Alcohols Phenols And Ethers – NEET UG ChemistryPractice Questions & PYQs

Aliphatic primary amines like $\mathrm{CH_3(CH_2)_3NH_2}$ undergo the carbylamine reaction with chloroform and alcoholic potassium hydroxide to yield foul-smelling isocyanides, while terminal alkynes react with ammoniacal $\mathrm{AgNO_3}$ to form characteristic white precipitates. Esters such as $\mathrm{CH_3CH_2COOCH_3}$ undergo alkaline hydrolysis to produce carboxylate salts and alcohols, and secondary alcohols like $\mathrm{CH_3CH(OH)CH_3}$ react with Lucas reagent to form cloudiness within five minutes due to moderate carbocation stability.

Acid strength directly correlates with the stability of the resulting conjugate base, which is enhanced by electron-withdrawing substituents and resonance delocalization. Aliphatic alcohols like cyclohexanol are the least acidic because alkyl groups increase electron density on the oxygen atom and destabilize the alkoxide ion. Phenol displays greater acidity than cyclohexanol due to aromatic ring resonance, and substituting it with three strongly electron-withdrawing nitro groups yields 2,4,6-trinitrophenol, which surpasses carboxylic acids in proton-donating ability. Acetic acid sits between picric acid and regular phenol because its carbonyl oxygen exerts a powerful electron-withdrawing effect that stabilizes the carboxylate anion more effectively than a standard phenoxide ion.

The iodoform test is successfully given by compounds containing either a methyl ketone group or a secondary alcohol with a $\mathrm{CH_3CH(OH)-}$ structural unit, along with ethanol which readily oxidizes to a methyl carbonyl intermediate. Examining the given choices, ethanol, acetone, and propan-2-ol all possess these necessary reactive sites capable of reacting with iodine and sodium hydroxide to precipitate iodoform. In contrast, methanol contains only a single carbon atom and lacks the required methyl group attached to the functional carbon, rendering it completely unresponsive to this halogen-based oxidation test.

The reaction of polyhydric alcohols with excess hydroiodic acid involves the sequential substitution of hydroxyl groups by halogen atoms followed by elimination and addition steps. When glycerol is subjected to this treatment, its three hydroxyl groups are initially replaced to form an unstable triiodo intermediate. This species spontaneously eliminates an iodine molecule to produce allyl iodide, which then adds hydrogen iodide across its double bond and rearranges to form 2-iodopropane as the ultimate stable product.

The acidity of a compound relies on how easily it releases a proton, a process directly enhanced by electron-withdrawing groups that polarize the $-\text{OH}$ bond. In phenol, the directly attached phenyl ring acts as an electron-withdrawing group via resonance, facilitating the liberation of $\text{H}^+$. Conversely, other structures feature intervening $-\text{CH}_2-$ units that dampen this electron-withdrawing influence or contain alkyl groups exerting a $+\text{I}$ effect to reduce $-\text{OH}$ bond polarity, thereby lowering acidic strength.

Substituents on an aromatic ring dictate acidity by either intensifying or dispersing the negative charge of the resulting phenoxide ion through inductive and resonance mechanisms. Methyl phenol contains an electron-donating methyl group that increases electron density and destabilizes the conjugate base, rendering it less acidic than unsubstituted phenol. Conversely, nitro groups act as strong electron-withdrawing substituents that stabilize the phenoxide ion, thereby enhancing proton release. Because the para position permits direct resonance delocalization into the nitro substituent, para-nitrophenol achieves greater stabilization and higher acidity than meta-nitrophenol.

The reduction of phenol with zinc dust yields benzene ($X$), which undergoes Friedel-Crafts alkylation with methyl chloride in the presence of anhydrous aluminum chloride to form toluene ($Y$). Subsequent oxidation of toluene using alkaline potassium permanganate converts the methyl side chain into a carboxylic acid group, ultimately producing benzoic acid ($Z$).

Periodic acid specifically cleaves vicinal diols by breaking the carbon-carbon bond between adjacent hydroxyl-bearing carbons and oxidizing each hydroxyl group into a carbonyl group. Because ethylene glycol possesses terminal adjacent hydroxyl groups, this oxidative cleavage splits the molecule symmetrically into two identical fragments. Consequently, heating $\mathrm{HOCH_2\cdot CH_2OH}$ with $\mathrm{HIO}_4$ successfully yields two molecules of formaldehyde, expressed as $2\mathrm{HCHO}$.

The cleavage of asymmetric ethers by hydrogen iodide involves protonation of the ether oxygen followed by a nucleophilic substitution pathway. Because the iodide ion targets the less sterically hindered alkyl component, reaction of $\mathrm{CH}_3-\overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{CH}}}-\mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3$ with heated $\mathrm{HI}$ selectively cleaves the carbon-oxygen bond next to the ethyl group. This process results in the formation of $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{I}$ and $\mathrm{CH}_3-\overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{CH}}} -\mathrm{CH}_2-\mathrm{OH}$.

The cleavage of unsymmetrical alkyl aryl or dialkyl ethers by hydrogen iodide is governed by steric hindrance and carbocation stability considerations during nucleophilic attack. Because the methyl group offers minimal steric resistance compared to the bulky isopropyl group, the incoming iodide ion preferentially attacks the methyl carbon via an associative pathway. Consequently, the reaction yields methyl iodide alongside isopropyl alcohol as the major organic products.

Aliphatic monohydric alcohols are oxygenated derivatives of alkanes where a terminal hydrogen atom is substituted by a hydroxyl functional group. Because the parent alkane series follows the general stoichiometry $\mathrm{C}_n\mathrm{H}_{2n+2}$, introducing a single oxygen atom alongside the hydroxyl hydrogen results in the molecular representation $\mathrm{C}_n\mathrm{H}_{2n+2}\mathrm{O}$. This expression accurately accounts for all acyclic saturated alcohols, spanning from methanol where $n=1$ up to higher homologues while maintaining single-bonded carbon chains.

The reaction of a Grignard reagent with ethylene oxide initiates through a nucleophilic ring-opening mechanism where the carbanion equivalent attacks one of the equivalent, unhindered carbons of the strained three-membered cyclic ether. This ring cleavage extends the carbon chain by two units and generates an intermediate magnesium alkoxide. Upon subsequent acid hydrolysis, this intermediate is protonated to form a straight-chain primary alcohol as the final product.

Acidity in these organic compounds depends directly on the stability of the conjugate base formed after the loss of an $\mathrm{H}^{+}$ ion. Phenols show significantly greater acidity than standard alcohols due to resonance stabilization of the phenoxide ion, a property that is further amplified by the presence of a strongly electron-withdrawing $-\mathrm{NO}_2$ group. This substituent successfully disperses the negative charge on the oxygen atom to pull the dissociation equilibrium forward, contrasting sharply with electron-donating groups like $-\mathrm{CH}_3$ which increase electron density and destabilize the product.

The iodoform reaction specifically targets compounds that feature a methyl group bonded to a carbonyl carbon or to a carbinol carbon bearing a hydrogen atom. Among the given choices, propan-2-ol, butan-2-ol, and ethanol all satisfy this structural requirement either directly or through preliminary oxidation. Methanol possesses only a single carbon atom and lacks the necessary adjacent methyl configuration, failing to yield the characteristic yellow precipitate when treated with alkaline iodine.

Treating phenol with $\mathrm{CHCl}_3$ in the presence of an aqueous $\mathrm{NaOH}$ base generates a reactive dichlorocarbene intermediate that attacks the aromatic ring via electrophilic substitution. This specific transformation, known as the Reimer-Tiemann reaction, selectively introduces a formyl group at the ortho position relative to the hydroxyl group on the benzene ring. Consequently, this reaction directly yields salicylaldehyde as the primary organic product.

Primary alcohols such as $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$ possess two hydrogens on the carbinol carbon, allowing them to oxidize first into aldehydes and subsequently into carboxylic acids upon treatment with $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. In contrast, secondary alcohols like isopropyl alcohol possess only one hydrogen on the carbinol carbon, yielding a stable ketone rather than undergoing further oxidation. Because the two alcohols produce distinct functional groups under this treatment, potassium dichromate serves as an effective chemical distinguishing reagent.

Ethanol and certain other secondary alcohols or methyl ketones react with a mixture of iodine and $\mathrm{NaOH}$ through a halogenation and cleavage sequence. This classic haloform reaction results in the rapid precipitation of iodoform as a distinct pale yellow solid consisting of hexagonal crystals. The other choices fail because reducing aldehydes yields primary rather than secondary alcohols, reacting vegetable oil with $\mathrm{H}_2\mathrm{SO}_4$ does not produce glycerine, and sucrose combined with $\mathrm{NaCl}$ does not form invert sugar.

Conversion of an alcohol to an alkyl halide using phosphorus tribromide yields bromoethane ($\mathrm{CH_3CH_2Br}$), which subsequently undergoes dehydrohalogenation with alcoholic potassium hydroxide to produce ethene ($\mathrm{CH_2=CH_2}$) as product $Y$. Subsequent treatment of ethene with cold concentrated sulfuric acid followed by acid-catalyzed hydrolysis yields the original reactant, regenerating ethanol ($\mathrm{CH_3CH_2OH}$) as the final product $Z$.

Converting terminal alkenes into primary alcohols requires an anti-Markovnikov hydration method that avoids unwanted skeletal rearrangements. Hydroboration-oxidation achieves this by adding borane across the double bond in a concerted step, placing the boron atom selectively at the less hindered carbon. Subsequent treatment with alkaline hydrogen peroxide smoothly replaces the carbon-boron bond with a hydroxyl group to deliver the primary alcohol.