Alcohols & phenols & ethers practice · Mixed difficulty

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

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

Option B ($C{H_2} = CH - C{H_2}OH$) is correct because $LiAlH_4$ reduces the carboxylic acid group to a primary alcohol without affecting the alkene.

Option A (Ether) is correct because sodium ethoxide reacts with ethyl chloride via the Williamson ether synthesis to form diethyl ether.

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

Option B ($\mathrm{C}{\mathrm{H}_3}-\mathrm{O}-\mathrm{C}{\mathrm{H}_3}$) is correct because ethers do not have acidic hydrogens and therefore do not react with sodium metal.

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

Option B is correct because both the assertion and reason are correct, but the reason (sodium methoxide being a strong nucleophile) doesn't fully explain why tert-butyl methyl ether isn't formed. The primary reason is steric hindrance around the tert-butyl group, which favors elimination over substitution.

Option C (Thioethers) is correct because dialkyl sulfides are also known as thioethers, analogous to ethers where oxygen is replaced by sulfur.

Option A ($R - OH + SOC{l_2} \to R - Cl + S{O_2} + HCl$) is correct because the reaction with thionyl chloride ($SOC{l_2}$) proceeds with retention of configuration and produces gaseous byproducts ($S{O_2}$ and $HCl$) that easily escape, leading to a cleaner product.

Option C (${B}_6 {H}_6$ and alk. ${H}_2 {O}_2$) is correct because this represents hydroboration-oxidation, which converts alkenes to alcohols with anti-Markovnikov addition, resulting in 1-propanol from propene.

Option C (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$) has the highest boiling point because it has stronger intermolecular hydrogen bonding compared to the other options.

Option C (S-butyl alcohol) is correct because it is the smallest alcohol with a chiral carbon atom, meaning it has four different groups attached to one carbon.

Option C ($\mathrm{C}_6 \mathrm{H}_5-\mathrm{CO}_2-\mathrm{CH}_2 \mathrm{C}_6 \mathrm{H}_5$) is correct because the ester is derived from benzoic acid and benzyl alcohol, both of which will be reduced to benzyl alcohol upon treatment with $\mathrm{LiAlH}_4$ followed by aqueous workup.

Option A ($10\%$) is correct because the percentage of ethanol formed during the fermentation of sugar molasses is approximately 10%.

Option A ($\mathrm{KMn}{\mathrm{O}_4}$) is correct because glycerol reacts vigorously with potassium permanganate, leading to a fire.

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.

Option B ($1/2$ mole of ${\mathrm{H}_2}$) is correct because 2 moles of sodium react with 2 moles of methanol to produce 1 mole of hydrogen gas.

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.

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.

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.

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

Option C is correct because primary and secondary alcohols can be oxidized to carboxylic acids, while tertiary alcohols are resistant to oxidation under these conditions.

Option D (If both Assertion and Reason are false) is correct because the addition of water to but-1-ene in acidic medium yields butan-2-ol, and the reaction proceeds via the formation of a secondary carbocation.

Option C is correct because ethanol has a boiling point of 78°C, a pleasant odor, and produces ethene upon reaction with concentrated sulfuric acid, which decolorizes bromine water and alkaline KMnO4.

Option A (${C_2}{H_5}OH$) is correct because ethanol reacts with $PCl_5$ to form ethyl chloride, which reacts with KCN to form ethyl cyanide, which hydrolyzes to propionic acid.

Option A (3) is correct because electron-withdrawing groups like $-\mathrm{NO}_2$, $-\mathrm{Br}$, and $-\mathrm{CHO}$ increase acidity.

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

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

Option B (III $>$ II $>$ IV $>$ I) is correct because the reactivity in Williamson's ether synthesis depends on the substrate's ability to undergo SN2 reaction. Allylic halides (III) are most reactive, followed by primary alkyl halides (II and IV), and then sterically hindered halides (I).