401. Ethanenitrile labelled with \({}^{13}\mathrm{C}\) at the nitrile carbon, \(\mathrm{CH_3{}^{13}CN}\), is completely hydrolysed. Where is the isotope found in the organic product?
ⓐ. In the carboxyl carbon of ethanoic acid
ⓑ. In carbon dioxide released during hydrolysis
ⓒ. In the methyl carbon of ethanoic acid
ⓓ. In the methyl carbon of ethanol
Correct Answer: In the carboxyl carbon of ethanoic acid
Explanation: Hydrolysis does not break the bond between the methyl group and the nitrile carbon. Instead, the nitrile carbon is progressively converted into the carbonyl carbon of an amide and then of a carboxylic acid. The \({}^{13}\mathrm{C}\) label therefore remains attached to the original methyl group. In ethanoic acid, it occupies the \(\mathrm{-COOH}\) carbon. The nitrogen atom is removed separately as ammonia or an ammonium species.
402. Which row correctly matches a nitrile with the medium and principal products of complete hydrolysis?
| Row | Nitrile | Medium | Principal products |
|---|
| P | Ethanenitrile | Acidic | Ethanoic acid and an ammonium salt |
| Q | Propanenitrile | Alkaline | Propanoic acid and ammonium ion directly |
| R | Benzonitrile | Acidic | Sodium benzoate and ammonia |
| S | Methanenitrile | Alkaline | Methanol and nitrogen gas |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Acidic hydrolysis of ethanenitrile produces ethanoic acid. The nitrogen-containing product is protonated and appears as an ammonium salt. Alkaline hydrolysis produces a carboxylate rather than the free carboxylic acid before acidification. Acidic hydrolysis of benzonitrile does not directly produce sodium benzoate or free ammonia. Nitrile hydrolysis gives carboxylic-acid derivatives rather than alcohol and nitrogen gas.
403. A \(14.4\,\mathrm{g}\) mixture of ethanenitrile and benzonitrile is completely hydrolysed in alkaline medium and releases \(4.48\,\mathrm{L}\) of ammonia at STP. The ethanoic acid obtained after acidification is recovered in \(80.0\%\) yield, while benzoic acid is recovered in \(75.0\%\) yield. Which pair gives the mole percentage of benzonitrile and the combined recovered acid mass? Use \(22.4\,\mathrm{L\,mol^{-1}}\) at STP.
ⓐ. \(25.0\%\) and \(11.98\,\mathrm{g}\)
ⓑ. \(50.0\%\) and \(13.95\,\mathrm{g}\)
ⓒ. \(75.0\%\) and \(16.10\,\mathrm{g}\)
ⓓ. \(50.0\%\) and \(18.20\,\mathrm{g}\)
Correct Answer: \(50.0\%\) and \(13.95\,\mathrm{g}\)
Explanation: The total ammonia amount is
\[
n(\mathrm{NH_3})=\frac{4.48}{22.4}=0.200\,\mathrm{mol}
\]
Each nitrile molecule gives one ammonia molecule.
Let \(x\) be ethanenitrile moles and \(y\) be benzonitrile moles.
\[
x+y=0.200
\]
Using molar masses \(41.0\,\mathrm{g\,mol^{-1}}\) and \(103\,\mathrm{g\,mol^{-1}}\),
\[
41x+103y=14.4
\]
Substitute \(x=0.200-y\):
\[
41(0.200-y)+103y=14.4
\]
\[
8.20+62y=14.4
\]
\[
y=0.100\,\mathrm{mol}
\]
\[
x=0.100\,\mathrm{mol}
\]
The benzonitrile mole percentage is
\[
\frac{0.100}{0.200}\times100=50.0\%.
\]
Recovered ethanoic acid is
\[
m=0.100\times0.800\times60.0=4.80\,\mathrm{g}
\]
Recovered benzoic acid is
\[
m=0.100\times0.750\times122=9.15\,\mathrm{g}
\]
\[
m_{\mathrm{total}}=4.80+9.15=13.95\,\mathrm{g}
\]
404. During hydrolysis of a nitrile under constant heated conditions, three concentration curves are observed. Curve P decreases continuously. Curve Q rises to a maximum and then decreases. Curve R increases gradually and approaches a plateau. What do P, Q, and R most reasonably represent?
ⓐ. P: carboxylic acid; Q: nitrile; R: amide
ⓑ. P: water; Q: ammonia; R: nitrile
ⓒ. P: amide; Q: carboxylic acid; R: nitrile
ⓓ. P: nitrile; Q: amide; R: carboxylic acid
Correct Answer: P: nitrile; Q: amide; R: carboxylic acid
Explanation: The nitrile is the starting material and is consumed throughout the reaction. Its concentration therefore falls continuously. The amide is formed from the nitrile during the first hydrolysis stage. It initially accumulates but later decreases as it undergoes further hydrolysis. The carboxylic acid or carboxylate is the final organic product and accumulates toward a limiting concentration. The rise-and-fall shape is characteristic of a reaction intermediate.
405. Consider the following statements about amide hydrolysis.
Statement I: Amides generally hydrolyse less readily than acyl chlorides.
Statement II: Acidic hydrolysis gives a carboxylic acid and an ammonium species.
Statement III: Alkaline hydrolysis gives a carboxylate ion and ammonia.
Statement IV: Alkaline hydrolysis directly gives the free carboxylic acid without acidification.
ⓐ. Statements I and IV are correct
ⓑ. Statements II and III are correct
ⓒ. Statements I, II, and III are correct
ⓓ. Statements I, II, III, and IV are correct
Correct Answer: Statements I, II, and III are correct
Explanation: Amides are resonance stabilised and are relatively resistant to hydrolysis compared with more reactive acid derivatives such as acyl chlorides. Acidic hydrolysis produces the free carboxylic acid, while the nitrogen product is protonated. Alkaline hydrolysis produces a carboxylate salt and ammonia. The carboxylate must be acidified to obtain the neutral carboxylic acid. Statement IV therefore confuses the reaction mixture before and after acidic work-up.
406. An unknown primary amide, \(7.30\,\mathrm{g}\), is completely hydrolysed in alkaline medium and produces \(2.24\,\mathrm{L}\) of ammonia at STP. After acidification, \(7.40\,\mathrm{g}\) of a monocarboxylic acid is obtained. Which pair identifies the amide and the acid?
ⓐ. Propanamide and propanoic acid
ⓑ. Ethanamide and ethanoic acid
ⓒ. Butanamide and butanoic acid
ⓓ. Methanamide and methanoic acid
Correct Answer: Propanamide and propanoic acid
Explanation: The ammonia amount is
\[
n(\mathrm{NH_3})=\frac{2.24}{22.4}=0.100\,\mathrm{mol}
\]
One mole of a primary amide gives one mole of ammonia.
\[
n(\mathrm{amide})=0.100\,\mathrm{mol}
\]
The amide molar mass is
\[
M=\frac{7.30}{0.100}=73.0\,\mathrm{g\,mol^{-1}}
\]
A primary amide with this molar mass is propanamide, \(\mathrm{C_2H_5CONH_2}\).
The acid amount after acidification is also
\[
0.100\,\mathrm{mol}.
\]
Its molar mass is
\[
M=\frac{7.40}{0.100}=74.0\,\mathrm{g\,mol^{-1}}
\]
This identifies propanoic acid and confirms retention of the three-carbon skeleton.
407. Acetamide labelled with \({}^{15}\mathrm{N}\), \(\mathrm{CH_3CO{}^{15}NH_2}\), undergoes complete acidic hydrolysis. Where is the isotope found?
ⓐ. In the carbonyl oxygen of ethanoic acid
ⓑ. In the methyl carbon of ethanoic acid
ⓒ. In carbon dioxide
ⓓ. In the ammonium ion or ammonium salt
Correct Answer: In the ammonium ion or ammonium salt
Explanation: Hydrolysis breaks the bond between the acyl carbon and nitrogen. The carbonyl carbon and oxygen remain with the organic fragment and form ethanoic acid. The labelled nitrogen becomes ammonia during cleavage. Because the medium is acidic, that ammonia is protonated to \(\mathrm{{}^{15}NH_4^+}\). The isotope therefore appears in the ammonium product rather than in the carboxylic acid.
408. Benzamide is heated with aqueous sodium hydroxide. Ammonia is evolved, but no benzoic acid precipitate is initially visible. After the mixture is cooled and acidified, a white solid appears. Which interpretation is most appropriate?
ⓐ. Sodium hydroxide reduces benzamide to benzyl alcohol, which precipitates after acidification
ⓑ. Hydrolysis gives soluble benzoate and ammonia; acidification then gives benzoic acid
ⓒ. Benzamide remains unchanged until hydrochloric acid directly converts it into benzoic acid
ⓓ. Cooling forms solid ammonium hydroxide, while benzoate stays in the organic layer
Correct Answer: Hydrolysis gives soluble benzoate and ammonia; acidification then gives benzoic acid
Explanation: Heating benzamide with sodium hydroxide cleaves the amide bond. The organic product in the alkaline solution is sodium benzoate rather than free benzoic acid. Sodium benzoate remains soluble, so an acid precipitate is not expected at that stage. Ammonia is released as the nitrogen-containing product. Acidification protonates benzoate ion and produces benzoic acid, whose lower water solubility allows the white solid to separate.
409. The decreasing order of hydrolysis reactivity among common carboxylic-acid derivatives is:
ⓐ. Amide \(\gt\) ester \(\gt\) anhydride \(\gt\) acyl chloride
ⓑ. Ester \(\gt\) amide \(\gt\) acyl chloride \(\gt\) anhydride
ⓒ. Acyl chloride \(\gt\) acid anhydride \(\gt\) ester \(\gt\) amide
ⓓ. Acid anhydride \(\gt\) acyl chloride \(\gt\) amide \(\gt\) ester
Correct Answer: Acyl chloride \(\gt\) acid anhydride \(\gt\) ester \(\gt\) amide
Explanation: Hydrolysis of an acid derivative occurs by nucleophilic acyl substitution. Acyl chlorides react most rapidly because chlorine withdraws electron density and chloride ion is a good leaving group. Acid anhydrides also contain a good carboxylate leaving group and hydrolyse readily. Esters react more slowly because alkoxide is a poorer leaving group. Amides are least reactive because nitrogen donates electron density strongly into the carbonyl group and the amide leaving group would be highly basic.
410. Assertion: Alkaline hydrolysis of an ester is often treated as effectively irreversible.
Reason: The carboxylic acid formed during acyl substitution is converted into a resonance-stabilised carboxylate ion, which does not readily react with the alcohol to regenerate the ester.
ⓐ. Assertion is true, but Reason is false
ⓑ. Assertion is false, but Reason is true
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Hydroxide attack on an ester ultimately produces an alcohol and a carboxylate ion. The carboxylate ion is stabilised by delocalisation of negative charge over two oxygen atoms. It is also much less electrophilic than the original ester. The reverse reaction with the alcohol is therefore strongly disfavoured in alkaline solution. The Reason directly explains why saponification proceeds essentially to completion.
411. Methyl benzoate, \(13.6\,\mathrm{g}\), is heated with \(75.0\,\mathrm{mL}\) of \(1.00\,\mathrm{mol\,L^{-1}}\) sodium hydroxide:
\[
\mathrm{C_6H_5COOCH_3+NaOH\rightarrow C_6H_5COONa+CH_3OH}
\]
After acidification, benzoic acid is isolated in \(80.0\%\) yield. Which pair gives the isolated acid mass and the mass of unreacted methyl benzoate?
ⓐ. \(7.32\,\mathrm{g}\) and \(3.40\,\mathrm{g}\)
ⓑ. \(9.15\,\mathrm{g}\) and \(3.40\,\mathrm{g}\)
ⓒ. \(7.32\,\mathrm{g}\) and \(6.80\,\mathrm{g}\)
ⓓ. \(9.15\,\mathrm{g}\) and \(0\,\mathrm{g}\)
Correct Answer: \(7.32\,\mathrm{g}\) and \(3.40\,\mathrm{g}\)
Explanation: The initial methyl benzoate amount is
\[
n=\frac{13.6}{136}=0.100\,\mathrm{mol}
\]
The sodium hydroxide amount is
\[
n=MV
\]
\[
n=1.00\times0.0750=0.0750\,\mathrm{mol}
\]
The saponification ratio is \(1:1\).
Sodium hydroxide is therefore limiting.
\[
n_{\mathrm{ester,hydrolysed}}=0.0750\,\mathrm{mol}
\]
The isolated benzoic acid amount is
\[
n=0.0750\times0.800=0.0600\,\mathrm{mol}
\]
\[
m=0.0600\times122=7.32\,\mathrm{g}
\]
The unreacted ester amount is
\[
n=0.100-0.0750=0.0250\,\mathrm{mol}
\]
\[
m=0.0250\times136=3.40\,\mathrm{g}
\]
Acidification changes benzoate into benzoic acid but does not hydrolyse the remaining ester under the stated calculation.
412. Which row correctly gives the products formed under the stated conditions?
| Row | Acid derivative and condition | Products before any additional work-up |
|---|
| P | Ethyl ethanoate with aqueous \(\mathrm{NaOH}\) | Ethanoic acid and sodium ethoxide |
| Q | Ethyl ethanoate with dilute acid and water | Sodium ethanoate and ethanol |
| R | Ethanoic anhydride with water | Ethanoic acid and ethanol |
| S | Acetyl chloride with water | Ethanoic acid and hydrogen chloride |
ⓐ. Row P
ⓑ. Row S
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: Acetyl chloride reacts rapidly with water to give ethanoic acid and hydrogen chloride. Alkaline hydrolysis of ethyl ethanoate gives sodium ethanoate and ethanol rather than the free acid. Acidic ester hydrolysis gives ethanoic acid and ethanol, not a sodium salt. Ethanoic anhydride gives two molecules of ethanoic acid on hydrolysis. Row S is the only product mapping that matches the stated medium.
413. Ethanoic anhydride, \(20.4\,\mathrm{g}\), is mixed with \(2.70\,\mathrm{g}\) of water:
\[
\mathrm{(CH_3CO)_2O+H_2O\rightarrow2CH_3COOH}
\]
Ethanoic acid is recovered in \(80.0\%\) yield. Which pair gives the recovered acid mass and the mass of unreacted anhydride?
ⓐ. \(18.0\,\mathrm{g}\) and \(0\,\mathrm{g}\)
ⓑ. \(14.4\,\mathrm{g}\) and \(10.2\,\mathrm{g}\)
ⓒ. \(14.4\,\mathrm{g}\) and \(5.10\,\mathrm{g}\)
ⓓ. \(12.0\,\mathrm{g}\) and \(5.10\,\mathrm{g}\)
Correct Answer: \(14.4\,\mathrm{g}\) and \(5.10\,\mathrm{g}\)
Explanation: The initial anhydride amount is
\[
n=\frac{20.4}{102}=0.200\,\mathrm{mol}
\]
The water amount is
\[
n=\frac{2.70}{18.0}=0.150\,\mathrm{mol}
\]
The reactants combine in a \(1:1\) ratio.
Water is the limiting reactant.
\[
n_{\mathrm{anhydride,hydrolysed}}=0.150\,\mathrm{mol}
\]
Two moles of acid form per mole of anhydride.
\[
n_{\mathrm{acid,theoretical}}=2(0.150)=0.300\,\mathrm{mol}
\]
The recovered acid amount is
\[
n=0.300\times0.800=0.240\,\mathrm{mol}
\]
\[
m=0.240\times60.0=14.4\,\mathrm{g}
\]
The anhydride left is
\[
n=0.200-0.150=0.0500\,\mathrm{mol}
\]
\[
m=0.0500\times102=5.10\,\mathrm{g}
\]
414. Acetyl chloride, \(7.85\,\mathrm{g}\), is completely hydrolysed in excess water. The mixture is then treated with \(150\,\mathrm{mL}\) of \(1.00\,\mathrm{mol\,L^{-1}}\) sodium hydroxide. The base neutralises hydrogen chloride before neutralising ethanoic acid. Which pair gives the mass of ethanoic acid remaining and the mass of sodium chloride formed? Use \(M(\mathrm{acetyl\ chloride})=78.5\,\mathrm{g\,mol^{-1}}\) and \(M(\mathrm{NaCl})=58.5\,\mathrm{g\,mol^{-1}}\).
ⓐ. \(3.00\,\mathrm{g}\) and \(5.85\,\mathrm{g}\)
ⓑ. \(6.00\,\mathrm{g}\) and \(2.93\,\mathrm{g}\)
ⓒ. \(3.00\,\mathrm{g}\) and \(8.78\,\mathrm{g}\)
ⓓ. \(0\,\mathrm{g}\) and \(5.85\,\mathrm{g}\)
Correct Answer: \(3.00\,\mathrm{g}\) and \(5.85\,\mathrm{g}\)
Explanation: The acetyl chloride amount is
\[
n=\frac{7.85}{78.5}=0.100\,\mathrm{mol}
\]
Hydrolysis produces
\[
0.100\,\mathrm{mol}
\]
of ethanoic acid and
\[
0.100\,\mathrm{mol}
\]
of hydrogen chloride.
The sodium hydroxide amount is
\[
n=1.00\times0.150=0.150\,\mathrm{mol}
\]
Neutralising hydrogen chloride consumes
\[
0.100\,\mathrm{mol}
\]
of sodium hydroxide.
The base remaining for ethanoic acid is
\[
0.150-0.100=0.0500\,\mathrm{mol}
\]
Thus, ethanoic acid left is
\[
0.100-0.0500=0.0500\,\mathrm{mol}
\]
\[
m=0.0500\times60.0=3.00\,\mathrm{g}
\]
Hydrogen chloride neutralisation produces
\[
0.100\,\mathrm{mol}
\]
of sodium chloride.
\[
m(\mathrm{NaCl})=0.100\times58.5=5.85\,\mathrm{g}
\]
415. A student claims that an ester and an acyl chloride should hydrolyse at similar rates because both contain a carbonyl group. The best correction is:
ⓐ. Esters react faster because alkoxide leaves more readily and their carbonyl is more electrophilic
ⓑ. Both derivatives react at the same negligible rate because water is weak
ⓒ. Hydrolysis rate depends only on molar mass, not leaving-group ability
ⓓ. Acyl chlorides react faster: chloride leaves readily and the carbonyl is more electrophilic
Correct Answer: Acyl chlorides react faster: chloride leaves readily and the carbonyl is more electrophilic
Explanation: Possession of a carbonyl group alone does not determine the rate of nucleophilic acyl substitution. Chlorine withdraws electron density strongly and increases the partial positive charge on the acyl carbon. Chloride ion is also a relatively stable leaving group. In an ester, the alkoxy group donates electron density by resonance and alkoxide is a poorer leaving group. These electronic and leaving-group differences make acyl-chloride hydrolysis far more rapid.
416. A concentration-time graph is recorded during acid-catalysed hydrolysis of an ester in a closed vessel. Curve P decreases and then becomes constant above zero. Curves Q and R increase from zero and reach equal plateaus. What do P, Q, and R represent?
ⓐ. P: water; Q and R: acid catalyst and unreacted ester
ⓑ. P: ester; Q and R: carboxylic acid and alcohol
ⓒ. P: carboxylic acid; Q and R: ester and water
ⓓ. P: alcohol; Q and R: ester and acid catalyst
Correct Answer: P: ester; Q and R: carboxylic acid and alcohol
Explanation: The ester is consumed as hydrolysis proceeds, so its concentration decreases. Because acidic hydrolysis is reversible, a non-zero amount of ester remains when equilibrium is reached. One mole of ester produces one mole each of carboxylic acid and alcohol. Their concentrations therefore rise together and reach equal increases when they start from zero. The acid catalyst changes the rate of reaching equilibrium but is not consumed according to this stoichiometry.
417. A \(14.69\,\mathrm{g}\) mixture contains ethyl ethanoate and acetyl chloride. Complete alkaline hydrolysis consumes \(0.250\,\mathrm{mol}\) of sodium hydroxide. Use the net equations:
\[
\mathrm{CH_3COOC_2H_5+OH^-\rightarrow CH_3COO^-+C_2H_5OH}
\]
\[
\mathrm{CH_3COCl+2OH^-\rightarrow CH_3COO^-+Cl^-+H_2O}
\]
After acidification, ethanoic acid is recovered in \(80.0\%\) yield. Which pair gives the mole percentage of acetyl chloride and the recovered acid mass?
ⓐ. \(25.0\%\) and \(10.5\,\mathrm{g}\)
ⓑ. \(50.0\%\) and \(8.40\,\mathrm{g}\)
ⓒ. \(42.9\%\) and \(8.40\,\mathrm{g}\)
ⓓ. \(42.9\%\) and \(10.5\,\mathrm{g}\)
Correct Answer: \(42.9\%\) and \(8.40\,\mathrm{g}\)
Explanation: Let \(x\) be moles of ethyl ethanoate and \(y\) be moles of acetyl chloride.
The mass equation is
\[
88x+78.5y=14.69
\]
The sodium hydroxide equation is
\[
x+2y=0.250
\]
From the second equation,
\[
x=0.250-2y
\]
Substitute into the mass equation:
\[
88(0.250-2y)+78.5y=14.69
\]
\[
22.0-97.5y=14.69
\]
\[
y=0.0750\,\mathrm{mol}
\]
\[
x=0.250-2(0.0750)=0.100\,\mathrm{mol}
\]
The acetyl-chloride mole percentage is
\[
\frac{0.0750}{0.100+0.0750}\times100=42.9\%.
\]
Both derivatives produce one acetate ion per mole.
\[
n_{\mathrm{acid,theoretical}}=0.100+0.0750=0.175\,\mathrm{mol}
\]
The recovered acid amount is
\[
n=0.175\times0.800=0.140\,\mathrm{mol}
\]
\[
m=0.140\times60.0=8.40\,\mathrm{g}
\]
418. The preparation of a carboxylic acid from a Grignard reagent is represented by:
ⓐ. \(\mathrm{RMgX\xrightarrow{H_2O}ROH\xrightarrow{CO_2}RCOOH}\)
ⓑ. \(\mathrm{RMgX\xrightarrow{CO_2}RCOOMgX\xrightarrow{H_3O^+}RCOOH}\)
ⓒ. \(\mathrm{RMgX\xrightarrow{O_2}ROOR\xrightarrow{H^+}RCOOH}\)
ⓓ. \(\mathrm{RMgX\xrightarrow{NH_3}RNH_2\xrightarrow{CO_2}RCOOH}\)
Correct Answer: \(\mathrm{RMgX\xrightarrow{CO_2}RCOOMgX\xrightarrow{H_3O^+}RCOOH}\)
Explanation: The carbon bonded to magnesium in a Grignard reagent behaves as a strongly nucleophilic carbon centre. It attacks the electrophilic carbon atom of carbon dioxide. This forms a magnesium carboxylate salt, \(\mathrm{RCOOMgX}\). Acidic work-up protonates the carboxylate and gives \(\mathrm{RCOOH}\). The carbon dioxide carbon becomes the new carboxyl carbon, so the product contains one more carbon atom than the organic group \(\mathrm{R}\).
419. Assertion: A Grignard reagent must react with carbon dioxide before water or dilute acid is added.
Reason: Water and acids protonate the carbon-magnesium bond and destroy the Grignard reagent.
ⓐ. 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
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A Grignard reagent contains a highly polar carbon-magnesium bond. The carbon behaves as a strong base as well as a nucleophile. Contact with water or an acid protonates this carbon and produces the corresponding hydrocarbon. The reagent would then no longer be available to attack carbon dioxide. Acidic hydrolysis must therefore be performed only after the magnesium carboxylate has formed.
420. Methylmagnesium bromide reacts with \({}^{13}\mathrm{CO_2}\), followed by acidic hydrolysis. Where is the isotope located in the product?
ⓐ. In the methyl carbon of methanol
ⓑ. In the carbon atom bonded to magnesium before reaction
ⓒ. In the carboxyl carbon of ethanoic acid
ⓓ. In carbon dioxide released during work-up
Correct Answer: In the carboxyl carbon of ethanoic acid
Explanation: The methyl carbon of the Grignard reagent attacks the labelled carbon of carbon dioxide. A new carbon-carbon bond forms between these two atoms. The \({}^{13}\mathrm{C}\) atom retains its bonds to oxygen and becomes the carbon of the carboxylate group. Acidic work-up changes the magnesium carboxylate into ethanoic acid. The isotope is therefore found at the carboxyl carbon rather than at the original methyl carbon.
