301. Which product pair is obtained from reductive ozonolysis of \((CH_3)_2C=CH_2\)?
ⓐ. \(CH_3CHO\) and \(CH_3CHO\)
ⓑ. \(CH_3COOH\) and \(HCOOH\)
ⓒ. \(CH_3CH_2CHO\) and \(HCHO\)
ⓓ. \(CH_3COCH_3\) and \(HCHO\)
Correct Answer: \(CH_3COCH_3\) and \(HCHO\)
Explanation: \(\textbf{Starting alkene:}\)
\((CH_3)_2C=CH_2\) has one double-bond carbon attached to two \(CH_3\) groups and the other as \(CH_2\).
\(\textbf{Cleavage rule:}\)
Each carbon of \(C=C\) becomes a carbonyl carbon.
\(\textbf{Ketone fragment:}\)
The \((CH_3)_2C=\) carbon becomes \(CH_3COCH_3\), propanone.
\(\textbf{Aldehyde fragment:}\)
The \(=CH_2\) carbon becomes \(HCHO\), methanal.
\(\textbf{Conclusion:}\)
The products are \(CH_3COCH_3\) and \(HCHO\).
302. Which alkene gives propanone and ethanal on reductive ozonolysis?
ⓐ. \(CH_2=CHCH_2CH_3\)
ⓑ. \(CH_3CH=CHCH_3\)
ⓒ. \((CH_3)_2C=CHCH_3\)
ⓓ. \((CH_3)_2C=C(CH_3)_2\)
Correct Answer: \((CH_3)_2C=CHCH_3\)
Explanation: \(\textbf{Target products:}\)
Propanone is \(CH_3COCH_3\), and ethanal is \(CH_3CHO\).
\(\textbf{Reverse mapping:}\)
Join the carbonyl carbons of the two products with a double bond.
\(\textbf{From propanone:}\)
The carbonyl carbon has two \(CH_3\) groups, giving \((CH_3)_2C=\).
\(\textbf{From ethanal:}\)
The carbonyl carbon has \(CH_3\) and \(H\), giving \(=CHCH_3\).
\(\textbf{Alkene:}\)
The original alkene is \((CH_3)_2C=CHCH_3\).
303. Which ozonolysis result indicates that the original alkene was symmetrical?
ⓐ. Two different aldehydes are formed.
ⓑ. Only one carbonyl compound is formed.
ⓒ. A ketone and an alcohol are formed.
ⓓ. Carbon dioxide is the only product.
Correct Answer: Only one carbonyl compound is formed.
Explanation: A symmetrical alkene has identical environments on both sides of the double bond. When such an alkene undergoes reductive ozonolysis, the two fragments can be identical. This gives only one type of carbonyl compound, though two molecules may form. For example, but-\(2\)-ene gives only ethanal.
304. Which alkene gives only propanone on reductive ozonolysis?
ⓐ. \((CH_3)_2C=C(CH_3)_2\)
ⓑ. \(CH_3CH_2C(CH_3)=CH_2\)
ⓒ. \(CH_3CH=CH_2\)
ⓓ. \(CH_2=CH_2\)
Correct Answer: \((CH_3)_2C=C(CH_3)_2\)
Explanation: \(\textbf{Target product:}\)
Only propanone, \(CH_3COCH_3\), should form.
\(\textbf{Reverse idea:}\)
Each double-bond carbon in the alkene must have two \(CH_3\) groups attached.
\(\textbf{Alkene structure:}\)
\((CH_3)_2C=C(CH_3)_2\) has identical substituted double-bond carbons.
\(\textbf{Cleavage result:}\)
Both sides become \(CH_3COCH_3\).
\(\textbf{Conclusion:}\)
\[(CH_3)_2C=C(CH_3)_2 \xrightarrow[Zn/H_2O]{O_3} 2CH_3COCH_3\]
305. An alkene gives \(HCHO\) and \(CH_3CH_2CHO\) on reductive ozonolysis. Which alkene is most suitable?
ⓐ. \(CH_3CH=CHCH_3\)
ⓑ. \(CH_3CH_2CH=CHCH_3\)
ⓒ. \(CH_3C(CH_3)=CH_2\)
ⓓ. \(CH_3CH_2CH=CH_2\)
Correct Answer: \(CH_3CH_2CH=CH_2\)
Explanation: \(\textbf{Products given:}\)
The products are \(HCHO\) and \(CH_3CH_2CHO\).
\(\textbf{Reverse mapping principle:}\)
Join the carbonyl carbons of the two products by a double bond.
\(\textbf{From }HCHO\textbf{:}\)
The carbonyl carbon contributes a \(CH_2\) end.
\(\textbf{From }CH_3CH_2CHO\textbf{:}\)
The carbonyl carbon contributes \(CHCH_2CH_3\).
\(\textbf{Alkene formed:}\)
The alkene is \(CH_3CH_2CH=CH_2\), but-\(1\)-ene.
306. Which statement is correct for the ozonolysis product from a double-bond carbon attached to two carbon groups?
ⓐ. It must become methanal.
ⓑ. It must become a carboxylic acid under reductive work-up.
ⓒ. It forms a ketone fragment.
ⓓ. It forms an ether fragment.
Correct Answer: It forms a ketone fragment.
Explanation: During reductive ozonolysis, each carbon of the original double bond becomes a carbonyl carbon. If that carbon was attached to two carbon groups, the product carbonyl carbon is also attached to two carbon groups. Such a carbonyl compound is a ketone. If the double-bond carbon had hydrogen attached, the product could be an aldehyde.
307. Which product is expected from reductive ozonolysis of \(CH_2=C(CH_3)CH_2CH_3\)?
ⓐ. \(CH_3CHO\) and \(CH_3COCH_3\)
ⓑ. \(HCHO\) and \(CH_3COCH_2CH_3\)
ⓒ. \(HCOOH\) and \(CH_3CH_2COOH\)
ⓓ. \(CH_3COCH_3\) and \(CH_3OH\)
Correct Answer: \(HCHO\) and \(CH_3COCH_2CH_3\)
Explanation: \(\textbf{Starting alkene:}\)
\(CH_2=C(CH_3)CH_2CH_3\) has one terminal \(CH_2\) carbon and one substituted double-bond carbon.
\(\textbf{Terminal fragment:}\)
The \(CH_2\) carbon becomes \(HCHO\).
\(\textbf{Substituted fragment:}\)
The other double-bond carbon is attached to \(CH_3\) and \(CH_2CH_3\), so it becomes \(CH_3COCH_2CH_3\).
\(\textbf{Final product pair:}\)
Reductive ozonolysis gives methanal and butan-\(2\)-one.
308. Which equation correctly represents reductive ozonolysis of propene?
ⓐ. \[CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3CHO + HCHO\]
ⓑ. \[CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3COCH_3\]
ⓒ. \[CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3CH_2CH_2OH\]
ⓓ. \[CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3COOH + CO_2\]
Correct Answer: \[CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3CHO + HCHO\]
Explanation: \(\textbf{Cleavage rule:}\)
Ozonolysis breaks the \(C=C\) bond and converts each alkene carbon into a carbonyl carbon.
\(\textbf{Left side of propene:}\)
The \(CH_3CH=\) portion becomes \(CH_3CHO\).
\(\textbf{Right side of propene:}\)
The \(=CH_2\) portion becomes \(HCHO\).
\(\textbf{Work-up condition:}\)
\(Zn/H_2O\) indicates reductive work-up, so aldehydes are retained.
\(\textbf{Conclusion:}\)
\[CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3CHO + HCHO\]
309. Which carbonyl compound is obtained by hydration of ethyne followed by tautomerisation?
ⓐ. Methanal
ⓑ. Propanone
ⓒ. Ethanoic acid
ⓓ. Ethanal
Correct Answer: Ethanal
Explanation: Ethyne undergoes hydration in the presence of \(Hg^{2+}\), \(H_2SO_4\), and water to form an unstable enol. This enol rapidly rearranges by keto-enol tautomerism. The stable product is ethanal, \(CH_3CHO\). This reaction is an important preparation route for ethanal from ethyne.
310. Which reagent condition is used for hydration of ethyne to ethanal?
ⓐ. \(Pd/BaSO_4,\ H_2\)
ⓑ. \(SnCl_2/HCl\), followed by water
ⓒ. \(Hg^{2+}, H_2SO_4, H_2O\)
ⓓ. \(Zn(Hg)/HCl\)
Correct Answer: \(Hg^{2+}, H_2SO_4, H_2O\)
Explanation: Hydration of ethyne is carried out using water in the presence of \(Hg^{2+}\) and dilute \(H_2SO_4\). The initial addition gives an enol-type intermediate. This intermediate tautomerises to ethanal. The other reagent sets correspond to Rosenmund reduction, Stephen reduction-type conditions, or Clemmensen reduction, not alkyne hydration.
311. Which product is mainly obtained when ethyne is hydrated in the presence of \(Hg^{2+}\), \(H_2SO_4\), and \(H_2O\)?
ⓐ. Methanal
ⓑ. Ethanal
ⓒ. Propanone
ⓓ. Ethanoic acid
Correct Answer: Ethanal
Explanation: \(\textbf{Starting alkyne:}\)
Ethyne is \(HC \equiv CH\).
\(\textbf{Reaction condition:}\)
Hydration occurs in the presence of \(Hg^{2+}\), \(H_2SO_4\), and \(H_2O\).
\(\textbf{Initial change:}\)
Water adds across the triple bond to give an unstable enol.
\(\textbf{Tautomerisation:}\)
The enol rearranges into a more stable carbonyl compound.
\(\textbf{Product:}\)
\[HC \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} CH_3CHO\]
\(\textbf{Conclusion:}\)
The product is ethanal.
312. Which statement best describes the role of tautomerisation in hydration of alkynes?
ⓐ. It converts the carbonyl compound into an ether salt.
ⓑ. It changes the alkyne directly into an alkyl halide.
ⓒ. It removes the oxygen atom from the hydrated product.
ⓓ. It changes the first enol into a carbonyl compound.
Correct Answer: It changes the first enol into a carbonyl compound.
Explanation: Hydration of an alkyne first gives an enol, a compound containing \(C=C\) and \(OH\) attached to a double-bond carbon. Enols are often unstable under these conditions. They rearrange into the more stable carbonyl form through keto-enol tautomerism. This is why alkyne hydration usually gives aldehydes or ketones rather than isolable enols.
313. Hydration of a higher terminal alkyne such as propyne generally gives which type of major carbonyl product?
ⓐ. Aldehyde
ⓑ. Ether
ⓒ. Ketone
ⓓ. Carboxylic acid
Correct Answer: Ketone
Explanation: Higher alkynes generally give ketones on hydration under \(Hg^{2+}/H_2SO_4\) conditions. Water first adds across the triple bond to form an enol intermediate. The enol then undergoes tautomerisation to a carbonyl compound. For terminal alkynes higher than ethyne, Markovnikov hydration followed by tautomerisation gives a methyl ketone.
314. What is the main product of hydration of propyne, \(CH_3C \equiv CH\), under \(Hg^{2+}/H_2SO_4\) conditions?
ⓐ. Propanal
ⓑ. Propanone
ⓒ. Ethanal
ⓓ. Propanoic acid
Correct Answer: Propanone
Explanation: \(\textbf{Starting alkyne:}\)
Propyne is \(CH_3C \equiv CH\).
\(\textbf{Condition:}\)
Hydration is carried out with \(Hg^{2+}\), \(H_2SO_4\), and \(H_2O\).
\(\textbf{Addition pattern:}\)
Water adds across the triple bond to form an enol intermediate.
\(\textbf{Tautomerisation:}\)
The enol rearranges to the more stable ketone form.
\(\textbf{Product:}\)
\[CH_3C \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} CH_3COCH_3\]
\(\textbf{Conclusion:}\)
The main product is propanone.
315. Which equation correctly represents hydration of ethyne followed by tautomerisation?
ⓐ. \[HC \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} CH_3CHO\]
ⓑ. \[HC \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} CH_3COCH_3\]
ⓒ. \[HC \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} HCOOH\]
ⓓ. \[HC \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} CH_3CH_2OH\]
Correct Answer: \[HC \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} CH_3CHO\]
Explanation: \(\textbf{Substrate:}\)
The substrate is ethyne, \(HC \equiv CH\).
\(\textbf{Reaction family:}\)
Alkyne hydration gives an enol first.
\(\textbf{Tautomerisation step:}\)
The enol changes into the corresponding carbonyl compound.
\(\textbf{Special case:}\)
Ethyne gives ethanal, not a ketone.
\(\textbf{Final equation:}\)
\[HC \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+},\ H_2O} CH_3CHO\]
316. Which intermediate type is formed before ethanal appears during hydration of ethyne?
ⓐ. Acetal
ⓑ. Enol
ⓒ. Cyanohydrin
ⓓ. Carboxylate
Correct Answer: Enol
Explanation: Hydration of ethyne first places \(H\) and \(OH\) across the triple bond. The immediate product has an \(OH\) group attached to a carbon-carbon double bond, so it is an enol. This enol is not the final stable product under the reaction conditions. It rearranges by tautomerisation to ethanal, \(CH_3CHO\).
317. Which alkyne is most directly converted into ethanal by acid-catalysed hydration with \(Hg^{2+}\)?
ⓐ. Propyne
ⓑ. But-\(1\)-yne
ⓒ. But-\(2\)-yne
ⓓ. Ethyne
Correct Answer: Ethyne
Explanation: Ethyne is the alkyne that gives ethanal on hydration under \(Hg^{2+}/H_2SO_4\) conditions. The reaction proceeds through an enol intermediate. Tautomerisation then gives \(CH_3CHO\). Higher alkynes usually give ketones rather than ethanal under these conditions.
318. Which statement is incorrect about hydration of alkynes under \(Hg^{2+}/H_2SO_4\) conditions?
ⓐ. Ethyne gives ethanal after tautomerisation.
ⓑ. Higher alkynes can give ketones.
ⓒ. The first formed enol may rearrange.
ⓓ. It directly gives carboxylate salts only.
Correct Answer: It directly gives carboxylate salts only.
Explanation: Hydration of alkynes gives carbonyl compounds after enol formation and tautomerisation. Ethyne gives ethanal, while many higher alkynes give ketones. The reaction does not directly produce carboxylate salts as the usual product. Carboxylate salts are more closely associated with acid-base reactions of carboxylic acids.
319. Which product is obtained from hydration of but-\(2\)-yne followed by tautomerisation?
ⓐ. Butan-\(1\)-ol
ⓑ. Butanal
ⓒ. Butan-\(2\)-one
ⓓ. Butanoic acid
Correct Answer: Butan-\(2\)-one
Explanation: \(\textbf{Starting alkyne:}\)
But-\(2\)-yne is \(CH_3C \equiv CCH_3\).
\(\textbf{Hydration step:}\)
Water adds across the triple bond under acidic mercuric ion conditions.
\(\textbf{Intermediate:}\)
An enol is formed first.
\(\textbf{Tautomerisation:}\)
The enol rearranges into a ketone.
\(\textbf{Product:}\)
For symmetrical but-\(2\)-yne, the carbonyl product is butan-\(2\)-one.
\(\textbf{Conclusion:}\)
The product is butan-\(2\)-one.
320. Which reaction route gives ethanal from ethyne?
ⓐ. Ozonolysis with \(O_3/Zn/H_2O\)
ⓑ. Hydration with \(Hg^{2+}/H_2SO_4/H_2O\)
ⓒ. Rosenmund reduction with \(Pd/BaSO_4\)
ⓓ. Clemmensen reduction with \(Zn(Hg)/HCl\)
Correct Answer: Hydration with \(Hg^{2+}/H_2SO_4/H_2O\)
Explanation: Ethyne can be converted into ethanal by hydration in the presence of \(Hg^{2+}\), \(H_2SO_4\), and water. The reaction forms an enol initially, which tautomerises to ethanal. Ozonolysis of alkenes is a different route to carbonyl compounds. Rosenmund reduction prepares aldehydes from acyl chlorides, while Clemmensen reduction removes the carbonyl oxygen.
