The compound A on treatment with Na gives ${\rm{B}}$, and with ${\rm{PC}}{{\rm{I}}_{\rm{5}}}$ gives ${\rm{C}}\,{\rm{.}}\,{\rm{B}}$ and ${\rm{C}}$ react together to give diethyl ether. ${\rm{A,}}\,\,{\rm{B}}\,\,{\rm{and}}\,\,{\rm{C}}$ are in the order.

${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH,}}\,\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}\,{\rm{ONa,}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{{\rm{4}}\,{\rm{ = }}\,{\rm{5}}}}\,{\rm{CI}}$

${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH,}}\,\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{,}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}\,{\rm{CI}}$

${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CI,}}\,\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{,}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}\,{\rm{OH}}$

${{\rm{C}}_{\rm{4}}}{{\rm{H}}_{\rm{5}}}{\rm{OH,}}\,\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CI,}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{ONa}}$

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NEET • CHEMISTRY • ALCOHOLS & PHENOLS & ETHERS

Alcohols & phenols & ethers practice · Medium

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61 Medium

61 of 64

Analyzing +4 | -1

Which of the following undergoes dehydration most readily?

View answer and explanation

Correct option: C

2-phenyl-2-butanol

Option C is correct because 2-phenyl-2-butanol forms the most stable carbocation (tertiary and benzylic) upon dehydration.

Solution approach

Correct Answer + Core Reason: 2-phenyl-2-butanol. It forms the most stable carbocation (tertiary and benzylic) upon dehydration.

Step-by-Step Solution:

Step 1: Dehydration of alcohols proceeds via a carbocation intermediate.

Step 2: The stability of the carbocation determines the ease of dehydration. More stable carbocations form more readily.

Step 3: Carbocation stability order: tertiary > secondary > primary.

Step 4: Consider the carbocations formed from each alcohol:
- 1-phenyl-1-butanol: forms a secondary carbocation.
- 1-phenyl-2-butanol: forms a secondary carbocation.
- 2-phenyl-2-butanol: forms a tertiary and benzylic carbocation (most stable).
- 2-phenyl-1-butanol: forms a primary carbocation.

Step 5: 2-phenyl-2-butanol forms the most stable carbocation, so it undergoes dehydration most readily.

Why other options are wrong:
- Option A: 1-phenyl-1-butanol forms a secondary carbocation, which is less stable than the tertiary carbocation formed by 2-phenyl-2-butanol.
- Option B: 1-phenyl-2-butanol forms a secondary carbocation, which is less stable than the tertiary carbocation formed by 2-phenyl-2-butanol.
- Option D: 2-phenyl-1-butanol forms a primary carbocation, which is the least stable.

Key Concepts:
- Dehydration of alcohols
- Carbocation stability

Memory Aids:
- Carbocation stability order: 3° > 2° > 1° > methyl

Quick Tips:
- Draw the carbocation intermediate to determine its stability.

62 Medium

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Application +4 | -1

The compound A on treatment with Na gives

\mathrm{B}

, and with

\mathrm{PC}{\mathrm{I}_\mathrm{5}}

gives

\mathrm{C}\,\mathrm{.}\,\mathrm{B}

and

\mathrm{C}

react together to give diethyl
ether.

\mathrm{A,}\,\,\mathrm{B}\,\,\mathrm{and}\,\,\mathrm{C}

are in the order.

View answer and explanation

Correct option: A

{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{OH,}\,\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\,\mathrm{ONa,}\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_{\mathrm{4}\,\mathrm{ = }\,\mathrm{5}}}\,\mathrm{CI}

Option A (

{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{OH}\,\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\,\mathrm{ONa}\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_{\mathrm{4}\,\mathrm{ = }\,\mathrm{5}}}\,\mathrm{CI}

) is correct because ethanol reacts with Na to form sodium ethoxide and with PCl5 to form chloroethane, which then react to form diethyl ether via Williamson ether synthesis.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{OH}\,\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\,\mathrm{ONa}\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_{\mathrm{4}\,\mathrm{ = }\,\mathrm{5}}}\,\mathrm{CI}$ . The reaction sequence is ethanol to sodium ethoxide and chloroethane, which react to form diethyl ether.

Step-by-Step Solution:

Step 1: Identify the reactions.\nReaction 1: A + Na -> B\nReaction 2: A + $PCl_5$ -> C\nReaction 3: B + C -> Diethyl ether

Step 2: Consider the reaction of ethanol with Na.
$2{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} + 2\mathrm{Na} \rightarrow 2{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa} + {\mathrm{H}_2}$ \nTherefore, A is ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ and B is ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa}$ .

Step 3: Consider the reaction of ethanol with $PCl_5$ .
${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} + \mathrm{PC}{\mathrm{I}_\mathrm{5}} \rightarrow {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{Cl} + \mathrm{POCl}_3 + \mathrm{HCl}$ \nTherefore, C is ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{Cl}$ .

Step 4: Consider the reaction of B and C to form diethyl ether (Williamson Ether Synthesis).
${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{Cl} \rightarrow {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5} + \mathrm{NaCl}$

Step 5: Summarize the compounds.\nA = ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ \nB = ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa}$ \nC = ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{Cl}$

Why other options are wrong:
- **Option B ( ${\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{OH}\,\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{6}}\mathrm{}\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\,\mathrm{CI}$ ): Ethanol does not react with Na to produce ethane.
- Option C ( ${\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{CI}\,\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{6}}\mathrm{}\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\,\mathrm{OH}$ ): The order is incorrect, and ethanol does not react with Na to produce ethane.
- Option D ( ${\mathrm{C}_\mathrm{4}}{\mathrm{H}_\mathrm{5}}\mathrm{OH}\,\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{CI}\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{ONa}$ ): The starting alcohol is incorrect.

Key Concepts:**
- Reactions of alcohols
- Williamson ether synthesis
- Reactions with sodium
- Reactions with $PCl_5$

Memory Aids:
- Na with alcohols forms alkoxides.
- $PCl_5$ replaces -OH with -Cl.
- Williamson ether synthesis: $R-O^- + R'-X \rightarrow R-O-R' + X^-$

Quick Tips:
- Recognize the Williamson ether synthesis as a key step.

63 Medium

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Understanding +4 | -1

Glycerol is highly viscous. It is due to the fact that:

View answer and explanation

Correct option: B

It forms extensive H-bonding

Option B (It forms extensive H-bonding) is correct because glycerol has three hydroxyl groups, allowing for extensive hydrogen bonding which increases viscosity.

Solution approach

Correct Answer + Core Reason: It forms extensive H-bonding. Glycerol has three hydroxyl groups, leading to extensive intermolecular hydrogen bonding.

Step-by-Step Solution:

Step 1: Glycerol structure: Glycerol has three hydroxyl groups ( $\mathrm{OH}$ ).

Step 2: Hydrogen bonding: Hydroxyl groups can form hydrogen bonds.

Step 3: Viscosity: Extensive hydrogen bonding increases intermolecular forces, leading to higher viscosity.

Why other options are wrong:
- Option A: It is highly polar: While glycerol is polar, the primary reason for its high viscosity is the extensive hydrogen bonding.
- Option C: It shows intramolecular H-bonding: Intramolecular hydrogen bonding would decrease intermolecular forces and thus decrease viscosity.
- Option D: It has high b.p.: High boiling point is a consequence of strong intermolecular forces, but the direct cause of viscosity is the hydrogen bonding itself.

Key Concepts:
- Hydrogen bonding
- Viscosity
- Intermolecular forces

Memory Aids:
- More OH groups = more H-bonding = higher viscosity.

Quick Tips:
- Remember that hydrogen bonding increases viscosity.

64 Medium

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Understanding +4 | -1

Which of the following ethers form peroxide readily?

View answer and explanation

Correct option: C

\mathrm{iPr}—\mathrm{O}—\mathrm{iPr}

Option C (

\mathrm{iPr}—\mathrm{O}—\mathrm{iPr}

) is correct because isopropyl ether is known to form peroxides more readily than other ethers due to the stability of the radical formed during the peroxide formation process.

Solution approach

Correct Answer + Core Reason: $\mathrm{iPr}—\mathrm{O}—\mathrm{iPr}$ . Isopropyl ether forms peroxides readily.

Step-by-Step Solution:

Step 1: Explain peroxide formation. Ethers can react with oxygen in the air to form explosive peroxides.

Step 2: Explain the ease of peroxide formation. Isopropyl ether forms peroxides more readily than other ethers due to the stability of the radical formed during the process.

Step 3: Conclude the peroxide formation. Isopropyl ether is the most likely to form peroxides readily among the given options.

Why other options are wrong:
- Option A: $\mathrm{Me}—\mathrm{O}—\mathrm{Me}$ (dimethyl ether) forms peroxides less readily.
- Option B: $\mathrm{Et}—\mathrm{O}—\mathrm{Et}$ (diethyl ether) forms peroxides less readily than isopropyl ether.
- Option D: $\mathrm{Me}—\mathrm{O}—\mathrm{Et}$ (ethyl methyl ether) forms peroxides less readily.

Key Concepts:
- Peroxide formation in ethers
- Stability of radicals

Memory Aids:
- Isopropyl ethers are known to be particularly prone to peroxide formation.

Quick Tips:
- Remember that isopropyl ether is more likely to form peroxides than other common ethers.

Revision notes and key takeaways

Review ALCOHOLS & PHENOLS & ETHERS in your textbook or class notes, then use this set to check your understanding.

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