Organic Compounds Containing Oxygen MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: The effect of heated copper at \(573\,\mathrm{K}\) depends on the structural class of the alcohol. Primary and secondary alcohols generally undergo dehydrogenation, whereas tertiary alcohols undergo dehydration under the characteristic conditions described in NCERT. Compound P is propan-1-ol, a primary alcohol. Dehydrogenation removes hydrogen from the hydroxyl group and the hydroxyl-bearing carbon, generating a carbonyl group. \[ \mathrm{CH_3CH_2CH_2OH\xrightarrow{Cu,\ 573\,K}CH_3CH_2CHO+H_2} \] The organic product is propanal, and hydrogen gas is liberated. Compound Q is propan-2-ol, a secondary alcohol. It undergoes the analogous removal of hydrogen to give propanone. \[ \mathrm{CH_3CH(OH)CH_3\xrightarrow{Cu,\ 573\,K}CH_3COCH_3+H_2} \] Thus, Q also liberates hydrogen gas, but its organic product is a ketone rather than an aldehyde. Compound R is 2-methylpropan-2-ol, a tertiary alcohol. Its hydroxyl-bearing carbon has no hydrogen atom available for ordinary dehydrogenation to a carbonyl compound. Consequently, R undergoes dehydration to produce 2-methylpropene and water. \[ \mathrm{(CH_3)_3COH\xrightarrow{Cu,\ 573\,K}(CH_3)_2C{=}CH_2+H_2O} \] This reaction does not liberate molecular hydrogen as the characteristic gaseous coproduct. The comparison therefore distinguishes dehydrogenation from dehydration using both the organic product class and the small molecule eliminated. Correct reasoning requires identifying the original alcohol class before predicting whether the reaction produces an aldehyde, ketone or alkene.

Explanation: Both starting compounds are secondary alcohols because their hydroxyl-bearing carbon atoms are attached to two other carbon atoms. PCC oxidises secondary alcohols to ketones without ordinarily changing their carbon skeletons. Butan-2-ol has the structure \(\mathrm{CH_3CH(OH)CH_2CH_3}\). Its oxidation produces butan-2-one. \[ \mathrm{CH_3CH(OH)CH_2CH_3+[O]\xrightarrow{PCC}CH_3COCH_2CH_3+H_2O} \] The resulting ketone P contains a methyl group directly attached to the carbonyl carbon. Thus, it possesses the structural unit \(\mathrm{CH_3CO-}\) required for the characteristic haloform reaction. Treatment with iodine and aqueous sodium hydroxide consequently produces a yellow precipitate of iodoform, \(\mathrm{CHI_3}\). \[ \mathrm{CH_3COCH_2CH_3+3I_2+4NaOH\longrightarrow CHI_3+CH_3CH_2COONa+3NaI+3H_2O} \] Pentan-3-ol has the structure \(\mathrm{CH_3CH_2CH(OH)CH_2CH_3}\). Oxidation produces pentan-3-one. \[ \mathrm{CH_3CH_2CH(OH)CH_2CH_3+[O]\xrightarrow{PCC}CH_3CH_2COCH_2CH_3+H_2O} \] The carbonyl carbon of pentan-3-one is attached to two ethyl groups rather than to a methyl group. It therefore lacks the methyl ketone structural requirement and does not undergo the characteristic iodoform reaction. The important conclusion is that all secondary alcohols can give ketones on suitable oxidation, but not every resulting ketone is iodoform-positive. The reaction identifies a particular structural arrangement rather than the entire ketone functional-group class.

Explanation: Lucas reagent consists of concentrated hydrochloric acid and anhydrous zinc chloride. Zinc chloride acts as a Lewis acid and facilitates replacement of the hydroxyl group by chlorine. Alcohols are initially soluble in the reagent, while the corresponding alkyl chlorides are comparatively immiscible and produce turbidity. The appearance of turbidity therefore provides information about the rate of alkyl chloride formation. Tertiary alcohols undergo the reaction most readily because the corresponding tertiary carbocations are comparatively stable. Compound P gives immediate turbidity and is therefore identified as 2-methylpropan-2-ol. This alcohol has three methyl groups attached directly to its hydroxyl-bearing carbon. Secondary alcohols usually react more slowly, producing turbidity after a few minutes under the specified conditions. Compound Q is therefore butan-2-ol, whose hydroxyl-bearing carbon is attached to two carbon groups. Primary alcohols ordinarily do not produce turbidity with Lucas reagent at room temperature within the usual observation period. Compound R is consequently butan-1-ol. The observations are consistent with the relative ease of substitution in the three alcohol classes. The molecular formula alone cannot distinguish the isomers because all three compounds contain four carbon atoms and one oxygen atom. The test depends primarily on the structure of the hydroxyl-bearing carbon and the reaction pathway made accessible under the strongly acidic conditions. Therefore, the turbidity sequence identifies the tertiary, secondary and primary alcohols in that order.

Explanation: The Assertion is correct because 2-methylpropan-2-ol is a tertiary alcohol and reacts rapidly with Lucas reagent. Butan-2-ol is a secondary alcohol and generally requires a longer time to produce observable turbidity at room temperature. Lucas reagent contains concentrated hydrochloric acid and anhydrous zinc chloride, which facilitate conversion of suitable alcohols into alkyl chlorides. The resulting alkyl chlorides have limited miscibility with the aqueous reagent, causing the characteristic turbidity. The Reason is also correct because formation of a tertiary carbocation is more favourable than formation of a comparable secondary carbocation. Alkyl groups stabilise positively charged carbon centres through electron-releasing effects and hyperconjugation. The tert-butyl carbocation is therefore comparatively stable relative to the secondary carbocation derived from butan-2-ol. This stability facilitates the reaction pathway leading to tert-butyl chloride. Butan-2-ol can also undergo substitution under Lucas-test conditions, but the process is slower. The rate difference is therefore directly connected to the substitution class of the alcohol and the stability of the relevant carbocation intermediate. The presence of the same hydroxyl functional group in both molecules does not imply identical reaction rates. Thus, both the Assertion and Reason are true, and the Reason correctly accounts for the faster appearance of turbidity from the tertiary alcohol.

Explanation: Butan-2-ol is a secondary alcohol containing the hydroxyl group on carbon 2 of a four-carbon chain. Under suitable strongly acidic dehydration conditions, the hydroxyl group is protonated, making water a better leaving group. Loss of water can generate a secondary carbocation intermediate in the usual acid-catalysed pathway. Formation of an alkene then requires removal of a hydrogen atom from a carbon adjacent to the positively charged carbon. Butan-2-ol has beta-hydrogen atoms available on both sides of the hydroxyl-bearing carbon. Removal of a hydrogen from the terminal side can produce but-1-ene, whereas removal from the other side produces but-2-ene. But-2-ene is the more substituted alkene and is generally thermodynamically more stable than but-1-ene. Consequently, it is favoured under the usual elimination conditions. But-2-ene also exists as cis and trans geometrical isomers because each carbon of the double bond is attached to two different groups. The trans isomer places the two methyl groups on opposite sides of the double bond and experiences less steric crowding than the cis isomer. Therefore, trans-but-2-ene is generally more stable and favoured within the but-2-ene product fraction. The elimination reaction does not oxidise the alcohol to a ketone because it involves loss of water rather than removal of hydrogen to form a carbonyl group. The correct prediction requires first selecting the more substituted alkene and then considering its geometrical stability.

Explanation: \(\textbf{Cyclohexanol dehydration:}\) Cyclohexanol is a secondary cyclic alcohol because its hydroxyl-bearing carbon is bonded to two other ring carbon atoms. Under suitable acid-catalysed conditions, the hydroxyl group is protonated and eliminated as water. Subsequent loss of a beta-hydrogen produces a double bond within the cyclohexane ring. Because the ring is unsubstituted, elimination towards either adjacent ring carbon gives the same constitutional alkene. \[ \mathrm{Cyclohexanol\xrightarrow{H^+,\ heat}Cyclohexene+H_2O} \] \(\textbf{1-Methylcyclohexan-1-ol dehydration:}\) The hydroxyl-bearing carbon in 1-methylcyclohexan-1-ol is attached to two ring carbon atoms and one methyl carbon. It is therefore tertiary. Protonation and loss of water can generate a tertiary carbocation at carbon 1. Beta-hydrogen removal from an adjacent ring carbon creates a double bond within the ring, giving 1-methylcyclohexene. Alternatively, removal of a hydrogen from the methyl substituent can form an exocyclic double bond, producing methylenecyclohexane. \(\textbf{Compare alkene substitution:}\) The double bond of 1-methylcyclohexene is trisubstituted because its two alkene carbons collectively carry three carbon substituents. The exocyclic double bond of methylenecyclohexane is disubstituted because its terminal methylene carbon carries no carbon substituent. The more substituted endocyclic alkene is generally favoured under the usual elimination conditions. \[ \mathrm{1\!-\!Methylcyclohexan\!-\!1\!-\!ol \xrightarrow{H^+,\ heat} 1\!-\!Methylcyclohexene+H_2O} \] Thus, cyclohexanol produces cyclohexene, whereas 1-methylcyclohexan-1-ol principally produces 1-methylcyclohexene. The comparison requires consideration of both alcohol class and the stability of the possible alkene products. The methyl substituent introduces an additional beta-hydrogen source and therefore creates a regioselectivity question absent from cyclohexanol itself.

Explanation: Statement I is correct because the characteristic ease of acid-catalysed dehydration follows tertiary alcohols greater than secondary alcohols greater than primary alcohols under comparable conditions. Compound R is tertiary, Q is secondary and P is primary. Their relative tendency towards dehydration therefore follows the stated order. Statement II is also correct because carbocation stability is important in the usual acid-catalysed dehydration pathways described in NCERT. Tertiary carbocations are comparatively more stable than secondary carbocations, which are generally more stable than corresponding primary carbocations. This stabilisation facilitates the formation of the reactive intermediate involved in many acid-catalysed eliminations. Statement III is correct because the hydroxyl group itself is a poor leaving group. Protonation converts it into an oxonium group from which neutral water can depart much more readily. Formation of water as the leaving group is therefore an important function of the acid catalyst. Statement IV is incorrect because the final elimination step involves removal of a beta-proton, not elimination of a hydroxide ion from the alcohol. The proton initially supplied by the acid is regenerated during the reaction sequence. Consequently, the acid functions as a catalyst rather than being consumed stoichiometrically in the ideal reaction cycle. The comparison also explains why merely increasing the total number of carbon atoms is not sufficient to predict dehydration ease. The substitution pattern of the hydroxyl-bearing carbon and the stability of intermediates must be considered. Thus, Statements I, II and III are correct, while Statement IV contains both an incorrect description of the leaving species and an incorrect conclusion about catalyst consumption.

Explanation: \(\textbf{Initial activation of ethanol:}\) The first stage of the NCERT dehydration mechanism involves protonation of the alcohol oxygen by the acid. This converts the hydroxyl-containing compound into a protonated alcohol. \[ \mathrm{CH_3CH_2OH+H^+\rightleftharpoons CH_3CH_2OH_2^+} \] Protonation is important because the original hydroxyl group is a poor leaving group, whereas water is a considerably better leaving group. \(\textbf{Formation of the reactive intermediate:}\) In the textbook mechanism, the protonated alcohol loses water to produce the corresponding positively charged carbon intermediate. \[ \mathrm{CH_3CH_2OH_2^+\longrightarrow CH_3CH_2^++H_2O} \] NCERT identifies this carbocation-forming stage as the slowest and rate-determining step in the represented mechanism. The energy required for the departure of water and formation of the high-energy intermediate is therefore crucial to the reaction rate. \(\textbf{Formation of the alkene:}\) Loss of a proton from a beta-carbon creates the carbon–carbon double bond. \[ \mathrm{CH_3CH_2^+\longrightarrow CH_2{=}CH_2+H^+} \] The proton is regenerated, completing the acid-catalysed reaction cycle. The overall transformation is dehydration of ethanol to ethene and water. \[ \mathrm{CH_3CH_2OH\xrightarrow[443\,K]{conc.\ H_2SO_4}CH_2{=}CH_2+H_2O} \] \(\textbf{Effect of removing the alkene:}\) Because dehydration is associated with an equilibrium, continuous removal of ethene decreases its concentration in the reaction system. According to Le Chatelier's principle, this favours further formation of the alkene. It does not prevent regeneration of the acid catalyst because proton release is part of the elimination stage. The stated mechanism does not require loss of hydroxide ions as the leaving group. The important distinction is between the kinetic role of the slowest elementary step and the equilibrium effect of continuously removing a product. Both ideas are needed to explain why the reaction is conducted under conditions favouring efficient alkene formation.

Explanation: \(\textbf{Interpret the Lucas observation:}\) The alcohol produces turbidity after several minutes rather than immediately. Under ordinary Lucas-test conditions, this behaviour is characteristic of a secondary alcohol. Tertiary alcohols generally produce immediate turbidity, while ordinary primary alcohols do not produce turbidity at room temperature within the usual observation period. Therefore, the first observation eliminates 2-methylbutan-2-ol and pentan-1-ol from consideration. \(\textbf{Interpret controlled oxidation:}\) The second observation establishes that oxidation produces a ketone. Secondary alcohols are oxidised to ketones because their hydroxyl-bearing carbon possesses one hydrogen and is attached to two other carbon groups. Both pentan-2-ol and pentan-3-ol satisfy this requirement. For pentan-2-ol: \[ \mathrm{CH_3CH(OH)CH_2CH_2CH_3+[O]\longrightarrow CH_3COCH_2CH_2CH_3+H_2O} \] The product is pentan-2-one. For pentan-3-ol: \[ \mathrm{CH_3CH_2CH(OH)CH_2CH_3+[O]\longrightarrow CH_3CH_2COCH_2CH_3+H_2O} \] The product is pentan-3-one. \(\textbf{Use the iodoform result to distinguish the ketones:}\) The iodoform reaction identifies compounds containing the methyl ketone unit \(\mathrm{CH_3CO-}\), among other eligible structures. Pentan-2-one contains this group because its carbonyl carbon is directly attached to a methyl group. It therefore gives a positive iodoform reaction. \[ \mathrm{CH_3COCH_2CH_2CH_3+3I_2+4NaOH} \] \[ \mathrm{\longrightarrow CHI_3+CH_3CH_2CH_2COONa+3NaI+3H_2O} \] Pentan-3-one contains ethyl groups on both sides of its carbonyl carbon and lacks the required methyl ketone structure. It does not give the characteristic iodoform reaction. The third observation therefore eliminates pentan-3-ol and identifies X as pentan-2-ol. This conclusion cannot be reached from the Lucas test alone because both remaining candidates are secondary alcohols. It also cannot be reached from the oxidation observation alone because both produce ketones. The three observations must be integrated in order to identify the correct positional isomer.

Explanation: \(\textbf{Identify the reaction pathways:}\) Primary and secondary alcohols undergo dehydrogenation over heated copper at \(573\,\mathrm{K}\), whereas tertiary alcohols undergo dehydration under the characteristic NCERT conditions. The gaseous products therefore depend on the structural class of each alcohol. \(\textbf{Sample I — Primary alcohol dehydrogenation:}\) Butan-1-ol is a primary alcohol and forms butanal with liberation of hydrogen. \[ \mathrm{CH_3CH_2CH_2CH_2OH\xrightarrow{Cu,\ 573\,K}CH_3CH_2CH_2CHO+H_2} \] One mole of butan-1-ol produces one mole of hydrogen. \[ n(\mathrm{H_2})_{\mathrm{I}}=0.24\times1 \] \[ n(\mathrm{H_2})_{\mathrm{I}}=0.24\,\mathrm{mol} \] \(\textbf{Sample II — Secondary alcohol dehydrogenation:}\) Butan-2-ol is secondary and forms butan-2-one with liberation of hydrogen. \[ \mathrm{CH_3CH(OH)CH_2CH_3\xrightarrow{Cu,\ 573\,K}CH_3COCH_2CH_3+H_2} \] The alcohol-to-hydrogen mole ratio is again \(1:1\). \[ n(\mathrm{H_2})_{\mathrm{II}}=0.18\times1 \] \[ n(\mathrm{H_2})_{\mathrm{II}}=0.18\,\mathrm{mol} \] \(\textbf{Total hydrogen evolved:}\) \[ n(\mathrm{H_2})_{\mathrm{total}}=0.24+0.18 \] \[ n(\mathrm{H_2})_{\mathrm{total}}=0.42\,\mathrm{mol} \] \(\textbf{Sample III — Tertiary alcohol dehydration:}\) 2-Methylpropan-2-ol is a tertiary alcohol. Its hydroxyl-bearing carbon has no directly attached hydrogen atom and therefore cannot undergo ordinary dehydrogenation to a carbonyl group. Instead, it loses water and produces 2-methylpropene. \[ \mathrm{(CH_3)_3COH\xrightarrow{Cu,\ 573\,K}(CH_3)_2C{=}CH_2+H_2O} \] One mole of tertiary alcohol produces one mole of the alkene. \[ n(\mathrm{C_4H_8})=0.12\times1 \] \[ n(\mathrm{C_4H_8})=0.12\,\mathrm{mol} \] Water formed during dehydration is removed by condensation, so it does not contribute to the collected gaseous mixture. \(\textbf{Calculate the requested mole ratio:}\) \[ \frac{n(\mathrm{H_2})}{n(\mathrm{C_4H_8})} =\frac{0.42}{0.12} \] \[ =\frac{42}{12} \] \[ =\frac{7}{2} \] \(\textbf{Chemical consistency check:}\) The first two samples produce hydrogen because their transformations involve dehydrogenation. The tertiary alcohol produces an alkene because its characteristic pathway is dehydration. The hydrocarbon gas amount must therefore be calculated only from Sample III, while hydrogen must be calculated from Samples I and II together. Treating all three alcohols as undergoing dehydrogenation would give an incorrect ratio. The carbon skeletons of the respective organic products are retained throughout the stated transformations. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{H_2}):n(\mathrm{C_4H_8})=7:2} \]

Explanation: Statement I is correct because compound P contains a hydroxyl group directly bonded to an \(sp^2\)-hybridised carbon of a carbon–carbon double bond. Such a structural arrangement is characteristic of an enol. P is therefore not an ordinary saturated primary alcohol merely because its formula contains an oxygen–hydrogen bond. Statement II is correct because Q contains the group \(\mathrm{-CH_2OH}\) attached to a carbon adjacent to a double bond. Its hydroxyl-bearing carbon is saturated and bonded directly to only one other carbon atom, making Q a primary alcohol. Since that carbon occupies an allylic position, Q is specifically a primary allylic alcohol. Statement III is correct because P can undergo keto–enol tautomerism, involving proton transfer and rearrangement of the double bond, to form ethanal. \[ \mathrm{CH_2=CHOH\rightleftharpoons CH_3CHO} \] Q has a different arrangement: its hydroxyl group is not bonded directly to either carbon of the double bond. Consequently, Q is not the enol tautomer of propanal. Statement IV is incorrect because alcohol classification depends on the number of carbon groups attached directly to the hydroxyl-bearing carbon, not the total number of carbon atoms in the molecule. Ethanol is a primary alcohol because its \(\mathrm{-CH_2OH}\) carbon is bonded to only one other carbon atom. The comparison illustrates why an oxygen-containing compound must first be classified from its precise connectivity before applying the usual rules for alcohol reactions.

Explanation: \(\textbf{Identify the alcohol classes:}\) Compound P is 2,2-dimethylpropan-1-ol, commonly called neopentyl alcohol. Its hydroxyl-bearing carbon is a terminal \(\mathrm{-CH_2OH}\) group attached to one other carbon atom. Therefore, P is a primary alcohol despite its highly branched carbon skeleton. Compound Q is 3-methylbutan-1-ol and also contains a terminal primary alcohol group. \(\textbf{Examine direct beta-elimination:}\) Ordinary dehydration to an alkene requires removal of water-related functionality from the hydroxyl-bearing carbon and a hydrogen from a neighbouring beta-carbon. In P, the carbon adjacent to the hydroxyl-bearing carbon is bonded to four carbon atoms. It has no hydrogen atom available for direct beta-elimination. Consequently, the unchanged carbon skeleton of P cannot produce an alkene through the ordinary direct elimination pathway. This does not establish that every possible acid-catalysed reaction of P is impossible, because rearrangement may require separate consideration under suitable conditions. In Q, the carbon adjacent to the hydroxyl-bearing carbon is a methylene group and possesses beta-hydrogens. Direct dehydration can therefore produce 3-methylbut-1-ene. \[ \mathrm{(CH_3)_2CHCH_2CH_2OH \longrightarrow (CH_3)_2CHCH=CH_2+H_2O} \] \(\textbf{Compare oxidation behaviour:}\) Both compounds contain a hydrogen atom on their hydroxyl-bearing carbon and are primary alcohols. Suitable controlled oxidation can therefore convert their \(\mathrm{-CH_2OH}\) groups into aldehyde groups. \[ \mathrm{(CH_3)_3CCH_2OH+[O]\longrightarrow (CH_3)_3CCHO+H_2O} \] \[ \mathrm{(CH_3)_2CHCH_2CH_2OH+[O]\longrightarrow (CH_3)_2CHCH_2CHO+H_2O} \] Thus, alcohol classification alone predicts similar oxidation behaviour but does not guarantee identical direct dehydration behaviour. The decisive additional structural check is the availability of a beta-hydrogen on the carbon adjacent to the hydroxyl-bearing carbon.

Explanation: Benzyl alcohol has the structure \(\mathrm{C_6H_5CH_2OH}\), and its hydroxyl-bearing carbon is attached directly to only one other carbon atom. It is therefore a primary alcohol, not a tertiary alcohol. Butan-1-ol is also primary because its terminal hydroxyl-bearing carbon has one carbon neighbour. However, the two compounds differ in the electronic environment surrounding the carbon–oxygen bond. In benzyl alcohol, the hydroxyl-bearing carbon occupies a benzylic position adjacent to an aromatic ring. Under suitable strongly acidic substitution conditions, protonation of the hydroxyl group can facilitate departure of water. The resulting benzylic carbocation is stabilised through delocalisation of positive charge into the aromatic system. \[ \mathrm{C_6H_5CH_2OH+H^+ \rightleftharpoons C_6H_5CH_2OH_2^+} \] \[ \mathrm{C_6H_5CH_2OH_2^+ \longrightarrow C_6H_5CH_2^++H_2O} \] This resonance stabilisation is not available to the ordinary primary carbocation that would result from direct ionisation of butan-1-ol. Consequently, the two primary alcohols need not exhibit identical substitution rates. The usual Lucas-test trend for ordinary aliphatic alcohols must therefore be applied with awareness of special benzylic and allylic stabilisation. The aromatic ring in benzyl alcohol remains attached to the side-chain carbon; substitution does not require conversion of the compound into phenol. The important distinction is between structural classification as primary and the additional electronic effects controlling substitution reactivity.

Explanation: Both reactions involve replacement of the hydroxyl group of an alcohol with a halogen-containing functionality. However, the identity of the halogen introduced depends on the reagent used. Thionyl chloride converts suitable alcohols into alkyl chlorides. With propan-1-ol, the organic product is 1-chloropropane. \[ \mathrm{CH_3CH_2CH_2OH+SOCl_2 \longrightarrow CH_3CH_2CH_2Cl+SO_2+HCl} \] Sulphur dioxide and hydrogen chloride are the characteristic byproducts of this conversion. In the presence of pyridine, hydrogen chloride may be captured as a pyridinium salt rather than escaping entirely as a gas. The formation of gaseous byproducts under suitable conditions helps in obtaining the alkyl chloride relatively free from certain inorganic contaminants. Phosphorus tribromide, in contrast, is employed to convert alcohols into alkyl bromides. \[ \mathrm{3CH_3CH_2CH_2OH+PBr_3 \longrightarrow 3CH_3CH_2CH_2Br+H_3PO_3} \] The principal organic product is therefore 1-bromopropane. Both reactions preserve the original three-carbon skeleton and replace the alcohol functional group without requiring carbon–carbon bond formation or cleavage. Formation of propene would involve elimination of water-related functionality and a beta-hydrogen, which is not the characteristic purpose of the specified reagents. The comparison demonstrates why identifying a reagent by its actual chemical role is more reliable than assuming that all reagents reacting with alcohols promote the same transformation.

Explanation: \(\textbf{Classify cyclohexanol:}\) In cyclohexanol, the carbon bearing the hydroxyl group is part of the six-membered ring. That carbon is directly attached to two neighbouring ring carbon atoms. Therefore, cyclohexanol is a secondary alcohol despite having only one hydroxyl group. PCC oxidises this secondary alcohol to the corresponding ketone. \[ \mathrm{Cyclohexanol+[O]\xrightarrow{PCC}Cyclohexanone+H_2O} \] Cyclohexanone is an ordinary ketone and does not give the characteristic silver-mirror reaction with Tollens' reagent. \(\textbf{Classify cyclohexylmethanol:}\) Cyclohexylmethanol contains a \(\mathrm{-CH_2OH}\) group attached to a cyclohexyl group. The hydroxyl-bearing carbon lies outside the ring and is bonded directly to only one carbon atom. Consequently, cyclohexylmethanol is a primary alcohol. Controlled oxidation converts its hydroxymethyl group into an aldehyde group. \[ \mathrm{C_6H_{11}CH_2OH+[O]\xrightarrow{PCC}C_6H_{11}CHO+H_2O} \] The product is cyclohexanecarbaldehyde. Because it contains an aldehyde group, it gives a positive Tollens' test through reduction of silver(I) to metallic silver. The cyclic carbon skeleton is retained during both oxidations. Neither reaction requires a carbon–carbon bond to be broken. The distinction arises because the hydroxyl-bearing carbon is part of the ring in one molecule and belongs to an external hydroxymethyl group in the other. This example illustrates why the classification of cyclic alcohols cannot be inferred solely from the presence of a ring.

Explanation: \(\textbf{Oxidation of the secondary alcohol:}\) Butan-2-ol is a secondary alcohol, so suitable oxidation produces butan-2-one. \[ \mathrm{CH_3CH(OH)CH_2CH_3 \longrightarrow CH_3COCH_2CH_3+2H^++2e^-} \] The displayed half-reaction shows that converting one alcohol molecule into the ketone involves loss of two electrons. \(\textbf{Reduction of dichromate in acidic medium:}\) \[ \mathrm{Cr_2O_7^{2-}+14H^++6e^- \longrightarrow 2Cr^{3+}+7H_2O} \] Each dichromate ion accepts six electrons. Therefore, one mole of dichromate can oxidise three moles of a secondary alcohol to the corresponding ketone. \(\textbf{Determine the oxidation capacity:}\) \[ n(\text{alcohol oxidisable})=3n(\mathrm{Cr_2O_7^{2-}}) \] \[ =3(0.080) \] \[ =0.240\,\mathrm{mol} \] The original alcohol quantity is \(0.30\,\mathrm{mol}\), which exceeds the oxidant capacity. Therefore, dichromate is the limiting reactant. \(\textbf{Amount of ketone produced:}\) \[ n(\text{butan-2-one})=0.240\,\mathrm{mol} \] \(\textbf{Alcohol remaining after oxidation:}\) \[ n(\text{unreacted alcohol})=0.300-0.240 \] \[ =0.060\,\mathrm{mol} \] \(\textbf{Derivative-formation stage:}\) A ketone reacts with 2,4-dinitrophenylhydrazine to produce the corresponding hydrazone derivative through addition followed by elimination of water. The stated stoichiometry is \(1:1\). \[ n(\text{ketone available})=0.240\,\mathrm{mol} \] \[ n(\text{2,4-DNPH available})=0.200\,\mathrm{mol} \] Therefore, the hydrazine reagent limits derivative formation. \[ n(\text{hydrazone derivative})=0.200\,\mathrm{mol} \] The derivative-forming reaction consumes ketone but does not convert the unreacted butan-2-ol into a carbonyl compound. Consequently, the remaining alcohol amount stays \(0.060\,\mathrm{mol}\). \(\textbf{Stoichiometric check:}\) The two reaction stages have different limiting reagents. Dichromate limits oxidation because its six-electron capacity is insufficient to oxidise all available alcohol. The hydrazine reagent limits the subsequent condensation because less than \(0.240\,\mathrm{mol}\) is supplied. Treating dichromate and alcohol as reacting in a \(1:1\) mole ratio would give an incorrect oxidation extent. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{derivative}},n_{\mathrm{alcohol\ remaining}}\right) =(0.20,\ 0.06)\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify both alcohol classes:}\) Butane-1,3-diol contains two hydroxyl groups attached to different carbon atoms. The hydroxyl group at carbon 1 is primary because its carbon is attached to only one other carbon. The hydroxyl group at carbon 3 is secondary because its carbon is attached to two other carbon atoms. \(\textbf{Predict oxidation of both groups:}\) PCC converts the primary hydroxyl group into an aldehyde and the secondary hydroxyl group into a ketone under the stated conditions. \[ \mathrm{HOCH_2CH_2CH(OH)CH_3+2[O] \xrightarrow{PCC} OHCCH_2COCH_3+2H_2O} \] The product is 3-oxobutanal. It contains both an aldehyde group and a methyl ketone group within the same molecule. \(\textbf{Tollens' test:}\) The aldehyde functionality is oxidised by Tollens' reagent, causing silver(I) ions to be reduced to metallic silver. Therefore, the compound gives a positive Tollens' test. \(\textbf{Iodoform test:}\) The ketone portion has the structure \(\mathrm{CH_3CO-}\), which satisfies the methyl ketone requirement for the iodoform reaction. Treatment with iodine and aqueous alkali therefore produces yellow iodoform. \(\textbf{2,4-Dinitrophenylhydrazine reaction:}\) Both the aldehyde and ketone functionalities contain carbonyl groups capable of reacting with 2,4-dinitrophenylhydrazine. Given sufficient reagent and complete conversion, each carbonyl group can form a hydrazone linkage. \[ \mathrm{R_2C=O+H_2NNHC_6H_3(NO_2)_2 \longrightarrow R_2C=NNHC_6H_3(NO_2)_2+H_2O} \] Accordingly, one molecule of the dicarbonyl compound can incorporate two hydrazone-forming groups. The three tests do not contradict one another because each detects a different structural feature. The aldehyde explains the Tollens' result, the methyl ketone explains the iodoform result, and both carbonyl groups contribute to hydrazone formation. This illustrates how a polyfunctional product can display the characteristic reactions of more than one carbonyl class simultaneously.

Explanation: \(\textbf{Sequence I — Reaction with an alcohol:}\) Butan-2-ol contains an oxygen–hydrogen bond and can donate its hydroxyl proton to a strongly basic Grignard reagent. Methylmagnesium bromide contains a polarised carbon–magnesium bond, and its methyl group behaves as a strong base towards protic substances. Direct treatment therefore produces methane through proton transfer rather than carbon–carbon bond formation. \[ \mathrm{CH_3CH(OH)CH_2CH_3+CH_3MgBr \longrightarrow CH_3CH(OMgBr)CH_2CH_3+CH_4} \] The reaction destroys the methylmagnesium bromide as a nucleophilic carbon reagent. Consequently, the tertiary alcohol expected from addition to a carbonyl compound is not formed in Sequence I. \(\textbf{Sequence II — Oxidation before organomagnesium addition:}\) Butan-2-ol is a secondary alcohol and is oxidised to butan-2-one under suitable controlled conditions. \[ \mathrm{CH_3CH(OH)CH_2CH_3+[O] \longrightarrow CH_3COCH_2CH_3+H_2O} \] Butan-2-one has an electrophilic carbonyl carbon capable of undergoing nucleophilic addition. The methyl group of methylmagnesium bromide attacks this carbon and forms a new carbon–carbon bond. \[ \mathrm{CH_3COCH_2CH_3+CH_3MgBr \longrightarrow (CH_3)_2C(OMgBr)CH_2CH_3} \] Aqueous work-up protonates the alkoxide intermediate. \[ \mathrm{(CH_3)_2C(OMgBr)CH_2CH_3 \xrightarrow{H_3O^+} (CH_3)_2C(OH)CH_2CH_3} \] The final product is 2-methylbutan-2-ol, a tertiary alcohol containing five carbon atoms. The two sequences therefore differ because the original alcohol possesses a proton that quenches the Grignard reagent, while the ketone produced after oxidation provides a carbonyl carbon for nucleophilic addition. The order of reactions is essential: direct exposure of the Grignard reagent to an alcohol does not produce the same outcome as first converting that alcohol into a carbonyl compound.

Explanation: \(\textbf{Formation of the tertiary alkyl bromide:}\) 2-Methylbutan-2-ol is a tertiary alcohol because its hydroxyl-bearing carbon is attached directly to three carbon groups. Under suitable acidic substitution conditions, the hydroxyl group becomes protonated and leaves as water, allowing bromide substitution. \[ \mathrm{(CH_3)_2C(OH)CH_2CH_3+HBr \longrightarrow (CH_3)_2C(Br)CH_2CH_3+H_2O} \] The product is 2-bromo-2-methylbutane. \(\textbf{Portion I — Aqueous hydroxide:}\) Aqueous potassium hydroxide provides hydroxide ions in a medium that favours hydrolysis of the tertiary alkyl bromide under suitable conditions. The carbon–bromine bond is replaced by a carbon–oxygen bond, regenerating the tertiary alcohol. \[ \mathrm{(CH_3)_2C(Br)CH_2CH_3+KOH(aq) \longrightarrow (CH_3)_2C(OH)CH_2CH_3+KBr} \] Thus, the principal product of Portion I is 2-methylbutan-2-ol. \(\textbf{Portion II — Elimination conditions:}\) Hot alcoholic potassium hydroxide favours removal of a beta-hydrogen together with bromide. Two constitutional alkenes are possible because beta-hydrogens occur at different neighbouring carbon positions. Formation of the more substituted alkene is generally favoured under the stated conditions. \[ \mathrm{(CH_3)_2C(Br)CH_2CH_3 \xrightarrow{KOH(alc.),\ heat} (CH_3)_2C=CHCH_3+KBr+H_2O} \] The principal alkene is 2-methylbut-2-ene. The comparison demonstrates that the same alkyl bromide can undergo substitution or elimination depending on the reaction medium and conditions. The first pathway regenerates the hydroxyl functional group, whereas the second removes a beta-hydrogen and establishes a carbon–carbon double bond. The tertiary carbon skeleton does not need to undergo rearrangement to account for either stated principal product.

Explanation: \(\textbf{Stereochemical structure of the starting alcohol:}\) Butan-2-ol has the structure \(\mathrm{CH_3CH(OH)CH_2CH_3}\). Its hydroxyl-bearing carbon is attached to four different groups: hydrogen, hydroxyl, methyl and ethyl. Therefore, butan-2-ol possesses a stereogenic centre and can exist as optical isomers. \(\textbf{Portion I — Oxidation of the stereogenic carbon:}\) Controlled oxidation converts the secondary alcohol group into a ketone. \[ \mathrm{CH_3CH(OH)CH_2CH_3+[O] \longrightarrow CH_3COCH_2CH_3+H_2O} \] The product is butan-2-one. The original tetrahedral hydroxyl-bearing carbon becomes a trigonal-planar carbonyl carbon. Consequently, that atom is no longer a stereogenic centre. Butan-2-one is therefore optically inactive. \(\textbf{Portion II — Dehydration and alkene selection:}\) Acid-catalysed dehydration removes water-related functionality and a beta-hydrogen to form an alkene. Butan-2-ol can yield but-1-ene and but-2-ene, but the more substituted but-2-ene is generally favoured. \[ \mathrm{CH_3CH(OH)CH_2CH_3 \xrightarrow{H^+,\ heat} CH_3CH=CHCH_3+H_2O} \] But-2-ene can exist as cis and trans geometrical isomers. The trans isomer has its methyl groups on opposite sides of the double bond and generally experiences less steric repulsion than the cis isomer. It is therefore more stable and ordinarily favoured within the but-2-ene product fraction. Neither cis-but-2-ene nor trans-but-2-ene contains the original stereogenic hydroxyl-bearing carbon. The trans isomer is geometrically distinct from the cis isomer, but geometrical isomerism does not by itself imply optical activity. Thus, both pathways eliminate the original optical activity through different structural changes: carbonyl formation in the first case and carbon–carbon double-bond formation in the second.