Alcohols Phenols And Ethers – JEE Main ChemistryPractice Questions & PYQs

The synthesis follows a logical progression of functional group transformations, utilizing the Grignard reaction to lengthen the carbon chain. First, ethanol is treated with phosphorus and iodine to generate iodoethane ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{I}$), identified here as compound A, which is then converted into the Grignard reagent ethylmagnesium iodide ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{MgI}$), compound B, through reaction with magnesium in ether. The nucleophilic carbon of this Grignard reagent subsequently attacks the carbonyl group of formaldehyde ($\mathrm{HCHO}$) to form an intermediate alkoxide, compound C. Upon final hydrolysis with water, this intermediate is protonated to produce the primary alcohol, n-propyl alcohol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$), which is compound D.

For an alcohol to yield a positive result in the iodoform test with $\mathrm{I}_2$ and $\mathrm{NaOH}$, it must possess the characteristic structural grouping $-\mathrm{CH(OH)CH}_3$. Upon reviewing the options, $\mathrm{PhCH(OH)CH}_3$ is the only molecule that contains this specific methyl secondary alcohol moiety, whereas the others are either primary alcohols or do not have a methyl group attached to the carbon bearing the hydroxyl group. Consequently, only this compound possesses the structural requirement to undergo the haloform reaction.

The reaction between methyl vinyl ether and hydrogen bromide is a classic electrophilic addition governed by the electronic effects of the substituent group. Under anhydrous conditions at room temperature, the electron-donating methoxy group stabilizes the developing carbocation through resonance, which dictates the regioselectivity of the reaction according to Markovnikov's rule. When the proton from $\mathrm{HBr}$ adds to the terminal methylene carbon of $\mathrm{CH}_2=\mathrm{CH}-\mathrm{OCH}_3$, a stable carbocation forms on the carbon atom directly bonded to the oxygen. Subsequently, the bromide ion attacks this stabilized cationic center to yield the final addition product, $\mathrm{H}_3\mathrm{C}-\mathrm{CHBr}-\mathrm{OCH}_3$.

The hydroxyl ($-OH$) group attached to the benzene ring acts as a powerful activator because it increases electron density within the ring through resonance, specifically directing this effect toward the ortho and para positions. This heightened reactivity makes these sites highly susceptible to electrophilic attack by bromine. Consequently, when phenol ($C_6H_5OH$) reacts with bromine water, the substitution occurs simultaneously at both ortho positions and the para position to yield a $C_6H_2Br_3OH$ derivative.

The Reimer-Tiemann reaction is defined by the generation of a highly reactive electrophile when phenol reacts with chloroform in the presence of a strong base. Through the process of dehydrohalogenation, chloroform (\text{CHCl}_3$) loses a proton and a chloride ion to form dichlorocarbene ($: \text{CCl}_2$), an electron-deficient neutral intermediate with an incomplete octet. This reactive species then functions as the electrophile by attacking the electron-rich phenoxide ion to facilitate the formylation process.

Grignard reagents like $\mathrm{C}_6\mathrm{H}_5\mathrm{MgBr}$ function as strong bases that readily undergo acid-base reactions with any compound possessing an active hydrogen atom, such as methanol. Because the hydroxyl proton in $\mathrm{CH}_3\mathrm{OH}$ is acidic, the phenyl group abstracts it to form the stable hydrocarbon benzene, denoted as $\mathrm{C}_6\mathrm{H}_6$. This transformation leaves behind the methoxide group to complex with the magnesium bromide moiety, yielding $\mathrm{Mg(OMe)Br}$ as the accompanying inorganic product.

The transformation of an allylic alcohol into an $\alpha,\beta$-unsaturated ketone requires a selective oxidizing agent capable of targeting the hydroxyl group without disrupting the sensitive carbon-carbon double bond. Stronger oxidizing agents such as acidic permanganate or acidic dichromate are non-specific and aggressive; they risk cleaving the alkene bond or over-oxidizing the structure through harsh reaction conditions. In contrast, Pyridinium chlorochromate functions under milder conditions, allowing it to efficiently oxidize the secondary alcohol $\mathrm{CH}_3-\mathrm{CH(OH)}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$ to the corresponding ketone $\mathrm{CH}_3-\mathrm{CO}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$ while preserving the integrity of the unsaturation elsewhere in the molecule.

Acid-catalyzed hydration proceeds according to Markovnikov's rule, where the electrophilic hydrogen attaches to the alkene carbon bearing more hydrogens, generating a carbocation intermediate that is subsequently attacked by water. Because the hydroxyl group bonds at the more substituted carbon, the reaction avoids forming primary alcohols and instead yields secondary or tertiary alcohols, as illustrated by propene forming $\mathrm{CH}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3$ and 2-methylpropene forming $\mathrm{CH}_3-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)_2$. This process ultimately results in the production of a mixture of secondary or tertiary alcohols depending on the degree of substitution in the starting alkene.

This transformation relies on the standard reactivity of acyl halides with organometallic reagents, where an initial nucleophilic acyl substitution is followed by a second addition step to produce a tertiary alcohol. When acetyl bromide reacts with the first equivalent of $\mathrm{CH}_3\mathrm{MgI}$, it generates acetone as an intermediate, which is rapidly consumed by the excess Grignard reagent present in the mixture. The subsequent nucleophilic attack by the second methyl group on the carbonyl carbon, followed by workup with a saturated $\mathrm{NH}_4\mathrm{Cl}$ solution, yields 2-methyl-2-propanol as the final product.

The acid-catalyzed dehydration of alcohols follows an E1 mechanism, where the reaction rate is determined by the stability of the intermediate carbocation, which follows the order tertiary > secondary > primary. Structure A, 3-methylpentan-3-ol, undergoes dehydration most readily because the loss of its hydroxyl group results in the formation of a stable tertiary carbocation represented by $\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}^+(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3$. Because the other provided compounds contain only primary or secondary hydroxyl groups, they would form significantly less stable carbocation intermediates, making them more resistant to the dehydration process.

The volatility of a substance is governed by the strength of the intermolecular forces holding its molecules together, and isomers like alcohol and ether with the same molecular formula, such as $\mathrm{C_2H_6O}$, exhibit distinct bonding characteristics. In alcohols like $\mathrm{CH_3CH_2OH}$, the presence of inter-molecular hydrogen bonding creates strong attractions between molecules, whereas ethers like $\mathrm{CH_3-O-CH_3}$ rely on weaker Van der Waals forces. Since hydrogen bonds are far more robust than Van der Waals forces, alcohol molecules are more difficult to pull apart into the vapor phase, causing the ether to be more volatile by comparison.

Concentrated $\mathrm{H}_2\mathrm{SO}_4$ acts as a crucial source of protons because the oxygen atom of an alcohol molecule possesses lone pairs of electrons that readily attract an $\mathrm{H}^+$ ion. This protonation process transforms the inherently poor hydroxyl leaving group ($-\mathrm{OH}$) into a significantly better leaving group, water ($-\mathrm{OH}_2^+$), which activates the molecule for the next stages of the reaction. As a result, the protonation of the alcohol molecule occurs as the vital initiation step, effectively preparing the substrate for the subsequent formation of a carbocation and the eventual elimination of a proton to yield an alkene.

This reaction sequence follows a standard synthetic strategy for replacing a hydroxyl group with a deuterium atom via an organometallic intermediate. Initially, treating the starting alcohol with $\mathrm{NaI}$ and $\mathrm{H}_3\mathrm{PO}_4$ facilitates the substitution of the -OH group with iodine, producing 2-iodobutane. This alkyl iodide subsequently reacts with magnesium in dry ether to generate the Grignard reagent $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{MgI})-\mathrm{CH}_3$. In the final step, exposing this reagent to $\mathrm{D}_2\mathrm{O}$ prompts the carbanionic center of the Grignard reagent to abstract a deuterium ion, resulting in the final product, $\mathrm{H}_3\mathrm{C}-\mathrm{CH}_2-\mathrm{CH}(\mathrm{D})-\mathrm{CH}_3$.

The reaction sequence initiates with the reduction of glucose by hydriodic acid to yield n-hexane, which is subsequently aromatized using vanadium pentoxide at 10-20 atm and 773 K to produce benzene. This benzene acts as the substrate for a Friedel-Crafts acylation involving benzoyl chloride and aluminum chloride, ultimately forming the major product benzophenone, $C_6H_5COC_6H_5$.

Evaluating the electron distribution in benzophenone requires identifying the $\pi$ electrons and lone pairs. The two aromatic benzene rings each contain six $\pi$ electrons, totaling 12 electrons from the aromatic systems. The carbonyl group contributes two $\pi$ electrons from its double bond, while the oxygen atom possesses two lone pairs, accounting for an additional four electrons. Summing these, the total number of $\pi$ electrons and lone pair electrons in the molecule equals 18.