textbf{Hydride Shift:} The secondary carbocation rearranges to a more stable tertiary carbocation through a 1,2-hydride shift from the 3rd carbon atom. \[ \text{CH}_3\text{CH}_2\text{C}^+(CH_3)\text{CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{C(CH}_3\text{)}^+-\text{CH}_2\text{CH}_3 \] \textbf{Structure:} \[ \chemfig{CH_3-CH_2-C^+(CH_3)-CH_2-CH_3} \] \item \textbf{Nucleophilic Attack:} The tertiary carbocation is then attacked by a bromide ion (\(\text{Br}^-\)) to form 2-bromo-2-methylbutane. \[ \text{CH}_3\text{CH}_2\text{C}^+(CH_3)\text{CH}_2\text{CH}_3 + \text{Br}^- \rightarrow \text{CH}_3\text{CH}_2\text{C(Br)(CH}_3\text{)CH}_2\text{CH}_3 \] \textbf{Structure:} \[ \chemfig{CH_3-CH_2-C(Br)(CH_3)-CH_2-CH_3} \] 

Question34. Classify the following as primary, secondary and tertiary alcohols:

(i) 

(ii)

(iii)

(iv) 

(v) 

(vi) 

Solution :
Primary alcohol → (i), (ii), (iii)

Secondary alcohol → (iv), (v)

Tertiary alcohol → (vi)

Question35. Identify allylic alcohols in the above examples.

Solution :
The alcohols given in (ii) and (vi) are allylic alcohols.

Question36. Name the following compounds according to IUPAC system.

(i) 

(ii) 

(iii) 

(iv) 

(v) 

Solution :
(i) 3-Chloromethyl-2-isopropylpentan-1-ol

(ii) 2, 5-Dimethylhexane-1, 3-diol

(iii) 3-Bromocyclohexanol

(iv) Hex-1-en-3-ol

(v) 2-Bromo-3-methylbut-2-en-1-ol

Question37. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?

(i) 

(ii) 

Solution :
(i)

(ii)

Question38. Write structures of the products of the following reactions:

(i) 

(ii) 

(iii) 

Solution :
(i)

(ii)

(iii)

Question39. Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl2 (b) HBr and (c) SOCl2. 

(i) Butan-1-ol 

(ii) 2-Methylbutan-2-ol

Solution :
(a)

(i)

Primary alcohols do not react appreciably with Lucas’ reagent (HCl-ZnCl2) at room temperature.

(ii)

Tertiary alcohols react immediately with Lucas’ reagent.

(b)

(i)

(ii)

(c)

(i)

(ii)

Question40. Predict the major product of acid catalysed dehydration of

(i) 1-Methylcyclohexanol and

(ii) Butan-1-ol

Solution :

(i)

(ii)

Question41. 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

It can be observed that the presence of nitro groups increases the stability of phenoxide ion.

Question42. Write the equations involved in the following reactions:

(i) Reimer-Tiemann reaction

(ii) Kolbe’s reaction

Solution :

1.Reimer-Tiemann reaction

2.Kolbe’s reaction

Question43. Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol. \section*

{Solution} Williamson ether synthesis involves the nucleophilic displacement of a halide ion or other good leaving group by an alkoxide ion. The synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol can be carried out through the following steps: \begin{enumerate} \item \textbf{Preparation of Ethyl Bromide from Ethanol:} Ethanol reacts with hydrogen bromide (HBr) to form ethyl bromide.\[ \chemfig{CH_3CH_2OH} + \chemfig{HBr} \rightarrow \chemfig{CH_3CH_2Br} + \chemfig{H_2O} \] \textbf{Reaction:} \[ \text{CH}_3\text{CH}_2\text{OH} + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br} + \text{H}_2\text{O} \] \item \textbf{Formation of Sodium 3-Methylpentan-2-oxide:} 3-Methylpentan-2-ol reacts with sodium to form sodium3-methylpentan-2-oxideandhydrogengas.\[\chemfig{CH_3CH_2CH(CH_3)CH(OH)CH_3} + \text{Na} \rightarrow \chemfig{CH_3CH_2CH(CH_3)CH(O^{-}Na^{+})CH_3}+ \frac{1}{2}\chemfig{H_2}\]\textbf{Reaction:}\[\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CH(OH)CH}_3 + \text{Na} \rightarrow \text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CH(O^{-}\text{Na}^{+})CH}_3 + \frac{1}{2}\text{H}_2 \] \item \textbf{Williamson Ether Synthesis:} Sodium 3-methylpentan-2-oxide reacts with ethyl bromide to form 2-ethoxy-3-methylpentane. \[ \chemfig{CH_3CH_2Br} + \chemfig{CH_3CH_2CH(CH_3)CH(O^{-}Na^{+})CH_3} \rightarrow \chemfig{CH_3CH_2CH(CH_3)CH(OCH_2CH_3)CH_3} + \text{NaBr} \] \textbf{Reaction:} \[ \text{CH}_3\text{CH}_2\text{Br} + \text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CH(O^{-}\text{Na}^{+})CH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CH(OCH}_2\text{CH}_3\text{)CH}_3 + \text{NaBr}

Question45. Predict the products of the following reactions:

(i)

(ii) 

(iii) 

(iv)

Solution :
(i)

(ii)

(iii)

(iv)