CBSE Class 12 Chemistry Alcohols Phenols and Ethers MCQs Set 09

Multiple Choice Questions

Question 1. Out of the following alkene, the one which will produce tertiary butyl alcohol on acid catalysed hydration is
(a) \( \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 \)
(b) \( \text{CH}_3\text{CH}=\text{CH}_2 \)
(c) \( \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \)
(d) \( (\text{CH}_3)_2\text{C}=\text{CH}_2 \)
Answer: (d) \( (\text{CH}_3)_2\text{C}=\text{CH}_2 \)
In simple words: We get tertiary butyl alcohol because water adds to the double bond of 2-methylpropene. This puts the alcohol group on the carbon that has three other carbons attached to it.

Exam Tip: Remember that acid-catalyzed hydration of branched alkenes follows Markovnikov's rule and forms highly stable tertiary carbocations.

Question 2. What would be the reactant and reagent used to obtain 2, 4-dimethyl pentan-3-ol?
(a) Propanal and propyl magnesium bromide
(b) 3-methylbutanal and 2-methyl magnesium iodide
(c) 2-dimethylpropanone and methyl magnesium iodide
(d) 2-methylpropanal and iso-propyl magnesium iodide
Answer: (d) 2-methylpropanal and iso-propyl magnesium iodide
In simple words: Mixing 2-methylpropanal with isopropyl magnesium iodide adds the carbon chains together. This creates the compound 2,4-dimethylpentan-3-ol after adding water.

Exam Tip: Grignard reactions involving branched aldehydes and branched Grignard reagents are prone to steric hindrance, so identify the starting fragments carefully by cleavage at the C-OH carbon.

Question 3. A tertiary alcohol is obtained by the reaction of Grignard reagent with
(a) butanone
(b) propanone
(c) acetone
(d) All of the options
Answer: (d) All of the options
In simple words: Any ketone reacts with a Grignard reagent to make a tertiary alcohol. Since all the options listed are ketones, they all do this.

Exam Tip: Keep in mind that formaldehyde yields primary alcohols, other aldehydes yield secondary alcohols, and ketones consistently yield tertiary alcohols upon reaction with Grignard reagents.

Question 4. Nucleophilic addition of Grignard reagent to ketones followed by hydrolysis with dilute acids forms
(a) alkene
(b) primary alcohol
(c) tertiary alcohol
(d) secondary alcohol
Answer: (c) tertiary alcohol
In simple words: When you mix a Grignard reagent with a ketone and then add water, you always get a tertiary alcohol.

Exam Tip: Clearly show the two-step mechanism in descriptive answers: first, nucleophilic addition to form an alkoxide intermediate, followed by acid hydrolysis.

Question 5. Select the incorrect statement about hydroboration oxidation reaction.
(a) Addition is against Markovnikov’s rule.
(b) Intermediate is a carbocation.
(c) It does not involve rearrangement.
(d) Alcohol is obtained in excellent yield.
Answer: (b) Intermediate is a carbocation.
In simple words: Hydroboration-oxidation does not form a carbocation intermediate because everything happens in a single, concerted step.

Exam Tip: This reaction proceeds through a concerted four-centered cyclic transition state, which explains why no skeletal rearrangements occur.

Question 6. For conversion of propene into 1-propanol, which of the following reagents and conditions should be used?
(a) Conc. \( \text{H}_2\text{SO}_4 \); \( \text{H}_2\text{O} \) and heat
(b) \( \text{B}_2\text{H}_6 \); \( \text{H}_2\text{O}_2 / \text{OH}^- \)
(c) Dil. \( \text{H}_2\text{SO}_4 \)
(d) \( \text{H}_2\text{O}/\text{H}^+ \)
Answer: (b) \( \text{B}_2\text{H}_6 \); \( \text{H}_2\text{O}_2 / \text{OH}^- \)
In simple words: Using diborane followed by hydrogen peroxide adds water in an anti-Markovnikov way. This turns propene into 1-propanol instead of 2-propanol.

Exam Tip: For preparing primary alcohols from terminal alkenes, always choose hydroboration-oxidation to achieve anti-Markovnikov regioselectivity.

Question 7. Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields
(a) o-cresol
(b) m-cresol
(c) p-cresol
(d) benzyl alcohol
Answer: (d) benzyl alcohol
In simple words: Sunlight helps chlorine attach to the side chain of toluene to make benzyl chloride. Then, NaOH replaces the chlorine with an OH group to make benzyl alcohol.

Exam Tip: Sunlight triggers free-radical substitution at the side-chain methyl group rather than electrophilic substitution on the aromatic ring.

Question 8. Arnav when reacted aniline with sodium nitrite and hydrochloric acid at certain temperature, a compound formed. This compound further gives phenol on reaction with
(a) \( \text{C}_2\text{H}_5\text{OH} \)
(b) \( \text{H}_2\text{O} \)
(c) \( \text{KOH} \)
(d) \( \text{Ca(OH)}_2 \)
Answer: (b) \( \text{H}_2\text{O} \)
In simple words: Aniline first turns into a diazonium salt using sodium nitrite and acid. Heating this salt with water replaces the nitrogen group with an OH group to make phenol.

Exam Tip: Diazotization must be performed at low temperatures (\( 0-5^\circ\text{C} \)) to prevent the highly unstable diazonium salt from decomposing prematurely.

Question 9. The boiling points of four compounds, an ether, an aldehyde, an alcohol and a haloalkane of comparable molecular weights, are given (not necessarily in the same order) in the table below.

Identify, which of the following compound is the alcohol?
(a) P
(b) Q
(c) R
(d) S
Answer: (d) S
In simple words: Alcohols can form strong hydrogen bonds between their molecules, which makes them boil at much higher temperatures than ethers, aldehydes, or haloalkanes.

Exam Tip: Alcohols consistently exhibit the highest boiling points among organic families of comparable molar mass due to extensive intermolecular hydrogen bonding.

Question 10. Which is the correct order of acid strength from the following?
(a) \( \text{C}_6\text{H}_5\text{OH} > \text{H}_2\text{O} > \text{ROH} \)
(b) \( \text{C}_6\text{H}_5\text{OH} > \text{ROH} > \text{H}_2\text{O} \)
(c) \( \text{ROH} > \text{C}_6\text{H}_5\text{OH} > \text{H}_2\text{O} \)
(d) \( \text{H}_2\text{O} > \text{C}_6\text{H}_5\text{OH} > \text{ROH} \)
Answer: (a) \( \text{C}_6\text{H}_5\text{OH} > \text{H}_2\text{O} > \text{ROH} \)
In simple words: Phenol is the strongest acid because its phenoxide ion is very stable. Water is slightly more acidic than alcohols because the alkyl groups in alcohols push electrons and weaken their acidity.

Exam Tip: Justify this order using the resonance stabilization of the phenoxide ion and the destabilizing +I effect of alkyl groups in alkoxide ions.

Question 11. Which of the following species can act as the strongest base?
(a) \( \text{OH}^- \)
(b) \( \text{C}_6\text{H}_5\text{O}^- \)
(c) \( \text{RO}^- \)
(d) \( \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{O}^- \)
Answer: (c) \( \text{RO}^- \)
In simple words: Since alcohols are the weakest acids, their conjugate bases are the strongest at grabbing protons.

Exam Tip: Base strength is inversely proportional to the acid strength of the corresponding conjugate acid; weaker acids yield stronger conjugate bases.

Question 13. On oxidation, an alcohol gave a product X which reduces Tollen’s reagent. Which of the following could the alcohols be?
I. \( \text{CH}_3-\text{CH}_2-\text{CH}_2\text{OH} \)
II. \( \text{CH}_3-\text{CH}_2-\text{CHOH}-\text{CH}_3 \)
III. \( \text{CH}_3-\text{CH}_2-\text{C(CH}_3)_2-\text{OH} \)
(a) Only I
(b) Only I and II
(c) Only II and III
(d) Any of the options
Answer: (a) Only I
In simple words: Only primary alcohols can be oxidized into aldehydes, which are the only ones that react with Tollen's reagent.

Exam Tip: Remember that only aldehydes reduce Tollen's reagent; therefore, the precursor must be a primary alcohol, as secondary alcohols oxidize to ketones.

Question 14. A compound (X) with the molecular formula \( \text{C}_3\text{H}_8\text{O} \) can be oxidised to another compound (Y) whose molecular formula is \( \text{C}_3\text{H}_6\text{O}_2 \). The compound (X) may be
(a) \( \text{CH}_3\text{CH}_2-\text{O}-\text{CH}_3 \)
(b) \( \text{CH}_3-\text{CH(OH)}-\text{CH}_3 \)
(c) \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{OH} \)
(d) \( \text{CH}_3-\text{CH}_2-\text{CHO} \)
Answer: (c) \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{OH} \)
In simple words: Propan-1-ol has three carbons and an OH group at the end. When oxidized, it easily turns into propanoic acid.

Exam Tip: The addition of one oxygen atom and loss of two hydrogen atoms (\( \text{C}_3\text{H}_8\text{O} \rightarrow \text{C}_3\text{H}_6\text{O}_2 \)) indicates oxidation of a primary alcohol to a carboxylic acid.

Question 15. Which of the following alcohols will not undergo oxidation?
(a) Butanol
(b) Butan-2-ol
(c) 2-methylbutan-2-ol
(d) 3-methylbutan-2-ol
Answer: (c) 2-methylbutan-2-ol
In simple words: Tertiary alcohols cannot be oxidized because the carbon holding the OH group does not have any hydrogen atoms to lose.

Exam Tip: Tertiary alcohols resist normal oxidation because they lack a hydrogen atom on the carbon bearing the hydroxyl group.

Question 16. Match the Column I with Column II and choose the correct option.

Codes:
(a) A-(ii), B-(iii), C-(i)
(b) A-(i), B-(ii), C-(iii)
(c) A-(iii), B-(ii), C-(i)
(d) A-(i), B-(iii), C-(ii)
Answer: (a) A-(ii), B-(iii), C-(i)
In simple words: This question matches reactants with their oxidized or dehydrated products under different conditions.

Exam Tip: Be mindful of reaction conditions: ethanol gives ethene at 443 K with \( \text{H}_2\text{SO}_4 \) but ethoxyethane at 413 K.

Question 17. Phenol does not undergo nucleophilic substitution reaction easily due to
(a) acidic nature of phenol.
(b) partial double bond character of C-OH bond.
(c) partial double bond character of C-C bond.
(d) instability of phenoxide ion.
Answer: (b) partial double bond character of C-OH bond.
In simple words: The bond between carbon and oxygen in phenol behaves like a double bond due to resonance, making it very strong and difficult to replace.

Exam Tip: Explain this by showing how resonance structures of phenol create positive charge on oxygen and partial double bond character in the carbon-oxygen bond.

Question 19. Phenol on being heated with concentrated \( \text{H}_2\text{SO}_4 \) and then with concentrated \( \text{HNO}_3 \) gives
(a) o-nitrophenol
(b) 2, 4, 6-trinitrophenol
(c) p-nitrophenol
(d) m-nitrophenol
Answer: (b) 2, 4, 6-trinitrophenol
In simple words: Mixing phenol with concentrated sulfuric and nitric acids adds three nitro groups to the ring, making picric acid.

Exam Tip: This sulfonation-nitration route minimizes oxidation of phenol, resulting in high yields of 2,4,6-trinitrophenol (picric acid).

Question 20. Consider the table given below.

Identify 'X', 'Y' and 'Z' in the above table.
(a) X = \( \text{C}_6\text{H}_5\text{O(CH}_2)_5-\text{CH}_3 \), Y = Phenyl iso-pentylether, Z = Methoxybenzene
(b) X = \( \text{C}_6\text{H}_5\text{O(CH}_2)_6-\text{CH}_3 \), Y = Methyl iso-pentylether, Z = Ethoxybenzene
(c) X = \( \text{C}_6\text{H}_5\text{O(CH}_2)_6-\text{CH}_3 \), Y = Phenyl iso-pentyl ether, Z = Methoxybenzene
(d) X = \( \text{C}_6\text{H}_5\text{O(CH}_2)_5-\text{CH}_3 \), Y = Phenyl iso-pentyl ether, Z = Ethoxybenzene
Answer: (c) X = \( \text{C}_6\text{H}_5\text{O(CH}_2)_6-\text{CH}_3 \), Y = Phenyl iso-pentyl ether, Z = Methoxybenzene
In simple words: This matches chemical formulas of ethers with their correct IUPAC and common names.

Exam Tip: Practice IUPAC nomenclature rules for mixed ethers, designating the smaller alkyl chain as the 'alkoxy' group and the larger one as the parent alkane.

Question 21. The major product formed when ethanol is dehydrated with Conc. \( \text{H}_2\text{SO}_4 \) at 413 K is
(a) ethoxyethane
(b) methoxyethane
(c) ethene
(d) methoxymethane
Answer: (a) ethoxyethane
In simple words: Heating ethanol at a lower temperature of 413 K with acid joins two ethanol molecules together to make ethoxyethane.

Exam Tip: Intermolecular dehydration of alcohols to form ethers takes place at lower temperatures, whereas intramolecular dehydration to alkenes occurs at higher temperatures.

Question 22. Williamson’s synthesis of preparing dimethyl ether is a/an
(a) \( \text{S}_{\text{N}}1 \) reaction
(b) elimination reaction
(c) \( \text{S}_{\text{N}}2 \) reaction
(d) nucleophilic addition reaction
Answer: (c) \( \text{S}_{\text{N}}2 \) reaction
In simple words: This reaction is an SN2 reaction because the alkoxide ion attacks the alkyl halide from the back in a single step to form an ether.

Exam Tip: Remember that Williamson's synthesis involves a bimolecular nucleophilic substitution mechanism, which requires primary alkyl halides to minimize steric hindrance.

Question 23. The reaction of \( \text{CH}_3\text{OC}_2\text{H}_5 \) with HI gives
(a) Only \( \text{CH}_3\text{I} \)
(b) Only \( \text{C}_2\text{H}_5\text{OH} \)
(c) \( \text{CH}_3\text{I} + \text{C}_2\text{H}_5\text{OH} \)
(d) \( \text{C}_2\text{H5}\text{I} + \text{CH}_3\text{OH} \)
Answer: (c) \( \text{CH}_3\text{I} + \text{C}_2\text{H}_5\text{OH} \)
In simple words: When reacting with HI, the smaller methyl group joins with iodine to make methyl iodide, leaving the rest as ethanol.

Exam Tip: For unsymmetrical ethers, the nucleophilic halide ion attacks the smaller alkyl group via an \( \text{S}_{\text{N}}2 \) pathway under standard conditions.

Question 24. When diethyl ether is heated with excess of HI, it produces
(a) ethanol
(b) iodoform
(c) methyl iodide
(d) ethyl iodide
Answer: (d) ethyl iodide
In simple words: Using a lot of HI splits the ether completely and turns both carbon chains into ethyl iodide.

Exam Tip: When excess HI is specified, any intermediate alcohols formed undergo further reaction to produce corresponding alkyl iodides.

Question 25. Which of the following will undergo electrophilic substitution reaction?
(a) \( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 \)
(b) \( \text{CH}_3\text{OC}_2\text{H}_5 \)
(c) \( \text{CH}_3\text{OCH}_3 \)
(d) \( \text{C}_6\text{H}_5\text{OCH}_3 \)
Answer: (d) \( \text{C}_6\text{H}_5\text{OCH}_3 \)
In simple words: Only anisole has a benzene ring that is highly activated by the oxygen atom, allowing it to undergo electrophilic substitution reactions.

Exam Tip: Ethers must have an aromatic ring, such as in anisole, to undergo electrophilic aromatic substitutions like nitration, halogenation, or Friedel-Crafts reactions.

Assertion-Reason

Directions In the following questions an Assertion (A) is followed by a corresponding Reason (R).
Use the following keys to choose the appropriate answer,
(a) Both (A) and (R) are true and (R) is the correct explanation of (A),
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A),
(c) (A) is true, but (R) is false,
(d) (A) is false, but (R) is true.

Question. Assertion (A) Carbon oxygen bond length of phenol is slightly less than that of methanol.
Reason (R) There exist a partial double bond character and \( sp^2 \)-hybridisation of carbon to which oxygen is attached in phenol.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: The bond in phenol is shorter because it shares electrons with the ring, making it act like a partial double bond.

Exam Tip: Highlight the partial double bond character resulting from resonance stabilization of the lone pairs on oxygen alongside the \( sp^2 \)-hybridized carbon in phenols.

Question. Assertion (A) Addition reaction of water to but-1-ene in acidic medium yields butan-2-ol.
Reason (R) Addition of water in acidic medium proceeds through the formation of primary carbanion.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: The reaction works by forming a stable carbocation intermediate, not a carbanion.

Exam Tip: Acid-catalyzed hydration of alkenes follows Markovnikov's rule and proceeds via a carbocation intermediate, not a carbanion.

Question. Assertion (A) Alcohols and phenols are soluble in water.
Reason (R) There occurs a dipole-dipole interaction of OH group of alcohol and phenol with water molecules which is responsible for their solubility in water.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: Alcohols dissolve in water because they can form strong hydrogen bonds with water molecules, not just simple dipole-dipole attractions.

Exam Tip: Clearly state that intermolecular hydrogen bonding, not simple dipole-dipole interaction, is the driving force behind the water solubility of alcohols and phenols.

Question. Assertion (A) Boiling point of p-nitrophenol is more than o-nitrophenol.
Reason (R) p-nitrophenol is steam volatile, due to intermolecular H-bonding.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: o-nitrophenol can form hydrogen bonds inside its own molecule, making it turn into steam easily, whereas p-nitrophenol links with other molecules and stays liquid longer.

Exam Tip: Understand that intramolecular hydrogen bonding in o-nitrophenol increases volatility, whereas intermolecular hydrogen bonding in p-nitrophenol elevates the boiling point.

Question. Assertion (A) The boiling point of ethanol is higher than that of methoxymethane.
Reason (R) There is intramolecular hydrogen bonding in ethanol.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: Ethanol molecules stick together tightly using intermolecular hydrogen bonds, which requires more heat to break than methoxymethane.

Exam Tip: Be careful not to confuse "intermolecular" (between separate molecules) and "intramolecular" (within the same molecule) hydrogen bonding.

Question. Assertion (A) \( \text{C}_2\text{H}_5\text{OH} \) is a weaker base than phenol but is a stronger nucleophile than phenol.
Reason (R) In phenol, the lone pair of electrons on oxygen is withdrawn towards the ring due to resonance.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: Phenol is a weaker base because its oxygen electrons are busy sharing with the benzene ring, unlike the free electrons in ethanol.

Exam Tip: Because of resonance, the oxygen lone pairs in phenol are delocalized into the ring, reducing base strength and nucleophilicity compared to ethanol.

Question. Assertion (A) p-methoxyphenol is a stronger acid than m-methoxyphenol.
Reason (R) Methoxy group exerts + R effect at both ortho and para position.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: The methoxy group at the para position pushes electrons toward the oxygen, making it harder to release the hydrogen proton.

Exam Tip: Resonance effects (+R) operate only at ortho and para positions, which decreases acidity at those positions compared to meta-substituted compounds.

Question. Assertion (A) Bromination of phenol can be carried out even in the absence of Lewis acid.
Reason (R) —OH group of phenol has the high activation effect.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Phenol is so active that it can polarize and react with bromine on its own without needing any extra catalyst.

Exam Tip: State that the strong resonance activation (+R effect) of the hydroxyl group in phenol increases electron density on the ring, enabling spontaneous halogenation.

Question. Assertion (A) Like bromination of benzene, bromination of phenol is also carried out in the presence of Lewis acid.
Reason (R) Lewis acid polarises the bromine molecule.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: Unlike benzene, phenol is extremely active and does not need a Lewis acid catalyst to react with bromine.

Exam Tip: Keep in mind that phenol reacts rapidly with bromine water without a Lewis acid catalyst to form a white precipitate of 2,4,6-tribromophenol.

Question. Assertion (A) IUPAC name of the compound \( \text{CH}_3-\text{CH(CH}_3)-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \) is 2-ethoxy-2-methylethane.
Reason (R) In IUPAC nomenclature, ether is regarded as hydrocarbon derivative in which a hydrogen atom is replaced by —OR or —OAr group [where, R = alkyl group and Ar = aryl group].
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: The correct name for the compound is 2-propoxypropane, which means the first statement is incorrect but the naming rule is correct.

Exam Tip: Ethers are systematically named as alkoxyalkanes by selecting the longest carbon chain as the parent alkane.

Question. Assertion (A) Bond angle in ethers is slightly less than the tetrahedral angle.
Reason (R) There is a repulsion between the two bulky (— R) groups.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: Ethers have a bond angle that is slightly wider than normal because the two bulky carbon chains push each other away.

Exam Tip: Due to steric hindrance between the alkyl groups, the C-O-C bond angle in ethers is slightly larger (about \( 111.7^\circ \)) than the tetrahedral angle of \( 109.5^\circ \).

Case-Study 37

Read the case given below and answer the questions that follow:
Alcohols are soluble in water due to their ability to form hydrogen bonds. The boiling point of alcohol generally increases with increase in molar mass and decreases with increased branching in the carbon chain. Alcohols on dehydration gives alkenes at 443 K. This reaction is commonly observed in compounds such as anisole and chlorobenzene. Phenol, on the other hand is used to prepare salicylaldehyde, salicylic acid, aspirin, methyl salicylate and p-benzoquinone. In addition, phenol undergoes electrophilic substitution reaction at o and p-position.
Observe the following table showing boiling points of alcohol, their respective molar mass. Study the table and answer the questions based on table and related studied concept.

Question (i) Why does tertiary butyl alcohol have lower boiling point than n-butyl alcohol?
Answer: Tertiary butyl alcohol exhibits a lower boiling point than n-butyl alcohol because its highly branched structure reduces its surface area, which decreases the strength of the intermolecular van der Waals forces.

Exam Tip: Increased branching makes the molecule more spherical, decreasing the contact area and reducing the strength of intermolecular dispersion forces.

Question (ii) Considering intermolecular forces, which alcohol would be expected to exhibit the strongest hydrogen bonding interactions?
(a) iso-propyl alcohol
(b) Methanol
(c) Ethanol
(d) n-butanol
Answer: (d) n-butanol
In simple words: The strength of hydrogen bonding increases with longer, more extended alkyl chains that support these interactions. Among the choices, n-butanol has the longest chain and exhibits the strongest hydrogen bonding.

Exam Tip: Longer, unbranched carbon chains provide greater surface area for cohesive intermolecular dispersion interactions alongside hydrogen bonding.

Question Or (ii) Which of the following alcohols would have the highest vapour pressure at room temperature?
(a) n-butanol
(b) n-propanol
(c) Ethanol
(d) Methanol
Answer: (d) Methanol
In simple words: Vapor pressure is inversely related to boiling temperature. Consequently, methanol, which has the lowest boiling point among the options, will possess the highest vapor pressure at room temperature.

Exam Tip: Compounds with weaker intermolecular forces and lower boiling points possess higher vapor pressures at any given temperature.

Question (iii) If Rahul have to mix an alcohol with water, which structural feature of the alcohol would likely make it less soluble?
(a) Low molecular weight with no branching.
(b) Presence of a hydroxyl group.
(c) Low molecular weight with branching.
(d) High molecular weight with no branching.
Answer: (d) High molecular weight with no branching.
In simple words: Alcohols with higher molecular weights and no branching are less soluble in water because their larger, unbranched hydrocarbon chains increase hydrophobicity, reducing their capacity to mix with water.

Exam Tip: Solubility of alcohols in water decreases as the size of the non-polar, hydrophobic alkyl part increases.

Case-Study 38

Read the case given below and answer the questions that follow:
Phenols undergo electrophilic substitution reactions readily due to the to strong activating effect of OH group attached the benzene ring. Since, the OH group increases the electron density more to o- and p- positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols.

Question (ii) Why phenol does not undergo protonation readily?
Answer: The lone pair of electrons on the oxygen of the hydroxyl group in phenol participates in resonance with the aromatic ring. Consequently, these electrons are less available to accept protons, preventing phenol from undergoing protonation easily.

Exam Tip: Show the resonance structures of phenol to illustrate that the positive charge acquired by oxygen makes it highly non-nucleophilic towards protonation.

Question (iii) Which is a stronger acid phenol or cresol? Give reason.
Answer: Phenol acts as a stronger acid than cresol because the methyl group in cresol exerts an electron-donating (+I) effect, which increases the electron density on the oxygen. This lowers the polarity of the O-H bond, making cresol less acidic.

Exam Tip: Electron-donating alkyl groups destabilize the phenoxide conjugate base, reducing its formation rate and overall acidity.