Alcohols & phenols & ethers practice · Hard

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 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 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 (${C_2}{H_5}OH > C{H_3}OC{H_3} > {C_2}{H_6}$) 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 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 D (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$) is correct because ethanol is a common solvent used to extract peppermint oil from plant sources.

Option A ($\mathrm{HI}$ and $\mathrm{AgN}{\mathrm{O}_3}$) is correct because HI cleaves methoxy groups to form methyl iodide, which then reacts with silver nitrate to form a precipitate of silver iodide.

Option C is correct because (CH3)3COH on dehydration gives (CH3)2C=CH2, which on ozonolysis gives acetone and formaldehyde.

Option D (All of these) is correct because ether, thiobarbiturates, and trichloromethane (chloroform) have all been used as anesthetic agents.

Option B (Glycerine trinitrate) is correct because glycerine trinitrate, also known as nitroglycerin, is the primary explosive component of dynamite.

Option B ((ii) and (iii)) is correct because cumene (isopropylbenzene) can be prepared by Friedel-Crafts alkylation of benzene with either 1-chloropropane or 2-chloropropane in the presence of AlCl3.

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 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.

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

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 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.