301. Acid-catalysed dehydration of 3,3-dimethylbutan-2-ol can involve a methyl shift before elimination. The major rearranged alkene is expected to be:
ⓐ. 3,3-dimethylbut-1-ene
ⓑ. 2,3-dimethylbut-2-ene
ⓒ. 2-methylpent-1-ene
ⓓ. 2,2-dimethylbut-2-ene
Correct Answer: 2,3-dimethylbut-2-ene
Explanation: Loss of water initially forms a secondary carbocation at carbon \(2\). A neighbouring methyl group can migrate from carbon \(3\) to carbon \(2\). This shift places the positive charge on carbon \(3\) and produces a more stable tertiary carbocation. Elimination of a hydrogen from the adjacent carbon then forms the highly substituted internal double bond. The resulting major product is 2,3-dimethylbut-2-ene rather than an unrearranged terminal alkene.
302. Identify the row containing the correct major-product prediction.
| Row | Alcohol | Major dehydration product |
|---|
| P | Butan-2-ol | But-1-ene only |
| Q | Pentan-2-ol | Pent-2-ene |
| R | 2-Methylbutan-2-ol | 2-Methylbut-1-ene only |
| S | Methanol | Methene |
ⓐ. Row P
ⓑ. Row R
ⓒ. Row Q
ⓓ. Row S
Correct Answer: Row Q
Explanation: Pentan-2-ol can form pent-1-ene and pent-2-ene, with pent-2-ene normally favoured as the more substituted alkene. Butan-2-ol does not form but-1-ene exclusively because but-2-ene is generally the major product. 2-Methylbutan-2-ol mainly forms 2-methylbut-2-ene rather than only the terminal alkene. Methene is not a valid alkene because a carbon-carbon double bond requires two carbon atoms. Row Q is the only prediction consistent with the usual dehydration trend.
303. Two ethanol molecules undergo intermolecular dehydration in the presence of concentrated sulfuric acid near \(413\,\mathrm{K}\). The organic product is:
ⓐ. ethoxyethane
ⓑ. ethene
ⓒ. ethanal
ⓓ. ethyl ethanoate
Correct Answer: ethoxyethane
Explanation: At a suitable lower temperature, two ethanol molecules can combine with loss of one water molecule. One ethanol molecule provides the ethoxy portion, while the second provides the ethyl group attached through oxygen. The product is \(\mathrm{C_2H_5OC_2H_5}\), named ethoxyethane. The overall reaction is \(\mathrm{2C_2H_5OH\rightarrow C_2H_5OC_2H_5+H_2O}\). Higher-temperature conditions favour intramolecular dehydration to ethene instead.
304. Which condition-product combination is most appropriate for ethanol in concentrated sulfuric acid?
ⓐ. Near \(413\,\mathrm{K}\): ethanoic acid; higher temperature: ethanal
ⓑ. Near \(413\,\mathrm{K}\): ethoxyethane; higher temperature: ethene
ⓒ. Near \(413\,\mathrm{K}\): ethene; higher temperature: ethoxyethane
ⓓ. Near \(413\,\mathrm{K}\): methane; higher temperature: methanol
Correct Answer: Near \(413\,\mathrm{K}\): ethoxyethane; higher temperature: ethene
Explanation: Suitable primary alcohols undergo competing dehydration pathways. At about \(413\,\mathrm{K}\), intermolecular reaction between two ethanol molecules favours ether formation. At a higher temperature, elimination within a single ethanol molecule becomes more important and produces ethene. Concentrated sulfuric acid acts as the acidic dehydrating medium in both cases. Temperature therefore controls whether the dominant pathway is intermolecular substitution or intramolecular elimination.
305. In the acid-catalysed formation of ethoxyethane from ethanol, the carbon-oxygen bond of the ether is formed when:
ⓐ. ethene attacks a hydroxide ion
ⓑ. an ethyl carbocation reacts with oxygen gas
ⓒ. ethanol attacks protonated ethanol
ⓓ. two ethyl radicals combine with an oxygen atom
Correct Answer: ethanol attacks protonated ethanol
Explanation: One ethanol molecule is first protonated, converting its hydroxyl group into a better leaving group. A second ethanol molecule uses an oxygen lone pair to attack the primary carbon of the protonated molecule. The displacement occurs through a bimolecular substitution-type step. Loss of a proton from the resulting oxonium ion gives neutral ethoxyethane. The pathway avoids formation of an unstable free primary carbocation.
306. Tertiary alcohols are generally unsuitable for preparing ethers by intermolecular dehydration because they:
ⓐ. cannot be protonated by acids
ⓑ. contain no carbon-oxygen bond
ⓒ. always form carboxylic acids
ⓓ. favour alkene elimination
Correct Answer: favour alkene elimination
Explanation: Protonation of a tertiary alcohol can be followed by rapid loss of water. The resulting tertiary carbocation is relatively stable. Removal of a \(\beta\)-hydrogen then forms an alkene efficiently. Steric crowding also makes bimolecular attack by another alcohol molecule less favourable. Elimination therefore competes strongly and usually prevents clean ether preparation by this method.
307. A mixture of methanol and ethanol is heated with concentrated sulfuric acid under ether-forming conditions. The likely organic product mixture contains:
ⓐ. dimethyl ether and methoxyethane, but no ethoxyethane
ⓑ. dimethyl ether and ethoxyethane, but no methoxyethane
ⓒ. dimethyl ether, methoxyethane, and ethoxyethane
ⓓ. methoxyethane and ethoxyethane, but no dimethyl ether
Correct Answer: dimethyl ether, methoxyethane, and ethoxyethane
Explanation: Methanol can react with another methanol molecule to form dimethyl ether. Ethanol can react with another ethanol molecule to form ethoxyethane. Cross-reaction between methanol and ethanol can form methoxyethane. Since several pairings are possible, the method does not selectively produce one unsymmetrical ether. Separation of the resulting mixture makes this route unattractive for preparing a pure mixed ether.
308. Assertion: Lower-temperature acid treatment of a suitable primary alcohol may favour ether formation, whereas higher temperature favours alkene formation.
Reason: Ether formation is an intermolecular substitution process, while alkene formation is an elimination process that becomes increasingly favoured at higher temperature.
ⓐ. 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: Primary alcohols can follow two competing acid-catalysed dehydration pathways. At a suitable lower temperature, attack by a second alcohol molecule produces an ether. Raising the temperature increases the relative importance of elimination and alkene formation. The two pathways differ in molecularity and product type. The Reason correctly connects those mechanistic alternatives with the temperature-dependent product change.
309. An \(18.4\,\mathrm{g}\) sample of ethanol is converted into ethoxyethane by intermolecular dehydration with a \(75\%\) yield. What mass of ethoxyethane is obtained? Use \(M(\mathrm{C_2H_5OH})=46\,\mathrm{g\,mol^{-1}}\) and \(M(\mathrm{C_2H_5OC_2H_5})=74\,\mathrm{g\,mol^{-1}}\).
ⓐ. \(7.40\,\mathrm{g}\)
ⓑ. \(11.1\,\mathrm{g}\)
ⓒ. \(14.8\,\mathrm{g}\)
ⓓ. \(22.2\,\mathrm{g}\)
Correct Answer: \(11.1\,\mathrm{g}\)
Explanation: \( \textbf{Amount of ethanol:} \)
\[
n(\mathrm{ethanol})=\frac{18.4\,\mathrm{g}}{46\,\mathrm{g\,mol^{-1}}}
\]
\[
n(\mathrm{ethanol})=0.400\,\mathrm{mol}
\]
\( \textbf{Reaction relation:} \)
\[
\mathrm{2C_2H_5OH\rightarrow C_2H_5OC_2H_5+H_2O}
\]
Two moles of ethanol give one mole of ethoxyethane.
\[
n_{\text{theoretical}}(\mathrm{ether})=\frac{0.400}{2}=0.200\,\mathrm{mol}
\]
\( \textbf{Theoretical ether mass:} \)
\[
m_{\text{theoretical}}=0.200\,\mathrm{mol}\times74\,\mathrm{g\,mol^{-1}}
\]
\[
m_{\text{theoretical}}=14.8\,\mathrm{g}
\]
\( \textbf{Apply the percentage yield:} \)
\[
m_{\text{actual}}=14.8\,\mathrm{g}\times\frac{75}{100}=11.1\,\mathrm{g}
\]
The factor of two in the ethanol-to-ether mole ratio must be applied before the yield calculation.
310. A chemist compares two experiments using ethanol and concentrated sulfuric acid.
Case 1: The mixture is maintained near \(413\,\mathrm{K}\).
Case 2: The mixture is heated more strongly under alkene-forming conditions.
The principal products expected in Case 1 and Case 2 are, respectively:
ⓐ. ethanal and ethanoic acid
ⓑ. ethene and ethoxyethane
ⓒ. ethyl ethanoate and ethanol
ⓓ. ethoxyethane and ethene
Correct Answer: ethoxyethane and ethene
Explanation: Near \(413\,\mathrm{K}\), two ethanol molecules undergo intermolecular dehydration and form ethoxyethane. The reaction removes one molecule of water from the pair of alcohol molecules. Under stronger heating, intramolecular elimination becomes dominant. A single ethanol molecule loses the elements of water and forms ethene. The product switch demonstrates that temperature is a reaction-control variable rather than merely a way to increase speed.
311. Acidified potassium permanganate is added to a suitable primary alcohol and the mixture is warmed. The expected colour change is:
ⓐ. orange to green
ⓑ. colourless to deep blue
ⓒ. yellow to brick red
ⓓ. purple to colourless
Correct Answer: purple to colourless
Explanation: Acidified permanganate contains purple \(\mathrm{MnO_4^-}\) ions. A primary alcohol is oxidised while manganese in the permanganate ion is reduced. In acidic medium, the principal manganese product is \(\mathrm{Mn^{2+}}\), which is very pale pink and often described as nearly colourless in dilute solution. The disappearance of the purple colour indicates that the oxidising agent has reacted. The observation alone does not establish whether the final organic product is an aldehyde or a carboxylic acid.
312. Samples P and Q are warmed separately with acidified potassium dichromate. P changes the reagent from orange to green, while Q shows no appreciable change under the mild conditions. Which interpretation is most reasonable?
ⓐ. P may be primary or secondary; Q may be tertiary
ⓑ. P must be a tertiary alcohol, while Q must be a primary alcohol
ⓒ. P must be an ether, while Q must be an aldehyde
ⓓ. P and Q must be the same class of alcohol
Correct Answer: P may be primary or secondary; Q may be tertiary
Explanation: Primary and secondary alcohols are commonly oxidised by acidified dichromate. During the reaction, orange chromium(VI) species are reduced to green chromium(III) species. Tertiary alcohols usually resist ordinary mild oxidation because the hydroxyl-bearing carbon has no hydrogen. Q may therefore be tertiary, while P may be primary or secondary. The colour test alone cannot distinguish a primary alcohol from a secondary alcohol because both can reduce the reagent.
313. Identify the row containing a consistent observation under ordinary mild oxidation conditions.
| Row | Substance and reagent | Expected observation |
|---|
| P | Tertiary alcohol with acidified dichromate | Immediate orange-to-green change in every case |
| Q | Secondary alcohol with acidified dichromate | Orange-to-green change as a ketone forms |
| R | Primary alcohol with acidified permanganate | Purple colour becomes more intense during oxidation |
| S | Secondary alcohol with acidified permanganate | No reaction because ketones contain oxygen |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row Q
Explanation: A secondary alcohol is oxidised to a ketone and can reduce acidified dichromate from chromium(VI) to chromium(III). The accompanying colour change is orange to green. A tertiary alcohol usually does not show the same rapid mild oxidation response. Acidified permanganate is consumed by suitable primary and secondary alcohols, so its purple colour fades rather than intensifies. Row Q correctly connects alcohol class, oxidation product, and reagent observation.
314. The general equation for Fischer esterification is:
ⓐ. \(\mathrm{RCOOH+R'OH\rightleftharpoons RCHO+R'OH_2^+}\)
ⓑ. \(\mathrm{RCOOH+R'OH\rightleftharpoons ROR'+CO_2}\)
ⓒ. \(\mathrm{RCOOH+R'OH\rightleftharpoons RCOONa+R'H}\)
ⓓ. \(\mathrm{RCOOH+R'OH\rightleftharpoons RCOOR'+H_2O}\)
Correct Answer: \(\mathrm{RCOOH+R'OH\rightleftharpoons RCOOR'+H_2O}\)
Explanation: Fischer esterification combines a carboxylic acid with an alcohol. The organic product is an ester, and water is formed as the small-molecule by-product. The double arrow is important because the reaction is reversible. An acid catalyst increases the rate of reaching equilibrium but does not make the reverse reaction impossible. The carbonyl-containing portion comes from the carboxylic acid, while the alkyl group attached through oxygen comes from the alcohol.
315. Ethanol reacts with propanoic acid in the presence of concentrated sulfuric acid. The ester formed is:
ⓐ. propyl ethanoate
ⓑ. ethyl propanoate
ⓒ. ethyl propanal
ⓓ. propoxyethane
Correct Answer: ethyl propanoate
Explanation: The alkyl part of an ester name is derived from the alcohol. Ethanol therefore contributes the ethyl group. The carboxylic acid contributes the alkanoate part of the name. Propanoic acid consequently gives propanoate. The product is \(\mathrm{CH_3CH_2COOCH_2CH_3}\), named ethyl propanoate.
316. Concentrated sulfuric acid catalyses Fischer esterification mainly by:
ⓐ. carbonyl protonation activates the acyl carbon
ⓑ. supplying the alkyl group that appears in the ester
ⓒ. permanently converting the alcohol into an alkane
ⓓ. reducing the carboxylic acid to an aldehyde
Correct Answer: carbonyl protonation activates the acyl carbon
Explanation: Protonation increases the electrophilic character of the carboxylic acid carbonyl carbon. The alcohol oxygen can then attack that carbon more readily. Subsequent proton transfers and loss of water lead to the ester. The acid catalyst is regenerated during the mechanism. It accelerates both forward and reverse reactions and does not supply the carbon skeleton of the product.
317. An equilibrium mixture contains a carboxylic acid, an alcohol, an ester, and water. Selective removal of water will:
ⓐ. shift the equilibrium toward the reactants
ⓑ. leave the equilibrium composition permanently unchanged
ⓒ. shift the equilibrium toward additional ester formation
ⓓ. stop both forward and reverse reactions immediately
Correct Answer: shift the equilibrium toward additional ester formation
Explanation: Water is one of the products of Fischer esterification. Removing a product disturbs the equilibrium composition. The system responds by favouring the forward reaction, which replaces some of the removed water. Additional carboxylic acid and alcohol are therefore converted into ester. This application of Le Chatelier’s principle can improve the equilibrium yield without changing the equilibrium constant at a fixed temperature.
318. Why is a reflux condenser commonly used while heating a Fischer esterification mixture?
ⓐ. It changes the equilibrium constant by removing the catalyst
ⓑ. permits prolonged heating while condensing volatile vapours
ⓒ. It converts the ester vapour into an alkene
ⓓ. It selectively destroys all water as soon as it forms
Correct Answer: permits prolonged heating while condensing volatile vapours
Explanation: Heating increases the reaction rate and helps the mixture approach equilibrium more quickly. Alcohols, esters, and other components may be volatile at the reaction temperature. A reflux condenser cools their vapours and returns the condensed liquids to the flask. This permits sustained heating without substantial loss of material. Reflux by itself does not shift equilibrium unless a component such as water is also removed selectively.
319. Direct acid-catalysed esterification is often slower with a highly hindered tertiary alcohol than with a comparable primary alcohol because:
ⓐ. steric crowding hinders nucleophilic attack
ⓑ. tertiary alcohols contain no oxygen lone pairs
ⓒ. primary alcohols are always stronger mineral acids
ⓓ. tertiary alcohols cannot form any bond to a carbonyl carbon
Correct Answer: steric crowding hinders nucleophilic attack
Explanation: Esterification requires interaction between the alcohol oxygen and the activated carboxylic acid carbonyl centre. Bulky alkyl groups around a tertiary alcohol can hinder effective approach and reaction. Under strongly acidic conditions, a tertiary alcohol may also undergo competing dehydration to an alkene. A primary alcohol is less crowded and generally participates more readily in the esterification pathway. The difference is kinetic and competitive rather than an absence of lone pairs on tertiary-alcohol oxygen.
320. Which naming rule correctly identifies the two parts of an ester name?
ⓐ. Both parts are taken from the alcohol
ⓑ. The first part comes from the acid, and the second part comes from the alcohol
ⓒ. the alkyl name comes from alcohol and alkanoate from acid
ⓓ. Both parts are taken from the carboxylic acid
Correct Answer: the alkyl name comes from alcohol and alkanoate from acid
Explanation: The group bonded directly to the ester oxygen is named first as an alkyl group. This group originates from the alcohol used in esterification. The carbonyl-containing portion is named as an alkanoate and originates from the carboxylic acid. Thus methanol and butanoic acid form methyl butanoate. Tracking the two starting molecules prevents reversal of names such as methyl ethanoate and ethyl methanoate.
