601. 2-Hydroxybenzaldehyde is subjected to the following controlled synthesis.
Stage I: The phenolic hydroxyl group is selectively acetylated using ethanoic anhydride, forming an aromatic acetate ester while leaving the aldehyde group unchanged.
Stage II: The aldehyde is oxidised selectively into a carboxylic acid without hydrolysing the aromatic acetate ester.
Stage III: The product is treated separately with aqueous sodium hydrogencarbonate and Tollens' reagent under conditions that do not hydrolyse the ester.
Which statement correctly identifies the final compound and its characteristic reactions?
ⓐ. The product is 2-hydroxybenzoic acid, which gives both a positive Tollens' test and hydrogencarbonate effervescence.
ⓑ. The product is methyl 2-hydroxybenzoate, which releases two moles of carbon dioxide per mole.
ⓒ. The product is 2-acetoxybenzoic acid, which reacts with hydrogencarbonate but lacks the original aldehyde required for a characteristic positive Tollens' reaction.
ⓓ. The product is 2-methoxybenzaldehyde, which is iodoform-positive but hydrogencarbonate-negative.
Correct Answer: The product is 2-acetoxybenzoic acid, which reacts with hydrogencarbonate but lacks the original aldehyde required for a characteristic positive Tollens' reaction.
Explanation: \(\textbf{Identify the starting functional groups:}\)
2-Hydroxybenzaldehyde contains an aromatic aldehyde and a phenolic hydroxyl group.
The two groups occupy adjacent positions on the benzene ring.
The synthesis modifies them in different stages.
\(\textbf{Stage I — Selective phenolic acetylation:}\)
Ethanoic anhydride supplies an acetyl group to the phenolic oxygen.
The product contains the linkage \(\mathrm{ArOCOCH_3}\).
The original aldehyde group remains intact under the stipulated conditions.
Therefore, the intermediate is 2-acetoxybenzaldehyde.
\(\textbf{Stage II — Oxidise the aldehyde:}\)
Selective oxidation converts the aromatic aldehyde group into a carboxylic acid group.
The aromatic acetate ester remains unchanged by assumption.
The product is 2-acetoxybenzoic acid, commonly called acetylsalicylic acid.
It contains one free carboxylic acid group and one aromatic acetate ester group.
\(\textbf{Stage III — Hydrogencarbonate reaction:}\)
The free carboxylic acid group protonates hydrogencarbonate.
Carbon dioxide is released through the characteristic acid–base reaction.
The ester group is not counted as a second free carboxylic acid group.
\(\textbf{Interpret the Tollens' result:}\)
The original aldehyde carbonyl has been oxidised into a carboxylic acid group.
Consequently, the final compound lacks the aldehyde functionality responsible for the characteristic positive Tollens' reaction of the starting material.
\(\textbf{Interpret the purpose of the sequence:}\)
Initial acetylation changes the phenolic hydroxyl into an ester.
The later oxidation changes the separate aldehyde into a carboxylic acid.
Both changes preserve the aromatic carbon skeleton and substitution positions.
The final identification therefore requires tracking two different functional groups through independent chemical transformations.
602. A chemist wishes to prepare ethyl tert-butyl ether, \(\mathrm{CH_3CH_2OC(CH_3)_3}\), through Williamson ether synthesis.
Two disconnections are proposed:
P: Sodium tert-butoxide with bromoethane.
Q: Sodium ethoxide with 2-bromo-2-methylpropane.
Assume ordinary Williamson substitution conditions and the usual competition between \(S_\mathrm{N}2\) substitution and elimination.
The ether obtained through the preferred route is subsequently treated with HI under conditions favouring cleavage at its tertiary alkyl–oxygen bond.
Which statement correctly describes the preferred preparation and principal initial cleavage products?
ⓐ. Q is preferred because a tertiary bromide undergoes \(S_\mathrm{N}2\) substitution most rapidly; cleavage gives iodoethane and tert-butanol.
ⓑ. P is preferred because bromoethane offers a primary electrophilic carbon; cleavage of the resulting ether can give tert-butyl iodide and ethanol.
ⓒ. Both disconnections are equally favourable because the oxygen atom has the same two carbon substituents in either product.
ⓓ. P is unsuitable because tert-butoxide cannot act as a nucleophile towards any primary alkyl halide; cleavage gives ethene and methanol.
Correct Answer: P is preferred because bromoethane offers a primary electrophilic carbon; cleavage of the resulting ether can give tert-butyl iodide and ethanol.
Explanation: \(\textbf{Identify the target ether connectivity:}\)
The target contains an oxygen atom connecting ethyl and tert-butyl groups.
A retrosynthetic disconnection can place the oxygen nucleophile on either side.
However, those formal alternatives are not necessarily equally feasible.
\(\textbf{Evaluate Route P:}\)
Sodium tert-butoxide is a strong alkoxide nucleophile and base.
Bromoethane contains a primary carbon bonded to bromine.
That carbon can undergo backside nucleophilic substitution under suitable conditions.
\[
\mathrm{(CH_3)_3CONa+CH_3CH_2Br}
\]
\[
\mathrm{\longrightarrow (CH_3)_3COCH_2CH_3+NaBr}
\]
The desired ether is formed through the stipulated Williamson pathway.
Although tert-butoxide is bulky and competing elimination may need to be controlled, the primary alkyl bromide provides a viable substitution centre.
\(\textbf{Evaluate Route Q:}\)
The proposed electrophile is 2-bromo-2-methylpropane.
Its carbon bearing bromine is tertiary.
Ordinary backside \(S_\mathrm{N}2\) attack at that carbon is strongly hindered.
Sodium ethoxide is also basic, making elimination an important competing pathway.
Therefore, this disconnection is unsuitable as the preferred Williamson route.
\(\textbf{Apply the specified HI cleavage:}\)
The resulting ether contains a tertiary alkyl–oxygen bond.
After protonation of oxygen, cleavage at that bond can produce a relatively stable tertiary carbocation.
Iodide capture gives tert-butyl iodide.
The remaining oxygen-containing fragment becomes ethanol.
\(\textbf{Interpret the asymmetry:}\)
Efficient synthesis and preferred acid cleavage need not occur at the same carbon–oxygen bond.
Williamson synthesis favours attack at an accessible primary carbon.
Acidic cleavage can instead favour departure of a tertiary carbon fragment through a carbocation pathway.
This difference arises from the mechanisms available under the two sets of reaction conditions.
603. Two independent routes are proposed for preparing aromatic carboxylic acids.
Route P: Benzyl bromide is converted into benzylmagnesium bromide in dry ether. The Grignard reagent reacts with carbon dioxide and undergoes acidic work-up.
Route Q: Ethylbenzene undergoes complete vigorous side-chain oxidation using hot alkaline potassium permanganate, followed by acidification.
Assume quantitative formation of the principal carboxylic acid products and no aromatic-ring cleavage.
Which statement correctly compares the carbon skeletons?
ⓐ. P gives benzoic acid with seven carbon atoms, while Q gives phenylacetic acid with eight carbon atoms.
ⓑ. Both routes give phenylacetic acid because their organic starting compounds each contain a benzene ring.
ⓒ. P gives eight-carbon phenylacetic acid by incorporation of carbon dioxide; Q gives seven-carbon benzoic acid by oxidation of the alkyl side chain.
ⓓ. Both routes give benzoic acid because carbon dioxide contributes oxygen atoms but never an additional carbon atom in Grignard reactions.
Correct Answer: P gives eight-carbon phenylacetic acid by incorporation of carbon dioxide; Q gives seven-carbon benzoic acid by oxidation of the alkyl side chain.
Explanation: \(\textbf{Examine the starting carbon skeleton in Route P:}\)
Benzyl bromide contains seven carbon atoms.
The bromine atom is attached to the saturated benzylic carbon.
Conversion into benzylmagnesium bromide preserves those seven carbon atoms.
\(\textbf{Apply Grignard carboxylation:}\)
The benzyl group attacks the electrophilic carbon of carbon dioxide.
The new carbon–carbon bond connects the benzylic carbon to the carbon dioxide carbon.
\[
\mathrm{C_6H_5CH_2MgBr+CO_2}
\]
\[
\mathrm{\longrightarrow C_6H_5CH_2COOMgBr}
\]
Acidic work-up gives:
\[
\mathrm{C_6H_5CH_2COOH}
\]
The product is phenylacetic acid.
It contains eight carbon atoms because one carbon was introduced from carbon dioxide.
\(\textbf{Examine the starting carbon skeleton in Route Q:}\)
Ethylbenzene also contains eight carbon atoms.
Its side chain is \(\mathrm{-CH_2CH_3}\), with two carbon atoms beyond the aromatic ring.
The benzylic carbon contains hydrogen and can undergo vigorous oxidation.
\(\textbf{Follow the oxidative transformation:}\)
Under the specified hot alkaline permanganate conditions, the side chain is oxidised to a carboxyl group.
The resulting product after acidification is:
\[
\mathrm{C_6H_5COOH}
\]
This is benzoic acid.
It contains seven carbon atoms.
The terminal carbon of the original ethyl side chain is removed during the vigorous oxidation process.
\(\textbf{Compare the two pathways:}\)
Route P increases the carbon count through carbon–carbon bond formation with carbon dioxide.
Route Q reduces the number of carbon atoms retained in the isolated aromatic organic acid through oxidative side-chain degradation.
Both routes produce aromatic carboxylic acids, but their carbon skeletons and products differ.
The comparison illustrates why identifying only the final functional group is insufficient for reaction prediction.
604. Three oxygen-containing aromatic compounds are treated separately with excess aqueous sodium hydroxide under suitable conditions.
P: Phenyl ethanoate, \(\mathrm{CH_3COOC_6H_5}\).
Q: Ethyl benzoate, \(\mathrm{C_6H_5COOCH_2CH_3}\).
R: Ethoxybenzene, \(\mathrm{C_6H_5OCH_2CH_3}\).
Assume complete alkaline hydrolysis of ester bonds, no cleavage of ordinary ether bonds and appropriate subsequent acidification of isolated aqueous fractions.
The regenerated organic products are examined for phenolic hydroxyl groups and free carboxylic acid groups.
Which statement correctly compares the transformations?
ⓐ. P yields phenol and ethanoic acid, Q yields benzoic acid and ethanol, and R remains unchanged under the stated alkaline hydrolysis conditions.
ⓑ. P yields benzyl alcohol and methanoic acid; Q yields phenol and ethanal; R yields phenol and ethene.
ⓒ. All three yield phenol because each contains oxygen directly connected to a phenyl-derived fragment.
ⓓ. Only R hydrolyses because ether oxygen is more basic than ester oxygen.
Correct Answer: P yields phenol and ethanoic acid, Q yields benzoic acid and ethanol, and R remains unchanged under the stated alkaline hydrolysis conditions.
Explanation: \(\textbf{Distinguish ester and ether connectivity:}\)
P and Q contain ester groups with an acyl carbon–oxygen linkage.
R contains an ether linkage connecting phenyl and ethyl groups without an ester carbonyl.
The specified alkaline conditions therefore do not affect all three compounds in the same manner.
\(\textbf{Evaluate P — Phenyl ethanoate:}\)
Phenyl ethanoate contains an ethanoic acid-derived acyl group and a phenoxy fragment.
Alkaline ester hydrolysis breaks the appropriate acyl–oxygen linkage.
The resulting species include ethanoate and phenol-derived material, present according to the alkaline conditions.
After suitable acidification, ethanoic acid and phenol can be recovered.
The regenerated phenol possesses a free phenolic oxygen–hydrogen bond.
\(\textbf{Evaluate Q — Ethyl benzoate:}\)
Ethyl benzoate contains a benzoic acid-derived acyl group and an ethoxy fragment.
Alkaline hydrolysis produces benzoate and ethanol.
Acidification regenerates benzoic acid from the benzoate.
The other organic product is ethanol, an ordinary primary alcohol.
Therefore, Q yields a free carboxylic acid but not phenol.
\(\textbf{Evaluate R — Ethoxybenzene:}\)
Ethoxybenzene contains an ordinary alkyl aryl ether linkage.
It lacks the ester carbonyl required for alkaline ester hydrolysis.
Under the specified conditions excluding ether cleavage, it remains unchanged.
\(\textbf{Interpret the characteristic reactions:}\)
The phenol recovered from P can undergo the characteristic phenolic reactions.
The benzoic acid recovered from Q releases carbon dioxide with aqueous sodium hydrogencarbonate.
The unchanged ether R lacks both a free phenolic hydroxyl group and a free carboxylic acid group.
The distinction depends on identifying whether oxygen is attached to an acyl carbon or connects two carbon groups in an ether.
605. Two aromatic aldehydes undergo nucleophilic addition of the same carbon nucleophile under otherwise comparable conditions:
P: Benzaldehyde.
Q: 4-Nitrobenzaldehyde.
An experimental study supplies the following idealised initial-rate law:
\[
r=k[\text{aldehyde}][\text{nucleophile}]
\]
At the same temperature:
\[
k_P=2.0\,\mathrm{L\,mol^{-1}\,s^{-1}}
\]
\[
k_Q=5.0\,\mathrm{L\,mol^{-1}\,s^{-1}}
\]
The initial aldehyde concentrations are \(0.20\,\mathrm{M}\) for P and \(0.050\,\mathrm{M}\) for Q. The nucleophile concentration is \(0.010\,\mathrm{M}\) in both experiments.
Which statement correctly compares the intrinsic carbonyl reactivity and the observed initial rates?
ⓐ. Q is intrinsically less reactive but reacts faster initially because its nitro group is electron withdrawing.
ⓑ. P and Q have identical intrinsic reactivity because both contain an aldehyde group attached to benzene.
ⓒ. Q has the larger intrinsic rate constant, but P has the larger observed initial rate under the supplied concentrations.
ⓓ. Q has both the larger intrinsic rate constant and the larger observed initial rate under the supplied concentrations.
Correct Answer: Q has the larger intrinsic rate constant, but P has the larger observed initial rate under the supplied concentrations.
Explanation: \(\textbf{Identify the intrinsic electronic comparison:}\)
Both compounds contain an aldehyde group attached directly to an aromatic ring.
However, Q also contains a para nitro substituent.
The nitro group withdraws electron density from the aromatic system.
Its electronic influence can increase the electrophilic character of the aldehyde carbonyl carbon.
The larger supplied rate constant for Q is consistent with greater intrinsic susceptibility towards the specified nucleophilic addition.
\(\textbf{Distinguish intrinsic reactivity from experimental rate:}\)
The initial reaction rate depends on the rate constant and the concentrations.
The question supplies:
\[
r=k[\text{aldehyde}][\text{nucleophile}]
\]
Therefore, different substrate concentrations can reverse the ordering of measured initial rates even when one substrate has the larger intrinsic rate constant.
\(\textbf{Calculate the rate for P:}\)
\[
r_P=2.0(0.20)(0.010)
\]
\[
r_P=0.0040\,\mathrm{mol\,L^{-1}\,s^{-1}}
\]
\(\textbf{Calculate the rate for Q:}\)
\[
r_Q=5.0(0.050)(0.010)
\]
\[
r_Q=0.0025\,\mathrm{mol\,L^{-1}\,s^{-1}}
\]
\(\textbf{Compare the observed rates:}\)
\[
\frac{r_P}{r_Q}=\frac{0.0040}{0.0025}
\]
\[
\frac{r_P}{r_Q}=1.6
\]
Thus, P reacts initially 1.6 times faster under the supplied experimental concentrations.
\(\textbf{Check the intrinsic rate constants:}\)
\[
\frac{k_Q}{k_P}=\frac{5.0}{2.0}
\]
\[
\frac{k_Q}{k_P}=2.5
\]
Q remains intrinsically more reactive by the criterion of the supplied rate constants.
Its smaller starting concentration accounts for its lower observed initial rate.
The result distinguishes substituent-controlled electrophilicity from concentration-controlled experimental kinetics.
\(\textbf{Final answer:}\)
\[
\boxed{k_Q\gt k_P,\qquad r_P=1.6r_Q}
\]
606. Two aromatic hydroxy acids are compared:
P: 2-Hydroxybenzoic acid.
Q: 4-Hydroxybenzoic acid.
Their approximate first-dissociation values are:
\[
\mathrm{p}K_{a1}(P)=2.98
\]
\[
\mathrm{p}K_{a1}(Q)=4.54
\]
Assume that only the first carboxyl-group dissociation is relevant at pH 3.50 and that the solutions exhibit ideal monoprotic behaviour for this comparison.
Which statement correctly compares the approximate fractions existing as the corresponding carboxylate forms at pH 3.50?
ⓐ. P is approximately \(23\%\) ionised and Q approximately \(92\%\) ionised.
ⓑ. Both are approximately \(50\%\) ionised because their functional groups are identical.
ⓒ. P is about \(77\%\) ionised and Q about \(8.4\%\); intramolecular interactions contribute to the ortho isomer's distinctive acidity.
ⓓ. P is approximately \(92\%\) ionised and Q approximately \(77\%\) ionised because both pH values are above their corresponding \(\mathrm{p}K_a\) values.
Correct Answer: P is about \(77\%\) ionised and Q about \(8.4\%\); intramolecular interactions contribute to the ortho isomer's distinctive acidity.
Explanation: \(\textbf{Identify the structural difference:}\)
Both acids contain a carboxylic acid group and a phenolic hydroxyl group.
In P, the two groups occupy adjacent aromatic positions.
In Q, they occupy para positions.
The ortho arrangement permits favourable intramolecular interactions, including hydrogen bonding involving the phenolic hydroxyl group and the neighbouring carboxyl-derived group.
Such interactions contribute to the relative stability of the relevant acid and conjugate-base forms.
\(\textbf{Use the supplied dissociation values:}\)
P has the smaller first \(\mathrm{p}K_a\).
It is therefore the stronger acid for the first dissociation.
At a common pH, it is expected to possess the larger carboxylate fraction.
\(\textbf{Apply the monoprotic ionisation relation:}\)
\[
r=\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}=10^{\mathrm{pH}-\mathrm{p}K_a}
\]
The ionised fraction is:
\[
f=\frac{r}{1+r}
\]
\(\textbf{Calculate the fraction for P:}\)
\[
r_P=10^{3.50-2.98}
\]
\[
r_P=10^{0.52}\approx3.31
\]
\[
f_P=\frac{3.31}{1+3.31}
\]
\[
f_P\approx0.768
\]
\[
f_P\approx76.8\%
\]
\(\textbf{Calculate the fraction for Q:}\)
\[
r_Q=10^{3.50-4.54}
\]
\[
r_Q=10^{-1.04}\approx0.0912
\]
\[
f_Q=\frac{0.0912}{1+0.0912}
\]
\[
f_Q\approx0.0836
\]
\[
f_Q\approx8.36\%
\]
\(\textbf{Interpret the positional effect:}\)
The differing ionisation fractions are much larger than would be predicted by treating ortho and para substitution as electronically identical.
The supplied first \(\mathrm{p}K_a\) values already incorporate the net effects of substituent position, electronic interactions and solvation.
It would be unjustified to attribute the exact numerical difference to only one effect without additional evidence.
Nevertheless, the calculations demonstrate that substitution position can strongly influence carboxylic acid speciation even when the two compounds possess the same functional groups.
\(\textbf{Final answer:}\)
\[
\boxed{f_P\approx76.8\%,\qquad f_Q\approx8.36\%}
\]
607. Two hypothetical monocarboxylic acids have the following dissociation constants:
\[
K_a(P)=1.0\times10^{-4}
\]
\[
K_a(Q)=1.0\times10^{-5}
\]
Equal analytical concentrations of P and Q are placed separately in aqueous buffers whose pH values can be adjusted.
An investigator seeks the pH at which the equilibrium concentration of the carboxylate form of P is exactly twice the equilibrium concentration of the carboxylate form of Q.
Assume ideal monoprotic acid–base behaviour and that the analytical concentrations of the two acids are equal.
Which pair gives the required pH and the approximate ionised fraction of P?
ⓐ. pH \(4.00\), with \(50.0\%\) P ionised
ⓑ. pH \(4.90\), with \(88.9\%\) P ionised
ⓒ. pH \(5.00\), with \(90.9\%\) P ionised
ⓓ. pH \(4.90\), with \(44.4\%\) P ionised
Correct Answer: pH \(4.90\), with \(88.9\%\) P ionised
Explanation: \(\textbf{Identify the condition imposed by equal analytical concentrations:}\)
Both solutions contain equal total analytical concentrations of their respective acids.
Therefore, the ratio of their equilibrium carboxylate concentrations equals the ratio of their ionised fractions.
The required condition is:
\[
\frac{f_P}{f_Q}=2
\]
\(\textbf{Write each ionised fraction:}\)
For a monoprotic weak acid:
\[
f=\frac{K_a}{K_a+[\mathrm{H^+}]}
\]
Let:
\[
h=[\mathrm{H^+}]
\]
Then:
\[
f_P=\frac{10^{-4}}{10^{-4}+h}
\]
\[
f_Q=\frac{10^{-5}}{10^{-5}+h}
\]
\(\textbf{Apply the required fraction ratio:}\)
\[
\frac{\dfrac{10^{-4}}{10^{-4}+h}}
{\dfrac{10^{-5}}{10^{-5}+h}}=2
\]
Simplifying:
\[
10\frac{10^{-5}+h}{10^{-4}+h}=2
\]
\[
10(10^{-5}+h)=2(10^{-4}+h)
\]
\[
10^{-4}+10h=2\times10^{-4}+2h
\]
\[
8h=10^{-4}
\]
\[
h=1.25\times10^{-5}\,\mathrm{mol\,L^{-1}}
\]
\(\textbf{Calculate the required pH:}\)
\[
\mathrm{pH}=-\log(1.25\times10^{-5})
\]
\[
\mathrm{pH}\approx4.90
\]
\(\textbf{Calculate the fraction of P ionised:}\)
\[
f_P=\frac{10^{-4}}{10^{-4}+1.25\times10^{-5}}
\]
\[
f_P=\frac{10^{-4}}{1.125\times10^{-4}}
\]
\[
f_P=\frac{1}{1.125}
\]
\[
f_P\approx0.8889
\]
\[
f_P\approx88.9\%
\]
\(\textbf{Verify the condition using Q:}\)
\[
f_Q=\frac{10^{-5}}{10^{-5}+1.25\times10^{-5}}
\]
\[
f_Q=\frac{1}{2.25}\approx0.4444
\]
\[
\frac{f_P}{f_Q}=\frac{0.8889}{0.4444}\approx2
\]
The required relationship is satisfied.
The stronger acid P is more extensively ionised, but both fractions approach unity at sufficiently high pH.
Therefore, their fraction ratio is not fixed at the ratio of the two dissociation constants.
The calculation must account for the common equilibrium hydrogen-ion concentration.
\(\textbf{Final answer:}\)
\[
\boxed{\mathrm{pH}\approx4.90,\qquad f_P\approx88.9\%}
\]
608. Ethanal reacts with excess ethanol in the presence of a suitable dry acid catalyst under conditions favouring complete acetal formation:
\[
\mathrm{CH_3CHO+2C_2H_5OH\rightleftharpoons CH_3CH(OC_2H_5)_2+H_2O}
\]
A reaction mixture initially contains \(0.20\,\mathrm{mol}\) ethanal and \(0.30\,\mathrm{mol}\) ethanol.
Assume that the limiting reactant is consumed completely and that no other reactions occur.
The acetal is isolated, and the remaining unreacted ethanal is separately recovered.
Which pair gives the amount of acetal formed and the amount of ethanal remaining, respectively?
ⓐ. \(0.20\,\mathrm{mol}\) and \(0.00\,\mathrm{mol}\)
ⓑ. \(0.15\,\mathrm{mol}\) and \(0.15\,\mathrm{mol}\)
ⓒ. \(0.10\,\mathrm{mol}\) and \(0.10\,\mathrm{mol}\)
ⓓ. \(0.15\,\mathrm{mol}\) and \(0.05\,\mathrm{mol}\)
Correct Answer: \(0.15\,\mathrm{mol}\) and \(0.05\,\mathrm{mol}\)
Explanation: \(\textbf{Identify the carbonyl derivative-forming reaction:}\)
Ethanal contains an aldehyde carbonyl group.
Under suitable acid-catalysed conditions, ethanol can react with its carbonyl carbon.
The overall conversion forms an acetal containing two ethoxy groups attached to the original aldehyde carbon.
The reaction also produces water.
\(\textbf{Write the governing stoichiometry:}\)
\[
\mathrm{CH_3CHO+2C_2H_5OH}
\]
\[
\mathrm{\longrightarrow CH_3CH(OC_2H_5)_2+H_2O}
\]
One mole of acetal requires one mole of ethanal and two moles of ethanol.
\(\textbf{Determine the theoretical product from ethanal:}\)
\[
n(\mathrm{acetal})_{\mathrm{from\ ethanal}}=0.20\,\mathrm{mol}
\]
\(\textbf{Determine the theoretical product from ethanol:}\)
\[
n(\mathrm{acetal})_{\mathrm{from\ ethanol}}=\frac{0.30}{2}
\]
\[
n(\mathrm{acetal})_{\mathrm{from\ ethanol}}=0.15\,\mathrm{mol}
\]
Therefore, ethanol is the limiting reactant.
\(\textbf{Calculate the acetal formed:}\)
\[
n(\mathrm{acetal})=0.15\,\mathrm{mol}
\]
\(\textbf{Calculate the ethanal consumed:}\)
\[
n(\mathrm{ethanal})_{\mathrm{consumed}}=0.15\,\mathrm{mol}
\]
\(\textbf{Calculate the remaining ethanal:}\)
\[
n(\mathrm{ethanal})_{\mathrm{remaining}}=0.20-0.15
\]
\[
n(\mathrm{ethanal})_{\mathrm{remaining}}=0.05\,\mathrm{mol}
\]
\(\textbf{Check water formation:}\)
The reaction produces one water molecule per acetal molecule.
\[
n(\mathrm{H_2O})=0.15\,\mathrm{mol}
\]
\(\textbf{Interpret the functional-group transformation:}\)
The isolated acetal no longer contains the original aldehyde carbonyl double bond.
Its two ethoxy oxygens originate from the alcohol reagent under the specified mechanism.
Acetal formation is reversible, so the assumption of complete limiting-reactant conversion depends on the explicitly controlled reaction conditions.
The calculation distinguishes the two-alcohol requirement of acetal formation from the one-reagent requirement of ordinary oxime or hydrazone formation.
\(\textbf{Final answer:}\)
\[
\boxed{n(\mathrm{acetal})=0.15\,\mathrm{mol},\qquad n(\mathrm{ethanal})_{\mathrm{remaining}}=0.05\,\mathrm{mol}}
\]
609. Three oxygen-containing compounds are examined under anhydrous conditions:
P: Ethane-1,2-diol, \(\mathrm{HOCH_2CH_2OH}\).
Q: 2-Methoxyethanol, \(\mathrm{CH_3OCH_2CH_2OH}\).
R: 1,2-Dimethoxyethane, \(\mathrm{CH_3OCH_2CH_2OCH_3}\).
Equal molar amounts of each compound are treated independently with excess sodium metal under conditions where every ordinary alcohol O–H group reacts completely but ether linkages remain unchanged.
The compounds are also examined for their ability to act as hydrogen-bond donors and acceptors towards water.
Which statement correctly compares their properties?
ⓐ. P and Q release equal hydrogen amounts because both contain two oxygen atoms; R releases twice as much hydrogen as P.
ⓑ. P evolves twice the hydrogen of Q; R evolves none through alcohol–sodium chemistry. All three can accept hydrogen bonds from water.
ⓒ. R cannot accept hydrogen bonds from water because it has no ordinary oxygen–hydrogen bond.
ⓓ. Q releases more hydrogen than P because its methoxy group increases the acidity of its remaining hydroxyl proton.
Correct Answer: P evolves twice the hydrogen of Q; R evolves none through alcohol–sodium chemistry. All three can accept hydrogen bonds from water.
Explanation: \(\textbf{Distinguish oxygen count from hydroxyl count:}\)
All three compounds contain two oxygen atoms.
However, the bonding environments of those oxygen atoms differ.
The amount of hydrogen released by sodium metal depends on the number of ordinary alcohol oxygen–hydrogen bonds.
It does not depend simply on the total number of oxygen atoms.
\(\textbf{Evaluate P:}\)
Ethane-1,2-diol contains two ordinary alcohol hydroxyl groups.
Both can react with excess sodium metal.
Therefore, one mole of P releases one mole of molecular hydrogen on complete reaction.
\(\textbf{Evaluate Q:}\)
2-Methoxyethanol contains one ether oxygen and one alcohol hydroxyl oxygen.
Only the hydroxyl group supplies the ordinary alcohol proton for sodium-metal hydrogen evolution.
Therefore, one mole of Q releases half a mole of molecular hydrogen.
\(\textbf{Evaluate R:}\)
1,2-Dimethoxyethane contains two ether oxygen atoms.
Neither oxygen is bonded to hydrogen.
Consequently, R does not undergo ordinary alcohol–sodium hydrogen evolution under the specified conditions.
\(\textbf{Compare hydrogen bonding with water:}\)
The oxygen atoms in all three compounds possess nonbonding electron pairs.
They can therefore accept hydrogen bonds from suitable water molecules.
P and Q can additionally donate hydrogen bonds through their alcohol hydroxyl groups.
R lacks an O–H bond and cannot donate conventional alcohol-type hydrogen bonds.
\(\textbf{Interpret the combined observations:}\)
A compound can contain several oxygen atoms and still possess no hydrogen capable of producing the ordinary alcohol–sodium reaction.
Likewise, absence of a hydrogen-bond donor does not imply absence of hydrogen-bond-accepting ability.
This distinction separates molecular polarity and oxygen count from the specific bonding features responsible for the two reactions.
610. Optically pure \((R)\)-1-phenylethanol is divided into two portions:
P: Treatment with Lucas reagent under conditions favouring ionisation to a resonance-stabilised benzylic carbocation followed by chloride capture from either face.
Q: Conversion into its sodium alkoxide, followed by reaction with iodomethane through Williamson ether synthesis.
Assume that P undergoes no competing elimination and that Q undergoes no reaction at the stereogenic carbon–oxygen bond.
Which statement correctly compares the stereochemical outcomes?
ⓐ. P gives optically pure \((R)\)-1-chloroethylbenzene, while Q gives a racemic ether because methylation occurs at the stereogenic carbon.
ⓑ. Both products must be racemic because any reaction involving a chiral alcohol necessarily destroys its stereogenic centre.
ⓒ. P can yield racemic 1-chloroethylbenzene via a planar benzylic intermediate; Q can preserve configuration while forming 1-methoxyethylbenzene.
ⓓ. P and Q both give achiral products because replacing an alcohol hydroxyl group always creates two identical carbon substituents.
Correct Answer: P can yield racemic 1-chloroethylbenzene via a planar benzylic intermediate; Q can preserve configuration while forming 1-methoxyethylbenzene.
Explanation: \(\textbf{Identify the stereogenic carbon:}\)
1-Phenylethanol contains a hydroxyl-bearing carbon attached to phenyl, methyl, hydrogen and oxygen.
These four substituent environments are different.
Therefore, the starting alcohol can exist as optically active enantiomers.
\(\textbf{Examine Route P:}\)
The question specifies ionisation to a benzylic carbocation under Lucas reagent conditions.
The benzylic carbocation is resonance stabilised by the aromatic ring.
Its positively charged carbon is approximately trigonal planar.
Consequently, chloride can approach from either face.
Under the stipulated equal-face conditions, both configurations of 1-chloroethylbenzene can form.
The product may therefore be racemic.
\(\textbf{Examine Route Q:}\)
Formation of sodium 1-phenylethoxide removes the hydrogen attached to oxygen.
It does not break the bond between oxygen and the stereogenic carbon.
The resulting alkoxide subsequently attacks the methyl carbon of iodomethane.
The new bond forms between the original alcohol oxygen and the methyl group.
The original stereogenic carbon is not the site of nucleophilic displacement.
Therefore, its spatial configuration can be preserved under the specified conditions.
\(\textbf{Interpret the stereochemical distinction:}\)
P involves formation of a planar carbocation at the originally stereogenic carbon.
This process can destroy the starting configurational information.
Q involves oxygen methylation at a different carbon.
It can preserve the stereogenic arrangement of the alcohol-derived carbon.
The two products differ not only in their functional groups but also in how their reaction pathways affect optical activity.
A common chiral alcohol starting material therefore does not imply a common stereochemical outcome.
611. Three compounds have the same molecular formula, \(\mathrm{C_4H_{10}O}\).
P: Butan-1-ol.
Q: 2-Methylpropan-2-ol.
R: Ethoxyethane.
Their approximate boiling points at atmospheric pressure are \(391\,\mathrm{K}\), \(355\,\mathrm{K}\) and \(308\,\mathrm{K}\), respectively.
The compounds are also subjected independently to PCC oxidation and reaction with sodium metal.
Consider the following statements.
Statement I: P has a higher boiling point than Q partly because the less-branched carbon skeleton permits more extensive intermolecular contact.
Statement II: R has the lowest boiling point because, unlike P and Q, it cannot form ordinary intermolecular O–H hydrogen bonds between its own molecules.
Statement III: Q gives a ketone on ordinary PCC oxidation because tertiary alcohols contain three carbon substituents at the hydroxyl-bearing carbon.
Statement IV: P and Q release hydrogen with sodium metal, whereas R lacks the ordinary alcohol O–H bond required for that reaction.
Which combination contains only the correct statements?
ⓐ. I, II and III only
ⓑ. I, II and IV only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: The three compounds are constitutional isomers, so their molecular masses are identical.
Their boiling-point differences must therefore be explained mainly through structural and intermolecular effects rather than molecular mass.
Statement I is correct because both P and Q possess alcohol hydroxyl groups and can form intermolecular hydrogen bonds.
However, P has a relatively extended carbon skeleton, whereas Q is highly branched.
Greater branching generally decreases intermolecular contact area and can lower the boiling point of comparable alcohol isomers.
Statement II is correct because ethoxyethane contains an ether oxygen but no O–H bond.
Its molecules can interact through dispersion and dipole forces, but they cannot donate conventional O–H hydrogen bonds to one another.
This helps explain why its boiling point is much lower than those of the two alcohols.
Statement III is incorrect because the hydroxyl-bearing carbon of Q has no directly attached hydrogen.
Ordinary oxidation of an alcohol to a carbonyl compound requires the appropriate hydrogen on that carbon.
Therefore, Q resists ordinary PCC oxidation rather than giving a ketone.
Statement IV is correct because P and Q possess alcohol hydroxyl groups capable of releasing hydrogen with sodium.
R contains no oxygen-bound hydrogen and does not undergo the corresponding alcohol–sodium reaction.
The evidence distinguishes hydrogen-bond donation, molecular branching and alcohol oxidation requirements.
612. A \(0.60\,\mathrm{mol}\) mixture contains only the following alcohols:
P: Butan-1-ol.
Q: Butan-2-ol.
R: 2-Methylpropan-2-ol.
The vapours are passed completely over heated copper at \(573\,\mathrm{K}\).
Assume the characteristic reactions proceed quantitatively:
I. P undergoes dehydrogenation to butanal, releasing one molecule of hydrogen.
II. Q undergoes dehydrogenation to butan-2-one, releasing one molecule of hydrogen.
III. R undergoes dehydration to 2-methylpropene and water without hydrogen evolution.
The collected hydrogen amounts to \(0.40\,\mathrm{mol}\).
The isolated carbonyl-containing fraction is treated with excess Tollens' reagent and produces \(0.50\,\mathrm{mol}\) metallic silver.
Assume each butanal molecule produces two silver atoms, and butan-2-one gives no characteristic Tollens' reaction.
What were the initial amounts of P, Q and R, respectively?
ⓐ. \(0.25,\ 0.15,\ 0.20\,\mathrm{mol}\)
ⓑ. \(0.20,\ 0.20,\ 0.20\,\mathrm{mol}\)
ⓒ. \(0.25,\ 0.20,\ 0.15\,\mathrm{mol}\)
ⓓ. \(0.15,\ 0.25,\ 0.20\,\mathrm{mol}\)
Correct Answer: \(0.25,\ 0.15,\ 0.20\,\mathrm{mol}\)
Explanation: \(\textbf{Assign the unknown alcohol amounts:}\)
Let the initial amounts of P, Q and R be \(x\), \(y\) and \(z\) moles.
The total amount gives:
\[
x+y+z=0.60
\]
\(\textbf{Use the different heated-copper reactions:}\)
A primary alcohol undergoes dehydrogenation to an aldehyde.
A secondary alcohol undergoes dehydrogenation to a ketone.
Under the specified conditions, the tertiary alcohol undergoes dehydration instead.
\[
\mathrm{CH_3CH_2CH_2CH_2OH
\xrightarrow{Cu,\ 573\,K}
CH_3CH_2CH_2CHO+H_2}
\]
\[
\mathrm{CH_3CH(OH)CH_2CH_3
\xrightarrow{Cu,\ 573\,K}
CH_3COCH_2CH_3+H_2}
\]
\[
\mathrm{(CH_3)_3COH
\xrightarrow{Cu,\ 573\,K}
(CH_3)_2C=CH_2+H_2O}
\]
\(\textbf{Use the hydrogen measurement:}\)
Only P and Q contribute hydrogen.
\[
x+y=0.40
\]
Combining this with the total amount:
\[
z=0.60-0.40
\]
\[
z=0.20\,\mathrm{mol}
\]
\(\textbf{Use the Tollens' result to identify P:}\)
Only butanal contributes to the stated silver formation.
Each butanal molecule reduces two silver(I) ions.
\[
2x=0.50
\]
\[
x=0.25\,\mathrm{mol}
\]
\(\textbf{Determine Q by difference:}\)
\[
y=0.40-0.25
\]
\[
y=0.15\,\mathrm{mol}
\]
\(\textbf{Verify the complete mixture:}\)
\[
x+y+z=0.25+0.15+0.20
\]
\[
x+y+z=0.60\,\mathrm{mol}
\]
The total amount, hydrogen evolution and silver yield are all satisfied.
\(\textbf{Interpret the diagnostic integration:}\)
Hydrogen evolution distinguishes the combined primary-and-secondary fraction from the tertiary fraction.
Tollens' reagent then separates the primary-alcohol contribution from the secondary-alcohol contribution after oxidation.
Neither measurement alone determines all three amounts.
\(\textbf{Final answer:}\)
\[
\boxed{(n_P,n_Q,n_R)=(0.25,\ 0.15,\ 0.20)\,\mathrm{mol}}
\]
613. Four oxygen-containing compounds are compared for dissolution in water.
P: Propan-1-ol.
Q: Propane-1,2-diol.
R: Propane-1,2,3-triol.
S: 1-Methoxypropane.
Assume ordinary dilute aqueous conditions without chemical reaction.
Consider the following statements.
Statement I: P, Q and R can both donate and accept hydrogen bonds involving water.
Statement II: S can accept hydrogen bonds from water despite lacking an ordinary O–H bond.
Statement III: Increasing the number of hydroxyl groups on the same three-carbon framework generally increases the capacity for hydrogen bonding with water.
Statement IV: S must be more water-soluble than R because ether oxygen has two lone pairs while an alcohol oxygen has only one.
Which combination contains only the correct statements?
ⓐ. I, II and IV only
ⓑ. II, III and IV only
ⓒ. I, III and IV only
ⓓ. I, II and III only
Correct Answer: I, II and III only
Explanation: The four compounds possess oxygen-containing functional groups, but their hydrogen-bonding capabilities differ.
Statement I is correct because P, Q and R contain alcohol hydroxyl groups.
The oxygen atoms of those groups can accept hydrogen bonds from water.
Their oxygen-bound hydrogens can also participate as hydrogen-bond donors.
Statement II is correct because the ether oxygen of S possesses nonbonding electron pairs.
Although S cannot donate an ordinary O–H hydrogen bond, it can accept one from a water molecule.
The absence of an O–H bond therefore does not imply complete inability to hydrogen-bond with water.
Statement III is correct because adding hydroxyl groups to the same carbon framework creates additional opportunities for interaction with water.
Q has two hydroxyl groups and R has three, compared with only one in P.
The hydrocarbon contribution remains based on the same three-carbon skeleton in this comparison.
However, hydrogen-bonding capacity should not be confused with an unrestricted numerical prediction of solubility under every set of conditions.
Statement IV is incorrect because an ordinary neutral ether oxygen and an ordinary neutral alcohol oxygen both possess lone pairs.
The proposed difference in lone-pair count is false.
Moreover, R possesses three hydroxyl groups and has extensive hydrogen-bonding opportunities.
The comparison demonstrates why counting oxygen atoms without identifying their bonding environments is an unreliable way to explain physical properties.
614. A liquid mixture contains benzoic acid and methyl benzoate.
The mixture is shaken with aqueous sodium hydrogencarbonate and an immiscible organic solvent under conditions where:
I. Benzoic acid is quantitatively converted into aqueous sodium benzoate.
II. Methyl benzoate undergoes negligible ester hydrolysis and remains in the organic layer.
III. Both layers are separated completely.
The aqueous layer is acidified strongly, while the organic layer is independently subjected to complete alkaline ester hydrolysis followed by acidification.
Which statement correctly predicts the products and explains the separation?
ⓐ. Only the original benzoic acid gives benzoic acid after the complete sequence; methyl benzoate produces methanol but no aromatic acid.
ⓑ. The initial separation is possible because methyl benzoate contains no aromatic ring, whereas benzoic acid does.
ⓒ. Both fractions ultimately yield benzoic acid, but through different transformations: protonation of benzoate in one fraction and ester hydrolysis followed by acidification in the other.
ⓓ. Both compounds immediately release carbon dioxide during the hydrogencarbonate extraction because each contains two oxygen atoms.
Correct Answer: Both fractions ultimately yield benzoic acid, but through different transformations: protonation of benzoate in one fraction and ester hydrolysis followed by acidification in the other.
Explanation: \(\textbf{Identify the original functional-group distinction:}\)
Benzoic acid contains a free carboxylic acid group.
Methyl benzoate contains an ester group formed from benzoic acid and methanol.
Although their aromatic carbon frameworks are closely related, their acid–base behaviour differs.
\(\textbf{Evaluate the initial extraction:}\)
Benzoic acid reacts with aqueous sodium hydrogencarbonate to form sodium benzoate.
The ionic sodium salt is preferentially accommodated in the aqueous layer under the stipulated conditions.
Carbon dioxide is also released during this neutralisation.
Methyl benzoate does not contain a free carboxylic acid proton.
It therefore remains predominantly unreacted in the organic layer.
\(\textbf{Follow the aqueous fraction:}\)
Strong acidification converts sodium benzoate into benzoic acid.
This transformation regenerates the original neutral acid through proton transfer.
No cleavage of a carbon–oxygen ester bond is required.
\(\textbf{Follow the organic fraction:}\)
Methyl benzoate undergoes alkaline ester hydrolysis.
The products include benzoate and methanol.
Subsequent acidification converts the benzoate into benzoic acid.
\(\textbf{Interpret the final convergence:}\)
Both fractions can ultimately furnish benzoic acid.
However, they reach that product through chemically distinct operations.
One route reverses an acid–base neutralisation.
The other first hydrolyses a covalent ester linkage.
Therefore, formation of the same final organic acid does not imply that the two original compounds had identical functional groups.
615. A \(0.40\,\mathrm{mol}\) sample of ethylmagnesium bromide is prepared in dry ether.
Before any further reaction, \(0.10\,\mathrm{mol}\) ethanol accidentally enters the reaction flask and reacts completely with an equal molar amount of the Grignard reagent.
The remaining ethylmagnesium bromide is then treated quantitatively with excess carbon dioxide, followed by acidic work-up.
Assume no competing reactions, complete recovery of the organic acid and complete subsequent neutralisation of that acid using excess aqueous sodium hydrogencarbonate.
What are the amounts of ethane produced during accidental quenching and carbon dioxide evolved during the final hydrogencarbonate reaction, respectively?
ⓐ. \(0.10\,\mathrm{mol}\) and \(0.40\,\mathrm{mol}\)
ⓑ. \(0.10\,\mathrm{mol}\) and \(0.30\,\mathrm{mol}\)
ⓒ. \(0.30\,\mathrm{mol}\) and \(0.10\,\mathrm{mol}\)
ⓓ. \(0.40\,\mathrm{mol}\) and \(0.30\,\mathrm{mol}\)
Correct Answer: \(0.10\,\mathrm{mol}\) and \(0.30\,\mathrm{mol}\)
Explanation: \(\textbf{Identify the incompatible functional group:}\)
A Grignard reagent contains a strongly basic carbon–magnesium bond.
Ordinary alcohols contain an oxygen-bound hydrogen capable of protonating that carbon fragment.
Therefore, ethanol destroys part of the active ethylmagnesium bromide through acid–base chemistry.
\(\textbf{Determine the accidental quenching reaction:}\)
\[
\mathrm{C_2H_5MgBr+C_2H_5OH}
\]
\[
\mathrm{\longrightarrow C_2H_6+C_2H_5OMgBr}
\]
The mole ratio between ethanol consumed and ethane produced is \(1:1\).
\[
n(\mathrm{C_2H_6})=0.10\,\mathrm{mol}
\]
\(\textbf{Calculate the remaining Grignard reagent:}\)
\[
n(\mathrm{C_2H_5MgBr})_{\mathrm{remaining}}=0.40-0.10
\]
\[
n(\mathrm{C_2H_5MgBr})_{\mathrm{remaining}}=0.30\,\mathrm{mol}
\]
\(\textbf{Apply carbon dioxide carboxylation:}\)
\[
\mathrm{C_2H_5MgBr+CO_2\longrightarrow C_2H_5COOMgBr}
\]
Acidic work-up yields propanoic acid.
The carbon dioxide carbon becomes the carboxyl carbon of the new acid.
\[
n(\mathrm{C_2H_5COOH})=0.30\,\mathrm{mol}
\]
\(\textbf{Calculate the subsequent carbon dioxide evolution:}\)
\[
\mathrm{C_2H_5COOH+NaHCO_3}
\]
\[
\mathrm{\longrightarrow C_2H_5COONa+CO_2+H_2O}
\]
One acid molecule produces one carbon dioxide molecule.
\[
n(\mathrm{CO_2})_{\mathrm{final}}=0.30\,\mathrm{mol}
\]
\(\textbf{Distinguish the two appearances of carbon dioxide:}\)
Carbon dioxide is consumed as a carbon-containing reagent during Grignard carboxylation.
It is subsequently produced during neutralisation of the resulting acid.
These are different chemical events.
The accidental ethanol quenching reduces the amount of carboxylic acid available for the final reaction.
\(\textbf{Final answer:}\)
\[
\boxed{n(\mathrm{C_2H_6})=0.10\,\mathrm{mol},\qquad n(\mathrm{CO_2})_{\mathrm{final}}=0.30\,\mathrm{mol}}
\]
616. Two compounds are constitutional isomers with molecular formula \(\mathrm{C_9H_{10}O_2}\).
P: Phenyl propanoate, \(\mathrm{CH_3CH_2COOC_6H_5}\).
Q: Benzyl ethanoate, \(\mathrm{CH_3COOCH_2C_6H_5}\).
Each undergoes complete alkaline ester hydrolysis, followed by suitable acidification and separation of the resulting organic products.
The alcohol- or phenol-derived fractions are then independently examined using neutral aqueous ferric chloride.
Which statement correctly compares the outcomes?
ⓐ. P gives phenol and propanoic acid; Q gives benzyl alcohol and ethanoic acid. Only P supplies the phenolic compound for the ferric chloride test.
ⓑ. P yields benzyl alcohol and ethanoic acid, while Q yields phenol and propanoic acid; both products are ordinary primary alcohols.
ⓒ. Both give phenol and ethanoic acid because their molecular formulae are identical.
ⓓ. P gives propan-1-ol and benzoic acid, while Q gives ethanal and phenol.
Correct Answer: P gives phenol and propanoic acid; Q gives benzyl alcohol and ethanoic acid. Only P supplies the phenolic compound for the ferric chloride test.
Explanation: \(\textbf{Distinguish the positions of the aromatic ring:}\)
P contains a phenoxy fragment directly connected to the ester oxygen.
Q contains a benzyl group in which a saturated methylene carbon separates the aromatic ring from oxygen.
This difference changes the identity of the oxygen-containing compound released during hydrolysis.
\(\textbf{Hydrolyse P:}\)
Phenyl propanoate contains a propanoyl group and a phenoxy group.
Alkaline hydrolysis produces propanoate and phenol-derived material.
After acidification, propanoic acid and phenol can be isolated.
\[
\mathrm{CH_3CH_2COOC_6H_5}
\]
\[
\mathrm{\xrightarrow[\text{work-up}]{NaOH,\ H_2O}
CH_3CH_2COOH+C_6H_5OH}
\]
\(\textbf{Hydrolyse Q:}\)
Benzyl ethanoate contains an ethanoyl group and a benzyl-derived alkoxy fragment.
Hydrolysis produces ethanoate and benzyl alcohol.
Acidification regenerates ethanoic acid.
\[
\mathrm{CH_3COOCH_2C_6H_5}
\]
\[
\mathrm{\xrightarrow[\text{work-up}]{NaOH,\ H_2O}
CH_3COOH+C_6H_5CH_2OH}
\]
\(\textbf{Interpret the ferric chloride test:}\)
Phenol contains hydroxyl directly attached to an aromatic carbon.
It can form the characteristic coloured complex with neutral ferric chloride under suitable conditions.
Benzyl alcohol contains its hydroxyl on a saturated carbon.
It does not exhibit the corresponding ordinary phenolic response.
\(\textbf{Integrate the structural evidence:}\)
Both starting compounds are aromatic esters with identical molecular formulae.
However, ester hydrolysis reveals different acid and hydroxyl-containing fragments.
The ferric chloride reaction then distinguishes the directly aromatic hydroxyl group from the benzylic alcohol hydroxyl group.
This makes hydrolysis followed by functional-group testing more informative than molecular formula comparison alone.
617. A phenolic compound X contains a methyl group at position 3 relative to its hydroxyl group.
It undergoes controlled aromatic mononitration.
Assume that the combined directing effects of the hydroxyl and methyl groups determine the electronically favoured positions, while steric crowding can affect the relative proportions of the individual products.
Which statement correctly identifies the positions simultaneously favoured by both substituents?
ⓐ. Positions 2, 4 and 6 are simultaneously ortho/para directed by the two substituents, although steric effects can influence the individual product proportions.
ⓑ. Only position 5 is favoured because the methyl group converts the hydroxyl group into a meta director.
ⓒ. Positions 3 and 5 are favoured because both substituents are electron withdrawing.
ⓓ. The substitution position cannot be predicted at all because hydroxyl and methyl exert opposite directing effects.
Correct Answer: Positions 2, 4 and 6 are simultaneously ortho/para directed by the two substituents, although steric effects can influence the individual product proportions.
Explanation: \(\textbf{Assign the ring positions:}\)
The compound is 3-methylphenol.
Hydroxyl occupies carbon 1, while methyl occupies carbon 3.
The aromatic hydrogens occupy positions 2, 4, 5 and 6.
\(\textbf{Apply the hydroxyl directing effect:}\)
The hydroxyl group donates electron density through resonance.
It activates the aromatic ring and favours substitution at ortho and para positions.
Relative to hydroxyl at position 1, these are positions 2, 4 and 6.
\(\textbf{Apply the methyl directing effect:}\)
The methyl group is also an ortho/para director through its electron-releasing influence.
Relative to methyl at position 3, the available favoured positions are 2, 4 and 6.
Therefore, the two directing groups reinforce substitution at the same three positions.
\(\textbf{Examine position 5:}\)
Position 5 is meta to both hydroxyl and methyl.
It is not electronically favoured in the same way as positions 2, 4 and 6.
\(\textbf{Consider steric selectivity:}\)
Position 2 lies between hydroxyl and methyl and may experience greater steric crowding.
Positions 4 and 6 have different local environments.
Therefore, identifying electronically favoured positions does not justify claiming equal product proportions.
The question separates positional direction from the quantitative distribution of nitration products.
Both effects must be considered before assigning a unique major isomer under specified experimental conditions.
618. An aromatic compound P contains a phenolic hydroxyl group and a carboxylic acid group at para positions.
A chemist separately performs the following reactions:
Route I: Methylation of the phenolic hydroxyl without changing the carboxylic acid group.
Route II: Esterification of the carboxylic acid group with methanol without changing the phenolic hydroxyl group.
The isolated products are treated independently with neutral aqueous ferric chloride and aqueous sodium hydrogencarbonate.
Assume neither reagent hydrolyses the methyl ether or methyl ester during testing.
Which statement correctly distinguishes the products?
ⓐ. Both products are ferric chloride-positive and hydrogencarbonate-positive because methylation does not change acid–base behaviour.
ⓑ. Route I gives a methyl ester that is ferric chloride-positive, while Route II gives an aromatic methyl ether that releases carbon dioxide.
ⓒ. Route I gives 4-methoxybenzoic acid, which retains the hydrogencarbonate response; Route II gives methyl 4-hydroxybenzoate, which retains the phenolic ferric chloride response.
ⓓ. Both routes produce methyl 4-methoxybenzoate because the same methyl group is introduced into the molecule.
Correct Answer: Route I gives 4-methoxybenzoic acid, which retains the hydrogencarbonate response; Route II gives methyl 4-hydroxybenzoate, which retains the phenolic ferric chloride response.
Explanation: \(\textbf{Identify the two independently reactive groups:}\)
The starting compound is 4-hydroxybenzoic acid.
It contains a phenolic oxygen–hydrogen group and a carboxylic acid group.
These two functionalities have different bonding environments and characteristic reactions.
\(\textbf{Analyse Route I:}\)
Selective methylation of phenolic oxygen converts the \(\mathrm{-OH}\) group into \(\mathrm{-OCH_3}\).
The aromatic carboxylic acid group is retained.
The product is 4-methoxybenzoic acid.
It no longer contains a free phenolic hydroxyl group.
Therefore, it lacks the ordinary characteristic ferric chloride response of the original phenol.
However, its free carboxylic acid group can still release carbon dioxide with sodium hydrogencarbonate.
\(\textbf{Analyse Route II:}\)
Esterification converts the carboxylic acid group into a methyl ester.
The phenolic hydroxyl remains unchanged.
The product is methyl 4-hydroxybenzoate.
It can exhibit the characteristic phenolic ferric chloride response.
However, the carboxylic acid proton is no longer present because that group has been esterified.
\(\textbf{Interpret the diagnostic pattern:}\)
Route I preserves the acid response while removing the free phenolic response.
Route II preserves the phenolic response while removing the free carboxylic acid response.
The two products each introduce one methyl group but have different connectivity.
The evidence therefore identifies the site of methylation rather than merely the increase in carbon count.
619. Phenol is dissolved in two separate solvents under conditions where no chemical reaction occurs.
Experiment P uses a nonpolar solvent in which intermolecular hydrogen-bonded association between phenol molecules is possible.
Experiment Q uses a sufficiently large amount of water, allowing phenol molecules to participate in hydrogen bonding with the solvent.
A researcher makes the following claims.
Assertion (A): The ability of phenol to form hydrogen bonds with water does not require phenol molecules to remain associated with one another in their original hydrogen-bonded arrangement.
Reason (R): Hydrogen bonds are intermolecular interactions that can be reorganised when the solvent changes, and phenol can both donate and accept hydrogen bonds involving water.
Which option is correct?
ⓐ. Both A and R are true, and R correctly explains A.
ⓑ. Both A and R are true, but R does not explain A.
ⓒ. A is true, but R is false.
ⓓ. A is false, but R is true.
Correct Answer: Both A and R are true, and R correctly explains A.
Explanation: Phenol contains a hydroxyl group directly attached to an aromatic carbon.
Its oxygen atom possesses nonbonding electron pairs, while its oxygen–hydrogen bond can act as a hydrogen-bond donor.
Phenol molecules can therefore participate in intermolecular hydrogen bonding.
However, a hydrogen bond is not a permanent covalent connection fixing two phenol molecules together.
Hydrogen-bonded arrangements can change as molecules move and interact with their surroundings.
In a suitable nonpolar solvent, interactions among phenol molecules may contribute significantly to association.
In water, phenol can form hydrogen bonds with water molecules.
Its hydroxyl hydrogen can interact with water oxygen, while phenol oxygen can accept hydrogen bonds from water.
These interactions can occur without preserving the original phenol–phenol associations.
Therefore, the Assertion is true.
The Reason is also true because it identifies the reversible intermolecular character of hydrogen bonding and the donor–acceptor capabilities of phenol.
It directly explains why solvent changes can reorganise the association pattern.
The example distinguishes the existence of hydrogen-bonding capability from one fixed molecular arrangement in every medium.
620. Two aqueous solutions contain equal analytical concentrations of phenol and ethanoic acid, respectively.
Both are separately treated with identical amounts of aqueous sodium hydroxide.
After complete reaction with the added base, the pH of each solution is adjusted to a value that is well below the \(\mathrm{p}K_a\) of phenol but well above the \(\mathrm{p}K_a\) of ethanoic acid.
Assume ideal acid–base equilibria and no other acid–base-active solutes.
Which statement correctly predicts the dominant chemical forms at the adjusted pH?
ⓐ. Both are predominantly present as their fully deprotonated sodium salts because they previously reacted with sodium hydroxide.
ⓑ. Phenol is mainly protonated and ethanoic acid mainly deprotonated; the final distribution is governed by equilibrium pH, not merely by adding base.
ⓒ. Phenol is predominantly phenoxide, whereas ethanoic acid is predominantly neutral because phenol is the stronger acid.
ⓓ. Both are predominantly neutral because adjusting pH reverses every previous acid–base reaction quantitatively.
Correct Answer: Phenol is mainly protonated and ethanoic acid mainly deprotonated; the final distribution is governed by equilibrium pH, not merely by adding base.
Explanation: \(\textbf{Identify the relevant acidity difference:}\)
Ethanoic acid is much stronger than phenol in aqueous solution.
Its carboxylate conjugate base is strongly stabilised by charge delocalisation over two oxygen atoms.
Phenoxide is resonance stabilised through the aromatic system, but phenol remains substantially weaker than an ordinary carboxylic acid.
\(\textbf{Apply the final pH condition to phenol:}\)
The solution pH is specified to be well below phenol's \(\mathrm{p}K_a\).
The Henderson–Hasselbalch relation therefore favours the protonated phenol form.
Even if some phenoxide formed during the earlier sodium hydroxide treatment, the final equilibrium conditions favour its protonation.
\(\textbf{Apply the same pH to ethanoic acid:}\)
The pH is simultaneously well above the \(\mathrm{p}K_a\) of ethanoic acid.
The ethanoate-to-ethanoic-acid ratio is therefore large.
Ethanoate is the dominant form.
\(\textbf{Interpret the apparent contradiction:}\)
Both compounds may react with sufficiently strong aqueous sodium hydroxide.
However, their final ionisation states need not remain identical after pH adjustment.
Acid–base equilibria respond to the current hydrogen-ion concentration.
They do not permanently record whether a molecule was previously neutralised.
\(\textbf{Apply the general principle:}\)
For any weak acid, the protonated form is favoured when pH is well below its \(\mathrm{p}K_a\).
The deprotonated form is favoured when pH is well above its \(\mathrm{p}K_a\).
This rule allows different functional groups to possess different dominant charge states at one common pH.
