The starting material for the preparation of ${\rm{C}}{{\rm{H}}_3}{\rm{I}}$ in one step reaction is:

${\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{CHO}}$

${\rm{C}}{{\rm{H}}_3}{\rm{COC}}{{\rm{H}}_3}$

In the reaction, the products formed are: ${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{C}}{{\rm{H}}_2} \cdot {\rm{C}}{{\rm{H}}_2} \cdot {\rm{O}} \cdot {\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3} + {\rm{HI}}\xrightarrow{{{\rm{Heated}}}}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{CHC}}{{\rm{H}}_2}{\rm{OH}} + {{\rm{C}}_2}{{\rm{H}}_5}{\rm{I}}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{CHC}}{{\rm{H}}_3} + {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{CH}} \cdot {\rm{C}}{{\rm{H}}_2}{\rm{OH}} + {{\rm{C}}_2}{{\rm{H}}_6}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{CH}} \cdot {\rm{C}}{{\rm{H}}_2}{\rm{I}} + {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{HBr}}$ reacts with${\rm{C}}{{\rm{H}}_2}$=${\rm{CH}}$—${\rm{OC}}{{\rm{H}}_3}$ under anhydrous conditions at room temperature to give:

${{\rm{H}}_3}{\rm{C}}$—${\rm{CHBr}}$—${\rm{OC}}{{\rm{H}}_3}$

${\rm{C}}{{\rm{H}}_3}{\rm{CHO}}$ and C${\rm{C}}{{\rm{H}}_3}{\rm{CHO}}$

${\rm{BrC}}{{\rm{H}}_2}{\rm{CHO}}$ and ${\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

${\rm{BrC}}{{\rm{H}}_2}$—${\rm{C}}{{\rm{H}}_2}$—${\rm{OC}}{{\rm{H}}_3}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{CONa}}$ on reaction with ${\rm{C}}{{\rm{H}}_3}{\rm{Br}}$ will give:

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{COC}}{{\rm{H}}_3}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{COC}}{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}$

${\rm{C}}{{\rm{H}}_3}{\rm{OC}}{{\rm{H}}_3}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OC}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}$

The reaction of ${\rm{C}}{{\rm{H}}_3}{\rm{O}}{{\rm{C}}_2}{{\rm{H}}_5}$ with ${\rm{HI}}$ gives:

${\rm{C}}{{\rm{H}}_3}{\rm{I}} + {{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{I}}$ only

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$ only

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{I}} + {\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

The toxicity order for ${\rm{C}}{{\rm{H}}_3}{\rm{OH}},{\rm{\;}}{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$ and ${{\rm{C}}_3}{{\rm{H}}_7}{\rm{OH}}$ is:

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} < {{\rm{C}}_3}{{\rm{H}}_7}{\rm{OH}} < C{{\rm{H}}_3}{\rm{OH}}$

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} < C{{\rm{H}}_3}{\rm{OH}} < {{\rm{C}}_3}{{\rm{H}}_7}{\rm{OH}}$

${{\rm{C}}_3}{{\rm{H}}_7}{\rm{OH}} < {{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} < C{{\rm{H}}_3}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{OH}} < {{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} < {{\rm{C}}_3}{{\rm{H}}_7}{\rm{OH}}$

ALCOHOLS & PHENOLS & ETHERS MCQs for NEET UG CHEMISTRY — Medium, Page 2

31 Medium

31 of 64

Analyzing +4 | -1

For which of the following parameters the structural isomers

{C_2}{H_5}OH

and

C{H_3}OC{H_3}

would be expected to have the same values? (Assume ideal behaviour)

View answer and explanation

Correct option: D

Gaseous densities at the same temperature and pressure

Option D is correct because gaseous density depends on molar mass, temperature, and pressure, and both isomers have the same molar mass.

Solution approach

Correct Answer + Core Reason: Gaseous densities at the same temperature and pressure. Density depends on molar mass, and both isomers have the same molar mass.

Step-by-Step Solution:

Step 1: The ideal gas law is $PV = nRT$ , where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is temperature.

Step 2: Density, $\rho$ , is defined as mass per unit volume, $\rho = \frac{m}{V}$ .

Step 3: Number of moles, $n = \frac{m}{M}$ , where $m$ is mass and $M$ is molar mass.

Step 4: Substituting $n$ into the ideal gas law: $PV = \frac{m}{M}RT$ .

Step 5: Rearranging for density: $\rho = \frac{m}{V} = \frac{PM}{RT}$ .

Step 6: At the same temperature and pressure, density is directly proportional to molar mass. Since both isomers have the same molar mass, their gaseous densities will be the same.

Why other options are wrong:
- Option A: Heat of vaporization depends on intermolecular forces, which are different for alcohols and ethers due to hydrogen bonding in alcohols.
- Option B: Vapor pressure depends on intermolecular forces, which are different for alcohols and ethers.
- Option C: Boiling points depend on intermolecular forces, which are different for alcohols and ethers.

Key Concepts:
- Ideal gas law
- Density
- Molar mass
- Intermolecular forces

Memory Aids:
- PV = nRT
- Density = mass/volume

Quick Tips:
- Remember the ideal gas law and its relationship to density.

32 Medium

32 of 64

Application +4 | -1

Which will not form yellow precipitate on heating with an alkaline solution of iodine?

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

Option C (

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

) is correct because it does not contain a methyl ketone or a secondary alcohol with a methyl group on the carbinol carbon, which are required for the iodoform test.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ . Propanol does not give a positive iodoform test.

Step-by-Step Solution:

Step 1: Understand the Iodoform Test: The iodoform test is given by compounds containing the $\mathrm{C}{\mathrm{H}_3}\mathrm{CO}$ group or compounds that can be oxidized to form this group (i.e., $\mathrm{C}{\mathrm{H}_3}\mathrm{CHOH}$ ).

Step 2: Analyze the given compounds:
- A: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CHOHC}{\mathrm{H}_3}$ can be oxidized to $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{COC}{\mathrm{H}_3}$ , which contains the $\mathrm{C}{\mathrm{H}_3}\mathrm{CO}$ group. So, it gives a positive iodoform test.
- B: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CHOHC}{\mathrm{H}_3}$ (same as A) gives a positive iodoform test.
- C: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ can be oxidized to $\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}$ , which does NOT contain the $\mathrm{C}{\mathrm{H}_3}\mathrm{CO}$ group. So, it does not give a positive iodoform test.
- D: $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ can be oxidized to $\mathrm{HCHO}$ , which does NOT contain the $\mathrm{C}{\mathrm{H}_3}\mathrm{CO}$ group, but methanol is an exception and gives positive iodoform test.

Step 3: Conclusion: Only $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ does not give a positive iodoform test.

Why other options are wrong:
- Option A: This compound gives a positive iodoform test.
- Option B: This compound gives a positive iodoform test.
- Option D: Methanol gives a positive iodoform test.

Key Concepts:
- Iodoform test
- Methyl ketones
- Secondary alcohols with a methyl group on the carbinol carbon

Memory Aids:
- Compounds with $\mathrm{C}{\mathrm{H}_3}\mathrm{CO}$ or $\mathrm{C}{\mathrm{H}_3}\mathrm{CHOH}$ groups give a positive iodoform test.

Quick Tips:
- Remember the structure of compounds that give a positive iodoform test.

33 Medium

33 of 64

Understanding +4 | -1

The starting material for the preparation of

\mathrm{C}{\mathrm{H}_3}\mathrm{I}

in one step reaction is:

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{H}_3}\mathrm{OH}

Option A (

\mathrm{CH}_3\mathrm{OH}

) is correct because methyl alcohol (

\mathrm{CH}_3\mathrm{OH}

) can be directly converted to methyl iodide (

\mathrm{CH}_3\mathrm{I}

) in a single step using a reagent like

\mathrm{HI}

or

\mathrm{P}{\mathrm{I}_3}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH}_3\mathrm{OH}$ . Methyl alcohol can be directly converted to methyl iodide in one step.

Step-by-Step Solution:

Step 1: Understand the conversion of alcohols to alkyl halides.

Step 2: Methyl alcohol ( $\mathrm{CH}_3\mathrm{OH}$ ) can react with $\mathrm{HI}$ or $\mathrm{P}{\mathrm{I}_3}$ to directly form methyl iodide ( $\mathrm{CH}_3\mathrm{I}$ ).

Step 3: The reaction is: $\mathrm{CH}_3\mathrm{OH} + \mathrm{HI} \rightarrow \mathrm{CH}_3\mathrm{I} + \mathrm{H}_2\mathrm{O}$

Why other options are wrong:
- Option B: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ : Ethyl alcohol would produce ethyl iodide, not methyl iodide.
- Option C: $\mathrm{CH}_3\mathrm{CHO}$ : Acetaldehyde would require multiple steps to convert to methyl iodide.
- Option D: $\mathrm{CH}_3\mathrm{COC}{\mathrm{H}_3}$ : Acetone would require multiple steps to convert to methyl iodide.

Key Concepts:
- Conversion of alcohols to alkyl halides

Memory Aids:
- Remember that alcohols can be converted to alkyl halides using hydrogen halides or phosphorus halides.

Quick Tips:
- Identify the starting material that can be directly converted to the desired product in one step.

34 Medium

34 of 64

Application +4 | -1

Propan-1-ol can be prepared from propane by

View answer and explanation

Correct option: C

{\mathrm{B}_2}{\mathrm{H}_6}

followed by

{\mathrm{H}_2}{\mathrm{O}_2}

Option C (

{\mathrm{B}_2}{\mathrm{H}_6}

followed by

{\mathrm{H}_2}{\mathrm{O}_2}

) is correct because it represents hydroboration-oxidation, which converts an alkene to an alcohol with anti-Markovnikov regioselectivity.

Solution approach

Correct Answer + Core Reason: ${\mathrm{B}_2}{\mathrm{H}_6}$ followed by ${\mathrm{H}_2}{\mathrm{O}_2}$ . Hydroboration-oxidation adds water across a double bond with anti-Markovnikov regioselectivity, yielding propan-1-ol from propene.

Step-by-Step Solution:

Step 1: Propane needs to be converted to propene first, then propene to propan-1-ol. The direct conversion of propane to propan-1-ol is not feasible with the given options.

Step 2: Identify the reaction that converts an alkene to an alcohol with anti-Markovnikov addition. This is hydroboration-oxidation.

Step 3: Hydroboration-oxidation involves the addition of borane ( ${\mathrm{BH}_3}$ ) to the alkene, followed by oxidation with hydrogen peroxide ( ${\mathrm{H}_2}{\mathrm{O}_2}$ ) in a basic medium.
${\mathrm{CH}_3}\mathrm{CH=CH}_2 \xrightarrow{{{\mathrm{B}_2}{\mathrm{H}_6}}} {\mathrm{(CH}_3}\mathrm{CH}_2\mathrm{CH}_2\mathrm{)}_3}\mathrm{B} \xrightarrow{{{\mathrm{H}_2}{\mathrm{O}_2},\mathrm{OH}^-}} {\mathrm{CH}_3}\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$

Step 4: The overall reaction converts propene to propan-1-ol.

Why other options are wrong:
- **Option A ( ${\mathrm{H}_2}\mathrm{O}/{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ ): This would add water according to Markovnikov's rule, yielding propan-2-ol.
- Option B ( $\mathrm{Hg}{\left( {\mathrm{OAc}} \right)_2}{\mathrm{H}_2}\mathrm{O}$ followed by $\mathrm{NaB}{\mathrm{H}_4}$ ): This is oxymercuration-demercuration, which also adds water according to Markovnikov's rule, yielding propan-2-ol.
- Option D ( $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{O}_2}\mathrm{H}/{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ ): This would add acetic acid to the alkene, not water, and would also follow Markovnikov's rule.

Key Concepts:
- Hydroboration-oxidation
- Markovnikov's rule
- Anti-Markovnikov's rule
- Addition reactions of alkenes

Memory Aids:
- Hydroboration-oxidation: "BAH!" (Boron Adds Here - Anti-Markovnikov)

Quick Tips:**
- Remember that hydroboration-oxidation is the key reaction for anti-Markovnikov addition of water to alkenes.
- Markovnikov's rule: The hydrogen atom adds to the carbon with more hydrogen atoms already attached.

35 Medium

35 of 64

Application +4 | -1

In the reaction, the products formed are:

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{C}{\mathrm{H}_2} \cdot \mathrm{C}{\mathrm{H}_2} \cdot \mathrm{O} \cdot \mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3} + \mathrm{HI}\xrightarrow{{\mathrm{Heated}}}

View answer and explanation

Correct option: C

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHC}{\mathrm{H}_2}\mathrm{OH} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{I}

Option C is correct because HI cleaves the ether at the less substituted side via SN1 or SN2 mechanism, forming an alcohol and an alkyl iodide.

Solution approach

Correct Answer + Core Reason: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHC}{\mathrm{H}_2}\mathrm{OH} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{I}$ . HI cleaves ethers to form alcohols and alkyl halides.

Step-by-Step Solution:

Step 1: Identify the ether: The given ether is ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{C}{\mathrm{H}_2} - \mathrm{C}{\mathrm{H}_2} - \mathrm{O} - \mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}$ .

Step 2: Reaction with HI: Ethers react with HI to form alcohols and alkyl iodides. The cleavage occurs at the C-O bond.

Step 3: Determine the products: In this case, the ether will cleave to form an alcohol from the more substituted side and an alkyl iodide from the less substituted side due to steric hindrance and carbocation stability (if SN1 mechanism is favored). Here, the more substituted side is ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{C}{\mathrm{H}_2} - \mathrm{C}{\mathrm{H}_2} - \mathrm{O}$ and the less substituted side is ${\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3} - \mathrm{O}}$ .

Step 4: Write the products: The products are ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHC}{\mathrm{H}_2}\mathrm{OH}$ and ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{I}$ .

Why other options are wrong:
- Option A: This option is incorrect because it forms an incorrect alcohol and alkyl group.
- Option B: This option is incorrect because it forms an incorrect alcohol and alkane.
- Option D: This option is incorrect because it forms an incorrect alkyl iodide and alcohol.

Key Concepts:
- Cleavage of ethers by HI
- SN1 and SN2 mechanisms
- Stability of carbocations
- Steric hindrance

Memory Aids:
- HI cleaves ethers at the C-O bond.
- More substituted side forms alcohol (usually, but depends on reaction conditions).

Quick Tips:
- Consider the stability of carbocations when determining the products of ether cleavage.

36 Medium

36 of 64

Remembering +4 | -1

Lucas reagent is used to distinguish among primary, secondary and tertiary:

View answer and explanation

Correct option: B

Alcohols

Option B (Alcohols) is correct because Lucas reagent is specifically used to differentiate primary, secondary, and tertiary alcohols based on their reactivity with the reagent.

Solution approach

Correct Answer + Core Reason: Alcohols. Lucas reagent is used to distinguish between primary, secondary, and tertiary alcohols based on their reaction rates.

Step-by-Step Solution:

Step 1: Recall the definition and use of Lucas reagent.

Step 2: Understand that Lucas reagent reacts with alcohols to form alkyl halides.

Step 3: Know that the reaction rate varies depending on whether the alcohol is primary, secondary, or tertiary.

Why other options are wrong:
- Option A: Alkyl halides are the products of the reaction with Lucas reagent, not the reactants being distinguished.
- Option C: Aliphatic amines react with different reagents to be distinguished.
- Option D: Aromatic amines react with different reagents to be distinguished.

Key Concepts:
- Lucas reagent
- Reactions of alcohols
- Primary, secondary, and tertiary alcohols

Memory Aids:
- Lucas reagent is used to test \"alcohol solubility\" (think of the \"L\" in Lucas for \"Liquids\" which alcohols are).

Quick Tips:
- Remember that tertiary alcohols react fastest with Lucas reagent, followed by secondary, and then primary.

37 Medium

37 of 64

Remembering +4 | -1

The explosive nitroglycerine is:

View answer and explanation

Correct option: C

An ester

Option C (An ester) is correct because nitroglycerine is the trinitrate ester of glycerol.

Solution approach

Correct Answer + Core Reason: An ester. Nitroglycerine is the trinitrate ester of glycerol.

Step-by-Step Solution:

Step 1: Recall the structure of nitroglycerine.

Step 2: Recognize that it is formed by the esterification of glycerol with nitric acid.

Step 3: Identify it as an ester.

Why other options are wrong:
- Option A: A soap is a salt of a fatty acid.
- Option B: A salt is an ionic compound formed from an acid and a base.
- Option D: A complex compound is a compound containing a metal ion bonded to ligands.

Key Concepts:
- Esters
- Nitroglycerine
- Esterification

Memory Aids:
- Remember that nitroglycerine is formed from glycerol and nitric acid, which results in an ester.

Quick Tips:
- Recognize functional groups in organic compounds.

38 Medium

38 of 64

Analyzing +4 | -1

If there be a compound of the formula

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\left( {\mathrm{OH}} \right)_3}

, which one of the following compounds would be obtained form it without treatment with any reagent?

View answer and explanation

Correct option: C

Acetic acid

Option C is correct because a geminal triol is unstable and loses water to form a carboxylic acid.

Solution approach

Correct Answer + Core Reason: Acetic acid. A compound with three hydroxyl groups on the same carbon is unstable and will lose water to form a carboxylic acid.

Step-by-Step Solution:

Step 1: The compound $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\left( {\mathrm{OH}} \right)_3}$ is a geminal triol (three hydroxyl groups on the same carbon).

Step 2: Geminal triols are unstable and readily lose a water molecule to form a carboxylic acid.

Step 3: The reaction is: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\left( {\mathrm{OH}} \right)_3} \to \mathrm{C}{\mathrm{H}_3}\mathrm{COOH} + {\mathrm{H}_2}\mathrm{O}$

Step 4: The product is acetic acid.

Why other options are wrong:
- Option A: Methanol is $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ , which is not formed from this compound.
- Option B: Ethanol is ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ , which is not formed from this compound.
- Option D: Formaldehyde is $\mathrm{HCHO}$ , which is not formed from this compound.

Key Concepts:
- Geminal triols
- Stability of organic compounds
- Dehydration reactions

Memory Aids:
- Geminal diols and triols are generally unstable.

Quick Tips:
- Recognize unstable functional groups.

39 Medium

39 of 64

Understanding +4 | -1

Ethyl alcohol is industrially prepared from the ethylene by:

View answer and explanation

Correct option: C

Absorbing in sulphuric acid followed by hydrolysis

Option C (Absorbing in sulphuric acid followed by hydrolysis) is correct because ethylene reacts with sulfuric acid to form ethyl hydrogen sulfate, which upon hydrolysis yields ethyl alcohol.

Solution approach

Correct Answer + Core Reason: Absorbing in sulphuric acid followed by hydrolysis. This is the industrial method for producing ethanol from ethene.

Step-by-Step Solution:

Step 1: Understand the industrial preparation of ethanol.

Step 2: Ethene reacts with concentrated sulfuric acid to form ethyl hydrogen sulfate ( $\mathrm{C}_2\mathrm{H}_5\mathrm{OSO}_3\mathrm{H}$ ).

Step 3: Ethyl hydrogen sulfate is then hydrolyzed with water to produce ethanol and sulfuric acid.

Step 4: The reactions are:
- $\mathrm{C}_2\mathrm{H}_4 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{OSO}_3\mathrm{H}$
- $\mathrm{C}_2\mathrm{H}_5\mathrm{OSO}_3\mathrm{H} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{OH} + \mathrm{H}_2\mathrm{SO}_4$

Why other options are wrong:
- Option A: Permanganate oxidation: This would lead to diols or further oxidation products, not ethanol directly.
- Option B: Catalytic reduction: This would reduce ethene to ethane, not ethanol.
- Option D: Fermentation: Fermentation is used to produce ethanol from sugars, not directly from ethylene.

Key Concepts:
- Industrial preparation of ethanol
- Electrophilic addition
- Hydrolysis

Memory Aids:
- Remember the sulfuric acid and hydrolysis steps.

Quick Tips:
- Know the different methods for producing ethanol.

40 Medium

40 of 64

Remembering +4 | -1

Fusel oil is a mixture of:

View answer and explanation

Correct option: A

Alcohols

Option A (Alcohols) is correct because fusel oil is a mixture of higher alcohols, mainly produced during alcoholic fermentation.

Solution approach

Correct Answer + Core Reason: Alcohols. Fusel oil is a mixture of higher alcohols.

Step-by-Step Solution:

Step 1: Understand the definition of fusel oil.

Step 2: Recall that fusel oil is a byproduct of fermentation, containing higher alcohols.

Step 3: Identify the correct option based on the definition.

Step 4: Conclude that the correct answer is Alcohols.

Why other options are wrong:
- Option B: Ethers are not the primary components of fusel oil.
- Option C: While alcohols are present, ethers are not a significant component of fusel oil.
- Option D: Acetone is not a typical component of fusel oil; alcohols are.

Key Concepts:
- Composition of Fusel Oil
- Alcoholic Fermentation

Memory Aids:
- Remember that \"fusel\" sounds like \"fuel,\" and alcohols are used as fuel.

Quick Tips:
- When dealing with mixtures, remember to consider the most abundant components.

41 Medium

41 of 64

Remembering +4 | -1

The prospective fuel ‘gashol’ is a mixture of:

View answer and explanation

Correct option: C

Petrol and ethanol

Option C (Petrol and ethanol) is correct because 'gashol' is a fuel mixture consisting of gasoline (petrol) and ethanol, used as an alternative fuel source.

Solution approach

Correct Answer + Core Reason: Petrol and ethanol. Gashol is a mixture of gasoline and ethanol.

Step-by-Step Solution:

Step 1: Define gashol.

Step 2: Gashol is a blend of gasoline (petrol) and ethanol.

Step 3: It is used as an alternative fuel.

Why other options are wrong:
- Option A: Gaseous hydrocarbons and heavy water is not the composition of gashol.
- Option B: Petrol and phenol is not the composition of gashol. Phenol is toxic and not used as a fuel additive in this way.
- Option D: Radioactive substances are not components of gashol.

Key Concepts:
- Definition of gashol

Memory Aids:
- Gashol = Gas(oline) + hol(alcohol)

Quick Tips:
- Remember the composition of common alternative fuels.

42 Medium

42 of 64

Remembering +4 | -1

Anisole is the product obtained from phenol by the reaction known as

View answer and explanation

Correct option: B

Etherification

Option B (Etherification) is correct because anisole is formed from phenol through an etherification reaction, where a methyl group is added to the oxygen of phenol.

Solution approach

Correct Answer + Core Reason: Etherification. Anisole is formed from phenol via etherification.

Step-by-Step Solution:

Step 1: Define anisole. Anisole is a methyl phenyl ether.

Step 2: Describe the formation. It is formed by reacting phenol with a methylating agent, such as methyl halide, in the presence of a base.

Step 3: Identify the reaction type. This reaction is an etherification, specifically Williamson ether synthesis.

Why other options are wrong:
- Option A: Coupling is not the reaction that forms anisole from phenol.
- Option C: Oxidation would result in different products, not anisole.
- Option D: Esterification involves the formation of an ester, not an ether.

Key Concepts:
- Etherification reaction
- Williamson ether synthesis
- Structure of anisole

Memory Aids:
- Etherification creates an ether, and anisole is an ether derived from phenol.

Quick Tips:
- Remember that etherification involves the formation of an ether linkage.

43 Medium

43 of 64

Understanding +4 | -1

Which of the following can not be used to oxidise primary alcohols to aldehydes

View answer and explanation

Correct option: C

KMn{O_4}

in acidic medium

Option C (

KMn{O_4}

in acidic medium) is correct because

KMn{O_4}

in acidic medium is a strong oxidizing agent that typically oxidizes primary alcohols all the way to carboxylic acids, not just aldehydes.

Solution approach

Correct Answer + Core Reason: $KMn{O_4}$ in acidic medium. It oxidizes primary alcohols to carboxylic acids, not just aldehydes.

Step-by-Step Solution:

Step 1: Understand the different oxidizing agents and their reactivity.

Step 2: $Cr{O_3}$ in anhydrous medium, Pyridinium chlorochromate (PCC), and heating in the presence of Cu at 573K are all suitable for oxidizing primary alcohols to aldehydes.

Step 3: $KMn{O_4}$ in acidic medium is a strong oxidizing agent that will further oxidize the aldehyde to a carboxylic acid.

Why other options are wrong:
- **Option A ( $Cr{O_3}$ in anhydrous medium): Can be used to oxidize primary alcohols to aldehydes.
- Option B (Pyridinium chlorochromate): PCC is a selective oxidizing agent that stops at the aldehyde stage.
- Option D (Heating in presence of Cu at 573K): Copper acts as a catalyst for dehydrogenation, leading to aldehyde formation.

Key Concepts:
- Oxidation of alcohols
- Selective oxidizing agents

Memory Aids:**
- PCC: Primary alcohol to Aldehyde
- $KMn{O_4}$ : Primary alcohol to Carboxylic acid

Quick Tips:
- Know the selectivity of different oxidizing agents.

44 Medium

44 of 64

Analyzing +4 | -1

Which of the property given below is not associated with glycerol?

View answer and explanation

Correct option: A

Formation of water and

\mathrm{C}{\mathrm{O}_2}

on reduction

Option A is correct because glycerol does not form water and carbon dioxide upon reduction; it typically undergoes reactions like esterification or ether formation.

Solution approach

Correct Answer + Core Reason: Formation of water and $\mathrm{C}{\mathrm{O}_2}$ on reduction. Glycerol typically doesn't form water and carbon dioxide upon reduction.

Step-by-Step Solution:

Step 1: Glycerol is a triol (an alcohol with three hydroxyl groups).

Step 2: Reduction typically involves adding hydrogen or removing oxygen.

Step 3: Reduction of glycerol would not directly lead to the formation of water and carbon dioxide. It might lead to other alcohols or alkanes under drastic conditions, but not directly to $\mathrm{H}_2\mathrm{O}$ and $\mathrm{CO}_2$ .

Step 4: Consider the other options:
- Formation of tartronic acid on oxidation: Glycerol can be oxidized to tartronic acid.
- Formation of acrolein on dehydration: Glycerol undergoes dehydration to form acrolein.
- Formation of allyl iodide with $\mathrm{P}{\mathrm{I}_3}$ : Glycerol reacts with $\mathrm{P}{\mathrm{I}_3}$ to form allyl iodide.

Why other options are wrong:
- Option B: Glycerol can be oxidized to tartronic acid.
- Option C: Glycerol can be dehydrated to form acrolein.
- Option D: Glycerol can react with $\mathrm{P}{\mathrm{I}_3}$ to form allyl iodide.

Key Concepts:
- Glycerol structure and properties
- Reduction reactions
- Oxidation reactions
- Dehydration reactions

Memory Aids:
- Glycerol is a triol.

Quick Tips:
- Know the common reactions of alcohols.

45 Medium

45 of 64

Remembering +4 | -1

Ether fire can be extinguished by:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because sand, pyrene, and CO2 can all be used to extinguish ether fires by smothering the flames or inhibiting the combustion process.

Solution approach

Correct Answer + Core Reason: All of these. Sand, pyrene, and $CO_2$ can all extinguish ether fires.

Step-by-Step Solution:

Step 1: Explain the nature of ether fires. Ether is highly flammable.

Step 2: Describe the extinguishing methods.
- Sand smothers the fire by cutting off oxygen supply.
- Pyrene (carbon tetrachloride) inhibits the combustion process.
- $CO_2$ displaces oxygen, extinguishing the fire.

Step 3: Conclude the effectiveness. All three methods are effective in extinguishing ether fires.

Why other options are wrong:
- Option A: Sand is effective, but not the only option.
- Option B: Pyrene is effective, but not the only option.
- Option C: $CO_2$ is effective, but not the only option.

Key Concepts:
- Flammability of ethers
- Fire extinguishing methods

Memory Aids:
- Think of smothering the fire with sand or displacing oxygen with $CO_2$ .

Quick Tips:
- Remember that multiple methods can be used to extinguish ether fires.

46 Medium

46 of 64

Understanding +4 | -1

Which of the following gives immediate turbidity with Lucas reagent?

View answer and explanation

Correct option: A

2 – methyl – propan – 2 – ol

Option A (2 – methyl – propan – 2 – ol) is correct because tertiary alcohols react fastest with Lucas reagent (HCl and ZnCl2) due to the formation of a stable tertiary carbocation intermediate.

Solution approach

Correct Answer + Core Reason: 2 – methyl – propan – 2 – ol. Tertiary alcohols react fastest with Lucas reagent.

Step-by-Step Solution:

Step 1: Describe Lucas reagent. Lucas reagent is a solution of anhydrous $\mathrm{ZnC}{\mathrm{l}_2}$ in concentrated HCl.

Step 2: Explain the reaction with alcohols. Lucas reagent reacts with alcohols to form alkyl chlorides, which are insoluble and cause turbidity.

Step 3: Explain the reactivity order. The reactivity order is tertiary > secondary > primary.

Step 4: Analyze the given alcohols.
- 2 – methyl – propan – 2 – ol is a tertiary alcohol.
- 2 – methyl – propan – 1 - ol is a primary alcohol.
- 2 – butanol is a secondary alcohol.
- 1 – butanol is a primary alcohol.

Step 5: Conclude the fastest reaction. 2 – methyl – propan – 2 – ol reacts fastest.

Why other options are wrong:
- Option B: 2 – methyl – propan – 1 - ol is a primary alcohol and reacts slowly.
- Option C: 2 – butanol is a secondary alcohol and reacts at an intermediate rate.
- Option D: 1 – butanol is a primary alcohol and reacts slowly.

Key Concepts:
- Lucas reagent test
- Reactivity of alcohols
- Carbocation stability

Memory Aids:
- Tertiary alcohols react fastest with Lucas reagent.

Quick Tips:
- Remember the reactivity order of alcohols with Lucas reagent: tertiary > secondary > primary.

47 Medium

47 of 64

Analyzing +4 | -1

Which of the following is not isomeric with diethyl ether?

View answer and explanation

Correct option: D

Butan-2-one

Option D (Butan-2-one) is correct because it has a different molecular formula (

C_4H_8O

) compared to diethyl ether (

C_4H_{10}O

).

Solution approach

Correct Answer + Core Reason: Butan-2-one. It has a different molecular formula than diethyl ether.

Step-by-Step Solution:

Step 1: Determine the molecular formula of diethyl ether: ${C}_4 {H}_{10} {O}$ .

Step 2: Determine the molecular formula of each option:
- A. Methyl $n -$ propyl ether: ${C}_4 {H}_{10} {O}$
- B. Butan-1-ol: ${C}_4 {H}_{10} {O}$
- C. 2-methyl propan-2-ol: ${C}_4 {H}_{10} {O}$
- D. Butan-2-one: ${C}_4 {H}_{8} {O}$

Step 3: Identify the compound with a different molecular formula.

Step 4: Butan-2-one has a different molecular formula ( ${C}_4 {H}_{8} {O}$ ) and is therefore not an isomer of diethyl ether.

Why other options are wrong:
- Option A: Methyl $n -$ propyl ether is an isomer of diethyl ether as it has the same molecular formula.
- Option B: Butan-1-ol is an isomer of diethyl ether as it has the same molecular formula.
- Option C: 2-methyl propan-2-ol is an isomer of diethyl ether as it has the same molecular formula.

Key Concepts:
- Isomers
- Molecular formula

Memory Aids:
- Isomers have the same molecular formula but different structural arrangements.

Quick Tips:
- Always determine the molecular formula first when identifying isomers.

48 Medium

48 of 64

Application +4 | -1

\mathrm{HBr}

reacts with

\mathrm{C}{\mathrm{H}_2}

=

\mathrm{CH}

—

\mathrm{OC}{\mathrm{H}_3}

under anhydrous conditions at room temperature to give:

View answer and explanation

Correct option: D

{\mathrm{H}_3}\mathrm{C}

—

\mathrm{CHBr}

—

\mathrm{OC}{\mathrm{H}_3}

The correct answer is D (

{\mathrm{H}_3}\mathrm{C}

—

\mathrm{CHBr}

—

\mathrm{OC}{\mathrm{H}_3}

) because the reaction follows Markovnikov's rule, where the hydrogen adds to the carbon with more hydrogens and the bromine adds to the other carbon.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_3}\mathrm{C}$ — $\mathrm{CHBr}$ — $\mathrm{OC}{\mathrm{H}_3}$ . The reaction follows Markovnikov's rule due to the electron-donating effect of the methoxy group.

Step-by-Step Solution:

Step 1: The given reactant is an ether with an alkene group: $\mathrm{C}{\mathrm{H}_2}=\mathrm{CH}$ — $\mathrm{OC}{\mathrm{H}_3}$ .

Step 2: $\mathrm{HBr}$ adds to alkenes according to Markovnikov's rule. This rule states that the hydrogen atom adds to the carbon atom with more hydrogen atoms already attached, and the halogen adds to the carbon atom with fewer hydrogen atoms.

Step 3: The methoxy group ( $-\mathrm{OC}{\mathrm{H}_3}$ ) is electron-donating and stabilizes the carbocation formed on the carbon adjacent to it.

Step 4: Therefore, the hydrogen from $\mathrm{HBr}$ adds to the $\mathrm{C}{\mathrm{H}_2}$ group, and the bromine adds to the $\mathrm{CH}$ group.

Step 5: The product is ${\mathrm{H}_3}\mathrm{C}$ — $\mathrm{CHBr}$ — $\mathrm{OC}{\mathrm{H}_3}$ .

Why other options are wrong:
- Option A: This option suggests the formation of aldehydes, which is not expected in this reaction. The ether linkage remains intact.
- Option B: This option also suggests the formation of an aldehyde and methanol, which is not expected. The ether linkage remains intact.
- Option C: This option suggests the bromine adds to the carbon further away from the methoxy group, which violates Markovnikov's rule.

Key Concepts:
- Electrophilic addition
- Markovnikov's rule
- Effect of substituents on alkene reactivity

Memory Aids:
- Markovnikov's rule: "The rich get richer" (hydrogen adds to the carbon with more hydrogens).

Quick Tips:
- Identify the alkene and any substituents. Consider the electronic effects of the substituents when applying Markovnikov's rule.

49 Medium

49 of 64

Application +4 | -1

The compound

B\;

formed in the following sequence of reactions,

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}\xrightarrow{{\mathrm{PC}{\mathrm{l}_5}}}A\xrightarrow{{\mathrm{Alc}.\mathrm{\;NaOH}}}B

will be:

View answer and explanation

Correct option: B

Propene

Option B (Propene) is correct because the reaction sequence involves converting an alcohol to an alkyl halide, followed by elimination to form an alkene.

Solution approach

Correct Answer + Core Reason: Propene. The reaction sequence is alcohol to alkyl halide followed by elimination to form an alkene.

Step-by-Step Solution:

Step 1: Identify the starting material and reagents. 1-propanol reacts with $PCl_5$ followed by alcoholic NaOH.

Step 2: Recall the reaction of alcohols with $PCl_5$ . $PCl_5$ converts alcohols to alkyl halides.
$\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{OH} + \mathrm{PC}{\mathrm{l}_5} \rightarrow \mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{Cl} + \mathrm{POCl}_3 + \mathrm{HCl}$

Step 3: Recall the reaction of alkyl halides with alcoholic NaOH. Alcoholic NaOH promotes elimination reactions (E2) leading to the formation of alkenes.
$\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{Cl} \xrightarrow{{\mathrm{Alc}.\mathrm{\;NaOH}}} \mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{CH}_2 + \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{O}$

Step 4: Combine the two steps to get the overall reaction.
$\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}\xrightarrow{{\mathrm{PC}{\mathrm{l}_5}}} \mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{Cl}\xrightarrow{{\mathrm{Alc}.\mathrm{\;NaOH}}}{\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{CH}_2}$

Why other options are wrong:
- Option A (Propyne): Propyne is an alkyne, which is not formed in this reaction sequence.
- Option C (Propanal): Propanal is an aldehyde, which is not formed in this reaction sequence.
- Option D (Propane): Propane is an alkane, which is not formed in this reaction sequence. Elimination reactions favor the formation of alkenes.

Key Concepts:
- Alcohol reactions
- Alkyl halide reactions
- Elimination reactions (E2)

Memory Aids:
- $PCl_5$ replaces -OH with -Cl.
- Alcoholic NaOH favors elimination.

Quick Tips:
- Always consider the type of reaction (substitution vs. elimination) when using reagents like NaOH.

50 Medium

50 of 64

Application +4 | -1

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{CONa}

on reaction with

\mathrm{C}{\mathrm{H}_3}\mathrm{Br}

will give:

View answer and explanation

Correct option: D

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{COC}{\mathrm{H}_3}

The correct answer is D (

{{\left( {\mathrm{C}{\mathrm{H}_3}} \right)}_3}\mathrm{COC}{\mathrm{H}_3}}

) because tert-butoxide reacts with methyl bromide via an SN2 mechanism to form tert-butyl methyl ether.

Solution approach

Correct Answer + Core Reason: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{COC}{\mathrm{H}_3}$ . The reaction is a Williamson ether synthesis.

Step-by-Step Solution:

Step 1: The reaction is between tert-butoxide ( ${{\left( {\mathrm{C}{\mathrm{H}_3}} \right)}_3}\mathrm{CONa}$ ) and methyl bromide ( ${\mathrm{CH}_3}\mathrm{Br}$ ).

Step 2: The tert-butoxide acts as a nucleophile and attacks the methyl bromide, displacing the bromide ion. This is a Williamson ether synthesis.

Step 3: The product is tert-butyl methyl ether, which is represented as ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{COC}{\mathrm{H}_3}$ .

Why other options are wrong:
- **Option A ( ${{\left( {\mathrm{C}{\mathrm{H}_3}} \right)}_3}\mathrm{COC}{{\left( {\mathrm{C}{\mathrm{H}_3}} \right)}_3}}$ ): This would require two tert-butyl groups to combine, which is not possible with methyl bromide.
- Option B ( ${\mathrm{CH}_3}\mathrm{OCH}_3}$ ): This would require two methyl groups to combine, which is not possible with tert-butoxide.
- Option C ( ${\mathrm{CH}_3}\mathrm{CH}_2\mathrm{OCH}_2{\mathrm{CH}_3}}$ ): This would require two ethyl groups to combine, which is not possible with tert-butoxide and methyl bromide.

Key Concepts:
- Williamson ether synthesis
- SN2 reaction

Memory Aids:
- Bulky alkoxides favor elimination reactions, but methyl halides favor SN2.

Quick Tips:**
- Consider the steric hindrance of the reactants when predicting the product.

51 Medium

51 of 64

Understanding +4 | -1

Ethyl alcohol reacts with following to form a compound of fruity smell:

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

Option C (

\mathrm{CH}_3\mathrm{COOH}

) is correct because ethyl alcohol reacts with acetic acid to form ethyl acetate, an ester with a fruity smell.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH}_3\mathrm{COOH}$ . Ethyl alcohol reacts with acetic acid to form an ester, which has a fruity smell.

Step-by-Step Solution:

Step 1: Understand esterification.

Step 2: Esterification is the reaction between an alcohol and a carboxylic acid to form an ester and water.

Step 3: Ethyl alcohol ( $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$ ) reacts with acetic acid ( $\mathrm{CH}_3\mathrm{COOH}$ ) to form ethyl acetate ( $\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5$ ) and water.

Step 4: The reaction is: $\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + \mathrm{CH}_3\mathrm{COOH} \rightarrow \mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5 + \mathrm{H}_2\mathrm{O}$

Step 5: Esters typically have fruity smells.

Why other options are wrong:
- Option A: $\mathrm{PCl}_5$ : Reacts to form chloroethane, not a fruity smell.
- Option B: ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7} + {\mathrm{H}_2}\mathrm{SO}_4$ : Oxidizes ethanol to ethanoic acid, not a fruity smell directly.
- Option D: $\mathrm{CH}_3\mathrm{COC}{\mathrm{H}_3}$ : Acetone does not react with ethanol to form a fruity smell.

Key Concepts:
- Esterification
- Esters

Memory Aids:
- Esters have fruity smells.

Quick Tips:
- Remember the reaction between alcohols and carboxylic acids.

52 Medium

52 of 64

Understanding +4 | -1

Most viscous among the following is:

View answer and explanation

Correct option: D

Propane-1,2,3-triol

Option D (Propane-1,2,3-triol) is correct because viscosity increases with the number of hydroxyl groups due to increased hydrogen bonding.

Solution approach

Correct Answer + Core Reason: Propane-1,2,3-triol. More hydroxyl groups lead to stronger intermolecular hydrogen bonding and higher viscosity.

Step-by-Step Solution:

Step 1: Understand viscosity.

Step 2: Viscosity is a measure of a fluid's resistance to flow.

Step 3: Hydrogen bonding increases viscosity.

Step 4: Compare the number of hydroxyl groups in each option:
- A. Propan-1-ol: 1 OH group
- B. Propan-2-ol: 1 OH group
- C. Propane-1,2-diol: 2 OH groups
- D. Propane-1,2,3-triol: 3 OH groups

Step 5: Propane-1,2,3-triol (glycerol) has the most OH groups and therefore the highest viscosity.

Why other options are wrong:
- Option A: Propan-1-ol: Less viscous than propane-1,2,3-triol.
- Option B: Propan-2-ol: Less viscous than propane-1,2,3-triol.
- Option C: Propane-1, 2-diol: Less viscous than propane-1,2,3-triol.

Key Concepts:
- Viscosity
- Hydrogen bonding

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

Quick Tips:
- Consider the intermolecular forces when comparing viscosities.

53 Medium

53 of 64

Understanding +4 | -1

A simple method to remove peroxides from ethes is to treat them with an aqueous solution of

View answer and explanation

Correct option: A

KI

Option A (KI) is correct because KI reduces peroxides to ethers and itself gets oxidized to iodine.

Solution approach

Correct Answer + Core Reason: KI. Potassium iodide reduces peroxides in ethers to the corresponding ether and is itself oxidized to iodine.

Step-by-Step Solution:

Step 1: Identify the problem: Ethers can form explosive peroxides upon exposure to air.

Step 2: Removal of peroxides: Peroxides can be removed by reduction.

Step 3: KI as a reducing agent: KI reduces the peroxide to ether.
$\mathrm{R-O-O-R + 2KI + H_2O \rightarrow 2ROH + I_2 + 2KOH}$

Why other options are wrong:
- Option B: KCNS: Potassium thiocyanate is not a common reagent for removing peroxides.
- Option C: $\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}$ : Sodium thiosulfate can be used as a reducing agent, but KI is more commonly used for removing peroxides from ethers.
- Option D: $\mathrm{B}{\mathrm{r}_2}$ : Bromine would react with the ether and not selectively remove peroxides.

Key Concepts:
- Peroxide formation in ethers
- Reducing agents for peroxide removal

Memory Aids:
- KI is a good reducing agent.

Quick Tips:
- Remember that ethers can form peroxides and that KI is a common reagent for their removal.

54 Medium

54 of 64

Remembering +4 | -1

A solution of

Cr{O_5}

in amyl alcohol has a ______
colour

View answer and explanation

Correct option: A

Blue

Option A is correct because a solution of CrO5 in amyl alcohol has a blue color.

Solution approach

Correct Answer + Core Reason: Blue. A solution of $CrO_5$ in amyl alcohol has a blue color.

Step-by-Step Solution:

Step 1: Recall the color of $CrO_5$ in organic solvents.

Step 2: $CrO_5$ (chromium pentoxide) has a butterfly structure and is unstable.

Step 3: When dissolved in amyl alcohol or ether, it forms a blue-colored solution.

Why other options are wrong:
- Option B: Yellow is not the color of $CrO_5$ in amyl alcohol.
- Option C: Green is not the color of $CrO_5$ in amyl alcohol.
- Option D: Orange-Red is not the color of $CrO_5$ in amyl alcohol.

Key Concepts:
- Chromium pentoxide ( $CrO_5$ )
- Color of transition metal compounds

Memory Aids:
- $CrO_5$ in amyl alcohol = Blue

Quick Tips:
- Remember the colors of common transition metal compounds and their solutions.

55 Medium

55 of 64

Understanding +4 | -1

The reaction of

\mathrm{C}{\mathrm{H}_3}\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5}

with

\mathrm{HI}

gives:

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{I} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}

Option C (

\mathrm{C}{\mathrm{H}_3}\mathrm{I} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}

) is correct because HI cleaves the ether bond, with the smaller alkyl group forming the iodide and the larger forming the alcohol.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{I} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ . HI cleaves unsymmetrical ethers, forming alkyl iodide from the smaller alkyl group and alcohol from the larger.

Step-by-Step Solution:

Step 1: Identify the ether: $\mathrm{C}{\mathrm{H}_3}\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5}$ is an unsymmetrical ether.

Step 2: Reaction with HI: The ether will undergo cleavage at the C-O bond.
$\mathrm{CH_3OC_2H_5 + HI \rightarrow CH_3I + C_2H_5OH}$

Step 3: Products: Methyl iodide ( $\mathrm{C}{\mathrm{H}_3}\mathrm{I}$ ) and ethanol ( ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ ) are formed.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{H}_3}\mathrm{I}$ only: This is incorrect because the ethyl group also reacts to form ethanol.
- Option B: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ only: This is incorrect because the methyl group also reacts to form methyl iodide.
- Option D: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{I} + \mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ : This is incorrect because the smaller alkyl group (methyl) forms the iodide, and the larger alkyl group (ethyl) forms the alcohol.

Key Concepts:
- Ether cleavage by HI
- Reaction of unsymmetrical ethers

Memory Aids:
- Smaller alkyl group becomes iodide, larger becomes alcohol.

Quick Tips:
- Remember that HI cleaves ethers at the C-O bond. In unsymmetrical ethers, the smaller alkyl group usually forms the iodide.

56 Medium

56 of 64

Understanding +4 | -1

The toxicity order for

\mathrm{C}{\mathrm{H}_3}\mathrm{OH},\mathrm{\;}{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}

and

{\mathrm{C}_3}{\mathrm{H}_7}\mathrm{OH}

is:

View answer and explanation

Correct option: C

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} < {\mathrm{C}_3}{\mathrm{H}_7}\mathrm{OH} < C{\mathrm{H}_3}\mathrm{OH}

Option C (

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} < {\mathrm{C}_3}{\mathrm{H}_7}\mathrm{OH} < C{\mathrm{H}_3}\mathrm{OH}

) is correct because methanol is metabolized into formaldehyde and formic acid, which are highly toxic, while ethanol and propanol are less toxic in comparison.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} < {\mathrm{C}_3}{\mathrm{H}_7}\mathrm{OH} < C{\mathrm{H}_3}\mathrm{OH}$ . Methanol is more toxic due to its metabolites.

Step-by-Step Solution:

Step 1: Understand the metabolism of methanol, ethanol, and propanol in the body.

Step 2: Methanol is metabolized into formaldehyde and formic acid, which are highly toxic and can cause blindness and death.

Step 3: Ethanol is metabolized into acetaldehyde and acetic acid, which are less toxic than the metabolites of methanol.

Step 4: Propanol is metabolized into propanal and propanoic acid, which are also less toxic than the metabolites of methanol.

Step 5: Therefore, the toxicity order is: Ethanol < Propanol < Methanol.

Why other options are wrong:
- **Option A ( ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} < C{\mathrm{H}_3}\mathrm{OH} < {\mathrm{C}_3}{\mathrm{H}_7}\mathrm{OH}$ ): Incorrect; propanol is less toxic than methanol.
- Option B ( ${\mathrm{C}_3}{\mathrm{H}_7}\mathrm{OH} < {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} < C{\mathrm{H}_3}\mathrm{OH}$ ): Incorrect; ethanol is less toxic than propanol.
- Option D ( $\mathrm{C}{\mathrm{H}_3}\mathrm{OH} < {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} < {\mathrm{C}_3}{\mathrm{H}_7}\mathrm{OH}$ ): Incorrect; methanol is the most toxic among the three.

Key Concepts:
- Toxicity of alcohols
- Metabolism of alcohols

Memory Aids:
- Methanol = Most Toxic

Quick Tips:**
- Remember that methanol is more toxic than ethanol and propanol due to its metabolites.

57 Medium

57 of 64

Application +4 | -1

Phenol on treatment with diethyl sulphate in presence of NaOH gives

View answer and explanation

Correct option: A

Phenetole

Option A (Phenetole) is correct because phenol reacts with diethyl sulfate in the presence of NaOH via Williamson ether synthesis to form phenetole.

Solution approach

Correct Answer + Core Reason: Phenetole. Williamson ether synthesis.

Step-by-Step Solution:

Step 1: Understand the reaction between phenol, diethyl sulfate, and NaOH.

Step 2: NaOH deprotonates phenol to form sodium phenoxide.

Step 3: Sodium phenoxide reacts with diethyl sulfate via an SN2 mechanism, where the phenoxide ion attacks the ethyl group of diethyl sulfate, displacing the ethyl sulfate ion.

Step 4: The product of the reaction is phenetole (ethyl phenyl ether).

Step 5: The reaction is known as Williamson ether synthesis.

Why other options are wrong:
- Option B (Anisole): Anisole would be formed if the alkylating agent was methyl sulfate.
- Option C (Diphenyl ether): Diphenyl ether would require two phenyl groups attached to the oxygen.
- Option D (Diethyl ether): Diethyl ether would require two ethyl groups attached to the oxygen, which is not possible in this reaction.

Key Concepts:
- Williamson ether synthesis
- SN2 reaction

Memory Aids:
- Phenol + Diethyl sulfate + NaOH = Phenetole

Quick Tips:
- Recognize Williamson ether synthesis reactions.

58 Medium

58 of 64

Application +4 | -1

For the preparation ter-butylmethylether by Williamson’s method the correct choice of reagents is:

View answer and explanation

Correct option: D

Ter-butoxide and methylbromide

The correct answer is D (Ter-butoxide and methylbromide) because Williamson's method requires a primary alkyl halide to react with an alkoxide to avoid elimination reactions.

Solution approach

Correct Answer + Core Reason: Ter-butoxide and methylbromide. Williamson's method favors SN2 reactions with less hindered alkyl halides.

Step-by-Step Solution:

Step 1: Williamson's ether synthesis involves the reaction of an alkoxide with an alkyl halide to form an ether.

Step 2: To prepare tert-butyl methyl ether, we need a tert-butyl group and a methyl group attached to the oxygen.

Step 3: Using tert-butoxide and methyl bromide will result in the desired ether. The reaction is: ${{\left( {\mathrm{CH}_3} \right)}_3}\mathrm{CO}^ - \mathrm{Na}^ + + \mathrm{CH}_3\mathrm{Br} \to {\left( {\mathrm{CH}_3} \right)_3}\mathrm{COCH}_3 + \mathrm{NaBr}$

Step 4: Using methoxide and tert-butyl bromide would lead to elimination due to the bulky tert-butyl group.

Why other options are wrong:
- Option A (Methoxide and ter-butylbromide): This would lead to elimination (E2) as the major product due to the bulky tert-butyl group.
- Option B (Methanol and 2-bromobutane): This requires an acid catalyst and would not directly form the ether.
- Option C (2-butanol and methylbromide): This requires an acid catalyst and would not directly form the ether. Also, 2-butanol is not tert-butanol.

Key Concepts:
- Williamson ether synthesis
- SN2 vs E2 reactions

Memory Aids:
- Bulky groups favor elimination.

Quick Tips:
- Use primary alkyl halides in Williamson ether synthesis to minimize elimination.

59 Medium

59 of 64

Analyzing +4 | -1

Which of the following cannot be made by using Williamson’s synthesis

View answer and explanation

Correct option: D

Di – tert – butyl ether

Option D (Di – tert – butyl ether) is correct because Williamson's synthesis involves

S_N2

reactions, and bulky tert-butyl groups lead to steric hindrance, making the reaction very slow and favoring elimination.

Solution approach

Correct Answer + Core Reason: Di – tert – butyl ether. Steric hindrance prevents $S_N2$ reaction with bulky tert-butyl groups.

Step-by-Step Solution:

Step 1: Recall Williamson's ether synthesis: $R-O^- + R'-X \rightarrow R-O-R' + X^-$ .

Step 2: Williamson's synthesis proceeds via an $S_N2$ mechanism.

Step 3: $S_N2$ reactions are susceptible to steric hindrance.

Step 4: Bulky alkyl groups (like tert-butyl) hinder the nucleophilic attack in $S_N2$ reactions.

Step 5: Di – tert – butyl ether requires the reaction of two tert-butyl groups, which is highly sterically hindered.

Why other options are wrong:
- Option A: Methoxy benzene can be formed by reacting sodium methoxide with bromobenzene or methyl halide with sodium phenoxide.
- Option B: Benzyl p – Nitro phenyl ether can be formed by reacting benzyl halide with sodium p-nitrophenoxide or vice versa.
- Option C: Methyl tertiary butyl ether can be formed by reacting sodium methoxide with tert-butyl halide, although elimination is favored.

Key Concepts:
- Williamson's ether synthesis
- $S_N2$ reaction
- Steric hindrance

Memory Aids:
- $S_N2$ : Inversion, 1 step, favored by less substituted carbons.

Quick Tips:
- Consider steric hindrance when dealing with bulky alkyl groups in $S_N2$ reactions.

60 Medium

60 of 64

Understanding +4 | -1

Which is the correct statement?

View answer and explanation

Correct option: C

Phenol is more acidic than propanol

Option C (Phenol is more acidic than propanol) is correct because phenol's conjugate base (phenoxide ion) is stabilized by resonance, making it more acidic than propanol.

Solution approach

Correct Answer + Core Reason: Phenol is more acidic than propanol. The phenoxide ion is resonance stabilized.

Step-by-Step Solution:

Step 1: Understand the structures of phenol, benzyl alcohol, ethanol, propanol, and ethane.

Step 2: Consider the acidity of phenols and alcohols. Phenols are more acidic than aliphatic alcohols due to resonance stabilization of the phenoxide ion.

Step 3: Compare the acidity of phenol and propanol. Phenol is more acidic because its conjugate base (phenoxide ion) is stabilized by resonance.

Why other options are wrong:
- Option A (Benzyl alcohol is more acidic than phenol): Phenol is more acidic than benzyl alcohol because the negative charge in phenoxide ion is more delocalized than in benzyloxide ion.
- Option B (Ethanol is a powerful oxidizing agent): Ethanol is not a powerful oxidizing agent; it can be oxidized but is not itself a strong oxidizing agent.
- Option D (Ethane has high boiling point than ethanol): Ethanol has a higher boiling point than ethane due to hydrogen bonding.

Key Concepts:
- Acidity of phenols and alcohols
- Resonance stabilization
- Hydrogen bonding

Memory Aids:
- Resonance = Stability = Increased Acidity
- Hydrogen bonding increases boiling point.

Quick Tips:
- Compare the stability of the conjugate bases to determine relative acidity.

Your Progress

Alcohols & phenols & ethers practice · Medium

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice