501. A \(0.250\,\mathrm{mol}\) mixture of sodium ethanoate and sodium propanoate has a mass of \(22.6\,\mathrm{g}\). Complete soda-lime decarboxylation gives methane and ethane. Which pair gives the methane volume percentage in the gaseous hydrocarbon mixture and the total gas volume at STP?
ⓐ. \(40.0\%\) and \(5.60\,\mathrm{L}\)
ⓑ. \(60.0\%\) and \(5.60\,\mathrm{L}\)
ⓒ. \(40.0\%\) and \(3.36\,\mathrm{L}\)
ⓓ. \(50.0\%\) and \(4.48\,\mathrm{L}\)
Correct Answer: \(40.0\%\) and \(5.60\,\mathrm{L}\)
Explanation: Let \(x\) be sodium-ethanoate moles and \(y\) be sodium-propanoate moles.
\[
x+y=0.250.
\]
Using molar masses \(82.0\,\mathrm{g\,mol^{-1}}\) and \(96.0\,\mathrm{g\,mol^{-1}}\),
\[
82x+96y=22.6.
\]
Substitute
\[
x=0.250-y.
\]
\[
82(0.250-y)+96y=22.6.
\]
\[
20.5+14y=22.6.
\]
\[
y=0.150\,\mathrm{mol}.
\]
\[
x=0.100\,\mathrm{mol}.
\]
Sodium ethanoate gives methane, so
\[
n(\mathrm{CH_4})=0.100\,\mathrm{mol}.
\]
The total hydrocarbon amount is
\[
0.100+0.150=0.250\,\mathrm{mol}.
\]
Methane volume percentage equals mole percentage:
\[
\frac{0.100}{0.250}\times100=40.0\%.
\]
The total volume at STP is
\[
V=0.250\times22.4=5.60\,\mathrm{L}.
\]
502. Consider the sequence:
\[
\mathrm{CH_3CH_2CH_2Br\xrightarrow{KCN}P\xrightarrow{hydrolysis}Q
\xrightarrow{NaOH}R\xrightarrow[\Delta]{soda\ lime}S}.
\]
What is S?
ⓐ. Methane
ⓑ. Ethane
ⓒ. Propane
ⓓ. Butane
Correct Answer: Propane
Explanation: Cyanide ion introduces one additional carbon atom into the chain. \(1\)-Bromopropane therefore forms butanenitrile, P. Hydrolysis converts the nitrile carbon into the carboxyl carbon of butanoic acid, Q. Sodium hydroxide gives sodium butanoate, R. Soda-lime decarboxylation removes the newly introduced carboxyl carbon, leaving the original three-carbon chain as propane.
503. Which equation represents the principal anodic process during Kolbe electrolysis of a simple carboxylate ion?
ⓐ. \(\mathrm{2RCOO^-+2e^-\rightarrow R-R+2CO_2}\)
ⓑ. \(\mathrm{2RCOO^-\rightarrow R-R+2CO_2+2e^-}\)
ⓒ. \(\mathrm{RCOO^-+H_2O\rightarrow RCOOH+OH^-}\)
ⓓ. \(\mathrm{2RCOO^-\rightarrow RCOOR+CO_2+O_2}\)
Correct Answer: \(\mathrm{2RCOO^-\rightarrow R-R+2CO_2+2e^-}\)
Explanation: A carboxylate ion loses one electron at the anode. The resulting carboxyl radical rapidly releases carbon dioxide and forms an alkyl radical, \(\mathrm{R\boldsymbol{\cdot}}\). Two identical alkyl radicals then couple to form \(\mathrm{R-R}\). Two carboxylate ions are therefore required for one molecule of the coupled alkane. The overall anodic equation also shows the release of two electrons.
504. What is the principal cathodic reaction during electrolysis of an aqueous sodium carboxylate solution under ordinary Kolbe conditions?
ⓐ. \(\mathrm{2Cl^-\rightarrow Cl_2+2e^-}\)
ⓑ. \(\mathrm{2RCOO^-\rightarrow R-R+2CO_2+2e^-}\)
ⓒ. \(\mathrm{Na^++e^-\rightarrow Na}\)
ⓓ. \(\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}\)
Correct Answer: \(\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}\)
Explanation: In an aqueous solution, water is reduced more readily than sodium ions. Water molecules accept electrons at the cathode and form hydrogen gas. Hydroxide ions are produced simultaneously. Sodium metal is not deposited from the aqueous medium because it would require a much more difficult reduction. Kolbe decarboxylation occurs at the anode, while hydrogen evolution occurs at the cathode.
505. A saturated monocarboxylate ion contains \(n\) carbon atoms, including its carboxyl carbon. Simple Kolbe coupling of identical ions gives an alkane containing ______ carbon atoms.
ⓐ. \(n+1\)
ⓑ. \(2n\)
ⓒ. \(2n-2\)
ⓓ. \(2n-1\)
Correct Answer: \(2n-2\)
Explanation: Decarboxylation removes the carboxyl carbon from each carboxylate ion. Each resulting alkyl radical therefore contains \(n-1\) carbon atoms. Two identical radicals couple during Kolbe electrolysis. The total number of carbon atoms in the alkane is consequently
\[
(n-1)+(n-1)=2(n-1).
\]
For example, a three-carbon propanoate ion gives the four-carbon alkane butane.
506. Why is ordinary Kolbe electrolysis of a mixture of sodium ethanoate and sodium propanoate unsuitable for preparing pure propane?
ⓐ. Neither carboxylate ion can lose an electron to form an alkyl radical at the anode
ⓑ. Methyl and ethyl radicals self-couple and cross-couple, giving three alkanes
ⓒ. Propane is oxidised back to propanoate at the cathode
ⓓ. Mixed carboxylates produce only \(\mathrm{CO_2}\) and \(\mathrm{H_2}\)
Correct Answer: Methyl and ethyl radicals self-couple and cross-couple, giving three alkanes
Explanation: Sodium ethanoate produces methyl radicals, while sodium propanoate produces ethyl radicals. Two methyl radicals can form ethane, and two ethyl radicals can form butane. A methyl radical can also combine with an ethyl radical to form propane. All three coupling pathways are possible in the same reaction mixture. The desired unsymmetrical alkane is therefore accompanied by two symmetrical products.
507. Sodium propanoate labelled at its carboxyl carbon, \(\mathrm{CH_3CH_2{}^{13}COONa}\), undergoes Kolbe electrolysis. Where is the isotope found?
ⓐ. In the terminal carbons of butane
ⓑ. In the cathodic hydrogen gas
ⓒ. In the carbon dioxide evolved at the anode
ⓓ. In the central carbon-carbon bond of butane
Correct Answer: In the carbon dioxide evolved at the anode
Explanation: The labelled atom is the carboxyl carbon. During anodic oxidation, the carboxylate group loses carbon dioxide before radical coupling occurs. The remaining ethyl fragment forms an ethyl radical. Two unlabelled ethyl radicals then couple to produce butane. The \({}^{13}\mathrm{C}\) label therefore leaves the organic fragment as \(\mathrm{{}^{13}CO_2}\).
508. An aqueous solution containing excess sodium propanoate is electrolysed with a charge of \(9650\,\mathrm{C}\). Assuming \(100\%\) current efficiency and \(F=96500\,\mathrm{C\,mol^{-1}}\), which pair gives the amount of butane formed at the anode and hydrogen formed at the cathode?
ⓐ. \(0.100\,\mathrm{mol}\) butane and \(0.0500\,\mathrm{mol}\) hydrogen
ⓑ. \(0.0500\,\mathrm{mol}\) butane and \(0.100\,\mathrm{mol}\) hydrogen
ⓒ. \(0.100\,\mathrm{mol}\) butane and \(0.100\,\mathrm{mol}\) hydrogen
ⓓ. \(0.0500\,\mathrm{mol}\) butane and \(0.0500\,\mathrm{mol}\) hydrogen
Correct Answer: \(0.0500\,\mathrm{mol}\) butane and \(0.0500\,\mathrm{mol}\) hydrogen
Explanation: The amount of electrons transferred is
\[
n(e^-)=\frac{Q}{F}.
\]
\[
n(e^-)=\frac{9650}{96500}.
\]
\[
n(e^-)=0.100\,\mathrm{mol}.
\]
At the anode,
\[
\mathrm{2CH_3CH_2COO^-\rightarrow CH_3CH_2CH_2CH_3+2CO_2+2e^-}.
\]
Two moles of electrons correspond to one mole of butane.
\[
n(\mathrm{butane})=\frac{0.100}{2}=0.0500\,\mathrm{mol}.
\]
At the cathode,
\[
\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}.
\]
Two moles of electrons also produce one mole of hydrogen.
\[
n(\mathrm{H_2})=\frac{0.100}{2}=0.0500\,\mathrm{mol}.
\]
Equal electron requirements give equal molar amounts of butane and hydrogen in this ideal case.
509. A graph plots moles of alkane formed during Kolbe electrolysis against charge passed through a solution containing a fixed amount of one sodium carboxylate. Which description is correct?
ⓐ. Rises with slope \(\frac{1}{2F}\), then plateaus after carboxylate exhaustion
ⓑ. Falls with slope \(-\frac{1}{2F}\) because alkane is reduced at the cathode
ⓒ. Rises with slope \(\frac{2}{F}\) even after all carboxylate is consumed
ⓓ. Remains horizontal because current produces only cathodic hydrogen
Correct Answer: Rises with slope \(\frac{1}{2F}\), then plateaus after carboxylate exhaustion
Explanation: Two moles of electrons are released for each mole of coupled alkane formed. While carboxylate remains available,
\[
n(\mathrm{alkane})=\frac{Q}{2F}.
\]
The alkane amount therefore increases linearly with charge and has slope \(\frac{1}{2F}\). Once all carboxylate ions have been consumed, additional charge cannot produce more of that alkane. The curve then becomes horizontal.
510. Which acid does not undergo ordinary Hell-Volhard-Zelinsky bromination?
ⓐ. \(\mathrm{CH_3CH_2COOH}\)
ⓑ. \(\mathrm{(CH_3)_3CCOOH}\)
ⓒ. \(\mathrm{CH_3CH_2CH_2COOH}\)
ⓓ. \(\mathrm{(CH_3)_2CHCOOH}\)
Correct Answer: \(\mathrm{(CH_3)_3CCOOH}\)
Explanation: The alpha carbon of \(2,2\)-dimethylpropanoic acid is bonded to three methyl groups and the carboxyl carbon. It carries no hydrogen atom. Propanoic acid and butanoic acid each contain an alpha \(\mathrm{CH_2}\) group. \(2\)-Methylpropanoic acid contains an alpha \(\mathrm{CH}\) group and therefore has one alpha hydrogen. Only \(2,2\)-dimethylpropanoic acid fails the structural requirement.
511. Propanoic acid is treated with bromine in the presence of red phosphorus and then hydrolysed. What is the principal organic product?
ⓐ. \(3\)-Bromopropanoic acid
ⓑ. \(2\)-Bromopropanoic acid
ⓒ. \(1\)-Bromopropan-\(2\)-one
ⓓ. Propanoyl bromide as the final isolated product
Correct Answer: \(2\)-Bromopropanoic acid
Explanation: The reaction introduces bromine at the alpha carbon of the carboxylic acid. In propanoic acid, the alpha carbon is carbon \(2\). Red phosphorus reacts with bromine to generate a phosphorus bromide species in the reaction mixture. An acyl-bromide intermediate is formed and undergoes alpha bromination. Hydrolysis restores the carboxylic-acid group, giving \(2\)-bromopropanoic acid.
512. Which sequence best represents the Hell-Volhard-Zelinsky reaction at recognition level?
ⓐ. Acid \(\rightarrow\) acyl halide \(\rightarrow\) alpha-halo acyl halide \(\rightarrow\) alpha-halo acid
ⓑ. Carboxylic acid \(\rightarrow\) primary alcohol \(\rightarrow\) aldehyde \(\rightarrow\) alpha-halo acid
ⓒ. Carboxylic acid \(\rightarrow\) carboxylate salt \(\rightarrow\) alkyl radical \(\rightarrow\) alpha-halo acid
ⓓ. Carboxylic acid \(\rightarrow\) ester \(\rightarrow\) enolate ion \(\rightarrow\) alpha-halo acid
Correct Answer: Acid \(\rightarrow\) acyl halide \(\rightarrow\) alpha-halo acyl halide \(\rightarrow\) alpha-halo acid
Explanation: Phosphorus and halogen generate a phosphorus halide capable of converting the acid into an acyl halide. The acyl halide can form an enol more readily than the original acid under the reaction conditions. Halogenation then occurs at the alpha carbon. Subsequent hydrolysis replaces the acyl-halide chlorine or bromine with hydroxyl. The final product is an alpha-halo carboxylic acid.
513. Assertion: Benzoic acid does not undergo the ordinary Hell-Volhard-Zelinsky reaction.
Reason: Benzoic acid has no alpha carbon bearing an alpha hydrogen next to its carboxyl group.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: In benzoic acid, the carboxyl group is attached directly to an aromatic-ring carbon. There is no separate saturated alpha carbon carrying a removable hydrogen. The Hell-Volhard-Zelinsky reaction specifically requires an alpha hydrogen. Ring halogenation of benzoic acid, when it occurs under other conditions, is a different reaction. The structural fact in the Reason directly explains the absence of ordinary alpha halogenation.
514. Butanoic acid, \(14.8\,\mathrm{g}\), is treated with \(25.6\,\mathrm{g}\) of bromine under Hell-Volhard-Zelinsky conditions:
\[
\mathrm{CH_3CH_2CH_2COOH+Br_2\rightarrow CH_3CH_2CHBrCOOH+HBr}.
\]
The alpha-bromo acid is isolated in \(75.0\%\) yield. Which pair gives the isolated product mass and the mass of unreacted butanoic acid? Use \(M(\mathrm{Br_2})=160\,\mathrm{g\,mol^{-1}}\) and \(M(2\text{-}\mathrm{bromobutanoic\ acid})=167\,\mathrm{g\,mol^{-1}}\).
ⓐ. \(26.72\,\mathrm{g}\) and \(0.72\,\mathrm{g}\)
ⓑ. \(15.03\,\mathrm{g}\) and \(5.92\,\mathrm{g}\)
ⓒ. \(20.04\,\mathrm{g}\) and \(0\,\mathrm{g}\)
ⓓ. \(20.04\,\mathrm{g}\) and \(0.72\,\mathrm{g}\)
Correct Answer: \(20.04\,\mathrm{g}\) and \(0.72\,\mathrm{g}\)
Explanation: \( \textbf{Butanoic-acid amount:} \) The molar mass of \(\mathrm{C_4H_8O_2}\) is \(88.0\,\mathrm{g\,mol^{-1}}\).
\[
n_{\mathrm{acid}}=\frac{14.8}{88.0}=0.1682\,\mathrm{mol}
\]
\( \textbf{Bromine amount:} \)
\[
n_{\mathrm{Br_2}}=\frac{25.6}{160}=0.160\,\mathrm{mol}
\]
\( \textbf{Limiting reactant:} \) The reaction ratio is \(1:1\), so bromine is limiting and \(0.160\,\mathrm{mol}\) of butanoic acid reacts.
\( \textbf{Theoretical alpha-bromo acid:} \)
\[
n_{\mathrm{product,theoretical}}=0.160\,\mathrm{mol}
\]
\( \textbf{Apply the isolation yield:} \)
\[
n_{\mathrm{product,isolated}}=0.160\times0.750=0.120\,\mathrm{mol}
\]
\[
m_{\mathrm{product}}=0.120\times167=20.04\,\mathrm{g}
\]
\( \textbf{Unreacted butanoic acid:} \)
\[
n_{\mathrm{unreacted}}=0.1682-0.160=0.00818\,\mathrm{mol}
\]
\[
m_{\mathrm{unreacted}}=0.00818\times88.0\approx0.72\,\mathrm{g}
\]
The \(75.0\%\) isolation yield changes the recovered product mass, but it does not change the amount of acid consumed by the limiting bromine.
515. An acid X has molecular formula \(\mathrm{C_4H_8O_2}\). Its sodium salt gives \(2,3\)-dimethylbutane on Kolbe electrolysis. X also undergoes Hell-Volhard-Zelinsky bromination to give a single alpha-bromo acid. What is X?
ⓐ. \(2\)-Methylpropanoic acid
ⓑ. Butanoic acid
ⓒ. \(2,2\)-Dimethylpropanoic acid
ⓓ. Butanedioic acid
Correct Answer: \(2\)-Methylpropanoic acid
Explanation: The sodium salt of X loses carbon dioxide to form the radical derived from its hydrocarbon portion. Formation of \(2,3\)-dimethylbutane requires coupling of two isopropyl radicals:
\[
\mathrm{2(CH_3)_2CH\boldsymbol{\cdot}\rightarrow (CH_3)_2CHCH(CH_3)_2}.
\]
The corresponding carboxylic acid is \(\mathrm{(CH_3)_2CHCOOH}\), or \(2\)-methylpropanoic acid. Its alpha carbon contains one hydrogen, so Hell-Volhard-Zelinsky bromination is possible. Butanoic acid would give \(n\)-hexane during Kolbe electrolysis rather than the branched product.
516. Assertion: The carboxyl group is meta directing in electrophilic aromatic substitution.
Reason: Sigma complexes formed during ortho and para attack are especially destabilised by resonance interaction with the electron-withdrawing carboxyl group.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The \(\mathrm{-COOH}\) group removes electron density from the aromatic ring. During ortho or para attack, the intermediate sigma complex has resonance contributors in which positive charge is unfavourably positioned relative to the carbon bearing the electron-withdrawing substituent. The corresponding meta intermediate avoids the most strongly destabilised contributor. Meta substitution is therefore favoured relative to ortho and para substitution. The Reason directly accounts for the directing effect stated in the Assertion.
517. Which order best represents the relative rates of nitration under comparable conditions?
ⓐ. Benzoic acid \(\gt\) benzene \(\gt\) methylbenzene
ⓑ. Benzene \(\gt\) benzoic acid \(\gt\) methylbenzene
ⓒ. Benzoic acid \(\gt\) methylbenzene \(\gt\) benzene
ⓓ. Methylbenzene \(\gt\) benzene \(\gt\) benzoic acid
Correct Answer: Methylbenzene \(\gt\) benzene \(\gt\) benzoic acid
Explanation: A methyl group donates electron density to the aromatic ring through the positive inductive effect and hyperconjugation. Methylbenzene is therefore more reactive than benzene toward electrophilic substitution. The carboxyl group strongly withdraws electron density and deactivates the ring. Benzoic acid consequently undergoes nitration more slowly than benzene. The order reflects electronic activation and deactivation rather than molecular mass.
518. Sulphonation of benzoic acid principally introduces the \(\mathrm{-SO_3H}\) group at which position relative to \(\mathrm{-COOH}\)?
ⓐ. Ortho
ⓑ. Para
ⓒ. Meta
ⓓ. Directly on the carboxyl carbon
Correct Answer: Meta
Explanation: Sulphonation is an electrophilic aromatic-substitution reaction. The carboxyl group is strongly electron withdrawing and directs incoming electrophiles mainly to the meta position. Ortho and para intermediates are more strongly destabilised by the substituent. The sulfonic-acid group is introduced on the aromatic ring rather than on the carboxyl carbon. The principal product is therefore \(3\)-sulfobenzoic acid.
519. Why does benzoic acid generally resist an ordinary Friedel-Crafts alkylation or acylation?
ⓐ. \(\mathrm{-COOH}\) strongly activates the ring and causes rapid polysubstitution
ⓑ. \(\mathrm{-COOH}\) directs substitution to ortho and para positions
ⓒ. \(\mathrm{AlCl_3}\) converts the acid into an acyl chloride that activates the ring
ⓓ. The ring is deactivated and \(\mathrm{-COOH}\) can complex with the Lewis acid
Correct Answer: The ring is deactivated and \(\mathrm{-COOH}\) can complex with the Lewis acid
Explanation: Friedel-Crafts reactions require an aromatic ring sufficiently reactive toward an electrophile. The \(\mathrm{-COOH}\) group strongly withdraws electron density and makes the ring a poor nucleophile. Its oxygen atoms can also coordinate with a Lewis acid such as \(\mathrm{AlCl_3}\). This interaction further reduces the suitability of the substrate under ordinary Friedel-Crafts conditions. The limitation arises from both ring deactivation and catalyst interaction.
520. A student predicts para nitration of benzoic acid because the para position is less sterically crowded than the ortho position. What is the best correction?
ⓐ. Oxygen lone pairs make \(\mathrm{-COOH}\) a strong para-directing activator
ⓑ. Lower steric crowding at para always overrides electronic effects, so para substitution dominates
ⓒ. Ortho and para sigma complexes are electronically destabilised, so meta substitution dominates
ⓓ. Nitration replaces the hydroxyl group of \(\mathrm{-COOH}\) rather than ring hydrogen
Correct Answer: Ortho and para sigma complexes are electronically destabilised, so meta substitution dominates
Explanation: Steric crowding can influence the relative amounts of products, but it does not determine the directing effect by itself. The carboxyl group strongly withdraws electron density from the ring. Ortho and para attack produce sigma complexes containing particularly unfavourable resonance contributors. Meta attack avoids this strongest electronic destabilisation. The principal orientation is therefore controlled by the electronic nature of \(\mathrm{-COOH}\), not merely by available space.
