${\rm{CH}}$≡${\rm{CH}}\xrightarrow{{{{\rm{O}}_3}/{\rm{NaOH}}}}X\xrightarrow{{{\rm{Zn}}/{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}}}Y$ is:

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

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

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

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

Conc. ${\rm{HCI}}$ and anhydrous ${\rm{ZnC}}{{\rm{I}}_2}$

Conc. ${\rm{HNO}}$ and hydrous ${\rm{ZnC}}{{\rm{I}}_2}$

Conc. ${\rm{HCI}}$ and hydrous ${\rm{ZnC}}{{\rm{I}}_2}$

Conc. ${\rm{HNO}}$ and anhydrous ${\rm{ZnC}}{{\rm{I}}_2}$

ALCOHOLS & PHENOLS & ETHERS MCQs for NEET UG CHEMISTRY — Medium

1 Medium

1 of 64

Understanding +4 | -1

Physical properties of:

View answer and explanation

Correct option: A

Alcohols lie between alkanes and

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

Option A (Alcohols lie between alkanes and

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

) is correct because alcohols have intermediate polarity due to the presence of both alkyl and hydroxyl groups, resulting in physical properties between those of nonpolar alkanes and highly polar water.

Solution approach

Correct Answer + Core Reason: Alcohols lie between alkanes and ${\mathrm{H}_2}\mathrm{O}$ . Alcohols have properties intermediate between alkanes and water due to their structure.

Step-by-Step Solution:

Step 1: Understand the physical properties of alkanes, alcohols, and water.

Step 2: Alkanes are nonpolar and have low boiling points.

Step 3: Water is highly polar and has high boiling points due to strong hydrogen bonding.

Step 4: Alcohols have both a nonpolar alkyl group and a polar hydroxyl group, giving them intermediate properties.

Step 5: Therefore, the physical properties of alcohols lie between those of alkanes and water.

Why other options are wrong:
- Option B: ${\mathrm{H}_2}\mathrm{O}$ lie between alcohols and alkenes: Incorrect, water is more polar than both alcohols and alkenes.
- Option C: Alkenes lie between alcohols and ${\mathrm{H}_2}\mathrm{O}$ : Incorrect, alkenes are less polar than both alcohols and water.
- Option D: None of the above: Incorrect, the first option is correct.

Key Concepts:
- Physical properties of organic compounds
- Polarity
- Hydrogen bonding

Memory Aids:
- Remember the relative polarities of alkanes, alcohols, and water.

Quick Tips:
- Consider the intermolecular forces when comparing physical properties.

2 Medium

2 of 64

Analyzing +4 | -1

{CH}_3 {OC}_2 {H}_5

and

\left({CH}_3\right)_3 {C}-{OCH}_3

are treated with hydroiodic acid. The fragments obtained after reactions are

View answer and explanation

Correct option: A

{CH}_3 {I}+{HOC}_2 {H}_5 ;\left({CH}_3\right)_3 {Cl}+{HOCH}_3

Option A (

{CH}_3 {I}+{HOC}_2 {H}_5 ;\left({CH}_3\right)_3 {Cl}+{HOCH}_3

) is correct because HI cleaves ethers, with the iodide attacking the less hindered alkyl group. In the case of tert-butyl methyl ether, the reaction proceeds via

S_N1

mechanism due to the stability of the tert-butyl carbocation.

Solution approach

Correct Answer + Core Reason: ${CH}_3 {I}+{HOC}_2 {H}_5 ;\left({CH}_3\right)_3 {I}+{HOCH}_3$ . HI cleaves ethers, with iodide attacking the less hindered alkyl group. (Corrected the product of tert-butyl methyl ether cleavage)

Step-by-Step Solution:

Step 1: Recall that ethers are cleaved by hydroiodic acid ( $HI$ ).

Step 2: Consider the first ether: ${CH}_3 {OC}_2 {H}_5$ . The cleavage will occur at the C-O bonds. Since both alkyl groups are primary, the iodide will attack the less bulky methyl group, forming ${CH}_3 {I}$ and ${HOC}_2 {H}_5$ .

Step 3: Consider the second ether: $\left({CH}_3\right)_3 {C}-{OCH}_3$ . The cleavage will occur at the C-O bonds. Due to the tert-butyl group, the reaction proceeds via $S_N1$ mechanism, forming a stable tert-butyl carbocation. The iodide will then attack the tert-butyl carbocation, forming $\left({CH}_3\right)_3 {CI}$ and ${HOCH}_3$ .

Step 4: Write the products of the reactions.

Why other options are wrong:
- Option B: Incorrect products for both reactions.
- Option C: Incorrect products for both reactions. The tert-butyl group forms tert-butyl iodide, not tert-butyl alcohol.
- Option D: Incorrect products for the first reaction and the tert-butyl group forms tert-butyl iodide, not tert-butyl alcohol in the second reaction.

Key Concepts:
- Ether cleavage by hydroiodic acid
- $S_N1$ and $S_N2$ mechanisms
- Stability of carbocations

Memory Aids:
- $HI$ cleaves ethers to form alcohols and alkyl halides.

Quick Tips:
- Consider the steric hindrance of the alkyl groups when predicting the products of ether cleavage.

3 Medium

3 of 64

Understanding +4 | -1

Phenol,

p

-methylphenol,

m

-nitrophenol and

p

-nitrophenol follows order of increasing strength as

View answer and explanation

Correct option: B

p

-methylphenol, pheol,

m

-nitrophenol,

p

-nitrophenol

Option B (

p

-methylphenol, phenol,

m

-nitrophenol,

p

-nitrophenol) is correct because electron-donating groups decrease acidity, while electron-withdrawing groups increase acidity, and nitro groups are stronger electron-withdrawing groups than methyl groups are electron-donating.

Solution approach

Correct Answer + Core Reason: $p$ -methylphenol, phenol, $m$ -nitrophenol, $p$ -nitrophenol. Electron-donating groups decrease acidity, electron-withdrawing groups increase acidity.

Step-by-Step Solution:

Step 1: Understand the effect of substituents on the acidity of phenols.

Step 2: Electron-donating groups (e.g., methyl) decrease acidity by destabilizing the phenoxide ion.

Step 3: Electron-withdrawing groups (e.g., nitro) increase acidity by stabilizing the phenoxide ion.

Step 4: The nitro group is a stronger electron-withdrawing group than the methyl group is an electron-donating group.

Step 5: Therefore, the acidity order is: $p$ -methylphenol < phenol < $m$ -nitrophenol < $p$ -nitrophenol.

Why other options are wrong:
- **Option A (Phenol, $p$ -methylphenol, $p$ -nitrophenol, $m$ -nitrophenol):** Incorrect; $p$ -methylphenol is less acidic than phenol.
- **Option C ( $p$ -methylphenol, $m$ -nitrophenol, phenol, $p$ -nitrophenol):** Incorrect; phenol is more acidic than $p$ -methylphenol and $m$ -nitrophenol is more acidic than phenol.
- **Option D ( $m$ -nitrophenol, $p$ -nitrophenol, phenol, $p$ -methylphenol):** Incorrect; $p$ -methylphenol is the least acidic.

Key Concepts:
- Acidity of phenols
- Substituent effects
- Electron-donating groups
- Electron-withdrawing groups

Memory Aids:
- EDG decreases acidity
- EWG increases acidity

Quick Tips:
- Consider the electronic effects of substituents when determining the acidity of phenols.

4 Medium

4 of 64

Application +4 | -1

2 mole of ethanol are burnt. The amount of

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

obtained will be:

View answer and explanation

Correct option: C

176 g

Option C (176 g) is correct because 2 moles of ethanol produce 4 moles of CO2, which corresponds to 176 g.

Solution approach

Correct Answer + Core Reason: 176 g. The amount of CO2 is calculated based on the stoichiometry of the combustion reaction of ethanol.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the combustion of ethanol:
${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} + 3{\mathrm{O}_2} \rightarrow 2{\mathrm{CO}_2} + 3{\mathrm{H}_2}\mathrm{O}$

Step 2: From the balanced equation, 1 mole of ethanol produces 2 moles of CO2. Therefore, 2 moles of ethanol will produce 4 moles of CO2.

Step 3: Calculate the molar mass of CO2:\nMolar mass of CO2 = 12 (C) + 2 * 16 (O) = 12 + 32 = 44 g/mol

Step 4: Calculate the mass of 4 moles of CO2:\nMass = moles * molar mass = 4 moles * 44 g/mol = 176 g

Step 5: Final Answer: 176 g

Why other options are wrong:
- Option A (132 g): This corresponds to 3 moles of CO2, which is incorrect based on the stoichiometry.
- Option B (44 g): This corresponds to 1 mole of CO2, which is incorrect based on the stoichiometry.
- Option D (88 g): This corresponds to 2 moles of CO2, which is incorrect based on the stoichiometry.

Key Concepts:
- Stoichiometry
- Balanced chemical equations
- Molar mass

Memory Aids:
- Remember to balance the chemical equation before performing stoichiometric calculations.

Quick Tips:
- Always start with a balanced chemical equation.
- Pay attention to the mole ratios between reactants and products.

5 Medium

5 of 64

Analyzing +4 | -1

Assertion: Hydroquinone is more acidic than resorcinol.

Reason:

OH

shows

-I

effect

View answer and explanation

Correct option: D

If both the Assertion and Reason are incorrect.

Option D (If both the Assertion and Reason are incorrect) is correct because hydroquinone is less acidic than resorcinol, and the -I effect of OH is not the primary reason for the difference in acidity.

Solution approach

Correct Answer + Core Reason: If both the Assertion and Reason are incorrect. Hydroquinone is less acidic than resorcinol, and the $-I$ effect of $OH$ is not the primary reason.

Step-by-Step Solution:

Step 1: Define hydroquinone and resorcinol. Hydroquinone is 1,4-dihydroxybenzene, and resorcinol is 1,3-dihydroxybenzene.

Step 2: Consider the acidity of dihydroxybenzenes. The acidity depends on the stability of the conjugate base (diphenoxide ion).

Step 3: Analyze the assertion. Hydroquinone is *less* acidic than resorcinol due to intramolecular hydrogen bonding in hydroquinone which stabilizes it and makes it less likely to donate a proton. Therefore, the assertion is incorrect.

Step 4: Analyze the reason. The $OH$ group does exhibit a $-I$ (inductive) effect, but this effect is not the primary reason for the difference in acidity between hydroquinone and resorcinol. The primary reason is intramolecular hydrogen bonding. Therefore, the reason is also incorrect.

Step 5: Conclusion: Both the assertion and the reason are incorrect.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion: Both are incorrect.
- Option B: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion: Both are incorrect.
- Option C: If Assertion is correct but Reason is incorrect: Both are incorrect.

Key Concepts:
- Acidity of phenols
- Inductive effect ( $-I$ effect)
- Intramolecular hydrogen bonding

Memory Aids:
- Intramolecular hydrogen bonding can affect acidity.

Quick Tips:
- Consider both inductive and resonance effects when comparing acidity.
- Intramolecular hydrogen bonding can stabilize molecules and affect their properties.

6 Medium

6 of 64

Remembering +4 | -1

A diazonium chloride reacts with

\phi \mathrm{OH}

to give an azodye. The reaction is called

View answer and explanation

Correct option: C

Coupling

Option C is correct because the reaction of a diazonium chloride with phenol to form an azodye is called a coupling reaction.

Solution approach

Correct Answer + Core Reason: Coupling. The reaction of a diazonium chloride with phenol to form an azodye is called a coupling reaction.

Step-by-Step Solution:

Step 1: Recall the reactions of diazonium salts.

Step 2: Diazonium salts react with phenols and aromatic amines in a coupling reaction to form azo compounds.

Step 3: These azo compounds are often colored and are used as dyes.

Why other options are wrong:
- Option A: Diazotization is the process of converting an aromatic amine to a diazonium salt.
- Option B: Condensation is a reaction where two molecules combine with the loss of a small molecule.
- Option D: Reduction is the gain of electrons or a decrease in oxidation state.

Key Concepts:
- Diazonium salts
- Coupling reaction
- Azo dyes

Memory Aids:
- Diazonium + Phenol = Coupling -> Azo dye

Quick Tips:
- Remember the characteristic reactions of diazonium salts.

7 Medium

7 of 64

Remembering +4 | -1

The organic compound present in tincture of iodine is:

View answer and explanation

Correct option: A

Alcohol

Option A is correct because tincture of iodine is a solution of iodine in alcohol.

Solution approach

Correct Answer + Core Reason: Alcohol. Tincture of iodine is a solution of iodine in alcohol.

Step-by-Step Solution:

Step 1: Recall the composition of tincture of iodine.

Step 2: Tincture of iodine is a solution of iodine (I2) in ethanol (alcohol) and water.

Step 3: It is used as an antiseptic.

Why other options are wrong:
- Option B: CCl4 is carbon tetrachloride, a solvent.
- Option C: Acetone is a ketone solvent.
- Option D: CS2 is carbon disulfide, a solvent.

Key Concepts:
- Tincture of iodine
- Alcohol as a solvent

Memory Aids:
- Tincture of iodine = Iodine in Alcohol

Quick Tips:
- Remember the composition of common solutions.

8 Medium

8 of 64

Application +4 | -1

Which of the following compounds when heated with CO at 150

^ \circ C

and 500 atm pressure in presence of

\mathrm{B}{\mathrm{F}_3}

forms ethyl propionate?

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because diethyl ether reacts with CO in the presence of BF3 to form ethyl propionate.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5}$ . Diethyl ether reacts with CO to form ethyl propionate.

Step-by-Step Solution:

Step 1: Identify the reactants: The reactants are CO, diethyl ether ( ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5}$ ), and $\mathrm{B}{\mathrm{F}_3}$ .

Step 2: Reaction: Diethyl ether reacts with CO in the presence of $\mathrm{B}{\mathrm{F}_3}$ at 150 $^\circ C$ and 500 atm pressure to form ethyl propionate ( $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{COOC}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}$ ). This is a Koch-Haaf carboxylation reaction.

Step 3: Write the reaction:
${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5} + \mathrm{CO} \xrightarrow{{\mathrm{B}{\mathrm{F}_3}, 150^\circ C, 500 \text{{ atm}}}}\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{COOC}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}$

Why other options are wrong:
- Option A: Ethanol will not form ethyl propionate under these conditions.
- Option B: Dimethyl ether will not form ethyl propionate under these conditions.
- Option D: Methyl ethyl ether will not form ethyl propionate under these conditions.

Key Concepts:
- Koch-Haaf carboxylation reaction
- Reaction of ethers with CO

Memory Aids:
- Diethyl ether + CO $\rightarrow$ Ethyl propionate

Quick Tips:
- Remember the Koch-Haaf reaction conditions.

9 Medium

9 of 64

Understanding +4 | -1

The factor adversely affecting the process of fermentation is:

View answer and explanation

Correct option: B

High concentration of sugars

Option B (High concentration of sugars) is correct because high sugar concentrations can inhibit the growth and activity of the microorganisms responsible for fermentation.

Solution approach

Correct Answer + Core Reason: High concentration of sugars. High sugar concentrations can inhibit the fermentation process.

Step-by-Step Solution:

Step 1: Understand fermentation.

Step 2: Fermentation is the process of converting sugars to alcohol and carbon dioxide using microorganisms (e.g., yeast).

Step 3: Factors affecting fermentation:
- Low concentration of sugar: May slow down the process but not adversely affect it.
- High concentration of sugars: Can inhibit the growth of microorganisms due to osmotic stress.
- Presence of ammonium salts: Can act as nutrients for the microorganisms, promoting fermentation.
- Presence of air: Can inhibit anaerobic fermentation processes.

Step 4: High sugar concentrations can create an environment that is too concentrated for the microorganisms to thrive.

Why other options are wrong:
- Option A: Low concentration of sugar: Low sugar concentration would slow down the process but not adversely affect it as much as high concentration.
- Option C: Presence of ammonium salts: Ammonium salts generally promote fermentation by providing nutrients.
- Option D: Presence of air: The presence of air inhibits anaerobic fermentation, but the question asks for the factor adversely affecting the process, and high sugar concentration is a more direct adverse factor in many fermentation processes.

Key Concepts:
- Fermentation
- Factors affecting fermentation

Memory Aids:
- High sugar concentration can inhibit microbial growth.

Quick Tips:
- Consider the optimal conditions for fermentation.

10 Medium

10 of 64

Analyzing +4 | -1

Which of the following can be used to distinguish between phenol and cyclohexanol

View answer and explanation

Correct option: B

NaOH

Option B (

NaOH

) is correct because phenol reacts with NaOH to form sodium phenoxide, which is soluble in water, while cyclohexanol does not react.

Solution approach

Correct Answer + Core Reason: $NaOH$ . Phenol is more acidic than cyclohexanol and reacts with $NaOH$ to form a salt, while cyclohexanol does not react.

Step-by-Step Solution:

Step 1: Understand the properties of phenol and cyclohexanol. Phenol is acidic due to resonance stabilization of the phenoxide ion. Cyclohexanol is an alcohol and less acidic than phenol.

Step 2: Consider the reaction with $NaOH$ . Phenol reacts with $NaOH$ to form sodium phenoxide and water: $C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$ . Cyclohexanol does not react with $NaOH$ under normal conditions.

Step 3: Consider the reaction with $Na_2CO_3$ . Phenol is not acidic enough to react with $Na_2CO_3$ .

Step 4: Consider the reaction with $CH_3COCl$ . Both phenol and cyclohexanol will react with $CH_3COCl$ to form esters. Therefore, it cannot be used to distinguish between them.

Step 5: Conclusion: $NaOH$ can be used to distinguish between phenol and cyclohexanol.

Why other options are wrong:
- Option A: $N{a_2}C{O_3}$ : Phenol is not acidic enough to react with $Na_2CO_3$ to evolve $CO_2$ .
- Option C: $C{H_3}COCl$ : Both phenol and cyclohexanol will react with $CH_3COCl$ to form esters, so it cannot distinguish them.
- Option D: All: Not all options can distinguish between phenol and cyclohexanol.

Key Concepts:
- Acidity of phenols and alcohols
- Reactions of phenols and alcohols with bases and acyl chlorides

Memory Aids:
- Phenols are more acidic than alcohols.
- $NaOH$ reacts with phenols to form salts.

Quick Tips:
- Consider the relative acidity of the compounds when choosing a reagent to distinguish between them.

11 Medium

11 of 64

Remembering +4 | -1

Phenol have ____ acidic strength than alcohols:

View answer and explanation

Correct option: B

Greater

Option B is correct because phenol is more acidic than alcohols due to resonance stabilization of the phenoxide ion.

Solution approach

Correct Answer + Core Reason: Greater. Phenols have greater acidic strength than alcohols due to resonance stabilization of the phenoxide ion.

Step-by-Step Solution:

Step 1: Recall the relative acidity of phenols and alcohols.

Step 2: Phenols are more acidic than alcohols because the phenoxide ion is resonance stabilized, which delocalizes the negative charge and makes it more stable.

Step 3: Alcohols do not have this resonance stabilization, so they are less acidic.

Why other options are wrong:
- Option A: Phenols are more acidic, not less acidic, than alcohols.
- Option C: Phenols and alcohols do not have the same acidic strength.
- Option D: Phenols are more acidic than alcohols.

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

Memory Aids:
- Phenols > Alcohols (acidity)

Quick Tips:
- Remember the factors that affect the acidity of organic compounds.

12 Medium

12 of 64

Remembering +4 | -1

Alcohol is sometimes used in:

View answer and explanation

Correct option: C

Thermometers

Option C (Thermometers) is correct because alcohol, specifically ethanol, is commonly used in thermometers due to its low freezing point and thermal expansion properties.

Solution approach

Correct Answer + Core Reason: Thermometers. Alcohol is used in thermometers due to its low freezing point and thermal expansion.

Step-by-Step Solution:

Step 1: Consider the properties of alcohol.

Step 2: Recall that alcohol has a low freezing point and expands with temperature.

Step 3: Recognize that these properties make it suitable for use in thermometers.

Why other options are wrong:
- Option A: Baking powder uses chemical reactions for leavening, not alcohol.
- Option B: Paints may contain solvents, but not typically pure alcohol.
- Option D: Weighing involves measuring mass, and alcohol is not directly used in the process.

Key Concepts:
- Properties of alcohol
- Thermometers
- Thermal expansion

Memory Aids:
- Think of thermometers as needing a liquid that works well in \"cold\" temperatures, like alcohol.

Quick Tips:
- Consider the physical properties of substances when determining their uses.

13 Medium

13 of 64

Remembering +4 | -1

In the fermentation of sugar molasses, the percentage of ethanol formed is:

View answer and explanation

Correct option: A

10\mathrm{ }\%

Option A (

10\%

) is correct because the percentage of ethanol formed during the fermentation of sugar molasses is approximately 10%.

Solution approach

Correct Answer + Core Reason: $10\mathrm{ }\%$ . The percentage of ethanol formed in the fermentation of sugar molasses is around 10%.

Step-by-Step Solution:

Step 1: Recall the process of fermentation of sugar molasses.

Step 2: Understand that fermentation produces ethanol and carbon dioxide.

Step 3: Remember the approximate percentage of ethanol formed during this process.

Why other options are wrong:
- Option B: $40\mathrm{ }\%$ is too high for the ethanol percentage in sugar molasses fermentation.
- Option C: $95\mathrm{ }\%$ is the percentage of absolute alcohol, achieved after distillation.
- Option D: $70\mathrm{ }\%$ is also too high for the ethanol percentage in sugar molasses fermentation.

Key Concepts:
- Fermentation
- Ethanol production
- Sugar molasses

Memory Aids:
- Think of the process as producing a \"small\" amount of alcohol, around 10%.

Quick Tips:
- Remember the approximate percentages for common industrial processes.

14 Medium

14 of 64

Analyzing +4 | -1

In which of the following bond angles on

s{p^3}

-hybridized are not contracted due to lone pair of electron?

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the oxygen atom in dimethyl ether has two lone pairs, but the steric bulk of the methyl groups prevents significant contraction of the bond angle.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{OC}{\mathrm{H}_3}$ . Steric bulk of methyl groups prevents significant bond angle contraction.

Step-by-Step Solution:

Step 1: Recall that lone pairs on the central atom in a molecule cause bond angle contraction due to increased repulsion.

Step 2: Consider the number of lone pairs on the oxygen atom in each molecule:
- $\mathrm{O}{\mathrm{F}_2}$ : 2 lone pairs
- ${\mathrm{H}_2}\mathrm{O}$ : 2 lone pairs
- $\mathrm{C}{\mathrm{H}_3}\mathrm{OC}{\mathrm{H}_3}$ : 2 lone pairs
- $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ : 2 lone pairs

Step 3: In $\mathrm{O}{\mathrm{F}_2}$ and ${\mathrm{H}_2}\mathrm{O}$ , the bond angles are contracted due to lone pair repulsion.

Step 4: In $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ , the bond angle is slightly contracted due to lone pair repulsion.

Step 5: In $\mathrm{C}{\mathrm{H}_3}\mathrm{OC}{\mathrm{H}_3}$ , the steric bulk of the methyl groups counteracts the lone pair repulsion, preventing significant bond angle contraction.

Why other options are wrong:
- Option A: $\mathrm{O}{\mathrm{F}_2}$ has a contracted bond angle due to lone pair repulsion.
- Option B: ${\mathrm{H}_2}\mathrm{O}$ has a contracted bond angle due to lone pair repulsion.
- Option D: $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ has a slightly contracted bond angle due to lone pair repulsion.

Key Concepts:
- VSEPR theory
- Lone pair repulsion
- Bond angle contraction
- Steric hindrance

Memory Aids:
- Lone pairs repel more than bonding pairs.

Quick Tips:
- Consider both lone pair repulsion and steric hindrance when determining bond angles.

15 Medium

15 of 64

Application +4 | -1

Propane,

\mathrm{C}{\mathrm{H}_3}-\mathrm{CH} = \mathrm{C}{\mathrm{H}_2}

can be converted into 1-propanol by oxidation. Which set of reagents among the following is ideal to effect the conversion?

View answer and explanation

Correct option: B

{\mathrm{B}_2}{\mathrm{H}_6},\mathrm{\;}{\mathrm{H}_2}{\mathrm{O}_2}

Option B (

{\mathrm{B}_2}{\mathrm{H}_6},\mathrm{\;}{\mathrm{H}_2}{\mathrm{O}_2}

) is correct because it represents hydroboration-oxidation, which converts alkenes to alcohols with anti-Markovnikov addition.

Solution approach

Correct Answer + Core Reason: ${\mathrm{B}_2}{\mathrm{H}_6},\mathrm{\;}{\mathrm{H}_2}{\mathrm{O}_2}$ . Hydroboration-oxidation gives anti-Markovnikov addition of water to the alkene.

Step-by-Step Solution:

Step 1: Identify the starting material and desired product. Propene is converted to 1-propanol.

Step 2: Recall the hydroboration-oxidation reaction. ${\mathrm{B}_2}{\mathrm{H}_6}$ adds to the alkene in an anti-Markovnikov fashion, followed by oxidation with ${\mathrm{H}_2}{\mathrm{O}_2}$ to give the alcohol.

Step 3: Apply the reaction to the given alkene. Propene reacts with ${\mathrm{B}_2}{\mathrm{H}_6}$ to form a trialkylborane, which is then oxidized with ${\mathrm{H}_2}{\mathrm{O}_2}$ to give 1-propanol.

Step 4: Write the overall reaction:
$\mathrm{C}{\mathrm{H}_3}-\mathrm{CH} = \mathrm{C}{\mathrm{H}_2} \xrightarrow{{{\mathrm{B}_2}{\mathrm{H}_6},\mathrm{\;}{\mathrm{H}_2}{\mathrm{O}_2}}} \mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$

Why other options are wrong:
- **Option A ( ${\mathrm{H}_2}\mathrm{O}$ ): Acid-catalyzed hydration of propene would give 2-propanol (Markovnikov addition), not 1-propanol.
- Option C ( ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ ): Similar to option A, this would lead to Markovnikov addition and formation of 2-propanol.
- Option D (None of these): Option B is the correct answer.

Key Concepts:
- Hydroboration-oxidation
- Anti-Markovnikov addition
- Alkene reactions

Memory Aids:
- BH3 gives Anti-Markovnikov addition of H and OH.

Quick Tips:**
- Hydroboration-oxidation is the best method for anti-Markovnikov alcohol synthesis from alkenes.

16 Medium

16 of 64

Application +4 | -1

\mathrm{CH}

≡

\mathrm{CH}\xrightarrow{{{\mathrm{O}_3}/\mathrm{NaOH}}}X\xrightarrow{{\mathrm{Zn}/\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}}}Y

is:

View answer and explanation

Correct option: A

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

—

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

Option A (

{\mathrm{CH}_2}\mathrm{OH}

—

{\mathrm{CH}_2}\mathrm{OH}

) is correct because the reaction sequence involves oxidative cleavage of alkyne to glyoxal followed by reduction to ethylene glycol.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ — $\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ . Ozonolysis of alkyne followed by reduction gives glycol.

Step-by-Step Solution:

Step 1: Identify the starting material and reagents. Ethyne (acetylene) reacts with $O_3/NaOH$ followed by $Zn/CH_3COOH$ .

Step 2: Recall the reaction of alkynes with ozone. Ozonolysis of alkynes results in oxidative cleavage of the triple bond, forming dicarbonyl compounds.
$\mathrm{CH}≡\mathrm{CH} \xrightarrow{{{\mathrm{O}_3}/\mathrm{NaOH}}} \mathrm{CHO} - \mathrm{CHO}$ (Glyoxal)

Step 3: Recall the reaction of aldehydes with $Zn/CH_3COOH$ . This is a reducing agent that reduces aldehydes to alcohols.
$\mathrm{CHO} - \mathrm{CHO} \xrightarrow{{\mathrm{Zn}/\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}}} \mathrm{C}{\mathrm{H}_2}\mathrm{OH} - \mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ (Ethylene glycol)

Step 4: Combine the two steps to get the overall reaction.
$\mathrm{CH}≡\mathrm{CH}\xrightarrow{{{\mathrm{O}_3}/\mathrm{NaOH}}} \mathrm{CHO} - \mathrm{CHO}\xrightarrow{{\mathrm{Zn}/\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}}} \mathrm{C}{\mathrm{H}_2}\mathrm{OH} - \mathrm{C}{\mathrm{H}_2}\mathrm{OH}$

Why other options are wrong:
- **Option B ( ${\mathrm{CH}_3}\mathrm{CH}_2\mathrm{OH}$ ): This is ethanol, which is not formed in this reaction sequence.
- Option C ( ${\mathrm{CH}_3}\mathrm{COOH}$ ): This is acetic acid, which is not formed in this reaction sequence.
- Option D ( ${\mathrm{CH}_3}\mathrm{OH}$ ): This is methanol, which is not formed in this reaction sequence.

Key Concepts:
- Ozonolysis of alkynes
- Reduction of aldehydes

Memory Aids:
- Ozonolysis cleaves multiple bonds.
- Zn reduces aldehydes to alcohols.

Quick Tips:**
- Remember that ozonolysis of alkynes gives dicarbonyl compounds, which can be further reduced.

17 Medium

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Analyzing +4 | -1

Gives are (i) cyclohexanol; (ii) acetic acid; (iii) 2, 4, 6-trinitrophenol; and (iv) phenol. In these the order of decreasing acidic character will be:

View answer and explanation

Correct option: A

(iii) > (ii) > (iv) > (i)

Option A ((iii) > (ii) > (iv) > (i)) is correct because the acidity order is based on the stability of the conjugate base. 2,4,6-trinitrophenol is the most acidic due to the strong electron-withdrawing nitro groups, followed by acetic acid, then phenol, and finally cyclohexanol.

Solution approach

Correct Answer + Core Reason: (iii) > (ii) > (iv) > (i). The acidity order depends on the stability of the conjugate base.

Step-by-Step Solution:

Step 1: 2,4,6-Trinitrophenol (iii) is the most acidic due to the presence of three strong electron-withdrawing nitro groups, which stabilize the phenoxide ion through resonance and inductive effects.

Step 2: Acetic acid (ii) is a carboxylic acid, which is more acidic than phenol due to the greater electronegativity of the carbonyl oxygen and resonance stabilization of the carboxylate ion.

Step 3: Phenol (iv) is more acidic than cyclohexanol due to resonance stabilization of the phenoxide ion.

Step 4: Cyclohexanol (i) is an alcohol and is the least acidic among the given compounds.

Step 5: Therefore, the order of decreasing acidic character is: 2,4,6-trinitrophenol > acetic acid > phenol > cyclohexanol.

Why other options are wrong:
- Option B: Incorrect order. 2,4,6-trinitrophenol is more acidic than acetic acid.
- Option C: Incorrect order. Phenol is more acidic than cyclohexanol.
- Option D: Incorrect order. Acetic acid is more acidic than phenol.

Key Concepts:
- Acidity of Organic Compounds
- Effect of Substituents on Acidity
- Resonance Stabilization
- Inductive Effect

Memory Aids:
- EWG increases acidity; EDG decreases acidity.

Quick Tips:
- Consider the stability of the conjugate base when comparing acidity.

18 Medium

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Analyzing +4 | -1

Oxidation of allyl alcohol,

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

=

\mathrm{CH}

—

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

gives a mixture of oxalic acid and formic acid. If this oxidation is done in presence of bromine. One would expect only:

View answer and explanation

Correct option: D

Acrylic acid

Option D is correct because oxidation of allyl alcohol in the presence of bromine leads to the formation of acrylic acid.

Solution approach

Correct Answer + Core Reason: Acrylic acid. Bromine will react with the double bond first, and subsequent oxidation will lead to acrylic acid.

Step-by-Step Solution:

Step 1: Allyl alcohol is $\mathrm{C}{\mathrm{H}_2}=\mathrm{CH}$ — $\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ .

Step 2: In the presence of bromine, the double bond will undergo addition, forming $\mathrm{BrC}{\mathrm{H}_2}$ - $\mathrm{CHBr}$ — $\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ .

Step 3: Oxidation of this dibromo alcohol will lead to the formation of acrylic acid ( $\mathrm{CH_2=CHCOOH}$ ). The bromine atoms are likely to be eliminated during the oxidation process, reforming the double bond.

Step 4: If oxidation occurs without bromine, the double bond is cleaved, leading to a mixture of oxalic acid and formic acid.

Why other options are wrong:
- Option A: Oxalic acid is formed when the double bond is cleaved, which doesn't happen in the presence of bromine.
- Option B: Formic acid is formed when the double bond is cleaved, which doesn't happen in the presence of bromine.
- Option C: Succinic acid is not formed in this reaction.

Key Concepts:
- Oxidation of alcohols
- Addition of bromine to alkenes

Memory Aids:
- Bromine adds to double bonds.

Quick Tips:
- Consider the effect of reagents on different functional groups.

19 Medium

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Analyzing +4 | -1

Decreasing order of boiling points of

n

-pentanol (

A

),

n

-pentane (

B

), 3-pentanol (

C

) and 2,2-dimethyl propanol (

D

) is :

View answer and explanation

Correct option: A

A,\;C,D,\;B

Option A is correct because boiling point decreases with branching and increases with molecular weight and hydrogen bonding.

Solution approach

Correct Answer + Core Reason: $A,\;C,D,\;B$ . Boiling point increases with molecular weight and hydrogen bonding, and decreases with branching.

Step-by-Step Solution:

Step 1: Identify the compounds:
- A: $n$ -pentanol ( $\mathrm{CH_3CH_2CH_2CH_2CH_2OH}$ )
- B: $n$ -pentane ( $\mathrm{CH_3CH_2CH_2CH_2CH_3}$ )
- C: 3-pentanol ( $\mathrm{CH_3CH_2CHOHCH_2CH_3}$ )
- D: 2,2-dimethyl propanol ( $\mathrm{(CH_3)_3CCH_2OH}$ )

Step 2: Consider the factors affecting boiling point:
- Molecular weight: Higher molecular weight, higher boiling point.
- Hydrogen bonding: Alcohols have hydrogen bonding, increasing boiling point.
- Branching: Branching decreases boiling point.

Step 3: A and C and D are alcohols and have higher boiling points than B (an alkane) due to hydrogen bonding.

Step 4: A ( $n$ -pentanol) has the least branching among the alcohols, so it has the highest boiling point.

Step 5: C (3-pentanol) has less branching than D (2,2-dimethyl propanol), so it has a higher boiling point than D.

Step 6: Therefore, the decreasing order of boiling points is A > C > D > B.

Why other options are wrong:
- Option B: Incorrect order.
- Option C: Incorrect order.
- Option D: Incorrect order.

Key Concepts:
- Boiling point
- Intermolecular forces
- Hydrogen bonding
- Branching

Memory Aids:
- Hydrogen bonding increases boiling point.
- Branching decreases boiling point.

Quick Tips:
- Consider all factors affecting boiling point.

20 Medium

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Remembering +4 | -1

Chlorex which is a good solvent for aromatic impurities is:

View answer and explanation

Correct option: B

Dichlorodiethyl ether

Option B (Dichlorodiethyl ether) is correct because dichlorodiethyl ether (also known as Chlorex) is known for its solvent properties, particularly for aromatic compounds.

Solution approach

Correct Answer + Core Reason: Dichlorodiethyl ether. Chlorex is dichlorodiethyl ether.

Step-by-Step Solution:

Step 1: Recall the structure and properties of Chlorex.

Step 2: Chlorex is dichlorodiethyl ether.

Step 3: It is a good solvent for aromatic impurities.

Why other options are wrong:
- Option A: Dichloro dimethyl ether is not Chlorex.
- Option C: Mono chloro ether is not Chlorex.
- Option D: Diethyl ether is a common solvent, but not specifically known as Chlorex for aromatic impurities.

Key Concepts:
- Properties of solvents
- Structure of ethers

Memory Aids:
- Chlorex = Dichlorodiethyl ether

Quick Tips:
- Memorize the names and structures of common solvents.

21 Medium

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Analyzing +4 | -1

The correct order of acid strength of the following compounds is

V. Phenol

VI.

p

-cresol

VII.

m

-nitrophenol

VIII.

p

-nitrophenol

View answer and explanation

Correct option: B

IV > III > I > II

Option B (IV > III > I > II) is correct because electron-withdrawing groups increase acidity, and nitro groups are stronger electron-withdrawing groups than methyl groups.

Solution approach

Correct Answer + Core Reason: IV > III > I > II. The acidity is determined by the stability of the conjugate base, which is influenced by electron-withdrawing and electron-donating groups.

Step-by-Step Solution:

Step 1: Identify the compounds: Phenol (I), p-cresol (II), m-nitrophenol (III), p-nitrophenol (IV).

Step 2: Consider the effect of substituents on acidity. Electron-withdrawing groups (like nitro, $-NO_2$ ) increase acidity by stabilizing the phenoxide ion through resonance and inductive effects. Electron-donating groups (like methyl, $-CH_3$ ) decrease acidity by destabilizing the phenoxide ion.

Step 3: Compare the acidity of nitrophenols. Nitro group at para position has stronger effect than at meta position due to resonance.

Step 4: Arrange the compounds in order of decreasing acidity: p-nitrophenol > m-nitrophenol > phenol > p-cresol.

Step 5: Final Answer: IV > III > I > II

Why other options are wrong:
- Option A: IIII > II > I > IV: Incorrect order. p-cresol is the least acidic due to the electron-donating methyl group, and nitrophenols are more acidic than phenol.
- Option C: II > IV > I > III: Incorrect order. p-cresol is the least acidic, and m-nitrophenol is less acidic than p-nitrophenol.
- Option D: I > II > IV > III: Incorrect order. Phenol is less acidic than both nitrophenols, and p-cresol is the least acidic.

Key Concepts:
- Acidity of phenols
- Effect of substituents on acidity (electron-withdrawing and electron-donating groups)
- Resonance and inductive effects

Memory Aids:
- Electron-withdrawing groups increase acidity.
- Electron-donating groups decrease acidity.

Quick Tips:
- Consider the stability of the conjugate base when comparing acidity.
- Nitro groups are strong electron-withdrawing groups, while alkyl groups are electron-donating.

22 Medium

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Remembering +4 | -1

In fermentation by zymase, alcohol and

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

are obtained from

View answer and explanation

Correct option: B

Glucose

Option B is correct because glucose is fermented by zymase to produce alcohol and carbon dioxide.

Solution approach

Correct Answer + Core Reason: Glucose. Zymase ferments glucose into alcohol and carbon dioxide.

Step-by-Step Solution:

Step 1: Recall the fermentation process.

Step 2: Zymase is an enzyme complex that catalyzes the fermentation of sugars.

Step 3: Glucose is fermented by zymase to produce ethanol and carbon dioxide.

Why other options are wrong:
- Option A: Invert sugar is a mixture of glucose and fructose, but glucose is the direct substrate for zymase.
- Option C: Fructose can be fermented, but glucose is the more common substrate.
- Option D: While invert sugar contains glucose and fructose, glucose is the direct substrate for zymase.

Key Concepts:
- Fermentation
- Zymase
- Glucose

Memory Aids:
- Glucose -> Zymase -> Alcohol + CO2

Quick Tips:
- Remember the substrates and products of fermentation.

23 Medium

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Remembering +4 | -1

Purity of ether before using it as anaesthetic agent is tested by:

View answer and explanation

Correct option: A

\mathrm{KI} +

starch

Option A (

\mathrm{KI} +

starch ) is correct because the presence of peroxides in ether, which can form upon exposure to air, is detected by reaction with KI, liberating iodine which turns starch blue.

Solution approach

Correct Answer + Core Reason: $\mathrm{KI} +$ starch. This tests for peroxides in ether.

Step-by-Step Solution:

Step 1: Explain peroxide formation. Ethers can form explosive peroxides upon exposure to air.

Step 2: Describe the test. The presence of peroxides is detected by adding KI and starch.

Step 3: Explain the observation. Peroxides oxidize KI to iodine, which reacts with starch to produce a blue color.

Why other options are wrong:
- Option B: $\mathrm{CuS}{\mathrm{O}_4}$ is not used to test for peroxides in ether.
- Option C: ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ is not used to test for peroxides in ether.
- Option D: The KI + starch test is a standard method.

Key Concepts:
- Peroxide formation in ethers
- Detection of peroxides

Memory Aids:
- KI + starch turns blue in the presence of peroxides.

Quick Tips:
- Remember that ethers can form peroxides, and the KI + starch test is used to detect them.

24 Medium

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Understanding +4 | -1

Which one has highest boiling point?

View answer and explanation

Correct option: C

Butan-1-ol

Option C (Butan-1-ol) is correct because alcohols have higher boiling points than alkanes and alkenes of comparable molecular weight due to hydrogen bonding.

Solution approach

Correct Answer + Core Reason: Butan-1-ol. Alcohols have higher boiling points due to hydrogen bonding.

Step-by-Step Solution:

Step 1: Identify the compounds and their functional groups: Ethane (alkane), Ethene (alkene), Butan-1-ol (alcohol), Pentane (alkane).

Step 2: Consider the intermolecular forces present in each compound: Ethane and Pentane (van der Waals forces), Ethene (van der Waals forces), Butan-1-ol (hydrogen bonding and van der Waals forces).

Step 3: Hydrogen bonding is a stronger intermolecular force than van der Waals forces.

Step 4: Butan-1-ol exhibits hydrogen bonding, while the other compounds only exhibit van der Waals forces. Therefore, Butan-1-ol has the highest boiling point.

Why other options are wrong:
- Option A (Ethane): Ethane is an alkane with only van der Waals forces and a low molecular weight, resulting in a low boiling point.
- Option B (Ethene): Ethene is an alkene with only van der Waals forces and a low molecular weight, resulting in a low boiling point.
- Option D (Pentane): Pentane is an alkane with only van der Waals forces. While it has a higher molecular weight than ethane and ethene, it still has a lower boiling point than butan-1-ol due to the absence of hydrogen bonding.

Key Concepts:
- Boiling point
- Intermolecular forces
- Hydrogen bonding
- Van der Waals forces

Memory Aids:
- Hydrogen bonding increases boiling point significantly.

Quick Tips:
- Look for hydrogen bonding when comparing boiling points.

25 Medium

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Analyzing +4 | -1

A compound

X

with molecular formula

{\mathrm{C}_3}{\mathrm{H}_8}\mathrm{O}

can be oxidised to a compound

Y

with the molecular formula

{\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_2}

.

X

is most likely to be:

View answer and explanation

Correct option: A

Primary alcohol

Option A (Primary alcohol) is correct because primary alcohols can be oxidized to carboxylic acids, while secondary alcohols are oxidized to ketones, and aldehydes/ketones cannot be further oxidized to carboxylic acids with the same number of carbon atoms.

Solution approach

Correct Answer + Core Reason: Primary alcohol. Primary alcohols can be oxidized to carboxylic acids with the same number of carbon atoms.

Step-by-Step Solution:

Step 1: The molecular formula of X is $C_3H_8O$ , which indicates an alcohol or ether.

Step 2: The molecular formula of Y is $C_3H_6O_2$ , which indicates a carboxylic acid.

Step 3: Primary alcohols can be oxidized to carboxylic acids with the same number of carbon atoms.
$RCH_2OH \xrightarrow{{\text{Oxidation}}} RCOOH$

Step 4: Secondary alcohols are oxidized to ketones.

Step 5: Aldehydes can be oxidized to carboxylic acids, but the question states that X is oxidized to Y, not that an aldehyde is oxidized to Y.

Step 6: Ketones cannot be oxidized to carboxylic acids with the same number of carbon atoms without breaking carbon-carbon bonds.

Step 7: Therefore, X is most likely a primary alcohol.

Why other options are wrong:
- Option B: Secondary alcohols are oxidized to ketones, not carboxylic acids.
- Option C: Aldehydes can be oxidized to carboxylic acids, but the question specifies that X is oxidized to Y.
- Option D: Ketones cannot be oxidized to carboxylic acids with the same number of carbon atoms without breaking carbon-carbon bonds.

Key Concepts:
- Oxidation of Alcohols
- Oxidation of Aldehydes and Ketones
- Functional Groups

Memory Aids:
- Primary alcohol $\rightarrow$ Aldehyde $\rightarrow$ Carboxylic acid
- Secondary alcohol $\rightarrow$ Ketone

Quick Tips:
- Remember the oxidation products of different types of alcohols.

26 Medium

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Analyzing +4 | -1

An unknown compound dissolves in

{H}_2 {SO}_4

, but does not decolourise

{Br}_2

water and does not react with

{Na}

. Which of the following classes of molecules would behave in this manner?

View answer and explanation

Correct option: D

Ether

Option D (Ether) is correct because ethers dissolve in strong acids like

H_2SO_4

due to protonation of the oxygen atom, but they don't react with

Br_2

water (unlike alkenes/phenols) or Na (unlike alcohols/phenols).

Solution approach

Correct Answer + Core Reason: Ether. Ethers dissolve in ${H}_2 {SO}_4$ but do not react with ${Br}_2$ water or ${Na}$ .

Step-by-Step Solution:

Step 1: Analyze the given information: The unknown compound dissolves in ${H}_2 {SO}_4$ , does not decolourise ${Br}_2$ water, and does not react with ${Na}$ .

Step 2: Consider the reactions of each class of molecules:
- Alkene: Reacts with ${Br}_2$ water (decolourises it).
- Alcohol: Reacts with ${Na}$ to produce hydrogen gas.
- Phenol: Reacts with ${Na}$ and decolourises ${Br}_2$ water.
- Ether: Dissolves in ${H}_2 {SO}_4$ due to protonation but does not react with ${Br}_2$ water or ${Na}$ .

Step 3: Match the given information with the properties of each class of molecules.

Step 4: Ethers fit the given criteria.

Why other options are wrong:
- Option A: Alkenes decolourise ${Br}_2$ water.
- Option B: Alcohols react with ${Na}$ .
- Option C: Phenols react with both ${Na}$ and ${Br}_2$ water.

Key Concepts:
- Reactions of alkenes, alcohols, phenols, and ethers
- Acidity and basicity

Memory Aids:
- Alkenes + ${Br}_2$ water = decolourisation
- Alcohols + ${Na}$ = ${H}_2$ gas

Quick Tips:
- Know the characteristic reactions of different functional groups.

27 Medium

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Analyzing +4 | -1

Assertion: Anisole reacts with

HI

to give phenol and chloromethane.

Reason: The

-OCH_3

group in anisole enriches the electron density in the ring through Resonance.

View answer and explanation

Correct option: B

Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.

Option B is correct because both the assertion and reason are correct, but the reason (resonance) doesn't directly explain why anisole reacts with HI to give phenol and methyl iodide (not chloromethane). The reaction proceeds via cleavage of the ether bond by HI.

Solution approach

Correct Answer + Core Reason: Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. HI cleaves the ether bond, and resonance explains the stability of anisole.

Step-by-Step Solution:

Step 1: Analyze the assertion: Anisole reacts with $HI$ to give phenol and methyl iodide (corrected from chloromethane). This is a known reaction.

Step 2: Analyze the reason: The $-OCH_3$ group in anisole enriches the electron density in the ring through Resonance. This is also a correct statement.

Step 3: Determine if the reason explains the assertion: The reaction of anisole with $HI$ involves the cleavage of the ether bond. The mechanism involves protonation of the oxygen atom followed by nucleophilic attack by $I^-$ . The resonance effect of the $-OCH_3$ group does not directly explain this cleavage. The reaction proceeds via $S_N1$ or $S_N2$ mechanism depending on the conditions. Since methyl is a primary alkyl group, $S_N2$ is favored, leading to methyl iodide.

Step 4: Conclude that both statements are correct, but the reason is not the correct explanation.

Why other options are wrong:
- Option A: The reason is not the correct explanation of the assertion.
- Option C: The reason is a correct statement.
- Option D: The assertion is a correct statement (after correcting chloromethane to methyl iodide).

Key Concepts:
- Electrophilic aromatic substitution
- Resonance
- Ether cleavage by $HI$
- $S_N1$ and $S_N2$ reactions

Memory Aids:
- $HI$ cleaves ethers to form alcohols and alkyl halides.

Quick Tips:
- When analyzing assertion-reason questions, consider whether the reason directly explains the mechanism of the reaction in the assertion.

28 Medium

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Remembering +4 | -1

The main product obtained from phenol with

{PCl}_5

is

View answer and explanation

Correct option: D

Triphenyl phosphate

Option D is correct because phenol reacts with PCl5 to form triphenyl phosphate.

Solution approach

Correct Answer + Core Reason: Triphenyl phosphate. Phenol reacts with PCl5 to give triphenyl phosphate as the main product.

Step-by-Step Solution:

Step 1: Recall the reaction of phenol with PCl5.

Step 2: Phenol reacts with PCl5 to form triphenyl phosphate ( (C6H5O)3PO ).

Why other options are wrong:
- Option A: BHC (Benzene hexachloride) is formed by the reaction of benzene with chlorine in the presence of sunlight.
- Option B: Hexachlorobenzene is formed by chlorination of benzene under specific conditions.
- Option C: Chlorobenzene can be formed by reaction of benzene with chlorine in presence of Lewis acid catalyst.

Key Concepts:
- Reaction of phenol with PCl5
- Triphenyl phosphate

Memory Aids:
- Phenol + PCl5 = Triphenyl phosphate

Quick Tips:
- Remember the reactions of alcohols and phenols with PCl5.

29 Medium

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Remembering +4 | -1

Lucas reagent is

View answer and explanation

Correct option: A

Conc.

\mathrm{HCI}

and anhydrous

\mathrm{ZnC}{\mathrm{I}_2}

Option A (Conc.

\mathrm{HCI}

and anhydrous

\mathrm{ZnC}{\mathrm{I}_2}

) is correct because Lucas reagent is a solution of concentrated hydrochloric acid and anhydrous zinc chloride.

Solution approach

Correct Answer + Core Reason: Conc. $\mathrm{HCI}$ and anhydrous $\mathrm{ZnC}{\mathrm{I}_2}$ . Lucas reagent is a solution of concentrated hydrochloric acid and anhydrous zinc chloride.

Step-by-Step Solution:

Step 1: Recall the composition of Lucas reagent.

Step 2: Remember that it consists of concentrated hydrochloric acid (HCl) and anhydrous zinc chloride ( $\mathrm{ZnC}{\mathrm{I}_2}$ ).

Why other options are wrong:
- Option B: Conc. $\mathrm{HNO}$ and hydrous $\mathrm{ZnC}{\mathrm{I}_2}$ is incorrect because Lucas reagent uses HCl, not nitric acid, and anhydrous $\mathrm{ZnC}{\mathrm{I}_2}$ .
- Option C: Conc. $\mathrm{HCI}$ and hydrous $\mathrm{ZnC}{\mathrm{I}_2}$ is incorrect because Lucas reagent uses anhydrous $\mathrm{ZnC}{\mathrm{I}_2}$ .
- Option D: Conc. $\mathrm{HNO}$ and anhydrous $\mathrm{ZnC}{\mathrm{I}_2}$ is incorrect because Lucas reagent uses HCl, not nitric acid.

Key Concepts:
- Lucas reagent
- Composition of Lucas reagent

Memory Aids:
- Think of Lucas reagent as having \"Chlorine\" in it (HCl and ZnCl2).

Quick Tips:
- Memorize the composition of important reagents.

30 Medium

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Understanding +4 | -1

Which of the following compounds is resistant to nucleophilic attack by hydroxyl ions?

View answer and explanation

Correct option: C

Diethyl ether

Option C (Diethyl ether) is correct because ethers lack a good leaving group and are not susceptible to nucleophilic attack by hydroxyl ions under normal conditions.

Solution approach

Correct Answer + Core Reason: Diethyl ether. Ethers are generally unreactive towards nucleophilic attack due to the strong C-O bond and lack of a good leaving group.

Step-by-Step Solution:

Step 1: Nucleophilic attack: Hydroxyl ions are strong nucleophiles.

Step 2: Reactivity of compounds:
- Acetamide: Can undergo nucleophilic acyl substitution.
- Methyl acetate: Can undergo nucleophilic acyl substitution.
- Diethyl ether: Relatively unreactive towards nucleophiles.
- Acetonitrile: Can undergo nucleophilic addition to the nitrile group under specific conditions.

Step 3: Resistance to nucleophilic attack: Diethyl ether is the most resistant due to the strong C-O bond and lack of a good leaving group.

Why other options are wrong:
- Option A: Acetamide: Can undergo nucleophilic acyl substitution at the carbonyl carbon.
- Option B: Methyl acetate: Can undergo nucleophilic acyl substitution at the carbonyl carbon.
- Option D: Acetonitrile: The carbon atom of the nitrile group is electrophilic and can be attacked by nucleophiles under specific conditions.

Key Concepts:
- Nucleophilic attack
- Reactivity of functional groups

Memory Aids:
- Ethers are generally unreactive.

Quick Tips:
- Consider the leaving group ability when determining susceptibility to nucleophilic attack.

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