Alcohols Phenols And Ethers – JEE Main ChemistryPractice Questions & PYQs

The reactivity of phenol during electrophilic aromatic substitution is highly sensitive to the solvent, as the reaction medium governs the degree of ring activation. In a polar medium like water, phenol ionizes into the phenoxide ion, a species that increases electron density significantly and activates the ring so strongly that it facilitates rapid trisubstitution when excess bromine is present, yielding 2,4,6-tribromophenol. Conversely, utilizing a non-polar solvent like $\mathrm{CS}_2$ maintains phenol in its neutral form, which provides much less activation and restricts the bromination to a single position, resulting only in a monobrominated product.

Darzen's reaction facilitates the conversion of primary or secondary aliphatic alcohols into pure alkyl chlorides by following an $\mathrm{S}_{\mathrm{N}}2$ mechanism. This process is highly selective, as phenolic $-\mathrm{OH}$ groups do not participate due to the $+R$ effect, which strengthens the $\mathrm{C}-\mathrm{O}$ bond through resonance-induced double bond character. Recognizing these distinct structural requirements allows for the clear identification of the specific alkyl chloride product expected in this transformation.

The Williamson synthesis is a standard laboratory method used to prepare a variety of ethers, including unsymmetrical ones like ethyl phenyl ether ($C_6H_5OC_2H_5$), making the assertion fundamentally correct. However, this reaction requires a reactive alkyl halide to undergo nucleophilic substitution, which is not the case for bromobenzene. In bromobenzene, the $C-Br$ bond exhibits significant partial double bond character due to resonance with the aromatic benzene ring, making the bond far too stable to be dissociated by the ethoxide nucleophile. Therefore, the proposed reaction route between bromobenzene and sodium ethoxide is ineffective, rendering the reasoning provided for the assertion incorrect.

The boiling point of a substance is fundamentally dictated by the strength of its intermolecular forces, with hydrogen bonding acting as a potent attractive force between molecules. Compound A contains a hydroxyl group ($-\text{OH}$), enabling it to form intermolecular hydrogen bonds that require additional energy to overcome during the transition to the gaseous state. Conversely, compound B features a methoxy group that cannot facilitate such hydrogen bonding, resulting in weaker intermolecular forces and a lower boiling point than compound A, even though both substances share the molecular formula $C_9H_18O_3$.

Acid-catalyzed dehydration of neopentyl alcohol initiates with the formation of a primary carbocation that rapidly undergoes a 1,2-methyl shift to produce the more stable tertiary carbocation, $(CH_3)_2C^+CH_2CH_3$. Following Zaitsev's rule, the elimination of a proton from this intermediate favors the formation of the more substituted 2-methyl-2-butene as the major product and 2-methyl-1-butene as the minor product. The 85:15 ratio is established by the thermodynamic stability of the more highly substituted alkene, which dictates the distribution of the final products.

The reaction sequence relies on the sequential conversion of an aromatic substrate through electrophilic aromatic substitution and subsequent functional group transformations. Initially, the aromatic ring undergoes nitration to introduce a nitro group, which is then chemically reduced to an amine, specifically aniline, serving as the intermediate B. The final step involves the diazotization of this primary aromatic amine using nitrous acid at $0-5^{\circ}C$, which successfully transforms the amine into a diazonium salt, identifying product C as the major aromatic species in this pathway.

The cleavage of an ether by concentrated hydroiodic acid ($HI$) involves the protonation of the ethereal oxygen followed by the formation of an alkyl iodide and an alcohol or enol. For the compound $1-methoxy-1-phenylethylene$ ($C_9H_{10}O$), this reaction yields methyl iodide ($CH_3I$) and the enol $1-phenylvinyl$ alcohol, where the methyl iodide reacts with $AgNO_3$ to produce a yellow precipitate of silver iodide. The unstable enol $1-phenylvinyl$ alcohol subsequently tautomerizes to form acetophenone, which possesses the requisite $CH_3CO-$ group to yield a positive iodoform test. Thus, the identity of the original compound 'A' is $1-methoxy-1-phenylethylene$.

The regioselectivity of this electrophilic addition reaction is governed by Markovnikov's rule, which dictates that the proton adds to the carbon atom with the greater number of hydrogen substituents to generate the most stable carbocation intermediate. The subsequent nucleophilic attack on this more stable, highly substituted carbocation yields the most substituted alkyl halide as the major species. This pathway is energetically favored due to the inductive and hyperconjugative stabilization provided by adjacent alkyl groups, confirming that the structure in the selected option represents the primary reaction product.

Ether cleavage with concentrated $HI$ involves the protonation of the ether oxygen followed by nucleophilic attack on the less sterically hindered carbon, yielding an alcohol and an alkyl iodide. The alcohol, 2-methylbutan-1-ol, undergoes acid-catalyzed dehydration via an $E1$ mechanism where the initial primary carbocation rearranges to a more stable tertiary carbocation, $CH_3-CH_2-C^+(CH_3)-CH_3$, via a 1,2-hydride shift. Elimination of a proton from the beta-carbon then yields the trisubstituted alkene $CH_3-CH=C(CH_3)-CH_3$ as the major product because it is the most stable isomer according to Saytzeff's rule.

The stability of alkoxide anions depends on the ability of adjacent groups to disperse the negative charge on the $O^-$ atom through resonance and inductive effects. Structures exhibiting strong resonance stabilization, such as (C), are inherently more stable than those with limited delocalization like (B), while aliphatic structures (A) are the least stable due to the electron-donating inductive effect of alkyl groups that increases negative charge density. Since increased charge delocalization significantly lowers the internal energy of the anion, the order of stability follows the efficiency of electron dispersal, confirming that (C) > (B) > (A).

The reaction of phenol with chloroform in the presence of aqueous sodium hydroxide follows the Reimer-Tiemann mechanism, producing salicylaldehyde with the molecular formula $C_7H_6O_2$ as the major product $P$. Calculating the molar mass of this compound involves summing the masses of seven carbon atoms at $7 \times 12 = 84$, six hydrogen atoms at $6 \times 1 = 6$, and two oxygen atoms at $2 \times 16 = 32$, which totals 122 grams per mole. The mass percentage of carbon is found by dividing the total mass contributed by carbon by the molecular weight, calculating $\frac{84}{122} \times 100$. This operation results in a value of approximately 68.85, which rounds to the nearest integer of 69.

The Ceric Ammonium Nitrate test is a reliable qualitative method specifically designed to identify the presence of aliphatic alcoholic $-\mathrm{OH}$ groups, which react to form a characteristic red-colored complex. Upon examining the molecular structure of product 'P', only the hydroxyl group identified as (b) functions as an aliphatic alcohol capable of undergoing this distinct coordination reaction. Consequently, the positive test result observed is attributed exclusively to the presence of this specific functional group.

The C-OH bond length is determined by the hybridization state of the carbon atom and the extent of resonance involved in the C-O linkage. In methanol, the oxygen atom is attached to an $sp^3$ hybridized carbon atom, which forms a pure single bond with no resonance, resulting in the longest bond length among the three compounds. In both phenol and p-ethoxyphenol, the oxygen atom is bonded to an $sp^2$ hybridized carbon atom of the benzene ring, which imparts partial double bond character through resonance; however, the strong electron-donating ethoxy group at the para position in p-ethoxyphenol increases electron density on the ring, which diminishes the delocalization of the lone pair from the hydroxyl group, making the C-OH bond in p-ethoxyphenol slightly longer than the more strongly resonated bond in phenol.

The reaction sequence begins with the acid-catalyzed hydration of an alkyne using $\text{HgSO}_4$ and $\text{dil.H}_2\text{SO}_4$. This reagent system facilitates the addition of a water molecule across the triple bond, which generates an enol intermediate that rapidly tautomerizes to form a ketone. In the final step, the addition of $\text{NaBH}_4$ acts as a selective reducing agent, converting the carbonyl group of the intermediate ketone into a secondary alcohol, which represents the major product [R].

The regioselectivity of this reaction is determined by the formation of the most stable carbocation intermediate, which is stabilized by inductive effects and hyperconjugation from adjacent alkyl groups. The nucleophile preferentially attacks the more substituted carbon atom, following the pathway to the most stable species, represented by $R_3C^+$, which is consistent with Markovnikov's rule for electrophilic addition reactions.

The synthesis of 2-phenylbutan-2-ol requires a ketone substrate to form a tertiary alcohol through the nucleophilic addition of a Grignard reagent. While reactions involving the ketones $PhCOCH_2CH_3$, $PhCOCH_3$, and $CH_3CH_2COCH_3$ successfully provide the necessary branched structure to yield this tertiary alcohol, formaldehyde ($HCHO$) contains two hydrogen atoms that limit the product to a primary alcohol. Therefore, the reaction with formaldehyde cannot produce the desired 2-phenylbutan-2-ol.

Electrophilic aromatic substitution proceeds via the attack of an electrophile at the position of highest electron density, which is dictated by the directing effects of existing substituents on the aromatic ring. Strongly activating, ortho/para-directing groups such as the methoxy group ($-\text{OCH}_3$) prioritize attack at the para position because it is significantly less sterically hindered than the crowded ortho positions. This kinetic preference for the para-substituted isomer makes it the predominant product observed in the reaction mixture.

The $pKa$ value serves as an inverse measure of acidic strength, meaning that a compound with a stronger tendency to donate protons will consistently exhibit a lower $pKa$ value. To determine the correct increasing order, you must first rank the compounds based on their relative acidity from highest to lowest. Since the most acidic species corresponds to the minimum $pKa$ and the least acidic to the maximum, simply reversing the acidity sequence provides the correct order of $pKa$ values.

The major product 'X' is determined by the thermodynamic stability of the resulting alkene, which follows Zaitsev's rule during elimination reactions. This principle states that the most substituted alkene is formed preferentially because the alkyl groups stabilize the double bond through hyperconjugation and inductive effects, effectively lowering the activation energy, $\Delta G^\ddagger$. Consequently, the reaction path involving the abstraction of the proton from the $\beta$-carbon that yields the more substituted, and thus more stable, isomer is kinetically and thermodynamically favored.

Lithium aluminum hydride ($\mathrm{LiAlH}_4$) is a powerful nucleophilic reducing agent that selectively reduces esters to primary alcohols while generally leaving conjugated carbon-carbon double bonds intact. When methyl crotonate undergoes this reduction, the ester group is converted to a primary alcohol functionality, yielding the unsaturated allylic alcohol $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_2\mathrm{OH}$ as the major product.