301. A ketone is subjected to four different reaction mixtures.
| Mixture | Added reagents and conditions |
|---|
| P | HCN and a small suitable amount of cyanide-generating base |
| Q | HCN in a strongly acidic medium that greatly suppresses free \(\mathrm{CN^-}\) |
| R | Ethylene glycol under suitable dry acid-catalysed conditions, with water removal |
| S | Ethylene glycol and aqueous NaOH, without acid catalysis or water removal |
Assume the ketone is otherwise stable, and compare only the ordinary carbonyl transformations specified.
Which statement best predicts the favourable reaction pathways?
ⓐ. P favours cyanohydrin formation and R favours cyclic ketal formation; Q and S lack important conditions supporting those respective pathways.
ⓑ. Q favours cyanohydrin formation because excess acid produces more cyanide ions, while S favours ketal formation because hydroxide removes water.
ⓒ. P and S both favour cyclic ketal formation because both contain bases, while Q and R favour cyanohydrin formation because both contain acids.
ⓓ. All four mixtures favour the same tetrahedral addition product because every nucleophile attacks the carbonyl carbon through an identical complete mechanism.
Correct Answer: P favours cyanohydrin formation and R favours cyclic ketal formation; Q and S lack important conditions supporting those respective pathways.
Explanation: The carbonyl carbon of a ketone is electrophilic and can undergo nucleophilic attack.
However, the complete reaction pathway depends on the nucleophile and reaction medium.
\(\textbf{Mixture P — Cyanide-assisted addition:}\)
The effective nucleophile in ordinary cyanohydrin formation is cyanide ion.
A suitable base increases the availability of \(\mathrm{CN^-}\) in the HCN-containing system.
Cyanide attacks the carbonyl carbon, and subsequent protonation produces a cyanohydrin.
Therefore, P provides favourable conditions for this pathway.
\(\textbf{Mixture Q — Suppressed cyanide availability:}\)
In a strongly acidic medium, cyanide is extensively protonated.
The concentration of free cyanide nucleophile is consequently reduced.
Although acid can activate carbonyl compounds in other contexts, that fact does not guarantee efficient cyanohydrin formation when the effective nucleophile is unavailable.
Therefore, Q is less favourable for the specified cyanide-ion addition mechanism.
\(\textbf{Mixture R — Cyclic ketal formation:}\)
Ethylene glycol contains two hydroxyl groups capable of reacting with a ketone carbonyl under suitable acid catalysis.
Formation of the cyclic ketal involves addition, proton transfers and loss of water.
Removing water favours the ketal side of the reversible reaction.
Consequently, R provides the appropriate conditions for cyclic ketal formation.
\(\textbf{Mixture S — Missing ketal-forming conditions:}\)
An aqueous strongly alkaline environment without suitable acid catalysis does not provide the characteristic conditions used for ordinary acid-catalysed ketal formation.
The presence of glycol alone does not ensure complete cyclic ketal production.
Therefore, S cannot be treated as equivalent to R.
The comparison demonstrates that nucleophile availability, catalytic activation and equilibrium control are separate decisions.
Both cyanohydrin and ketal formation involve carbonyl electrophilicity, but their optimal conditions are not identical.
302. Benzaldehyde is subjected separately to three reactions under achiral conditions:
Reaction P: Addition of HCN to form the corresponding cyanohydrin.
Reaction Q: Reaction with ethylene glycol to form the corresponding five-membered cyclic acetal.
Reaction R: Reaction with hydroxylamine to form benzaldehyde oxime.
Assume complete conversion and ignore conformational isomerism.
Consider the following statements.
Statement I: P can be obtained as a racemic mixture because cyanohydrin formation creates a stereogenic carbon.
Statement II: Q has a stereogenic carbon at the former carbonyl centre because its two oxygen-containing ring paths are different.
Statement III: R can exhibit geometrical isomerism about its carbon–nitrogen double bond.
Statement IV: All three reactions retain the original benzaldehyde carbonyl oxygen in the isolated organic products.
Which combination contains only the correct statements?
ⓐ. I and II only
ⓑ. I and III only
ⓒ. II and III only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: \(\textbf{Reaction P — Cyanohydrin stereochemistry:}\)
Benzaldehyde contains a planar aldehyde carbonyl group.
Cyanide can attack from either face under achiral conditions.
The resulting cyanohydrin has the structure:
\[
\mathrm{C_6H_5CH(OH)CN}
\]
Its former carbonyl carbon is attached to hydrogen, phenyl, hydroxyl and cyanide groups.
These groups are all different, so the carbon is stereogenic.
Consequently, Statement I is correct.
\(\textbf{Reaction Q — Symmetry of the cyclic acetal:}\)
Reaction with ethylene glycol forms a five-membered cyclic acetal under appropriate acid-catalysed conditions.
The former carbonyl carbon is attached to hydrogen, phenyl and two ring oxygen atoms.
However, the two oxygen-containing ring paths are constitutionally equivalent because the cyclic acetal is formed from symmetrical ethylene glycol.
Therefore, the former carbonyl carbon does not possess four different substituent paths.
Statement II is incorrect.
\(\textbf{Reaction R — Oxime geometry:}\)
Hydroxylamine converts the aldehyde group into the linkage \(\mathrm{C=N-OH}\).
\[
\mathrm{C_6H_5CHO+NH_2OH
\longrightarrow C_6H_5CH=NOH+H_2O}
\]
The carbon of the carbon–nitrogen double bond bears hydrogen and phenyl groups.
These are different substituents, permitting geometrical isomerism.
Thus, Statement III is correct.
\(\textbf{Track the carbonyl oxygen in the three reactions:}\)
During cyanohydrin formation, the carbonyl oxygen remains attached to carbon and becomes the hydroxyl oxygen.
During acetal formation, the original carbonyl oxygen is eliminated through the overall condensation process.
During oxime formation, the original carbonyl oxygen is also lost as part of water.
Therefore, Statement IV is incorrect.
The comparison demonstrates that stereogenic-centre formation, geometrical isomerism and oxygen-atom retention depend on different features of the reaction products.
Only Statements I and III are correct.
303. Acetophenone is subjected to the following reaction sequence.
Stage I: Controlled aromatic mononitration under conditions favouring substitution directed by the acetyl group.
Stage II: Complete iodoform reaction of the isolated mononitrated ketone using excess iodine and aqueous sodium hydroxide.
Stage III: Acidification of the isolated organic carboxylate.
Assume the aromatic nitro group remains unchanged during Stages II and III.
Which compound is obtained after Stage III?
ⓐ. 2-Nitrobenzoic acid
ⓑ. 4-Nitrobenzoic acid
ⓒ. 3-Nitrobenzyl alcohol
ⓓ. 3-Nitrobenzoic acid
Correct Answer: 3-Nitrobenzoic acid
Explanation: \(\textbf{Identify the directing group before nitration:}\)
Acetophenone contains the acetyl substituent \(\mathrm{-COCH_3}\) directly attached to the aromatic ring.
The carbonyl group withdraws electron density from the aromatic system.
It is therefore deactivating and meta directing towards electrophilic aromatic substitution.
\(\textbf{Stage I — Aromatic nitration:}\)
Controlled nitration introduces the nitro group predominantly at a position meta to the acetyl substituent.
\[
\mathrm{C_6H_5COCH_3
\xrightarrow{controlled\ nitration}
m\!-\!NO_2C_6H_4COCH_3}
\]
The principal product is 3-nitroacetophenone.
The methyl ketone functionality remains present.
\(\textbf{Stage II — Recognise the haloform-reactive group:}\)
3-Nitroacetophenone contains the structural unit \(\mathrm{-COCH_3}\).
The methyl group attached directly to the ketone carbonyl undergoes exhaustive iodination.
Subsequent cleavage produces iodoform and an aromatic carboxylate.
\[
\mathrm{m\!-\!NO_2C_6H_4COCH_3+3I_2+4NaOH}
\]
\[
\mathrm{\longrightarrow m\!-\!NO_2C_6H_4COONa+CHI_3+3NaI+3H_2O}
\]
The original ketone carbonyl carbon becomes the carboxyl carbon.
The nitro group remains meta to this carbonyl-derived substituent.
\(\textbf{Stage III — Acidification:}\)
Acidification converts sodium 3-nitrobenzoate into 3-nitrobenzoic acid.
\[
\mathrm{m\!-\!NO_2C_6H_4COONa+H^+
\longrightarrow m\!-\!NO_2C_6H_4COOH+Na^+}
\]
The final product is therefore 3-nitrobenzoic acid.
The sequence preserves the aromatic substitution relationship while changing an acetyl group into a carboxyl group through haloform cleavage.
Predicting the product requires applying the directing effect before determining the carbon-skeleton change.
304. Two structurally related compounds undergo separate reactions with methanal under conditions suitable for forming aromatic substitution products.
P: Phenol, \(\mathrm{C_6H_5OH}\)
Q: Anisole, \(\mathrm{C_6H_5OCH_3}\)
Assume the reaction medium generates an appropriate methanal-derived electrophile and that the relevant process is electrophilic substitution at the aromatic ring rather than cleavage of either original carbon–oxygen bond.
Which statement correctly compares their ring reactivity and oxygen-containing groups?
ⓐ. Only phenol can undergo electrophilic substitution because anisole lacks an oxygen–hydrogen bond.
ⓑ. Both rings are activated by oxygen lone-pair donation, but P retains a phenolic OH while Q bears an ether linkage.
ⓒ. Both molecules must undergo ordinary aldol condensation because methanal is an aldehyde.
ⓓ. Anisole is necessarily meta directing because its hydroxyl hydrogen has been replaced by methyl.
Correct Answer: Both rings are activated by oxygen lone-pair donation, but P retains a phenolic OH while Q bears an ether linkage.
Explanation: Phenol contains a hydroxyl group directly attached to an aromatic carbon.
Anisole contains a methoxy group directly attached to an aromatic carbon.
Although their oxygen-containing functional groups differ, both oxygen atoms possess lone pairs capable of conjugation with the benzene ring.
This resonance donation increases electron density at suitable ortho and para positions.
\(\textbf{Interpret the phenol reaction:}\)
Phenol can participate in aromatic substitution with appropriately activated methanal-derived electrophilic species.
The new carbon–carbon bond forms at an activated ring carbon.
The oxygen–hydrogen bond need not be cleaved during the initial aromatic carbon–carbon bond-forming event.
Therefore, P retains a phenolic hydroxyl group in the specified substitution product.
\(\textbf{Interpret the anisole reaction:}\)
Anisole contains an aryl methyl ether linkage.
Methylation of the original phenolic oxygen removes the oxygen–hydrogen bond, but it does not remove the oxygen lone pairs.
Consequently, its methoxy group remains an ortho/para-directing aromatic substituent.
Under the conditions explicitly specified, Q retains its ether linkage during aromatic substitution.
\(\textbf{Distinguish the process from aldol addition:}\)
Ordinary aldol chemistry depends on an enolate-type nucleophile generated from a suitable carbonyl compound.
Neither the phenolic nor the anisole aromatic ring is functioning as an ordinary carbonyl enolate in this reaction.
The aromatic ring instead supplies the nucleophilic carbon centre.
Therefore, the presence of methanal does not automatically establish an aldol mechanism.
The comparison shows why a phenolic proton is important for acid–base behaviour but is not essential for oxygen-mediated activation of an aromatic ring.
The principal distinction between P and Q is their oxygen functional-group identity, not complete loss of aromatic electrophilic-substitution reactivity.
305. Two separate samples of 4-formylbenzoic acid, \(\mathrm{p\!-\!HOOC-C_6H_4-CHO}\), are reduced under different conditions.
Sample P: Sodium borohydride under conditions selectively reducing the aldehyde group while preserving the carboxylic acid group.
Sample Q: Excess lithium aluminium hydride in dry ether, followed by controlled aqueous work-up, under conditions reducing both functional groups completely.
The isolated products are treated separately with excess aqueous sodium hydrogencarbonate.
Which statement correctly predicts the organic products and hydrogencarbonate observations?
ⓐ. P gives benzene-1,4-dimethanol and produces carbon dioxide; Q gives 4-(hydroxymethyl)benzoic acid and produces no carbon dioxide.
ⓑ. Both P and Q give benzene-1,4-dimethanol, but only P produces carbon dioxide.
ⓒ. P gives 4-(hydroxymethyl)benzoic acid and produces carbon dioxide; Q gives benzene-1,4-dimethanol and does not produce the characteristic carboxylic-acid effervescence.
ⓓ. P gives 4-hydroxybenzaldehyde and produces no carbon dioxide; Q gives terephthalic acid and produces carbon dioxide.
Correct Answer: P gives 4-(hydroxymethyl)benzoic acid and produces carbon dioxide; Q gives benzene-1,4-dimethanol and does not produce the characteristic carboxylic-acid effervescence.
Explanation: \(\textbf{Identify both reducible functional groups:}\)
4-Formylbenzoic acid contains an aldehyde group and a carboxylic acid group at para positions on the benzene ring.
These groups are not equivalent in their susceptibility to ordinary hydride reducing agents.
The question specifies conditions that distinguish their reduction levels.
\(\textbf{Sample P — Selective aldehyde reduction:}\)
Sodium borohydride reduces the aldehyde group to a primary alcohol under the stated conditions.
The carboxylic acid functionality remains present.
\[
\mathrm{p\!-\!HOOCC_6H_4CHO
\xrightarrow{NaBH_4}
p\!-\!HOOCC_6H_4CH_2OH}
\]
The product is 4-(hydroxymethyl)benzoic acid.
It contains one primary alcohol group and one carboxylic acid group.
\(\textbf{Sample Q — Stronger hydride reduction:}\)
Excess lithium aluminium hydride can reduce both the aldehyde group and the carboxylic acid group.
After aqueous work-up, each becomes a primary alcohol group.
\[
\mathrm{p\!-\!HOOCC_6H_4CHO
\xrightarrow[\text{work-up}]{excess\ LiAlH_4}
p\!-\!HOCH_2C_6H_4CH_2OH}
\]
The product is benzene-1,4-dimethanol.
It contains two primary alcohol groups but no remaining carboxylic acid functionality.
\(\textbf{Hydrogencarbonate test of P:}\)
The carboxylic acid group of P is sufficiently acidic to react with sodium hydrogencarbonate.
The reaction liberates carbon dioxide.
The ordinary alcohol hydroxyl group does not account for this characteristic effervescence.
\(\textbf{Hydrogencarbonate test of Q:}\)
Q contains only ordinary alcohol hydroxyl groups.
Therefore, it does not produce the characteristic carboxylic-acid carbon dioxide evolution with aqueous sodium hydrogencarbonate.
The two reactions preserve the aromatic carbon skeleton but differ in how many oxygen-containing groups are converted into alcohols.
This comparison connects reagent strength, functional-group selectivity and a subsequent diagnostic acid–base reaction.
306. A bifunctional compound P has the structure:
\[
\mathrm{OHCCH_2COCH_3}
\]
It contains an aldehyde group and a methyl ketone group.
P is treated with one equivalent of HCN under conditions specified to favour complete selective addition at the aldehyde carbonyl only.
The cyanohydrin Q is isolated and its nitrile group is hydrolysed completely to a carboxylic acid under conditions that preserve the ketone group.
Which combination correctly identifies the final structure and its characteristic tests?
ⓐ. \(\mathrm{HOOCCH(OH)CH_2COCH_3}\); positive iodoform test and carbon dioxide evolution with aqueous sodium hydrogencarbonate.
ⓑ. \(\mathrm{HOOCCH(OH)CH_2COCH_3}\); positive Tollens' test and negative iodoform test.
ⓒ. \(\mathrm{HOOCCH_2CH(OH)COCH_3}\); positive iodoform test and no carbon dioxide evolution with aqueous sodium hydrogencarbonate.
ⓓ. \(\mathrm{HOCH_2CH(OH)CH_2COCH_3}\); positive Tollens' test and carbon dioxide evolution with aqueous sodium hydrogencarbonate.
Correct Answer: \(\mathrm{HOOCCH(OH)CH_2COCH_3}\); positive iodoform test and carbon dioxide evolution with aqueous sodium hydrogencarbonate.
Explanation: \(\textbf{Assign the two carbonyl environments:}\)
P contains a terminal aldehyde and an internal methyl ketone.
The question specifies selective cyanide addition at the aldehyde carbonyl, so the ketone carbonyl remains unchanged during the first stage.
\(\textbf{Cyanohydrin formation:}\)
Cyanide attacks the aldehyde carbonyl carbon.
The carbonyl oxygen becomes a hydroxyl group after protonation.
\[
\mathrm{OHCCH_2COCH_3+HCN}
\]
\[
\mathrm{\longrightarrow NCCH(OH)CH_2COCH_3}
\]
The cyanide reagent contributes one additional carbon atom.
The product still contains the original methyl ketone group.
\(\textbf{Hydrolyse the nitrile group:}\)
Complete acidic hydrolysis converts the nitrile carbon into a carboxylic acid carbon.
The hydroxyl group on the neighbouring carbon remains intact.
\[
\mathrm{NCCH(OH)CH_2COCH_3}
\]
\[
\mathrm{\xrightarrow{H_3O^+,\ heat}
HOOCCH(OH)CH_2COCH_3}
\]
The product contains five carbon atoms.
It possesses a carboxylic acid group, an ordinary alcohol hydroxyl group and a methyl ketone group.
\(\textbf{Predict the iodoform test:}\)
The ketone functionality retains the structure \(\mathrm{-COCH_3}\).
Therefore, the final compound remains eligible for the characteristic iodoform reaction.
A positive test produces yellow iodoform.
\(\textbf{Predict the hydrogencarbonate test:}\)
The newly formed carboxylic acid group reacts with aqueous sodium hydrogencarbonate and releases carbon dioxide.
The ordinary alcohol hydroxyl group does not produce the corresponding effervescence.
\(\textbf{Interpret the Tollens' requirement:}\)
The original aldehyde group has been converted into a cyanohydrin and subsequently into a hydroxycarboxylic acid fragment.
It is no longer present as a free aldehyde.
Therefore, the original ordinary aldehyde Tollens' response should not be assigned to the final product.
The problem requires tracing selective carbonyl addition, retention of a different carbonyl group and conversion of the cyanide-derived carbon into a carboxyl carbon.
307. 4-Methylbenzaldehyde undergoes controlled aromatic mononitration under conditions favouring the position jointly directed by its formyl and methyl substituents.
The isolated mononitrated aldehyde is subsequently treated with sodium borohydride under conditions preserving both the nitro group and the aromatic methyl substituent.
Number the starting benzaldehyde ring so that the aldehyde-bearing carbon is position 1 and the methyl-bearing carbon is position 4.
Which structure represents the principal final product?
ⓐ. 2-Nitro-4-methylbenzyl alcohol
ⓑ. 4-Nitro-3-methylbenzyl alcohol
ⓒ. 3-Nitro-4-methylbenzyl alcohol
ⓓ. 3-Nitro-4-methylbenzoic acid
Correct Answer: 3-Nitro-4-methylbenzyl alcohol
Explanation: \(\textbf{Locate the two original substituents:}\)
4-Methylbenzaldehyde contains a formyl group at carbon 1 and a methyl group at carbon 4.
The formyl group withdraws electron density and is meta directing.
The methyl group releases electron density and is ortho/para directing.
\(\textbf{Determine the positions favoured by formyl:}\)
With the formyl group at carbon 1, the meta positions are carbons 3 and 5.
Both positions are initially available for electrophilic substitution.
\(\textbf{Determine the positions favoured by methyl:}\)
With the methyl group at carbon 4, carbons 3 and 5 are its ortho positions.
The para position relative to methyl is carbon 1, which is already occupied by the formyl group.
Thus, the available positions favoured by methyl are also carbons 3 and 5.
\(\textbf{Combine the directing effects:}\)
Both substituents favour the same pair of available aromatic positions.
Carbons 3 and 5 are equivalent by symmetry in the starting para-disubstituted compound.
Therefore, mononitration at either position gives the same constitutional product.
\[
\mathrm{p\!-\!CH_3C_6H_4CHO
\xrightarrow{controlled\ nitration}
3\!-\!NO_2\!-\!4\!-\!CH_3C_6H_3CHO}
\]
\(\textbf{Reduce the aldehyde group selectively:}\)
Sodium borohydride converts the formyl group into a hydroxymethyl group under the conditions specified.
\[
\mathrm{3\!-\!NO_2\!-\!4\!-\!CH_3C_6H_3CHO}
\]
\[
\mathrm{\xrightarrow{NaBH_4}
3\!-\!NO_2\!-\!4\!-\!CH_3C_6H_3CH_2OH}
\]
The final product is 3-nitro-4-methylbenzyl alcohol.
The positions of nitro and methyl remain unchanged during carbonyl reduction.
The product is not a carboxylic acid because sodium borohydride reduces the aldehyde rather than oxidising it.
The key reasoning combines two reinforcing directing effects with a subsequent functional-group transformation.
308. A reaction mixture contains benzaldehyde and propanone in equimolar amounts.
A carefully controlled quantity of ethylmagnesium bromide is introduced in dry ether under conditions specified to give complete preferential 1,2-addition to benzaldehyde, while propanone remains unchanged.
After aqueous acidic work-up, the two organic compounds are separated.
The alcohol product is examined for optical isomerism, while the unchanged carbonyl compound is tested with iodine and aqueous sodium hydroxide.
Which statement correctly describes the outcomes?
ⓐ. The alcohol is 2-phenylpropan-2-ol and is achiral; the unchanged carbonyl compound gives a negative iodoform test.
ⓑ. The alcohol is 2-phenylpropan-1-ol and can form a racemic mixture; the unchanged benzaldehyde gives a positive iodoform test.
ⓒ. The alcohol is 1-phenylpropan-1-ol and is necessarily optically pure; the unchanged propanone gives a negative iodoform test.
ⓓ. The alcohol is 1-phenylpropan-1-ol and can form a racemic mixture; the unchanged propanone gives a positive iodoform test.
Correct Answer: The alcohol is 1-phenylpropan-1-ol and can form a racemic mixture; the unchanged propanone gives a positive iodoform test.
Explanation: \(\textbf{Identify the selectively reacting carbonyl compound:}\)
Benzaldehyde is an aldehyde containing a phenyl group and hydrogen attached to its carbonyl carbon.
Propanone is a ketone containing two methyl groups attached to its carbonyl carbon.
Aldehydes are generally less sterically hindered and more susceptible to ordinary nucleophilic addition than comparable ketones.
The question explicitly specifies conditions under which ethylmagnesium bromide reacts completely with benzaldehyde while leaving propanone unchanged.
\(\textbf{Grignard addition to benzaldehyde:}\)
The ethyl group supplied by the organomagnesium reagent attacks the aldehyde carbonyl carbon.
The original phenyl group and hydrogen remain attached to that carbon.
Aqueous acidic work-up converts the resulting magnesium alkoxide into the secondary alcohol.
\[
\mathrm{C_6H_5CHO+C_2H_5MgBr}
\]
\[
\mathrm{\xrightarrow[\text{2. }H_3O^+]{\text{1. dry ether}}
C_6H_5CH(OH)CH_2CH_3}
\]
The product is 1-phenylpropan-1-ol.
\(\textbf{Evaluate stereogenicity:}\)
The hydroxyl-bearing carbon is bonded to hydrogen, hydroxyl, phenyl and ethyl groups.
These are four different substituents.
Consequently, this carbon is stereogenic.
Because the original carbonyl group is planar and the reaction conditions are achiral, addition can occur from either face.
The isolated alcohol can therefore be obtained as a racemic mixture.
\(\textbf{Identify the unchanged carbonyl compound:}\)
Propanone remains present according to the stipulated chemoselectivity.
Its structure is \(\mathrm{CH_3COCH_3}\), which contains the methyl ketone functionality required for a positive iodoform test.
Treatment with excess iodine and aqueous sodium hydroxide therefore produces yellow iodoform.
\(\textbf{Check the combined inference:}\)
The stereogenic alcohol originates from Grignard addition to benzaldehyde.
The positive iodoform observation originates from the separate, unreacted propanone.
The tests must not be assigned to the wrong product simply because both carbonyl compounds were initially present in the same mixture.
The problem therefore integrates comparative carbonyl reactivity, Grignard product connectivity, stereochemistry and structural identification of the residual ketone.
309. A polyhydric alcohol X has the structure:
\[
\mathrm{HOCH_2CH(OH)C(OH)(CH_3)_2}
\]
X is treated with sufficient PCC under anhydrous conditions to oxidise every primary and secondary alcohol group while leaving the tertiary alcohol group unchanged.
The resulting compound P is isolated and subjected separately to Tollens' reagent, 2,4-dinitrophenylhydrazine and the iodoform test.
Which statement correctly describes P and its expected behaviour?
ⓐ. P contains one aldehyde and one ketone group; it gives positive Tollens' and iodoform tests because it contains two methyl groups.
ⓑ. P contains two aldehyde groups; it gives a positive Tollens' test and consumes two equivalents of 2,4-dinitrophenylhydrazine.
ⓒ. P contains one aldehyde and one ketone group; it gives a positive Tollens' test, can consume two equivalents of 2,4-dinitrophenylhydrazine and gives a negative iodoform test.
ⓓ. P contains one ketone and two tertiary alcohol groups; it gives a negative Tollens' test but a positive iodoform test.
Correct Answer: P contains one aldehyde and one ketone group; it gives a positive Tollens' test, can consume two equivalents of 2,4-dinitrophenylhydrazine and gives a negative iodoform test.
Explanation: \(\textbf{Classify the three hydroxyl-bearing carbons:}\)
The terminal \(\mathrm{-CH_2OH}\) group is a primary alcohol because its hydroxyl-bearing carbon is attached to only one other carbon.
The neighbouring \(\mathrm{-CH(OH)-}\) group is a secondary alcohol because its hydroxyl-bearing carbon is attached to two carbon atoms.
The carbon bearing hydroxyl and two methyl groups is tertiary because it is attached to three carbon atoms.
These classifications depend on the carbon directly bonded to hydroxyl rather than on the overall branching of the molecule.
\(\textbf{Apply the specified oxidation conditions:}\)
PCC converts the primary alcohol group into an aldehyde and the secondary alcohol group into a ketone.
The tertiary alcohol group remains unchanged under the stated conditions.
\[
\mathrm{HOCH_2CH(OH)C(OH)(CH_3)_2}
\]
\[
\mathrm{\xrightarrow{PCC}
OHCCOC(OH)(CH_3)_2}
\]
The product can be written more explicitly as \(\mathrm{OHC-C(=O)-C(OH)(CH_3)_2}\).
\(\textbf{Interpret the Tollens' test:}\)
P contains an aldehyde group, which can undergo characteristic oxidation with Tollens' reagent.
Therefore, a positive silver-mirror test is expected.
\(\textbf{Interpret carbonyl derivative formation:}\)
P also contains two carbonyl groups: one aldehyde and one ketone.
Under complete derivative-forming conditions, each can react with one molecule of 2,4-dinitrophenylhydrazine.
Consequently, two equivalents of the reagent can be consumed per molecule of P.
\(\textbf{Check the iodoform requirement:}\)
Although P contains two methyl groups, neither is attached directly to its ketone carbonyl carbon.
Both methyl groups are attached to the tertiary alcohol carbon.
Therefore, the ketone does not possess the required \(\mathrm{CH_3CO-}\) arrangement.
The characteristic iodoform test is consequently negative.
This example shows why methyl-group presence alone is insufficient to establish methyl ketone character.
310. Butane-2,3-dione, \(\mathrm{CH_3COCOCH_3}\), is completely reduced with excess sodium borohydride under achiral conditions, followed by aqueous work-up.
The resulting butane-2,3-diol is considered in all its possible configurational forms.
A separate \(0.20\,\mathrm{mol}\) sample of the fully reduced diol is treated with excess sodium metal, assuming complete reaction of both hydroxyl groups.
Which option correctly gives the total number of distinct stereoisomers of the diol and the amount of hydrogen gas evolved?
ⓐ. Four stereoisomers and \(0.20\,\mathrm{mol}\) hydrogen
ⓑ. Three stereoisomers and \(0.20\,\mathrm{mol}\) hydrogen
ⓒ. Two stereoisomers and \(0.10\,\mathrm{mol}\) hydrogen
ⓓ. Three stereoisomers and \(0.40\,\mathrm{mol}\) hydrogen
Correct Answer: Three stereoisomers and \(0.20\,\mathrm{mol}\) hydrogen
Explanation: \(\textbf{Identify the carbonyl reduction product:}\)
Butane-2,3-dione contains two adjacent ketone carbonyl groups.
Each carbonyl group is reduced by sodium borohydride to a secondary alcohol group.
The carbon skeleton remains unchanged.
\[
\mathrm{CH_3COCOCH_3
\xrightarrow[\text{work-up}]{excess\ NaBH_4}
CH_3CH(OH)CH(OH)CH_3}
\]
The product is butane-2,3-diol.
\(\textbf{Analyse the stereogenic centres:}\)
Carbons 2 and 3 each become bonded to hydrogen, hydroxyl, methyl and the remaining substituted carbon chain.
Both are therefore stereogenic.
Two stereogenic centres would ordinarily suggest a maximum of four configurational arrangements.
However, the molecule has a symmetrical carbon skeleton.
The \((2R,3S)\) and \((2S,3R)\) descriptions refer to the same meso compound because of its internal symmetry.
The other configurations, \((2R,3R)\) and \((2S,3S)\), form an enantiomeric pair.
Therefore, only three distinct stereoisomers exist.
\(\textbf{Calculate hydrogen evolution:}\)
Each butane-2,3-diol molecule contains two oxygen–hydrogen bonds.
Both react with sodium metal under the stated complete-reaction conditions.
\[
\mathrm{HOCH(CH_3)CH(CH_3)OH+2Na}
\]
\[
\mathrm{\longrightarrow NaOCH(CH_3)CH(CH_3)ONa+H_2}
\]
One mole of diol therefore releases one mole of hydrogen gas.
For the specified sample:
\[
n(\mathrm{H_2})=0.20\times1
\]
\[
n(\mathrm{H_2})=0.20\,\mathrm{mol}
\]
The stereoisomer count depends on molecular symmetry, whereas hydrogen evolution depends on the number of hydroxyl groups.
All three stereoisomeric forms possess two hydroxyl groups and therefore have the same ideal sodium-reaction stoichiometry.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{stereoisomers}}=3,\qquad n(\mathrm{H_2})=0.20\,\mathrm{mol}}
\]
311. Two optically active alcohols are subjected separately to controlled oxidation with PCC under anhydrous conditions.
P: \((R)\)-2-Methylbutan-1-ol, \(\mathrm{HOCH_2CH(CH_3)CH_2CH_3}\)
Q: \((R)\)-Butan-2-ol, \(\mathrm{CH_3CH(OH)CH_2CH_3}\)
Assume neither sample undergoes racemisation or carbon-skeleton rearrangement during oxidation.
The isolated products are examined for the presence of their original stereogenic centres and are separately treated with Tollens' reagent.
Which statement is correct?
ⓐ. P loses its original stereogenic centre and gives a positive Tollens' test; Q retains its stereogenic centre and gives a negative Tollens' test.
ⓑ. Both products become achiral because oxidation of any alcohol destroys every stereogenic centre in the molecule.
ⓒ. Both products retain their original stereogenic centres, but only Q gives a positive Tollens' test.
ⓓ. P retains its stereogenic centre and gives a positive Tollens' test; Q loses its original stereogenic centre and gives a negative Tollens' test.
Correct Answer: P retains its stereogenic centre and gives a positive Tollens' test; Q loses its original stereogenic centre and gives a negative Tollens' test.
Explanation: \(\textbf{Identify the hydroxyl-bearing carbon in P:}\)
P is a primary alcohol because its hydroxyl group is attached to a terminal methylene carbon.
Its stereogenic centre is the neighbouring carbon, which is attached to hydrogen, methyl, ethyl and hydroxymethyl groups.
Therefore, the stereogenic centre is not the carbon undergoing oxidation.
\(\textbf{Oxidise P:}\)
PCC converts the primary alcohol group into an aldehyde.
\[
\mathrm{HOCH_2CH(CH_3)CH_2CH_3}
\]
\[
\mathrm{\xrightarrow{PCC}
OHCCH(CH_3)CH_2CH_3}
\]
The product is 2-methylbutanal.
Its carbon at position 2 remains tetrahedral and retains four different substituents.
Under the stated no-racemisation assumption, the original stereogenic centre is preserved.
The aldehyde group gives a positive Tollens' test.
\(\textbf{Identify the hydroxyl-bearing carbon in Q:}\)
Q is a secondary alcohol.
Its hydroxyl-bearing carbon is itself the stereogenic centre because it is attached to hydrogen, hydroxyl, methyl and ethyl groups.
\(\textbf{Oxidise Q:}\)
PCC converts the secondary alcohol into butan-2-one.
\[
\mathrm{CH_3CH(OH)CH_2CH_3
\xrightarrow{PCC}
CH_3COCH_2CH_3}
\]
The original stereogenic carbon becomes a trigonal-planar ketone carbonyl carbon.
Its tetrahedral stereogenicity is therefore destroyed.
Butan-2-one is an ordinary ketone and does not give the characteristic positive Tollens' reaction.
The comparison demonstrates that oxidation does not automatically remove every stereogenic centre.
The result depends on whether the stereogenic centre is the hydroxyl-bearing carbon or another carbon in the molecule.
312. An alcohol X has the structure \(\mathrm{p\!-\!CH_3C_6H_4CH(OH)CH_3}\).
It undergoes the following sequence:
Stage I: Controlled oxidation with PCC.
Stage II: Complete reaction of the isolated product with excess iodine and aqueous sodium hydroxide.
Stage III: Acidification of the organic carboxylate, followed by vigorous benzylic oxidation with hot alkaline potassium permanganate and final acidification.
Assume the aromatic ring remains intact and each specified transformation is complete.
Which pair correctly identifies the organic acids formed immediately after Stage II acidification and after Stage III?
ⓐ. 4-Methylbenzoic acid and benzene-1,4-dicarboxylic acid
ⓑ. Benzoic acid and benzene-1,4-dicarboxylic acid
ⓒ. 4-Methylbenzoic acid and benzene-1,3-dicarboxylic acid
ⓓ. 4-Ethylbenzoic acid and benzene-1,4-dicarboxylic acid
Correct Answer: 4-Methylbenzoic acid and benzene-1,4-dicarboxylic acid
Explanation: \(\textbf{Classify the starting alcohol:}\)
The hydroxyl-bearing carbon of X is attached to a phenyl group and a methyl group.
It also possesses one hydrogen atom.
Therefore, X is a secondary benzylic alcohol.
The aromatic ring contains a separate methyl substituent para to the alcohol-containing side chain.
\(\textbf{Stage I — Oxidation of the secondary alcohol:}\)
PCC converts the secondary alcohol into a ketone.
\[
\mathrm{p\!-\!CH_3C_6H_4CH(OH)CH_3}
\]
\[
\mathrm{\xrightarrow{PCC}
p\!-\!CH_3C_6H_4COCH_3}
\]
The product is 4-methylacetophenone.
Its ketone group contains the methyl ketone arrangement \(\mathrm{-COCH_3}\).
\(\textbf{Stage II — Haloform cleavage:}\)
Iodine and aqueous alkali convert the methyl ketone into iodoform and an aromatic carboxylate.
\[
\mathrm{p\!-\!CH_3C_6H_4COCH_3+3I_2+4NaOH}
\]
\[
\mathrm{\longrightarrow p\!-\!CH_3C_6H_4COONa+CHI_3+3NaI+3H_2O}
\]
Acidification gives 4-methylbenzoic acid.
The methyl group originally attached directly to the aromatic ring remains unchanged during this selected haloform reaction.
\(\textbf{Stage III — Oxidation of the remaining benzylic methyl group:}\)
4-Methylbenzoic acid contains a methyl substituent with benzylic hydrogen atoms.
Vigorous permanganate oxidation converts this methyl group into another carboxyl group.
\[
\mathrm{p\!-\!CH_3C_6H_4COOH}
\]
\[
\mathrm{\xrightarrow[\text{acidification}]{hot\ alkaline\ KMnO_4}
p\!-\!HOOCC_6H_4COOH}
\]
The final product is benzene-1,4-dicarboxylic acid, also called terephthalic acid.
The two carboxyl groups remain para to each other because none of the reactions changes their aromatic positions.
The sequence requires distinguishing oxidation of an alcohol, haloform cleavage of a methyl ketone and strong oxidation of a separate benzylic side chain.
313. Two alcohols are separately passed over heated copper at approximately \(573\,\mathrm{K}\).
P: 1-Phenylpropan-1-ol, \(\mathrm{C_6H_5CH(OH)CH_2CH_3}\)
Q: 2-Phenylpropan-2-ol, \(\mathrm{C_6H_5C(OH)(CH_3)_2}\)
Assume that P undergoes the characteristic secondary-alcohol dehydrogenation and Q undergoes the characteristic tertiary-alcohol dehydration.
The principal product from Q is subsequently subjected to reductive ozonolysis.
Which statement correctly identifies the products?
ⓐ. P gives acetophenone; Q gives styrene, which produces methanal and benzaldehyde on ozonolysis.
ⓑ. P gives propiophenone; Q gives alpha-methylstyrene, which produces benzaldehyde and ethanal on ozonolysis.
ⓒ. P gives propiophenone; Q gives alpha-methylstyrene, which produces acetophenone and methanal on ozonolysis.
ⓓ. P gives 1-phenylprop-1-ene; Q gives acetophenone, which produces benzaldehyde on ozonolysis.
Correct Answer: P gives propiophenone; Q gives alpha-methylstyrene, which produces acetophenone and methanal on ozonolysis.
Explanation: \(\textbf{Determine the alcohol classes:}\)
P contains a hydroxyl-bearing carbon attached to a phenyl group and an ethyl group.
It also possesses a hydrogen atom, making it a secondary alcohol.
Q contains a hydroxyl-bearing carbon attached to one phenyl group and two methyl groups.
It has no hydrogen attached directly to that carbon and is therefore tertiary.
\(\textbf{Reaction of P over heated copper:}\)
Suitable secondary alcohols undergo dehydrogenation to ketones at approximately \(573\,\mathrm{K}\).
\[
\mathrm{C_6H_5CH(OH)CH_2CH_3}
\]
\[
\mathrm{\xrightarrow{Cu,\ 573\,K}
C_6H_5COCH_2CH_3+H_2}
\]
The product is 1-phenylpropan-1-one, commonly called propiophenone.
\(\textbf{Reaction of Q over heated copper:}\)
The tertiary alcohol cannot undergo ordinary carbonyl-forming dehydrogenation because its hydroxyl-bearing carbon lacks hydrogen.
Under the stated conditions, it undergoes dehydration.
\[
\mathrm{C_6H_5C(OH)(CH_3)_2}
\]
\[
\mathrm{\xrightarrow{Cu,\ 573\,K}
C_6H_5C(CH_3)=CH_2+H_2O}
\]
The product is 2-phenylprop-1-ene, commonly called alpha-methylstyrene.
\(\textbf{Interpret reductive ozonolysis:}\)
Ozonolysis cleaves the carbon–carbon double bond of alpha-methylstyrene.
The alkene carbon attached to phenyl and methyl becomes a ketone carbonyl carbon.
The terminal methylene carbon becomes the carbonyl carbon of methanal.
\[
\mathrm{C_6H_5C(CH_3)=CH_2}
\]
\[
\mathrm{\xrightarrow[\text{reductive work-up}]{O_3}
C_6H_5COCH_3+HCHO}
\]
The products are acetophenone and methanal.
The two starting alcohols therefore undergo different transformations because of the number of carbon substituents on their hydroxyl-bearing carbons.
The later ozonolysis also distinguishes propiophenone from the ketone obtained by cleavage of the Q-derived alkene.
314. Two unsaturated alcohols are examined:
P: Prop-2-en-1-ol, \(\mathrm{CH_2=CHCH_2OH}\)
Q: But-3-en-1-ol, \(\mathrm{CH_2=CHCH_2CH_2OH}\)
Both are classified as primary alcohols. Under comparable conditions, their rates of conversion into organic chlorides by concentrated HCl and anhydrous zinc chloride are compared, without relying on turbidity as the sole measurement.
Separate samples are oxidised with PCC under conditions preserving their carbon–carbon double bonds.
Which statement is correct?
ⓐ. P is secondary because its hydroxyl group is close to an alkene; Q is primary, and only Q forms an aldehyde with PCC.
ⓑ. Both are primary; P has an allylic hydroxyl-bearing carbon and can show enhanced substitution reactivity, while PCC gives prop-2-enal from P and but-3-enal from Q.
ⓒ. Both are primary, but Q is allylic whereas P is not; PCC gives a ketone from P and an aldehyde from Q.
ⓓ. P and Q must have identical chloride-formation rates because primary classification completely determines substitution kinetics.
Correct Answer: Both are primary; P has an allylic hydroxyl-bearing carbon and can show enhanced substitution reactivity, while PCC gives prop-2-enal from P and but-3-enal from Q.
Explanation: \(\textbf{Classify the hydroxyl-bearing carbons:}\)
In P, the hydroxyl-bearing carbon is a terminal methylene carbon attached to one other carbon.
In Q, the hydroxyl-bearing carbon is also attached to one other carbon.
Therefore, both molecules are primary alcohols.
Their different relationships to the alkene linkage do not change this classification.
\(\textbf{Identify the allylic arrangement:}\)
In P, the hydroxyl-bearing carbon lies directly next to a carbon–carbon double bond.
It is therefore an allylic primary alcohol.
In Q, one additional saturated methylene group separates the hydroxyl-bearing carbon from the alkene.
Q is a homoallylic primary alcohol rather than an allylic one.
\(\textbf{Compare chloride-formation behaviour:}\)
The allylic arrangement in P can stabilise appropriate substitution intermediates or transition states through interaction with the adjacent pi system.
Consequently, P can undergo substitution more readily than an ordinary primary alcohol under suitable conditions.
Q does not possess the same direct allylic stabilisation at its hydroxyl-bearing carbon.
Thus, identical primary classification does not necessarily imply identical substitution rates.
\(\textbf{Oxidise P with PCC:}\)
\[
\mathrm{CH_2=CHCH_2OH
\xrightarrow{PCC}
CH_2=CHCHO}
\]
The product is prop-2-enal.
\(\textbf{Oxidise Q with PCC:}\)
\[
\mathrm{CH_2=CHCH_2CH_2OH
\xrightarrow{PCC}
CH_2=CHCH_2CHO}
\]
The product is but-3-enal.
Both products retain the alkene linkage because the oxidation conditions are explicitly selective for the primary alcohol group.
The comparison demonstrates that alcohol class, allylic stabilisation and oxidation product must be assessed as related but distinct structural properties.
315. 1-Methylcyclopentan-1-ol is heated with a suitable acid catalyst under conditions favouring dehydration to the more substituted alkene, without skeletal rearrangement.
The principal alkene P is isolated and subjected to reductive ozonolysis.
The resulting single open-chain organic compound Q is examined separately with Tollens' reagent and iodine in aqueous sodium hydroxide.
Which statement correctly identifies Q and its test behaviour?
ⓐ. Q is \(\mathrm{OHC(CH_2)_3CHO}\); it gives a positive Tollens' test but a negative iodoform test.
ⓑ. Q is \(\mathrm{CH_3CO(CH_2)_2CHO}\); it gives both tests positive.
ⓒ. Q is \(\mathrm{CH_3CO(CH_2)_3CHO}\); it gives a negative Tollens' test and a positive iodoform test.
ⓓ. Q is \(\mathrm{CH_3CO(CH_2)_3CHO}\); it gives positive Tollens' and iodoform tests.
Correct Answer: Q is \(\mathrm{CH_3CO(CH_2)_3CHO}\); it gives positive Tollens' and iodoform tests.
Explanation: \(\textbf{Classify the starting cyclic alcohol:}\)
The hydroxyl-bearing ring carbon is attached to two neighbouring ring carbons and one methyl group.
It is therefore a tertiary alcohol.
Under the specified acid-catalysed conditions, dehydration can form an alkene.
\(\textbf{Identify the more substituted elimination product:}\)
Elimination involving a hydrogen on an adjacent ring carbon produces 1-methylcyclopentene.
Elimination involving the methyl substituent could instead give methylenecyclopentane.
The endocyclic alkene is more highly substituted and is favoured under the specified conditions.
Thus, P is 1-methylcyclopentene.
\(\textbf{Apply reductive ozonolysis:}\)
Ozonolysis cleaves the carbon–carbon double bond and opens the ring at that bond.
The alkene carbon carrying the methyl substituent has no attached hydrogen.
It therefore forms a ketone carbonyl.
The other alkene carbon possesses hydrogen and therefore forms an aldehyde carbonyl.
The three remaining ring methylene groups connect the two new carbonyl groups.
\[
\mathrm{1\!-\!Methylcyclopentene}
\]
\[
\mathrm{\xrightarrow[\text{reductive work-up}]{O_3}
CH_3CO(CH_2)_3CHO}
\]
The product Q is 5-oxohexanal.
\(\textbf{Interpret the Tollens' test:}\)
Q contains an aldehyde group and therefore gives the characteristic positive silver-mirror reaction.
\(\textbf{Interpret the iodoform test:}\)
Q also contains a methyl ketone group, \(\mathrm{CH_3CO-}\).
It therefore gives yellow iodoform with iodine and aqueous alkali.
Both tests can be positive for the same compound because they act on different functional groups.
The original tertiary alcohol contains neither of these carbonyl groups; they arise after dehydration followed by alkene cleavage.
The correct prediction requires identifying the major alkene and independently determining the carbonyl group formed at each end of its cleaved double bond.
316. Ethanol labelled specifically at two positions has the structure:
\[
\mathrm{{}^{13}CH_3CH_2{}^{18}OH}
\]
The methyl carbon contains \({}^{13}\mathrm{C}\), and the hydroxyl oxygen contains \({}^{18}\mathrm{O}\).
The sample undergoes complete intramolecular dehydration with concentrated sulphuric acid at approximately \(443\,\mathrm{K}\).
Assume the acid acts catalytically and that no oxygen-isotope exchange, carbon-skeleton rearrangement or other isotope scrambling occurs.
Which pair correctly identifies the labelled alkene and the oxygen-containing small-molecule product?
ⓐ. \(\mathrm{{}^{13}CH_2=CH_2}\) and \(\mathrm{H_2{}^{18}O}\)
ⓑ. \(\mathrm{CH_2={}^{13}CH_2}\) and unlabelled \(\mathrm{H_2O}\), with the oxygen label retained in sulphuric acid.
ⓒ. \(\mathrm{{}^{13}CH_3CH=O}\) and \(\mathrm{H_2{}^{18}O}\)
ⓓ. \(\mathrm{{}^{13}CH_2=CH_2}\) and unlabelled \(\mathrm{H_2O}\), with \({}^{18}\mathrm{O}\) entering the alkene.
Correct Answer: \(\mathrm{{}^{13}CH_2=CH_2}\) and \(\mathrm{H_2{}^{18}O}\)
Explanation: \(\textbf{Identify the atoms involved in dehydration:}\)
Ethanol contains two carbon atoms and a terminal oxygen–hydrogen bond.
The carbon bearing hydroxyl is the alpha-carbon for the elimination.
The neighbouring methyl carbon is the beta-carbon.
The \({}^{13}\mathrm{C}\) label is therefore located at the beta-carbon.
The \({}^{18}\mathrm{O}\) label belongs to the alcohol hydroxyl group.
\(\textbf{Protonate the alcohol oxygen:}\)
In acid-catalysed dehydration, protonation converts the poor leaving group \(\mathrm{-OH}\) into a better leaving group.
The resulting protonated alcohol contains an oxygen atom originally supplied by ethanol.
Under the stated no-exchange assumption, this oxygen retains its \({}^{18}\mathrm{O}\) label.
\(\textbf{Departure of water:}\)
The protonated hydroxyl group can leave as water.
The oxygen atom of this departing water is therefore the original labelled alcohol oxygen.
Consequently, the oxygen-containing small-molecule product is \(\mathrm{H_2{}^{18}O}\).
\(\textbf{Formation of the alkene:}\)
Elimination of an appropriate beta-hydrogen forms the carbon–carbon double bond.
The beta-carbon remains part of the two-carbon alkene skeleton.
Therefore, the \({}^{13}\mathrm{C}\) label is retained in ethene.
\[
\mathrm{{}^{13}CH_3CH_2{}^{18}OH}
\]
\[
\mathrm{\xrightarrow{conc.\ H_2SO_4,\ 443\,K}
{}^{13}CH_2=CH_2+H_2{}^{18}O}
\]
The displayed equation tracks the net isotope locations rather than every intermediate proton transfer.
The acid catalyst is regenerated during the overall reaction.
The carbon label cannot enter water because dehydration does not cleave the carbon skeleton.
The oxygen label cannot enter ethene because the final alkene contains no oxygen.
This atom-tracking problem therefore distinguishes carbon–carbon double-bond formation from departure of the original hydroxyl oxygen.
317. Separate \(0.20\,\mathrm{mol}\) samples of methanol and ethanol are subjected to acid-catalysed dehydration.
Sample P: Methanol undergoes complete intermolecular dehydration to dimethyl ether under suitable conditions.
Sample Q: Ethanol undergoes complete intramolecular dehydration to ethene using concentrated sulphuric acid at approximately \(443\,\mathrm{K}\).
Assume both organic products are collected as dry gases at STP, where one mole of gas occupies \(22.4\,\mathrm{L}\).
Which statement correctly gives the respective gas volumes and explains why methanol cannot undergo ordinary alkene-forming dehydration?
ⓐ. Both produce \(4.48\,\mathrm{L}\) of gas because each alcohol molecule loses one water molecule.
ⓑ. Methanol produces \(4.48\,\mathrm{L}\) of dimethyl ether and ethanol produces \(2.24\,\mathrm{L}\) of ethene; methanol has no alpha-hydrogen.
ⓒ. Methanol gives \(2.24\,\mathrm{L}\) dimethyl ether; ethanol gives \(4.48\,\mathrm{L}\) ethene. Methanol lacks an adjacent beta-carbon for elimination.
ⓓ. Both produce \(2.24\,\mathrm{L}\) of gas because two alcohol molecules are always required for dehydration.
Correct Answer: Methanol gives \(2.24\,\mathrm{L}\) dimethyl ether; ethanol gives \(4.48\,\mathrm{L}\) ethene. Methanol lacks an adjacent beta-carbon for elimination.
Explanation: \(\textbf{Distinguish the two dehydration pathways:}\)
Intermolecular dehydration combines two alcohol molecules to form an ether and water.
Intramolecular dehydration removes water from one suitable alcohol molecule to form an alkene.
The stoichiometries are therefore different.
\(\textbf{Sample P — Methanol to dimethyl ether:}\)
\[
\mathrm{2CH_3OH
\xrightarrow{H^+,\ suitable\ conditions}
CH_3OCH_3+H_2O}
\]
Two moles of methanol produce one mole of dimethyl ether.
\[
n(\mathrm{CH_3OH})=0.20\,\mathrm{mol}
\]
\[
n(\mathrm{CH_3OCH_3})=\frac{0.20}{2}
\]
\[
n(\mathrm{CH_3OCH_3})=0.10\,\mathrm{mol}
\]
\(\textbf{Calculate the dimethyl ether volume:}\)
\[
V_P=nV_m
\]
\[
V_P=0.10(22.4)
\]
\[
V_P=2.24\,\mathrm{L}
\]
\(\textbf{Sample Q — Ethanol to ethene:}\)
\[
\mathrm{CH_3CH_2OH
\xrightarrow{conc.\ H_2SO_4,\ 443\,K}
CH_2=CH_2+H_2O}
\]
One mole of ethanol produces one mole of ethene.
\[
n(\mathrm{CH_2=CH_2})=0.20\,\mathrm{mol}
\]
\[
V_Q=0.20(22.4)
\]
\[
V_Q=4.48\,\mathrm{L}
\]
\(\textbf{Explain the structural limitation of methanol:}\)
Ordinary alkene-forming dehydration requires formation of a carbon–carbon double bond involving the hydroxyl-bearing carbon and an adjacent beta-carbon.
Methanol contains only one carbon atom.
It therefore has no neighbouring beta-carbon and cannot produce a carbon–carbon double bond through ordinary intramolecular dehydration.
However, two methanol molecules can combine through an ether-forming dehydration pathway.
The difference in gas volumes reflects both the reaction stoichiometry and the different structures of the starting alcohols.
\(\textbf{Final answer:}\)
\[
\boxed{V_P=2.24\,\mathrm{L},\qquad V_Q=4.48\,\mathrm{L}}
\]
318. Two constitutional isomers are considered:
P: 2,4-Dimethylpentan-3-ol, \(\mathrm{CH_3CH(CH_3)CH(OH)CH(CH_3)CH_3}\)
Q: 2,4-Dimethylpentan-2-ol, \(\mathrm{(CH_3)_2C(OH)CH_2CH(CH_3)CH_3}\)
Each undergoes acid-catalysed dehydration under conditions favouring the more substituted alkene without carbon-skeleton rearrangement.
Separate samples are also passed over heated copper at approximately \(573\,\mathrm{K}\).
Which statement correctly compares the reactions?
ⓐ. Both alcohols are tertiary and therefore give identical ketones over heated copper.
ⓑ. Both can give 2,4-dimethylpent-2-ene as a principal dehydration product, but heated copper converts P into a ketone whereas Q undergoes dehydration.
ⓒ. P gives 2,4-dimethylpent-1-ene while Q gives exclusively 2,4-dimethylpent-3-ene; both form aldehydes over heated copper.
ⓓ. Both can give 2,4-dimethylpent-2-ene, but heated copper converts P into an aldehyde and Q into a ketone.
Correct Answer: Both can give 2,4-dimethylpent-2-ene as a principal dehydration product, but heated copper converts P into a ketone whereas Q undergoes dehydration.
Explanation: \(\textbf{Classify P and Q:}\)
In P, the hydroxyl-bearing carbon is attached to two carbon atoms and one hydrogen.
Therefore, P is a secondary alcohol.
In Q, the hydroxyl-bearing carbon is attached to three carbon atoms and has no directly attached hydrogen.
Therefore, Q is a tertiary alcohol.
The compounds have the same molecular formula but different hydroxyl positions.
\(\textbf{Dehydration of P:}\)
P contains equivalent beta-carbons on either side of its hydroxyl-bearing carbon.
Removal of a suitable beta-hydrogen and the hydroxyl functionality forms a carbon–carbon double bond.
Because the starting skeleton is symmetrical about the hydroxyl-bearing carbon, elimination towards either side produces the same constitutional alkene.
\[
\mathrm{CH_3CH(CH_3)CH(OH)CH(CH_3)CH_3}
\]
\[
\mathrm{\xrightarrow{H^+,\ heat}
CH_3C(CH_3)=CHCH(CH_3)CH_3+H_2O}
\]
The product is 2,4-dimethylpent-2-ene.
\(\textbf{Dehydration of Q:}\)
Q can eliminate a beta-hydrogen from the methylene group adjacent to its tertiary hydroxyl-bearing carbon.
This gives the more substituted alkene:
\[
\mathrm{(CH_3)_2C(OH)CH_2CH(CH_3)CH_3}
\]
\[
\mathrm{\xrightarrow{H^+,\ heat}
(CH_3)_2C=CHCH(CH_3)CH_3+H_2O}
\]
This is also 2,4-dimethylpent-2-ene.
\(\textbf{Compare heated-copper behaviour:}\)
Secondary alcohol P possesses a hydrogen on its hydroxyl-bearing carbon.
It can therefore undergo dehydrogenation over heated copper to form 2,4-dimethylpentan-3-one.
Tertiary alcohol Q lacks that carbon-bound hydrogen and does not undergo ordinary ketone-forming dehydrogenation.
Under the stated heated-copper conditions, its characteristic transformation is dehydration.
Thus, two alcohol classes can converge on the same alkene under one reagent system while giving different reaction classes under another.
319. 3,3-Dimethylbutan-2-ol has the structure:
\[
\mathrm{CH_3CH(OH)C(CH_3)_3}
\]
It undergoes acid-catalysed dehydration under conditions that permit a carbocation pathway.
Assume that water departure initially produces a carbocation at carbon 2 and that a favourable 1,2-methyl shift occurs from the adjacent quaternary carbon before the final elimination step.
Which principal rearrangement-derived alkene is expected?
ⓐ. 2,3-Dimethylbut-2-ene, formed after migration of a methyl group and elimination from the rearranged carbocation.
ⓑ. 3,3-Dimethylbut-1-ene, formed without any change in the carbon skeleton.
ⓒ. 2,2-Dimethylbut-2-ene, formed by removal of a hydrogen directly from the original hydroxyl-bearing carbon.
ⓓ. 2-Methylpent-2-ene, formed by migration of the original hydroxyl oxygen along the carbon chain.
Correct Answer: 2,3-Dimethylbut-2-ene, formed after migration of a methyl group and elimination from the rearranged carbocation.
Explanation: \(\textbf{Classify the starting alcohol:}\)
The hydroxyl-bearing carbon is attached to two carbon groups and one hydrogen.
Therefore, the substrate is a secondary alcohol.
The adjacent carbon bearing three methyl groups is quaternary and contains no hydrogen.
\(\textbf{Protonation and water departure:}\)
The acid protonates the alcohol oxygen, converting hydroxyl into a better leaving group.
Departure of water produces the carbocation specified in the question.
\[
\mathrm{CH_3CH(OH)C(CH_3)_3}
\]
\[
\mathrm{\xrightarrow[\text{water departure}]{H^+}
CH_3CH^+C(CH_3)_3}
\]
The initial positive charge resides at carbon 2.
\(\textbf{Identify the possible rearrangement:}\)
The neighbouring quaternary carbon has no hydrogen available for a 1,2-hydride shift.
However, one of its methyl groups can migrate to the positively charged carbon.
The migrating methyl group moves with the bonding electron pair.
The positive charge is consequently transferred to the carbon from which methyl migrated.
\(\textbf{Write the rearranged carbocation:}\)
The resulting carbon skeleton can be represented as:
\[
\mathrm{(CH_3)_2CH-C^+(CH_3)_2}
\]
This carbocation is tertiary because the positively charged carbon is attached to three carbon groups.
Its greater substitution provides the stability advantage specified by the rearrangement pathway.
\(\textbf{Select the final elimination:}\)
Removal of the hydrogen on the adjacent carbon bearing two methyl groups forms a carbon–carbon double bond between the two central carbons.
\[
\mathrm{(CH_3)_2CH-C^+(CH_3)_2}
\]
\[
\mathrm{\longrightarrow (CH_3)_2C=C(CH_3)_2+H^+}
\]
The product is 2,3-dimethylbut-2-ene.
This alkene is tetrasubstituted and is more highly substituted than the direct terminal alkene available before rearrangement.
The question explicitly specifies that methyl migration occurs, so the correct answer must preserve the migrated carbon skeleton.
The example distinguishes a 1,2-methyl shift from an unavailable hydride shift at a quaternary neighbouring carbon.
320. An optically pure sample of 3-methylbutan-2-ol is subjected to the following sequence.
Stage I: Complete oxidation with PCC to the corresponding ketone.
Stage II: Addition of HCN to the isolated ketone under achiral conditions, giving the corresponding cyanohydrin.
Stage III: Complete acidic hydrolysis of the nitrile group without rearrangement or racemisation of any newly formed stereogenic centre.
Which statement correctly describes the stereochemical consequences?
ⓐ. The original alcohol remains configurationally unchanged through all three stages because the carbon skeleton is retained.
ⓑ. PCC creates a second stereogenic carbon, and HCN addition eliminates both centres.
ⓒ. Ketone formation erases the original stereogenic centre; HCN addition creates a new centre, giving racemic cyanohydrin and alpha-hydroxy acid.
ⓓ. The ketone remains optically active because its carbonyl carbon retains tetrahedral geometry, and HCN addition gives an achiral product.
Correct Answer: Ketone formation erases the original stereogenic centre; HCN addition creates a new centre, giving racemic cyanohydrin and alpha-hydroxy acid.
Explanation: \(\textbf{Identify the original stereogenic centre:}\)
3-Methylbutan-2-ol has the structure \(\mathrm{CH_3CH(OH)CH(CH_3)_2}\).
Its hydroxyl-bearing carbon is attached to hydrogen, hydroxyl, methyl and isopropyl groups.
These four substituents are different.
Therefore, the carbon is stereogenic in the optically pure starting alcohol.
\(\textbf{Stage I — Oxidation of the secondary alcohol:}\)
PCC oxidises the secondary alcohol to a ketone.
\[
\mathrm{CH_3CH(OH)CH(CH_3)_2}
\]
\[
\mathrm{\xrightarrow{PCC}
CH_3COCH(CH_3)_2}
\]
The product is 3-methylbutan-2-one.
The original stereogenic carbon becomes a trigonal-planar carbonyl carbon.
Its original stereochemical configuration is therefore lost.
\(\textbf{Stage II — Creation of a new stereogenic centre:}\)
Cyanide attacks the planar ketone carbonyl carbon.
After protonation, the product is:
\[
\mathrm{CH_3C(OH)(CN)CH(CH_3)_2}
\]
The former carbonyl carbon is now attached to hydroxyl, cyanide, methyl and isopropyl groups.
These are four different substituents, making this carbon stereogenic.
Under achiral conditions, attack from the two faces of the planar ketone produces enantiomeric cyanohydrins.
A racemic mixture can therefore form.
\(\textbf{Stage III — Nitrile hydrolysis:}\)
Complete acidic hydrolysis converts the nitrile group into a carboxylic acid group.
\[
\mathrm{CH_3C(OH)(CN)CH(CH_3)_2}
\]
\[
\mathrm{\xrightarrow{H_3O^+,\ heat}
CH_3C(OH)(COOH)CH(CH_3)_2}
\]
The hydroxyl-bearing carbon remains tetrahedral and retains four different substituents.
Under the stated no-racemisation assumption, the stereochemical mixture remains racemic.
Thus, oxidation destroys the original stereogenic centre, but subsequent carbonyl addition creates a new one.
The final optical behaviour depends on the mechanism and reaction environment, not simply on whether the carbon skeleton remains unchanged.
