Organic Compounds Containing Oxygen MCQs with Answers – Part 9 (NEET Chemistry)

Explanation: \(\textbf{Identify the reacting carbon atoms:}\) The original molecule contains eight carbon atoms arranged in an open chain. Its ketone carbonyl groups occupy positions 2 and 7. The question specifies that carbon 7 acts as the electrophilic acceptor in both routes. The position of the enolate carbon therefore determines which new carbon–carbon bond forms. \(\textbf{Evaluate Route P:}\) The nucleophilic carbon is carbon 3. It attacks the carbonyl carbon at position 7. The new carbon–carbon bond joins carbons 3 and 7. The resulting cyclic path contains: \[ \mathrm{C_3-C_4-C_5-C_6-C_7-C_3} \] The ring contains five carbon atoms. \(\textbf{Identify the product functionality in P:}\) The attacked ketone carbonyl at carbon 7 becomes an alcohol-containing tetrahedral carbon after protonation. The carbonyl group at carbon 2 remains available in the beta-hydroxy ketone product. Thus, ring formation does not require loss of the original ketone carbonyl carbon. \(\textbf{Evaluate Route Q:}\) The nucleophilic carbon is the terminal methyl carbon at position 1. Attack occurs at carbon 7. The new carbon–carbon bond joins carbons 1 and 7. The resulting cyclic path contains: \[ \mathrm{C_1-C_2-C_3-C_4-C_5-C_6-C_7-C_1} \] The ring contains seven carbon atoms. \(\textbf{Compare the proposed pathways:}\) Both reactions introduce a new carbon–carbon bond within the same eight-carbon molecule. Neither adds an external carbon atom. However, selecting different alpha-carbons changes how many original chain atoms become enclosed in the ring. \(\textbf{Interpret the ring-size preference:}\) Five- and six-membered rings are generally favourable targets for many intramolecular cyclisations. The actual competition between these proposed routes would depend on enolate formation and cyclisation conditions. The question avoids assuming that either product must be experimentally dominant by specifying which enolate is generated in each experiment. The correct ring sizes follow from the explicit carbon connectivity.

Explanation: \(\textbf{Identify the carbon introduced during cyanohydrin formation:}\) HCN adds to a carbonyl group through formation of a new carbon–carbon bond. The cyanide carbon remains attached to the former carbonyl carbon. Hydrolysis of the nitrile later converts that introduced carbon into a carboxylic acid carbon. \(\textbf{Follow the methanal pathway:}\) Methanal forms the cyanohydrin: \[ \mathrm{HCHO+HCN\longrightarrow HOCH_2CN} \] Complete nitrile hydrolysis produces glycolic acid: \[ \mathrm{HOCH_2CN\xrightarrow{hydrolysis}HOCH_2COOH} \] Its hydroxyl group belongs to a primary alcohol carbon. The stipulated complete oxidation produces: \[ \mathrm{HOCH_2COOH\xrightarrow{[O]}HOOCCOOH} \] The final product is ethanedioic acid, commonly called oxalic acid. \(\textbf{Count acidic groups in P's final product:}\) Oxalic acid contains two free carboxylic acid groups. For: \[ n(P)=0.10\,\mathrm{mol} \] Complete hydrogencarbonate neutralisation gives: \[ n(\mathrm{CO_2})_P=2(0.10) \] \[ n(\mathrm{CO_2})_P=0.20\,\mathrm{mol} \] \(\textbf{Follow the ethanal pathway:}\) Ethanal forms: \[ \mathrm{CH_3CH(OH)CN} \] Nitrile hydrolysis gives 2-hydroxypropanoic acid: \[ \mathrm{CH_3CH(OH)COOH} \] Its alcohol-bearing carbon is secondary. Selective oxidation therefore produces: \[ \mathrm{CH_3CH(OH)COOH\xrightarrow{[O]}CH_3COCOOH} \] The final product is 2-oxopropanoic acid. \(\textbf{Count acidic groups in Q's final product:}\) 2-Oxopropanoic acid contains one free carboxylic acid group. Its ketone carbonyl is not another free acid group. Therefore: \[ n(\mathrm{CO_2})_Q=1(0.10) \] \[ n(\mathrm{CO_2})_Q=0.10\,\mathrm{mol} \] \(\textbf{Interpret the difference:}\) Both routes introduce one new carbon through cyanohydrin formation. However, the original carbonyl substituents determine whether the resulting hydroxy acid contains a primary or secondary alcohol group. Subsequent oxidation therefore produces a second carboxylic acid group only in P. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{CO_2})_P=0.20\,\mathrm{mol},\qquad n(\mathrm{CO_2})_Q=0.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the two requirements for condensation:}\) Benzaldehyde contains an electrophilic carbonyl carbon. Hydroxylamine supplies a nucleophilic nitrogen atom. Successful oxime formation requires carbonyl attack followed by proton transfers and dehydration. The reaction rate consequently depends on both carbonyl activation and nucleophile availability. \(\textbf{Explain the effect of moderate acidity:}\) Protonation of the carbonyl oxygen can increase the electrophilic character of the carbonyl carbon. This can facilitate attack by the nitrogen nucleophile. Suitable acid catalysis also assists proton-transfer steps associated with dehydration. The rising part of the graph is consistent with these catalytic contributions. \(\textbf{Examine excessive acidity:}\) Hydroxylamine is a basic nitrogen-containing reagent. In a strongly acidic medium, a larger fraction becomes protonated. Protonated hydroxylamine has reduced availability of the nitrogen lone pair required for ordinary nucleophilic attack. Consequently, increasing acidity may eventually reduce the concentration of the effective attacking species. \(\textbf{Interpret the graph maximum:}\) At intermediate acidity, the benefits of carbonyl activation and the availability of unprotonated hydroxylamine can be favourably balanced. At very high acidity, nucleophile protonation can outweigh further carbonyl activation. The initial rate can therefore decrease. \(\textbf{Evaluate the assertion and reason:}\) Assertion A is correct because the supplied graph shows that the rate does not increase indefinitely. Reason R is also correct because it identifies competing acid-dependent influences on the reaction. The reason directly explains why a maximum can occur. Therefore, both statements are correct, and R is the correct explanation of A. This illustrates why identifying a reaction as acid-catalysed does not imply that unlimited addition of acid always increases its rate.

Explanation: \(\textbf{Identify the two parts of an ester:}\) An ester contains an acyl-derived carbon fragment and an alkoxy-derived carbon fragment. Complete lithium aluminium hydride reduction can convert the acyl-derived portion into a primary alcohol. The ester alkoxy-derived portion is recovered as its corresponding alcohol during work-up. The two resulting alcohols therefore need not remain joined within one molecule. \(\textbf{Follow methyl ethanoate reduction:}\) Methyl ethanoate contains a two-carbon acetyl-derived fragment and a one-carbon methoxy-derived fragment. Its acyl carbonyl carbon becomes the alcohol-bearing carbon of ethanol. The methoxy-derived fragment gives methanol. With the label at the ester carbonyl carbon: \[ \mathrm{CH_3{}^{13}COOCH_3} \] \[ \mathrm{\xrightarrow[\text{work-up}]{LiAlH_4}CH_3{}^{13}CH_2OH+CH_3OH} \] Therefore, the labelled product is ethanol. \(\textbf{Follow ethyl methanoate reduction:}\) Ethyl methanoate contains a one-carbon methanoyl-derived fragment and a two-carbon ethoxy-derived fragment. The acyl carbon becomes the alcohol-bearing carbon of methanol. The ethoxy-derived fragment gives ethanol. \[ \mathrm{H{}^{13}COOCH_2CH_3} \] \[ \mathrm{\xrightarrow[\text{work-up}]{LiAlH_4}{}^{13}CH_3OH+CH_3CH_2OH} \] Therefore, the labelled product is methanol. \(\textbf{Compare the unlabelled product compositions:}\) Both starting esters ultimately produce one mole of methanol and one mole of ethanol per mole of ester. Without isotope information, the two complete-reduction product mixtures have the same constitutional alcohol components in the same molar ratio. However, the isotope identifies which alcohol originated from the ester acyl carbon. \(\textbf{Evaluate sodium-metal hydrogen evolution:}\) Each methanol molecule contains one O–H bond. Each ethanol molecule also contains one O–H bond. Thus, one mole of either ester ultimately supplies two moles of alcohol O–H groups. \[ n(\mathrm{O-H})=2n(\mathrm{ester}) \] Since two alcohol O–H hydrogens form one hydrogen molecule: \[ n(\mathrm{H_2})=\frac{2n(\mathrm{ester})}{2} \] \[ n(\mathrm{H_2})=n(\mathrm{ester}) \] \(\textbf{Interpret the reverse identification:}\) The complete-reduction products cannot distinguish these ester isomers merely through the identities or amounts of their unlabelled alcohols. The isotope label provides additional structural information by revealing the origin of the acyl carbon. This illustrates how two different starting connectivities can converge on the same ordinary product mixture while retaining distinguishable atom histories. \(\textbf{Final answer:}\) \[ n(\mathrm{H_2})=n(\mathrm{ester})\text{ for either starting isomer} \]

Explanation: \(\textbf{Distinguish the heated-copper reactions:}\) Ethanol is a primary alcohol and undergoes dehydrogenation to an aldehyde. Propan-2-ol is a secondary alcohol and undergoes dehydrogenation to a ketone. The tertiary alcohol has no hydrogen on its hydroxyl-bearing carbon and undergoes dehydration under the stipulated conditions. \(\textbf{Represent the three transformations:}\) \[ \mathrm{CH_3CH_2OH\xrightarrow{Cu,\ 573\,K}CH_3CHO+H_2} \] \[ \mathrm{CH_3CH(OH)CH_3\xrightarrow{Cu,\ 573\,K}CH_3COCH_3+H_2} \] \[ \mathrm{(CH_3)_3COH\xrightarrow{Cu,\ 573\,K}(CH_3)_2C=CH_2+H_2O} \] \(\textbf{Assign the unknown amounts:}\) Let the initial amounts of the primary, secondary and tertiary alcohols be \(x\), \(y\) and \(z\), respectively. \[ x+y+z=0.50 \] \(\textbf{Use the hydrogen measurement:}\) Each primary or secondary alcohol molecule produces one hydrogen molecule. The tertiary alcohol contributes no hydrogen through its stipulated dehydration. \[ x+y=0.35 \] Therefore: \[ z=0.50-0.35 \] \[ z=0.15\,\mathrm{mol} \] \(\textbf{Use the Fehling's measurement:}\) Ethanal is the only aldehyde among the stipulated heated-copper products. It gives the characteristic positive Fehling's reaction. Propanone is an ordinary ketone and does not contribute to the measured copper(I) oxide. \[ n(\mathrm{Cu_2O})=x \] \[ x=0.22\,\mathrm{mol} \] \(\textbf{Determine the secondary alcohol amount:}\) \[ y=0.35-0.22 \] \[ y=0.13\,\mathrm{mol} \] \(\textbf{Check the chemical balance:}\) \[ 0.22+0.13+0.15=0.50\,\mathrm{mol} \] The two dehydrogenating alcohols contribute: \[ 0.22+0.13=0.35\,\mathrm{mol\ H_2} \] Thus, the mixture composition satisfies both independent measurements. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} n(\mathrm{ethanol})=0.22\,\mathrm{mol}\\ n(\mathrm{propan\!-\!2\!-\!ol})=0.13\,\mathrm{mol}\\ n(\mathrm{2\!-\!methylpropan\!-\!2\!-\!ol})=0.15\,\mathrm{mol} \end{gathered}} \]

Explanation: \(\textbf{Distinguish activation from orientation:}\) Phenolic hydroxyl increases aromatic electron density through resonance donation. It ordinarily favours electrophilic attack at ortho and para positions. Bromine withdraws electron density through induction and deactivates the ring overall. However, its lone-pair resonance contribution makes bromine an ortho/para director. Therefore, the two groups can reinforce the same positions even though their overall activating effects differ. \(\textbf{Analyse 3-bromophenol:}\) Place hydroxyl at carbon 1 and bromine at carbon 3. The hydroxyl-directed positions are carbons 2, 4 and 6. Relative to bromine at carbon 3, carbons 2 and 4 are ortho, while carbon 6 is para. All three positions are therefore favoured by both directing effects. The actual proportions need not be equal because positions differ in steric accessibility. \(\textbf{Analyse 4-bromophenol:}\) Place hydroxyl at carbon 1 and bromine at carbon 4. Hydroxyl directs towards carbons 2 and 6 because its para position is occupied. Bromine directs towards carbons 3 and 5 because its para position is occupied by hydroxyl. The two sets contain no common vacant position. \(\textbf{Interpret the competition:}\) The absence of reinforcing positions in Q does not prohibit nitration. Its more strongly activating hydroxyl group can still exert an important influence on substitution. However, the two substituents do not jointly direct attack at one vacant position. The comparison demonstrates that directing effects must be mapped onto actual ring positions rather than inferred solely from the names of the substituents. In particular, a halogen's deactivating nature does not make it a meta director.

Explanation: \(\textbf{Identify the first Grignard reagents:}\) Bromoethane reacts with magnesium in dry ether to produce ethylmagnesium bromide. Benzyl bromide produces benzylmagnesium bromide under the corresponding stipulated conditions. Both organomagnesium compounds contain a carbon–magnesium bond capable of attacking carbon dioxide. \(\textbf{Follow carboxylation in Route P:}\) \[ \mathrm{CH_3CH_2MgBr+CO_2\longrightarrow CH_3CH_2COOMgBr} \] Acidic work-up converts the magnesium carboxylate into propanoic acid. \[ \mathrm{CH_3CH_2COOH} \] The original two-carbon ethyl fragment now belongs to a three-carbon acid. \(\textbf{Follow carboxylation in Route Q:}\) \[ \mathrm{C_6H_5CH_2MgBr+CO_2\longrightarrow C_6H_5CH_2COOMgBr} \] Acidic work-up gives phenylacetic acid. \[ \mathrm{C_6H_5CH_2COOH} \] The new carboxyl carbon is attached to the original benzyl carbon. \(\textbf{Apply soda-lime decarboxylation:}\) Sodium propanoate produces ethane: \[ \mathrm{CH_3CH_2COONa+NaOH \xrightarrow{CaO,\ \Delta}CH_3CH_3+Na_2CO_3} \] Sodium phenylacetate produces toluene: \[ \mathrm{C_6H_5CH_2COONa+NaOH \xrightarrow{CaO,\ \Delta}C_6H_5CH_3+Na_2CO_3} \] \(\textbf{Track the carbon introduced by carbon dioxide:}\) The carboxylation reaction adds one carbon atom to each original organic fragment. That introduced carbon becomes the carboxyl carbon. Soda-lime decarboxylation removes the same carboxyl-derived carbon into an inorganic carbonate product. The original ethyl and benzyl carbon frameworks are therefore recovered as their corresponding hydrocarbons. This is a carbon-skeleton accounting problem rather than a simple recognition of a Grignard reagent.

Explanation: \(\textbf{Classify the original alcohol centres:}\) In P, the hydroxyl-bearing carbon is attached to phenyl and methyl groups. It also possesses one hydrogen. Therefore, P is a secondary alcohol. In Q, the hydroxyl-bearing carbon is terminal and attached to only one other carbon. Therefore, Q is a primary alcohol. \(\textbf{Follow oxidation of P:}\) PCC converts the secondary alcohol into a ketone: \[ \mathrm{C_6H_5CH(OH)CH_3\xrightarrow{PCC}C_6H_5COCH_3} \] The product is acetophenone. Its carbonyl carbon is attached directly to a methyl group. Therefore, it contains the structural arrangement required for the characteristic iodoform reaction. \(\textbf{Apply iodoform cleavage:}\) \[ \mathrm{C_6H_5COCH_3\xrightarrow{I_2/OH^-}C_6H_5COO^-+CHI_3} \] The methyl carbon becomes the carbon of iodoform. The original ketone carbonyl carbon remains attached to the aromatic ring in the benzoate ion. Acidification therefore produces benzoic acid. \(\textbf{Follow oxidation of Q:}\) PCC converts the primary alcohol into phenylacetaldehyde: \[ \mathrm{C_6H_5CH_2CH_2OH\xrightarrow{PCC}C_6H_5CH_2CHO} \] The aldehyde carbon remains separated from the aromatic ring by one methylene group. \(\textbf{Apply Tollens' oxidation:}\) The aldehyde carbonyl is oxidised to a carboxylate. After acidification: \[ \mathrm{C_6H_5CH_2CHO\xrightarrow{[O]}C_6H_5CH_2COOH} \] The recovered acid is phenylacetic acid. \(\textbf{Interpret the contrasting carbon histories:}\) Both starting compounds contain two carbon atoms outside the aromatic ring. However, iodoform cleavage of P removes one of those carbons from the acid product. Tollens' oxidation of Q preserves both side-chain carbon atoms in the acid product. Consequently, positional isomerism in the starting alcohols leads to carboxylic acids with different carbon skeletons.

Explanation: \(\textbf{Identify the condition for benzylic oxidation:}\) Vigorous oxidation of an alkylbenzene side chain with hot alkaline potassium permanganate generally requires at least one hydrogen attached to the benzylic carbon. A suitable alkyl side chain is converted into a carboxyl group attached directly to the aromatic ring. \(\textbf{Oxidation of P:}\) 1-Ethyl-4-methylbenzene contains an ethyl group and a methyl group at para-related positions. The ethyl group contains a benzylic \(\mathrm{-CH_2-}\) carbon, while the methyl group contains three benzylic hydrogens. Both substituents therefore undergo the specified oxidation. The product contains two carboxylic acid groups at para positions and is benzene-1,4-dicarboxylic acid, commonly called terephthalic acid. \(\textbf{Oxidation of Q:}\) 1-tert-Butyl-4-methylbenzene contains a methyl group that possesses benzylic hydrogens. This group is oxidised to a carboxylic acid group. However, the tert-butyl substituent is attached to the aromatic ring through a carbon bonded to three methyl groups. This benzylic carbon has no hydrogen and is preserved under the stated conditions. Therefore, Q produces 4-tert-butylbenzoic acid. \(\textbf{Calculate carbon dioxide from P:}\) Benzene-1,4-dicarboxylic acid contains two carboxylic acid groups. Each group reacts with sodium hydrogencarbonate to release one molecule of carbon dioxide. \[ n(\mathrm{CO_2})_P=2n(\mathrm{dicarboxylic\ acid}) \] \[ n(\mathrm{CO_2})_P=2(0.10) \] \[ n(\mathrm{CO_2})_P=0.20\,\mathrm{mol} \] \(\textbf{Calculate carbon dioxide from Q:}\) 4-tert-Butylbenzoic acid contains only one carboxylic acid group. \[ n(\mathrm{CO_2})_Q=n(\mathrm{monocarboxylic\ acid}) \] \[ n(\mathrm{CO_2})_Q=0.10\,\mathrm{mol} \] \(\textbf{Final comparison:}\) Although both starting compounds contain two alkyl substituents, only P has benzylic hydrogens in both side chains. Therefore, oxidation of P produces a dicarboxylic acid, whereas oxidation of Q produces a monocarboxylic acid. The corresponding carbon dioxide yields are: \[ \boxed{n(\mathrm{CO_2})_P=0.20\,\mathrm{mol},\qquad n(\mathrm{CO_2})_Q=0.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the sulphonation intermediate:}\) Phenol-2,4-disulphonic acid contains one original phenolic hydroxyl group and two aromatic sulphonic acid groups. Its sulphonic acid groups occupy positions 2 and 4 relative to hydroxyl. These groups are substantially stronger acids than an ordinary phenolic hydroxyl. \(\textbf{Calculate the neutralisation capacity in P:}\) The question specifies complete neutralisation of both sulphonic acid groups. Each sulphonic acid proton consumed by hydrogencarbonate generates one carbon dioxide molecule. \[ n(\mathrm{substrate})_P=0.10\,\mathrm{mol} \] \[ n(\mathrm{acidic\ -SO_3H\ groups})_P=2(0.10) \] \[ n(\mathrm{acidic\ -SO_3H\ groups})_P=0.20\,\mathrm{mol} \] Therefore: \[ n(\mathrm{CO_2})_P=0.20\,\mathrm{mol} \] \(\textbf{Follow the characteristic nitration sequence:}\) Concentrated nitric acid converts the specified disulphonated phenol into 2,4,6-trinitrophenol under the stated conditions. The final aromatic product contains nitro groups at positions 2, 4 and 6. It is commonly called picric acid. The original sulphonic acid groups are not retained in the final product. \(\textbf{Identify the acidic functionality in Q:}\) The three strongly electron-withdrawing nitro groups substantially increase the acidity of the remaining phenolic O–H bond. However, the final molecule contains only one such acidic hydroxyl group. It no longer contains the original two sulphonic acid groups. \(\textbf{Calculate Q's gas yield:}\) \[ n(\mathrm{picric\ acid})_Q=0.10\,\mathrm{mol} \] Under the stipulated complete hydrogencarbonate reaction: \[ n(\mathrm{CO_2})_Q=1(0.10) \] \[ n(\mathrm{CO_2})_Q=0.10\,\mathrm{mol} \] \(\textbf{Interpret the structural change:}\) The final nitration product is a much stronger phenolic acid than unsubstituted phenol. Nevertheless, it contains fewer groups capable of supplying the stipulated hydrogencarbonate-reactive protons than the disulphonic acid intermediate. Acid strength and the number of acidic functional groups therefore determine different aspects of the observations. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{CO_2})_P=0.20\,\mathrm{mol},\qquad n(\mathrm{CO_2})_Q=0.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the alpha-carbon requirement:}\) The characteristic Hell–Volhard–Zelinsky reaction brominates the carbon directly adjacent to the carboxyl carbon in suitable aliphatic acid structures. An appropriate alpha-hydrogen is necessary for the ordinary reaction pathway. The presence of hydrogens elsewhere in a molecule is not sufficient. \(\textbf{Evaluate phenylacetic acid:}\) Phenylacetic acid contains the group \(\mathrm{C_6H_5CH_2COOH}\). Its saturated alpha-carbon is the methylene carbon between phenyl and carboxyl. That carbon bears two hydrogens. Therefore, P can undergo alpha-bromination. \[ \mathrm{C_6H_5CH_2COOH \xrightarrow{Br_2/red\ P} C_6H_5CHBrCOOH} \] The product is 2-bromo-2-phenylethanoic acid. \(\textbf{Follow hydroxyl substitution:}\) Under the stipulated substitution conditions, bromine at the alpha-carbon is replaced by hydroxyl. \[ \mathrm{C_6H_5CHBrCOOH \xrightarrow{OH^-,\ work-up} C_6H_5CH(OH)COOH} \] The product is 2-hydroxy-2-phenylethanoic acid, commonly called mandelic acid. Its hydroxyl-bearing carbon is attached to hydrogen, hydroxyl, phenyl and carboxyl groups. These four substituents are different, so that carbon is stereogenic. Without a chiral influence, the reaction need not produce one enantiomer exclusively. \(\textbf{Evaluate benzoic acid:}\) Benzoic acid contains its carboxyl group attached directly to an aromatic carbon. It has no saturated alpha-carbon bearing the hydrogen required for the stipulated Hell–Volhard–Zelinsky pathway. Therefore, Q does not undergo the specified alpha-bromination. \(\textbf{Evaluate 2,2-dimethylpropanoic acid:}\) The carbon directly adjacent to its carboxyl group is attached to three methyl groups. It carries no hydrogen. Although the terminal methyl groups possess hydrogens, those hydrogens are not attached to the alpha-carbon. Therefore, R also lacks the required alpha-hydrogen. \(\textbf{Interpret the comparison:}\) P and Q both contain a phenyl ring, but their carboxyl groups have different local carbon environments. P and R both have saturated alpha-carbons, but only P has an alpha-hydrogen. The reaction depends on this precise structural requirement rather than aromatic character or overall hydrogen content.

Explanation: \(\textbf{Identify the Cannizzaro-reactive functionality:}\) 4-Hydroxybenzaldehyde contains a phenolic hydroxyl group and an aromatic aldehyde group. Its aldehyde group lacks an enolisable alpha-hydrogen and can undergo the Cannizzaro disproportionation under sufficiently concentrated alkaline conditions. In this reaction, two aldehyde molecules participate: one is oxidised into a carboxylate, while the other is reduced into a primary alcohol. \(\textbf{Account for phenoxide formation:}\) During the alkaline reaction, the phenolic hydroxyl group may be deprotonated by sodium hydroxide to form phenoxide. This acid–base reaction does not eliminate the phenolic oxygen or change the position of the substituent on the aromatic ring. After acidification, the phenolic hydroxyl group is regenerated in both organic products. \(\textbf{Identify the reduction product:}\) One aldehyde group is reduced into a primary alcohol group. The recovered compound has the structure: \[ \mathrm{p\!-\!HOC_6H_4CH_2OH} \] This compound is 4-hydroxybenzyl alcohol. \(\textbf{Identify the oxidation product:}\) The second aldehyde group is oxidised into a carboxylate group. Acidification subsequently produces: \[ \mathrm{p\!-\!HOC_6H_4COOH} \] This compound is 4-hydroxybenzoic acid. \(\textbf{Apply Cannizzaro stoichiometry:}\) Two molecules of the starting aldehyde give one molecule of the alcohol-derived product and one molecule of the acid-derived product. The initial amount is: \[ n(\mathrm{starting\ aldehyde})=0.20\,\mathrm{mol} \] Therefore: \[ n(\mathrm{4\!-\!hydroxybenzyl\ alcohol})=\frac{0.20}{2} \] \[ n(\mathrm{4\!-\!hydroxybenzyl\ alcohol})=0.10\,\mathrm{mol} \] Similarly: \[ n(\mathrm{4\!-\!hydroxybenzoic\ acid})=0.10\,\mathrm{mol} \] \(\textbf{Calculate carbon dioxide evolution:}\) The acid-containing product has one carboxylic acid group and one phenolic hydroxyl group. Under the stipulated aqueous sodium hydrogencarbonate conditions, only the carboxylic acid group liberates carbon dioxide appreciably. One mole of 4-hydroxybenzoic acid therefore produces one mole of carbon dioxide. \[ n(\mathrm{CO_2})=n(\mathrm{4\!-\!hydroxybenzoic\ acid}) \] \[ n(\mathrm{CO_2})=0.10\,\mathrm{mol} \] \(\textbf{Final interpretation:}\) The Cannizzaro reaction produces equal molar amounts of 4-hydroxybenzyl alcohol and 4-hydroxybenzoic acid. The phenolic hydroxyl group remains present in both products after acidification. Only the separated carboxylic acid product is used for the final hydrogencarbonate reaction. \(\textbf{Final numerical result:}\) \[ \boxed{n(\mathrm{CO_2})=0.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify glycerol's three alcohol groups:}\) Glycerol contains two terminal primary alcohol groups and one central secondary alcohol group. The two terminal \(\mathrm{-CH_2OH}\) groups are constitutionally equivalent in the original molecule. Its central carbon is not initially stereogenic because the two groups attached on either side are identical. \(\textbf{Follow Route P:}\) Selective oxidation of one terminal primary alcohol produces: \[ \mathrm{OHCCH(OH)CH_2OH} \] The product is 2,3-dihydroxypropanal, commonly called glyceraldehyde. It contains an aldehyde group. Therefore, it gives the characteristic positive Tollens' reaction. \(\textbf{Evaluate the stereogenic centre in P:}\) The central carbon of glyceraldehyde is attached to hydrogen, hydroxyl, an aldehyde-derived group and a hydroxymethyl group. These four substituents are different. Therefore, the central carbon is stereogenic. Under the stated achiral and nonselective conditions, formation of one enantiomer exclusively is not implied. \(\textbf{Follow Route Q:}\) Oxidation of the central secondary alcohol produces: \[ \mathrm{HOCH_2COCH_2OH} \] The product is 1,3-dihydroxypropan-2-one, commonly called dihydroxyacetone. It contains an ordinary ketone carbonyl. It does not give the characteristic positive Tollens' reaction under the specified ordinary conditions. \(\textbf{Evaluate the stereochemical structure of Q:}\) The central ketone carbon is trigonal planar. The terminal hydroxymethyl groups are identical. No stereogenic carbon is present in the stated product. \(\textbf{Interpret the relationship:}\) Glyceraldehyde and dihydroxyacetone have the same molecular formula, \(\mathrm{C_3H_6O_3}\). They are functional-group isomers containing an aldehyde or ketone group, respectively. Their different stereochemical properties arise because terminal oxidation destroys the equivalence of glycerol's two ends, whereas central oxidation preserves it. The reaction therefore connects selective alcohol oxidation, carbonyl classification and the creation or absence of molecular chirality.

Explanation: \(\textbf{Distinguish aldehydic hydrogen from alpha-hydrogen:}\) Both compounds contain a terminal aldehyde group. The hydrogen attached directly to the aldehyde carbonyl carbon is not an alpha-hydrogen. The alpha-carbon is the carbon immediately adjacent to that carbonyl carbon. Its hydrogen content determines eligibility for ordinary enolate formation. \(\textbf{Examine P:}\) In 2,2-dimethylpropanal, the alpha-carbon is attached to the aldehyde carbon and three methyl groups. It possesses no hydrogen. Therefore, P cannot form an ordinary alpha-enolate through deprotonation at that carbon. Under the specified concentrated alkaline conditions, it can undergo Cannizzaro disproportionation. \(\textbf{Follow P's disproportionation:}\) One aldehyde molecule is reduced to an alcohol. The other is oxidised to a carboxylate. \[ \mathrm{2(CH_3)_3CCHO+NaOH} \] \[ \mathrm{\longrightarrow (CH_3)_3CCH_2OH+(CH_3)_3CCOONa} \] The alcohol is 2,2-dimethylpropan-1-ol. The salt contains the 2,2-dimethylpropanoate ion. \(\textbf{Examine Q:}\) In 2-methylpropanal, the alpha-carbon is bonded to two methyl groups, an aldehyde carbon and one hydrogen. That hydrogen can be removed under suitable enolate-forming conditions. Therefore, Q can function as an aldol enolate donor. \(\textbf{Follow Q's aldol addition:}\) The enolate carbon attacks the carbonyl carbon of another Q molecule. A new carbon–carbon bond forms. Protonation produces a beta-hydroxy aldehyde. The question explicitly stops the reaction at this addition stage. \(\textbf{Interpret the distinction:}\) Both aldehydes are branched near their carbonyl groups. However, the additional methyl substituent in P removes the final alpha-hydrogen. This changes its eligibility for enolate chemistry. Therefore, structural branching must be assessed at the precise alpha-carbon rather than treated as a general prohibition on aldol reactions.

Explanation: \(\textbf{Identify the carbon source in Route P:}\) The Gattermann–Koch reaction introduces a formyl group into a suitable aromatic ring. The carbon atom of the formyl group originates from carbon monoxide. Using \({}^{13}\mathrm{CO}\) therefore places the isotope at the carbonyl carbon of the introduced aldehyde group. \(\textbf{Follow the first intermediate:}\) \[ \mathrm{C_6H_6+{}^{13}CO+HCl} \] \[ \mathrm{\xrightarrow{AlCl_3/CuCl}C_6H_5{}^{13}CHO} \] The organic product is labelled benzaldehyde. It contains an aldehyde group and can give a positive characteristic Tollens' reaction. \(\textbf{Apply aldehyde oxidation:}\) Oxidation changes the aldehyde group into a carboxylic acid group. It does not remove the aldehyde-derived carbon. Therefore: \[ \mathrm{C_6H_5{}^{13}CHO\xrightarrow{[O]}C_6H_5{}^{13}COOH} \] The final benzoic acid is labelled at its carboxyl carbon. \(\textbf{Identify the carbon source in Route Q:}\) Friedel–Crafts methylation introduces the methyl carbon supplied by methyl chloride. \[ \mathrm{C_6H_6+{}^{13}CH_3Cl} \] \[ \mathrm{\xrightarrow{AlCl_3}C_6H_5{}^{13}CH_3+HCl} \] The intermediate is labelled toluene. It does not contain an aldehyde or carboxylic acid group. \(\textbf{Follow vigorous side-chain oxidation:}\) Toluene contains benzylic hydrogens, so its methyl group undergoes the stipulated oxidation. The carbon directly attached to benzene becomes the carboxyl carbon. \[ \mathrm{C_6H_5{}^{13}CH_3\xrightarrow{KMnO_4/OH^-,\ \Delta;\ H^+}C_6H_5{}^{13}COOH} \] \(\textbf{Interpret the convergence:}\) Both routes introduce exactly one new carbon atom into the benzene framework. In P, that carbon enters as an aldehyde carbon. In Q, it enters as a methyl carbon. Their subsequent oxidations converge on the same labelled carboxylic acid. The isotope therefore reveals the origin of the new carboxyl carbon without implying that the two preparative mechanisms are identical.

Explanation: \(\textbf{Identify the two reducible functional groups:}\) Methyl 3-oxobutanoate contains a ketone carbonyl and an ester carbonyl. The carbonyl groups are chemically distinguishable. The question specifies different reduction extents in the two routes. \(\textbf{Follow the selective reduction in P:}\) Sodium borohydride reduces the ketone carbonyl to a secondary alcohol. The methyl ester remains unchanged. \[ \mathrm{CH_3COCH_2COOCH_3} \] \[ \mathrm{\xrightarrow{NaBH_4}CH_3CH(OH)CH_2COOCH_3} \] The product is methyl 3-hydroxybutanoate. Each product molecule contains one free alcohol O–H group. Its ester oxygen atoms do not provide additional ordinary alcohol hydrogens. \(\textbf{Calculate hydrogen from P:}\) \[ n(\mathrm{product})_P=0.10\,\mathrm{mol} \] \[ n(\mathrm{O-H})_P=1(0.10)=0.10\,\mathrm{mol} \] Two alcohol O–H hydrogens form one hydrogen molecule. \[ n(\mathrm{H_2})_P=\frac{0.10}{2} \] \[ n(\mathrm{H_2})_P=0.05\,\mathrm{mol} \] \(\textbf{Follow the complete reduction in Q:}\) Excess lithium aluminium hydride reduces the ketone to a secondary alcohol. It also reduces the ester acyl carbon into a primary alcohol-bearing carbon. The ester methoxy-derived fragment becomes methanol after work-up. \[ \mathrm{CH_3COCH_2COOCH_3} \] \[ \mathrm{\xrightarrow[\text{work-up}]{excess\ LiAlH_4} CH_3CH(OH)CH_2CH_2OH+CH_3OH} \] The principal four-carbon product is butane-1,3-diol. It contains two alcohol O–H groups. Methanol supplies one additional O–H group per original ester molecule. \(\textbf{Count all alcohol hydrogens in Q:}\) \[ n(\mathrm{diol})=0.10\,\mathrm{mol} \] \[ n(\mathrm{methanol})=0.10\,\mathrm{mol} \] \[ n(\mathrm{O-H})_Q=2(0.10)+1(0.10) \] \[ n(\mathrm{O-H})_Q=0.30\,\mathrm{mol} \] \(\textbf{Calculate hydrogen from Q:}\) \[ n(\mathrm{H_2})_Q=\frac{0.30}{2} \] \[ n(\mathrm{H_2})_Q=0.15\,\mathrm{mol} \] \(\textbf{Check the difference:}\) The additional hydrogen evolved in Q reflects formation of an alcohol from the original ester acyl fragment and recovery of methanol from its alkoxy fragment. Counting only the four-carbon diol would underestimate the total O–H content of the isolated organic product fraction. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{H_2})_P=0.05\,\mathrm{mol},\qquad n(\mathrm{H_2})_Q=0.15\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the independent reactive functionalities:}\) Each starting molecule contains one methyl ester and one aromatic aldehyde. Saponification converts the ester into a carboxylate. Cannizzaro disproportionation converts aldehyde groups into alcohol-derived and carboxylate-derived groups. The question specifies complete operation of both processes. \(\textbf{Count the products of the Cannizzaro reaction:}\) Two aldehyde molecules participate in one ordinary self-Cannizzaro event. One aldehyde carbon is reduced to a hydroxymethyl group. The other is oxidised to a carboxylate group. \[ n(\mathrm{starting\ molecules})=0.20\,\mathrm{mol} \] Therefore: \[ n(\mathrm{Cannizzaro\ reduced\ products})=\frac{0.20}{2} \] \[ n(\mathrm{Cannizzaro\ reduced\ products})=0.10\,\mathrm{mol} \] Similarly: \[ n(\mathrm{Cannizzaro\ oxidised\ products})=0.10\,\mathrm{mol} \] \(\textbf{Account for ester saponification:}\) Every starting molecule contains one methyl ester. Complete saponification converts that ester into a carboxylate, irrespective of whether the aldehyde group was oxidised or reduced. After acidification, the reduced Cannizzaro product is: \[ \mathrm{p\!-\!HOCH_2{-}C_6H_4{-}COOH} \] This is 4-(hydroxymethyl)benzoic acid. It contains one carboxylic acid group. \(\textbf{Identify the oxidised product:}\) The oxidised Cannizzaro product has its original methyl ester converted into one acid group. Its aldehyde has been converted into a second acid group. After acidification, it is: \[ \mathrm{HOOC{-}C_6H_4{-}COOH} \] This is benzene-1,4-dicarboxylic acid, commonly called terephthalic acid. It contains two carboxylic acid groups. \(\textbf{Calculate the gas from the reduced fraction:}\) \[ n(\mathrm{CO_2})_{\mathrm{reduced}}=1(0.10) \] \[ n(\mathrm{CO_2})_{\mathrm{reduced}}=0.10\,\mathrm{mol} \] \(\textbf{Calculate the gas from the oxidised fraction:}\) \[ n(\mathrm{CO_2})_{\mathrm{oxidised}}=2(0.10) \] \[ n(\mathrm{CO_2})_{\mathrm{oxidised}}=0.20\,\mathrm{mol} \] \(\textbf{Combine the gas yields:}\) \[ n(\mathrm{CO_2})_{\mathrm{total}}=0.10+0.20 \] \[ n(\mathrm{CO_2})_{\mathrm{total}}=0.30\,\mathrm{mol} \] \(\textbf{Check the functional-group count:}\) All \(0.20\,\mathrm{mol}\) starting molecules contribute one acid group through ester hydrolysis. Only half of the aldehyde groups additionally become acid groups through Cannizzaro oxidation. Thus: \[ n(\mathrm{COOH\ groups})=0.20+0.10 \] \[ n(\mathrm{COOH\ groups})=0.30\,\mathrm{mol} \] The hydroxybenzyl alcohol functionality does not contribute an additional hydrogencarbonate-reactive proton. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{CO_2})_{\mathrm{total}}=0.30\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the two bromine-reactive sites:}\) 4-Vinylphenol contains an alkene side chain and an activated phenolic aromatic ring. The two sites can react with bromine through different mechanisms. The question deliberately separates alkene addition from aromatic substitution. \(\textbf{Follow Stage I:}\) The vinyl group contains one carbon–carbon double bond. One bromine molecule adds across that bond. \[ \mathrm{HOC_6H_4CH=CH_2+Br_2} \] \[ \mathrm{\longrightarrow HOC_6H_4CHBrCH_2Br} \] The product contains a saturated dibrominated side chain. For the original sample: \[ n(\mathrm{Br_2})_I=1(0.20) \] \[ n(\mathrm{Br_2})_I=0.20\,\mathrm{mol} \] \(\textbf{Locate the available aromatic positions:}\) Place the phenolic hydroxyl group at aromatic position 1. The side chain occupies position 4. The para position relative to hydroxyl is therefore unavailable for ordinary replacement of ring hydrogen. The two ortho positions, carbons 2 and 6, remain available. \(\textbf{Follow Stage II:}\) The question specifies complete aromatic substitution at both ortho positions. Each substitution consumes one bromine molecule. Thus, two additional bromine molecules react per organic molecule. \[ n(\mathrm{Br_2})_{II}=2(0.20) \] \[ n(\mathrm{Br_2})_{II}=0.40\,\mathrm{mol} \] \(\textbf{Combine the bromine requirements:}\) \[ n(\mathrm{Br_2})_{\mathrm{total}}=0.20+0.40 \] \[ n(\mathrm{Br_2})_{\mathrm{total}}=0.60\,\mathrm{mol} \] \(\textbf{Determine the sodium-reactive O–H count:}\) The original phenolic hydroxyl remains intact throughout both stages. The bromination reactions introduce no new ordinary alcohol hydroxyl group. Therefore, each final organic molecule contains one free phenolic O–H bond. \[ n(\mathrm{O-H})=0.20\,\mathrm{mol} \] Two oxygen-bound hydrogens produce one hydrogen molecule. \[ n(\mathrm{H_2})=\frac{0.20}{2} \] \[ n(\mathrm{H_2})=0.10\,\mathrm{mol} \] \(\textbf{Interpret the stoichiometric distinction:}\) Alkene bromination consumes bromine through addition without replacing hydrogen. Aromatic bromination consumes bromine through substitution of ring hydrogens. The sodium-metal reaction probes the remaining phenolic hydroxyl rather than the number of bromine atoms introduced. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Br_2})_{\mathrm{total}}=0.60\,\mathrm{mol},\qquad n(\mathrm{H_2})=0.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify both carbonyl environments:}\) X contains an aldehyde group at one end of its four-carbon skeleton. It also contains a methyl ketone group at the other end. The two carbonyl groups have different diagnostic reactivities. Their characteristic responses must be counted independently. \(\textbf{Experiment P — Determine silver formation:}\) The question specifies oxidation of the aldehyde group with Tollens' reagent. Each X molecule contains one aldehyde group. Two silver(I) ions are reduced per reacting aldehyde group. \[ n(\mathrm{aldehyde\ groups})=0.15\,\mathrm{mol} \] Therefore: \[ n(\mathrm{Ag})=2(0.15) \] \[ n(\mathrm{Ag})=0.30\,\mathrm{mol} \] \(\textbf{Experiment Q — Locate the haloform-reactive group:}\) X contains the fragment \(\mathrm{CH_3CO-}\). Its methyl carbon is directly attached to a ketone carbonyl carbon. This is the characteristic methyl ketone arrangement required for the iodoform reaction. The question specifies one iodoform molecule per reacting methyl ketone group. \[ n(\mathrm{CHI_3})=1(0.15) \] \[ n(\mathrm{CHI_3})=0.15\,\mathrm{mol} \] \(\textbf{Experiment R — Reduce both carbonyl groups:}\) Sodium borohydride converts the aldehyde group into a primary alcohol. It converts the ketone group into a secondary alcohol. \[ \mathrm{CH_3COCH_2CHO} \] \[ \mathrm{\xrightarrow{NaBH_4}CH_3CH(OH)CH_2CH_2OH} \] The resulting compound is butane-1,3-diol. Each product molecule contains two free alcohol hydroxyl groups. \(\textbf{Count hydrogen-producing groups:}\) \[ n(\mathrm{diol})=0.15\,\mathrm{mol} \] \[ n(\mathrm{O-H})=2(0.15) \] \[ n(\mathrm{O-H})=0.30\,\mathrm{mol} \] \(\textbf{Apply sodium-metal stoichiometry:}\) \[ \mathrm{2ROH+2Na\longrightarrow 2RONa+H_2} \] Therefore: \[ n(\mathrm{H_2})=\frac{0.30}{2} \] \[ n(\mathrm{H_2})=0.15\,\mathrm{mol} \] \(\textbf{Check the independent functional-group counts:}\) Each X molecule has one Tollens-reactive aldehyde group. It also has one methyl ketone group capable of producing iodoform under the stipulated conditions. After complete hydride reduction, those two carbonyl groups become two ordinary alcohol hydroxyl groups. The final measurements therefore reveal different consequences of the same original multifunctional structure. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} n(\mathrm{Ag})=0.30\,\mathrm{mol}\\ n(\mathrm{CHI_3})=0.15\,\mathrm{mol}\\ n(\mathrm{H_2})=0.15\,\mathrm{mol} \end{gathered}} \]

Explanation: \(\textbf{Distinguish molecular amount from particle amount:}\) The original sample contains a fixed number of ethanoic acid formula units. Hydrogen-bonded dimer formation does not change this amount of acid. However, every dimer replaces two independent monomer particles with one associated particle. Therefore, association reduces the total number of independently behaving solute particles. \(\textbf{Calculate the mass of acid present:}\) \[ m=nM \] \[ m=0.50\times60 \] \[ m=30\,\mathrm{g} \] \(\textbf{Use the apparent molar mass:}\) \[ M_{\mathrm{app}}=\frac{m}{n_{\mathrm{particles}}} \] Therefore: \[ n_{\mathrm{particles}}=\frac{30}{100} \] \[ n_{\mathrm{particles}}=0.30\,\mathrm{mol} \] \(\textbf{Assign monomer and dimer amounts:}\) Let \(x\) be the amount of dimers in moles. Each dimer contains two original ethanoic acid molecules. The amount of unassociated monomer is therefore: \[ n_{\mathrm{monomer}}=0.50-2x \] The equilibrium particle amount is: \[ n_{\mathrm{particles}}=(0.50-2x)+x \] \[ 0.30=0.50-x \] \[ x=0.20\,\mathrm{mol} \] \(\textbf{Calculate the fraction associated:}\) The dimers contain: \[ n_{\mathrm{acid\ units\ in\ dimers}}=2(0.20) \] \[ n_{\mathrm{acid\ units\ in\ dimers}}=0.40\,\mathrm{mol} \] Therefore: \[ \alpha=\frac{0.40}{0.50} \] \[ \alpha=0.80=80\% \] \(\textbf{Determine the effect of complete dissociation:}\) Before dilution, there are \(0.30\,\mathrm{mol}\) independent solute particles. After complete dimer dissociation, all \(0.50\,\mathrm{mol}\) acid formula units behave as separate particles. \[ \text{Particle-number factor}=\frac{0.50}{0.30} \] \[ =\frac53 \] \(\textbf{Interpret the chemical origin:}\) Carboxylic acids can form cyclic hydrogen-bonded dimers in suitable nonpolar environments. This association contributes to their characteristic intermolecular behaviour. A change in solvent can alter molecular association without changing the number of acid molecules originally introduced. Consequently, an apparent molar mass derived from particle count need not equal the formula molar mass. \(\textbf{Final answer:}\) \[ \boxed{\alpha=80\%,\qquad \frac{n_{\mathrm{particles,\ after}}}{n_{\mathrm{particles,\ before}}}=\frac53} \]