401. Ethanol vapour participates in two reversible acid-catalysed reactions under a hypothetical controlled system where all the species listed below are in the gaseous phase.
Reaction I:
\[
\mathrm{2C_2H_5OH(g)\rightleftharpoons C_2H_5OC_2H_5(g)+H_2O(g)}
\]
Reaction II:
\[
\mathrm{C_2H_5OH(g)\rightleftharpoons C_2H_4(g)+H_2O(g)}
\]
The temperature is held constant, and the ideal-gas approximation is applicable.
Which statement correctly predicts the direct equilibrium effect of increasing the total pressure by decreasing the vessel volume?
ⓐ. Both equilibria shift towards products because dehydration always removes water.
ⓑ. Reaction I shifts towards reactants, while Reaction II is unaffected because both produce one molecule of water.
ⓒ. Reaction I is unaffected by the change in the number of gaseous molecules, whereas Reaction II is favoured in the reverse direction.
ⓓ. Reaction I shifts towards products, while Reaction II shifts towards products because the acid catalyst is regenerated.
Correct Answer: Reaction I is unaffected by the change in the number of gaseous molecules, whereas Reaction II is favoured in the reverse direction.
Explanation: \(\textbf{Separate chemical equilibrium from catalyst action:}\)
Both reactions are acid catalysed, but the catalyst does not determine the direction of the equilibrium shift caused by a pressure change.
The relevant principle is the difference in the number of gaseous molecules between reactant and product sides.
The question explicitly places every species in the gaseous phase.
\(\textbf{Evaluate Reaction I:}\)
Two ethanol molecules appear on the reactant side.
The product side contains one ether molecule and one water molecule.
Therefore, both sides contain two gaseous molecules per stoichiometric reaction event.
\[
\Delta n_{\mathrm{gas},I}=2-2=0
\]
For ideal gases, compression at constant temperature does not favour one side of this equilibrium solely through a difference in gaseous mole count.
\(\textbf{Evaluate Reaction II:}\)
One ethanol molecule appears on the reactant side.
The product side contains one ethene molecule and one water molecule.
\[
\Delta n_{\mathrm{gas},II}=2-1=+1
\]
The forward reaction increases the number of gaseous molecules.
Compression therefore favours the side containing fewer gaseous molecules, namely the ethanol side.
\(\textbf{Interpret the selectivity consequence:}\)
In this idealised comparison, increasing pressure directly opposes alkene-forming dehydration more than ether formation through the mole-count effect.
This does not establish that ether formation must become the only reaction.
Actual product distribution may also depend on temperature, reaction rates and the activity of the species involved.
The question isolates the equilibrium contribution from pressure while holding those other conditions fixed.
402. A mixture contains only ethanol and propan-2-ol, with a total amount of \(0.45\,\mathrm{mol}\).
Both alcohols are oxidised completely using acidified potassium dichromate under conditions converting ethanol into ethanoic acid and propan-2-ol into propanone.
Exactly \(0.24\,\mathrm{mol}\) dichromate ions are reduced to chromium(III), and no other reducing species are present.
The organic products are separated. The carboxylic acid fraction is treated with excess sodium hydrogencarbonate, while the ketone fraction undergoes a complete iodoform reaction.
Assume quantitative transformations.
What are the respective amounts of carbon dioxide and iodoform produced?
ⓐ. \(0.18\,\mathrm{mol}\) and \(0.27\,\mathrm{mol}\)
ⓑ. \(0.24\,\mathrm{mol}\) and \(0.21\,\mathrm{mol}\)
ⓒ. \(0.30\,\mathrm{mol}\) and \(0.15\,\mathrm{mol}\)
ⓓ. \(0.27\,\mathrm{mol}\) and \(0.18\,\mathrm{mol}\)
Correct Answer: \(0.27\,\mathrm{mol}\) and \(0.18\,\mathrm{mol}\)
Explanation: \(\textbf{Identify the different oxidation requirements:}\)
Complete oxidation of ethanol to ethanoic acid is a four-electron oxidation per alcohol molecule.
Oxidation of propan-2-ol to propanone is a two-electron oxidation per alcohol molecule.
Each dichromate ion accepts six electrons when reduced from chromium(VI) to chromium(III).
\(\textbf{Represent the unknown alcohol amounts:}\)
Let:
\[
x=n(\text{ethanol})
\]
\[
y=n(\text{propan-2-ol})
\]
The mixture composition gives:
\[
x+y=0.45
\]
\(\textbf{Establish the electron balance:}\)
Electrons accepted by dichromate:
\[
n(e^-)_{\mathrm{accepted}}=6(0.24)
\]
\[
n(e^-)_{\mathrm{accepted}}=1.44\,\mathrm{mol}
\]
Electrons released by the alcohols:
\[
n(e^-)_{\mathrm{released}}=4x+2y
\]
Therefore:
\[
4x+2y=1.44
\]
\(\textbf{Solve for the alcohol composition:}\)
Using:
\[
y=0.45-x
\]
Substitute:
\[
4x+2(0.45-x)=1.44
\]
\[
4x+0.90-2x=1.44
\]
\[
2x=0.54
\]
\[
x=0.27\,\mathrm{mol}
\]
Consequently:
\[
y=0.45-0.27
\]
\[
y=0.18\,\mathrm{mol}
\]
\(\textbf{Calculate carbon dioxide formation:}\)
Ethanol produces ethanoic acid in a \(1:1\) molar ratio.
Each ethanoic acid molecule gives one carbon dioxide molecule with excess sodium hydrogencarbonate.
\[
n(\mathrm{CO_2})=x
\]
\[
n(\mathrm{CO_2})=0.27\,\mathrm{mol}
\]
\(\textbf{Calculate iodoform formation:}\)
Propan-2-ol produces propanone in a \(1:1\) molar ratio.
Propanone contains a methyl ketone group and yields one iodoform molecule per ketone molecule under complete haloform conditions.
\[
n(\mathrm{CHI_3})=y
\]
\[
n(\mathrm{CHI_3})=0.18\,\mathrm{mol}
\]
\(\textbf{Check the overall balances:}\)
The recovered alcohol amounts sum to the original \(0.45\,\mathrm{mol}\).
Their electron contributions also reproduce the measured dichromate requirement.
The two final tests distinguish the carboxylic acid and methyl ketone products independently.
\(\textbf{Final answer:}\)
\[
\boxed{\left(n(\mathrm{CO_2}),n(\mathrm{CHI_3})\right)
=(0.27,\ 0.18)\,\mathrm{mol}}
\]
403. Two isotopically labelled samples of propan-2-ol are passed separately over heated copper at approximately \(573\,\mathrm{K}\).
P: \(\mathrm{CH_3CH(OH)CH_2D}\), with deuterium on a terminal methyl carbon.
Q: \(\mathrm{CH_3CD(OH)CH_3}\), with deuterium directly attached to the hydroxyl-bearing carbon.
Assume complete ordinary secondary-alcohol dehydrogenation, no isotope exchange and no skeletal rearrangement.
Which statement correctly identifies the isotopic locations after reaction?
ⓐ. P gives deuterated propanone and ordinary \(\mathrm{H_2}\); Q gives unlabelled propanone and \(\mathrm{HD}\).
ⓑ. P gives unlabelled propanone and \(\mathrm{HD}\); Q gives deuterated propanone and ordinary \(\mathrm{H_2}\).
ⓒ. Both P and Q give deuterated propanone because heated copper removes only the hydroxyl hydrogen.
ⓓ. Both give ordinary propanone and \(\mathrm{D_2}\) because the carbon-bound deuterium atoms are removed preferentially.
Correct Answer: P gives deuterated propanone and ordinary \(\mathrm{H_2}\); Q gives unlabelled propanone and \(\mathrm{HD}\).
Explanation: \(\textbf{Identify the bonds broken during dehydrogenation:}\)
A secondary alcohol can undergo dehydrogenation over heated copper to form a ketone.
The net transformation removes one hydrogen isotope from the hydroxyl group and another from the hydroxyl-bearing carbon.
The neighbouring methyl carbons remain attached to the carbonyl carbon.
\(\textbf{Track the label in P:}\)
P contains deuterium at a terminal carbon rather than at the carbon bearing hydroxyl.
The hydroxyl-bearing carbon still possesses ordinary hydrogen.
Therefore, ketone formation removes ordinary hydrogen from that carbon and ordinary hydrogen from oxygen.
\[
\mathrm{CH_3CH(OH)CH_2D}
\]
\[
\mathrm{\xrightarrow{Cu,\ 573\,K}
CH_3COCH_2D+H_2}
\]
The deuterium label remains in the ketone product.
The evolved gas is ordinary molecular hydrogen.
\(\textbf{Track the label in Q:}\)
Q contains deuterium directly attached to the hydroxyl-bearing carbon.
Formation of the ketone requires removal of that carbon-bound deuterium.
The hydroxyl group contains ordinary hydrogen.
\[
\mathrm{CH_3CD(OH)CH_3}
\]
\[
\mathrm{\xrightarrow{Cu,\ 573\,K}
CH_3COCH_3+HD}
\]
The carbonyl product is unlabelled propanone.
The gaseous product contains one hydrogen atom and one deuterium atom.
\(\textbf{Interpret the experiment:}\)
Both samples undergo the same functional-group transformation.
Their different isotope locations reveal which carbon–hydrogen bond is broken.
The experiment therefore distinguishes a hydrogen on the hydroxyl-bearing carbon from hydrogen atoms on neighbouring carbons.
A carbon-bound isotope does not necessarily leave during alcohol oxidation simply because it belongs to the same molecule.
404. Consider the NCERT-described mechanism of acid-catalysed ethanol dehydration.
Under a simplified kinetic model, protonation of ethanol is a rapid pre-equilibrium:
\[
\mathrm{C_2H_5OH+H^+\rightleftharpoons C_2H_5OH_2^+}
\]
The subsequent departure of water from protonated ethanol is the rate-determining step:
\[
\mathrm{C_2H_5OH_2^+\xrightarrow{slow}C_2H_5^++H_2O}
\]
Assume that the protonation equilibrium constant and the rate constant for water departure remain fixed, that protonated ethanol is present only in a small equilibrium fraction, and that the simplified model applies under both compared conditions.
In Experiment P, the concentrations are \([\mathrm{C_2H_5OH}]=0.20\,\mathrm{mol\,L^{-1}}\) and \([\mathrm{H^+}]=0.050\,\mathrm{mol\,L^{-1}}\).
In Experiment Q, the corresponding concentrations are \(0.30\,\mathrm{mol\,L^{-1}}\) and \(0.150\,\mathrm{mol\,L^{-1}}\).
What is the ratio of the initial dehydration rates, \(r_Q/r_P\), predicted by this model?
ⓐ. \(1.5\)
ⓑ. \(3.0\)
ⓒ. \(4.5\)
ⓓ. \(9.0\)
Correct Answer: \(4.5\)
Explanation: \(\textbf{Relate the slow step to its reacting intermediate:}\)
The specified rate-determining step is departure of water from protonated ethanol.
Under the stated elementary-step model:
\[
r=k[\mathrm{C_2H_5OH_2^+}]
\]
The concentration of protonated ethanol is not independently supplied.
It must be related to the preceding protonation equilibrium.
\(\textbf{Apply the rapid pre-equilibrium:}\)
For the protonation equilibrium:
\[
K=\frac{[\mathrm{C_2H_5OH_2^+}]}{[\mathrm{C_2H_5OH}][\mathrm{H^+}]}
\]
Rearranging:
\[
[\mathrm{C_2H_5OH_2^+}]
=K[\mathrm{C_2H_5OH}][\mathrm{H^+}]
\]
Substituting into the slow-step rate expression:
\[
r=kK[\mathrm{C_2H_5OH}][\mathrm{H^+}]
\]
\(\textbf{Form the experiment-to-experiment ratio:}\)
The constants \(k\) and \(K\) are specified to remain unchanged.
Therefore:
\[
\frac{r_Q}{r_P}
=\frac{[\mathrm{C_2H_5OH}]_Q[\mathrm{H^+}]_Q}
{[\mathrm{C_2H_5OH}]_P[\mathrm{H^+}]_P}
\]
\(\textbf{Insert the supplied concentrations:}\)
\[
\frac{r_Q}{r_P}
=\frac{(0.30)(0.150)}{(0.20)(0.050)}
\]
\[
=\frac{0.0450}{0.0100}
\]
\[
=4.5
\]
\(\textbf{Interpret the kinetic conclusion:}\)
Ethanol concentration increases by a factor of \(1.5\).
Hydrogen-ion concentration increases by a factor of \(3\).
The product of these factors is \(4.5\).
This rate dependence follows from the explicitly supplied pre-equilibrium and rate-determining-step assumptions.
It should not be treated as an experimentally universal rate law for every alcohol dehydration mechanism.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{r_Q}{r_P}=4.5}
\]
405. Two isomeric secondary alcohols are subjected independently to oxidation with PCC and acid-catalysed dehydration.
P: Pentan-2-ol.
Q: Pentan-3-ol.
Assume that dehydration of each produces pent-2-ene as the principal alkene and that skeletal rearrangements do not occur.
The principal alkenes are separately subjected to reductive ozonolysis.
The PCC oxidation products are independently examined using the characteristic iodoform test.
Which statement correctly compares the outcomes?
ⓐ. Both alcohols give the same principal alkene and the same ozonolysis products, but only the PCC oxidation product from P gives a positive iodoform test.
ⓑ. Both alcohols give the same ketone after PCC oxidation because their dehydration products are identical.
ⓒ. P gives methanal and butanone on ozonolysis, whereas Q gives ethanal and propanal; both oxidation products are iodoform-positive.
ⓓ. Only Q can form pent-2-ene because P has a hydroxyl group at carbon 2 rather than carbon 3.
Correct Answer: Both alcohols give the same principal alkene and the same ozonolysis products, but only the PCC oxidation product from P gives a positive iodoform test.
Explanation: \(\textbf{Compare the alcohol structures:}\)
Pentan-2-ol and pentan-3-ol are both secondary alcohols.
Their hydroxyl groups occupy different positions in the same five-carbon chain.
Consequently, oxidation preserves different carbonyl positions.
\(\textbf{PCC oxidation of P:}\)
\[
\mathrm{CH_3CH(OH)CH_2CH_2CH_3}
\]
\[
\mathrm{\xrightarrow{PCC}
CH_3COCH_2CH_2CH_3}
\]
The product is pentan-2-one.
Its carbonyl carbon is bonded directly to a methyl group.
Therefore, it is a methyl ketone and gives a positive iodoform test.
\(\textbf{PCC oxidation of Q:}\)
\[
\mathrm{CH_3CH_2CH(OH)CH_2CH_3}
\]
\[
\mathrm{\xrightarrow{PCC}
CH_3CH_2COCH_2CH_3}
\]
The product is pentan-3-one.
Both groups attached directly to its carbonyl carbon are ethyl groups.
It therefore does not give the characteristic positive iodoform reaction.
\(\textbf{Dehydration of the two alcohols:}\)
P can form pent-2-ene by elimination towards carbon 3.
Q can form pent-2-ene by elimination towards either constitutionally equivalent side of carbon 3.
The question specifies this common alkene as the principal product in each experiment.
\(\textbf{Apply reductive ozonolysis:}\)
Pent-2-ene has the structure:
\[
\mathrm{CH_3CH=CHCH_2CH_3}
\]
Cleavage of its carbon–carbon double bond gives:
\[
\mathrm{CH_3CHO+CH_3CH_2CHO}
\]
The products are ethanal and propanal.
Therefore, the identical principal dehydration products cannot distinguish P from Q.
Their PCC-derived ketones, however, remain constitutionally distinct.
The iodoform test provides the additional structural evidence needed to distinguish the starting alcohols.
406. Two diols undergo complete intramolecular dehydration under suitably selected acid-catalysed conditions favouring cyclic ether formation without rearrangement.
P: Butane-1,4-diol, \(\mathrm{HO(CH_2)_4OH}\).
Q: Pentane-1,5-diol, \(\mathrm{HO(CH_2)_5OH}\).
The cyclic products are isolated before further reactions.
Separate equivalent samples of the original diols and their corresponding cyclic ethers are treated with excess sodium metal under suitable anhydrous conditions.
Which statement correctly compares ring formation and the sodium-metal test?
ⓐ. P forms a four-membered cyclic ether and Q forms a five-membered cyclic ether; both cyclic products retain one hydroxyl group.
ⓑ. Both products retain two hydroxyl groups because intramolecular dehydration changes only carbon–carbon bonds.
ⓒ. Both P and Q form six-membered cyclic ethers because oxygen is excluded from ring-size counting.
ⓓ. P gives a five-membered ether ring and Q a six-membered ring; the diols evolve \(\mathrm{H_2}\) with sodium, unlike the cyclised ethers.
Correct Answer: P gives a five-membered ether ring and Q a six-membered ring; the diols evolve \(\mathrm{H_2}\) with sodium, unlike the cyclised ethers.
Explanation: \(\textbf{Identify the two terminal hydroxyl groups:}\)
Both starting compounds contain primary alcohol groups at opposite ends of a saturated carbon chain.
Under the specified intramolecular dehydration conditions, one terminal oxygen can become the ether oxygen connecting the two ends of the chain.
Formation of the new carbon–oxygen bond is accompanied by overall elimination of water.
\(\textbf{Determine the ring obtained from P:}\)
Butane-1,4-diol contains four carbon atoms between its terminal oxygen-bearing positions.
Cyclisation incorporates all four carbon atoms and one oxygen atom into a ring.
The product is tetrahydrofuran, a five-membered cyclic ether.
\[
\mathrm{HO(CH_2)_4OH
\xrightarrow{H^+,\ heat}
(CH_2)_4O+H_2O}
\]
The ring formula represents the oxygen as one member of the five-membered ring.
\(\textbf{Determine the ring obtained from Q:}\)
Pentane-1,5-diol contains five carbon atoms.
Its corresponding cyclic ether contains those five carbon atoms and one oxygen atom.
The product is tetrahydropyran, a six-membered cyclic ether.
\(\textbf{Compare sodium-metal reactivity:}\)
Each original diol contains two oxygen–hydrogen bonds.
Both can react with sodium metal to form the corresponding alkoxide functionalities and release hydrogen.
One mole of either diol can therefore release one mole of hydrogen under complete reaction.
In contrast, complete cyclic ether formation removes the original free hydroxyl groups.
The isolated cyclic ethers contain ether oxygen but no ordinary hydroxyl hydrogen.
They therefore do not undergo the characteristic alcohol–sodium hydrogen-evolution reaction.
This example connects ring-size determination with the change from alcohol functionality to ether functionality.
407. Two cyclic alcohols have the molecular formula \(\mathrm{C_5H_{10}O}\).
P: Cyclopentanol.
Q: 1-Methylcyclobutan-1-ol.
They are separately subjected to controlled PCC oxidation and heated-copper treatment at approximately \(573\,\mathrm{K}\).
For the heated-copper experiments, assume characteristic secondary-alcohol dehydrogenation for P and tertiary-alcohol dehydration for Q, without skeletal rearrangement.
Which statement correctly compares the products?
ⓐ. P and Q both form cyclopentanone under PCC oxidation because their molecular formulae are identical.
ⓑ. P resists oxidation because its hydroxyl group is cyclic, while Q forms methylcyclobutanone because it contains a methyl substituent.
ⓒ. P yields cyclopentanone by dehydrogenation; Q cannot form a ketone by dehydrogenation at its OH-bearing carbon and instead dehydrates.
ⓓ. Both undergo ordinary carbonyl-forming dehydrogenation because their hydroxyl-bearing carbons are bonded to ring atoms.
Correct Answer: P yields cyclopentanone by dehydrogenation; Q cannot form a ketone by dehydrogenation at its OH-bearing carbon and instead dehydrates.
Explanation: \(\textbf{Identify the oxygen-bearing carbon in P:}\)
Cyclopentanol contains hydroxyl on a saturated ring carbon.
That carbon is bonded to two neighbouring ring carbons and one hydrogen.
It is therefore a secondary alcohol.
\(\textbf{Predict carbonyl formation from P:}\)
Secondary alcohols can undergo oxidation to ketones.
PCC oxidation converts cyclopentanol into cyclopentanone.
Under suitable heated-copper conditions, secondary-alcohol dehydrogenation also produces cyclopentanone with liberation of molecular hydrogen.
The five-membered ring is preserved.
\(\textbf{Identify the oxygen-bearing carbon in Q:}\)
In 1-methylcyclobutan-1-ol, the hydroxyl-bearing ring carbon is bonded to two neighbouring ring carbons and a methyl group.
It therefore has three carbon attachments and no directly attached hydrogen.
Consequently, Q is a tertiary alcohol.
\(\textbf{Evaluate Q under PCC conditions:}\)
Ordinary tertiary-alcohol oxidation to a ketone would require a carbon-bound hydrogen at the hydroxyl-bearing carbon.
Q lacks this hydrogen.
Therefore, it resists the characteristic carbonyl-forming oxidation under the stated conditions.
\(\textbf{Evaluate Q over heated copper:}\)
The question specifies tertiary-alcohol dehydration without skeletal rearrangement.
The reaction removes water and introduces a carbon–carbon double bond using an available beta-hydrogen.
This transformation differs from secondary-alcohol dehydrogenation.
\(\textbf{Interpret the structural comparison:}\)
The identical molecular formulae do not establish identical alcohol classes.
The ring carbon bearing hydroxyl in P is secondary, whereas that in Q is tertiary because it carries an additional methyl substituent.
The decisive classification therefore comes from counting carbon neighbours of the hydroxyl-bearing carbon.
408. A mixture contains a secondary alcohol P and a tertiary alcohol Q.
Both undergo rapid formation of a common family of carbocation-derived intermediates under suitably strong acidic conditions.
In one experiment, the reaction medium contains a high effective concentration of chloride nucleophile and the temperature is kept relatively low.
In a second experiment, the medium contains a poorly nucleophilic acid catalyst and is heated sufficiently to favour alkene formation.
Assume each reaction follows a pathway permitting competition between nucleophilic capture and beta-proton elimination.
Which statement is most defensible?
ⓐ. Increasing chloride availability must increase the intrinsic rate of every initial carbocation-forming event by the same factor.
ⓑ. The first conditions favour substitution and the hotter, weakly nucleophilic conditions favour elimination, even if initial carbocation formation is similar.
ⓒ. Both experiments must produce only alkyl chlorides because protonated alcohols always react exclusively by substitution.
ⓓ. Both experiments must produce only alkenes because the presence of a beta-hydrogen prevents chloride attack.
Correct Answer: The first conditions favour substitution and the hotter, weakly nucleophilic conditions favour elimination, even if initial carbocation formation is similar.
Explanation: \(\textbf{Identify the shared activation step:}\)
An alcohol hydroxyl group is a poor leaving group under ordinary conditions.
Suitable acid protonation converts it into a group capable of departing as water.
If a carbocation-like intermediate is generated, its subsequent reaction depends on the available pathways.
\(\textbf{Nucleophile-rich conditions:}\)
A high effective chloride concentration provides opportunities for nucleophilic capture.
Under suitable conditions, chloride attack produces an alkyl chloride.
Relatively low temperature can also help suppress some competing elimination pathways.
The question therefore describes conditions compatible with substitution.
\(\textbf{Heated poorly nucleophilic conditions:}\)
A medium containing a suitable acid catalyst but lacking an effective external nucleophile can favour loss of a beta-proton.
This generates a carbon–carbon double bond.
Heating can favour dehydration under the stated reaction environment.
\(\textbf{Distinguish reaction rate from product distribution:}\)
The rate of initial ionisation or water departure and the relative rates of later intermediate-capture pathways are not automatically the same quantity.
Changing chloride availability may significantly affect which product forms without requiring an identical proportional change in the initial carbocation-forming rate.
\(\textbf{Recognise the limitations of the comparison:}\)
The exact competition also depends on substrate structure, solvent and experimental conditions.
The question does not establish a universal product ratio.
It asks which qualitative interpretation is consistent with the specified pathways.
The defensible conclusion is that substitution and elimination may compete, and changing the reaction environment can alter their relative importance.
409. A mixture contains \(0.20\,\mathrm{mol}\) propan-1-ol and \(0.30\,\mathrm{mol}\) propan-2-ol.
The alcohols undergo separate quantitative reactions over heated copper at approximately \(573\,\mathrm{K}\).
The carbonyl products are combined and divided into two equivalent samples, each containing the entire product composition stated above.
Sample I undergoes complete reaction with excess 2,4-dinitrophenylhydrazine.
Sample II undergoes complete characteristic Tollens' oxidation.
Assume one carbonyl group consumes one molecule of 2,4-dinitrophenylhydrazine and each aldehyde reduces two silver(I) ions.
What are the amounts of 2,4-dinitrophenylhydrazine consumed and metallic silver produced, respectively?
ⓐ. \(0.50\,\mathrm{mol}\) and \(0.40\,\mathrm{mol}\)
ⓑ. \(0.50\,\mathrm{mol}\) and \(1.00\,\mathrm{mol}\)
ⓒ. \(0.20\,\mathrm{mol}\) and \(0.40\,\mathrm{mol}\)
ⓓ. \(0.30\,\mathrm{mol}\) and \(0.60\,\mathrm{mol}\)
Correct Answer: \(0.50\,\mathrm{mol}\) and \(0.40\,\mathrm{mol}\)
Explanation: \(\textbf{Identify the heated-copper products:}\)
Propan-1-ol is primary and undergoes dehydrogenation to propanal.
Propan-2-ol is secondary and undergoes dehydrogenation to propanone.
\[
\mathrm{CH_3CH_2CH_2OH
\xrightarrow{Cu,\ 573\,K}
CH_3CH_2CHO+H_2}
\]
\[
\mathrm{CH_3CH(OH)CH_3
\xrightarrow{Cu,\ 573\,K}
CH_3COCH_3+H_2}
\]
\(\textbf{Determine carbonyl amounts:}\)
Each alcohol produces one carbonyl compound molecule.
Therefore:
\[
n(\text{propanal})=0.20\,\mathrm{mol}
\]
\[
n(\text{propanone})=0.30\,\mathrm{mol}
\]
The total carbonyl amount is:
\[
n(\text{carbonyl groups})=0.20+0.30
\]
\[
n(\text{carbonyl groups})=0.50\,\mathrm{mol}
\]
\(\textbf{Sample I — Carbonyl derivative formation:}\)
Both aldehyde and ketone carbonyl groups form suitable 2,4-dinitrophenylhydrazone derivatives.
Each carbonyl group consumes one molecule of the reagent.
\[
n(\text{2,4-DNPH})=0.50\,\mathrm{mol}
\]
\(\textbf{Sample II — Silver formation:}\)
Propanal is an aldehyde and reduces Tollens' reagent.
Propanone is an ordinary ketone and does not undergo the characteristic positive Tollens' reaction.
One mole of aldehyde reduces two moles of silver(I) ions.
\[
n(\mathrm{Ag})=2n(\text{propanal})
\]
\[
n(\mathrm{Ag})=2(0.20)
\]
\[
n(\mathrm{Ag})=0.40\,\mathrm{mol}
\]
\(\textbf{Compare the analytical information:}\)
The 2,4-dinitrophenylhydrazine reaction counts all suitable carbonyl groups in the mixture.
The Tollens' reaction selectively measures the aldehyde contribution under the stated conditions.
The two samples are independent, so conversion in one does not consume the carbonyl compounds used in the other.
\(\textbf{Final answer:}\)
\[
\boxed{\left(n(\text{2,4-DNPH}),n(\mathrm{Ag})\right)
=(0.50,\ 0.40)\,\mathrm{mol}}
\]
410. Two primary alcohols are subjected to acid-catalysed dehydration:
P: 2,2-Dimethylpropan-1-ol, \(\mathrm{(CH_3)_3CCH_2OH}\).
Q: 3,3-Dimethylbutan-1-ol, \(\mathrm{(CH_3)_3CCH_2CH_2OH}\).
The question concerns only direct ordinary beta-elimination without skeletal rearrangement.
Which statement correctly compares their structural eligibility?
ⓐ. Both are ineligible because every primary alcohol with a quaternary carbon elsewhere in its structure lacks beta-hydrogen.
ⓑ. P is eligible but Q is not, because a beta-carbon must be bonded to exactly three carbon atoms.
ⓒ. Neither can undergo ordinary dehydration because primary alcohols cannot form carbon–carbon double bonds.
ⓓ. P lacks a beta-hydrogen on its carbon adjacent to the hydroxyl-bearing carbon, whereas Q possesses beta-hydrogens and can form a terminal alkene through direct elimination.
Correct Answer: P lacks a beta-hydrogen on its carbon adjacent to the hydroxyl-bearing carbon, whereas Q possesses beta-hydrogens and can form a terminal alkene through direct elimination.
Explanation: \(\textbf{Classify both alcohols:}\)
P and Q contain terminal hydroxyl-bearing methylene carbons.
Both are therefore primary alcohols.
Their different dehydration behaviour does not result from one of them being secondary or tertiary.
\(\textbf{Locate the beta-carbon in P:}\)
The hydroxyl-bearing carbon is directly attached to a quaternary carbon.
That adjacent carbon is bonded to three methyl groups and the hydroxymethyl carbon.
It possesses no hydrogen atom.
Therefore, there is no ordinary beta-hydrogen available for direct alkene-forming elimination from P.
\(\textbf{Locate the beta-carbon in Q:}\)
Q contains the sequence \(\mathrm{-C(CH_3)_3-CH_2-CH_2OH}\).
The hydroxyl-bearing carbon is the terminal \(\mathrm{CH_2OH}\) carbon.
Its neighbouring beta-carbon is a methylene group containing two hydrogen atoms.
Direct elimination can therefore form a double bond between these two terminal side-chain carbons.
\[
\mathrm{(CH_3)_3CCH_2CH_2OH}
\]
\[
\mathrm{\xrightarrow{H^+,\ heat}
(CH_3)_3CCH=CH_2+H_2O}
\]
The product is a terminal alkene.
\(\textbf{Interpret the structural distinction:}\)
Both compounds contain a quaternary carbon.
However, only P has that carbon directly adjacent to the hydroxyl-bearing carbon.
Q contains an intervening methylene group that supplies the required beta-hydrogen.
Therefore, the relevant structural criterion is the substitution and hydrogen content of the immediate beta-carbon.
The question excludes rearrangement, so indirect alkene formation through carbon-skeleton migration is not included.
411. Three proposed alcohol-to-alkene transformations are examined.
P: Ethanol to ethene.
Q: Cyclohexanol to cyclohexene.
R: Benzyl alcohol to styrene.
All are proposed to occur through ordinary intramolecular dehydration without introduction of additional carbon atoms or carbon–carbon skeletal rearrangement.
Which statement correctly evaluates the proposals?
ⓐ. P and Q are feasible; R is not, because dehydration of benzyl alcohol cannot supply the additional carbon atom required for styrene.
ⓑ. Only P is feasible because cyclic alcohols cannot undergo beta-elimination.
ⓒ. All are feasible because phenyl groups can supply an additional carbon atom during dehydration.
ⓓ. Q and R are feasible, but P is not because ethanol is a primary alcohol.
Correct Answer: P and Q are feasible; R is not, because dehydration of benzyl alcohol cannot supply the additional carbon atom required for styrene.
Explanation: \(\textbf{Apply carbon-atom conservation:}\)
Ordinary dehydration removes the elements of water from an alcohol.
It does not introduce an additional carbon atom.
Therefore, the carbon count of the organic product must match that of the starting alcohol unless some other carbon-containing reagent or skeletal transformation is explicitly involved.
\(\textbf{Evaluate P:}\)
Ethanol contains two carbon atoms and a suitable beta-carbon bearing hydrogen.
It can undergo acid-catalysed dehydration to ethene.
\[
\mathrm{CH_3CH_2OH
\xrightarrow{H^+,\ heat}
CH_2=CH_2+H_2O}
\]
Both reactant and organic product contain two carbon atoms.
\(\textbf{Evaluate Q:}\)
Cyclohexanol contains six ring carbon atoms.
The hydroxyl-bearing carbon has adjacent ring carbons containing beta-hydrogens.
Dehydration can form a carbon–carbon double bond within the ring.
\[
\mathrm{Cyclohexanol
\xrightarrow{H^+,\ heat}
Cyclohexene+H_2O}
\]
The six-carbon ring is retained.
\(\textbf{Evaluate R:}\)
Benzyl alcohol has the molecular structure \(\mathrm{C_6H_5CH_2OH}\), containing seven carbon atoms.
Styrene has the structure \(\mathrm{C_6H_5CH=CH_2}\), containing eight carbon atoms.
The proposed conversion therefore requires an additional carbon atom.
Ordinary intramolecular dehydration cannot accomplish this increase.
Moreover, benzyl alcohol does not possess the ordinary saturated beta-carbon arrangement needed to form a two-carbon side-chain alkene directly.
The proposal is consequently invalid under the stated restrictions.
The question requires checking both beta-hydrogen availability and conservation of the organic carbon skeleton.
412. Two experiments are performed on the same optically active secondary alcohol, \((R)\)-hexan-3-ol.
Experiment I: Conversion to hexan-3-one using PCC, followed by addition of methylmagnesium bromide and aqueous work-up.
Experiment II: Direct conversion to the corresponding alkyl bromide using a pathway specified to involve ideal backside displacement at the hydroxyl-bearing carbon, without rearrangement.
Assume all reagents are achiral and that no additional stereogenic centres are introduced elsewhere.
Which statement correctly compares the products?
ⓐ. Experiment I necessarily retains the original absolute configuration, whereas Experiment II gives racemic bromide.
ⓑ. Experiment I gives a tertiary alcohol whose hydroxyl-bearing carbon is not stereogenic because it carries two identical groups; Experiment II retains the starting tetrahedral arrangement.
ⓒ. Experiment I gives a tertiary alcohol with a newly formed stereogenic carbon and can produce a racemic mixture; Experiment II gives a substitution product with inversion at the original stereogenic carbon.
ⓓ. Both products are achiral because oxidation or substitution of any secondary alcohol removes its carbon skeleton.
Correct Answer: Experiment I gives a tertiary alcohol with a newly formed stereogenic carbon and can produce a racemic mixture; Experiment II gives a substitution product with inversion at the original stereogenic carbon.
Explanation: \(\textbf{Identify the starting stereogenic carbon:}\)
Hexan-3-ol contains a hydroxyl-bearing carbon attached to hydrogen, hydroxyl, ethyl and propyl groups.
These four substituents are different.
Therefore, the specified starting alcohol is optically active.
\(\textbf{Experiment I — Destroy the original centre:}\)
PCC oxidises the secondary alcohol to hexan-3-one.
\[
\mathrm{CH_3CH_2CH(OH)CH_2CH_2CH_3}
\]
\[
\mathrm{\xrightarrow{PCC}
CH_3CH_2COCH_2CH_2CH_3}
\]
The original stereogenic carbon becomes a trigonal-planar carbonyl carbon.
Its original configuration is consequently lost.
\(\textbf{Create a tertiary alcohol with Grignard reagent:}\)
Methylmagnesium bromide introduces a methyl group at the ketone carbonyl carbon.
After aqueous work-up, the former carbonyl carbon is bonded to hydroxyl, methyl, ethyl and propyl groups.
\[
\mathrm{CH_3CH_2COCH_2CH_2CH_3}
\]
\[
\mathrm{\xrightarrow[\text{work-up}]{CH_3MgBr}
CH_3CH_2C(OH)(CH_3)CH_2CH_2CH_3}
\]
These four substituents are different.
The carbon is therefore newly stereogenic.
Because the ketone is planar and the addition conditions are achiral, attack can occur from either face.
A racemic tertiary alcohol can consequently form.
\(\textbf{Experiment II — Direct backside displacement:}\)
The question specifically selects a direct substitution pathway without a free carbocation.
Attack from the backside replaces the original hydroxyl-derived leaving group at the stereogenic carbon.
Its tetrahedral arrangement undergoes inversion.
The alkyl bromide retains a stereogenic carbon because bromine replaces hydroxyl without making two substituents identical.
\(\textbf{Compare the stereochemical mechanisms:}\)
Experiment I destroys and then recreates a stereogenic centre through a planar carbonyl intermediate.
Experiment II preserves the existence of the stereogenic centre but reverses its spatial arrangement through substitution.
The identical original alcohol therefore gives fundamentally different stereochemical outcomes under the two routes.
413. Consider the following Assertion and Reason.
Assertion (A): Ethanoic acid is considerably more acidic than phenol, although the conjugate bases of both compounds exhibit resonance.
Reason (R): The two principal resonance contributors of the ethanoate ion are equivalent, with negative charge delocalised over oxygen atoms, whereas the resonance contributors of phenoxide are non-equivalent and include structures with negative charge on ring carbon atoms.
Which option is correct?
ⓐ. Both A and R are true, and R is the correct explanation of A.
ⓑ. Both A and R are true, but R is not the correct explanation of A.
ⓒ. A is true, but R is false.
ⓓ. A is false, but R is true.
Correct Answer: Both A and R are true, and R is the correct explanation of A.
Explanation: \(\textbf{Compare the conjugate bases:}\)
Ethanoic acid loses its carboxylic oxygen-bound hydrogen to form the ethanoate ion.
Phenol loses its phenolic oxygen-bound hydrogen to form the phenoxide ion.
Both conjugate bases exhibit resonance, but the nature of that resonance differs.
The ethanoate ion has two equivalent principal resonance contributors.
In these contributors, the negative charge is associated with either of the two oxygen atoms.
Consequently, the negative charge is effectively delocalised over two electronegative atoms.
The phenoxide ion also possesses resonance contributors, but they are not equivalent in the same way.
Some phenoxide contributors place negative charge on aromatic carbon atoms, which are less electronegative than oxygen.
Therefore, the stabilising effect of resonance is greater for the carboxylate conjugate base in the comparison described.
Greater conjugate-base stabilisation favours proton donation by ethanoic acid.
This explains why ordinary carboxylic acids are generally stronger acids than simple phenols.
Both Assertion and Reason are correct, and the Reason directly explains the relative acidity.
414. Three compounds contain oxygen and have the following structures:
P: \(\mathrm{CH_3COOH}\)
Q: \(\mathrm{HCOOCH_3}\)
R: \(\mathrm{HOCH_2CHO}\)
Each compound is treated separately with aqueous sodium hydrogencarbonate under ordinary conditions. Independent samples are examined for the presence of an aldehyde group, and Q is subjected to complete alkaline hydrolysis.
Which statement correctly integrates the observations?
ⓐ. P and Q both release carbon dioxide because each contains a carbonyl group bonded to oxygen; R is an ordinary carboxylic acid.
ⓑ. Only Q releases carbon dioxide; its alkaline hydrolysis produces ethanoate and water.
ⓒ. P and R release carbon dioxide because both contain oxygen-bound hydrogen, while Q remains unchanged on alkaline hydrolysis.
ⓓ. P releases carbon dioxide; R contains an aldehyde group; alkaline hydrolysis of Q produces methanol and methanoate.
Correct Answer: P releases carbon dioxide; R contains an aldehyde group; alkaline hydrolysis of Q produces methanol and methanoate.
Explanation: \(\textbf{Identify the complete carboxyl group:}\)
P contains a carbonyl group and a hydroxyl group attached to the same carbon atom.
This arrangement constitutes the carboxylic acid functionality.
Therefore, P can donate a proton to aqueous hydrogencarbonate and liberate carbon dioxide.
Q contains an ester group rather than a free carboxylic acid group.
Although it contains two oxygen atoms, neither is present as a free carboxylic hydroxyl group.
Consequently, Q does not show the characteristic carboxylic-acid hydrogencarbonate reaction.
R contains an aldehyde group and a separate alcohol hydroxyl group.
Its ordinary alcohol proton is not sufficiently acidic to produce the characteristic hydrogencarbonate effervescence.
The aldehyde group of R is relevant to characteristic carbonyl reactions such as the Tollens' test.
\(\textbf{Trace ester hydrolysis:}\)
\[
\mathrm{HCOOCH_3+NaOH\longrightarrow HCOONa+CH_3OH}
\]
The ester Q therefore produces sodium methanoate and methanol.
The different reactions depend on the connectivity of the oxygen atoms rather than simply their number.
This distinction prevents confusion between carboxylic acids, esters and hydroxyaldehydes.
415. Two monoprotic carboxylic acids P and Q are dissolved separately in buffers maintained at \(\mathrm{pH}=4.00\).
The following data are supplied.
| Acid | Analytical concentration | \(\mathrm{p}K_a\) |
|---|
| P | \(0.20\,\mathrm{mol\,L^{-1}}\) | 3.00 |
| Q | \(0.50\,\mathrm{mol\,L^{-1}}\) | 5.00 |
Assume ideal monoprotic acid behaviour, constant buffer pH and negligible changes in solution volume.
What is the ratio of equilibrium carboxylate-ion concentrations, \([\mathrm{P^-}]/[\mathrm{Q^-}]\)?
ⓐ. \(1:4\)
ⓑ. \(4:1\)
ⓒ. \(10:1\)
ⓓ. \(40:1\)
Correct Answer: \(4:1\)
Explanation: \(\textbf{Relate pH to carboxylate formation:}\)
Each acid undergoes dissociation according to:
\[
\mathrm{RCOOH\rightleftharpoons RCOO^-+H^+}
\]
The Henderson–Hasselbalch relation is:
\[
\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\mathrm{RCOO^-}]}{[\mathrm{RCOOH}]}
\]
Let the ionised-to-unionised concentration ratio be \(r\).
\[
r=10^{\mathrm{pH}-\mathrm{p}K_a}
\]
The fraction of the analytical acid concentration present as carboxylate is:
\[
f=\frac{r}{1+r}
\]
\(\textbf{Calculate carboxylate concentration for P:}\)
\[
r_P=10^{4.00-3.00}=10
\]
\[
f_P=\frac{10}{11}
\]
\[
[\mathrm{P^-}]=0.20\times\frac{10}{11}
\]
\[
[\mathrm{P^-}]=0.1818\,\mathrm{mol\,L^{-1}}
\]
\(\textbf{Calculate carboxylate concentration for Q:}\)
\[
r_Q=10^{4.00-5.00}=0.10
\]
\[
f_Q=\frac{0.10}{1.10}=\frac{1}{11}
\]
\[
[\mathrm{Q^-}]=0.50\times\frac{1}{11}
\]
\[
[\mathrm{Q^-}]=0.04545\,\mathrm{mol\,L^{-1}}
\]
\(\textbf{Compare the equilibrium concentrations:}\)
\[
\frac{[\mathrm{P^-}]}{[\mathrm{Q^-}]}
=\frac{0.1818}{0.04545}
\]
\[
=4
\]
Although Q has the greater total analytical concentration, its larger \(\mathrm{p}K_a\) gives a considerably smaller ionised fraction.
The result therefore requires considering both acid strength and analytical concentration.
The ratio of ionised-to-unionised forms alone is not the requested concentration ratio.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{[\mathrm{P^-}]}{[\mathrm{Q^-}]}=4:1}
\]
416. Four acids are represented by the following structures:
P: \(\mathrm{CH_3COOH}\)
Q: \(\mathrm{ClCH_2COOH}\)
R: \(\mathrm{Cl_2CHCOOH}\)
S: \(\mathrm{Cl_3CCOOH}\)
The compounds are compared at the same temperature in water.
Consider the following statements.
Statement I: Their acidity increases in the order P < Q < R < S.
Statement II: The chlorine substituents strengthen these acids primarily by stabilising the corresponding carboxylate ions through their electron-withdrawing inductive effects.
Statement III: The increasing number of chlorine atoms creates additional oxygen atoms over which the carboxylate charge can resonate.
Statement IV: If their measured \(\mathrm{p}K_a\) values follow the expected trend, the strongest acid has the smallest \(\mathrm{p}K_a\).
Which combination contains only the correct statements?
ⓐ. I, II and III only
ⓑ. II, III and IV only
ⓒ. I, II and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: The four compounds have the same carboxylic acid functional group.
They differ in the number of chlorine atoms attached to the carbon adjacent to the carboxyl carbon.
Statement I is correct because each additional chlorine substituent contributes an electron-withdrawing inductive influence.
The combined withdrawing effect increases the stability of the corresponding carboxylate ion.
Consequently, trichloroethanoic acid is stronger than dichloroethanoic acid, which is stronger than chloroethanoic acid.
Ethanoic acid, lacking these chlorine substituents, is weakest among the four.
Statement II correctly identifies the principal electronic reason for the trend.
The carboxylate anion retains the same two oxygen atoms, but its negative charge is stabilised more effectively by the neighbouring electron-withdrawing substituents.
Statement III is incorrect because replacing hydrogen with chlorine does not introduce additional oxygen atoms into the carboxylate group.
The two principal carboxylate resonance contributors remain associated with the original two oxygen atoms.
Statement IV is correct because:
\[
\mathrm{p}K_a=-\log K_a
\]
A larger acid dissociation constant corresponds to a smaller \(\mathrm{p}K_a\).
Therefore, the strength order and \(\mathrm{p}K_a\) order run in opposite directions.
The comparison requires distinguishing substituent-induced stabilisation from the intrinsic resonance arrangement of the carboxylate group.
417. A researcher compares three aromatic and aryl-substituted carboxylic acids using the following supplied data.
| Compound | Structure | Approximate \(\mathrm{p}K_a\) |
|---|
| P | \(\mathrm{C_6H_5COOH}\) | 4.19 |
| Q | \(\mathrm{C_6H_5CH_2COOH}\) | 4.31 |
| R | \(\mathrm{CH_3CH_2COOH}\) | 4.87 |
Which interpretation correctly accounts for their acidity order without treating every phenyl-containing acid as electronically identical?
ⓐ. Q is stronger than P because the intervening methylene group permits stronger direct resonance between the aromatic ring and the carboxylate oxygens.
ⓑ. P is stronger than Q, and both are stronger than R; the position and hybridisation of the carbon group adjacent to the carboxyl carbon help explain the difference.
ⓒ. R is strongest because alkyl groups always stabilise negatively charged conjugate bases more effectively than phenyl groups.
ⓓ. P and Q must have identical acidity because they contain the same phenyl ring and each possesses one carboxyl group.
Correct Answer: P is stronger than Q, and both are stronger than R; the position and hybridisation of the carbon group adjacent to the carboxyl carbon help explain the difference.
Explanation: \(\textbf{Read the acidity data correctly:}\)
A smaller \(\mathrm{p}K_a\) corresponds to a stronger acid.
The supplied values establish that P is stronger than Q and Q is stronger than R.
\(\textbf{Examine benzoic acid:}\)
In P, the carboxyl carbon is directly attached to an aromatic \(sp^2\)-hybridised carbon.
NCERT explains that the greater electronegativity associated with the \(sp^2\)-hybridised carbon contributes to increased acidity.
This helps account for the relatively high acidity of benzoic acid.
\(\textbf{Examine phenylacetic acid:}\)
Q contains an intervening methylene group between the phenyl ring and the carboxyl carbon.
The carboxyl group is therefore not attached directly to an aromatic carbon.
The electronic influence of the phenyl group is modified by this intervening saturated carbon.
It is incorrect to assume that adding methylene automatically strengthens direct conjugation with the carboxyl group.
\(\textbf{Examine propanoic acid:}\)
R contains an ordinary alkyl group attached to the carboxyl carbon.
Alkyl groups generally exert an electron-releasing inductive influence that tends to decrease carboxylic acid strength.
The supplied \(\mathrm{p}K_a\) values are consistent with this effect.
\(\textbf{Integrate the distinctions:}\)
All three acids form resonance-stabilised carboxylate ions upon deprotonation.
Their different acid strengths arise from the electronic character and connectivity of the carbon substituent attached to the carboxyl group.
Thus, merely counting phenyl rings or carboxyl groups is insufficient to explain the observed trend.
418. A sample contains three positional isomers:
P: 2-Chlorobutanoic acid, \(\mathrm{CH_3CH_2CH(Cl)COOH}\)
Q: 3-Chlorobutanoic acid, \(\mathrm{CH_3CH(Cl)CH_2COOH}\)
R: 4-Chlorobutanoic acid, \(\mathrm{ClCH_2CH_2CH_2COOH}\)
Assume that the electron-withdrawing inductive influence of chlorine decreases with increasing carbon-chain distance from the carboxyl group.
Which combination correctly predicts their relative acidity and the structural reason?
ⓐ. P > Q > R, because the chlorine atom is closest to the carboxyl group in P and farthest from it in R.
ⓑ. R > Q > P, because the inductive effect increases as the chlorine atom moves farther from the carboxyl group.
ⓒ. P = Q = R, because all three contain the same number of chlorine atoms.
ⓓ. Q > P > R, because chlorine on a secondary carbon necessarily exerts a stronger inductive effect than chlorine on an alpha-carbon.
Correct Answer: P > Q > R, because the chlorine atom is closest to the carboxyl group in P and farthest from it in R.
Explanation: All three compounds have the same molecular formula and contain one chlorine substituent.
They also possess the same carboxylic acid functional group.
The important difference is the distance between chlorine and the carboxyl carbon.
\(\textbf{Evaluate P:}\)
In 2-chlorobutanoic acid, chlorine is attached to the alpha-carbon.
This is the carbon immediately adjacent to the carboxyl group.
Its electron-withdrawing inductive effect can therefore influence the carboxylate conjugate base relatively strongly.
\(\textbf{Evaluate Q:}\)
In 3-chlorobutanoic acid, chlorine is attached to the beta-carbon.
The effect must be transmitted through a greater number of sigma bonds.
Therefore, its influence on the carboxyl group is weaker than in P.
\(\textbf{Evaluate R:}\)
In 4-chlorobutanoic acid, chlorine is attached to the gamma-carbon.
Its distance from the carboxyl group is greatest among the three.
Consequently, its acid-strengthening inductive contribution is weakest under the stated comparison.
\(\textbf{Combine the effects:}\)
The resulting acidity order is P > Q > R.
The number and identity of chlorine substituents remain unchanged.
Therefore, their positions, rather than substituent count or molecular formula, determine the predicted order.
This is an example of distance-dependent transmission of an inductive effect through saturated carbon–carbon bonds.
419. Two acids P and Q are compared.
P: Trifluoroethanoic acid, \(\mathrm{CF_3COOH}\), with \(\mathrm{p}K_a=0.23\).
Q: Ethanoic acid, \(\mathrm{CH_3COOH}\), with \(\mathrm{p}K_a=4.76\).
Assume both acid dissociation constants are measured under the same conditions.
How many times greater is \(K_a\) for P than for Q, and which substituent effect explains the difference?
ⓐ. Approximately \(3.4\times10^4\); the three fluorine atoms exert a strong electron-withdrawing inductive effect.
ⓑ. Approximately \(4.53\); fluorine increases acidity by donating electrons towards the carboxyl group.
ⓒ. Approximately \(3.4\times10^4\); fluorine strengthens the acid by introducing three additional carboxylate resonance contributors involving fluorine–oxygen bonds.
ⓓ. Approximately \(3.0\times10^{-5}\); fluorine destabilises the conjugate base by withdrawing electron density.
Correct Answer: Approximately \(3.4\times10^4\); the three fluorine atoms exert a strong electron-withdrawing inductive effect.
Explanation: \(\textbf{Relate acid strength to the dissociation constant:}\)
The acid dissociation constant measures the extent of proton donation under specified conditions.
The logarithmic relation is:
\[
\mathrm{p}K_a=-\log_{10}K_a
\]
Therefore:
\[
K_a=10^{-\mathrm{p}K_a}
\]
\(\textbf{Form the required ratio:}\)
\[
\frac{K_{a,P}}{K_{a,Q}}
=\frac{10^{-0.23}}{10^{-4.76}}
\]
\[
=10^{-0.23+4.76}
\]
\[
=10^{4.53}
\]
\(\textbf{Evaluate the logarithmic difference:}\)
\[
10^{4.53}=10^4\times10^{0.53}
\]
Using:
\[
10^{0.53}\approx3.39
\]
Therefore:
\[
\frac{K_{a,P}}{K_{a,Q}}\approx3.39\times10^4
\]
\(\textbf{Explain the structural cause:}\)
The methyl group of ethanoic acid has an electron-releasing inductive influence.
In trifluoroethanoic acid, that methyl group is replaced by a strongly electron-withdrawing trifluoromethyl group.
The fluorine atoms withdraw electron density through sigma bonds.
This stabilises the corresponding carboxylate conjugate base and increases the tendency of the acid to donate a proton.
The effect does not require introducing new oxygen atoms or changing the two principal carboxylate resonance contributors.
\(\textbf{Interpret the numerical result:}\)
The difference of \(4.53\) \(\mathrm{p}K_a\) units represents a multiplicative, rather than additive, difference in acid dissociation constants.
The stronger acid has the smaller \(\mathrm{p}K_a\) and the larger \(K_a\).
\(\textbf{Final answer:}\)
\[
\boxed{\frac{K_{a,P}}{K_{a,Q}}\approx3.4\times10^4}
\]
420. Three aromatic carboxylic acids have the following NCERT-style data:
| Acid | Structure | \(\mathrm{p}K_a\) |
|---|
| P | 4-Nitrobenzoic acid | 3.41 |
| Q | Benzoic acid | 4.19 |
| R | 4-Methoxybenzoic acid | 4.46 |
Equal analytical concentrations of the three acids are maintained in separate ideal buffers at \(\mathrm{pH}=4.19\).
What are the approximate percentages of P, Q and R present as carboxylate ions?
ⓐ. \(35\%,\ 50\%,\ 86\%\)
ⓑ. \(50\%,\ 86\%,\ 35\%\)
ⓒ. \(86\%,\ 35\%,\ 50\%\)
ⓓ. \(86\%,\ 50\%,\ 35\%\)
Correct Answer: \(86\%,\ 50\%,\ 35\%\)
Explanation: \(\textbf{Relate aromatic substituents to acidity:}\)
The nitro group strongly withdraws electron density and increases the acidity of benzoic acid.
The para-methoxy group can donate electron density through resonance and decreases acidity in the supplied comparison.
Therefore, P has the smallest \(\mathrm{p}K_a\), while R has the largest.
\(\textbf{Use the buffer relation:}\)
For each monoprotic acid:
\[
\frac{[\mathrm{RCOO^-}]}{[\mathrm{RCOOH}]}
=10^{\mathrm{pH}-\mathrm{p}K_a}
\]
Let this concentration ratio be \(r\).
The ionised fraction is:
\[
f=\frac{r}{1+r}
\]
\(\textbf{Calculate the fraction for P:}\)
\[
r_P=10^{4.19-3.41}
\]
\[
r_P=10^{0.78}\approx6.03
\]
\[
f_P=\frac{6.03}{1+6.03}
\]
\[
f_P\approx0.858
\]
Thus, P is approximately \(86\%\) ionised.
\(\textbf{Calculate the fraction for Q:}\)
\[
r_Q=10^{4.19-4.19}=1
\]
\[
f_Q=\frac{1}{1+1}=0.500
\]
Thus, Q is \(50\%\) ionised.
\(\textbf{Calculate the fraction for R:}\)
\[
r_R=10^{4.19-4.46}
\]
\[
r_R=10^{-0.27}\approx0.537
\]
\[
f_R=\frac{0.537}{1+0.537}
\]
\[
f_R\approx0.349
\]
Thus, R is approximately \(35\%\) ionised.
\(\textbf{Interpret the results:}\)
The common buffer pH does not produce identical ionised fractions.
Instead, stronger acids have greater carboxylate fractions under the same conditions.
The numerical ordering is consistent with the electron-withdrawing and electron-releasing effects of the aromatic substituents.
\(\textbf{Final answer:}\)
\[
\boxed{(f_P,f_Q,f_R)\approx(86\%,\,50\%,\,35\%)}
\]
