401. Assertion: Phenol reacts with both sodium metal and aqueous sodium hydroxide, whereas ethanol reacts with sodium metal but not appreciably with aqueous sodium hydroxide.
Reason: Phenoxide ion is resonance stabilised, making phenol more acidic than ethanol.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Sodium metal reacts with compounds that possess sufficiently reactive hydroxyl hydrogens and releases hydrogen gas. Both phenol and ethanol can therefore react with sodium. Aqueous sodium hydroxide deprotonates phenol because the resulting phenoxide ion is stabilised by resonance. Ethoxide lacks comparable delocalisation, so ethanol remains largely unreacted with aqueous hydroxide. The conjugate-base stability described in the Reason explains the different base reactions.
402. Phenol undergoes electrophilic aromatic substitution more readily than benzene because the hydroxyl group:
ⓐ. withdraws all electron density from the ring
ⓑ. destroys aromaticity before the reaction starts
ⓒ. donates ring electron density by resonance
ⓓ. converts every electrophile into a nucleophile
Correct Answer: donates ring electron density by resonance
Explanation: A lone pair on the phenolic oxygen overlaps with the aromatic \(\pi\)-system. This resonance interaction increases electron density in the ring, especially at the ortho and para positions. A more electron-rich ring attacks electrophiles more readily than benzene. The hydroxyl group therefore activates the aromatic ring. Its resonance donation dominates over its electron-withdrawing inductive effect during electrophilic substitution.
403. Ortho and para substitution are favoured in phenol because the corresponding sigma complexes:
ⓐ. oxygen donation adds resonance stabilisation
ⓑ. contain no positive charge at any stage
ⓒ. are formed only after the hydroxyl group leaves
ⓓ. convert the electrophile into water
Correct Answer: oxygen donation adds resonance stabilisation
Explanation: Electrophilic attack temporarily produces a non-aromatic positively charged sigma complex. For ortho and para attack, resonance structures can be drawn in which oxygen donates a lone pair into the ring. This donation helps delocalise and stabilise the positive charge. The meta sigma complex does not receive the same direct oxygen-assisted stabilisation. Ortho and para pathways therefore have lower-energy intermediates and are favoured.
404. Which statement about meta substitution in phenol is most accurate?
ⓐ. It is impossible because the meta carbon contains no electrons
ⓑ. It is always the only product
ⓒ. It is favoured because oxygen withdraws by resonance
ⓓ. meta attack lacks extra oxygen-derived resonance
Correct Answer: meta attack lacks extra oxygen-derived resonance
Explanation: Every ring position contains electron density and can in principle be attacked under suitable conditions. The important difference lies in the stability of the sigma complexes. Ortho and para attack produce intermediates that can be stabilised by oxygen lone-pair donation. Meta attack lacks this particular stabilising resonance contributor. Meta substitution is therefore disfavoured rather than absolutely forbidden.
405. Phenol usually does not require a Lewis acid catalyst such as \(\mathrm{FeBr_3}\) for bromination because:
ⓐ. bromine is already negatively charged
ⓑ. the hydroxyl group strongly activates the aromatic ring
ⓒ. phenol first converts into benzene
ⓓ. the reaction proceeds by a free-radical chain mechanism only
Correct Answer: the hydroxyl group strongly activates the aromatic ring
Explanation: Benzene normally requires a catalyst to generate a sufficiently strong brominating electrophile. Phenol is substantially more reactive because oxygen donates electron density into the ring. The activated ring can polarise and react with bromine under much milder conditions. A Lewis acid catalyst is therefore often unnecessary. The reaction remains electrophilic aromatic substitution rather than radical side-chain bromination.
406. The phenoxide ion is generally even more activating toward electrophilic attack than neutral phenol because:
ⓐ. its oxygen has no lone pairs
ⓑ. it has lost aromaticity permanently
ⓒ. negative charge activates the ring
ⓓ. it contains a carbon-halogen bond
Correct Answer: negative charge activates the ring
Explanation: Phenoxide carries a formal negative charge on oxygen in an important contributing structure. This charge is delocalised into the aromatic ring through resonance. The ring therefore becomes highly electron rich. Electrophiles are strongly attracted to such an activated system. Practical reaction conditions must still be considered because some electrophiles may react with the strongly basic phenoxide ion in other ways.
407. When both ortho and para products are possible in the substitution of phenol by a bulky electrophile, the para product may predominate because:
ⓐ. the para site is less sterically crowded
ⓑ. oxygen directs exclusively to the meta position
ⓒ. the para carbon carries no hydrogen
ⓓ. ortho substitution destroys the carbon-oxygen bond
Correct Answer: the para site is less sterically crowded
Explanation: Resonance activation makes both ortho and para positions favourable. The ortho positions lie directly beside the hydroxyl group. A bulky electrophile experiences greater crowding when approaching these positions. The para position is farther from the hydroxyl substituent and is more accessible. Steric effects can therefore make the para product more abundant even though both orientations are electronically favoured.
408. A reaction-coordinate graph compares electrophilic substitution of benzene and phenol by the same electrophile. Which feature is expected for phenol?
ⓐ. A higher activation barrier because oxygen removes ring electrons
ⓑ. An identical activation barrier in every case
ⓒ. No energy maximum because the reaction is instantaneous
ⓓ. a lower barrier from a resonance-stabilised sigma complex
Correct Answer: a lower barrier from a resonance-stabilised sigma complex
Explanation: The slow step in electrophilic aromatic substitution involves formation of the sigma complex. In phenol, oxygen lone-pair donation stabilises the developing positive charge. This lowers the energy of the transition state and intermediate relative to the corresponding benzene pathway. The activation barrier is therefore reduced. A lower barrier accounts for the faster reaction of phenol under comparable conditions.
409. Bromination of phenol in a nonpolar solvent such as carbon disulfide at low temperature mainly gives:
ⓐ. \(m\)-bromophenol only
ⓑ. 2,4,6-tribromophenol exclusively
ⓒ. bromobenzene and water
ⓓ. ortho- and para-bromophenols
Correct Answer: ortho- and para-bromophenols
Explanation: The hydroxyl group activates the aromatic ring and directs bromine toward ortho and para positions. A nonpolar solvent and controlled conditions limit the extent of substitution. Monobromination therefore predominates rather than immediate replacement of three ring hydrogens. Both ortho and para products are formed. The para isomer is commonly favoured because it experiences less steric crowding.
410. Phenol reacts with excess bromine water to form:
ⓐ. meta-bromophenol
ⓑ. bromobenzene product
ⓒ. 2,4,6-tribromophenol
ⓓ. para-bromophenol only
Correct Answer: 2,4,6-tribromophenol
Explanation: The hydroxyl group strongly activates the aromatic ring through resonance donation. It also directs electrophilic substitution toward both ortho positions and the para position. In bromine water, the reaction is sufficiently rapid for all three activated positions to undergo substitution. The product is therefore 2,4,6-tribromophenol. Controlled monobromination requires milder conditions such as bromine in a nonpolar solvent.
411. The characteristic observation when bromine water is added to an aqueous solution of phenol is:
ⓐ. bromine decolourises and a white precipitate forms
ⓑ. evolution of a colourless gas without any precipitate
ⓒ. formation of a blue solution
ⓓ. no visible change at room temperature
Correct Answer: bromine decolourises and a white precipitate forms
Explanation: Phenol rapidly consumes bromine because its aromatic ring is strongly activated. The brown or orange colour of bromine water disappears as bromine is incorporated into the ring. The product 2,4,6-tribromophenol is only sparingly soluble in water. It consequently separates as a white precipitate. The combined colour loss and precipitate formation provide a useful qualitative observation for phenol.
412. Phenol undergoes tribromination in bromine water without requiring \(\mathrm{FeBr_3}\) because:
ⓐ. water converts bromine completely into bromide ions
ⓑ. phenol reacts by a free-radical mechanism
ⓒ. the hydroxyl group leaves before bromination begins
ⓓ. oxygen donation activates the ring
Correct Answer: oxygen donation activates the ring
Explanation: A lone pair on the phenolic oxygen enters conjugation with the benzene ring. This increases electron density particularly at the ortho and para positions. The activated ring can polarise bromine and attack the electrophilic end without an added Lewis acid. Repeated substitution occurs because the remaining activated positions continue to react. The process is electrophilic aromatic substitution rather than radical bromination.
413. Bromination of phenol in bromine water differs from bromination in cold carbon disulfide mainly because:
ⓐ. water: meta monobromination; carbon disulfide: no reaction
ⓑ. water: 2,4,6-tribromination; carbon disulfide: ortho and para monobromination
ⓒ. water: ortho and para monobromination; carbon disulfide: 2,4,6-tribromination
ⓓ. both media: identical ortho and para monobromination
Correct Answer: water: 2,4,6-tribromination; carbon disulfide: ortho and para monobromination
Explanation: The solvent and reaction conditions influence how far substitution proceeds. In bromine water, the highly activated phenol ring undergoes rapid substitution at both ortho positions and the para position. In a nonpolar solvent under controlled conditions, the reaction can be limited mainly to monobromination. The hydroxyl group directs both reactions toward ortho and para positions. The difference is primarily the extent of substitution rather than a change to meta direction.
414. Which equation correctly represents bromination of phenol with excess bromine water?
ⓐ. \(\mathrm{C_6H_5OH+Br_2\rightarrow C_6H_5Br+H_2O}\)
ⓑ. \(\mathrm{C_6H_5OH+2Br_2\rightarrow C_6H_3Br_2OH+2HBr}\)
ⓒ. \(\mathrm{C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr}\)
ⓓ. \(\mathrm{C_6H_5OH+3HBr\rightarrow C_6H_2Br_3OH+3H_2}\)
Correct Answer: \(\mathrm{C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr}\)
Explanation: Three ring hydrogen atoms are replaced by bromine atoms. These substitutions occur at positions \(2\), \(4\), and \(6\) relative to the hydroxyl group. Each substitution consumes one molecule of bromine and releases one molecule of hydrogen bromide. Three molecules of bromine are therefore required per molecule of phenol. The molecular formula of the organic product is \(\mathrm{C_6H_3Br_3O}\), equivalently written as \(\mathrm{C_6H_2Br_3OH}\).
415. What mass of bromine is required for complete tribromination of \(9.4\,\mathrm{g}\) of phenol? Use \(M(\mathrm{phenol})=94\,\mathrm{g\,mol^{-1}}\) and \(M(\mathrm{Br_2})=160\,\mathrm{g\,mol^{-1}}\).
ⓐ. \(48.0\,\mathrm{g}\)
ⓑ. \(16.0\,\mathrm{g}\)
ⓒ. \(32.0\,\mathrm{g}\)
ⓓ. \(64.0\,\mathrm{g}\)
Correct Answer: \(48.0\,\mathrm{g}\)
Explanation: \( \textbf{Amount of phenol:} \)
\[
n(\mathrm{phenol})=\frac{9.4\,\mathrm{g}}{94\,\mathrm{g\,mol^{-1}}}
\]
\[
n(\mathrm{phenol})=0.100\,\mathrm{mol}
\]
\( \textbf{Reaction relation:} \)
\[
\mathrm{C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr}
\]
One mole of phenol consumes three moles of bromine.
\[
n(\mathrm{Br_2})=0.100\times3=0.300\,\mathrm{mol}
\]
\( \textbf{Mass of bromine:} \)
\[
m=nM
\]
\[
m=0.300\,\mathrm{mol}\times160\,\mathrm{g\,mol^{-1}}
\]
\[
m=48.0\,\mathrm{g}
\]
The factor of three arises from substitution at the two ortho positions and the para position.
416. Which result best distinguishes phenol from ethanol using bromine water?
ⓐ. ethanol decolourises bromine water with a white precipitate; phenol does not
ⓑ. phenol decolourises bromine water with a white precipitate; ethanol does not
ⓒ. both compounds form white tribromo precipitates
ⓓ. both release hydrogen gas from bromine water
Correct Answer: phenol decolourises bromine water with a white precipitate; ethanol does not
Explanation: Phenol contains an activated aromatic ring that rapidly undergoes electrophilic bromination. The reaction consumes bromine and forms insoluble 2,4,6-tribromophenol. Ethanol has no aromatic ring and cannot undergo the same substitution. Under ordinary test conditions, it does not produce the characteristic white tribromo precipitate. The contrast identifies the phenolic structure rather than merely the presence of an oxygen atom.
417. Phenol reacts with excess bromine water to form 2,4,6-tribromophenol:
[
\mathrm{C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr}
]
A (4.70,\mathrm{g}) sample of phenol gives the tribromo product in an (80%) yield. What mass of 2,4,6-tribromophenol is obtained? Use (M(\text{phenol})=94,\mathrm{g,mol^{-1}}) and (M(\text{2,4,6-tribromophenol})=331,\mathrm{g,mol^{-1}}).
ⓐ. (10.08,\mathrm{g})
ⓑ. (16.55,\mathrm{g})
ⓒ. (13.24,\mathrm{g})
ⓓ. (33.10,\mathrm{g})
Correct Answer: (13.24,\mathrm{g})
Explanation: ( \textbf{Amount of phenol:} )
[
n(\text{phenol})=\frac{4.70,\mathrm{g}}{94,\mathrm{g,mol^{-1}}}
]
[
n(\text{phenol})=0.0500,\mathrm{mol}
]
( \textbf{Product relation:} )
The equation shows that one mole of phenol forms one mole of 2,4,6-tribromophenol.
[
n_{\text{theoretical}}(\text{product})=0.0500,\mathrm{mol}
]
( \textbf{Apply the percentage yield:} )
[
n_{\text{actual}}=0.0500\times\frac{80}{100}
]
[
n_{\text{actual}}=0.0400,\mathrm{mol}
]
( \textbf{Product mass:} )
[
m=0.0400,\mathrm{mol}\times331,\mathrm{g,mol^{-1}}
]
[
m=13.24,\mathrm{g}
]
( \textbf{Final answer:} ) The mass of 2,4,6-tribromophenol obtained is (13.24,\mathrm{g}). The three molecules of bromine required per phenol molecule do not change the (1:1) mole relation between phenol and the organic product.
418. Treatment of phenol with dilute nitric acid at a controlled temperature gives mainly:
ⓐ. \(m\)-nitrophenol only
ⓑ. a mixture of \(o\)-nitrophenol and \(p\)-nitrophenol
ⓒ. 2,4,6-trinitrophenol only
ⓓ. nitrobenzene after loss of the hydroxyl group
Correct Answer: a mixture of \(o\)-nitrophenol and \(p\)-nitrophenol
Explanation: The hydroxyl group activates the aromatic ring and directs electrophilic substitution toward ortho and para positions. Under dilute and controlled conditions, mononitration predominates. Both ortho and para positions are available, so a mixture of the two nitrophenols is obtained. Meta substitution is minor because its sigma complex lacks the extra resonance stabilisation provided by oxygen. Extensive nitration requires stronger conditions.
419. A mixture of \(o\)-nitrophenol and \(p\)-nitrophenol obtained from dilute nitration can be separated by steam distillation because:
ⓐ. \(p\)-nitrophenol is the only compound containing oxygen
ⓑ. both isomers have identical intermolecular attractions
ⓒ. the ortho isomer has intramolecular hydrogen bonding
ⓓ. \(p\)-nitrophenol decomposes completely in steam
Correct Answer: the ortho isomer has intramolecular hydrogen bonding
Explanation: The adjacent hydroxyl and nitro groups in \(o\)-nitrophenol form an intramolecular hydrogen bond. This reduces association between separate ortho-isomer molecules. The compound is consequently more volatile and can distil with steam. \(p\)-Nitrophenol forms stronger intermolecular hydrogen-bonded associations and is less volatile. Steam distillation therefore exploits a physical-property difference created by the substituent positions.
420. An \(18.8\,\mathrm{g}\) sample of phenol undergoes dilute nitration. Only \(50\%\) of the phenol reacts, and \(60\%\) of the nitrated product is \(p\)-nitrophenol. What mass of \(p\)-nitrophenol is formed? Use \(M(\mathrm{phenol})=94\,\mathrm{g\,mol^{-1}}\) and \(M(p\text{-nitrophenol})=139\,\mathrm{g\,mol^{-1}}\).
ⓐ. \(13.9\,\mathrm{g}\)
ⓑ. \(5.56\,\mathrm{g}\)
ⓒ. \(16.7\,\mathrm{g}\)
ⓓ. \(8.34\,\mathrm{g}\)
Correct Answer: \(8.34\,\mathrm{g}\)
Explanation: \( \textbf{Initial amount of phenol:} \)
\[
n(\mathrm{phenol})=\frac{18.8\,\mathrm{g}}{94\,\mathrm{g\,mol^{-1}}}
\]
\[
n(\mathrm{phenol})=0.200\,\mathrm{mol}
\]
\( \textbf{Phenol that undergoes nitration:} \)
\[
n_{\text{reacted}}=0.200\times\frac{50}{100}
\]
\[
n_{\text{reacted}}=0.100\,\mathrm{mol}
\]
Mononitration gives one mole of nitrophenol per mole of phenol consumed.
\( \textbf{Amount of para isomer:} \)
\[
n(p\text{-nitrophenol})=0.100\times\frac{60}{100}
\]
\[
n(p\text{-nitrophenol})=0.0600\,\mathrm{mol}
\]
\( \textbf{Mass of para isomer:} \)
\[
m=0.0600\,\mathrm{mol}\times139\,\mathrm{g\,mol^{-1}}
\]
\[
m=8.34\,\mathrm{g}
\]
Both the percentage conversion and the product-distribution percentage must be applied.
