Alcohols Phenols And Ethers – JEE Main ChemistryPractice Questions & PYQs

The cleavage of a mixed ether by excess hot concentrated hydrogen iodide splits the molecule at the oxygen linkage, yielding two alkyl iodides that convert into alcohols upon treatment with aqueous sodium hydroxide. For these resulting alcohols to produce a positive result in the iodoform test, they must feature the $CH_3CH(OH)-$ structural unit, which is characteristic of secondary alcohols with a methyl substituent. This requirement dictates that the alkyl groups comprising the original ether must be structured such that, once the oxygen bridge is replaced by hydroxyl groups, they form these specific secondary alcohols. Consequently, the ether must possess groups like isopropyl or sec-butyl to generate products (Q) and (R) capable of producing the characteristic yellow precipitate with $NaOI$.

The Reimer-Tiemann reaction involves treating phenol with $CHCl_3$ and aqueous $KOH$ to introduce a formyl group onto the benzene ring. Contrary to the claim in the first statement, this transformation yields $o$-hydroxybenzaldehyde as the major product rather than the para-isomer. The ortho-isomer forms preferentially due to the specific mechanism involving the dichlorocarbene intermediate, making the first statement incorrect.

Regarding the separation of these products, steam distillation serves as an effective technique. The ortho-isomer exhibits intramolecular hydrogen bonding, which stabilizes the molecule and lowers its effective intermolecular attractive forces, thereby rendering it steam volatile. In contrast, the para-isomer forms extensive intermolecular hydrogen bonds, resulting in a higher boiling point and negligible steam volatility. Consequently, this significant difference in physical properties allows for their straightforward separation, confirming the validity of the second statement.

Acidity is primarily determined by the stability of the conjugate base, which increases with the presence of electron-withdrawing groups that distribute negative charge via resonance or inductive effects. Compound III exerts the most profound stabilization on the conjugate base due to its specific substituent arrangement, placing it at the top of the acidity scale, followed by I and II, while the configuration of compound IV provides the least stabilization. Therefore, the relative acidity decreases in the sequence $III > I > II > IV$ based on the total cumulative electronic contribution of the substituents in each case.

A substance will dissolve in aqueous sodium hydroxide if it possesses a level of acidity sufficient to donate a proton to the hydroxide ion, resulting in the formation of a water-soluble salt. This transformation is spontaneous whenever the molecule exhibits a $pK_a$ value lower than that of water, establishing the necessary conditions for deprotonation by $NaOH$ at room temperature. Upon examining the chemical properties of the listed structures, it becomes clear that compounds II, III, IV, V, and VII are acidic enough to react in this manner. Since these five specific compounds are capable of undergoing this process, the total number of soluble substances is 5.

The reaction between benzyl phenyl ether ($\mathrm{C_6H_5-O-CH_2-C_6H_5}$) and $\mathrm{HI}$ begins with the protonation of the ether oxygen, followed by a nucleophilic attack of the iodide ion on the benzylic carbon. This selective cleavage occurs because the phenyl-oxygen bond possesses partial double bond character and is resistant to breakage, while the $\mathrm{O-CH_2}$ bond is susceptible to rupture, leading to the formation of phenol and benzyl iodide. Consequently, the assertion that the products are benzyl alcohol and iodobenzene is false, whereas the description of the $\mathrm{O-CH_2}$ bond cleavage is correct.

Acid-catalyzed dehydration of alcohols involves the removal of a hydroxyl group and an adjacent hydrogen atom to form a double bond, requiring strong acids like concentrated $\mathrm{H}_2\mathrm{SO}_4$ or $\mathrm{H}_3\mathrm{PO}_4$ as the reagent (D). During the transformation of compound A to the alkene B, these acidic conditions protonate the hydroxyl group to generate a good leaving group, which subsequently departs to form a carbocation that stabilizes by releasing a proton to establish the double bond. When compound C undergoes this same elimination reaction using the reagent (D), it results in the unsaturated product E with the structural formula $\mathrm{CH}_2 = \mathrm{CH} - \mathrm{CH}(\mathrm{OH}) - \mathrm{CH}_3$.

Synthesizing 3,3-dimethyl-2-butanol is challenging because its structure imposes significant steric hindrance that undermines common synthetic approaches. The nucleophilic addition of a bulky tert-butyl Grignard reagent to an aldehyde is inefficient because the extreme spatial crowding forces the reaction toward elimination or reduction pathways rather than the intended addition. Furthermore, attempting an electrophilic hydration of 3,3-dimethyl-1-butene fails to yield the target alcohol because the carbocation intermediate undergoes a rapid 1,2-hydride shift to form a more stable tertiary carbocation, ultimately resulting in 2,3-dimethyl-2-butanol.

The molar mass of compound (x) is determined to be 94 g/mol based on the provided vapour density, and the empirical data identifies the substance as phenol. This phenol undergoes the Kolbe-Schmidt reaction with $\mathrm{CO}_2$ and $\mathrm{NaOH}$ followed by acidification to produce salicylic acid as compound (y). Salicylic acid is a stronger acid than phenol due to the presence of the electron-withdrawing carboxyl group, making the statement that phenol is more acidic inherently incorrect. Furthermore, salicylic acid reacts with $\mathrm{NaHCO}_3$ to evolve $\mathrm{CO}_2$, while both compounds are aromatic and yield a sooty flame during combustion, and both successfully dissolve in $\mathrm{NaOH}$ by forming their respective salts.

The reaction sequence initiates with the catalytic dehydrogenation of an alcohol using copper metal at 573 K, which removes hydrogen to yield an aldehyde or a ketone based on the alcohol's structural classification. Subsequently, the molecule reacts with $\text{PhCOOH}$ in an acidic medium, facilitating an esterification pathway to append the benzoate moiety to the framework. This two-part transformation effectively converts the original hydroxyl group into the final benzoate ester product $P$ through a combination of controlled oxidation and acid-catalyzed condensation.

Nitrating phenol with dilute nitric acid at 298 K yields a mixture where only ortho-nitrophenol possesses significant steam volatility due to its internal hydrogen bonding, which minimizes intermolecular forces. Para-nitrophenol, hindered by intermolecular hydrogen bonding, does not volatilize under the same conditions. Identifying the steam-volatile compound as ortho-nitrophenol allows for a direct comparison of oxygen content between the starting material and the product.

Phenol has a molecular formula of $C_6H_6O$ and a molar mass of 94 $\mathrm{g \; mol^{-1}}$, meaning its 16 $\mathrm{g \; mol^{-1}}$ oxygen content accounts for 17.021 percent of the total mass. Ortho-nitrophenol, represented by $C_6H_5NO_3$, has a molar mass of 139 $\mathrm{g \; mol^{-1}}$ and contains three oxygen atoms totaling 48 $\mathrm{g \; mol^{-1}}$, resulting in an oxygen percentage of $\frac{48}{139} \times 100$, or 34.532 percent. Subtracting the initial percentage from the final percentage gives an increase of 17.511 percent. Writing this value as $175.11 \times 10^{-1}$ leads to the required numerical result of 175.

Identifying a secondary alcohol requires finding a hydroxyl group ($-OH$) attached to a carbon atom that is directly bonded to two other carbon atoms, which fits the general structural formula $R_2CH-OH$. Applying this structural assessment to the provided set of molecules confirms that exactly three of these compounds meet the criteria, while the remaining structures are excluded because they contain primary or tertiary alcohols or different functional groups entirely.

Heating phenol with zinc dust reduces it to benzene, following a simple stoichiometry where one mole of phenol produces one mole of benzene. The molar mass of phenol, $C_6H_5OH$, is determined by the sum of its atoms as $(6 \times 12) + (6 \times 1) + 16 = 94$ g/mol. Dividing the starting mass of 4.7 g by this molar mass indicates that 0.05 mol of phenol are available for the reaction. Because the process occurs with a 60% yield, the final amount of benzene obtained is $0.05 \times 0.60 = 0.03$ mol. Expressing this quantity as $3 \times 10^{-2}$ mol reveals that the integer required is 3.

The synthesis of alcohols from hydrocarbons relies on achieving partial oxidation rather than complete combustion or oxidative cleavage, a process that depends entirely on the specific catalyst and reaction conditions employed. Reaction C serves as a clear example of this, where the oxidation of isobutane using $\mathrm{KMnO}_4$ yields tert-butyl alcohol. Similarly, reaction D achieves the selective oxidation of methane to produce methanol ($\mathrm{CH}_3\mathrm{OH}$) when carried out under the controlled conditions of 523 K and 100 atm using copper as a catalyst.

Conversely, the other reactions result in different functional groups based on the reagents provided. Reaction A uses $\mathrm{Mo}_2\mathrm{O}_3$ to transform methane into formaldehyde, while reaction B utilizes $(\mathrm{CH}_3\mathrm{COO})_2\mathrm{Mn}$ to oxidize ethane into acetic acid. Finally, reaction E involves the treatment of an alkene with acidic $\mathrm{KMnO}_4$, which causes oxidative cleavage and results in the formation of carboxylic acids rather than alcohols.

Acetylation effectively replaces the hydrogen atom of a hydroxyl group with an acetyl group, resulting in a predictable mass increase because the incoming acetyl moiety has a higher molar mass than the hydrogen atom it displaces. Specifically, replacing a hydrogen atom with a mass of 1 g mol$^{-1}$ with an acetyl group possessing a mass of 43 g mol$^{-1}$ adds a net weight of 42 g mol$^{-1}$ for every hydroxyl group present. The total mass added to the compound during the reaction is the difference between the final molar mass of 290 g mol$^{-1}$ and the initial molar mass of 122 g mol$^{-1}$, which results in a total gain of 168 g mol$^{-1}$. Dividing this total increase by the 42 g mol$^{-1}$ added per individual site determines that there are $n = \frac{168}{42} = 4$ hydroxyl groups in the original compound.

The stoichiometry of the bromination of phenol dictates that one mole of $C_6H_5OH$ reacts with three moles of $Br_2$ to yield 2,4,6-tribromophenol. To determine the required mass of bromine, start by finding the number of moles of phenol in a 2 g sample by using its molar mass of 94 g/mol. Since the balanced reaction requires three times as many moles of bromine as phenol, you must multiply the moles of phenol by 3 to calculate the total moles of bromine needed. Completing the calculation by multiplying this molar quantity by the molar mass of $Br_2$, which is 160 g/mol, results in a final required mass of 10.22 g.

When an aryl alkyl ether reacts with concentrated hydrobromic acid, the process begins with the protonation of the ether oxygen. This step activates the molecule, setting the stage for a nucleophilic substitution where the bromide ion selectively attacks the aliphatic alkyl carbon. The cleavage occurs specifically at the alkyl-oxygen bond because the bond linking the aromatic ring to the oxygen possesses partial double-bond character due to resonance stabilization, which makes it far too stable to break under these conditions. This reaction, represented as $Ar-O-R + HBr \rightarrow Ar-OH + R-Br$, ensures that the aryl group remains attached to the oxygen, successfully yielding phenol as a product while the alkyl group is converted into an alkyl halide.

The solubility of ethers in water is governed by the balance between the ability of the oxygen lone pairs to form hydrogen bonds and the hydrophobic, non-polar nature of the alkyl groups attached to them. Dimethyl ether ($CH_3OCH_3$) is completely soluble because its small methyl groups offer minimal hydrophobic hindrance, while the larger ethyl groups in diethyl ether ($C_2H_5OC_2H_5$) create significant hydrophobic interference that limits its solubility to a very small extent. Because metallic sodium reacts readily with the active hydrogen in the hydroxyl group of ethyl alcohol ($C_2H_5OH$), it cannot be used to dry alcohols, yet it functions effectively as a drying agent for diethyl ether since ethers lack active hydrogens and remain unreactive toward the metal.

The reaction sequence is driven by the susceptibility of aromatic substituents to systematic modification through sulfonation, nucleophilic substitution, and reduction. Benzene first undergoes sulfonation with oleum to form benzenesulfonic acid, which then converts into phenol upon alkali fusion with molten sodium hydroxide followed by acidification. The concluding step utilizes zinc dust to perform a reductive deoxygenation on the phenol, effectively stripping the hydroxyl group and regenerating benzene ($C_6H_6$) as the final product.

The enhanced reaction rate observed during alkaline hydrolysis is fundamentally attributed to neighboring group participation, where an internal nucleophile facilitates a more efficient reaction pathway. Specifically, the lone pair of electrons on the nitrogen atom initiates the process by undergoing an intramolecular substitution, resulting in the formation of a highly reactive cyclic intermediate. This internal mechanism is kinetically favored over direct substitution by an external hydroxide ion, which explains why the reaction proceeds at a faster rate. Because the intramolecular step effectively accelerates the hydrolysis, both the claim regarding the faster rate and the mechanistic description are entirely correct.

The boiling points of alcohols and phenols trend upward as the number of carbon atoms in the chain increases, primarily because larger molecular sizes yield greater surface area and subsequently intensify the van der Waals forces between molecules. Furthermore, these compounds possess a characteristic $-OH$ group that enables strong intermolecular hydrogen bonding, a structural feature absent in ethers and haloalkanes. Since breaking these hydrogen bonds demands significant energy, alcohols and phenols naturally exhibit higher boiling points than those other classes of organic compounds.