Alcohols & phenols & ethers practice · Mixed difficulty

Option C ($\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{6} \mathrm{H}_{5}$) is correct because the C-O bond in diphenyl ether has partial double bond character due to resonance with the benzene rings, making it more difficult to cleave.

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.

Option A ($\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$) is correct because wood spirit is the common name for methanol.

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.

Option D (2, 4, 6-trinitrophenol) is correct because 2,4,6-trinitrophenol, also known as picric acid, is a well-known explosive due to the presence of multiple nitro groups.

Option D (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$) is correct because ethanol is a small, polar molecule capable of forming hydrogen bonds with water, making it miscible in all proportions.

Option C (O - Nitro phenol > P - chloro phenol > Phenol > O - cresol) is correct because the acidity of phenols is affected by the substituents. Electron-withdrawing groups increase acidity, while electron-donating groups decrease acidity. Ortho-nitrophenol is more acidic than para-nitrophenol due to intramolecular hydrogen bonding making it easier to lose a proton.

Option A (${1^ \circ }$ alcohol) is correct because aldehydes are reduced to primary alcohols by $\mathrm{Zn}/\mathrm{HCl}$.

Option B is correct because carbolic acid is a common name for phenol.

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

Option D (Comparatively inert) is correct because the oxygen atom in ethers is relatively unreactive due to the electron-donating nature of the alkyl groups attached to it, which reduces its electrophilicity.

Option A (Thio-alcohols) is correct because mercaptans are the common name for thio-alcohols, which are compounds with an -SH group.

Option A (Both the statements are true) is correct because the +I effect of the tertiary butyl group increases the electron density on the oxygen atom, making it a stronger Lewis base.

Option C (Benzoic acid) is correct because the reaction sequence involves reduction of phenol to benzene, followed by Friedel-Crafts alkylation to toluene, and finally oxidation to benzoic acid.

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.

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.

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.

Option B (If both Assertion & reason are True but Reason is not a correct explanation of the Assertion) is correct because while both the assertion and reason are true, the stabilization of the phenoxide ion by resonance is not the *only* reason for phenol's higher acidity compared to alcohols; inductive effects also play a role.

Option A (2-chloro-2-methylbutane) is correct because the reaction proceeds via a carbocation intermediate, and rearrangement occurs to form the more stable tertiary carbocation, which then reacts with chloride.

Option D ($C{H_3} - \mathop {\mathop {C}\limits^{\mathop {C{H_3}}\limits_|} }\limits_{\mathop {OH}\limits^| } - C{H_3}$) is correct because Lucas reagent reacts fastest with tertiary alcohols due to the formation of a more stable tertiary carbocation intermediate.

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

Option D (Tert-Butyl alcohol) is correct because branching in alcohols increases solubility in water due to reduced hydrophobic interactions.

Option C (${\mathrm{S}_\mathrm{N}}2$ mechanism) is correct because Williamson's synthesis proceeds through an SN2 mechanism, where an alkoxide ion attacks a primary alkyl halide.

Option A (${\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\mathrm{OH}\,\,\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_\mathrm{5}}\,\mathrm{ONa}\,{\mathrm{C}_\mathrm{2}}{\mathrm{H}_{\mathrm{4}\,\mathrm{ = }\,\mathrm{5}}}\,\mathrm{CI}$) is correct because ethanol reacts with Na to form sodium ethoxide and with PCl5 to form chloroethane, which then react to form diethyl ether via Williamson ether synthesis.

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

Option C is correct because 1-propanol and 2-propanol give different products upon oxidation with copper, which can be distinguished by Fehling's solution.

Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because acetophenone has alpha hydrogens and the carbonyl group is necessary for aldol condensation to occur.

Option A (Resonance of phenoxide ion) is correct because the resonance stabilization of the phenoxide ion contributes to the acidic character of phenol by delocalizing the negative charge.

Option C ($\mathrm{iPr}—\mathrm{O}—\mathrm{iPr}$) is correct because isopropyl ether is known to form peroxides more readily than other ethers due to the stability of the radical formed during the peroxide formation process.

Option B (1-hydroxy-1-phenylethane) is correct because the hydration of styrene follows Markovnikov's rule, where the hydroxyl group attaches to the more substituted carbon.