501. Which product is formed when a primary alcohol is oxidised mildly?
ⓐ. Carboxylic acid
ⓑ. Aldehyde
ⓒ. Ether
ⓓ. Alkene
Correct Answer: Aldehyde
Explanation: A primary alcohol has the general structure \(R-CH_2OH\). Mild oxidation removes hydrogen from the \(O-H\) group and from the carbon bearing \(OH\), forming a carbonyl group. This gives an aldehyde, \(R-CHO\). On stronger oxidation, the aldehyde can further oxidise to a carboxylic acid.
502. Which product is obtained when a secondary alcohol is oxidised?
ⓐ. Aldehyde
ⓑ. Ether
ⓒ. Alkene
ⓓ. Ketone
Correct Answer: Ketone
Explanation: A secondary alcohol has the general structure \(R-CH(OH)-R'\). Oxidation converts the \(C-OH\) part into a carbonyl group, \(C=O\). Since the carbonyl carbon remains attached to two carbon groups, the product is a ketone. Secondary alcohols do not normally oxidise to carboxylic acids without breaking carbon-carbon bonds.
503. Which alcohol gives ethanal on controlled oxidation?
ⓐ. \(CH_3CH_2OH\)
ⓑ. \(CH_3OH\)
ⓒ. \(CH_3CH(OH)CH_3\)
ⓓ. \(CH_3OCH_3\)
Correct Answer: \(CH_3CH_2OH\)
Explanation: \(\textbf{Starting alcohol needed:}\)
Ethanal is \(CH_3CHO\).
\(\textbf{Reverse mapping:}\)
The aldehyde group \(CHO\) comes from oxidation of a terminal \(CH_2OH\) group.
\(\textbf{Alcohol:}\)
\(CH_3CH_2OH\) has the required terminal \(CH_2OH\) group.
\(\textbf{Reaction:}\)
\[
CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O
\]
\(\textbf{Final Answer:}\)
Ethanol gives ethanal on controlled oxidation.
504. Which product is formed when propan-\(2\)-ol is oxidised?
ⓐ. Propanal
ⓑ. Propanoic acid
ⓒ. Propanone
ⓓ. Propene
Correct Answer: Propanone
Explanation: Propan-\(2\)-ol, \(CH_3CH(OH)CH_3\), is a secondary alcohol. Oxidation of a secondary alcohol gives a ketone. The carbon bearing \(OH\) becomes the carbonyl carbon, while both methyl groups remain attached. Therefore the product is propanone, \(CH_3COCH_3\).
505. Which equation represents oxidation of ethanol to ethanal?
ⓐ. \[CH_3CH_2OH + HBr \rightarrow CH_3CH_2Br + H_2O\]
ⓑ. \[CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O\]
ⓒ. \[CH_3CHO + 2[H] \rightarrow CH_3CH_2OH\]
ⓓ. \[2CH_3CH_2OH \rightarrow CH_3CH_2OCH_2CH_3 + H_2O\]
Correct Answer: \[CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O\]
Explanation: \(\textbf{Alcohol type:}\)
Ethanol is a primary alcohol.
\(\textbf{Oxidation rule:}\)
Controlled oxidation of a primary alcohol gives an aldehyde.
\(\textbf{Product:}\)
The terminal \(CH_2OH\) group becomes \(CHO\).
\(\textbf{Equation:}\)
\[
CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O
\]
\(\textbf{Final Answer:}\)
Ethanol is oxidised to ethanal.
506. Which reagent can oxidise a primary alcohol to a carboxylic acid under strong oxidising conditions?
ⓐ. Acidified \(KMnO_4\)
ⓑ. Neutral \(KMnO_4\) at room temperature
ⓒ. Dry ether
ⓓ. \(NaCl\)
Correct Answer: Acidified \(KMnO_4\)
Explanation: Acidified \(KMnO_4\) is a strong oxidising agent. A primary alcohol can first oxidise to an aldehyde and then further to a carboxylic acid. Under strong oxidising conditions, the aldehyde stage is usually not the final product. \(NaBH_4\) is a reducing agent and does not carry out this oxidation.
507. What is the final oxidation product of ethanol under strong oxidising conditions?
ⓐ. Methanol
ⓑ. Ethene
ⓒ. Diethyl ether
ⓓ. Ethanoic acid
Correct Answer: Ethanoic acid
Explanation: \(\textbf{Starting compound:}\)
Ethanol is \(CH_3CH_2OH\), a primary alcohol.
\(\textbf{First oxidation:}\)
It can form ethanal, \(CH_3CHO\).
\(\textbf{Further oxidation:}\)
The aldehyde is further oxidised to ethanoic acid, \(CH_3COOH\).
\(\textbf{Overall change:}\)
\[
CH_3CH_2OH \xrightarrow{[O]} CH_3CHO \xrightarrow{[O]} CH_3COOH
\]
\(\textbf{Final Answer:}\)
Strong oxidation gives ethanoic acid.
508. Which alcohol is resistant to oxidation under ordinary conditions because it has no hydrogen on the carbon bearing \(OH\)?
ⓐ. \(CH_3CH_2OH\)
ⓑ. \(CH_3CH(OH)CH_3\)
ⓒ. \((CH_3)_3COH\)
ⓓ. \(CH_3CH_2CH_2OH\)
Correct Answer: \((CH_3)_3COH\)
Explanation: Tertiary alcohols have the \(OH\)-bearing carbon attached to three carbon groups. This carbon has no hydrogen atom attached to it. Oxidation to a carbonyl compound normally requires removal of hydrogen from the \(OH\)-bearing carbon. Therefore tertiary alcohols resist oxidation under ordinary conditions unless harsh conditions break carbon-carbon bonds.
509. Which product is obtained when \(CH_3CH_2CH_2OH\) undergoes controlled oxidation?
ⓐ. \(CH_3CH_2OCH_3\)
ⓑ. \(CH_3COCH_3\)
ⓒ. \(CH_3CH_2COOH\)
ⓓ. \(CH_3CH_2CHO\)
Correct Answer: \(CH_3CH_2CHO\)
Explanation: \(CH_3CH_2CH_2OH\) is propan-\(1\)-ol, a primary alcohol. Controlled oxidation of a primary alcohol gives the corresponding aldehyde. The terminal \(CH_2OH\) group changes into \(CHO\). Therefore propan-\(1\)-ol gives propanal, \(CH_3CH_2CHO\).
510. Which product is formed when butan-\(2\)-ol is oxidised?
ⓐ. Butanal
ⓑ. Butanoic acid
ⓒ. But-\(2\)-ene
ⓓ. Butan-\(2\)-one
Correct Answer: Butan-\(2\)-one
Explanation: Butan-\(2\)-ol is a secondary alcohol because the \(OH\)-bearing carbon is attached to two carbon atoms. Oxidation converts the \(CHOH\) unit into a carbonyl group. The carbon skeleton remains unchanged, giving \(CH_3COCH_2CH_3\). This compound is butan-\(2\)-one.
511. Which statement correctly compares oxidation of primary and secondary alcohols?
ⓐ. Primary alcohols give only ketones, while secondary alcohols give aldehydes.
ⓑ. Primary alcohols give aldehydes or acids; secondary alcohols give ketones.
ⓒ. Primary and secondary alcohols both give only ethers.
ⓓ. Secondary alcohols always give carboxylic acids without bond breaking.
Correct Answer: Primary alcohols give aldehydes or acids; secondary alcohols give ketones.
Explanation: Primary alcohols contain a terminal \(CH_2OH\) group. Controlled oxidation can give an aldehyde, while stronger oxidation can give a carboxylic acid. Secondary alcohols contain \(CHOH\) between two carbon groups and oxidise to ketones. This difference is a major reaction-based method for distinguishing alcohol classes.
512. Which alcohol gives acetone on oxidation?
ⓐ. \(CH_3CH_2OH\)
ⓑ. \(CH_3CH_2CH_2OH\)
ⓒ. \(CH_3CH(OH)CH_3\)
ⓓ. \((CH_3)_3COH\)
Correct Answer: \(CH_3CH(OH)CH_3\)
Explanation: Acetone is the common name for propanone, \(CH_3COCH_3\). The alcohol that gives propanone on oxidation must have \(OH\) on the middle carbon of a three-carbon chain. \(CH_3CH(OH)CH_3\), propan-\(2\)-ol, has this structure. Oxidation changes \(CH(OH)\) into \(C=O\), giving acetone.
513. Which equation correctly represents oxidation of propan-\(2\)-ol?
ⓐ. \[CH_3CH(OH)CH_3 + [O] \rightarrow CH_3COCH_3 + H_2O\]
ⓑ. \[CH_3CH_2CH_2OH + [O] \rightarrow CH_3COCH_3 + H_2O\]
ⓒ. \[(CH_3)_3COH + [O] \rightarrow CH_3COOH + CH_4\]
ⓓ. \[CH_3OCH_3 + [O] \rightarrow CH_3CH_2OH\]
Correct Answer: \[CH_3CH(OH)CH_3 + [O] \rightarrow CH_3COCH_3 + H_2O\]
Explanation: \(\textbf{Alcohol type:}\)
Propan-\(2\)-ol is a secondary alcohol.
\(\textbf{Oxidation rule:}\)
Secondary alcohols are oxidised to ketones.
\(\textbf{Product:}\)
The \(CH(OH)\) group becomes \(CO\), giving propanone.
\(\textbf{Equation:}\)
\[
CH_3CH(OH)CH_3 + [O] \rightarrow CH_3COCH_3 + H_2O
\]
\(\textbf{Final Answer:}\)
Propan-\(2\)-ol gives propanone.
514. Which alcohol cannot be oxidised to a carbonyl compound without breaking a carbon-carbon bond?
ⓐ. \(CH_3CH_2OH\)
ⓑ. \(CH_3CH_2CH_2OH\)
ⓒ. \(CH_3CH(OH)CH_3\)
ⓓ. \((CH_3)_3COH\)
Correct Answer: \((CH_3)_3COH\)
Explanation: \((CH_3)_3COH\) is a tertiary alcohol. The carbon bearing \(OH\) is attached to three methyl groups and has no hydrogen attached to it. Formation of a carbonyl compound by oxidation normally needs removal of a hydrogen from the \(OH\)-bearing carbon. Without such a hydrogen, ordinary oxidation to a carbonyl compound is not possible without carbon-carbon bond cleavage.
515. Which conversion is an oxidation reaction?
ⓐ. \(CH_3CHO \rightarrow CH_3CH_2OH\)
ⓑ. \(CH_2=CH_2 \rightarrow CH_3CH_2OH\)
ⓒ. \(CH_3CH_2OH \rightarrow CH_3CHO\)
ⓓ. \(CH_3CH_2OH \rightarrow CH_3CH_2Cl\)
Correct Answer: \(CH_3CH_2OH \rightarrow CH_3CHO\)
Explanation: Oxidation of ethanol removes hydrogen and forms a carbonyl compound, ethanal. The conversion \(CH_3CHO \rightarrow CH_3CH_2OH\) is reduction, not oxidation. Hydration of ethene forms ethanol by addition of water. Conversion of ethanol to chloroethane is substitution, not oxidation.
516. Which test reagent gives an orange-to-green colour change with oxidisable alcohols?
ⓐ. Acidified \(KMnO_4\) solution
ⓑ. Acidified \(K_2Cr_2O_7\)
ⓒ. Dry ether
ⓓ. \(NaHCO_3\)
Correct Answer: Acidified \(K_2Cr_2O_7\)
Explanation: Acidified potassium dichromate, \(K_2Cr_2O_7\), is a common oxidising agent for alcohols. During oxidation, dichromate ion is reduced, producing a characteristic colour change from orange to green. Primary and secondary alcohols show this oxidation behaviour. Tertiary alcohols generally resist oxidation under ordinary conditions.
517. Which alcohol would give a positive oxidation test with acidified \(K_2Cr_2O_7\) most readily?
ⓐ. \(CH_3CH_2OH\)
ⓑ. \((CH_3)_3COH\)
ⓒ. \(CH_3OCH_3\)
ⓓ. \(C_6H_6\)
Correct Answer: \(CH_3CH_2OH\)
Explanation: Ethanol is a primary alcohol and can be oxidised by acidified \(K_2Cr_2O_7\). The oxidising agent changes colour as the alcohol is oxidised. tert-Butyl alcohol is tertiary and resists ordinary oxidation. Dimethyl ether and benzene are not simple oxidisable alcohols in this test context.
518. Which product is expected when \(CH_3CH_2OH\) is passed over heated copper at about \(573\,\text{K}\)?
ⓐ. \(CH_3COOH\)
ⓑ. \(CH_2=CH_2\)
ⓒ. \(CH_3CHO\)
ⓓ. \(CH_3OCH_3\)
Correct Answer: \(CH_3CHO\)
Explanation: Passing a primary alcohol over heated copper at about \(573\,\text{K}\) causes dehydrogenation. Ethanol loses hydrogen to form ethanal. This is different from acid-catalysed dehydration, which gives ether or alkene depending on conditions. The reaction can be represented as \(CH_3CH_2OH \rightarrow CH_3CHO + H_2\).
519. Which equation represents dehydrogenation of ethanol over heated copper?
ⓐ. \[CH_3CH_2OH \xrightarrow{Cu,573\,\text{K}} CH_2=CH_2 + H_2O\]
ⓑ. \[2CH_3CH_2OH \xrightarrow{Cu,573\,\text{K}} C_2H_5OC_2H_5 + H_2O\]
ⓒ. \[CH_3CH_2OH + Na \rightarrow CH_3CH_2ONa + \frac{1}{2}H_2\]
ⓓ. \[CH_3CH_2OH \xrightarrow{Cu,573\,\text{K}} CH_3CHO + H_2\]
Correct Answer: \[CH_3CH_2OH \xrightarrow{Cu,573\,\text{K}} CH_3CHO + H_2\]
Explanation: \(\textbf{Reaction condition:}\)
Heated copper at about \(573\,\text{K}\) causes dehydrogenation.
\(\textbf{Alcohol type:}\)
Ethanol is a primary alcohol.
\(\textbf{Product rule:}\)
Primary alcohols give aldehydes on dehydrogenation.
\(\textbf{Equation:}\)
\[
CH_3CH_2OH \xrightarrow{Cu,573\,\text{K}} CH_3CHO + H_2
\]
\(\textbf{Final Answer:}\)
Ethanol gives ethanal and hydrogen.
520. Which product forms when propan-\(2\)-ol is passed over heated copper at about \(573\,\text{K}\)?
ⓐ. Propene
ⓑ. Propanone
ⓒ. Propanal
ⓓ. Propanoic acid
Correct Answer: Propanone
Explanation: Propan-\(2\)-ol is a secondary alcohol. On dehydrogenation over heated copper, secondary alcohols give ketones. The \(CH(OH)\) group in \(CH_3CH(OH)CH_3\) becomes \(C=O\). Therefore the product is propanone, \(CH_3COCH_3\).
