Alcohols Phenols And Ethers – JEE Main ChemistryPractice Questions & PYQs

The reaction involves the nucleophilic attack of the phenyl carbanion from $\mathrm{PhMgBr}$ onto the electrophilic carbonyl carbon of a ketone, resulting in an intermediate alkoxide that yields a tertiary alcohol after acidic hydrolysis. Specifically, acetophenone ($\mathrm{Ph}-\text{CO}-\text{CH}_3$) reacts with one mole of $\mathrm{PhMgBr}$ to introduce the second phenyl group at the carbonyl center, which upon protonation produces $\mathrm{Ph}-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)(\mathrm{Ph})$.

Carbolic acid is the common name for phenol, which is synthesized by the hydrolysis of benzenediazonium chloride upon warming with water. The reaction proceeds according to the equation $C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 \uparrow + HCl$, where the diazonium group is replaced by a hydroxyl group.

The industrial synthesis of phenol via the cumene process relies on the controlled air oxidation of isopropylbenzene to form cumene hydroperoxide, which corresponds to compound A. This intermediate subsequently undergoes acid-catalyzed cleavage to yield phenol and acetone, where the latter is identified as compound B. The overall transformation can be summarized as $C_6H_5CH(CH_3)_2 \xrightarrow{O_2} C_6H_5C(CH_3)_2OOH \xrightarrow{H^+} C_6H_5OH + CH_3COCH_3$, highlighting the conversion of the alkyl-substituted aromatic into the target alcohol and ketone.

Pyridinium chlorochromate (PCC) selectively oxidizes the secondary alcohol in hex-4-ene-2-ol to the corresponding ketone, producing hex-4-en-2-one ($CH_3CH=CHCH_2COCH_3$). Because this ketone contains a methyl ketone group, it undergoes the haloform reaction with sodium hypoiodite ($NaOI$) to produce iodoform ($CHI_3$) and the sodium salt of pent-3-enoic acid ($CH_3CH=CHCH_2COO^-Na^+$). Finally, decarboxylation of this carboxylic acid salt upon heating with soda lime replaces the carboxylate group with a hydrogen atom to yield but-2-ene ($CH_3CH=CHCH_3$) as the final product.

The chemical name 4-iodomethylphenol specifies a phenol ring where the para position-the carbon directly opposite the hydroxyl group-bears a methyl group that is substituted with an iodine atom. Recognizing this nomenclature allows us to isolate the molecular structure, which must contain an $-OH$ group on the aromatic ring and a $-CH_2I$ group at the 4-position. This specific arrangement of functional groups confirms the identity of the target molecule formed in the reaction.

The cleavage of ethers using concentrated hydroiodic acid (\text{HI}$) involves the protonation of the ether oxygen followed by nucleophilic attack, yielding an alkyl iodide and a hydroxy compound, such as a phenol. Treating this phenolic hydroxy compound with zinc dust (\text{Zn}$) triggers a reduction reaction, which effectively removes the oxygen atom and replaces the hydroxyl group with a hydrogen atom to generate a hydrocarbon as compound B. Consequently, compound A is the intermediate hydroxy derivative, and compound B is the corresponding hydrocarbon resulting from the deoxygenation of A.

The diazotization reaction of primary aromatic amines, such as aniline, involves treatment with nitrous acid-generated in situ from sodium nitrite and hydrochloric acid-at low temperatures between 273 K and 278 K. This process converts the amino group into a diazonium group, resulting in the formation of benzenediazonium chloride as the key intermediate $X$. This species is highly reactive and functions as an electrophile in subsequent coupling reactions with electron-rich aromatic compounds to produce azo dyes.

The distinct bromination outcomes for phenol are fundamentally governed by the solvent's polarity, which determines the degree of phenol ionization. In water, a highly polar medium, phenol readily dissociates into the phenoxide ion, a species with significantly higher electron density that directs electrophilic substitution to all ortho and para positions, yielding 2,4,6-tribromophenol. Conversely, chloroform acts as a low-polarity solvent where phenol remains largely unionized, leaving the aromatic ring less activated and restricting the bromination to a single position.

Birch reduction functions as a chemoselective process that transforms aromatic rings into non-conjugated 1,4-cyclohexadienes using an alkali metal, such as sodium or lithium, dissolved in liquid ammonia in the presence of a proton donor. The reaction proceeds via the formation of a radical anion intermediate, where the site of protonation and subsequent electron transfer are governed by the inductive and resonance effects of substituents attached to the aromatic core. Consequently, electron-donating groups direct the reduction differently than electron-withdrawing groups, leading to the specific substitution pattern found in the major product of this transformation.

Markovnikov's rule dictates that in the electrophilic addition of hydrogen halides to unsymmetrical alkenes, the electrophile $H^+$ adds to the carbon atom containing the greater number of hydrogen atoms to form the more stable carbocation, yielding the Markovnikov product. Conversely, when $HBr$ is added in the presence of organic peroxides, a free-radical mechanism is initiated, which places the bromine atom on the less substituted carbon, known as the anti-Markovnikov addition or peroxide effect.

$\text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{ionic}} \text{CH}_3-\text{CH(Br)}-\text{CH}_3$

$\text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br}$

The reaction 'A' proceeds via the standard ionic route to form the Markovnikov product, while reaction 'B', occurring in the presence of peroxide, follows the free-radical route to form the anti-Markovnikov product.

The reduction of a nitrile group with catalytic hydrogenation using $H_2/Ni$ transforms the cyanide functional group into a primary amine, which subsequently serves as the substrate for deamination via nitrous acid treatment. In the reaction sequence, the nitrile $R-CN$ is reduced to the primary amine $R-CH_2NH_2$ (Component A), which then reacts with $HNO_2$ (prepared in situ from $NaNO_2/HCl$) to form a highly unstable aliphatic diazonium salt that decomposes instantly to release nitrogen gas and generate a carbocation. This carbocation is rapidly captured by a nucleophilic water molecule to yield the corresponding primary alcohol $R-CH_2OH$ (Component B).

Phenol serves as a versatile starting material where specific chemical transformations are dictated by the chosen reagents. Bromination with $\mathrm{Br}_2$ in a non-polar solvent like $\mathrm{CS}_2$ promotes monobromination to yield p-bromophenol, while oxidation with $\mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7/\mathrm{H}_2\mathrm{SO}_4$ converts phenol into p-benzoquinone. The application of $\mathrm{Zn}$ dust facilitates the reduction of the phenol hydroxyl group to produce benzene, and the Reimer-Tiemann reaction employing $\mathrm{CHCl}_3/\mathrm{NaOH}$ introduces a formyl group to the aromatic ring to form salicylaldehyde.

Phenols are inherently weakly acidic because the negative charge on the phenoxide ion, which results from deprotonation, is stabilized by resonance within the aromatic ring. Although phenols are sufficiently acidic to dissolve in an aqueous $\mathrm{NaOH}$ solution, they are actually stronger acids than water and aliphatic alcohols rather than weaker. Consequently, the assertion in the second statement regarding their relative acidic strength is false, even though the first statement accurately identifies their acidic nature.

Although the presence of an active methylene group in acetylacetone typically favors an acid-base reaction with a Grignard reagent ($RMgX$), this specific problem requires focusing on the carbonyl group as the reactive electrophilic center. By proceeding through a nucleophilic addition reaction at the carbonyl carbon rather than undergoing deprotonation, the Grignard reagent attacks the electrophile to yield the addition product corresponding to option A.

The overall efficiency of a sequential chemical process depends on the cumulative product of the percentage yields of each individual step, as the output of the first stage serves as the input for the second. If the first step produces an intermediate with a yield of 60%, the amount of available substance is reduced to 0.6 of the starting quantity. When this intermediate undergoes the second step with a 50% yield, the final amount of product is determined by the mathematical relationship $0.6 \times 0.5 = 0.3$. Consequently, the total yield for the entire process is 30% of the initial reactant quantity.

Carbon atoms participating in aromatic conjugation or double bonds, specifically $C=C$ or $C=O$ groups, undergo $sp^2$ hybridization to maintain a trigonal planar geometry. In the structure of compound 'X', the six carbon atoms within the aromatic benzene ring contribute 6 to the total count of $sp^2$ hybridized carbons. The carbonyl functional group accounts for 1 $sp^2$ hybridized carbon, and the presence of an additional alkenyl double bond ($C=C$) within the molecular framework contributes the final 1, resulting in a total of 8 $sp^2$ hybridized carbon atoms.

Compound 'P' is identified as phenol ($C_6H_5OH$) because its nitration with dilute $\mathrm{HNO}_3$ produces ortho-nitrophenol and para-nitrophenol, which are separable by steam distillation due to the intramolecular hydrogen bonding present in the ortho-isomer and intermolecular hydrogen bonding in the para-isomer. The reaction of phenol with concentrated $\mathrm{HNO}_3$ results in the formation of 2,4,6-trinitrophenol, commonly known as picric acid, which is compound 'C'. The molecular formula for picric acid is $C_6H_2(OH)(NO_2)_3$, which can be expressed as $C_6H_3N_3O_7$. This molecule contains one hydroxyl group and three nitro groups, so the total number of oxygen atoms is calculated as $1 + (3 \times 2) = 7$.

The reaction begins with the dehydration of ethanol in the presence of heated concentrated $\mathrm{H}_2\mathrm{SO}_4$, which eliminates water to produce the alkene gas ethene, $\mathrm{CH}_2=\mathrm{CH}_2$. When this gas interacts with a cold dilute aqueous solution of Baeyer's reagent, the double bond undergoes a hydroxylation process that adds two hydroxyl groups across the carbon atoms. This specific reaction sequence ultimately forms ethane-1,2-diol, commonly referred to as glycol.

Acid-catalyzed dehydration of butan-2-ol proceeds via the E1 mechanism, where the elimination of water from the carbocation intermediate follows Saytzeff's rule to produce the most substituted alkene, but-2-ene, as major product $A$. Subsequent electrophilic addition of $HBr$ to this alkene occurs according to Markovnikov's rule, proceeding through a stable secondary carbocation intermediate to yield 2-bromobutane as major product $B$.

According to Zaitsev's rule, in dehydrohalogenation reactions involving strong bases, the major product is the most substituted and thermodynamically stable alkene. In the reaction of 2-bromo-3-methylbutane with $NaOEt/EtOH$, the ethoxide ion acts as a strong base to facilitate an E2 elimination by abstracting a proton from the $\beta$-carbon. The pathway that yields the more substituted 2-methyl-2-butene is kinetically favored and results in a significantly more stable product compared to the alternative 3-methyl-1-butene, thereby establishing it as the major product.