Organic Compounds Containing Oxygen MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: The classification of an oxygen-containing organic compound depends on the connectivity of its functional group, not merely on the presence of an oxygen atom. In phenols, the hydroxyl group is bonded directly to an aromatic-ring carbon. Therefore, \(\mathrm{C_6H_5OH}\) is phenol because its oxygen-bearing carbon belongs to the benzene ring. In \(\mathrm{C_6H_5CH_2OH}\), the hydroxyl group is attached to a saturated carbon outside the aromatic ring, making the compound benzyl alcohol. The additional \(\mathrm{-CH_2-}\) group changes the functional-group classification even though the molecule still contains a benzene ring. In \(\mathrm{C_6H_5OCH_3}\), oxygen connects phenyl and methyl groups, so the compound is an ether. The compound \(\mathrm{C_6H_5CHO}\) contains an aldehyde group attached to the aromatic ring. This distinction is important because phenols and benzyl alcohols exhibit different acidity and chemical reactivity. Direct attachment of \(\mathrm{-OH}\) to an aromatic carbon is the decisive structural feature.

Explanation: Sodium borohydride is a reducing agent that converts aldehydes and ketones into their corresponding alcohols. During reduction, the carbonyl group undergoes addition of hydrogen equivalents, and the carbon skeleton remains unchanged. Propanal has the structure \(\mathrm{CH_3CH_2CHO}\), in which the carbonyl carbon is bonded to one alkyl group and one hydrogen atom. Reduction converts its terminal \(\mathrm{-CHO}\) group into \(\mathrm{-CH_2OH}\), producing propan-1-ol. Propanone has the structure \(\mathrm{CH_3COCH_3}\), in which the carbonyl carbon is bonded to two methyl groups. Its reduction gives \(\mathrm{CH_3CH(OH)CH_3}\), which is propan-2-ol. The resulting hydroxyl-bearing carbon is primary in the first product and secondary in the second. Neither reduction adds a carbon atom or rearranges the carbon skeleton. Formation of carboxylic acids would require oxidation rather than reduction. Thus, identifying whether the starting carbonyl compound is an aldehyde or ketone determines the alcohol class obtained.

Explanation: Both treatments convert the carbon–carbon double bond of propene into an alcohol, but the position at which the hydroxyl group appears depends on the reaction pathway. In acid-catalysed hydration, the alkene is protonated to produce the more stable carbocation. For propene, protonation that gives a secondary carbocation is favoured over formation of a primary carbocation. Water subsequently attacks the secondary carbocation, and deprotonation produces propan-2-ol. The addition therefore follows Markovnikov's rule. Hydroboration–oxidation proceeds differently and does not involve the formation of the same free carbocation intermediate. During hydroboration, boron becomes attached preferentially to the less substituted carbon of the double bond. Subsequent oxidation replaces the carbon–boron bond with a carbon–oxygen bond. Consequently, propan-1-ol is obtained from Treatment II. Both products retain the original three-carbon skeleton. The difference arises from regioselectivity rather than a change in the number of carbon atoms or the overall addition of water.

Explanation: Grignard reagents contain a carbon–magnesium bond in which the carbon behaves as a nucleophilic centre. Ethylmagnesium bromide contributes an ethyl group, \(\mathrm{-CH_2CH_3}\), to an electrophilic carbonyl carbon. The first stage produces a magnesium alkoxide, which yields an alcohol only after hydrolysis. Methanal has two hydrogen atoms attached to its carbonyl carbon. Addition of the ethyl group gives an alkoxide whose carbon bearing oxygen is attached to only one carbon group. Hydrolysis therefore produces the primary alcohol propan-1-ol. The carbon-skeleton change can be represented as: \[ \mathrm{HCHO \xrightarrow[\text{2. }H_3O^+]{\text{1. }C_2H_5MgBr} CH_3CH_2CH_2OH} \] Propanone contains two methyl groups attached to the carbonyl carbon. Addition of an ethyl group creates a carbon bonded to three alkyl groups, giving a tertiary alcohol after hydrolysis. \[ \mathrm{CH_3COCH_3 \xrightarrow[\text{2. }H_3O^+]{\text{1. }C_2H_5MgBr} (CH_3)_2C(OH)CH_2CH_3} \] The first product contains three carbon atoms, whereas the second contains five. This difference reflects the carbon atoms already present in the two carbonyl compounds. The defining principle is that methanal gives a primary alcohol, while a ketone gives a tertiary alcohol on Grignard addition followed by hydrolysis.

Explanation: \(\textbf{Required product structure:}\) Pentan-3-ol has its hydroxyl group on the middle carbon of a five-carbon chain. \[ \mathrm{CH_3CH_2CH(OH)CH_2CH_3} \] The carbon bearing the hydroxyl group is attached to two ethyl groups and one hydrogen atom. It is therefore a secondary alcohol. A secondary alcohol can be prepared by adding a Grignard reagent to an aldehyde other than methanal, followed by hydrolysis. \(\textbf{Reverse carbonyl identification:}\) One ethyl group must already be attached to the aldehyde carbonyl carbon, while the other can be supplied by the Grignard reagent. Propanal has an ethyl group attached to its carbonyl carbon. \[ \mathrm{CH_3CH_2CHO+C_2H_5MgBr \longrightarrow CH_3CH_2CH(OMgBr)CH_2CH_3} \] \(\textbf{Hydrolysis:}\) \[ \mathrm{CH_3CH_2CH(OMgBr)CH_2CH_3 \xrightarrow{H_3O^+} CH_3CH_2CH(OH)CH_2CH_3} \] Butanal with methylmagnesium bromide instead gives pentan-2-ol because the two alkyl groups attached to the alcohol carbon are unequal. Propanone with ethylmagnesium bromide gives a tertiary alcohol because the ketone already has two carbon substituents. Methanal with a butyl Grignard reagent gives a primary alcohol. Thus, both the carbonyl class and the distribution of carbon atoms around its carbonyl carbon must be checked. Matching the total carbon count alone is insufficient to identify the required alcohol.

Explanation: \(\textbf{Reaction I — Carbonyl addition:}\) Ethylmagnesium bromide contains a two-carbon ethyl group. Methanal contributes one additional carbon atom through its carbonyl carbon. Their reaction, followed by hydrolysis, produces the three-carbon primary alcohol propan-1-ol. \[ \mathrm{CH_3CH_2MgBr+HCHO \xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}} CH_3CH_2CH_2OH} \] \(\textbf{Complete oxidation:}\) A primary alcohol undergoes oxidation through the aldehyde oxidation level to the corresponding carboxylic acid under sufficiently strong oxidising conditions. \[ \mathrm{CH_3CH_2CH_2OH+2[O] \xrightarrow{acidified\ KMnO_4} CH_3CH_2COOH+H_2O} \] The final product of Reaction I is propanoic acid. \(\textbf{Reaction II — Carbon dioxide addition:}\) A Grignard reagent reacts with carbon dioxide to form a magnesium carboxylate. Acidification converts this intermediate into the corresponding carboxylic acid. \[ \mathrm{CH_3CH_2MgBr+CO_2 \longrightarrow CH_3CH_2COOMgBr} \] \[ \mathrm{CH_3CH_2COOMgBr \xrightarrow{H_3O^+} CH_3CH_2COOH} \] Carbon dioxide contributes one carbon atom, just as methanal contributes one carbon atom in Reaction I. The difference is that carbon dioxide directly establishes the carboxyl oxidation level, whereas methanal initially produces an alcohol. Both pathways therefore contain three carbon atoms in their final products and produce the same carboxylic acid. This is an example of distinct preparation routes converging on an identical functional-group product.

Explanation: Chlorobenzene and benzyl chloride contain the same aromatic ring but differ in the carbon atom to which chlorine is attached. In chlorobenzene, chlorine is bonded directly to an aromatic \(sp^2\)-hybridised carbon. The aryl carbon–chlorine bond has partial double-bond character due to resonance and does not readily undergo ordinary nucleophilic substitution under mild conditions. Therefore, the specified severe conditions are required to convert chlorobenzene into sodium phenoxide. Acidification of sodium phenoxide produces phenol, in which the hydroxyl group is directly attached to the benzene ring. In benzyl chloride, chlorine is attached to a saturated carbon adjacent to the aromatic ring. Aqueous hydroxide can replace this chlorine under much milder conditions, producing benzyl alcohol. The resulting hydroxyl group remains attached to the side-chain carbon rather than the ring carbon. Both reactions replace a chlorine-containing functionality with a hydroxyl-containing functionality, but they produce different functional-group classes. The important distinction is the original connectivity of the carbon–chlorine bond. Similar molecular components do not imply identical substitution behaviour or identical final functional groups.

Explanation: Benzenesulphonic acid and its sodium salt provide a preparation route for phenol through alkaline fusion. During fusion with molten sodium hydroxide, the sulphonic-acid-derived group attached to the aromatic ring is replaced by an oxygen-containing group. Because the reaction mixture is strongly alkaline, the product exists as sodium phenoxide rather than free phenol. Sodium phenoxide contains the phenoxide ion, \(\mathrm{C_6H_5O^-}\), paired with sodium ions. Subsequent acidification protonates the phenoxide ion and releases phenol. This acidification step is necessary for obtaining the neutral hydroxyl compound from its salt. The aromatic six-carbon skeleton is retained throughout the conversion. Benzyl alcohol cannot be the product because the route does not introduce a \(\mathrm{-CH_2OH}\) side chain. Benzoic acid would require formation of a carboxyl group, which is not accomplished by this transformation. Anisole requires an additional methyl group bonded to oxygen, and no methylating reagent is present. The defining sequence is therefore alkaline fusion to sodium phenoxide followed by acidification to phenol.

Explanation: \(\textbf{Initial industrial conversion:}\) Cumene is isopropylbenzene, containing an isopropyl substituent attached to the aromatic ring. Oxidation with air produces cumene hydroperoxide. Acid-catalysed decomposition of this hydroperoxide yields two important organic products: phenol and acetone. The aromatic portion of the starting compound gives phenol, while the three-carbon side-chain portion gives acetone. This transformation therefore involves rearrangement and cleavage rather than a simple conversion of the side-chain carbon into an alcohol. \(\textbf{Subsequent selective reduction:}\) Acetone is propanone, a ketone with the structure \(\mathrm{CH_3COCH_3}\). Sodium borohydride reduces its carbonyl group to a secondary alcohol. \[ \mathrm{CH_3COCH_3 \xrightarrow[\text{aqueous work-up}]{NaBH_4} CH_3CH(OH)CH_3} \] The product is propan-2-ol because the carbon bearing the hydroxyl group is attached to two methyl groups. Phenol does not contain an aldehyde or ketone carbonyl group and is not converted into benzyl alcohol by sodium borohydride. The reduction therefore changes the acetone component while leaving the phenolic component unchanged. The final mixture contains phenol and propan-2-ol. This example illustrates how the products of an initial preparation reaction can respond differently to a subsequent reagent.

Explanation: \(\textbf{Given:}\) Each ethanol sample contains \(0.60\,\mathrm{mol}\). The samples undergo different dehydration pathways because the reaction temperatures are different. \(\textbf{Stage 1 — Intermolecular dehydration:}\) At \(413\,\mathrm{K}\), ethanol undergoes intermolecular dehydration to form ethoxyethane. \[ 2\mathrm{C_2H_5OH} \xrightarrow[\ 413\,\mathrm{K}\ ]{\mathrm{conc.\ H_2SO_4}} \mathrm{C_2H_5OC_2H_5+H_2O} \] The balanced equation shows that two moles of ethanol produce one mole of ethoxyethane. \(\textbf{Stoichiometric relation:}\) \[ n(\text{ethoxyethane})=\frac{n(\text{ethanol})}{2} \] \(\textbf{Substitution:}\) \[ n(\text{ethoxyethane})=\frac{0.60}{2} \] \[ n(\text{ethoxyethane})=0.30\,\mathrm{mol} \] \(\textbf{Stage 2 — Intramolecular dehydration:}\) At \(443\,\mathrm{K}\), ethanol undergoes dehydration to form ethene. \[ \mathrm{C_2H_5OH} \xrightarrow[\ 443\,\mathrm{K}\ ]{\mathrm{conc.\ H_2SO_4}} \mathrm{CH_2{=}CH_2+H_2O} \] One mole of ethanol produces one mole of ethene. \[ n(\text{ethene})=n(\text{ethanol}) \] \[ n(\text{ethene})=0.60\,\mathrm{mol} \] \(\textbf{Required mole ratio:}\) \[ \frac{n(\text{ethoxyethane})}{n(\text{ethene})} =\frac{0.30}{0.60} \] \[ =\frac{1}{2} \] The ether pathway combines two ethanol molecules into one organic product molecule, whereas the alkene pathway converts each ethanol molecule into one ethene molecule. This difference explains the mole ratio even though the initial ethanol amounts are equal. The carbon balance is consistent in both reactions: four carbon atoms enter each ether molecule, while two carbon atoms enter each ethene molecule. \(\textbf{Final answer:}\) \[ \boxed{n(\text{ethoxyethane}):n(\text{ethene})=1:2} \]

Explanation: Ethoxybenzene has the structure \(\mathrm{C_6H_5OC_2H_5}\), in which oxygen connects a phenyl group to an ethyl group. Williamson ether synthesis proceeds through nucleophilic substitution between an alkoxide or phenoxide ion and a suitable alkyl halide. Sodium phenoxide provides the \(\mathrm{C_6H_5O^-}\) nucleophile required for constructing the phenyl–oxygen portion of the ether. Bromoethane provides the ethyl group and contains a primary carbon–bromine bond that is suitable for \(S_\mathrm{N}2\) substitution. The phenoxide oxygen attacks the ethyl carbon as bromide leaves, forming ethoxybenzene. Using sodium ethoxide with bromobenzene is not an equivalent practical route because the aryl carbon–bromine bond does not undergo ordinary \(S_\mathrm{N}2\) substitution. Sodium phenoxide and chlorobenzene likewise do not provide the required ethyl group. Sodium ethoxide with benzyl bromide instead forms benzyl ethyl ether, which contains an additional methylene group between the aromatic ring and oxygen. The correct retrosynthetic choice must preserve the desired connectivity and place the leaving group on a carbon suitable for substitution.

Explanation: \(\textbf{Required ether connectivity:}\) The target compound is 2-methoxy-2-methylpropane, containing a methyl group and a tert-butyl group attached to oxygen. A Williamson reaction requires a nucleophilic alkoxide and an alkyl halide capable of undergoing \(S_\mathrm{N}2\) substitution. The important structural decision is which alkyl group should come from the halide and which should come from the alkoxide. \(\textbf{Preferred substitution route:}\) Sodium tert-butoxide provides the tert-butyl–oxygen portion of the ether. Bromomethane provides an unhindered methyl carbon at which nucleophilic substitution is highly favourable. \[ \mathrm{(CH_3)_3CONa+CH_3Br \longrightarrow (CH_3)_3COCH_3+NaBr} \] \(\textbf{Why the reversed route fails:}\) If tert-butyl bromide is used with sodium methoxide, the leaving group is attached to a tertiary carbon. Ordinary \(S_\mathrm{N}2\) substitution is strongly hindered at this carbon. Methoxide is also a strong base and favours elimination, giving 2-methylpropene instead of the required ether. \[ \mathrm{(CH_3)_3CBr+NaOCH_3 \longrightarrow (CH_3)_2C{=}CH_2+CH_3OH+NaBr} \] The combination containing sodium ethoxide and tert-butyl bromide suffers from the same elimination preference. Sodium methoxide with bromoethane can undergo substitution, but it produces methoxyethane rather than the target ether. Thus, the correct synthesis places the bulky tert-butyl group in the alkoxide and the unhindered methyl group in the halide. The ability to exchange the two nominal ether fragments does not mean the two Williamson routes are chemically equivalent.

Explanation: Acid-catalysed intermolecular dehydration involves reaction between suitable alcohol molecules to form an ether and water. When only one primary alcohol is used, the corresponding symmetrical ether can be obtained under suitable conditions. A mixture of methanol and ethanol contains two different alcohols capable of participating in intermolecular dehydration. Consequently, methanol can react with methanol to form dimethyl ether, and ethanol can react with ethanol to form diethyl ether. Methanol can also react with ethanol to form the desired unsymmetrical ether, methoxyethane. The resulting reaction therefore lacks the selectivity needed to produce only one ether. Williamson synthesis provides greater control because the alkoxide and alkyl halide partners are selected independently. Sodium methoxide supplies the methoxy group, while bromoethane supplies the ethyl group through substitution at a primary carbon. The desired transformation is: \[ \mathrm{CH_3ONa+CH_3CH_2Br \longrightarrow CH_3OCH_2CH_3+NaBr} \] The specified partners provide a direct route to the required connectivity without the three competing alcohol-pair combinations. The key reasoning is not simply that both methods can form ethers, but that the preparation of an unsymmetrical ether requires control over which carbon fragments become connected through oxygen.

Explanation: \(\textbf{Stage 1 — Diazotisation:}\) Aniline is an aromatic primary amine. Treatment with nitrous acid generated from sodium nitrite and hydrochloric acid at \(273-278\,\mathrm{K}\) converts it into benzenediazonium chloride. \[ \mathrm{C_6H_5NH_2 \xrightarrow[\ 273-278\,K\ ]{NaNO_2/HCl} C_6H_5N_2^+Cl^-} \] \(\textbf{Stage 2 — Hydrolysis:}\) On warming with water, the diazonium group is replaced by a hydroxyl group, forming phenol. \[ \mathrm{C_6H_5N_2^+Cl^-+H_2O \xrightarrow{warm} C_6H_5OH+N_2+HCl} \] Thus, compound P is phenol. \(\textbf{Stage 3 — Formation of phenoxide:}\) Phenol reacts with sodium hydroxide to form sodium phenoxide because phenol is sufficiently acidic to undergo deprotonation by hydroxide ions. \[ \mathrm{C_6H_5OH+NaOH \longrightarrow C_6H_5ONa+H_2O} \] Compound Q is therefore sodium phenoxide. \(\textbf{Stage 4 — Ether formation:}\) Sodium phenoxide attacks the primary carbon of bromoethane through Williamson ether synthesis. \[ \mathrm{C_6H_5ONa+C_2H_5Br \longrightarrow C_6H_5OC_2H_5+NaBr} \] Compound R is ethoxybenzene. The aromatic carbon framework remains unchanged throughout the sequence, while the nitrogen-containing group is first replaced by hydroxyl and then converted into an ether linkage. No methylene group is inserted between the benzene ring and oxygen. Therefore, the final product is an alkyl aryl ether rather than benzyl alcohol or a benzyl ether.

Explanation: \(\textbf{Initial structural identification:}\) Propan-1-ol is a primary alcohol with three carbon atoms. Primary alcohols can be oxidised to aldehydes or further to carboxylic acids depending on the oxidising reagent and reaction conditions. \(\textbf{Portion I — Controlled oxidation:}\) PCC is a reagent used to oxidise primary alcohols to aldehydes without the usual further oxidation associated with strong aqueous oxidising conditions. \[ \mathrm{CH_3CH_2CH_2OH+[O] \xrightarrow{PCC} CH_3CH_2CHO+H_2O} \] The product P is propanal. Propanal contains a polar carbonyl group whose carbon atom is electrophilic. Cyanide ions generated in the HCN addition system can attack this carbonyl carbon to form a cyanohydrin after protonation. This supports the identification of P as a carbonyl compound. \(\textbf{Portion II — Complete oxidation:}\) Acidified potassium permanganate is sufficiently strong to oxidise the primary alcohol to its corresponding carboxylic acid. \[ \mathrm{CH_3CH_2CH_2OH+2[O] \xrightarrow{acidified\ KMnO_4} CH_3CH_2COOH+H_2O} \] Compound Q is therefore propanoic acid. Carboxylic acids react with sodium hydrogencarbonate to liberate carbon dioxide because they are sufficiently acidic to protonate hydrogencarbonate ions. \[ \mathrm{CH_3CH_2COOH+NaHCO_3 \longrightarrow CH_3CH_2COONa+CO_2+H_2O} \] The supplementary reactions confirm the functional-group identities inferred from the preparation conditions. Both P and Q retain three carbon atoms, but their oxidation levels and characteristic reactions differ. Therefore, P is propanal and Q is propanoic acid.

Explanation: \(\textbf{Stage 1 — Locate the double bond:}\) The starting compound is 2-methylbut-2-ene, containing five carbon atoms. One carbon of its double bond is attached to two methyl groups, while the other double-bond carbon is attached to a methyl group and a hydrogen atom. \(\textbf{Stage 2 — Ozonolysis:}\) Reductive ozonolysis cleaves the carbon–carbon double bond and converts each original double-bond carbon into a carbonyl carbon. The carbon carrying two methyl groups becomes the carbonyl carbon of propanone. The carbon carrying one methyl group and one hydrogen becomes the carbonyl carbon of ethanal. \[ \mathrm{CH_3C(CH_3){=}CHCH_3 \xrightarrow[\text{2. }Zn/H_2O]{\text{1. }O_3} CH_3COCH_3+CH_3CHO} \] \(\textbf{Stage 3 — Carbonyl reduction:}\) Propanone is a ketone, and sodium borohydride reduces its carbonyl group to a secondary alcohol. \[ \mathrm{CH_3COCH_3 \xrightarrow[\text{aqueous work-up}]{NaBH_4} CH_3CH(OH)CH_3} \] This product is propan-2-ol. Ethanal is an aldehyde, and reduction converts it into a primary alcohol. \[ \mathrm{CH_3CHO \xrightarrow[\text{aqueous work-up}]{NaBH_4} CH_3CH_2OH} \] This product is ethanol. \(\textbf{Carbon-skeleton check:}\) The original five carbon atoms are divided into a three-carbon fragment and a two-carbon fragment during ozonolysis. Sodium borohydride changes the carbonyl groups into hydroxyl groups without altering the carbon count of either fragment. The product classes are also consistent: the ketone gives a secondary alcohol and the aldehyde gives a primary alcohol. Hence, the final mixture contains propan-2-ol and ethanol.

Explanation: \(\textbf{Starting functional group:}\) Ethanenitrile has the structure \(\mathrm{CH_3CN}\). The nitrile carbon is part of the carbon skeleton and can be converted into a carbonyl carbon through suitable reactions. However, different reagents produce different carbonyl classes. \(\textbf{Sequence I — Stephen reduction:}\) Stannous chloride in hydrochloric acid reduces the nitrile to an imine-derived intermediate. Subsequent hydrolysis converts this intermediate into an aldehyde. \[ \mathrm{CH_3CN \xrightarrow[\text{2. hydrolysis}]{\text{1. }SnCl_2/HCl} CH_3CHO} \] The product is ethanal, which contains two carbon atoms. This route is useful because it converts a nitrile into an aldehyde without introducing an additional carbon atom. The aldehyde is obtained after hydrolysis of the reduction intermediate rather than by complete reduction to an amine. \(\textbf{Sequence II — Grignard addition:}\) Methylmagnesium bromide supplies a nucleophilic methyl group. This methyl group adds to the electrophilic carbon of the nitrile, producing a magnesium-containing imine intermediate. Acidic hydrolysis of that intermediate gives a ketone. \[ \mathrm{CH_3CN \xrightarrow[\text{2. }H_3O^+]{\text{1. }CH_3MgBr,\ dry\ ether} CH_3COCH_3} \] The product is propanone, which contains three carbon atoms. One carbon atom has been added from the Grignard reagent, while the original nitrile carbon becomes the ketone carbonyl carbon. Thus, Sequence I changes the oxidation and bonding arrangement without increasing the carbon count, whereas Sequence II constructs a new carbon–carbon bond. The final functional-group distinction is equally important: Stephen reduction followed by hydrolysis gives an aldehyde, while reaction with a Grignard reagent followed by hydrolysis gives a ketone. Therefore, ethanal and propanone are the correct final products.

Explanation: Propanoyl chloride is an acyl chloride containing a three-carbon skeleton and the functional group \(\mathrm{-COCl}\). Its reduction product depends on the reducing agent and the extent of reduction. In Portion I, hydrogenation over palladium supported on barium sulphate is the Rosenmund reduction. The specially controlled catalyst permits conversion of the acyl chloride into an aldehyde without ordinary further hydrogenation to the alcohol. Consequently, propanoyl chloride produces propanal. \[ \mathrm{CH_3CH_2COCl+H_2 \xrightarrow{Pd/BaSO_4} CH_3CH_2CHO+HCl} \] Propanal can subsequently be reduced with a suitable carbonyl-reducing agent to produce propan-1-ol. Statement I is therefore correct. Lithium aluminium hydride is a considerably stronger reducing agent and reduces an acyl chloride to a primary alcohol under the stated work-up conditions. The aldehyde oxidation level is passed during this reduction, but an aldehyde need not be isolated as the final product. \[ \mathrm{CH_3CH_2COCl \xrightarrow[\text{2. aqueous work-up}]{\text{1. }LiAlH_4} CH_3CH_2CH_2OH} \] Statement II is also correct. Both reactions preserve the three-carbon skeleton, but the final oxidation levels differ. The important principle is that reagent selection determines whether reduction stops at the aldehyde or proceeds to the primary alcohol.

Explanation: \(\textbf{Sequence I — Controlled side-chain oxidation:}\) Toluene contains a methyl group directly attached to the benzene ring. Chromyl chloride selectively oxidises this benzylic methyl group through formation of a chromium-containing intermediate. Subsequent hydrolysis produces benzaldehyde. This transformation is known as the Etard reaction. \[ \mathrm{C_6H_5CH_3 \xrightarrow[\text{2. hydrolysis}]{\text{1. }CrO_2Cl_2} C_6H_5CHO} \] The product retains the carbon atom of the original methyl substituent as the aldehyde carbonyl carbon. \(\textbf{Sequence II — Strong oxidation:}\) Hot alkaline potassium permanganate oxidises an alkyl side chain possessing a benzylic hydrogen to a carboxylate group. Subsequent acidification gives benzoic acid. \[ \mathrm{C_6H_5CH_3 \xrightarrow[\text{2. }H_3O^+]{\text{1. hot alkaline }KMnO_4} C_6H_5COOH} \] The benzene ring remains intact, and the side-chain carbon is converted into the carboxyl carbon. \(\textbf{Chemical identification:}\) Benzaldehyde does not possess the acidic carboxyl group required to liberate carbon dioxide from sodium hydrogencarbonate. Benzoic acid does possess this group and undergoes an acid–base reaction. \[ \mathrm{C_6H_5COOH+NaHCO_3 \longrightarrow C_6H_5COONa+CO_2+H_2O} \] Thus, only the product of Sequence II produces carbon dioxide effervescence. The combined evidence distinguishes controlled oxidation to an aldehyde from stronger oxidation to a carboxylic acid. The different products arise from the oxidising conditions, not from differences in the starting aromatic compound.

Explanation: \(\textbf{Reaction I — Aromatic formylation:}\) Benzene reacts with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride and cuprous chloride. This is the Gattermann–Koch reaction, which introduces a formyl group into the aromatic ring. The carbon atom of carbon monoxide becomes the carbonyl carbon of benzaldehyde. \[ \mathrm{C_6H_6+CO \xrightarrow{HCl,\ AlCl_3/CuCl} C_6H_5CHO} \] Benzaldehyde is an aldehyde because its carbonyl carbon is attached to one phenyl group and one hydrogen atom. Reduction converts the aldehyde group into a primary alcohol group. \[ \mathrm{C_6H_5CHO \xrightarrow[\text{aqueous work-up}]{NaBH_4} C_6H_5CH_2OH} \] This product is benzyl alcohol. \(\textbf{Reaction II — Aromatic acylation:}\) Benzene reacts with ethanoyl chloride through Friedel–Crafts acylation to produce acetophenone. \[ \mathrm{C_6H_6+CH_3COCl \xrightarrow{anhydrous\ AlCl_3} C_6H_5COCH_3+HCl} \] Acetophenone is a ketone because the carbonyl carbon is attached to phenyl and methyl groups. Reduction converts this carbonyl carbon into the secondary alcohol carbon of 1-phenylethanol. \[ \mathrm{C_6H_5COCH_3 \xrightarrow[\text{aqueous work-up}]{NaBH_4} C_6H_5CH(OH)CH_3} \] The products therefore differ in both carbon count and alcohol classification. Reaction I introduces one carbon atom through formylation, whereas Reaction II introduces a two-carbon acyl fragment. Neither reduction changes the carbon skeleton formed in the preceding aromatic substitution step. Hence, the alcohols are benzyl alcohol and 1-phenylethanol, respectively.