Organic Compounds Containing Oxygen MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: \(\textbf{Identify the two reducible functionalities:}\) Cinnamaldehyde contains an aldehyde carbonyl group and an alkene linkage. The two double bonds are conjugated, but they are not chemically identical. Their selective reduction therefore depends on the reagent and conditions. \(\textbf{Route P — Carbonyl-selective hydride reduction:}\) The question specifies that sodium borohydride reduces the aldehyde group while leaving the carbon–carbon double bond unchanged. \[ \mathrm{C_6H_5CH=CHCHO} \] \[ \mathrm{\xrightarrow{NaBH_4} C_6H_5CH=CHCH_2OH} \] The product is cinnamyl alcohol, also called 3-phenylprop-2-en-1-ol. It contains a primary alcohol group and retains the alkene linkage. It can therefore undergo ordinary bromine addition at the carbon–carbon double bond. It no longer contains the original aldehyde carbonyl group and does not give the characteristic ordinary aldehyde Tollens' response. \(\textbf{Route Q — Alkene-selective hydrogenation:}\) Under the explicitly supplied conditions, one equivalent of hydrogen saturates the carbon–carbon double bond. \[ \mathrm{C_6H_5CH=CHCHO+H_2} \] \[ \mathrm{\longrightarrow C_6H_5CH_2CH_2CHO} \] The product is 3-phenylpropanal. Its aldehyde group remains intact and can reduce Tollens' reagent. However, the carbon–carbon double bond originally present in cinnamaldehyde has been consumed. \(\textbf{Interpret the diagnostic distinction:}\) Route P retains the alkene but removes the aldehyde functionality. Route Q retains the aldehyde but removes the alkene functionality. The two independent tests therefore provide complementary evidence of successful chemoselectivity. The question requires tracking which double bond is reduced rather than assuming every reduction of a conjugated system gives complete saturation.

Explanation: \(\textbf{Interpret the first three observations:}\) A positive 2,4-dinitrophenylhydrazine reaction indicates a suitable aldehyde or ketone carbonyl group. Positive Tollens' and Fehling's reactions under ordinary conditions are consistent with an aliphatic aldehyde. Thus, butanal and 2-methylpropanal remain plausible after the diagnostic-test evidence. Butan-2-one is a ketone and does not satisfy the characteristic aldehyde tests. 2-Methylpropan-2-ol contains no aldehyde or ketone carbonyl group. \(\textbf{Examine the alpha-carbon of butanal:}\) Butanal contains the structure \(\mathrm{CH_3CH_2CH_2CHO}\). Its alpha-carbon is a methylene group containing two hydrogen atoms. During self-aldol addition, one alpha-hydrogen is removed to form the donor nucleophile. The resulting donor-derived alpha-carbon retains another hydrogen. Therefore, its beta-hydroxy aldehyde can undergo ordinary subsequent dehydration. \(\textbf{Examine the alpha-carbon of 2-methylpropanal:}\) 2-Methylpropanal contains the structure \(\mathrm{(CH_3)_2CHCHO}\). Its alpha-carbon has only one hydrogen. Self-aldol addition uses that hydrogen during enolate formation. After the carbon–carbon bond forms, the donor-derived alpha-carbon becomes bonded to four carbon substituent connections and carries no hydrogen. The retained aldehyde is therefore adjacent to an alpha-carbon lacking the hydrogen required for ordinary aldol dehydration. \(\textbf{Use the fourth observation:}\) Only 2-methylpropanal matches this post-addition structural condition. The diagnostic tests establish the carbonyl class, while the aldol product's alpha-hydrogen count distinguishes the positional isomers. The final identification therefore depends on analysing the product of an initial reaction rather than simply noting that the starting aldehyde contains an alpha-hydrogen.

Explanation: \(\textbf{Identify the two distinct carbonyl-containing groups:}\) 3-Oxobutanoic acid contains a carboxylic acid group and a separate ketone carbonyl. Its structure can also be written as: \[ \mathrm{CH_3COCH_2COOH} \] The ketone carbonyl is directly attached to a methyl group. Therefore, the molecule contains the characteristic methyl ketone arrangement required for the iodoform reaction. \(\textbf{Identify the carbon entering iodoform:}\) During the haloform reaction, the methyl group adjacent to the ketone carbonyl undergoes exhaustive iodination. That original methyl carbon becomes the carbon of \(\mathrm{CHI_3}\). The ketone carbonyl carbon remains with the other portion of the organic molecule. \(\textbf{Track the remaining carbon skeleton:}\) Cleavage of the methyl ketone converts its carbonyl-containing fragment into a carboxylate. The original carboxylic acid functionality also exists predominantly as carboxylate in the alkaline medium. The resulting organic skeleton is the propanedioate framework: \[ \mathrm{{^-}OOCCH_2COO^-} \] The two carboxylate groups are separated by one methylene carbon. \(\textbf{Apply mild acidification:}\) Protonation of the carboxylate groups gives: \[ \mathrm{HOOCCH_2COOH} \] This compound is propanedioic acid, also called malonic acid. \(\textbf{Check carbon conservation:}\) The starting compound contains four carbon atoms. The haloform product contains one carbon atom. The remaining dicarboxylic acid contains three carbon atoms. The specified conditions exclude subsequent decarboxylation, so this three-carbon acid remains the final organic acid. The inference requires distinguishing the carbon that becomes iodoform from the ketone carbonyl carbon that becomes an additional carboxyl group.

Explanation: \(\textbf{Assign the unknown compound amounts:}\) Let: \[ x=n(\mathrm{CH_3CHO}) \] \[ y=n(\mathrm{C_6H_5CHO}) \] \[ z=n(\mathrm{CH_3COCH_3}) \] \(\textbf{Use the Tollens' result:}\) Ethanal and benzaldehyde are aldehydes. Both reduce Tollens' reagent under the stated conditions. Propanone is an ordinary ketone and is Tollens-negative. Each aldehyde produces two silver atoms. \[ 2(x+y)=0.80 \] \[ x+y=0.40 \] \(\textbf{Use the Fehling result:}\) Ethanal is an aliphatic aldehyde and gives the characteristic positive Fehling reaction. Benzaldehyde does not show the corresponding ordinary positive reaction. Propanone is also negative. \[ x=n(\mathrm{Cu_2O}) \] \[ x=0.25\,\mathrm{mol} \] \(\textbf{Determine benzaldehyde:}\) \[ y=0.40-0.25 \] \[ y=0.15\,\mathrm{mol} \] \(\textbf{Use the iodoform result:}\) Ethanal contains the structural arrangement required for the characteristic iodoform reaction. Propanone is a methyl ketone and is also iodoform-positive. Benzaldehyde lacks the corresponding structural arrangement. Therefore: \[ x+z=0.40 \] \(\textbf{Determine propanone:}\) \[ z=0.40-0.25 \] \[ z=0.15\,\mathrm{mol} \] \(\textbf{Verify all three observations:}\) Tollens' silver: \[ 2(0.25+0.15)=0.80\,\mathrm{mol} \] Fehling's copper(I) oxide: \[ 0.25\,\mathrm{mol} \] Iodoform: \[ 0.25+0.15=0.40\,\mathrm{mol} \] All three independently supplied experimental results are satisfied. The solution depends on recognising that ethanal is positive in both Fehling's and iodoform tests, whereas benzaldehyde and propanone have different diagnostic patterns. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{benzaldehyde}},n_{\mathrm{propanone}}\right) =(0.15,\ 0.15)\,\mathrm{mol}} \]

Explanation: \(\textbf{Examine the two carbonyl environments:}\) Butane-2,3-dione has the structure: \[ \mathrm{CH_3COCOCH_3} \] Each carbonyl carbon is attached directly to a methyl group. Therefore, the molecule contains two methyl ketone reaction sites. The question specifies complete haloform transformation at both ends. \(\textbf{Apply the first haloform cleavage:}\) One terminal methyl group undergoes exhaustive iodination. Cleavage produces one molecule of iodoform. The remaining three-carbon organic fragment contains a ketone and a newly formed carboxylate group. Its connectivity corresponds to a pyruvate-type fragment: \[ \mathrm{CH_3COCOO^-} \] \(\textbf{Apply the second haloform cleavage:}\) The remaining ketone still contains a methyl group directly attached to its carbonyl carbon. It can therefore undergo another complete haloform transformation. Its methyl carbon becomes a second molecule of iodoform. The remaining two original ketone carbonyl carbons now form the two carboxylate carbons of the final organic fragment. \[ \mathrm{{^-}OOCCOO^-} \] \(\textbf{Apply the specified acidification:}\) Mild acidification gives: \[ \mathrm{HOOCCOOH} \] This compound is ethanedioic acid, commonly called oxalic acid. \(\textbf{Verify the carbon balance:}\) The starting diketone contains four carbon atoms. Two original methyl carbons become the carbon atoms of two iodoform molecules. The remaining two carbon atoms form ethanedioic acid. Therefore, all four carbon atoms are accounted for. \(\textbf{Distinguish the intermediate from the final product:}\) Stopping after the first cleavage would leave a three-carbon carbonyl carboxylate. The question explicitly requires complete reaction of both methyl ketone sites. Consequently, the final organic acid contains only two carbon atoms. The reasoning combines repeated haloform eligibility with carbon tracing through two successive carbon–carbon bond cleavages.

Explanation: Sodium hydrogensulphite undergoes nucleophilic addition with suitable aldehydes and ketones. The carbonyl carbon acts as the electrophilic centre during this reaction. Statement I is correct because the sulphur-containing nucleophile forms a bond with the original carbonyl carbon. The carbonyl oxygen becomes part of a hydroxyl group after the necessary proton transfers. For benzaldehyde, the addition product can be represented as: \[ \mathrm{C_6H_5CHO+NaHSO_3\rightleftharpoons C_6H_5CH(OH)SO_3Na} \] Statement II is correct because treatment of the isolated addition compound with suitable dilute acid or alkali can regenerate benzaldehyde. This reversibility makes hydrogensulphite addition useful for separating and purifying certain carbonyl compounds. NCERT explains that the equilibrium generally favours addition more strongly for aldehydes than for many ketones because of steric effects. Statement III is correct because acetophenone contains the methyl ketone group \(\mathrm{-COCH_3}\). It therefore undergoes the characteristic iodoform reaction under suitable conditions. As an ordinary ketone, acetophenone does not give the characteristic positive Tollens' test. Statement IV is incorrect because ketones can undergo nucleophilic addition. Their less favourable hydrogensulphite-addition equilibria in many cases do not establish absolute chemical inability. The separation therefore relies on differing equilibrium behaviour rather than an exclusive reaction available only to aldehydes.

Explanation: \(\textbf{Identify the electrophilic carbon in the nitrile:}\) Ethanenitrile contains a polar carbon–nitrogen triple bond. The nitrile carbon can undergo nucleophilic attack by a suitable Grignard reagent. Ethylmagnesium bromide supplies a two-carbon ethyl group. \(\textbf{Stage I — Carbon–carbon bond formation:}\) The ethyl group attacks the nitrile carbon. A magnesium-containing imine derivative is formed during the initial addition. The nitrile carbon remains part of the organic skeleton. Therefore, the intermediate contains four carbon atoms. \(\textbf{Stage II — Hydrolysis to a ketone:}\) Controlled hydrolysis of the Grignard–nitrile addition intermediate produces a ketone. \[ \mathrm{CH_3CN \xrightarrow[\text{2. }H_3O^+]{\text{1. }C_2H_5MgBr} CH_3COCH_2CH_3} \] The carbonyl carbon originates from the original nitrile carbon. One group attached to it is methyl, and the other is ethyl. Consequently, X is butan-2-one. \(\textbf{Stage III — Grignard addition to X:}\) Butan-2-one contains a ketone carbonyl group capable of ordinary nucleophilic addition. Methylmagnesium bromide introduces a methyl group at its carbonyl carbon. \[ \mathrm{CH_3COCH_2CH_3 \xrightarrow[\text{2. }H_3O^+]{\text{1. }CH_3MgBr} CH_3C(OH)(CH_3)CH_2CH_3} \] The final product is 2-methylbutan-2-ol. Its hydroxyl-bearing carbon is attached to three carbon groups, making it a tertiary alcohol. \(\textbf{Compare the two Grignard reactions:}\) A nitrile followed by hydrolysis yields a ketone through the specified pathway. A ketone followed by Grignard addition and work-up yields a tertiary alcohol. The original nitrile carbon therefore passes through two different functional-group environments without being removed. The sequence introduces three carbon atoms overall: two from ethylmagnesium bromide and one from methylmagnesium bromide.

Explanation: \(\textbf{Distinguish the two original carbonyl groups:}\) The underlying compound contains one terminal aldehyde and one methyl ketone. These carbonyl groups would ordinarily both be susceptible to suitable hydride reduction. However, the question specifies that the ketone has already been protected as a cyclic ketal. This protection removes its ordinary ketone carbonyl reactivity during Stage I. \(\textbf{Stage I — Selective aldehyde reduction:}\) Sodium borohydride reduces the unprotected aldehyde into a primary alcohol. The terminal \(\mathrm{-CHO}\) group becomes \(\mathrm{-CH_2OH}\). The protected ketone remains unchanged under the stipulated conditions. \(\textbf{Stage II — Regenerate the ketone:}\) Controlled aqueous acid hydrolysis reverses cyclic ketal formation. The ketone carbonyl is regenerated without oxidising the newly formed primary alcohol. The isolated product consequently has the structure: \[ \mathrm{HOCH_2CH_2CH_2COCH_3} \] Numbering from the end that gives the ketone the appropriate lower locant identifies the compound as 5-hydroxypentan-2-one. \(\textbf{Stage III — Interpret the functional-group tests:}\) The compound contains a free primary alcohol hydroxyl group. It can therefore react with sodium metal to release hydrogen. It also contains the methyl ketone arrangement \(\mathrm{CH_3CO-}\). That arrangement gives a positive characteristic iodoform reaction. \(\textbf{Explain the synthetic advantage:}\) Protection prevents unwanted reduction of one carbonyl group while another undergoes the desired transformation. Subsequent deprotection restores the original ketone functionality. The final compound therefore possesses both an alcohol and a ketone even though the intermediate reduction reagent could ordinarily react with an unprotected ketone.

Explanation: \(\textbf{Identify the three carbon groups at the target alcohol carbon:}\) The hydroxyl-bearing carbon of 3-methylhexan-3-ol is attached to a methyl group, an ethyl group and a propyl group. These three carbon groups remain attached after formation of the tertiary alcohol. Grignard addition to a ketone contributes one new carbon-containing group. The other two carbon groups originate from the ketone. \(\textbf{Evaluate Combination I:}\) Butan-2-one has methyl and ethyl groups attached to its carbonyl carbon. Propylmagnesium bromide supplies the third required group. \[ \mathrm{CH_3COCH_2CH_3+C_3H_7MgBr} \] After appropriate work-up, the product contains methyl, ethyl and propyl groups around the hydroxyl-bearing carbon. Therefore, I is valid. \(\textbf{Evaluate Combination II:}\) Pentan-2-one contains methyl and propyl groups attached to its carbonyl carbon. Ethylmagnesium bromide supplies the missing ethyl group. The same tertiary alcohol is produced. Therefore, II is valid. \(\textbf{Evaluate Combination III:}\) Hexan-3-one contains ethyl and propyl groups attached to its carbonyl carbon. Methylmagnesium bromide supplies the methyl group. Therefore, III is also valid. \(\textbf{Evaluate Combination IV:}\) Pentan-3-one contains two identical ethyl groups attached to the carbonyl carbon. Ethylmagnesium bromide introduces a third ethyl group. The product consequently contains three ethyl substituents at the hydroxyl-bearing carbon. It is 3-ethylpentan-3-ol rather than the target compound. \(\textbf{Retrosynthetic conclusion:}\) A tertiary alcohol with three different carbon substituents can sometimes be disconnected into three distinct ketone–Grignard combinations. However, every proposed combination must reproduce the complete set of substituents around the hydroxyl-bearing carbon. Carbon count alone is insufficient because Combination IV also produces a seven-carbon tertiary alcohol but with different connectivity.

Explanation: \(\textbf{Compare the intermolecular forces:}\) Butane is a hydrocarbon whose molecules interact predominantly through dispersion forces. Methoxyethane contains polar carbon–oxygen bonds, but it has no ordinary oxygen–hydrogen bond. Therefore, it cannot donate conventional O–H hydrogen bonds to other ether molecules. Propanal and propanone contain polar carbonyl groups. The carbonyl bond contributes a substantial molecular dipole. Consequently, dipole–dipole attractions help raise their boiling points above those of comparable hydrocarbons and ethers. \(\textbf{Examine the alcohol:}\) Propan-1-ol contains a free hydroxyl group. One alcohol molecule can donate a hydrogen bond to the oxygen atom of another alcohol molecule. This produces stronger intermolecular association and contributes to the considerably higher boiling point of propan-1-ol. \(\textbf{Examine hydrogen bonding with water:}\) The oxygen of an ether can accept a hydrogen bond from water. The oxygen of an aldehyde or ketone carbonyl can also accept hydrogen bonds from water. Thus, absence of a hydrogen-bond-donating O–H group does not imply inability to interact through hydrogen bonding with water. \(\textbf{Interpret the complete trend:}\) The supplied boiling-point ordering is consistent with increasing contributions from intermolecular polarity and hydrogen bonding. However, molecular mass, surface area and dispersion forces also influence boiling points. Therefore, the numbers should not be explained by hydrogen bonding alone. The decisive distinction is between a compound's ability to accept hydrogen bonds from water and its ability to donate O–H hydrogen bonds to molecules of its own kind.

Explanation: Both compounds contain ordinary ketone carbonyl groups. Their carbonyl carbons are electrophilic and can undergo suitable nucleophilic addition reactions. Cyanide attacks the carbonyl carbon and produces an alkoxide intermediate. Protonation then yields a cyanohydrin. \(\textbf{Compare the environments around the two carbonyls:}\) Cyclohexanone has two ring methylene groups adjacent to its carbonyl carbon. 2,6-Dimethylcyclohexanone has additional methyl substituents on those adjacent ring carbons. These groups increase steric congestion near the site approached by the nucleophile. The tetrahedral cyanohydrin product can also experience steric interactions not present to the same extent in the starting carbonyl compound. Therefore, addition can be less favourable for the more crowded ketone. \(\textbf{Examine the catalyst change:}\) Cyanide ions are effective nucleophiles for cyanohydrin formation. Increasing their availability may influence the rate at which carbonyl addition occurs. However, a catalyst does not automatically remove structural steric interactions or make an unfavourable product equilibrium quantitatively complete. The equilibrium position also depends on substrate and product stability and the activities of the reacting species. \(\textbf{Interpret the observation:}\) The lower yield from Q is consistent with the influence of steric congestion. It is not evidence that Q lacks a carbonyl group. Nor does it establish an absolute prohibition on cyanohydrin formation. The comparison distinguishes an experimentally less favourable addition reaction from one that is chemically impossible.

Explanation: \(\textbf{Identify the starting carbonyl class:}\) Benzophenone is a ketone whose carbonyl carbon is attached to two phenyl groups. It contains no ordinary alpha-hydrogen because its adjacent carbon atoms are aromatic ring carbons. However, the absence of alpha-hydrogen does not prevent suitable carbonyl reduction or carbonyl derivative formation. \(\textbf{Route I — Hydride reduction:}\) Sodium borohydride supplies hydride to the electrophilic ketone carbonyl carbon. Aqueous work-up protonates the resulting alkoxide. \[ \mathrm{(C_6H_5)_2CO \xrightarrow[\text{work-up}]{NaBH_4} (C_6H_5)_2CHOH} \] The product is diphenylmethanol, commonly called benzhydrol. It is a secondary alcohol because the hydroxyl-bearing carbon is attached to two carbon groups. \(\textbf{Route II — Wolff–Kishner reduction:}\) Hydrazine first forms the corresponding hydrazone. Further treatment under the specified strongly basic heated conditions converts the original carbonyl carbon into a methylene carbon. \[ \mathrm{(C_6H_5)_2CO \xrightarrow[\text{KOH,\ heat}]{NH_2NH_2} (C_6H_5)_2CH_2} \] The product is diphenylmethane. It contains neither the original carbonyl group nor a newly formed alcohol hydroxyl group. \(\textbf{Route III — Carbonyl derivative formation:}\) 2,4-Dinitrophenylhydrazine reacts through addition followed by dehydration. The product contains the characteristic carbon–nitrogen double bond of a hydrazone derivative. It is not the same as the hydrocarbon obtained after complete Wolff–Kishner reduction. \(\textbf{Compare the three reaction levels:}\) Route I reduces the ketone to an alcohol. Route II deoxygenates the ketone to a methylene-containing hydrocarbon. Route III converts the carbonyl into a nitrogen-containing derivative. The reactions therefore differ in the bonds retained at the original carbonyl carbon and must not be treated as equivalent simply because nitrogen or hydrogen-containing reagents are involved.

Explanation: \(\textbf{Identify the carbonyl arrangement:}\) Pentane-2,4-dione contains two ketone carbonyl groups separated by one methylene carbon. Its central carbon therefore occupies an alpha position relative to both carbonyl groups. Statement I is correct. \(\textbf{Examine the conjugate base:}\) Removal of one central methylene hydrogen produces a carbon-centred conjugate base capable of resonance delocalisation. The electron distribution can interact with the carbonyl group on either side. Consequently, more than one carbonyl oxygen contributes to stabilisation of the resulting enolate system. Statement II is correct. \(\textbf{Compare propanone:}\) Propanone contains alpha-hydrogens on its two methyl groups. Those hydrogens are more acidic than ordinary saturated alkane hydrogens because the adjacent carbonyl group stabilises an enolate. Therefore, Statement III is incorrect. Alpha-hydrogen acidity is not restricted to compounds containing a carboxylic acid group. \(\textbf{Compare the extent of stabilisation:}\) A propanone enolate is stabilised through interaction with one carbonyl system. The central enolate of pentane-2,4-dione can delocalise negative charge through two carbonyl systems. This additional stabilisation generally favours removal of a central methylene hydrogen from the diketone. Statement IV is therefore correct. \(\textbf{Interpret the distinction:}\) The relevant comparison concerns the stability of the species formed after proton removal. It is not sufficient to count hydrogen atoms or carbonyl groups without considering their relative positions. Two carbonyl groups separated by a suitable alpha-carbon can influence one another through the shared enolate framework.

Explanation: \(\textbf{Identify the two enolate donors:}\) Ethanal contains alpha-hydrogens on its methyl group. Propanal contains alpha-hydrogens on its methylene group. Therefore, either aldehyde can supply an enolate under suitable dilute alkaline conditions. Both also contain electrophilic aldehyde carbonyl groups. \(\textbf{Count the self-aldol pathways:}\) An ethanal enolate can attack another ethanal molecule. This gives 3-hydroxybutanal as the initial addition product. A propanal enolate can attack another propanal molecule. This gives 3-hydroxy-2-methylpentanal. The self-reactions therefore contribute two constitutional products. \(\textbf{Identify the first crossed pathway:}\) An ethanal-derived enolate can attack propanal. The product has the structure: \[ \mathrm{CH_3CH_2CH(OH)CH_2CHO} \] This is 3-hydroxypentanal. \(\textbf{Identify the reverse crossed pathway:}\) A propanal-derived enolate can attack ethanal. The product has the structure: \[ \mathrm{CH_3CH(OH)CH(CH_3)CHO} \] This is 3-hydroxy-2-methylbutanal. The two crossed products possess different carbon connectivity. They are therefore constitutional isomers rather than alternative descriptions of the same compound. \(\textbf{Check the complete product set:}\) There are two distinct choices of enolate donor. For each donor, there are two possible aldehyde acceptors. This gives four distinct donor–acceptor combinations. The question excludes dehydration and stereoisomer counting, so neither operation increases the requested constitutional-product count. The result illustrates why crossed aldol reactions involving two enolate-forming carbonyl compounds can produce complicated mixtures.

Explanation: \(\textbf{Identify the aromatic substituents:}\) Benzaldehyde contains a formyl group directly attached to the aromatic ring. Acetophenone contains an acetyl group directly attached to the ring. Both substituents contain carbonyl carbon bonded to the aromatic carbon. \(\textbf{Determine their directing effects:}\) Aromatic aldehyde and ketone carbonyl groups withdraw electron density from the ring. They are deactivating and meta directing in ordinary electrophilic aromatic substitution. Therefore, the specified controlled nitration gives the corresponding meta-nitro products. The presence of oxygen in the carbonyl group does not make these substituents equivalent to phenolic hydroxyl or methoxy groups. \(\textbf{Follow the benzaldehyde carbonyl:}\) Nitration changes an aromatic ring hydrogen into a nitro substituent. It does not remove the original aldehyde group under the stated conditions. The benzaldehyde-derived product is 3-nitrobenzaldehyde. It retains an aldehyde carbonyl group and is eligible for the characteristic Tollens' reaction. \(\textbf{Follow the acetophenone carbonyl:}\) The acetophenone-derived product is 3-nitroacetophenone. Its original \(\mathrm{-COCH_3}\) group is unchanged. It consequently retains the methyl ketone arrangement required for the characteristic iodoform reaction. \(\textbf{Integrate the two reaction types:}\) Nitration position depends on the electronic influence of the existing aromatic substituent. The diagnostic carbonyl reaction depends on the connectivity retained within that substituent. Thus, two compounds can exhibit the same meta-directing behaviour while retaining different aldehyde-versus-ketone properties.

Explanation: \(\textbf{Count the original carbonyl groups:}\) Cyclohexane-1,4-dione contains two ketone carbonyl groups within one six-membered ring. The sample amount is: \[ n(\mathrm{diketone})=0.18\,\mathrm{mol} \] Therefore, the total amount of carbonyl groups is: \[ n(\mathrm{C=O})=2(0.18) \] \[ n(\mathrm{C=O})=0.36\,\mathrm{mol} \] \(\textbf{Follow the hydride reduction:}\) Sodium borohydride converts each ketone group into a secondary alcohol group. The six-membered carbon ring is preserved. The product contains two oxygen–hydrogen bonds per molecule. \(\textbf{Calculate hydrogen evolution:}\) The general sodium-metal reaction is: \[ \mathrm{2ROH+2Na\longrightarrow 2RONa+H_2} \] Two hydroxyl groups therefore produce one hydrogen molecule on complete reaction. Each molecule of cyclohexane-1,4-diol contains exactly two hydroxyl groups. Hence: \[ n(\mathrm{H_2})=n(\mathrm{diol}) \] \[ n(\mathrm{H_2})=0.18\,\mathrm{mol} \] \(\textbf{Evaluate the independent derivative-forming experiment:}\) The original diketone contains two ketone carbonyl groups. Each reacts with one molecule of 2,4-dinitrophenylhydrazine under the stipulated complete derivatisation conditions. Therefore: \[ n(\mathrm{2,4\!-\!DNPH})=2n(\mathrm{diketone}) \] \[ n(\mathrm{2,4\!-\!DNPH})=2(0.18) \] \[ n(\mathrm{2,4\!-\!DNPH})=0.36\,\mathrm{mol} \] \(\textbf{Check the functional-group transformation:}\) The original molecule contains two ketone carbonyl groups but no alcohol hydroxyl groups. Reduction changes both carbonyl groups into secondary alcohol functionalities. Therefore, the sodium-metal reaction measures the hydroxyl groups formed during reduction. The derivative-forming experiment separately counts the original carbonyl groups. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2}),n(\mathrm{2,4\!-\!DNPH})\right) =(0.18,\ 0.36)\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the carbonyl structure:}\) Butan-2-one contains a ketone carbonyl group attached to methyl and ethyl groups. Its carbon skeleton contains four carbon atoms. The two proposed sequences differ in the reduction level established during their first transformations. \(\textbf{Sequence P — Reduce to an alcohol:}\) Sodium borohydride converts the ketone carbonyl group into a secondary alcohol. \[ \mathrm{CH_3COCH_2CH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_2CH_3} \] The product is butan-2-ol. Under suitable acid-catalysed heating, the alcohol can undergo dehydration. Removal of a beta-hydrogen gives a carbon–carbon double bond. \[ \mathrm{CH_3CH(OH)CH_2CH_3 \xrightarrow{H^+,\ heat} CH_3CH=CHCH_3+H_2O} \] But-2-ene is the more substituted alkene relative to but-1-ene. \(\textbf{Sequence Q — Deoxygenating reduction:}\) Wolff–Kishner reduction converts the original ketone carbonyl group into a methylene group. \[ \mathrm{CH_3COCH_2CH_3 \xrightarrow[\text{KOH,\ heat}]{NH_2NH_2} CH_3CH_2CH_2CH_3} \] The product is butane. It contains no hydroxyl group available for ordinary alcohol dehydration. Treatment with ordinary aqueous acid does not reproduce the missing alcohol intermediate or directly create the required alkene. \(\textbf{Synthetic conclusion:}\) Reducing a ketone to an alcohol preserves an oxygen-containing functionality that can later undergo dehydration. Reducing the same ketone directly to a hydrocarbon removes that opportunity. The reaction sequence must therefore be planned according to the intended final oxidation level and functional-group transformation.

Explanation: \(\textbf{Identify the target functional group:}\) 4-Methylbenzaldehyde contains a formyl group attached directly to an aromatic ring. The methyl substituent occupies the para position relative to this formyl group. The question specifies that the ring methyl group remains unchanged. \(\textbf{Route P — Controlled nitrile reduction:}\) A nitrile can undergo controlled reduction with a suitable hydride reagent such as DIBAL-H. With appropriate conditions and work-up, the nitrile carbon becomes the aldehyde carbon. Therefore, 4-methylbenzonitrile can produce 4-methylbenzaldehyde through this route. \(\textbf{Route Q — Rosenmund reduction:}\) An acyl chloride can be hydrogenated using a suitably modified palladium catalyst under Rosenmund conditions. The reaction can stop at the aldehyde stage. Consequently, 4-methylbenzoyl chloride is a suitable precursor to the target aromatic aldehyde. \(\textbf{Route R — Grignard addition to a nitrile:}\) Methylmagnesium bromide contributes a methyl group through nucleophilic attack at the nitrile carbon. Hydrolysis of the resulting intermediate produces a ketone rather than an aldehyde. \[ \mathrm{p\!-\!CH_3C_6H_4CN} \] \[ \mathrm{\xrightarrow[\text{2. hydrolysis}]{\text{1. }CH_3MgBr} p\!-\!CH_3C_6H_4COCH_3} \] The product is 4-methylacetophenone. Its carbonyl carbon is attached to the aromatic ring and to a newly introduced methyl group. \(\textbf{Compare the synthetic pathways:}\) Routes P and Q reach the aldehyde oxidation level through controlled reduction. Route R forms a new carbon–carbon bond and yields a ketone. Therefore, possession of the correct aromatic precursor skeleton does not guarantee the correct carbonyl class. The identity and chemical role of the reagent determine whether the product is an aldehyde or ketone.

Explanation: \(\textbf{Compare aldehyde and ketone oxidation:}\) Aldehydes are generally oxidised more readily than ordinary ketones. Their carbonyl carbon bears an aldehydic hydrogen and can be converted into a carboxyl-derived functionality. Statement I is correct because benzaldehyde undergoes oxidation in Tollens' reagent. The alkaline medium favours the corresponding benzoate form during the oxidation. Silver(I) ions are reduced to metallic silver in the same reaction. Statement II is correct because butan-2-one is an ordinary ketone. It does not undergo the characteristic positive Tollens' reaction under normal testing conditions. \(\textbf{Interpret vigorous ketone oxidation:}\) Ketones are relatively resistant to mild oxidation. However, sufficiently powerful oxidising agents at elevated temperatures can oxidise ketones through cleavage of carbon–carbon bonds. Therefore, Statement III is incorrect. The expression that ketones resist oxidation must not be expanded into absolute resistance under every possible condition. \(\textbf{Examine the possible cleavage products:}\) Vigorous ketone oxidation can produce carboxylic acid fragments containing fewer carbon atoms than the parent ketone. Actual product distribution depends on substrate structure and reaction conditions. The question does not specify a unique cleavage pattern. Consequently, assigning one definite acid as the exclusive product would introduce unsupported selectivity. Statement IV is correct. The comparison distinguishes a reliable qualitative oxidation-class test from the potentially complex products of vigorous oxidative cleavage. It also explains why a negative Tollens' result cannot be interpreted as proof that a compound is incapable of every oxidation reaction.

Explanation: \(\textbf{Identify the two carbonyl environments:}\) X contains a terminal aldehyde group and a methyl ketone group. The aldehyde carbon is at one end of the five-carbon chain. The ketone contains the \(\mathrm{-COCH_3}\) arrangement at the opposite end. Their different reactivities permit the stipulated sequential transformations. \(\textbf{Stage I — Selective aldehyde oxidation:}\) Tollens' reagent oxidises the aldehyde functionality into a carboxylate. The question specifies preservation of the ketone carbonyl. After accounting for the alkaline medium, the organic product has the connectivity: \[ \mathrm{{^-}OOCCH_2CH_2COCH_3} \] No carbon atom has been lost. The original aldehyde carbon is now a carboxylate carbon. \(\textbf{Stage II — Identify the haloform-reactive group:}\) The remaining ketone contains a terminal methyl group directly attached to its carbonyl carbon. This methyl group undergoes exhaustive iodination. Subsequent cleavage forms iodoform. The original terminal methyl carbon becomes the carbon atom of \(\mathrm{CHI_3}\). \(\textbf{Trace the organic fragment after cleavage:}\) The original ketone carbonyl carbon becomes an additional carboxylate carbon. The two methylene groups between the original carbonyl functionalities remain intact. Therefore, the final organic carboxylate framework is: \[ \mathrm{{^-}OOCCH_2CH_2COO^-} \] \(\textbf{Stage III — Regenerate the neutral dicarboxylic acid:}\) Mild acidification protonates the two carboxylate groups. \[ \mathrm{HOOCCH_2CH_2COOH} \] The product is butanedioic acid, also called succinic acid. \(\textbf{Check carbon conservation:}\) The starting ketoaldehyde contains five carbon atoms. The haloform product contains one original carbon atom. The remaining dicarboxylic acid contains four carbon atoms. The specified mild conditions prevent loss of either carboxyl carbon by subsequent decarboxylation. Thus, the final answer is fixed by selective aldehyde oxidation followed by carbon-specific methyl ketone cleavage.