RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
A single –OH group turns a hydrocarbon into an alcohol or a phenol, and the two behave nothing alike once that group sits on an sp³ carbon versus directly on a benzene ring. This chapter builds the preparation and reaction map of alcohols, phenols and ethers, and pins down exactly why phenol is acidic while ethanol is barely so.
Which compound gives an immediate turbidity with the Lucas reagent at room temperature?
Answer
2-Methylpropan-2-ol. This tertiary alcohol ionises rapidly in the Lucas reagent (conc. HCl/ZnCl₂) because the resulting 3° carbocation is highly stabilised, so the insoluble alkyl chloride separates as cloudiness within seconds. Secondary alcohols need a few minutes and primary alcohols react only on warming, since their carbocations are progressively less stable.
Phenol is a stronger acid than ethanol mainly because:
Answer
The phenoxide ion is resonance-stabilised by delocalisation into the ring. Once phenol loses its proton, the negative charge on oxygen can spread onto the ortho and para carbons through resonance, lowering the energy of the phenoxide ion and pulling the equilibrium towards ionisation. The ethoxide ion formed from ethanol has no such delocalisation, so it is a much stronger, less stable base, and ethanol is correspondingly a far weaker acid.
The Williamson synthesis of tert-butyl methyl ether should be carried out using:
Answer
Methyl bromide and sodium tert-butoxide. The alkyl halide must be attacked by the alkoxide in an SN2 step, so it should be primary (methyl here) to avoid steric hindrance and elimination. Using tert-butyl bromide with sodium methoxide would instead favour E2 elimination to isobutylene, because the bulky base finds the hindered tertiary carbon difficult to attack but easily removes a β-hydrogen.
Treating phenol with bromine water at room temperature gives:
Answer
2,4,6-Tribromophenol as a white precipitate. The –OH group donates its lone pair into the ring, making phenol so strongly activated towards electrophilic substitution that bromine water reacts directly, without the FeBr₃ catalyst benzene would need, and substitutes at all three positions activated by resonance — both ortho positions and the para position.
Assertion (A): Phenol is more acidic than cyclohexanol. Reason (R): The oxygen of phenol is attached to an sp² carbon, which is more electronegative than the sp³ carbon of cyclohexanol.
Answer
Both A and R are true but R is not the correct explanation of A. Phenol is indeed far more acidic than cyclohexanol, and it is true that the sp² carbon of the ring is somewhat more electronegative than an sp³ carbon (a real but minor effect). However, the dominant reason for the huge acidity gap is not this inductive difference but resonance stabilisation of the phenoxide ion by delocalisation of its negative charge into the aromatic ring, an effect unavailable to cyclohexoxide.
Write the major product and name the mechanism when 2-methylpropan-2-ol is heated with concentrated H₂SO₄.
Answer
Product: 2-methylprop-1-ene, (CH₃)₂C=CH₂, formed by E1 elimination. Protonation of the –OH gives a good leaving group (water), which departs to form the stable tertiary carbocation (CH₃)₃C⁺; loss of a β-hydrogen by a base (HSO₄⁻ or another alcohol molecule) then generates the alkene. Because only one type of β-hydrogen is available here, isobutylene is the sole alkene product.
Describe the Kolbe reaction: reagents, conditions and product starting from phenol, and explain why phenoxide reacts with carbon dioxide when phenol itself does not.
Answer
Phenol is first converted to sodium phenoxide with NaOH. The phenoxide is then heated with CO₂ at about 400 K under 4–7 atm pressure, and the resulting solid is acidified with dilute acid. C₆H₅O⁻Na⁺ + CO₂ → sodium salicylate → (H⁺) → salicylic acid (2-hydroxybenzoic acid). Phenoxide reacts because the negatively charged oxygen pushes extra electron density into the ring by resonance, making the ortho carbon nucleophilic enough to attack the weak electrophile CO₂. Neutral phenol has only a lone pair, not a full negative charge, so its ring is not nucleophilic enough to attack CO₂ under these mild conditions.
Arrange the following in decreasing order of boiling point and justify: propan-1-ol, propane, diethyl ether (all roughly similar molar mass) — wait, use propan-1-ol, methoxyethane and butane.
Answer
Order: propan-1-ol > methoxyethane > butane. Propan-1-ol has an O–H bond and forms intermolecular hydrogen bonds, which need considerable extra energy to break, giving it the highest boiling point. Methoxyethane has polar C–O bonds and a net dipole but no O–H hydrogen, so it experiences only weaker dipole–dipole attraction. Butane is essentially non-polar and held together only by weak van der Waals forces, giving it the lowest boiling point of the three despite comparable molar mass.
(i) Starting from phenol, write the reagents and equation for the Reimer–Tiemann reaction and name the product. (ii) Explain, with resonance structures in words, why the –OH group of phenol is ortho/para-directing. (iii) Explain why anisole (methoxybenzene) undergoes electrophilic substitution faster than benzene.
Answer
(i) Phenol is treated with chloroform and aqueous NaOH at about 340 K. NaOH first deprotonates chloroform-derived species to generate dichlorocarbene, :CCl₂, which attacks the electron-rich ring ortho to the –OH group. The resulting dichloromethyl intermediate is hydrolysed by aqueous NaOH to an aldehyde group. Overall: C₆H₅OH + CHCl₃ + 3NaOH → (2-hydroxyphenyl)methanal (salicylaldehyde) + 3NaCl + 2H₂O. (ii) The oxygen lone pair on phenol's –OH can delocalise into the ring: one resonance structure places extra electron density (and formal negative charge) at the ortho carbon, another at the para carbon, and a third at the other ortho carbon, while the ipso and meta carbons never receive this extra density. An incoming electrophile therefore finds the ortho and para positions specifically electron-rich, and the carbocation intermediate formed by attack there is further resonance-stabilised by the oxygen lone pair, lowering the activation energy for substitution at those positions only. (iii) In anisole the methoxy oxygen similarly donates a lone pair into the ring by resonance, activating it towards electrophiles exactly as in phenol, though slightly less strongly since the methyl group is a poorer electron donor by induction than a free O–H can be balanced by hydrogen bonding effects. This electron donation outweighs the inductive electron withdrawal by the electronegative oxygen, so the net effect is still a ring more electron-rich than benzene, and anisole undergoes bromination and nitration considerably faster than benzene itself, substituting mainly at the ortho and para positions.
(i) An organic compound A (C₄H₁₀O) does not react with sodium metal. On treatment with excess hot concentrated HI it gives one equivalent of CH₃I and one equivalent of a compound B. Identify A and B and explain the mechanism of ether cleavage that gives this product distribution. (ii) Distinguish phenol from ethanol using two simple chemical tests.
Answer
(i) A does not react with sodium, so it has no free –OH; it is an ether of formula C₄H₁₀O, i.e. either diethyl ether or methyl propyl ether/methyl isopropyl ether. Producing exactly one equivalent of CH₃I on cleavage identifies A as methoxypropane variants only if a methyl group is present — matching methyl propyl ether, CH₃–O–CH₂CH₂CH₃; excess HI cleaves it completely. Mechanism: protonation of the ether oxygen gives an oxonium ion; iodide then attacks the less hindered carbon (SN2, backside attack on the methyl carbon, which is far less hindered than the propyl carbon) displacing propan-1-ol as the leaving fragment, giving CH₃I. The liberated propan-1-ol, B, is itself attacked by more HI (also SN2, since it is primary) to give 1-iodopropane as the final organic product, so B taken as the first-formed alcohol is propan-1-ol, CH₃CH₂CH₂OH. (ii) Test 1 — neutral FeCl₃: phenol gives a violet colouration with neutral FeCl₃ solution due to complex formation, while ethanol gives no colour change. Test 2 — bromine water: phenol instantly gives a white precipitate of 2,4,6-tribromophenol at room temperature without any catalyst, while ethanol shows no such precipitate because it has no aromatic ring to brominate.
A chemist has three unlabelled bottles P, Q, R, each containing one of: propan-1-ol, propan-2-ol, phenol. With Lucas reagent, P stays clear even on gentle warming; Q turns cloudy within about 5 minutes; R is insoluble in Lucas reagent but dissolves in NaOH solution. With neutral FeCl₃, R gives a violet colour and P, Q give none. (a) Identify P, Q and R with reasons. (b) Write the equation for R reacting with NaOH. (c) Suggest one chemical test, other than FeCl₃, that would separately confirm R is phenol and not an alcohol. (d) Explain why P reacts so slowly with the Lucas reagent compared with Q.
Answer
(a) R dissolves in NaOH but not in Lucas reagent and gives a violet FeCl₃ colour, so R is phenol (mildly acidic, forms water-soluble sodium phenoxide, and gives the characteristic iron complex). Q turns cloudy within minutes, matching a secondary alcohol, so Q is propan-2-ol. P stays clear even warm, the slowest Lucas response, so P is propan-1-ol, a primary alcohol. (b) C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O. (c) Treat R with bromine water: phenol gives an immediate white precipitate of 2,4,6-tribromophenol at room temperature, a reaction alcohols do not give since they lack an activated aromatic ring. (d) In the Lucas mechanism the alcohol must first form a carbocation (SN1). A primary carbocation from P is highly unstable and forms only slowly, so the reaction needs heating and still proceeds sluggishly, whereas the secondary carbocation from Q is considerably more stable and forms fast enough to give visible cloudiness within minutes at room temperature.
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