Alcohols Phenols And Ethers – JEE Main ChemistryPractice Questions & PYQs

The decolorization of bromine water is a definitive test for carbon-carbon unsaturation, identifying the presence of $C=C$ or $C \equiv C$ bonds that undergo electrophilic addition with $Br_2$. Ethylmagnesium bromide, a strong nucleophile and base, reacts with compounds containing acidic hydrogen atoms or electrophilic carbonyl centers. The compound in option D possesses both the unsaturated structural component required for bromine addition and the necessary functional group or acidic proton to undergo reaction with the Grignard reagent.

The reaction proceeds via a Williamson ether synthesis mechanism where $K_2CO_3$ acts as a base to deprotonate the phenolic hydroxyl group of m-cresol, generating a nucleophilic phenoxide anion. This anion subsequently performs an $S_N2$ attack on the methylene carbon of propargyl bromide, displacing the bromide ion to form 3-methylphenyl propargyl ether.

Reacting aniline with sodium nitrite in dilute $HCl$ at $0^\circ C$ generates benzenediazonium chloride, which functions as a key electrophile for the subsequent coupling step. When added to the mixture of aniline and phenol, the reaction selectively proceeds with aniline because the slightly acidic medium favors its higher reactivity over that of phenol. This results in an electrophilic aromatic substitution at the para-position of the aniline ring, yielding $p$-aminoazobenzene as the major product.

The reaction proceeds via an $E2$ elimination mechanism, where the ethoxide base abstracts a proton from the $\beta$-carbon to form a carbon-carbon double bond. The major product is the most substituted alkene, 2-methyl-2-butene, because it is thermodynamically more stable due to the greater degree of hyperconjugation and reduced steric strain compared to the less substituted isomer.

The reaction of phenol with $CO_2$ in the presence of $NaOH$ is the Kolbe-Schmitt reaction, which yields salicylic acid ($2$-hydroxybenzoic acid). Subsequent treatment of salicylic acid with acetic anhydride, $(CH_3CO)_2O$, in the presence of a catalytic amount of $H_2SO_4$ results in the acetylation of the phenolic hydroxyl group, producing acetylsalicylic acid, commonly known as aspirin. This transformation effectively converts the hydroxyl group into an acetoxy group, $-OCOCH_3$, while preserving the ortho-carboxylic acid substituent.

Phenol undergoes a Schotten-Baumann acylation reaction with methyl chloroformate in the presence of $NaOH$ to yield phenyl methyl carbonate as product A, where the phenolic oxygen atom is acylated. This ester group exerts an ortho/para-directing effect due to the resonance contribution of the oxygen atom attached to the benzene ring, causing subsequent electrophilic bromination by $Br_2$ to occur preferentially at the para position to produce p-bromophenyl methyl carbonate as product B.

The formation of the major product in this reaction is governed by the relative stability of the resulting intermediate or final product, which dictates the pathway with the lowest activation energy. According to the principle of thermodynamic stability, the reaction preferentially proceeds to form the most substituted species-whether it is a carbocation intermediate during electrophilic addition or an alkene product in an elimination reaction-because these species are significantly stabilized by hyperconjugation and inductive effects. This stabilization lowers the energy of the transition state, ensuring that the most substituted, and thus most stable, isomer is obtained as the major product.

The formation of optically active compounds requires the creation of at least one chiral center, which occurs in this reaction when the chlorine atom replaces a hydrogen atom on the carbon-2 position of the n-pentane chain. While 1-chloropentane and 3-chloropentane are achiral and do not rotate plane-polarized light, the produced 2-chloropentane possesses a chiral carbon atom attached to four unique groups: a hydrogen, a methyl group, an ethyl group, and a chlorine atom. This chiral center allows for the existence of two distinct enantiomers, the $d$ and $l$ forms, resulting in a total of two optically active compounds.

Boiling points are determined by the strength of intermolecular forces, where hydrogen bonding is the strongest, followed by dipole-dipole interactions, while London dispersion forces are the weakest. Ethanol ($C_2H_5OH$) forms strong intermolecular hydrogen bonds and possesses the highest boiling point, whereas propane ($C_2H_5CH_3$) relies only on weak dispersion forces and exhibits the lowest boiling point. Among the intermediate compounds, chloroethane ($C_2H_5Cl$) possesses stronger dipole-dipole interactions than the ether ethyl methyl ether ($C_2H_5OCH_3$), leading to the increasing order (III) < (IV) < (II) < (I).

Free radical allylic bromination introduces a bromine atom at the carbon adjacent to the double bond by exploiting the stability of the resulting resonance-stabilized allylic radical. This initial step is followed by hydrolysis, where the bromine atom ($-\text{Br}$) is displaced by a hydroxyl group ($-\text{OH}$) through a nucleophilic substitution process. This two-step sequence effectively converts an allylic hydrocarbon into an allylic alcohol by specifically modifying the carbon position next to the unsaturation.

The Bouveault-Blanc reduction is a specific organic transformation defined by the conversion of esters into primary alcohols using metallic sodium and ethanol. This process relies on sodium as the electron donor to effect the reduction, while ethanol functions as the proton source, summarized as $RCOOR' + 4[H] \xrightarrow{Na/C_2H_5OH} RCH_2OH + R'OH$. Because this unique system specifically targets the ester functional group to generate alcohols, it stands apart from other reduction methods involving different reagents like hydride donors or acidic metallic catalysts.

Tertiary alkyl halides such as 2-chloro-2-methylpentane possess significant steric hindrance around the electrophilic carbon center, which effectively prevents nucleophilic substitution and favors E2 elimination in the presence of strong bases like sodium methoxide. Under these conditions, the methoxide ion functions as a base by abstracting a $\beta$-hydrogen to yield an alkene through the formation of the most stable, highly substituted double bond, in accordance with Zaitsev's rule. This mechanism suppresses both substitution and the formation of less substituted isomers, resulting exclusively in 2-methylpent-2-ene, which is represented by the formula $\mathrm{C_2H_5CH}=\mathrm{C(CH_3)_2}$.

Grignard reagents behave as strong bases because of the high electron density on the alkyl group, making them react vigorously with any source of acidic protons. When methylmagnesium bromide ($\mathrm{CH_3MgBr}$) is treated with methanol ($\mathrm{CH_3OH}$), the methyl group rapidly abstracts the acidic hydroxyl hydrogen to generate methane. This acid-base reaction proceeds as
$\mathrm{CH_3MgBr} + \mathrm{CH_3OH} \rightarrow \mathrm{CH_3OMgBr} + \mathrm{CH_4} \uparrow$
and confirms that methane is the gas evolved in the process.

Achieving the partial oxidation of a primary alcohol ($\mathrm{R-CH_2-OH}$) into an aldehyde ($\mathrm{R-CHO}$) requires a reagent capable of stopping the reaction before it progresses further into a carboxylic acid. Strong oxidizing agents like $\mathrm{KMnO_4}$, $\mathrm{K_2Cr_2O_7}$, or $\mathrm{CrO_3}$ are typically too aggressive and will drive the reaction to completion, whereas Pyridinium Chlorochromate (PCC) functions as a mild, selective oxidant specifically formulated for this precise transformation. By utilizing PCC under controlled, anhydrous conditions, the oxidation is halted effectively at the aldehyde stage, as represented by the reaction $\mathrm{RCH_2OH} \xrightarrow{\mathrm{PCC}} \mathrm{RCHO}$.

The reaction between an alcohol and the Lucas reagent relies on the generation of a carbocation intermediate, with the reaction rate being directly proportional to the stability of this intermediate. Tertiary alcohols react fastest because they efficiently form a highly stable carbocation, allowing the conversion to proceed through the $S_N1$ mechanism. In contrast, primary and secondary alcohols react more slowly as their corresponding carbocation intermediates are significantly less stable and harder to achieve under these conditions.

The solubility of organic compounds in water is fundamentally determined by their capacity to form hydrogen bonds with the solvent. Within the $o$-Nitrophenol molecule, the hydrogen atom of the $-\mathrm{OH}$ group is stabilized through an intra-molecular hydrogen bond with the oxygen atom of the adjacent nitro group, effectively shielding it from interacting with the surrounding medium. Since the hydrogen is already engaged in this internal bonding, it remains unavailable to form the hydrogen bonds with $H_2\mathrm{O}$ necessary for dissolution. In contrast, $p$- and $m$-Nitrophenols lack this internal stabilization, allowing them to readily form inter-molecular hydrogen bonds with water molecules and thereby exhibit greater solubility.

The reaction between potassium bromide and potassium bromate functions as a reagent to generate molecular bromine in situ through the chemical pathway $5\,\mathrm{KBr} + \mathrm{KBrO}_3 + 3\,\mathrm{H}_2\mathrm{O} \longrightarrow 3\,\mathrm{Br}_2 + 6\,\mathrm{KOH}$. This molecular bromine acts as an electrophile, and because the hydroxyl group in phenol is a strongly activating group, it drastically increases the electron density of the benzene ring at the ortho and para positions. This electronic activation makes the ring highly susceptible to electrophilic aromatic substitution, causing bromine to replace the hydrogen atoms at all two ortho positions and the single para position simultaneously to yield 2,4,6-tribromophenol as the final product.

The reaction of alcohols with the Lucas reagent, which consists of concentrated $\mathrm{HCl}$ and anhydrous $\mathrm{ZnCl}_2$, proceeds via an $\mathrm{S}_{\mathrm{N}}1$ mechanism where the rate of reaction depends directly on the stability of the resulting carbocation intermediate. Since the formation of a tertiary carbocation is energetically favorable compared to primary or secondary options, tertiary alcohols exhibit the fastest reactivity under these conditions. Among the provided choices, 2-methylpropan-2-ol is the only tertiary alcohol, allowing it to readily lose water and form a stable $(\mathrm{CH}_3)_3\mathrm{C}^+$ ion, which subsequently reacts with the chloride ion to yield the final product.

The interaction of phenol with sodium hydroxide followed by carbon dioxide constitutes the Kolbe-Schmidt reaction, an electrophilic aromatic substitution where the nucleophilic phenoxide ion attacks carbon dioxide to introduce a carboxyl group specifically at the ortho position. The reaction proceeds according to the pathway $C_6H_5OH \xrightarrow{NaOH} C_6H_5ONa \xrightarrow{CO_2, \text{pressure}} C_6H_4(OH)COONa \xrightarrow{H^+} C_6H_4(OH)COOH$, yielding salicylic acid as the major product.

When phenol reacts with concentrated $H_2SO_4$, it undergoes electrophilic substitution to produce a mixture of ortho- and para-phenolsulfonic acids. Because the ortho-isomer is more stable at room temperature, it facilitates the subsequent reaction with concentrated $HNO_3$ to yield $o$-nitrophenol as the final product.