Organic Compounds Containing Oxygen MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: \(\textbf{Identify the carbonyl classes:}\) P is propanal, an aldehyde containing one hydrogen and one ethyl group attached to its carbonyl carbon. Q is propanone, a ketone containing two methyl groups attached to its carbonyl carbon. Both compounds contain three carbon atoms before the reaction. \(\textbf{Cyanohydrin formation from propanal:}\) Cyanide attacks the electrophilic carbonyl carbon, and subsequent protonation gives the corresponding cyanohydrin. \[ \mathrm{CH_3CH_2CHO+HCN \longrightarrow CH_3CH_2CH(OH)CN} \] One additional carbon atom is introduced through the cyanide group. \(\textbf{Hydrolysis of the propanal-derived nitrile:}\) Complete acidic hydrolysis converts the nitrile carbon into a carboxyl carbon. \[ \mathrm{CH_3CH_2CH(OH)CN+2H_2O+H^+ \longrightarrow CH_3CH_2CH(OH)COOH+NH_4^+} \] Numbering begins at the carboxyl carbon, so the hydroxyl group occupies carbon 2. The product is 2-hydroxybutanoic acid. \(\textbf{Cyanohydrin formation from propanone:}\) The same type of nucleophilic addition occurs at the carbonyl carbon of propanone. \[ \mathrm{CH_3COCH_3+HCN \longrightarrow (CH_3)_2C(OH)CN} \] The original two methyl groups remain attached to the former carbonyl carbon. \(\textbf{Hydrolysis of the propanone-derived nitrile:}\) \[ \mathrm{(CH_3)_2C(OH)CN+2H_2O+H^+ \longrightarrow (CH_3)_2C(OH)COOH+NH_4^+} \] The product is 2-hydroxy-2-methylpropanoic acid. \(\textbf{Compare the product skeletons:}\) Both products contain four carbon atoms because HCN addition introduces exactly one carbon atom. However, their carbon skeletons differ. The aldehyde-derived product has a four-carbon chain, whereas the ketone-derived product has a three-carbon parent chain with an additional methyl substituent. No carbon–carbon bond migration or rearrangement is required. The distinct products arise from the different groups originally attached to the carbonyl carbons of the two functional isomers. Identical molecular formulae of the starting compounds therefore do not imply identical products after nucleophilic addition and subsequent hydrolysis.

Explanation: \(\textbf{Identify the two reactive functional groups:}\) But-2-enal is an unsaturated aldehyde containing a carbonyl group and an alkene linkage. The carbonyl carbon is electrophilic because its carbon–oxygen double bond is polarised towards oxygen. The carbon–carbon double bond is a separate reactive site capable of undergoing suitable addition reactions. The question specifies selective 1,2-addition at the carbonyl group during the first stage, so other possible pathways of conjugated carbonyl chemistry are excluded. \(\textbf{Stage I — Cyanohydrin formation:}\) Cyanide attacks the carbonyl carbon, and the carbonyl pi electrons move towards oxygen. Subsequent protonation produces a hydroxyl group attached to the original carbonyl carbon. \[ \mathrm{CH_3CH=CHCHO+HCN \longrightarrow CH_3CH=CHCH(OH)CN} \] The product contains one additional carbon atom supplied by cyanide. The original carbon–carbon double bond remains intact because the reaction conditions specify selective carbonyl addition. \(\textbf{Stage II — Bromine addition:}\) Bromine adds across the remaining alkene double bond. Each of the two original alkene carbon atoms receives one bromine atom. \[ \mathrm{CH_3CH=CHCH(OH)CN+Br_2 \longrightarrow CH_3CHBrCHBrCH(OH)CN} \] The cyanohydrin hydroxyl group and nitrile group remain present under the specified conditions. The final product therefore contains two bromine atoms, one hydroxyl group and one nitrile group. The newly introduced cyanide carbon remains directly attached to the carbon that originally belonged to the aldehyde group. This sequence demonstrates that two different unsaturated bonds within one molecule can undergo chemically distinct additions. The correct structure follows from tracking the reaction site and carbon connectivity at each stage.

Explanation: All three reactions begin with a compound containing an electrophilic carbonyl carbon. However, the reactions differ in the nature of the nucleophile and the transformations following initial attack. Statement I is correct because hydrogensulphite ion can add across the carbonyl group of suitable aldehydes. The addition product contains a hydroxyl group and a hydrogensulphite-derived substituent on the former carbonyl carbon. Under suitable conditions, this addition compound can decompose to regenerate the original aldehyde. This reversibility makes hydrogensulphite addition useful in separating and purifying certain carbonyl compounds. Statement II is correct because ethylene glycol can react with a ketone under acid-catalysed conditions to form a cyclic ketal. The overall transformation involves formation of carbon–oxygen bonds and elimination of water. The carbonyl group is consequently converted into an acetal-type carbon bearing two oxygen substituents. Statement III is incorrect because hydroxylamine forms an oxime through addition followed by dehydration. The final characteristic linkage is \(\mathrm{C=N-OH}\), not an unchanged carbonyl group. The original carbonyl oxygen is removed during the dehydration stage. Statement IV is correct because all three processes depend on carbonyl electrophilicity, but the hydrogensulphite adduct, cyclic ketal and oxime are structurally different products. These reactions must therefore not be treated as identical simply because their first mechanistic event involves nucleophilic attack.

Explanation: \(\textbf{Identify the carbonyl substrate:}\) Propanal contains a three-carbon skeleton with a terminal aldehyde group. Its carbonyl carbon is electrophilic and can undergo acid-catalysed addition of alcohols. The products depend on whether the reaction stops after initial addition or proceeds to complete acetal formation. \(\textbf{Condition I — Hemiacetal formation:}\) Methanol adds to the carbonyl group under suitable acid-catalysed conditions. The resulting hemiacetal contains one hydroxyl group and one methoxy group attached to the same carbon atom. \[ \mathrm{CH_3CH_2CHO+CH_3OH \rightleftharpoons CH_3CH_2CH(OH)OCH_3} \] The product is 1-methoxypropan-1-ol. The question explicitly specifies isolation of this intermediate before further reaction. \(\textbf{Condition II — Complete acetal formation:}\) Under dry acid-catalysed conditions with sufficient methanol, the hemiacetal can undergo further reaction. Its hydroxyl functionality is replaced through the acetal-forming process, producing a carbon bonded to two methoxy groups. \[ \mathrm{CH_3CH_2CHO+2CH_3OH \rightleftharpoons CH_3CH_2CH(OCH_3)_2+H_2O} \] The resulting compound is 1,1-dimethoxypropane, an acetal. Unlike the hemiacetal, it has no hydroxyl group on the original carbonyl carbon. \(\textbf{Condition III — Acidic hydrolysis:}\) Acetals undergo hydrolysis with aqueous mineral acids to regenerate the corresponding carbonyl compound. \[ \mathrm{CH_3CH_2CH(OCH_3)_2+H_2O \xrightarrow{H^+} CH_3CH_2CHO+2CH_3OH} \] The regenerated carbonyl compound is propanal, not propanone or propan-1-ol. The original carbonyl carbon remains the terminal carbon of the three-carbon skeleton throughout the sequence. The distinction between hemiacetal, acetal and regenerated aldehyde is therefore established by the number of oxygen substituents and the reaction medium.

Explanation: \(\textbf{Determine the starting carbonyl structure:}\) Propanone has the structure \(\mathrm{CH_3COCH_3}\). Its carbonyl carbon is attached to two methyl groups and therefore belongs to a ketone. Ethylene glycol contains two hydroxyl groups on adjacent carbon atoms. These groups can react with the carbonyl functionality under suitable acid-catalysed conditions. \(\textbf{Form the cyclic ketal:}\) Initial attack by one hydroxyl oxygen produces an addition intermediate. Subsequent reaction involving the second hydroxyl group and elimination of water produces a cyclic ketal. The former carbonyl carbon becomes bonded to both glycol oxygen atoms. The resulting ring contains the former carbonyl carbon, two oxygen atoms and two methylene carbon atoms from ethylene glycol. Therefore, the newly formed ring has five members. The cyclic product is commonly called 2,2-dimethyl-1,3-dioxolane. The overall reaction is: \[ \mathrm{CH_3COCH_3+HOCH_2CH_2OH \rightleftharpoons (CH_3)_2C(OCH_2CH_2O)+H_2O} \] In the condensed product notation, the two oxygen atoms are connected through the ethylene glycol fragment and the former carbonyl carbon to form the ring. \(\textbf{Hydrolysis of the ketal:}\) Aqueous mineral acid promotes cleavage of the ketal linkages and regenerates the original carbonyl compound. \[ \mathrm{(CH_3)_2C(OCH_2CH_2O)+H_2O \xrightarrow{H^+} CH_3COCH_3+HOCH_2CH_2OH} \] The carbonyl carbon returns to its trigonal-planar ketone environment. Neither the carbon skeleton of propanone nor that of ethylene glycol is changed. The product is therefore a five-membered cyclic ketal that regenerates propanone and ethylene glycol on hydrolysis.

Explanation: \(\textbf{Identify the functional groups requiring different treatment:}\) 4-Oxopentanal contains a terminal aldehyde group and an internal ketone group. Both carbonyl groups would ordinarily be susceptible to reduction by sodium borohydride. However, the question specifies that the aldehyde group is selectively converted into an acetal before the reducing agent is introduced. \(\textbf{Stage I — Protect the aldehyde functionality:}\) Acetal formation converts the aldehyde carbonyl into a carbon bonded to two alkoxy oxygen atoms. This removes the original carbonyl pi bond from the protected aldehyde centre. The ketone carbonyl at carbon 4 remains available for subsequent reduction. \(\textbf{Stage II — Reduce the unprotected ketone:}\) Sodium borohydride reduces the ketone carbonyl to a secondary alcohol group. The relevant transformation is: \[ \mathrm{-CH_2COCH_3 \xrightarrow[\text{work-up}]{NaBH_4} -CH_2CH(OH)CH_3} \] Under the stated conditions, the acetal remains intact during this stage. Therefore, the original aldehyde carbon is not reduced to a primary alcohol. \(\textbf{Stage III — Regenerate the aldehyde:}\) Aqueous acid hydrolyses the acetal and restores the terminal aldehyde functionality. The secondary alcohol produced at carbon 4 remains unchanged under the specified conditions. The final organic structure is: \[ \mathrm{OHCCH_2CH_2CH(OH)CH_3} \] \(\textbf{Nomenclature and structural verification:}\) The aldehyde is the principal functional group and its carbon is assigned position 1. The hydroxyl group is therefore located at carbon 4. The product is 4-hydroxypentanal. Pentane-1,4-diol would be produced if the original aldehyde were reduced instead of being protected. 4-Oxopentan-1-ol would require the opposite selectivity, namely reduction of the aldehyde while preserving the ketone. Thus, selective acetal formation permits a different reduction outcome from direct treatment of the original dicarbonyl compound.

Explanation: Benzaldehyde contains a carbonyl group conjugated with the aromatic ring. This conjugation influences the distribution of electron density around the carbonyl group and therefore its susceptibility to nucleophilic attack. Statement I is correct because a nitro group is strongly electron withdrawing. At the para position, its electronic influence can be transmitted through the conjugated aromatic system. This tends to increase the electrophilic character of the aldehyde carbonyl carbon relative to unsubstituted benzaldehyde under suitable comparable conditions. Statement II is correct because the methoxy group can donate electron density into the aromatic ring through resonance. When placed para to the formyl group, this electron-donating influence can reduce the electron deficiency of the carbonyl carbon. Consequently, 4-methoxybenzaldehyde is generally less electrophilic than the corresponding unsubstituted aromatic aldehyde in the relevant comparison. Statement III is incorrect because it reverses the normal electronic effect of an electron-withdrawing nitro substituent. Nitro substitution does not increase aromatic electron density through ordinary resonance donation. Statement IV is correct because molecular mass alone does not determine carbonyl electrophilicity. The type of substituent, its position and its electronic interaction with the aromatic carbonyl system are essential. Steric factors and reaction conditions can also influence actual addition rates. The correct comparison is therefore based on electronically meaningful structures rather than treating every aromatic aldehyde as equally reactive. Statements I, II and IV correctly identify the relevant distinctions.

Explanation: The reaction of aldehydes and ketones with hydrogen cyanide produces cyanohydrins. However, NCERT notes that reaction with pure HCN occurs very slowly. The chemically effective nucleophile in the standard cyanohydrin mechanism is cyanide ion, \(\mathrm{CN^-}\). This ion donates an electron pair to the electrophilic carbonyl carbon. The addition step can be represented as: \[ \mathrm{R_2C=O+CN^- \longrightarrow R_2C(O^-)CN} \] Subsequent protonation produces the cyanohydrin. \[ \mathrm{R_2C(O^-)CN+H^+ \longrightarrow R_2C(OH)CN} \] Condition II supplies a suitable base that increases the availability of cyanide ions from the HCN-containing system. The greater availability of the effective nucleophile facilitates attack at the carbonyl carbon. Consequently, this condition is favourable for the standard cyanohydrin-forming pathway. Condition I provides comparatively little cyanide ion because pure HCN is weakly ionised. Condition III also provides little free cyanide because a strongly acidic environment shifts cyanide towards its protonated form. Although suitable protonation can activate a carbonyl compound towards other reactions, this does not mean that strongly acidic conditions automatically favour a pathway requiring free cyanide ion. The relevant comparison must therefore include both electrophilic carbonyl activation and nucleophile availability. The addition tendency is not determined by total carbon count or simply by the presence of hydrogen ions. The correct interpretation identifies cyanide-ion generation as the principal catalytic role in the NCERT cyanohydrin reaction.

Explanation: \(\textbf{Relate addition-product concentration to the free carbonyl compound:}\) The problem defines the equilibrium constant for reversible hydrogensulphite addition. \[ K=\frac{[Q]}{[P][\mathrm{HSO_3^-}]} \] Rearranging: \[ \frac{[Q]}{[P]}=K[\mathrm{HSO_3^-}] \] The ratio of addition product to free carbonyl compound therefore depends on the free hydrogensulphite concentration. \(\textbf{Experiment I — Higher hydrogensulphite concentration:}\) \[ [\mathrm{HSO_3^-}]_{\mathrm{I}}=0.50\,\mathrm{mol\,L^{-1}} \] Substitution gives: \[ \frac{[Q]}{[P]}=6.0(0.50) \] \[ \frac{[Q]}{[P]}=3.0 \] Thus: \[ [Q]=3[P] \] The total carbonyl-derived concentration is: \[ C_{\mathrm{total}}=[P]+[Q] \] \[ C_{\mathrm{total}}=[P]+3[P]=4[P] \] The fraction present as Q is: \[ f_{\mathrm{I}}=\frac{[Q]}{C_{\mathrm{total}}} =\frac{3[P]}{4[P]} \] \[ f_{\mathrm{I}}=0.75 \] Therefore: \[ \%\ Q_{\mathrm{I}}=75.0\% \] \(\textbf{Experiment II — Lower hydrogensulphite concentration:}\) \[ [\mathrm{HSO_3^-}]_{\mathrm{II}}=0.10\,\mathrm{mol\,L^{-1}} \] \[ \frac{[Q]}{[P]}=6.0(0.10) \] \[ \frac{[Q]}{[P]}=0.60 \] Thus: \[ [Q]=0.60[P] \] The total concentration becomes: \[ C_{\mathrm{total}}=[P]+0.60[P] \] \[ C_{\mathrm{total}}=1.60[P] \] The addition-product fraction is: \[ f_{\mathrm{II}}=\frac{0.60[P]}{1.60[P]} \] \[ f_{\mathrm{II}}=0.375 \] Therefore: \[ \%\ Q_{\mathrm{II}}=37.5\% \] \(\textbf{Equilibrium interpretation:}\) The higher hydrogensulphite concentration favours formation of the addition product. Reducing the free hydrogensulphite concentration decreases the product-to-reactant concentration ratio. However, a fivefold decrease in hydrogensulphite concentration does not imply a fivefold decrease in the percentage present as Q. The percentage depends on the ratio of Q to the total amount of P and Q, not simply on \([Q]/[P]\). \(\textbf{Final answer:}\) \[ \boxed{(\%\ Q_{\mathrm{I}},\%\ Q_{\mathrm{II}})=(75.0\%,\,37.5\%)} \]

Explanation: \(\textbf{Identify the original stereogenic centre:}\) 2-Methylbutanal has the structure \(\mathrm{CH_3CH_2CH(CH_3)CHO}\). The carbon adjacent to the aldehyde group is attached to hydrogen, methyl, ethyl and formyl groups. These four substituents are different, making this carbon stereogenic. The question specifies an optically pure starting material and excludes racemisation of this centre. \(\textbf{Effect of cyanide addition:}\) The aldehyde carbonyl carbon is approximately trigonal planar before reaction. Cyanide can attack this carbon from either face. Subsequent protonation gives the cyanohydrin: \[ \mathrm{CH_3CH_2CH(CH_3)CHO+HCN} \] \[ \mathrm{\longrightarrow CH_3CH_2CH(CH_3)CH(OH)CN} \] The former carbonyl carbon is now attached to hydrogen, hydroxyl, cyanide and the substituted butyl-derived fragment. It therefore becomes a new stereogenic centre. \(\textbf{Relationship between the products:}\) Because the original stereogenic centre retains its configuration, the two products obtained by attack from opposite carbonyl faces differ at the newly created centre but not at the pre-existing centre. Such stereoisomers are diastereomers rather than enantiomers. The formation of diastereomers does not require the two facial attacks to occur at equal rates. Indeed, an existing stereogenic centre can make the two carbonyl faces diastereotopic. The question does not provide sufficient information to determine their relative yields. The products have the same atom connectivity because cyanide attacks the same carbonyl carbon in both pathways. Therefore, two diastereomeric cyanohydrins are possible under the stated assumptions. The result illustrates why stereochemical outcomes must account for stereogenic centres already present in the reactant.

Explanation: Propanal contains three carbon atoms and an aldehyde group. Its carbonyl carbon is electrophilic and can undergo different nucleophilic reactions depending on the reagent. \(\textbf{Reaction I — Addition of HCN:}\) Cyanide ion attacks the carbonyl carbon, producing an alkoxide intermediate that becomes protonated. The product contains hydroxyl and nitrile groups attached to the same former carbonyl carbon. \[ \mathrm{CH_3CH_2CHO+HCN \longrightarrow CH_3CH_2CH(OH)CN} \] This compound is a cyanohydrin. The cyanide reagent contributes one additional carbon atom, so P contains four carbon atoms. The original carbonyl oxygen remains in P as the hydroxyl oxygen. \(\textbf{Reaction II — Reaction with hydroxylamine:}\) Hydroxylamine attacks the aldehyde carbonyl group under suitable conditions. The initial addition intermediate subsequently loses water to produce an oxime. \[ \mathrm{CH_3CH_2CHO+NH_2OH \longrightarrow CH_3CH_2CH=NOH+H_2O} \] The product Q is propanal oxime. No carbon atom is introduced because hydroxylamine contains no carbon. Consequently, Q retains the original three-carbon skeleton. The final \(\mathrm{C=N-OH}\) linkage also differs from the tetrahedral carbon containing hydroxyl and nitrile groups in P. The original carbonyl oxygen is lost as water during oxime formation. Thus, cyanohydrin formation is an ordinary addition process, while oxime formation involves addition followed by dehydration. The two reactions begin at the same electrophilic carbonyl carbon but lead to different functional groups and carbon counts.

Explanation: Oxime formation begins with a carbonyl compound and hydroxylamine. The nitrogen atom of hydroxylamine acts as a nucleophilic centre and attacks the electrophilic carbonyl carbon. This generates an addition intermediate that can undergo suitable proton transfers and dehydration to form a carbon–nitrogen double bond. The overall transformation is: \[ \mathrm{R_2C=O+NH_2OH \rightleftharpoons R_2C=NOH+H_2O} \] A suitable acid catalyst can facilitate activation of the carbonyl group towards nucleophilic addition. Acidic conditions can also assist the later proton-transfer and dehydration processes. However, hydroxylamine itself is a basic nitrogen-containing compound and can become protonated. In a strongly acidic medium, substantial protonation reduces the availability of its unprotonated nucleophilic form. Therefore, increasing acid concentration indefinitely does not guarantee a faster or more complete oxime-forming reaction. Condition I provides a favourable balance between carbonyl activation and availability of the nitrogen nucleophile. Condition II can suppress effective nucleophilic attack if hydroxylamine is excessively protonated. Condition III cannot yield the required oxime because hydroxylamine, the nitrogen-containing reactant, is absent. An acid catalyst alone cannot supply the nitrogen atom of the final \(\mathrm{C=N-OH}\) linkage. The key distinction is that acid catalysis can support several mechanistic stages while excessive protonation of the nucleophile can oppose the initial attack. This reasoning also applies broadly to suitable carbonyl reactions involving ammonia derivatives.

Explanation: \(\textbf{Identify the oxygen sources:}\) The starting propanone contains a labelled oxygen atom in its carbonyl group. Hydroxylamine supplies a second oxygen atom that is initially bonded to nitrogen. The product of the reaction is an oxime, which contains a carbon–nitrogen double bond and an oxygen–hydrogen bond attached to nitrogen. \(\textbf{Initial nucleophilic addition:}\) Hydroxylamine attacks through its nitrogen atom at the electrophilic carbonyl carbon. The carbonyl pi electrons move towards the labelled oxygen atom. An addition intermediate containing the original carbonyl oxygen and the incoming hydroxylamine fragment is formed. \(\textbf{Dehydration of the intermediate:}\) Suitable proton transfers allow the oxygen atom originally belonging to the carbonyl group to leave as part of a water molecule. The nitrogen atom of hydroxylamine remains attached to the former carbonyl carbon. Formation of the carbon–nitrogen double bond gives the oxime. \(\textbf{Track the hydroxylamine oxygen:}\) The oxygen originally bonded to hydroxylamine nitrogen remains bonded to that nitrogen in the final \(\mathrm{N-OH}\) group. The labelled reaction can be represented schematically as: \[ \mathrm{(CH_3)_2C={}^{18}O+NH_2{}^{16}OH} \] \[ \mathrm{\longrightarrow (CH_3)_2C=N{}^{16}OH+H_2{}^{18}O} \] The oxime therefore contains the unlabelled hydroxylamine-derived oxygen. The water contains the carbonyl-derived labelled oxygen. The assumption excluding isotope exchange ensures that the observed label distribution reflects the reaction's bond changes rather than subsequent exchange processes. This is an important distinction from ordinary hydride reduction or cyanohydrin formation, in which the carbonyl oxygen is retained in the alcohol product. Oxime formation instead replaces the carbonyl oxygen-containing functionality with a nitrogen-containing derivative through dehydration.

Explanation: Oxime formation converts an aldehyde or ketone carbonyl group into a carbon–nitrogen double bond. The characteristic linkage is \(\mathrm{C=N-OH}\). Geometrical isomerism about this linkage can arise when the carbon atom of the double bond carries two different substituents. The restricted rotation associated with the carbon–nitrogen double bond allows different relative arrangements of suitable substituents. \(\textbf{Propanone oxime:}\) Propanone has two identical methyl groups attached to its carbonyl carbon. Its oxime is: \[ \mathrm{(CH_3)_2C=NOH} \] The carbon of the double bond still carries two identical methyl groups. Therefore, propanone oxime does not exhibit distinct geometrical isomers arising from interchange of these identical carbon substituents. \(\textbf{Butan-2-one oxime:}\) Butan-2-one contains methyl and ethyl groups attached to its carbonyl carbon. Its oxime is: \[ \mathrm{CH_3C(=NOH)CH_2CH_3} \] The carbon–nitrogen double-bond carbon therefore bears two different groups. Distinct geometrical arrangements are possible. \(\textbf{Benzaldehyde oxime:}\) Benzaldehyde contains hydrogen and phenyl groups attached to the carbonyl carbon. Its oxime is: \[ \mathrm{C_6H_5CH=NOH} \] Hydrogen and phenyl are different substituents, so geometrical isomerism about the carbon–nitrogen double bond is also possible. Thus, Q and R can exhibit geometrical isomerism, whereas P cannot. The decisive criterion is substituent identity at the carbonyl-derived carbon, not whether the starting compound is an aldehyde or a ketone. The stereochemical consequence becomes apparent only after recognising the structure of the oxime produced.

Explanation: Aldehydes and ketones react with ammonia derivatives through nucleophilic addition followed by elimination of water. The final products contain a carbon–nitrogen double bond. The group attached to the imine-like nitrogen depends on the structure of the nitrogen-containing reagent. Reaction P is correctly represented because hydroxylamine gives an oxime. \[ \mathrm{(CH_3)_2CO+NH_2OH \longrightarrow (CH_3)_2C=NOH+H_2O} \] The hydroxyl group remains bonded to nitrogen in the oxime. Reaction Q is also correct because hydrazine forms a hydrazone. \[ \mathrm{(CH_3)_2CO+NH_2NH_2 \longrightarrow (CH_3)_2C=NNH_2+H_2O} \] The product retains a terminal \(\mathrm{-NH_2}\) group attached through the nitrogen–nitrogen bond. Reaction R is incorrect because semicarbazide contains the group \(\mathrm{-NHCONH_2}\), not \(\mathrm{-NHCOCH_3}\), attached to the nitrogen that participates in derivative formation. The correct semicarbazone linkage is: \[ \mathrm{(CH_3)_2C=NNHCONH_2} \] The claimed structure instead contains a different acyl substituent and does not preserve the nitrogen-containing fragment supplied by semicarbazide. Reaction S is correct because phenylhydrazine produces a phenylhydrazone. Its phenyl group remains attached to the terminal nitrogen of the hydrazone linkage. \[ \mathrm{(CH_3)_2C=NNHC_6H_5} \] The important distinction is between the common carbonyl-derived \(\mathrm{C=N}\) linkage and the different substituents retained from each reagent. The reagent's nitrogen-containing fragment must be traced accurately rather than inferred from the derivative name alone.

Explanation: The Assertion is correct because 2,4-dinitrophenylhydrazine is used to characterise suitable aldehydes and ketones. Its reaction with a carbonyl compound can produce a coloured solid derivative. NCERT describes 2,4-dinitrophenylhydrazone derivatives as yellow, orange or red solids. The actual colour depends on the compound and its structural features. A positive result therefore establishes the presence of a suitably reactive carbonyl functionality but does not by itself determine whether the original compound is an aldehyde or a ketone. Ethanal and propanone both possess carbonyl groups and can form the corresponding derivatives. The Reason is also correct because the carbonyl carbon is electrophilic in both classes of compounds. The nitrogen-containing reagent initially attacks this carbonyl centre through its nucleophilic nitrogen atom. The addition intermediate then undergoes appropriate proton transfers and dehydration to produce a carbon–nitrogen double bond. The common derivative-forming reaction is represented schematically as: \[ \mathrm{R_2C=O+H_2NNHC_6H_3(NO_2)_2} \] \[ \mathrm{\longrightarrow R_2C=NNHC_6H_3(NO_2)_2+H_2O} \] The original aldehyde or ketone substituents remain attached to the carbonyl-derived carbon. Consequently, the same broad mechanism explains why both aldehydes and ketones can give this observation. Distinguishing between the two classes requires additional chemical evidence from an appropriate characteristic reaction. Thus, both statements are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Identify the derivative-forming transformation:}\) Hydroxylamine reacts with aldehydes and ketones to produce oximes. The reaction involves initial nucleophilic addition followed by dehydration. Each molecule containing one carbonyl group produces one molecule of the corresponding oxime under the stated quantitative conditions. The general reaction is: \[ \mathrm{R_2C=O+NH_2OH \longrightarrow R_2C=NOH+H_2O} \] \(\textbf{Sample I — Formation of ethanal oxime:}\) Ethanal has the structure \(\mathrm{CH_3CHO}\). Its reaction with hydroxylamine produces ethanal oxime. \[ \mathrm{CH_3CHO+NH_2OH \longrightarrow CH_3CH=NOH+H_2O} \] The molecular formula of ethanal oxime is \(\mathrm{C_2H_5NO}\). Its molar mass is: \[ M_{\mathrm{ethanal\ oxime}} =2(12)+5(1)+14+16 \] \[ =24+5+14+16 \] \[ M_{\mathrm{ethanal\ oxime}}=59\,\mathrm{g\,mol^{-1}} \] The reaction has a \(1:1\) mole ratio between ethanal and its oxime. \[ n_{\mathrm{ethanal\ oxime}}=0.10\,\mathrm{mol} \] Therefore: \[ m_{\mathrm{ethanal\ oxime}}=nM \] \[ =0.10(59) \] \[ m_{\mathrm{ethanal\ oxime}}=5.90\,\mathrm{g} \] \(\textbf{Sample II — Formation of propanone oxime:}\) Propanone has the structure \(\mathrm{CH_3COCH_3}\). Its reaction with hydroxylamine gives propanone oxime. \[ \mathrm{CH_3COCH_3+NH_2OH \longrightarrow (CH_3)_2C=NOH+H_2O} \] The molecular formula of propanone oxime is \(\mathrm{C_3H_7NO}\). Its molar mass is: \[ M_{\mathrm{propanone\ oxime}} =3(12)+7(1)+14+16 \] \[ =36+7+14+16 \] \[ M_{\mathrm{propanone\ oxime}}=73\,\mathrm{g\,mol^{-1}} \] The amount of oxime produced is: \[ n_{\mathrm{propanone\ oxime}}=0.15\,\mathrm{mol} \] Its mass is: \[ m_{\mathrm{propanone\ oxime}}=0.15(73) \] \[ m_{\mathrm{propanone\ oxime}}=10.95\,\mathrm{g} \] \(\textbf{Total isolated oxime mass:}\) \[ m_{\mathrm{total}}=5.90+10.95 \] \[ m_{\mathrm{total}}=16.85\,\mathrm{g} \] \(\textbf{Chemical consistency check:}\) Oxime formation introduces nitrogen and retains the oxygen originally present in hydroxylamine. The original carbonyl oxygen is eliminated as part of water during the reaction. Consequently, the oxime molar mass cannot be calculated by simply adding the entire molar mass of hydroxylamine to that of the carbonyl compound. The eliminated water must be accounted for in the molecular formula. The two samples are reacted separately with excess reagent, so neither carbonyl compound competes with the other for hydroxylamine. The calculated total mass therefore equals the sum of the independently isolated dry oxime masses. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{total\ oximes}}=16.85\,\mathrm{g}} \]

Explanation: \(\textbf{Identify the carbon groups attached to the alcohol carbon:}\) The carbon bearing the hydroxyl group is attached to three different alkyl groups: methyl, ethyl and propyl. Therefore, X is 3-methylhexan-3-ol, a tertiary alcohol containing seven carbon atoms. A tertiary alcohol can be prepared by adding a suitable Grignard reagent to a ketone. During this reaction, the two carbon groups originally attached to the ketone carbonyl carbon remain attached. The third carbon group is supplied by the Grignard reagent. \(\textbf{First retrosynthetic disconnection:}\) The methyl group may be supplied by methylmagnesium bromide. The ketone must then contain ethyl and propyl groups attached to its carbonyl carbon. \[ \mathrm{CH_3CH_2COCH_2CH_2CH_3+CH_3MgBr} \] \[ \mathrm{\xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}} CH_3CH_2C(OH)(CH_3)CH_2CH_2CH_3} \] \(\textbf{Second retrosynthetic disconnection:}\) The ethyl group may instead come from ethylmagnesium bromide. The ketone must then contain methyl and propyl groups. \[ \mathrm{CH_3COCH_2CH_2CH_3+C_2H_5MgBr} \] \[ \mathrm{\xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}} CH_3CH_2C(OH)(CH_3)CH_2CH_2CH_3} \] \(\textbf{Third retrosynthetic disconnection:}\) The propyl group may be supplied by propylmagnesium bromide. The remaining methyl and ethyl groups must belong to the ketone. \[ \mathrm{CH_3COCH_2CH_3+CH_3CH_2CH_2MgBr} \] \[ \mathrm{\xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}} CH_3CH_2C(OH)(CH_3)CH_2CH_2CH_3} \] These are three constitutionally distinct ketone–Grignard combinations. Interchanging the two groups already attached to the carbonyl carbon does not create another ketone structure. \(\textbf{Evaluate subsequent oxidation:}\) The hydroxyl-bearing carbon in X has no directly attached hydrogen atom. Therefore, X resists ordinary alcohol-to-carbonyl oxidation without carbon–carbon bond cleavage. The correct answer combines three valid synthetic disconnections with the oxidation behaviour of a tertiary alcohol.

Explanation: \(\textbf{Route I — Addition to an aromatic ketone:}\) Acetophenone contains a carbonyl carbon attached to a phenyl group and a methyl group. Ethylmagnesium bromide supplies an ethyl nucleophile, which attacks the carbonyl carbon. \[ \mathrm{C_6H_5COCH_3+C_2H_5MgBr} \] \[ \mathrm{\xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}} C_6H_5C(OH)(CH_3)CH_2CH_3} \] The product is 2-phenylbutan-2-ol. Its hydroxyl-bearing carbon is attached to phenyl, methyl and ethyl groups, making it tertiary. These three carbon groups are different, and the fourth substituent is hydroxyl. Therefore, the former carbonyl carbon becomes stereogenic. Under achiral conditions, nucleophilic attack on the planar ketone can occur from either face, producing a racemic mixture. \(\textbf{Route II — Addition to an aromatic aldehyde:}\) Benzaldehyde has a phenyl group and hydrogen attached to its carbonyl carbon. The propyl group supplied by the Grignard reagent becomes attached to that carbon. \[ \mathrm{C_6H_5CHO+CH_3CH_2CH_2MgBr} \] \[ \mathrm{\xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}} C_6H_5CH(OH)CH_2CH_2CH_3} \] The product is 1-phenylbutan-1-ol. Its hydroxyl-bearing carbon is bonded to hydrogen, phenyl and propyl groups, so it is secondary. Because these groups are different, this carbon is also stereogenic. Formation from an achiral planar aldehyde under achiral conditions gives a racemic mixture. \(\textbf{Distinguish the oxidation outcomes:}\) The tertiary alcohol obtained in Route I has no hydrogen on its hydroxyl-bearing carbon and resists ordinary oxidation with PCC. The secondary alcohol from Route II possesses the required carbon-bound hydrogen and can be oxidised to 1-phenylbutan-1-one. Both alcohols have the molecular formula \(\mathrm{C_{10}H_{14}O}\), but their structures and oxidation behaviour differ. Thus, identical molecular formulae and the presence of stereogenic centres do not imply identical alcohol classification or reactivity.

Explanation: \(\textbf{Stage I — Grignard reaction with a nitrile:}\) Benzonitrile has the structure \(\mathrm{C_6H_5CN}\). Its nitrile carbon is directly bonded to the phenyl ring. Reaction with methylmagnesium bromide introduces a methyl group at this carbon through nucleophilic addition. Complete acidic hydrolysis of the resulting imine-derived intermediate forms the corresponding ketone. \[ \mathrm{C_6H_5CN \xrightarrow[\text{2. }H_3O^+]{\text{1. }CH_3MgBr,\ dry\ ether} C_6H_5COCH_3} \] The product is acetophenone. The nitrile carbon has become the ketone carbonyl carbon. \(\textbf{Stage II — Preserve the required reaction conditions:}\) The ketone is isolated and dried before another Grignard reagent is introduced. This is essential because the acidic aqueous medium used in Stage I would destroy phenylmagnesium bromide through proton transfer. The reaction sequence therefore requires a separate dry organomagnesium-addition stage. \(\textbf{Stage III — Grignard addition to the ketone:}\) Acetophenone contains phenyl and methyl groups attached to its carbonyl carbon. Phenylmagnesium bromide supplies a second phenyl group. \[ \mathrm{C_6H_5COCH_3+C_6H_5MgBr} \] \[ \mathrm{\xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}} (C_6H_5)_2C(OH)CH_3} \] The final product is 1,1-diphenylethan-1-ol. Its hydroxyl-bearing carbon is tertiary because it is attached to three carbon groups. \(\textbf{Check for optical activity:}\) The same carbon bears two identical phenyl substituents. Therefore, it does not have four different groups and is not a stereogenic centre. Formation of a tetrahedral tertiary alcohol carbon does not automatically produce a chiral molecule. The sequence illustrates two distinct Grignard applications: conversion of a nitrile into a ketone and conversion of that ketone into a tertiary alcohol.