701. Ethanal and propanone are subjected to two separately controlled crossed aldol reactions.
Route P: The enolate is generated exclusively from propanone and reacts with ethanal as the carbonyl acceptor.
Route Q: The enolate is generated exclusively from ethanal and reacts with propanone as the carbonyl acceptor.
In each route, the isolated beta-hydroxy carbonyl compound undergoes complete dehydration under conditions producing the corresponding alpha,beta-unsaturated carbonyl compound.
Assume that the specified donor–acceptor roles are maintained, with no self-condensation or skeletal rearrangement.
Which statement correctly compares the dehydrated products?
ⓐ. Both routes produce pent-3-en-2-one because the same two carbonyl compounds are used.
ⓑ. P produces 3-methylbut-2-enal, whereas Q produces pent-3-en-2-one; neither product can exhibit geometrical isomerism.
ⓒ. P gives pent-3-en-2-one, capable of \(E/Z\) isomerism; Q gives 3-methylbut-2-enal, which lacks \(E/Z\) isomerism because one double-bond carbon bears two methyl groups.
ⓓ. P produces a saturated ketone, whereas Q produces a saturated aldehyde because dehydration cannot introduce a carbon–carbon double bond.
Correct Answer: P gives pent-3-en-2-one, capable of \(E/Z\) isomerism; Q gives 3-methylbut-2-enal, which lacks \(E/Z\) isomerism because one double-bond carbon bears two methyl groups.
Explanation: \(\textbf{Identify the significance of the donor:}\)
An aldol enolate forms through removal of an alpha-hydrogen from a suitable carbonyl compound.
The enolate carbon forms a new carbon–carbon bond by attacking another carbonyl carbon.
Changing which compound supplies the enolate can therefore change the product connectivity.
\(\textbf{Route P — Propanone supplies the enolate:}\)
The propanone-derived enolate attacks the aldehyde carbonyl of ethanal.
After protonation, the beta-hydroxy ketone is:
\[
\mathrm{CH_3COCH_2CH(OH)CH_3}
\]
Dehydration gives:
\[
\mathrm{CH_3COCH_2CH(OH)CH_3}
\]
\[
\mathrm{\longrightarrow CH_3COCH=CHCH_3+H_2O}
\]
The product is pent-3-en-2-one.
Each carbon of its carbon–carbon double bond carries two different substituents.
Therefore, \(E/Z\) geometrical isomerism is possible.
\(\textbf{Route Q — Ethanal supplies the enolate:}\)
The ethanal-derived enolate attacks the carbonyl carbon of propanone.
The beta-hydroxy aldehyde is:
\[
\mathrm{OHCCH_2C(OH)(CH_3)_2}
\]
Dehydration gives:
\[
\mathrm{OHCCH=C(CH_3)_2+H_2O}
\]
The product is 3-methylbut-2-enal.
Its beta alkene carbon is attached to two identical methyl groups.
Therefore, the double bond cannot exhibit \(E/Z\) isomerism.
\(\textbf{Compare the two products:}\)
Both products contain five carbon atoms.
Both contain one carbonyl group and one newly formed carbon–carbon double bond.
However, one retains a ketone carbonyl, while the other retains an aldehyde carbonyl.
The difference arises from the original donor–acceptor assignment, not from the number of carbon atoms contributed by either reactant.
This illustrates why crossed aldol reactions require explicit control or identification of the enolate donor when both starting carbonyl compounds possess alpha-hydrogens.
702. An unknown acyclic alkene X undergoes complete reductive ozonolysis, producing methanal and butan-2-one as its only organic products.
Two independent portions of X are then hydrated.
Route P: Acid-catalysed hydration under ordinary Markovnikov conditions without skeletal rearrangement.
Route Q: Hydroboration–oxidation under ordinary anti-Markovnikov conditions.
The isolated alcohols are separately treated with PCC.
Assume complete regioselectivity in each hydration and preservation of the carbon skeleton during oxidation.
Which statement correctly identifies X and compares the oxidation products?
ⓐ. X is pent-2-ene; P gives pentan-3-one and Q gives pentanal.
ⓑ. X is 2-methylbut-1-ene; P gives a tertiary alcohol resistant to PCC, whereas Q gives a primary alcohol oxidisable to 2-methylbutanal.
ⓒ. X is 3-methylbut-1-ene; P gives 3-methylbutanal and Q gives 3-methylbutan-2-one.
ⓓ. X is 2-methylbut-2-ene; both hydration routes give the same secondary alcohol and subsequently the same ketone.
Correct Answer: X is 2-methylbut-1-ene; P gives a tertiary alcohol resistant to PCC, whereas Q gives a primary alcohol oxidisable to 2-methylbutanal.
Explanation: \(\textbf{Reconstruct the original double bond:}\)
Reductive ozonolysis replaces the two carbons of an alkene double bond with carbonyl carbons.
Methanal must originate from an alkene carbon bearing two hydrogen atoms.
Therefore, X contains a terminal \(\mathrm{CH_2=}\) group.
Butan-2-one contains a carbonyl carbon bonded to methyl and ethyl groups.
The other original alkene carbon must therefore carry those two substituents.
Reconnecting the two carbonyl carbons through a double bond gives:
\[
\mathrm{CH_2=C(CH_3)CH_2CH_3}
\]
Thus, X is 2-methylbut-1-ene.
\(\textbf{Follow Markovnikov hydration:}\)
In Route P, hydroxyl becomes attached to the more substituted alkene carbon.
\[
\mathrm{CH_2=C(CH_3)CH_2CH_3}
\]
\[
\mathrm{\xrightarrow{H_2O/H^+}CH_3C(OH)(CH_3)CH_2CH_3}
\]
The product is 2-methylbutan-2-ol.
Its hydroxyl-bearing carbon is tertiary and has no directly attached hydrogen.
It therefore resists ordinary PCC oxidation to a carbonyl compound.
\(\textbf{Follow hydroboration–oxidation:}\)
The hydroxyl group becomes attached to the less substituted terminal carbon.
\[
\mathrm{CH_2=C(CH_3)CH_2CH_3}
\]
\[
\mathrm{\xrightarrow[\mathrm{H_2O_2/OH^-}]{\mathrm{BH_3}}
HOCH_2CH(CH_3)CH_2CH_3}
\]
The product is 2-methylbutan-1-ol.
It is a primary alcohol.
PCC can oxidise it to:
\[
\mathrm{OHCCH(CH_3)CH_2CH_3}
\]
This is 2-methylbutanal.
\(\textbf{Interpret the integrated evidence:}\)
Ozonolysis identifies the original alkene connectivity.
The two hydration pathways introduce hydroxyl at different carbons.
The subsequent PCC reaction distinguishes the tertiary and primary alcohol products.
The reasoning therefore requires reconstructing the precursor before predicting either hydration outcome.
703. Two isotopically labelled ethylbenzene samples undergo separate vigorous benzylic oxidations using hot alkaline potassium permanganate, followed by acidification.
P contains \({}^{13}\mathrm{C}\) at its benzylic methylene carbon:
\[
\mathrm{C_6H_5{}^{13}CH_2CH_3}
\]
Q contains \({}^{13}\mathrm{C}\) at its terminal methyl carbon:
\[
\mathrm{C_6H_5CH_2{}^{13}CH_3}
\]
Assume complete characteristic side-chain oxidation to benzoic acid, no isotope exchange and no rearrangement of the aromatic carbon skeleton.
Which statement correctly identifies the location of the isotope label in the isolated benzoic acid?
ⓐ. P gives unlabelled benzoic acid, whereas Q gives benzoic acid labelled at the carboxyl carbon.
ⓑ. Both give benzoic acid labelled at the aromatic carbon originally attached to the side chain.
ⓒ. Both give benzoic acid labelled at the carboxyl carbon because the two side-chain carbons become equivalent during oxidation.
ⓓ. P gives benzoic acid labelled at the carboxyl carbon, whereas Q gives benzoic acid without a \({}^{13}\mathrm{C}\) label.
Correct Answer: P gives benzoic acid labelled at the carboxyl carbon, whereas Q gives benzoic acid without a \({}^{13}\mathrm{C}\) label.
Explanation: \(\textbf{Identify the carbon retained during benzylic oxidation:}\)
Ethylbenzene contains a two-carbon side chain.
The benzylic carbon is the carbon directly attached to the benzene ring.
It contains hydrogen and therefore satisfies the characteristic structural requirement for vigorous side-chain oxidation.
\(\textbf{Track the benzylic carbon:}\)
During conversion into benzoic acid, the carbon directly bonded to the aromatic ring becomes the carboxyl carbon.
The aromatic carbon–benzylic carbon connection is retained.
Therefore, a label initially placed at that benzylic carbon remains in the isolated aromatic acid.
For P:
\[
\mathrm{C_6H_5{}^{13}CH_2CH_3}
\]
\[
\mathrm{\xrightarrow[\mathrm{H^+}]{KMnO_4/OH^-,\ heat}
C_6H_5{}^{13}COOH}
\]
The labelled carbon is now the carboxyl carbon.
\(\textbf{Track the terminal methyl carbon:}\)
The terminal methyl carbon is not directly attached to the aromatic ring.
The characteristic vigorous oxidation degrades the extended side chain while retaining only the benzylic carbon within the isolated benzoic acid.
Therefore, the terminal labelled carbon of Q does not remain in its isolated benzoic acid product.
\(\textbf{Interpret the atom-tracing distinction:}\)
The two starting samples have identical ordinary chemical structures.
Their labels occupy different positions within the same side chain.
The oxidation products are chemically benzoic acid in both experiments, but their isotope compositions differ.
The experiment shows that removal of side-chain carbons does not imply removal of the carbon directly connecting the original side chain to the aromatic ring.
It also distinguishes carbon-skeleton tracing from a simple count of carbon atoms before and after oxidation.
704. Anisole undergoes a controlled three-stage reaction sequence.
Stage I: Friedel–Crafts acylation with benzoyl chloride and anhydrous aluminium chloride under conditions favouring substitution para to the methoxy group.
Stage II: Complete Wolff–Kishner reduction of the isolated aromatic ketone, converting its carbonyl group into a methylene group while preserving the aromatic methoxy ether.
Stage III: Cleavage of the aromatic methyl ether using concentrated HI under conditions preserving the newly formed carbon–carbon skeleton.
Assume no additional aromatic substitution or rearrangement.
Which compound is obtained after Stage III?
ⓐ. 4-Hydroxybenzophenone
ⓑ. 4-Methoxybenzophenone
ⓒ. 4-Benzylphenol
ⓓ. 4-Methoxybenzoic acid
Correct Answer: 4-Benzylphenol
Explanation: \(\textbf{Identify the directing group in Stage I:}\)
Anisole contains a methoxy group directly attached to an aromatic ring.
The oxygen lone pairs can donate electron density into the aromatic system through resonance.
The methoxy group activates the ring and directs electrophilic substitution towards ortho and para positions.
Under the stipulated para-selective Friedel–Crafts conditions, the benzoyl group enters the para position.
\(\textbf{Identify the acylation product:}\)
Benzoyl chloride introduces the group \(\mathrm{-COC_6H_5}\).
The resulting compound is 4-methoxybenzophenone.
It contains an aromatic ketone carbonyl between two phenyl-derived fragments.
\(\textbf{Apply Wolff–Kishner reduction:}\)
Hydrazine and strongly basic heating conditions convert the ketone carbonyl into a methylene group.
The two aromatic carbon groups remain attached to that carbon.
The relevant transformation is:
\[
\mathrm{ArCOC_6H_5\longrightarrow ArCH_2C_6H_5}
\]
The aromatic methoxy group is preserved by the question's assumptions.
The Stage II product is therefore 4-methoxydiphenylmethane, also describable as 4-methoxybenzylbenzene.
\(\textbf{Apply concentrated HI cleavage:}\)
An aromatic methyl ether contains an aryl–oxygen bond and a methyl–oxygen bond.
Under the specified cleavage conditions, the methyl–oxygen bond is broken.
The aromatic carbon–oxygen bond remains intact.
The methoxy group therefore becomes a phenolic hydroxyl group.
\(\textbf{Identify the final product:}\)
The benzyl substituent remains para to the newly regenerated phenolic hydroxyl.
The product is 4-benzylphenol.
The sequence integrates aromatic substitution orientation, carbonyl deoxygenation and selective cleavage of an aromatic methyl ether.
It also shows that removing the carbonyl oxygen and removing the methyl group of an ether are chemically different transformations.
705. A mixture contains equal analytical concentrations of benzoic acid and 4-nitrophenol.
Their approximate acid dissociation values are:
\[
\mathrm{p}K_a(\text{benzoic acid})=4.20
\]
\[
\mathrm{p}K_a(\text{4-nitrophenol})=7.20
\]
The mixture is equilibrated at pH 5.20 in a suitable aqueous–organic extraction system.
Assume:
I. Ionised forms preferentially remain in the aqueous layer.
II. Neutral forms preferentially remain in the organic layer.
III. Each compound obeys ideal monoprotic acid–base equilibrium.
IV. No chemical decomposition or additional acid–base-active species interfere.
Which statement correctly identifies the approximate ionised fractions and the compound preferentially recovered in the aqueous phase?
ⓐ. Benzoic acid is approximately \(90.9\%\) ionised, 4-nitrophenol approximately \(1.0\%\) ionised, and benzoate is preferentially enriched in the aqueous phase.
ⓑ. Benzoic acid is approximately \(1.0\%\) ionised, 4-nitrophenol approximately \(90.9\%\) ionised, and nitrophenoxide is preferentially enriched in the aqueous phase.
ⓒ. Both compounds are approximately \(50\%\) ionised because they are present at equal analytical concentrations.
ⓓ. Both compounds are approximately \(99\%\) ionised because the solution pH exceeds the \(\mathrm{p}K_a\) of benzoic acid.
Correct Answer: Benzoic acid is approximately \(90.9\%\) ionised, 4-nitrophenol approximately \(1.0\%\) ionised, and benzoate is preferentially enriched in the aqueous phase.
Explanation: \(\textbf{Connect ionic form with extraction behaviour:}\)
The problem distinguishes neutral and ionised molecules through their stipulated distribution between two liquid phases.
The relative amount in each acid–base form depends on the pH and the compound's \(\mathrm{p}K_a\).
Equal analytical concentrations do not imply equal ionisation.
\(\textbf{Use the monoprotic dissociation relation:}\)
For an acid HA:
\[
\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}
=10^{\mathrm{pH}-\mathrm{p}K_a}
\]
Its ionised fraction is:
\[
f=\frac{10^{\mathrm{pH}-\mathrm{p}K_a}}
{1+10^{\mathrm{pH}-\mathrm{p}K_a}}
\]
\(\textbf{Calculate the benzoic acid fraction:}\)
\[
r_{\mathrm{benzoic}}=10^{5.20-4.20}
\]
\[
r_{\mathrm{benzoic}}=10
\]
Therefore:
\[
f_{\mathrm{benzoic}}=\frac{10}{11}
\]
\[
f_{\mathrm{benzoic}}\approx0.909
\]
\[
f_{\mathrm{benzoic}}\approx90.9\%
\]
\(\textbf{Calculate the 4-nitrophenol fraction:}\)
\[
r_{\mathrm{nitrophenol}}=10^{5.20-7.20}
\]
\[
r_{\mathrm{nitrophenol}}=10^{-2}=0.0100
\]
Therefore:
\[
f_{\mathrm{nitrophenol}}=\frac{0.0100}{1.0100}
\]
\[
f_{\mathrm{nitrophenol}}\approx0.00990
\]
\[
f_{\mathrm{nitrophenol}}\approx0.990\%
\]
\(\textbf{Apply the stipulated extraction preference:}\)
Benzoic acid is predominantly present as benzoate.
Its ionic form is consequently preferentially enriched in the aqueous layer.
4-Nitrophenol remains predominantly neutral and is preferentially associated with the organic layer.
\(\textbf{Interpret the separation principle:}\)
The two compounds are both oxygen-containing aromatic acids, but their dissociation constants differ substantially.
A carefully chosen intermediate pH can therefore favour different charge states.
The chemical basis for the separation is selective ionisation rather than the presence or absence of an aromatic ring.
\(\textbf{Final answer:}\)
\[
\boxed{f_{\mathrm{benzoate}}\approx90.9\%,\qquad
f_{\mathrm{nitrophenoxide}}\approx0.990\%}
\]
706. A mixture contains propanal and propanone dissolved in a suitable organic solvent.
The mixture is treated with aqueous sodium hydrogensulphite, \(\mathrm{NaHSO_3}\), under conditions that permit reversible carbonyl addition.
The experiment is specifically adjusted so that the propanal-derived hydrogensulphite addition compound is recovered quantitatively in an aqueous fraction, whereas propanone remains predominantly in the organic fraction.
The isolated aqueous addition compound is then treated with dilute mineral acid to regenerate its parent carbonyl compound.
Which statement correctly explains the separation and identifies the recovered carbonyl?
ⓐ. The separated addition compound is an oxime, and acid hydrolysis produces propanone.
ⓑ. Sodium hydrogensulphite irreversibly oxidises propanal to propanoic acid, leaving propanone unchanged.
ⓒ. Propanal forms a reversible carbonyl-addition derivative from which it can be regenerated; separation uses differences in equilibrium and phase partitioning.
ⓓ. The recovered compound must be propanone because ketones always have a more favourable hydrogensulphite addition equilibrium than corresponding aldehydes.
Correct Answer: Propanal forms a reversible carbonyl-addition derivative from which it can be regenerated; separation uses differences in equilibrium and phase partitioning.
Explanation: \(\textbf{Identify the reaction centre:}\)
Both propanal and propanone contain a carbonyl group.
Its carbon atom is electrophilic because oxygen withdraws electron density.
Suitable nucleophiles can therefore attack that carbon.
\(\textbf{Apply hydrogensulphite addition:}\)
Hydrogensulphite can form reversible addition compounds with aldehydes and ketones.
The carbonyl carbon changes from an approximately trigonal-planar environment to a tetrahedral environment.
The original carbonyl oxygen becomes part of a hydroxyl-containing addition product.
For propanal, the connectivity is represented by:
\[
\mathrm{CH_3CH_2CH(OH)SO_3Na}
\]
\(\textbf{Compare the relevant equilibrium tendencies:}\)
Ordinary aldehydes generally possess less steric hindrance at the carbonyl carbon than corresponding ketones.
This can favour nucleophilic addition to aldehydes.
NCERT describes sodium hydrogensulphite addition compounds as useful in the separation and purification of suitable aldehydes.
\(\textbf{Interpret the specified extraction:}\)
The question supplies conditions under which the propanal-derived addition product is recovered quantitatively in the aqueous fraction.
Propanone remains predominantly in the organic fraction.
The separation is therefore based on reversible addition and partitioning rather than permanent destruction of either carbonyl framework.
\(\textbf{Regenerate the original aldehyde:}\)
Treatment with suitable dilute mineral acid can decompose the hydrogensulphite addition compound and regenerate the corresponding carbonyl group.
The carbon skeleton of propanal is unchanged.
The recovered compound is therefore propanal.
\(\textbf{Resolve the mechanistic misconception:}\)
Hydrogensulphite addition does not mean that a sulphur-containing group permanently replaces the aldehyde oxygen.
Nor is it equivalent to formation of an oxime through nitrogen-containing condensation.
The reversibility of the addition is essential to its usefulness for carbonyl-compound purification.
707. Two hydroxyaldehydes undergo separate intramolecular nucleophilic additions under controlled conditions:
P: 4-Hydroxybutanal, \(\mathrm{HOCH_2CH_2CH_2CHO}\).
Q: 3-Hydroxypropanal, \(\mathrm{HOCH_2CH_2CHO}\).
Assume that the alcohol oxygen attacks the aldehyde carbonyl carbon within the same molecule to form a cyclic hemiacetal, without cleavage of the carbon skeleton or elimination of water.
The comparison concerns the ring sizes and their ordinary geometrical favourability, rather than a claim of quantitative ring closure.
Which statement is correct?
ⓐ. P forms a four-membered cyclic hemiacetal, whereas Q forms a five-membered cyclic hemiacetal; Q is necessarily more favourable.
ⓑ. Both form six-membered rings because the aldehyde oxygen must remain in the ring.
ⓒ. Both form acetals containing two new ether bonds because intramolecular addition always eliminates water.
ⓓ. P can form a five-membered cyclic hemiacetal, whereas Q would form a four-membered cyclic hemiacetal; the five-membered arrangement is ordinarily more favourable.
Correct Answer: P can form a five-membered cyclic hemiacetal, whereas Q would form a four-membered cyclic hemiacetal; the five-membered arrangement is ordinarily more favourable.
Explanation: \(\textbf{Identify the reacting atoms:}\)
Each molecule contains an alcohol hydroxyl group and an aldehyde carbonyl group.
Under suitable conditions, the alcohol oxygen can act as a nucleophile.
It attacks the aldehyde carbonyl carbon within the same molecule.
This creates a new carbon–oxygen bond.
\(\textbf{Count the ring atoms in P:}\)
4-Hydroxybutanal has the connectivity:
\[
\mathrm{HO-C_4H_2-C_3H_2-C_2H_2-C_1HO}
\]
The hydroxyl oxygen is connected initially to carbon 4.
Its attack on aldehyde carbon 1 creates a ring following:
\[
\mathrm{O-C_4-C_3-C_2-C_1-O}
\]
There are five atoms in this cycle: one oxygen and four carbons.
Therefore, P forms a five-membered cyclic hemiacetal.
\(\textbf{Count the ring atoms in Q:}\)
3-Hydroxypropanal contains only three carbon atoms between the ring-closing positions.
The resulting cyclic connectivity contains one oxygen and three carbon atoms.
Therefore, the proposed hemiacetal ring is four-membered.
\(\textbf{Compare geometrical favourability:}\)
Five-membered rings can generally accommodate favourable bond angles with less strain than corresponding four-membered rings.
The five-membered cyclic hemiacetal of P is consequently the more favourable ordinary ring-size arrangement.
\(\textbf{Distinguish hemiacetal formation from acetal formation:}\)
Intramolecular addition to an aldehyde carbonyl produces a tetrahedral carbon bearing both an oxygen-derived ring linkage and a hydroxyl group.
The product is a hemiacetal.
Formation of a complete acetal would require an additional transformation replacing the remaining hydroxyl-derived functionality under suitable conditions.
The specified intramolecular addition does not eliminate water.
The problem therefore requires tracing the newly formed bond and counting the atoms in the actual ring.
708. Acetophenone and benzophenone are subjected to a controlled crossed aldol reaction.
The enolate is generated exclusively from acetophenone, \(\mathrm{C_6H_5COCH_3}\).
Benzophenone, \(\mathrm{(C_6H_5)_2CO}\), acts exclusively as the carbonyl acceptor.
The resulting beta-hydroxy ketone is isolated and subsequently dehydrated completely.
Assume that the bulky ketone acceptor reacts through the stipulated pathway, that no rearrangement occurs and that the product double bond forms by ordinary alpha,beta elimination.
Which statement correctly identifies the dehydrated product and its stereochemical character?
ⓐ. \(\mathrm{C_6H_5COCH=C(C_6H_5)_2}\), with no \(E/Z\) isomerism because one double-bond carbon bears two identical phenyl groups.
ⓑ. \(\mathrm{C_6H_5COCH=C(C_6H_5)_2}\), exhibiting \(E/Z\) isomerism because the product contains three aromatic rings.
ⓒ. \(\mathrm{(C_6H_5)_2C=CHCOCH_3}\), containing a new aldehyde group produced by ketone dehydration.
ⓓ. \(\mathrm{C_6H_5CH=CHC_6H_5}\), because the original acetophenone carbonyl carbon is eliminated during the condensation.
Correct Answer: \(\mathrm{C_6H_5COCH=C(C_6H_5)_2}\), with no \(E/Z\) isomerism because one double-bond carbon bears two identical phenyl groups.
Explanation: \(\textbf{Identify the enolate donor:}\)
Acetophenone contains a methyl group directly attached to its ketone carbonyl.
That methyl group possesses alpha-hydrogens.
Therefore, acetophenone can generate the stipulated enolate.
\(\textbf{Examine the acceptor:}\)
Benzophenone contains a carbonyl carbon attached directly to two phenyl groups.
It lacks an ordinary enolisable alpha-hydrogen.
Under the question's controlled conditions, it functions as the electrophilic carbonyl acceptor.
\(\textbf{Construct the aldol-addition product:}\)
The alpha-carbon of the acetophenone enolate attacks the benzophenone carbonyl carbon.
The new carbon–carbon bond connects the acetophenone-derived alpha-carbon with the former benzophenone carbonyl carbon.
After protonation, the product is:
\[
\mathrm{C_6H_5COCH_2C(OH)(C_6H_5)_2}
\]
The acetophenone ketone carbonyl remains intact.
The original benzophenone carbonyl oxygen becomes a hydroxyl group.
\(\textbf{Apply dehydration:}\)
Removal of the elements of water from the beta-hydroxy ketone gives:
\[
\mathrm{C_6H_5COCH=C(C_6H_5)_2}
\]
The retained carbonyl and newly formed carbon–carbon double bond form an alpha,beta-unsaturated ketone system.
\(\textbf{Examine geometrical isomerism:}\)
The beta carbon of the new double bond carries two identical phenyl substituents.
The presence of a double bond alone does not establish \(E/Z\) isomerism.
Each double-bond carbon must possess two different substituents.
Because this condition fails at the beta carbon, the product has no \(E/Z\) geometrical isomers about that double bond.
\(\textbf{Interpret the distinction:}\)
The product contains three phenyl rings and a conjugated carbonyl system.
Neither feature automatically creates geometrical isomerism.
The decisive structural information is the substituent identity at each individual double-bond carbon.
709. Ethanoic anhydride undergoes complete reaction with ethanol according to:
\[
\mathrm{(CH_3CO)_2O+C_2H_5OH
\longrightarrow CH_3COOC_2H_5+CH_3COOH}
\]
A reaction mixture initially contains \(0.30\,\mathrm{mol}\) ethanoic anhydride and \(0.20\,\mathrm{mol}\) ethanol.
Assume that ethanol reacts completely through the specified pathway, that the unreacted anhydride is separated without hydrolysis, and that no other reactions occur.
The ethanoic acid product is then completely neutralised with excess aqueous sodium hydrogencarbonate.
The isolated ethyl ethanoate is independently hydrolysed completely in aqueous sodium hydroxide, and its resulting carboxylate is acidified.
What are the amount of carbon dioxide evolved from the initially formed acid and the amount of additional ethanoic acid recovered from ester hydrolysis, respectively?
ⓐ. \(0.10\,\mathrm{mol}\) and \(0.20\,\mathrm{mol}\)
ⓑ. \(0.20\,\mathrm{mol}\) and \(0.10\,\mathrm{mol}\)
ⓒ. \(0.20\,\mathrm{mol}\) and \(0.20\,\mathrm{mol}\)
ⓓ. \(0.30\,\mathrm{mol}\) and \(0.20\,\mathrm{mol}\)
Correct Answer: \(0.20\,\mathrm{mol}\) and \(0.20\,\mathrm{mol}\)
Explanation: \(\textbf{Determine the limiting reactant:}\)
One mole of ethanoic anhydride reacts with one mole of ethanol under the specified acyl-substitution reaction.
The initial amounts are:
\[
n(\mathrm{anhydride})=0.30\,\mathrm{mol}
\]
\[
n(\mathrm{ethanol})=0.20\,\mathrm{mol}
\]
Therefore, ethanol is the limiting reactant.
\(\textbf{Calculate the initial organic products:}\)
The reaction equation gives one mole of ethyl ethanoate and one mole of ethanoic acid per mole of consumed ethanol.
Thus:
\[
n(\mathrm{ethyl\ ethanoate})=0.20\,\mathrm{mol}
\]
\[
n(\mathrm{ethanoic\ acid})_{\mathrm{initial}}=0.20\,\mathrm{mol}
\]
The remaining ethanoic anhydride is:
\[
n(\mathrm{anhydride})_{\mathrm{remaining}}=0.30-0.20
\]
\[
n(\mathrm{anhydride})_{\mathrm{remaining}}=0.10\,\mathrm{mol}
\]
It is separated without hydrolysis by assumption.
\(\textbf{Determine carbon dioxide from the acid fraction:}\)
\[
\mathrm{CH_3COOH+NaHCO_3
\longrightarrow CH_3COONa+CO_2+H_2O}
\]
One mole of ethanoic acid produces one mole of carbon dioxide.
Therefore:
\[
n(\mathrm{CO_2})=0.20\,\mathrm{mol}
\]
\(\textbf{Follow the independent ester hydrolysis:}\)
\[
\mathrm{CH_3COOC_2H_5+NaOH
\longrightarrow CH_3COONa+C_2H_5OH}
\]
One mole of ester produces one mole of ethanoate.
After acidification:
\[
n(\mathrm{CH_3COOH})_{\mathrm{from\ ester}}
=0.20\,\mathrm{mol}
\]
\(\textbf{Interpret the two acid quantities:}\)
The first acid fraction is produced directly during reaction of the anhydride with ethanol.
The second acid fraction becomes available only after hydrolysis of the ester.
They arise through different bond-cleavage processes and are recovered from independently treated fractions.
The unreacted anhydride does not contribute to either result because the question explicitly excludes its hydrolysis.
\(\textbf{Final answer:}\)
\[
\boxed{n(\mathrm{CO_2})=0.20\,\mathrm{mol},\qquad
n(\mathrm{CH_3COOH})_{\mathrm{from\ ester}}=0.20\,\mathrm{mol}}
\]
710. Acetylsalicylic acid contains one free carboxylic acid group and one aromatic acetate ester group.
A \(0.15\,\mathrm{mol}\) sample is divided into two equal portions.
Experiment P: One portion is treated directly with excess aqueous sodium hydrogencarbonate.
Experiment Q: The other portion undergoes complete hydrolysis of its aromatic acetate ester linkage. After suitable acidification, all resulting organic acids are recovered without loss and treated with excess aqueous sodium hydrogencarbonate.
Assume:
I. Each free carboxylic acid group produces one carbon dioxide molecule on complete neutralisation.
II. Ordinary phenolic hydroxyl groups do not produce the characteristic hydrogencarbonate response under the specified conditions.
III. Hydrolysis produces salicylic acid and ethanoic acid quantitatively.
What is the ratio of the carbon dioxide amounts obtained in Experiments Q and P?
ⓐ. \(1:1\)
ⓑ. \(2:1\)
ⓒ. \(3:1\)
ⓓ. \(1:2\)
Correct Answer: \(2:1\)
Explanation: \(\textbf{Determine the sample size in each experiment:}\)
The original acetylsalicylic acid amount is:
\[
n_{\mathrm{total}}=0.15\,\mathrm{mol}
\]
Two equal portions contain:
\[
n_P=n_Q=\frac{0.15}{2}
\]
\[
n_P=n_Q=0.075\,\mathrm{mol}
\]
\(\textbf{Experiment P — Count the initial free acid groups:}\)
Each acetylsalicylic acid molecule contains one free carboxylic acid group.
Its other oxygen-containing substituent is an aromatic acetate ester.
That ester is not a second free carboxylic acid group.
Therefore:
\[
n(\mathrm{CO_2})_P=0.075\,\mathrm{mol}
\]
\(\textbf{Experiment Q — Follow complete ester hydrolysis:}\)
Hydrolysis of the aromatic acetate ester produces salicylic acid and ethanoic acid.
One starting molecule gives one molecule of each acid.
\[
n(\mathrm{salicylic\ acid})=0.075\,\mathrm{mol}
\]
\[
n(\mathrm{ethanoic\ acid})=0.075\,\mathrm{mol}
\]
Both products contain one free carboxylic acid group.
The salicylic acid also contains a phenolic hydroxyl group, but that group is excluded from the stated hydrogencarbonate response.
\(\textbf{Calculate the second gas amount:}\)
\[
n(\mathrm{CO_2})_Q=0.075+0.075
\]
\[
n(\mathrm{CO_2})_Q=0.150\,\mathrm{mol}
\]
\(\textbf{Calculate the requested ratio:}\)
\[
\frac{n(\mathrm{CO_2})_Q}{n(\mathrm{CO_2})_P}
=\frac{0.150}{0.075}
\]
\[
\frac{n(\mathrm{CO_2})_Q}{n(\mathrm{CO_2})_P}=2
\]
\(\textbf{Interpret the structural change:}\)
Hydrolysis does not simply strengthen the existing acid group.
It separates an ester into two organic products, each containing its own carboxylic acid group.
Therefore, the total number of free carboxylic acid functionalities available for neutralisation increases.
The comparison distinguishes acid-strength changes from changes in the number of acidic functional groups produced by hydrolysis.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{n(\mathrm{CO_2})_Q}{n(\mathrm{CO_2})_P}=2}
\]
711. Two saturated, acyclic carbonyl compounds have molecular formula \(\mathrm{C_5H_{10}O}\).
P: Pentanal.
Q: Pentan-2-one.
Each is subjected independently to two oxidation experiments.
Experiment I: Treatment with Tollens' reagent under ordinary diagnostic conditions.
Experiment II: Treatment with a sufficiently strong oxidising agent at elevated temperature to permit vigorous carbonyl-related oxidation.
For Experiment II, assume that the aldehyde carbon is oxidised without cleavage of the five-carbon skeleton, whereas oxidation of the ketone proceeds through carbon–carbon bond cleavage as described in NCERT.
Consider the following statements.
Statement I: P can give a positive Tollens' reaction because its aldehyde carbon bears a hydrogen.
Statement II: Q ordinarily gives a negative Tollens' reaction but can undergo oxidation under sufficiently vigorous conditions.
Statement III: Vigorous oxidation of Q must retain all five carbon atoms in one monocarboxylic acid because its ketone carbonyl is polar.
Statement IV: Oxidation of P to pentanoic acid can preserve its five-carbon skeleton, whereas vigorous oxidation of Q can give carboxylic acid products with fewer carbon atoms than the starting ketone.
Which combination contains only the correct statements?
ⓐ. I, II and III only
ⓑ. I, III and IV only
ⓒ. II, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: \(\textbf{Compare the original carbonyl environments:}\)
Pentanal contains a terminal aldehyde carbonyl group.
Pentan-2-one contains an internal ketone carbonyl group.
Although both have molecular formula \(\mathrm{C_5H_{10}O}\), their carbonyl carbons have different substituents.
\(\textbf{Evaluate the mild oxidation experiment:}\)
The aldehyde carbon of P carries hydrogen.
It can undergo oxidation to a carboxylate-derived functionality using Tollens' reagent.
The silver(I) ions are reduced to metallic silver.
Statement I is correct.
Q is an ordinary ketone.
It does not give the characteristic positive Tollens' response under normal diagnostic conditions.
Statement II correctly recognises this distinction while allowing vigorous ketone oxidation.
\(\textbf{Examine aldehyde oxidation:}\)
Controlled oxidation of pentanal converts its terminal aldehyde group into a carboxylic acid group.
\[
\mathrm{CH_3CH_2CH_2CH_2CHO
\xrightarrow{[O]}
CH_3CH_2CH_2CH_2COOH}
\]
The product is pentanoic acid.
The five-carbon skeleton is retained.
\(\textbf{Examine vigorous ketone oxidation:}\)
Ketones ordinarily resist mild oxidation more strongly than aldehydes.
Under vigorous conditions, carbon–carbon bonds adjacent to the ketone carbonyl can be cleaved.
This produces carboxylic acid products containing fewer carbon atoms than the original ketone.
Statement III is incorrect because carbonyl polarity does not prevent such cleavage.
Statement IV is correct.
\(\textbf{Interpret the distinction:}\)
The negative Tollens' result of a ketone does not prove absolute resistance to every oxidising agent.
Oxidation behaviour depends on both functional-group structure and reagent strength.
The comparison also demonstrates that oxidation may preserve the carbon skeleton in one substrate while breaking carbon–carbon bonds in another.
712. Three ketones are compared for the constitutionally distinct positions from which an alpha-hydrogen may be removed.
P: Propanone, \(\mathrm{CH_3COCH_3}\).
Q: Pentane-2,4-dione, \(\mathrm{CH_3COCH_2COCH_3}\).
R: Pentan-3-one, \(\mathrm{CH_3CH_2COCH_2CH_3}\).
Here, distinct alpha sites mean sites giving constitutionally different enolates; equivalent positions related by molecular symmetry are counted once.
Consider the following statements.
Statement I: Q has two constitutionally different alpha-carbon environments, namely a central methylene environment and a terminal methyl environment.
Statement II: Deprotonation at the central methylene group of Q gives an enolate stabilised by two neighbouring carbonyl groups.
Statement III: R must form two constitutionally distinct enolates because it has an ethyl group on each side of its carbonyl carbon.
Statement IV: P has only one constitutionally distinct alpha-carbon environment because its two methyl groups are equivalent.
Which combination contains only the correct statements?
ⓐ. I, II and IV only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: \(\textbf{Identify distinct alpha environments:}\)
An alpha-carbon is directly bonded to a carbonyl carbon.
Removing an alpha-hydrogen can produce an enolate because the resulting negative charge can be delocalised towards the carbonyl oxygen.
However, a molecule with alpha-carbon atoms on both sides of its carbonyl does not necessarily give two constitutionally different enolates.
Molecular symmetry must be examined before counting the possible enolate connectivities.
\(\textbf{Evaluate the diketone:}\)
In Q, the central \(\mathrm{-CH_2-}\) group lies between two carbonyl groups, whereas the two terminal methyl groups each lie next to only one carbonyl.
The central and terminal alpha environments are therefore constitutionally different, making Statement I correct.
Deprotonation at the central methylene allows resonance stabilisation involving both adjacent carbonyl groups.
That additional delocalisation explains why this site is especially favourable and confirms Statement II.
\(\textbf{Check symmetry in the ordinary ketones:}\)
P has two equivalent methyl groups, so deprotonation on either side gives the same constitutional enolate; Statement IV is correct.
R also has identical ethyl groups on its two sides.
Its two methylene alpha-carbons are symmetry-related and therefore give the same constitutional enolate, not two different ones.
Statement III is consequently incorrect.
The decisive distinction is between counting available alpha-hydrogens and counting non-equivalent sites for enolate formation.
Statements I, II and IV are the correct combination.
713. A chemist wishes to synthesise the tertiary alcohol 2-methylbutan-2-ol:
\[
\mathrm{(CH_3)_2C(OH)CH_2CH_3}
\]
Two Grignard disconnections are proposed.
Route P: Propanone reacts with ethylmagnesium bromide, followed by aqueous acidic work-up.
Route Q: Butan-2-one reacts with methylmagnesium bromide, followed by aqueous acidic work-up.
The products are isolated separately and examined for stereogenicity at the hydroxyl-bearing carbon and ordinary PCC oxidation.
Assume complete nucleophilic addition without rearrangement.
Which statement correctly compares the two routes?
ⓐ. P forms a secondary alcohol, whereas Q forms a tertiary alcohol; only P is oxidised by PCC.
ⓑ. Both routes form the same tertiary alcohol with two identical methyl groups at its hydroxyl-bearing carbon; neither product is stereogenic at that carbon or undergoes ordinary PCC oxidation.
ⓒ. Both routes form different tertiary alcohols because the carbon group introduced by a Grignard reagent must remain distinguishable from every original substituent.
ⓓ. P forms an optically active alcohol, whereas Q produces an aldehyde because methylmagnesium bromide transfers hydrogen instead of methyl.
Correct Answer: Both routes form the same tertiary alcohol with two identical methyl groups at its hydroxyl-bearing carbon; neither product is stereogenic at that carbon or undergoes ordinary PCC oxidation.
Explanation: \(\textbf{Identify the carbon groups required at the target alcohol centre:}\)
The hydroxyl-bearing carbon in the target is attached to two methyl groups and one ethyl group.
It is therefore tertiary.
Any suitable Grignard synthesis must establish these three carbon substituents at the former carbonyl carbon.
\(\textbf{Evaluate Route P:}\)
Propanone has two methyl groups attached to its ketone carbonyl carbon.
Ethylmagnesium bromide supplies an ethyl group.
Following aqueous work-up:
\[
\mathrm{(CH_3)_2CO+C_2H_5MgBr}
\]
\[
\mathrm{\xrightarrow{work-up}(CH_3)_2C(OH)CH_2CH_3}
\]
The product is 2-methylbutan-2-ol.
\(\textbf{Evaluate Route Q:}\)
Butan-2-one contains methyl and ethyl substituents at its carbonyl carbon.
Methylmagnesium bromide introduces another methyl group.
\[
\mathrm{CH_3COCH_2CH_3+CH_3MgBr}
\]
\[
\mathrm{\xrightarrow{work-up}(CH_3)_2C(OH)CH_2CH_3}
\]
The product is again 2-methylbutan-2-ol.
\(\textbf{Examine stereogenicity:}\)
The hydroxyl-bearing carbon has two identical methyl substituents.
Therefore, it does not possess four different substituents.
It is not stereogenic, regardless of which methyl group originated in the Grignard reagent.
\(\textbf{Examine PCC oxidation:}\)
The hydroxyl-bearing carbon is bonded to three carbon groups.
It carries no hydrogen.
Therefore, the product resists ordinary alcohol-to-carbonyl PCC oxidation.
\(\textbf{Interpret the synthetic design:}\)
Different carbonyl compounds and different Grignard reagents can produce the same alcohol if their combined carbon substituents are identical.
The reaction must be analysed from the final carbon connectivity, not simply from the identities of the individual starting reagents.
The same structural result also determines the alcohol class, stereogenicity and ordinary oxidation behaviour.
714. An unknown saturated, acyclic ketone X has molecular formula \(\mathrm{C_5H_{10}O}\).
It gives a positive 2,4-dinitrophenylhydrazine test and a negative characteristic Tollens' test.
It also gives a negative iodoform test.
A separate sample of X is reduced completely with sodium borohydride, followed by aqueous work-up.
The isolated alcohol is dehydrated under conditions favouring the more substituted alkene and excluding skeletal rearrangement.
The principal alkene can exhibit \(E/Z\) geometrical isomerism.
Which statement correctly identifies X and the transformation sequence?
ⓐ. X is pentan-2-one; its reduction gives pentan-2-ol, and dehydration produces only pent-1-ene.
ⓑ. X is pentanal; its reduction gives pentan-1-ol, and direct dehydration produces pent-2-ene.
ⓒ. X is pentan-3-one; its reduction gives pentan-3-ol, and dehydration can produce pent-2-ene with \(E/Z\) geometrical isomerism.
ⓓ. X is 3-methylbutan-2-one; its reduction gives 3-methylbutan-2-ol, and dehydration produces 2-methylbut-2-ene with \(E/Z\) geometrical isomerism.
Correct Answer: X is pentan-3-one; its reduction gives pentan-3-ol, and dehydration can produce pent-2-ene with \(E/Z\) geometrical isomerism.
Explanation: \(\textbf{Interpret the initial carbonyl tests:}\)
A positive 2,4-dinitrophenylhydrazine reaction indicates a suitable aldehyde or ketone carbonyl group.
The negative characteristic Tollens' test supports an ordinary ketone rather than an aldehyde.
Therefore, pentanal is excluded.
\(\textbf{Use the negative iodoform test:}\)
The characteristic positive iodoform reaction requires an appropriate methyl ketone arrangement or another recognised positive structural class.
Pentan-2-one contains the group \(\mathrm{CH_3CO-}\).
3-Methylbutan-2-one also contains this methyl ketone group.
Both would give the characteristic positive iodoform reaction.
Pentan-3-one instead has the structure:
\[
\mathrm{CH_3CH_2COCH_2CH_3}
\]
Its ketone carbonyl carbon is attached to two ethyl groups.
It contains no methyl group directly attached to the carbonyl carbon.
Therefore, its negative iodoform result is consistent with the observations.
\(\textbf{Follow sodium borohydride reduction:}\)
Pentan-3-one undergoes ordinary ketone reduction to the corresponding secondary alcohol.
\[
\mathrm{CH_3CH_2COCH_2CH_3}
\]
\[
\mathrm{\xrightarrow{NaBH_4}CH_3CH_2CH(OH)CH_2CH_3}
\]
The product is pentan-3-ol.
\(\textbf{Follow controlled dehydration:}\)
The hydroxyl-bearing carbon has suitable beta-hydrogens on either side.
Elimination towards either constitutionally equivalent side produces pent-2-ene.
\[
\mathrm{CH_3CH_2CH(OH)CH_2CH_3}
\]
\[
\mathrm{\longrightarrow CH_3CH=CHCH_2CH_3+H_2O}
\]
Each carbon of the resulting double bond has two different substituents.
Therefore, pent-2-ene can exhibit \(E/Z\) geometrical isomerism.
\(\textbf{Interpret the combined identification:}\)
The carbonyl tests first distinguish a non-methyl ketone from aldehydes and methyl ketones.
Reduction then determines the alcohol class.
Dehydration provides an additional structural check through the constitutional alkene and its geometrical-isomerism capability.
The identity of X is therefore supported by three connected stages rather than by a single positive or negative diagnostic test.
715. Two independent \(0.50\,\mathrm{mol}\) samples of the same mixture, containing only propanal and benzaldehyde in identical proportions, are prepared.
Experiment P: Complete oxidation of the first sample with excess Tollens' reagent produces \(1.00\,\mathrm{mol}\) metallic silver.
Experiment Q: Complete reaction of the second sample with excess Fehling's solution produces \(0.20\,\mathrm{mol}\) copper(I) oxide, \(\mathrm{Cu_2O}\).
Assume that both aldehydes reduce two silver(I) ions per molecule, whereas only propanal gives the characteristic positive Fehling's reaction, producing one \(\mathrm{Cu_2O}\) formula unit per molecule.
The aldehydes are subsequently reduced separately and completely using sodium borohydride.
Which statement correctly gives the composition of each original sample and identifies the alcohol obtained from the Fehling-positive component?
ⓐ. \(0.20\,\mathrm{mol}\) propanal and \(0.30\,\mathrm{mol}\) benzaldehyde; propan-1-ol
ⓑ. \(0.30\,\mathrm{mol}\) propanal and \(0.20\,\mathrm{mol}\) benzaldehyde; propan-1-ol
ⓒ. \(0.20\,\mathrm{mol}\) propanal and \(0.30\,\mathrm{mol}\) benzaldehyde; propan-2-ol
ⓓ. \(0.10\,\mathrm{mol}\) propanal and \(0.40\,\mathrm{mol}\) benzaldehyde; propan-1-ol
Correct Answer: \(0.20\,\mathrm{mol}\) propanal and \(0.30\,\mathrm{mol}\) benzaldehyde; propan-1-ol
Explanation: \(\textbf{Identify the aldehyde reactions:}\)
Both propanal and benzaldehyde give positive Tollens' tests under suitable conditions because both contain an aldehyde group. Each aldehyde molecule reduces two silver(I) ions to metallic silver.
Fehling's solution distinguishes the selected compounds because propanal gives a positive characteristic reaction, whereas benzaldehyde ordinarily does not.
\(\textbf{Assign the mixture composition:}\)
Let \(x\) denote the moles of propanal and \(y\) the moles of benzaldehyde in each sample.
The total amount is:
\[
x+y=0.50
\]
\(\textbf{Apply the Tollens' result:}\)
Each aldehyde molecule produces two silver atoms.
\[
n(\mathrm{Ag})=2(x+y)
\]
Substitution gives:
\[
1.00=2(0.50)
\]
The result confirms the total aldehyde content but does not distinguish the individual amounts.
\(\textbf{Apply the Fehling's result:}\)
Only propanal contributes to the stipulated formation of copper(I) oxide.
\[
n(\mathrm{Cu_2O})=x
\]
Given:
\[
x=0.20\,\mathrm{mol}
\]
\(\textbf{Calculate benzaldehyde:}\)
\[
y=0.50-0.20
\]
\[
y=0.30\,\mathrm{mol}
\]
\(\textbf{Identify the alcohol from propanal:}\)
Sodium borohydride reduces the aldehyde carbonyl group of propanal to a primary alcohol group.
\[
\mathrm{CH_3CH_2CHO\xrightarrow[\text{aqueous work-up}]{NaBH_4}CH_3CH_2CH_2OH}
\]
The product is propan-1-ol because the hydroxyl group occupies the terminal carbon of the three-carbon chain.
\(\textbf{Verify the composition:}\)
The \(0.20\,\mathrm{mol}\) propanal and \(0.30\,\mathrm{mol}\) benzaldehyde together produce \(1.00\,\mathrm{mol}\) silver. The propanal alone accounts for \(0.20\,\mathrm{mol}\) copper(I) oxide. Therefore, each sample contains \(0.20\,\mathrm{mol}\) propanal and \(0.30\,\mathrm{mol}\) benzaldehyde, and the Fehling-positive component yields propan-1-ol.
\(\textbf{Final numerical results:}\)
\[
\boxed{n(\mathrm{propanal})=0.20\,\mathrm{mol},\qquad n(\mathrm{benzaldehyde})=0.30\,\mathrm{mol}}
\]
716. Two synthetic routes begin with benzene.
Route P:
Benzene undergoes Friedel–Crafts acylation with ethanoyl chloride and anhydrous aluminium chloride. The isolated aromatic ketone is mononitrated under conditions favouring its ordinary directing effect. The ketone carbonyl is subsequently reduced completely by the Wolff–Kishner method without changing the nitro group.
Route Q:
Benzene undergoes Friedel–Crafts ethylation. The isolated ethylbenzene is then subjected to controlled aromatic mononitration without side-chain oxidation.
Assume that in each route aromatic substitution follows the principal directing influence of the substituent already attached to the ring.
Which statement correctly compares the principal nitroethylbenzene substitution patterns?
ⓐ. Both routes give predominantly 1-ethyl-3-nitrobenzene because nitration always follows the meta rule when the original aromatic starting material is benzene.
ⓑ. Route P favours 1-ethyl-2-nitrobenzene, whereas Route Q favours 1-ethyl-3-nitrobenzene because acyl groups are ortho directors.
ⓒ. Both routes give the same para-substituted product because carbonyl reduction removes the effect of the original ketone before nitration.
ⓓ. Route P favours 1-ethyl-3-nitrobenzene, whereas Route Q favours ortho- and para-nitroethylbenzene products; the order of substitution and carbonyl reduction changes the directing environment during nitration.
Correct Answer: Route P favours 1-ethyl-3-nitrobenzene, whereas Route Q favours ortho- and para-nitroethylbenzene products; the order of substitution and carbonyl reduction changes the directing environment during nitration.
Explanation: \(\textbf{Identify the first intermediate in Route P:}\)
Friedel–Crafts acylation of benzene with ethanoyl chloride produces acetophenone.
\[
\mathrm{C_6H_6\xrightarrow{CH_3COCl/AlCl_3}C_6H_5COCH_3}
\]
The acetyl substituent contains a carbonyl group directly attached to the aromatic ring.
That group withdraws electron density and is meta directing.
\(\textbf{Apply nitration before reduction:}\)
Controlled nitration of acetophenone therefore favours substitution meta to the acetyl group.
The principal nitrated intermediate is 3-nitroacetophenone.
The question specifies that the nitro group survives the subsequent Wolff–Kishner reduction.
\(\textbf{Follow the carbonyl reduction:}\)
\[
\mathrm{ArCOCH_3\xrightarrow{NH_2NH_2/KOH,\ heat}ArCH_2CH_3}
\]
The original acetyl group becomes an ethyl substituent.
Its aromatic attachment position does not change.
Therefore, Route P gives predominantly 1-ethyl-3-nitrobenzene.
\(\textbf{Evaluate Route Q:}\)
Friedel–Crafts ethylation introduces an ethyl group directly.
An ethyl substituent is ordinarily an activating ortho/para director in electrophilic aromatic substitution.
Subsequent nitration therefore favours the corresponding ortho and para positions, with their relative proportions influenced by reaction conditions and steric effects.
\(\textbf{Identify the controlling difference:}\)
In Route P, nitration occurs while the substituent contains a deactivating carbonyl group.
In Route Q, nitration occurs after introduction of the electron-releasing alkyl group.
The subsequent reduction in P does not move the already introduced nitro substituent.
This explains why routes ending with similar functional-group identities can give different aromatic positional isomers.
The key reasoning is the order in which electrophilic substitution and carbonyl deoxygenation occur.
717. Benzaldehyde and acetophenone are subjected to the following independent reactions.
Experiment P: Benzaldehyde alone is treated with concentrated aqueous sodium hydroxide under ordinary Cannizzaro conditions.
Experiment Q: Benzaldehyde reacts with the selectively generated enolate of acetophenone under controlled crossed aldol conditions, followed by complete dehydration.
The isolated product of Q is subsequently subjected to the characteristic iodoform test.
Assume that the crossed aldol reaction proceeds exclusively through acetophenone as the enolate donor and that no self-condensation occurs.
Which statement correctly compares the pathways?
ⓐ. P gives benzyl alcohol and benzoate; Q gives \(\mathrm{C_6H_5COCH=CHC_6H_5}\), which lacks the methyl ketone group originally present in acetophenone.
ⓑ. P gives benzyl alcohol and methanoate; Q gives acetophenone oxime, which is iodoform-positive.
ⓒ. P gives benzoate only; Q gives \(\mathrm{C_6H_5CH_2COCH_3}\), which remains a saturated methyl ketone.
ⓓ. P and Q both give benzyl alcohol because benzaldehyde lacks alpha-hydrogens and cannot accept an enolate.
Correct Answer: P gives benzyl alcohol and benzoate; Q gives \(\mathrm{C_6H_5COCH=CHC_6H_5}\), which lacks the methyl ketone group originally present in acetophenone.
Explanation: \(\textbf{Examine benzaldehyde acting alone:}\)
Benzaldehyde lacks an ordinary enolisable alpha-hydrogen.
It therefore cannot function as the normal enolate donor in a self-aldol reaction.
In concentrated aqueous alkali, it can undergo Cannizzaro disproportionation.
One benzaldehyde-derived carbon is oxidised to benzoate.
Another is reduced to benzyl alcohol.
\[
\mathrm{2C_6H_5CHO+NaOH
\longrightarrow C_6H_5COONa+C_6H_5CH_2OH}
\]
\(\textbf{Identify the enolate donor in Q:}\)
Acetophenone contains an alpha-methyl group adjacent to its ketone carbonyl.
That group can supply an enolate under suitable conditions.
Benzaldehyde still lacks an alpha-hydrogen, but this does not prevent its carbonyl carbon from acting as an electrophilic acceptor.
\(\textbf{Construct the crossed aldol product:}\)
The acetophenone enolate attacks benzaldehyde.
After protonation, the beta-hydroxy ketone is:
\[
\mathrm{C_6H_5COCH_2CH(OH)C_6H_5}
\]
Dehydration produces:
\[
\mathrm{C_6H_5COCH=CHC_6H_5}
\]
The product is commonly called benzalacetophenone or chalcone.
\(\textbf{Evaluate the iodoform structural requirement:}\)
Acetophenone initially contains \(\mathrm{-COCH_3}\).
The methyl carbon participates in forming the new carbon–carbon bond during the aldol reaction.
After dehydration, the carbonyl group is attached to an alkenyl carbon rather than a methyl group.
Consequently, the isolated unsaturated ketone lacks the original characteristic methyl ketone arrangement.
It does not give the ordinary positive iodoform reaction on that basis.
\(\textbf{Interpret the reaction selection:}\)
The absence of alpha-hydrogen makes benzaldehyde a Cannizzaro substrate when reacting alone under suitable conditions.
However, it can still accept an enolate generated from another carbonyl compound.
Therefore, absence of alpha-hydrogen does not prohibit crossed aldol addition as an electrophilic carbonyl partner.
The reaction outcome depends on the complete set of reactants and the specified conditions.
718. Three phenolic compounds are examined:
P: Phenol, \(\mathrm{C_6H_5OH}\).
Q: 4-Nitrophenol.
R: 2,4,6-Trinitrophenol.
For this question, use:
\[
\mathrm{p}K_a(P)=10.0,\qquad
\mathrm{p}K_a(Q)=7.2,\qquad
\mathrm{p}K_a(R)=0.4
\]
\[
\mathrm{p}K_{a1}(\mathrm{H_2CO_3})=6.35
\]
Each compound is separately treated with excess aqueous sodium hydrogencarbonate.
Assume that characteristic carbon dioxide evolution is appreciable when the reacting phenolic acid is substantially stronger than carbonic acid and that no competing reaction occurs.
Which statement correctly predicts the characteristic observations and explains the distinction?
ⓐ. All three release comparable carbon dioxide amounts because each contains one phenolic O–H bond.
ⓑ. Only P releases carbon dioxide because phenoxide is more resonance stabilised than nitrophenoxide.
ⓒ. R readily liberates carbon dioxide under the stated test; P and Q do not, because R's additional nitro substituents greatly increase phenolic acidity.
ⓓ. Q and R release carbon dioxide, whereas P does not, because any single nitro group automatically makes a phenol stronger than carbonic acid.
Correct Answer: R readily liberates carbon dioxide under the stated test; P and Q do not, because R's additional nitro substituents greatly increase phenolic acidity.
Explanation: \(\textbf{Identify the relevant acid–base competition:}\)
Hydrogencarbonate can accept a proton from an acid and produce carbonic acid.
Carbonic acid can then yield carbon dioxide and water.
Whether proton transfer is favourable depends on the relative acid strengths.
\(\textbf{Compare ordinary phenol:}\)
Phenol has a substantially larger \(\mathrm{p}K_a\) than carbonic acid.
It is therefore the weaker acid in the stated comparison.
Its proton transfer to hydrogencarbonate is not appreciably favoured under the stipulated diagnostic criterion.
\(\textbf{Examine 4-nitrophenol:}\)
The para nitro group withdraws electron density and stabilises the phenoxide conjugate base.
Consequently, Q is stronger than ordinary phenol.
However, its supplied \(\mathrm{p}K_a\) remains larger than that of carbonic acid.
Therefore, Q is still the weaker acid in this comparison.
It does not satisfy the stipulated criterion for characteristic hydrogencarbonate effervescence.
\(\textbf{Examine 2,4,6-trinitrophenol:}\)
R contains three strongly electron-withdrawing nitro groups.
The corresponding conjugate base receives substantial electronic stabilisation.
Its very small \(\mathrm{p}K_a\) shows that it is much stronger than carbonic acid.
Proton transfer to hydrogencarbonate is consequently strongly favoured.
\(\textbf{Interpret the conceptual exception:}\)
The statement that phenol does not ordinarily release carbon dioxide with sodium hydrogencarbonate is useful for simple qualitative tests.
It must not be converted into an absolute rule applying to every substituted phenol.
Sufficiently strong electron-withdrawing substituents can substantially increase phenolic acidity.
The correct conclusion depends on comparing the particular acid dissociation constants rather than merely recognising a phenolic hydroxyl group.
719. Three structurally different alcohols undergo dehydration:
P: Butan-1-ol.
Q: Butan-2-ol.
R: 2-Methylpropan-2-ol.
For each alcohol, consider the possibility of a direct elimination pathway that forms a double bond between the original hydroxyl-bearing carbon and an adjacent beta-carbon.
Assume that no carbon-skeleton rearrangement occurs.
The products are examined for their constitutional identities and their capacity to exhibit \(E/Z\) geometrical isomerism.
Which statement correctly evaluates the possible alkene products?
ⓐ. P directly produces but-2-ene, Q produces only but-1-ene, and R produces but-2-ene.
ⓑ. P can directly form but-1-ene; Q can form but-1-ene or but-2-ene; R forms 2-methylpropene. Only but-2-ene among these product structures can exhibit \(E/Z\) isomerism.
ⓒ. P and Q each necessarily produce the same single alkene because they contain four carbon atoms, whereas R cannot dehydrate.
ⓓ. Every alkene formed from the three alcohols can exhibit \(E/Z\) isomerism because dehydration creates a carbon–carbon double bond.
Correct Answer: P can directly form but-1-ene; Q can form but-1-ene or but-2-ene; R forms 2-methylpropene. Only but-2-ene among these product structures can exhibit \(E/Z\) isomerism.
Explanation: \(\textbf{Identify the beta-hydrogens in P:}\)
Butan-1-ol contains hydroxyl on carbon 1.
The directly adjacent beta-carbon is carbon 2.
Removal of water through direct elimination therefore creates a double bond between carbons 1 and 2.
The constitutional product is but-1-ene.
\(\textbf{Examine Q:}\)
Butan-2-ol contains hydroxyl on carbon 2.
Both neighbouring carbons can provide beta-hydrogens.
Elimination towards carbon 1 gives but-1-ene.
Elimination towards carbon 3 gives but-2-ene.
Under suitable conditions, the more substituted but-2-ene is generally favoured.
\(\textbf{Examine R:}\)
2-Methylpropan-2-ol is a tertiary alcohol.
Its hydroxyl-bearing carbon is attached to three equivalent methyl groups.
Elimination involving any suitable adjacent methyl group produces the same constitutional alkene.
That product is 2-methylpropene.
\(\textbf{Evaluate geometrical isomerism:}\)
But-1-ene contains a terminal double-bond carbon bearing two identical hydrogen substituents.
2-Methylpropene contains a double-bond carbon bearing two identical methyl substituents.
Neither can exhibit \(E/Z\) isomerism.
In but-2-ene, each double-bond carbon bears hydrogen and methyl groups.
Those substituents permit distinct geometrical arrangements.
\(\textbf{Interpret the relationship:}\)
The number and location of beta-hydrogens determine which constitutional alkene structures are accessible.
The identities of substituents attached to the final double-bond carbons determine geometrical-isomerism eligibility.
These are related but separate structural questions.
720. An unknown oxygen-containing organic compound X has molecular formula \(\mathrm{C_4H_8O_2}\).
It undergoes complete alkaline hydrolysis to produce an alcohol P and a sodium carboxylate Q.
After acidification, Q gives a carboxylic acid containing three carbon atoms.
The alcohol P does not undergo ordinary PCC oxidation to an aldehyde or ketone.
Assume that X is a simple acyclic ester and that no carbon-skeleton rearrangement occurs.
Which statement correctly assesses the observations?
ⓐ. X must be ethyl ethanoate because it gives a three-carbon acid and a tertiary alcohol.
ⓑ. X must be methyl propanoate because methanol resists ordinary PCC oxidation.
ⓒ. X must be propyl methanoate because methanoic acid contains three carbon atoms after acidification.
ⓓ. The data conflict: a three-carbon acid requires a methanol fragment, which is ordinarily susceptible to oxidation.
Correct Answer: The data conflict: a three-carbon acid requires a methanol fragment, which is ordinarily susceptible to oxidation.
Explanation: \(\textbf{Use the ester carbon balance:}\)
The starting ester contains four carbon atoms.
Its complete alkaline hydrolysis separates an acid-derived fragment from an alcohol-derived fragment.
The acid recovered after acidification is specified to contain three carbon atoms.
Therefore, the alcohol fragment must contain the remaining one carbon atom.
\(\textbf{Identify the necessary ester connectivity:}\)
A three-carbon saturated monocarboxylic acid is propanoic acid.
Its methyl ester is methyl propanoate:
\[
\mathrm{CH_3CH_2COOCH_3}
\]
Alkaline hydrolysis gives:
\[
\mathrm{CH_3CH_2COOCH_3+NaOH}
\]
\[
\mathrm{\longrightarrow CH_3CH_2COONa+CH_3OH}
\]
Acidification of the carboxylate gives propanoic acid.
The alcohol is methanol.
\(\textbf{Check the oxidation observation:}\)
Methanol is a primary alcohol.
It contains hydrogen atoms directly attached to the carbon bearing hydroxyl.
Under suitable controlled oxidation conditions, it can be converted into methanal.
Therefore, methanol does not satisfy the stipulated resistance to ordinary PCC oxidation.
\(\textbf{Eliminate the alternative esters:}\)
Ethyl ethanoate produces a two-carbon acid and a two-carbon alcohol.
Propyl methanoate produces a one-carbon acid and a three-carbon alcohol.
Neither matches the specified acid fragment.
\(\textbf{Identify the inconsistency:}\)
The carbon balance and hydrolysis products uniquely require methyl propanoate among the relevant simple saturated esters.
However, its alcohol fragment contradicts the negative oxidation observation.
The complete experimental description therefore cannot be satisfied under the stated assumptions.
The problem illustrates why an identification must satisfy every reported transformation, not merely one molecular formula and one hydrolysis product.
