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NEET Chemistry: Alcohols, Phenols and Ethers questions with solutions

33 questions with worked solutions.

Questions

Q1. Common name for methanol is:

  1. volemitol
  2. rubbing alcohol
  3. wood spirit
  4. xylitol

Answer: wood spirit

Methanol is commonly called wood spirit because it was originally produced from wood distillation. The other options refer to different compounds: rubbing alcohol is usually isopropyl alcohol, and xylitol/volemitol are sugar alcohols.

Q2. When glycerol is treated with excess of HI, it produces:

  1. glycerol triiodide
  2. 2–iodopropane
  3. allyl iodide
  4. propene

Answer: 2–iodopropane

When glycerol reacts with excess HI, it undergoes a series of substitution reactions where hydroxyl groups are replaced by iodine, ultimately forming 2-iodopropane as the major product.

Q3. The rates of reaction of different alcohols with Lucas reagent follows the order: 3° alcohol > 2° alcohol > 1° alcohol.

  1. Since carbocations are formed as intermediate, more stable the carbocation, higher will be the reactivity of the parent compound (alcohol).
  2. 2-methylpropan-2-ol generates a 3° carbocation, so it will react fastest; other three generate either 1° or 2° carbocations.
  3. CH3CH2OH reacts slower than CH3CH(OH)CH3.
  4. CH3CH2OH reacts faster than CH3CH(OH)CH3.

Answer: Since carbocations are formed as intermediate, more stable the carbocation, higher will be the reactivity of the parent compound (alcohol).

The reactivity order with Lucas reagent is determined by the stability of the carbocation intermediate. Tertiary carbocations are the most stable, followed by secondary, and then primary, which explains the given order.

Q4. Lucas reagent is conc. HCl + anhyd. ZnCl2.

  1. CH3CH2OH reacts with Lucas reagent to form CH3CH2Cl.
  2. CH3CH2OH reacts faster than CH3CH(OH)CH3.
  3. CH3CH2OH reacts slower than CH3CH(OH)CH3.
  4. CH3CH2OH reacts with Lucas reagent to form CH3CH2Br.

Answer: CH3CH2OH reacts slower than CH3CH(OH)CH3.

Lucas reagent distinguishes alcohols based on their reactivity. Secondary alcohols like CH3CH(OH)CH3 react faster than primary alcohols like CH3CH2OH due to the stability of the carbocation intermediate.

Q5. Phenols are more acidic than alcohols as they are resonance stabilized whereas alcohols are not.

  1. Phenols are less acidic than alcohols.
  2. Alcohols are more acidic than phenols.
  3. Phenols are more acidic than alcohols as they are resonance stabilized whereas alcohols are not.
  4. Alcohols and phenols have the same acidity.

Answer: Phenols are more acidic than alcohols as they are resonance stabilized whereas alcohols are not.

Phenols are more acidic than alcohols because the phenoxide ion formed after losing a proton is resonance stabilized, whereas the alkoxide ion from alcohols lacks such stabilization.

Q6. Given are cyclohexanol (I), acetic acid (II), 2,4,6-trinitrophenol (III), and phenol (IV). In these, the order of decreasing acidic character will be:

  1. (a) III > II > IV > I
  2. (b) II > III > IV > I
  3. (c) II > III > I > IV
  4. (d) III > IV > I > II

Answer: (a) III > II > IV > I

The acidic strength depends on the stability of the conjugate base formed after losing a proton. 2,4,6-Trinitrophenol (III) is the most acidic due to strong electron-withdrawing nitro groups. Acetic acid (II) follows due to resonance stabilization of its carboxylate ion. Phenol (IV) is less acidic than acetic acid but more than cyclohexanol (I), which lacks resonance stabilization.

Q7. When 3,3-dimethyl-2-butanol is heated with H2SO4, the major product obtained is:

  1. 2,3-dimethyl-2-butene
  2. 3,3-dimethyl-1-butene
  3. 2,3-dimethyl-1-butene
  4. cis & trans isomers of 2,3-dimethyl-2-butene

Answer: 2,3-dimethyl-2-butene

When 3,3-dimethyl-2-butanol is heated with H2SO4, dehydration occurs, leading to the formation of the most stable alkene. The major product is 2,3-dimethyl-2-butene due to its higher stability from hyperconjugation and alkyl substitution.

Q8. Number of isomeric alcohols of molecular formula C6H14O which give positive iodoform test is

  1. two
  2. three
  3. four
  4. five

Answer: three

The iodoform test is given by alcohols with the structure R-CH(OH)-CH3. For C6H14O, three such isomeric alcohols can be formed: 2-pentanol, 3-methyl-2-butanol, and 2-methyl-2-pentanol.

Q9. When vapours of a secondary alcohol is passed over heated copper at 573 K, the product formed is

  1. an alkene
  2. a carboxylic acid
  3. an aldehyde
  4. a ketone

Answer: a ketone

When a secondary alcohol is passed over heated copper at 573 K, it undergoes dehydrogenation to form a ketone. This is a characteristic reaction of secondary alcohols.

Q10. n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?

  1. PCl5
  2. Reduction
  3. Oxidation with potassium dichromate
  4. Ozonolysis

Answer: Oxidation with potassium dichromate

n-Propyl alcohol and isopropyl alcohol can be distinguished by oxidation with potassium dichromate. n-Propyl alcohol (a primary alcohol) oxidizes to an aldehyde, while isopropyl alcohol (a secondary alcohol) oxidizes to a ketone.

Q11. Propan-1-ol may be prepared by the reaction of propene with

  1. H3BO3
  2. H2SO4/H2O
  3. B2H6/NaOH/H2O2
  4. CH3C-O-H

Answer: B2H6/NaOH/H2O2

Propan-1-ol is prepared by the hydroboration-oxidation of propene using B2H6 followed by treatment with NaOH and H2O2, which adds -OH to the less substituted carbon (anti-Markovnikov addition).

Q12. Which one of the following on oxidation gives a ketone?

  1. Primary alcohol
  2. Secondary alcohol
  3. Tertiary alcohol
  4. All of these

Answer: Secondary alcohol

Secondary alcohols oxidize to form ketones, while primary alcohols form aldehydes and tertiary alcohols resist oxidation under normal conditions.

Q13. HBr reacts fastest with

  1. 2-Methylpropan-1-ol
  2. 2-Methylpropan-2-ol
  3. propan-2-ol
  4. propan-1-ol

Answer: 2-Methylpropan-2-ol

HBr reacts fastest with 2-Methylpropan-2-ol because it forms a tertiary carbocation during the reaction, which is the most stable and facilitates the reaction rate.

Q14. The compound which reacts fastest with Lucas reagent at room temperature is

  1. Butan-1-ol
  2. Butan-2-ol
  3. 2-Methylpropan-1-ol
  4. 2-Methylpropan-2-ol

Answer: 2-Methylpropan-2-ol

Lucas reagent reacts fastest with tertiary alcohols due to the formation of a stable tertiary carbocation. Among the options, 2-Methylpropan-2-ol is a tertiary alcohol and reacts the fastest.

Q15. What is formed when a primary alcohol undergoes catalytic dehydrogenation?

  1. Aldehyde
  2. Ketone
  3. Alkene
  4. Acid

Answer: Aldehyde

When a primary alcohol undergoes catalytic dehydrogenation, it loses hydrogen atoms to form an aldehyde. This is a common reaction in organic chemistry.

Q16. Lucas reagent is

  1. Conc. HCl and anhydrous ZnCl2
  2. Conc. HNO3 and hydrous ZnCl2
  3. Conc. HCl and hydrous ZnCl2
  4. Conc. HNO3 and anhydrous ZnCl2

Answer: Conc. HCl and anhydrous ZnCl2

Lucas reagent is a mixture of concentrated HCl and anhydrous ZnCl2, used to distinguish between primary, secondary, and tertiary alcohols based on their reactivity.

Q17. Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional group?

  1. -CH2Cl
  2. -COOH
  3. -CHCl2
  4. -CHO

Answer: -CHO

Phenol reacts with chloroform in the presence of dilute sodium hydroxide to form salicylaldehyde, introducing the -CHO (aldehyde) group. This is known as the Reimer-Tiemann reaction.

Q18. Among the following four compounds: (i) phenol (ii) methylphenol (iii) meta-nitrophenol (iv) para-nitrophenol The acidity order is:

  1. ii > i > iii > iv
  2. iv > iii > i > ii
  3. iii > iv > i > ii
  4. iv > i > iii > ii

Answer: iv > i > iii > ii

The acidity of phenols is influenced by the substituents on the benzene ring. Electron-withdrawing groups like nitro (-NO2) increase acidity, and their position matters. Para-nitrophenol (iv) is more acidic than meta-nitrophenol (iii) due to better resonance stabilization. Phenol (i) is less acidic than nitrophenols, and methylphenol (ii) is the least acidic due to the electron-donating methyl group.

Q19. Which one of the following compounds is most acidic?

  1. CH2=CH-CH2-OH
  2. OH
  3. NO2
  4. CH3

Answer: OH

The hydroxyl group (OH) is inherently acidic due to its ability to donate a proton (H+), making it the most acidic among the given options.

Q20. When phenol is treated with CHCl3 and NaOH, the product formed is:

  1. benzaldehyde
  2. salicylaldehyde
  3. salicylic acid
  4. benzoic acid

Answer: salicylaldehyde

When phenol reacts with CHCl3 and NaOH, it undergoes the Reimer-Tiemann reaction, forming salicylaldehyde as the product.

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