Alcohols & phenols & ethers practice · Mixed difficulty

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

Option A (Phenetole) is correct because sodium phenoxide reacts with ethyl iodide via Williamson ether synthesis to form phenetole (ethyl phenyl ether).

Option A (2) is correct because there are two optically active primary alcohols with the formula $C_5H_{12}O$: 2-methylbutan-1-ol and 3-methylbutan-2-ol. However, only 2-methylbutan-1-ol is a primary alcohol.

Option D ($100\mathrm{\% \;}{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$) is correct because absolute alcohol is defined as pure ethanol without any water content.

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.

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 $\mathrm{P}{\mathrm{I}_3}$.

Option B ($o -$nitrophenol) is correct because it exhibits intramolecular hydrogen bonding, which lowers its boiling point compared to other isomers.

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.

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.

Option C (Higher) is correct because thioethers have higher boiling points than ethers due to the larger size and greater polarizability of sulfur compared to oxygen, leading to stronger London dispersion forces.

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.

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

Option B (If both Assertion & reason are True but Reason is not a correct explanation of the Assertion) is correct because while both statements are true, the reason (p-Nitrophenol has intermolecular H-bonding) does not explain why m-Nitrophenol is less acidic than p-Nitrophenol. The difference in acidity is primarily due to the resonance and inductive effects of the nitro group at different positions.

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

Option A ($m$-nitrophenol) will show the highest pH value because it is the least acidic among the given nitrophenols, leading to a lower concentration of H+ ions and thus a higher pH.

Option D (2-methoxy propane) is correct because the methoxy group is attached to the second carbon of the propane chain.

Option C ($\mathrm{CHO}$—$\mathrm{CHO}$) is correct because glyoxal is a dialdehyde with the formula (CHO)₂.

Option A is correct because $\mathrm{PCl}_5$ readily replaces the hydroxyl group (OH) of alcohols with a chlorine atom.

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.

Option C (Ether) is correct because Williamson's synthesis is a method for preparing ethers by reacting an alkoxide with a haloalkane.

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

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.

Option B is correct because both the assertion and reason are true, but the reason doesn't explain why isopropyl alcohol is generally preferred over ethanol.

Option D (Alcohols) is correct because the Lucas test is used to differentiate between primary, secondary, and tertiary alcohols based on their reactivity with Lucas reagent (HCl and ZnCl₂).

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.

Option D (${C_2}{H_5}OH > C{H_3}OC{H_3} > {C_2}{H_6}$) is correct because alcohols have hydrogen bonding, ethers have dipole-dipole interactions, and alkanes have only London dispersion forces, with the strength of intermolecular forces determining the boiling point.

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.

Option A (${C_2}{H_5}OC{H_3}$) is correct because functional isomers have the same molecular formula but different functional groups. Propanol ($C{H_3}C{H_2}C{H_2}OH$) and ethyl methyl ether (${C_2}{H_5}OC{H_3}$) both have the formula ${C_3}{H_8}O$, but propanol is an alcohol and ethyl methyl ether is an ether.

Option B (${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{COOH}$) is correct because aromatic carboxylic acids have a high carbon to hydrogen ratio, leading to incomplete combustion and sooty flames.

Option D (On reduction it yields phenol) is correct because the reduction of benzaldehyde yields benzyl alcohol, not phenol.