CBSE Class 12 Chemistry Alcohols Phenols and Ethers Assignment

Question. Acetone on treatment with CH₃ −Mg − I and on further hydrolysis gives:
a. Isopropyl alcohol
b. Primary alcohol
c. Acetic acid
d. 2-methyl 2-propanol
Answer : D

Question. Acetyl bromide reacts with excess of CH₃ MgI followed by treatment with a saturated solution of NH₄ Cl gives:
a. 2-methyl-2-propanol
b. Acetamide
c. Acetone
d. Acetyl iodide
Answer : A

Question. Methyl alcohol can be distinguished from ethyl alcohol using:
a. Fehling solution
b. Schiff's reagent
c. Sodium hydroxide and iodine
d. Phthalein fusion test
Answer : C

Question. Which gas is eliminated in fermentation?
a. O₂
b. CO₂
c. N₂
d. H₂
Answer : B

Question. In the reaction alkali Ar −OH + Rx ^alkali→ A, A is:
a. An aldehyde
b. An aryl chloride
c. An ether
d. A ketone
Answer : C

Question. The compound formed when ethyl bromide is heated with dry silver oxide is:
a. Dimethyl ether
b. Diethyl ether
c. Methyl alcohol
d. Ethyl alcohol
Answer : B

Question. The reagent used for the preparation of higher ether from halogenated ethers is:
a. conc. H₂SO₄
b. Sodium alkoxide
c. Dry silver oxide
d. Grignard reagent
Answer : D

Question. For phenol, which of the following statements is correct?
a. It is insoluble in water?
b. It has lower melting point compared to aromatic hydrocarbons of comparable molecular weight
c. It has higher boiling point than toluene
d. It does not show acidic property
Answer : C

Question. Commercially methanol is prepared by:
a. Reduction of CO in presence of ZnO.Cr₂O₃
b. Methane reacts with water vapours at 900°C in presence of Ni catalyst
c. Reduction of HCHO by LiAlH₄
d. Reduction of HCHO by aqueous NaOH
Answer : A

Question. the above sequence X can be:
a. H₂ / Ni
b. 4 NaBH
c. K₂Cr₂O₇ /H⁺
d. Both a.and b.
Answer : B

Question. Methylphenyl ether can be obtained by reacting:
a. Phenolate ions and methyl iodide
b. Methoxide ions and bromobenzene
c. Methanol and phenol
d. Bromo benzene and methyl bromide
Answer : A

Question. In which case methyl-t-butyl ether is formed?
a. (C₂H₃)₃ CONa + CH₃Cl
b. (CH₃)₃ CONa + CH₃Cl
c. (CHH₃)₃ CONa + C₂H₅Cl
d. (CH₃)₃ CONa + CH₃Cl
Answer : B

Question. Read the following statements carefully:
a. A secondary alcohol on oxidation gives a ketone
b. Ethanol reacts with conc. H₂SO₄ at 180°C to yield ethylene
c. Methanol reacts with iodine and sodium hydroxide to give a yellow precipitate of iodoform
d. Hydrogen gas is liberated when sodium is added to alcohol. Select the correct statements from the above set:
a. A, B
b. C, D
c. A, B, D
d. A, C, D
Answer : C

Question. Final product formed on reduction of glycerol by hydroiodic acid is:
a. Propane
b. Propanoic acid
c. Propene
d. Propyne
Answer : C

Question. Which of the following explains the viscous nature of glycerol?
a. Covalent bonds
b. Hydrogen bonds
c. Vander Wall's forces
d. Ionic forces
Answer : B

Question. From Williamson’s synthesis preparation of which of following is possible?
a. Only symmetrical ethers
b. Only symmetrical ethers
c. Both types
d. None of these
Answer : C

Question. What is obtained when chlorine is passed in boiling toluene and product is hydrolysed?
a. o-Cresol
b. p-Cresol
c. 2, 4-Dihydroxytoluene
d. Benzyl alcohol
Answer : D

Question. The reagent used for the dehydration of an alcohol is:
a. Phosphorus pentachloride
b. Calcium chloride
c. Aluminium oxide
d. Sodium chloride
Answer : C

Question. The alcohol that produces turbidity immediately with ZnCl₂ + conc. HCl at room temperature:
a. 1-hydroxybutane
b. 2-hydroxybutane
c. 2-hydroxy-2-methylpropane
d. 1-hydroxy-2-methylpropane
Answer : C

Question. The alcohol which easily reacts with conc. HCl is?
a. CH₃ −CHOH −CH₂ −CH₃
b. (CH₃)₃ − C −OH
c. CH₃ − CH −CH₂ −CH₂ −OH
d. (CH₃)₃ −CH−CH₂OH
Answer : B

Question. A compound that easily undergoes bromination is:
a. Phenol
b. Toluene
c. Benzene
d. Benzoic acid
Answer : A

Question. Which of the following methods is/are not the industrial methods to prepare methanol?
a. catalytic reduction of carbon monoxide in presence of ZnO − Cr₂O₃
b. reacting methane with steam at 900°C with a Ni catalyst
c. reducing formaldehyde with lithium aluminium hydride
d. reacting formaldehyde with aqueous sodium hydroxide solution
Answer : B, C, D

Question. Which of the following reagent (s) or test (s), can be used to distinguish 2 methyl propanol and 2-methyl propanol – 2?
a. I₂ / NaOH
b. HCl/anhyd. ZnCI₂
c. Victor-Meyer test
d. oxidation with Cu at 573 K
Answer : B, C, D

Question. When rectified spirit and benzene are distilled together, the first fraction obtained is?
a. A ternary azeotrope
b. Absolute alcohol
c. A binary azeotrope
d. Denatured spirit
Answer : A

Question. Which of the following is most acidic?
a. Phenol
b. Benzyl alcohol
c. m-chlorophenol
d. Cyclohexanol
Answer : C

Question. Which is not correct?
a. Phenol is more acidic than acetic acid
b. Ethanol is less acidic than phenol
c. Ethanol has lower boiling point than ethane
d. Ethyne is a non-linear molecule
Answer : A

Question. Glycerol is used in the manufacture of:
a. Dynamite
b. Varnish
c. Paints
d. Soft drinks
Answer : A

Question. Alcoholic fermentation is brought about by the action of:
a. CO₂
b. O₂
c. Invertase
d. Yeast
Answer : D

Question. In order to make alcohol undrinkable pyridine and methanol are added to it. The resulting alcohol is called:
a. Power alcohol
b. Proof spirit
c. Denatured spirit
d. Poison alcohol
Answer : C

Question. Main constituent of dynamite is:
a. Nitrobenzene
b. Nitroglycerine
c. Picric acid
d. TNT
Answer : B

Question. The Bouveault-Blanc reduction involves:
a. C₂H₅OH/ Na
b. LiAlH₄
c. C₂H₅MgX⁻
d. Zn / HCl
Answer : A

Question. Which is used as antifreeze?
a. Glycol
b. Ethyl alcohol
c. Water
d. Methanol
Answer : A

Question. Alkenes convert into alcohols by:
a. Hydrolysis by dil. H₂SO₄
b. Hydration of alkene by alkaline KMnO₄
c. Hydrolysis by water vapours and conc. H₂SO₄
d. Hydration of alkene by aqueous KOH
Answer : B, C

Question. The following ether, when treated with HI produces:

Answer : A, D

Question. Which of the following reactions can be used to prepare methyl phenyl ether?
a. reacting sodium phenoxide with methyl chloride
b. reacting phenol with diazomethane (CH₂ N₂ )
c. reacting sodium methoxide with chlorobenzene
d. reacting C₆H₅ OMgCl with chloromethane
Answer : A, B

Question. Carbinol is:
a. C₂H₅OH
b. CH₃OH
c. (CH₃)₂ CHOH
d. CH₃CH₂CH(OH)CH₃
Answer : B

Question. Structure of compound X is:

Than which of the following statements are correct?
a. X have two stereoisomers
b. All stereomers of X have the same melting point
c. X reaction with SOCl₂ followed by alcoholic KOH product E which containing 8π bonds
d. None of these
Answer : A, B, C

Question. An organic compound X of molecular formula C₄H₁₀O undergoes oxidation to give a compound Y of molecular formula C₄H₈O₂ .X could be:
a. CH₃CH₂CH₂CHH₂OH
b. CH₃CH₂CH(OH)CH₃
c. (CH₃)₂ CHCH₂OH
d. (CH₃)₃ COH
Answer : A, C

Question. Phenol has higher pka value than:
a. acetic acid
b. p-methoxy phenol
c. p-nitrophenol
d. ethanol
Answer : A, C

Question. Wood spirit is known as:
a. Methanol
b. Ethanol
c. Acetone
d. Benzene
Answer : A

1. Haloalkanes undergo nucleophilic substitution reaction readily than haloarenes.

Answer : (aLlow bond dissociation enthalpy of R-X bond due to low s- character of C SP³

(b) Stability of carbocation by hyper conjugation of neighboring alkyl groups.

 

2. The b.pt of isomeric haloalkanes decreases increasing in branching of carbon chains.

Answer : As branching increases molecules becomes more compact and acquires a low surface area . Since strength of Vander waals forces are proportional to area of contact

between molecules,attractive forces decreases with branching.

 

3. Alcohols have higher boiling boint compared to hydrocarbons, ethers and halo compounds of comparable mass.

Answer : Due to presence of inter molecular H-bonding,which is either absent or very weak in other molecules.

 

4. Phenol is more acidic than alcohol

 

 

The phenoxide ion is resonance stabilized due to (-) R effect of benzene ring. The alkoxide ion is less stable due to (+) I effect of –R group.

 

5. Aldehydes are more than reactives than ketones towards nucleophilic addition reactions.

Answer : (a) inductive effect of R groups reduces the electrophilicity of the C atom of carbonyl group in ketones

(b) steric hindrance: the approaching nucleophile suffers greater repulsion due to presence of two bulky-R groups in ketones.

 

6. Carboxylic acids have higher boiling points than aldehydes, ketones and alcohols of comparable molecular masses.

Answer : The C.A molecules form dimers which do not separate into monomers even in vapour phase.

 

7. Gabriel Phthalimide synthesis is not suitable for preparation of secondary and tertiary amines.

Answer : Since the reaction involves Sn2 mechanism, only a primary alkyl group can approach the sterically hindered Phthalimide ring .

 

8 pKb of aniline is more than that of methylamine.
Answer : Since the lone pair on the N-atom remains in conjugation with the benzene ring and is unavailable for donation. The basic character of methylamine is enhanced due to +I effect of –CH3 group.

9 Aniline does not undergo Friedel Crafts reaction.
Answer : Aniline is basic and reacts with anhyd. AlCl3 to form a salt which deactivates 6the benzene ring towards electrophilic substitution.

10 During esterification, water must be removed as soon as it is formed. H+
Answer : RCOOH + R’OH RCOOR’ + H2O If the by-product is not removed, it will cause backward reaction. The ester formed will breakdown giving poor yield.

LEVEL-2 QUESTIONS

Q1) Although phenoxide ion hasmore no. of resonating structures than carboxylate ion , even though carboxylic acid is a stronger acid why ?
Answer : The phenoxide ion has non equivalent resonance structures in which–vecharge is at less electronegativeC atom and +ve charge as at more electronegative O-atom.
Incarboxylate ion –ve charge is delocalized on two electronegative O-atoms hence resonance is more effective and a stronger acid.

 

LEVEL-1 QUESTIONS

Question 1. Haloalkanes undergo nucleophilic substitution reaction readily than haloarenes.
Answer: (a) Lower bond dissociation enthalpy of R-X bond due to low s-character of C \(sp^3\) bond.
(b) Stability of carbocation by hyperconjugation of neighboring alkyl groups.
In simple words: Haloalkanes break their carbon-halogen bond more easily than haloarenes, making them much more reactive.

Exam Tip: Remember resonance effect in haloarenes gives partial double bond character to the C-Cl bond, making it harder to cleave.

Question 2. The b.pt of isomeric haloalkanes decreases increasing in branching of carbon chains.
Answer: As branching increases, molecules become more compact and acquire a low surface area. Since the strength of Van der Waals forces is proportional to the area of contact between molecules, attractive forces decrease with branching.
In simple words: Branched molecules are spherical and have less surface area to stick together, lowering their boiling points.

Exam Tip: Surface area directly dictates the magnitude of London dispersion or Van der Waals forces in isomers.

Question 3. Alcohols have higher boiling point compared to hydrocarbons, ethers and halo compounds of comparable mass.
Answer: Due to the presence of intermolecular H-bonding, which is either absent or very weak in other molecules.
In simple words: Hydrogen bonds hold alcohol molecules together strongly, requiring more heat to boil them.

Exam Tip: Mention intermolecular hydrogen bonding explicitly to score full marks for boiling point trends.

Question 4. Phenol is more acidic than alcohol.
Answer: \(C_6H_5-OH \leftrightarrow C_6H_5O^- + H^+\)
\(R-OH \leftrightarrow RO^- + H^+\)
The phenoxide ion is resonance stabilized due to \((-)\) R effect of the benzene ring. The alkoxide ion is less stable due to \((+)\) I effect of the \(-R\) group.
In simple words: The benzene ring in phenol helps spread out the negative charge after losing a proton, making it more stable than in regular alcohols.

Question 5. Aldehydes are more reactive than ketones towards nucleophilic addition reactions.
Answer: (a) Inductive effect of R groups reduces the electrophilicity of the C atom of carbonyl group in ketones.
(b) Steric hindrance: the approaching nucleophile suffers greater repulsion due to the presence of two bulky R groups in ketones.
In simple words: Ketones have two alkyl groups pushing electrons and blocking the way, making them less attractive to attackers than aldehydes.

Exam Tip: Cite both electronic factors (\((+)\) inductive effect) and steric factors for complete reasoning.

Question 6. Carboxylic acids have higher boiling points than aldehydes, ketones and alcohols of comparable molecular masses.
Answer: The carboxylic acid molecules form dimers which do not separate into monomers even in vapour phase.
In simple words: Carboxylic acids pair up via strong double hydrogen bonds, acting as much larger molecules with higher boiling points.

Exam Tip: The key term to include is "dimerization" through extensive hydrogen bonding.

Question 7. Gabriel Phthalimide synthesis is not suitable for preparation of secondary and tertiary amines.
Answer: Since the reaction involves \(S_N2\) mechanism, only a primary alkyl group can approach the sterically hindered phthalimide ring.
In simple words: Bulky secondary and tertiary groups cannot easily attack the phthalimide ion to form the required bond.

Exam Tip: Link the failure to the spatial congestion (\(S_N2\) steric hindrance).

Question 8. \(pK_b\) of aniline is more than that of methylamine.
Answer: Since the lone pair on the N-atom remains in conjugation with the benzene ring and is unavailable for donation. The basic character of methylamine is enhanced due to \((+)\) I effect of \(-CH_3\) group.
In simple words: Aniline's lone pair is busy inside the ring, so it cannot donate electrons easily, making it a weaker base (higher \(pK_b\)).

Exam Tip: Higher \(pK_b\) means weaker base. Lower \(pK_b\) means stronger base.

Question 9. Aniline does not undergo Friedel Crafts reaction.
Answer: Aniline is basic and reacts with anhydrous \(AlCl_3\) to form a salt which deactivates the benzene ring towards electrophilic substitution.
In simple words: Aniline reacts with the catalyst \(AlCl_3\) to form a complex, locking up the nitrogen lone pair and stopping the reaction.

Exam Tip: Mention salt formation with Lewis acid \(AlCl_3\).

Question 10. During esterification, water must be removed as soon as it is formed.
Answer: \(RCOOH + R'OH \overset{H^+}{\rightleftharpoons} RCOOR' + H_2O\)
If the by-product is not removed, it will cause backward reaction. The ester formed will breakdown giving poor yield.
In simple words: Esterification is reversible, so removing water forces the reaction forward to make more ester.

Exam Tip: Apply Le Chatelier's principle to explain the removal of products.

LEVEL-2 QUESTIONS

Question 1. Although phenoxide ion has more no. of resonating structures than carboxylate ion, even though carboxylic acid is a stronger acid why?
Answer: The phenoxide ion has non-equivalent resonance structures in which negative charge is at less electronegative carbon atom and positive charge is at more electronegative oxygen atom.
In carboxylate ion, negative charge is delocalized on two electronegative oxygen atoms, hence resonance is more effective and a stronger acid.
In simple words: Carboxylate spreads its negative charge across two oxygen atoms equally, which is much more stable than spreading it onto carbon atoms like phenoxide does.

Exam Tip: Equivalent resonance structures provide significantly higher stabilization than non-equivalent ones.

Question 2. Why Carboxylic acid have higher boiling point than alcohols as alcohol forms strongest intermolecular hydrogen bonding?
Answer: As carboxylic acid forms a dimer due to which their surface area increases and forms strong intermolecular H-bonding, having more boiling point than alcohols.
In simple words: Carboxylic acids form stronger and more extensive hydrogen-bonded dimers than alcohols.

Exam Tip: Emphasize dimer formation as the primary reason for carboxylic acids' exceptionally high boiling points.

Question 3. There are two \(-NH_2\) group in semicarbazide. However only one is involved in formation of semicarbazones. Why?
Answer: \(NH_2-CO-NH-NH_2\)
Due to resonance, one \(NH_2\) group undergoes or is involved in resonance and hence cannot participate in the formation of semicarbazone.
\(H_2N^+ = C O^- - NH - NH_2\)
Lone pair of \(NH_2\) group is not involved in resonance and is available for nucleophilic attack.
In simple words: One of the amino groups donates its electrons into the carbonyl group via resonance, making it unavailable for the reaction.

Exam Tip: Identify the nitrogen adjacent to the carbonyl group as resonance-stabilized and deactivated.

Question 4. Why does solubility decreases with increasing molecular mass in carboxylic acid?
Answer: Because of increase in alkyl chain length which is hydrophobic in nature. Hence solubility decreases.
In simple words: As the carbon chain gets longer, the oily non-polar part overpowers the polar acid part, making it less soluble in water.

Exam Tip: Balance between hydrophilic (\(-COOH\) group) and hydrophobic (alkyl chain) nature determines solubility.

Question 5. Why are aldehydes more reactive than ketones when undergo nucleophilic addition reaction?
Answer: (a) \((+)\) I effect: The alkyl group in ketones due to their electron-releasing character decrease the positive charge on C-Atom and thus reduce its reactivity.
(b) Steric hindrance: Due to steric hindrance in ketones they are less reactive.
In simple words: Ketones have two electron-pushing groups that crowd the central carbon and reduce its positive charge, slowing down attacks.

Exam Tip: Mention both inductive electron donation and steric crowding.

Question 6. Why PCC cannot oxidise methanol to methanoic acid while \(KMnO_4\) can?
Answer: This is because PCC is a mild oxidising agent and can oxidize methanol to methanal only, while \(KMnO_4\) being a strong oxidising agent oxidises it to methanoic acid.

In simple words: PCC stops at the aldehyde stage, whereas strong oxidizers like potassium permanganate push the reaction all the way to carboxylic acid.

Exam Tip: Classify PCC (Pyridinium chlorochromate) as a selective/mild reagent and \(KMnO_4\) as a vigorous/strong reagent.

Question 7. During preparation of esters from a carboxylic acid and an alcohol in the presence of acid catalyst water or ester formed should be removed as soon as it is formed.
Answer: The formation of esters from a carboxylic acid and an alcohol in the presence of acid catalyst is a reversible reaction.
\(R-COOH + R'OH \rightleftharpoons R-COOR' + H_2O\)
To shift the equilibrium in forward direction, the water or ester formed should be removed as fast as it is formed.
In simple words: Removing products forces a reversible equilibrium reaction to keep producing more ester.

Exam Tip: Use Le Chatelier's principle in your answer.

Question 8. Why HCOOH does not give HVZ reaction while \(CH_3COOH\) does?
Answer: \(CH_3COOH\) contains \(\alpha\)-hydrogens and hence gives HVZ reaction, but HCOOH does not contain an \(\alpha\)-hydrogen and hence does not give HVZ reaction.
In simple words: The Hell-Volhard-Zelinsky reaction requires an alpha-hydrogen atom next to the acid group, which formic acid lacks.

Exam Tip: Highlight the requirement of at least one \(\alpha\)-hydrogen for HVZhalogenation.

Question 9. Suggest a reason for the large difference in the boiling point of butanol and butanal although they have same solubility in water.
Answer: Because butanol has strong intermolecular H-bonding while butanal has weak dipole-dipole interaction. However, both of them form H-bonds with water and hence are soluble.
In simple words: Alcohols form strong self-hydrogen bonds (high boiling point), but both alcohols and aldehydes can form hydrogen bonds with water molecules (making them soluble).

Exam Tip: Differentiate between pure-state intermolecular forces (boiling point) and solute-solvent interactions (solubility).

Question 10. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reaction than propanal. Explain.
Answer: C-atom of carbonyl group of benzaldehyde is less electrophilic than C-atom of carbonyl group in propanal.
Polarity of carbonyl group in benzaldehyde is reduced due to resonance, making it less reactive in nucleophilic addition reactions.
In simple words: Resonance in benzaldehyde spreads out the positive charge on the carbonyl carbon, making it less attractive to incoming nucleophiles compared to propanal.

Exam Tip: Resonance stabilization in aromatic aldehydes reduces their carbonyl electrophilicity.

Question 11. Why does methanal not give aldol condensation while ethanol gives?
Answer: This is because only those compounds which have \(\alpha\)-hydrogen atoms can undergo aldol reaction. Ethanol possesses \(\alpha\)-hydrogen and undergoes aldol condensation. Methanal has no alpha hydrogen atoms, hence does not undergo aldol condensation.
In simple words: Aldol condensation needs acidic alpha-hydrogens, which methanal does not possess.

Exam Tip: Mention the necessity of \(\alpha\)-hydrogens for carbanion formation in aldol reactions.

Question 12. Why does methanal undergo Cannizzaro's reaction?
Answer: Because it does not possess an \(\alpha\)-hydrogen atom.
In simple words: Without alpha hydrogens, aldehydes undergo self-oxidation and reduction (Cannizzaro reaction) in the presence of concentrated alkali.

Exam Tip: Absence of \(\alpha\)-hydrogen is the hallmark condition for Cannizzaro reaction.

Question 13. Which acid is stronger and why? \(F_3C-C_6H_4COOH\) and \(CH_3C_6H_4COOH\)
Answer: \(F_3C^-\) has a strong \((-I)\) effect, whereas \(CH_3^-\) has a strong \((+I)\) effect. Due to greater stability of \(F_3CC_6H_4COO^-\) ion over \(CH_3C_6H_4COO^-\) ion, \(F_3C-C_6H_4COOH\) is a much stronger acid than \(CH_3-C_6H_4COOH\).
In simple words: The trifluoromethyl group pulls electrons away, stabilizing the negative charge and making the acid much stronger.

Exam Tip: Electron-withdrawing groups increase acidity; electron-donating groups decrease acidity.

Question 14. Explain why o-hydroxy benzaldehyde is a liquid at room temperature while p-hydroxy benzaldehyde is a high melting solid.
Answer: Due to intramolecular H-bonding in o-hydroxy benzaldehyde, it exists as discrete molecules, whereas due to intermolecular H-bonding, p-hydroxy benzaldehyde exists as associated molecules.
To break this intermolecular H-bond, a large amount of energy is needed. Consequently, p-isomer has a much higher m.p. and b.p. than that of o-isomer. As a result, o-hydroxy benzaldehyde is a liquid.
In simple words: Ortho-hydroxybenzaldehyde forms internal hydrogen bonds (staying separate as a liquid), while para-hydroxybenzaldehyde forms chains linking many molecules together (making it a solid).

Exam Tip: Clearly distinguish intramolecular (chelation/internal) vs. intermolecular (chain association) hydrogen bonding.

Question 15. Why is the boiling point of an acid anhydride higher than the acid from which it is derived?
Answer: Acid anhydrides are bigger in size than corresponding acids, have more surface area and more van der Waals forces of attraction, hence have higher boiling point.
In simple words: Anhydrides have larger molecular weights and larger surface areas, leading to stronger intermolecular forces.

Exam Tip: Size and molecular mass correlate directly with London dispersion forces.

Question 16. Why do Carboxylic acids not give the characteristic reactions of a carbonyl group?
Answer: Due to resonance, it doesn't give the characteristic reactions of carbonyl group. It does not have a free \(C=O\) group.
In simple words: The carbonyl carbon is involved in resonance with the adjacent hydroxyl group, locking up its double bond character.

Exam Tip: Resonance involvement prevents carboxylic acids from reacting with typical nucleophilic reagents like Grignard reagents (for addition) or carbonyl reagents like \(NaHSO_3\).

Question 17. Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. Why?
Answer: In 2,2,6-trimethylcyclohexanone there is steric hindrance of 3 methyl groups. It does not form cyanohydrin in good yield.
In simple words: The three methyl groups crowd the carbonyl carbon so tightly that cyanide ions cannot approach to react.

Exam Tip: Steric congestion prevents bulky or crowded ketones from forming addition products.

Question 18. Why is carboxyl group in benzoic acid meta directing?
Answer: In benzoic acid the carboxyl group is meta-directing because it is electron-withdrawing. There is positive charge on ortho and para positions. Electrophilic substitution takes place at meta-position.
In simple words: The carboxylic acid group pulls electron density away from the ring, leaving ortho and para positions positively charged and forcing incoming groups to the meta position.

Exam Tip: Electron-withdrawing groups deactivate ortho and para positions, making meta the preferred substitution site.

Question 19. Treatment of Benzaldehyde with HCN gives a mixture of two isomers which cannot be separated even by careful fractional distillation. Explain why?
Answer: It is because we get two optical isomers (enantiomers) which have same physical properties and cannot be separated by fractional distillation.
In simple words: The reaction creates a chiral center, producing a racemic mixture of enantiomers that share identical boiling points.

Exam Tip: Enantiomers possess identical physical properties except optical rotation, rendering distillation useless for separation.

Question 20. Sodium Bisulphite is used for the purification of aldehydes and ketones. Explain.
Answer: Aldehydes and ketones form addition compounds with \(NaHSO_3\) whereas impurities do not. On hydrolysis we get pure aldehydes and ketones back.
\(CH_3-CHO + NaHSO_3 \rightarrow CH_3-CH(OH)-SO_3Na\)
Treatment with \(H_2O\) regenerates \(CH_3-CHO + NaHSO_3\)
In simple words: Aldehydes and ketones form a solid bisulfite precipitate that can be separated from impurities and then converted back using acid or base.

Exam Tip: This addition reaction is reversible and serves as a classic purification technique.

Question 21. Why pH of reaction should be carefully controlled while preparing ammonia derivatives of carbonyl compound?
Answer: In strongly acidic medium, ammonia derivatives being basic will react with acids and will not react with carbonyl compound. In basic medium, \(OH^-\) will attack carbonyl group. pH of a reaction should be carefully controlled.
In simple words: If it is too acidic, the reagent gets protonated and stops reacting; if it is too basic, nucleophilic attack fails. A mildly acidic buffer is ideal.

Exam Tip: Optimal pH is generally maintained around 4.5 for nucleophilic addition of ammonia derivatives.

Question 22. Why formic acid is stronger acid than acetic acid?
Answer: Due to \((+)\) I effect, \(-CH_3\) group in acetic acid increases electron density on carbon atom which makes it a weak acid. While in formic acid no such pushing group is present, hence is more stronger acid than acetic acid.
In simple words: The methyl group in acetic acid pushes electrons towards the carboxyl group, destabilizing the negative charge and lowering acidity.

Question 23. Why is oxidation of alcohols to get aldehydes carried out under controlled conditions?
Answer: It is because aldehydes get further oxidised to acids. Oxidation of alcohols to aldehydes needs to be controlled.
In simple words: Without control, the reaction will continue past the aldehyde stage and turn into a carboxylic acid.

Exam Tip: Use special reagents like PCC or anhydrous media to stop oxidation at the aldehyde stage.

Question 24. Why the oxidation of toluene to benzaldehyde with \(CrO_3\) is carried out in the presence of acetic anhydride?
Answer: If acetic anhydride is not used we will get benzoic acid. Acetic anhydride is used to prevent oxidation of benzaldehyde to benzoic acid.
In simple words: Acetic anhydride traps the intermediate so it cannot oxidize further into benzoic acid (Etard reaction modification).

Exam Tip: This complex formation protects the intermediate aldehyde from over-oxidation.

Question 25. Melting point of an acid with even no. of carbon atoms is higher than those of its neighbour with odd no. of carbon atoms.
Answer: They fit into crystal lattice more readily than odd ones, that is why they have higher lattice energy and higher melting point.
In simple words: Even-carbon acid chains pack together more efficiently in a crystal structure, requiring more heat to melt.

Question 26. Why do aldehydes have lower boiling point corresponding alcohols?
Answer: Alcohols are associated with intermolecular H-bonding whereas aldehydes are associated with weaker dipole-dipole interactions. Alcohols have higher b.p.
In simple words: Alcohols have strong hydrogen bonds holding them together, whereas aldehydes only have weaker dipole attractions.

Exam Tip: Hydrogen bonding is always stronger than dipole-dipole attractions.

Question 27. Why do aldehydes behave like polar compounds?
Answer: Due to presence of \(C=O\) group which is polar.
In simple words: Oxygen pulls electrons away from carbon in the carbonyl double bond, creating a permanent dipole moment.

Exam Tip: Electronegativity difference between oxygen and carbon creates polarity.

Question 28. Most aromatic acids are solids while acetic acid and others of this series are liquids. Explain why?
Answer: Aromatic acids have higher molecular weight, more Van der Waals force of attraction as compared to aliphatic acids. They are solids.
In simple words: Aromatic acids are heavier and larger, so intermolecular forces hold them firmly together as solids at room temperature.

Exam Tip: Molecular weight and surface area dictate physical state (solid vs. liquid).

Question 29. Ethers possess a dipole moment even if the alkyl radicals in the molecule are identical. Why?
Answer: It is because ethers are bent molecules, dipoles do not get cancelled.
\(R-O-R\)
In simple words: The C-O-C bond angle is not linear (it is bent), so individual bond dipoles do not cancel each other out.

Exam Tip: Molecular geometry (bent shape) accounts for net dipole moments in symmetrical ethers.

Question 30. Why does acyl chlorides have lower boiling point than corresponding acids?
Answer: Acyl chlorides are not associated with intermolecular H-bonding. They have lower boiling point.
In simple words: Acyl chlorides lack the hydrogen bonding found in carboxylic acids, making them easier to boil.

Exam Tip: Absence of O-H groups rules out hydrogen bonding in acid chlorides.

Question 31. Why ethers are stored in coloured bottles?
Answer: They are stored in coloured bottles because in presence of sunlight they react with oxygen to form peroxides which may cause explosion.
In simple words: Colored glass blocks light and prevents the formation of explosive organic peroxides.

Exam Tip: Always add antioxidant or store in dark bottles to prevent auto-oxidation of ethers.

Question 32. Why formaldehyde cannot be prepared by Rosenmund's reduction?
Answer: Because the formyl chloride (\(H-COCl\)) thus formed is unstable at room temperature and cannot be prepared by Rosenmund reduction.
In simple words: Formyl chloride decomposes instantly at room temperature, making this method unusable for formaldehyde.

Exam Tip: Instability of formyl chloride is the reason behind this limitation.

LEVEL-3 QUESTIONS

Question 1. A compound 'A' dissolves in water and its 40% aqueous solution is used as a preservation for zoological specimen. When its aq. solution is allowed to stand, it forms trimer 'B'. 6 moles of it react to give 'C' in presence of \(Ca(OH)_2\). 'A' undergoes disproportionation reaction in presence of conc. alkali to form 'D' and 'E'. 'D' liberates \(H_2(g)\) with Na metal. Identify 'A' to E writing the chemical equations involved.
Answer: A = Formaldehyde (\(HCHO\))
B = Trioxane (\(C_3H_6O_3\))
C = Glucose (\(C_6H_{12}O_6\))
D = Methanol (\(CH_3OH\))
E = Sodium formate (\(HCOONa\))
Reactions:
1) \(3HCHO \rightarrow \text{Trioxane (B)}\)
2) \(6HCHO \xrightarrow{Ca(OH)_2} C_6H_{12}O_6 \text{ (Glucose, C)}\)
3) \(2HCHO \xrightarrow{\text{Conc. }NaOH} CH_3OH \text{ (D)} + HCOONa \text{ (E)}\)
4) \(2CH_3OH + 2Na \rightarrow 2CH_3ONa + H_2 \uparrow\)
In simple words: Formaldehyde undergoes polymerization, Cannizzaro disproportionation, and formose condensation to yield various carbon compounds.

Exam Tip: Formalin (40% aqueous formaldehyde) is a classic biological preservative.

Question 2. Complete the following and identify from A to E.
\(CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} A \xrightarrow{OH^-} B \xrightarrow{\text{Heat}} C \xrightarrow{\text{Tollen's reagent}} D \xrightarrow{Br_2/H_2O} E\)
Answer: A = \(CH_3CHO\) (Acetaldehyde)
B = \(CH_3-CH(OH)-CH_2-CHO\) (3-hydroxybutanal)
C = \(CH_3-CH=CH-CHO\) (But-2-enal)
D = \(CH_3-CH=CH-COOH\) (But-2-enoic acid)
E = \(CH_3-CH(Br)-CH(Br)-COOH\) (2,3-dibromobutanoic acid)
In simple words: Acyl chloride reduces to aldehyde, undergoes aldol condensation, dehydration, mild oxidation of the aldehyde group, and bromine addition across the double bond.

Exam Tip: Rosenmund reduction followed by aldol condensation and Tollen's oxidation creates a multi-functional chain pathway.

Question 3. An organic compound (A) having MF \(C_8H_8O\) forms an orange red precipitate (B) with 2,4-DNP. Compound (A) gives yellow ppt ('C') when heated in presence of \(I_2\) and NaOH along with a colorless compound 'D'. 'A' does not reduce Fehling's solution and does not decolorize bromine water. On drastic oxidation of 'A' with chromic acid, a carboxylic acid 'E' with MF \(C_7H_6O_2\) is formed. Deduce the structures of A to E. Also, write the reactions involved.
Answer: A = Acetophenone (\(C_6H_5COCH_3\))
B = Acetophenone 2,4-dinitrophenylhydrazone
C = Iodoform (\(CHI_3\))
D = Sodium benzoate (\(C_6H_5COONa\))
E = Benzoic acid (\(C_7H_6O_2\))
Reactions:
1) \(C_6H_5COCH_3 + 2,4-DNP \rightarrow \text{Orange-red ppt (B)}\)
2) \(C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 \text{ (yellow ppt, C)} + C_6H_5COONa \text{ (D)}\)
3) \(C_6H_5COCH_3 \xrightarrow{\text{Chromic acid}} C_6H_5COOH \text{ (E)} + CO_2 + H_2O\)
In simple words: Acetophenone reacts with 2,4-DNP, gives an iodoform test due to the \(CH_3CO\) group, and oxidizes into benzoic acid.

Exam Tip: Positive iodoform test combined with a negative Fehling test confirms a methyl ketone.

Question 4. Complete the following and identify the compounds A to E.
1) Cyclohexene derivative + \(O_3 / Zn-H_2O \rightarrow A + B\)
2) \(A + B \xrightarrow{dil. NaOH} C + H_2O\)
3) \(C \xrightarrow{O_3 / Zn-H_2O} A + D\)
4) \(D \xrightarrow{H_2/Ni} E\)
Answer: A = Benzaldehyde (\(C_6H_5CHO\))
B = Pentanedione / Keto-aldehyde intermediate
C = Aldol condensation product / cyclic derivative
D = Diketone / Keto-aldehyde cleavage product
E = Corresponding reduced alcohol/diol derivative
In simple words: Ozonolysis splits cyclic alkenes into dialdehydes/diones, which then undergo aldol reactions, further ozonolysis, and catalytic hydrogenation.

Exam Tip: Ozonolysis cleavage breaks unsaturated carbon bonds into respective carbonyl fragments.

Question 5. An organic compound A with MF \(C_5H_8O_2\) is reduced to n-pentane on treatment with \(Zn-Hg/HCl\). 'A' forms di-oxine with \(NH_2OH\) and gives a positive iodoform test and Tollen's reagent test. Identify the compound 'A' and deduce its structure.
Answer: Structure of A = \(OHC-CH_2-CH_2-CO-CH_3\) (Pentane-1,4-dione / containing both aldehyde and ketone groups)
Reactions:
1) \(OHC-CH_2-CH_2-CO-CH_3 \xrightarrow{Zn-Hg/HCl} CH_3-CH_2-CH_2-CH_2-CH_3\) (n-pentane)
2) Forms dioxime with \(2NH_2OH\)
3) Gives positive iodoform test (due to \(CH_3CO-\)) and Tollen's test (due to \(-CHO\)).
In simple words: Compound A contains both an aldehyde and a ketone group along a straight 5-carbon chain.

Question 6. An organic compound 'A' with MF \(C_8H_8O\) gives positive iodoform and 2,4-DNP tests. It does not reduce Fehling's solution and does not decolorize \(Br_2/H_2O\). On oxidation with chromic oxide, it gives a carboxylic acid 'B' with MF \(C_7H_6O_2\). Deduce the structures of A and B.
Answer: A = Acetophenone (\(C_6H_5COCH_3\))
B = Benzoic acid (\(C_6H_5COOH\))
Reactions:
\(C_6H_5COCH_3 \xrightarrow{2,4-DNP} \text{Derivative (positive test)}\)
\(C_6H_5COCH_3 \xrightarrow{I_2/NaOH} CHI_3 + C_6H_5COONa\)
\(C_6H_5COCH_3 \xrightarrow{\text{Chromic oxide}} C_6H_5COOH \text{ (B)}\)
In simple words: Acetophenone gives a positive iodoform test because of its methyl ketone group and oxidizes into benzoic acid.

Exam Tip: Match the molecular formula \(C_8H_8O\) directly with aromatic methyl ketones. Benzene ring accounts for \(C_6H_5-\), leaving \(-COCH_3\).

Question 7. A compound 'X' with MF \(C_2H_4O\) on oxidation gives Y with MF \(C_2H_4O_2\). X undergoes haloform reaction. On treatment with HCN, X gives Z which on hydrolysis gives 2-hydroxy propanoic acid. Write down the structures of X, Y and Z. Name the product when X reacts with diluted NaOH.
Answer: X = Acetaldehyde (\(CH_3CHO\))
Y = Acetic acid (\(CH_3COOH\))
Z = Acetaldehyde cyanohydrin (\(CH_3CH(OH)CN\))
Product with dilute NaOH: 3-hydroxybutanal (Aldol product)
Reactions:
1) \(CH_3CHO \xrightarrow{[O]} CH_3COOH \text{ (Y)}\)
2) \(CH_3CHO + HCN \rightarrow CH_3CH(OH)CN \text{ (Z)}\)
3) \(CH_3CH(OH)CN \xrightarrow{H_3O^+} CH_3CH(OH)COOH\) (2-hydroxypropanoic acid)
4) \(2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO\) (3-hydroxybutanal)
In simple words: X is acetaldehyde, which oxidizes to acetic acid, reacts with HCN to form cyanohydrin, and undergoes aldol addition with dilute alkali.

Question 8. An unknown aldehyde A on reaction with alkali gives \(\beta\)-hydroxy aldehyde which loses \(H_2O\) forming an unsaturated aldehyde 2-butenal. Another aldehyde B undergoes disproportionation reaction in the presence of conc. alkali to produce C and D. C is an aryl alcohol with MF \(C_7H_8O\). Write the sequence of reactions involved. Identify A to D. Name the product when B reacts with \(Zn-Hg/HCl\).
Answer: A = Acetaldehyde (\(CH_3CHO\))
B = Benzaldehyde (\(C_6H_5CHO\))
C = Benzyl alcohol (\(C_6H_5CH_2OH\))
D = Sodium benzoate (\(C_6H_5COONa\))
Product when B reacts with \(Zn-Hg/HCl\): Toluene (\(C_6H_5CH_3\))
Reactions:
1) \(2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO \xrightarrow{\text{heat}, -H_2O} CH_3-CH=CH-CHO\) (2-butenal)
2) \(2C_6H_5CHO \xrightarrow{\text{Conc. }NaOH} C_6H_5CH_2OH \text{ (C)} + C_6H_5COONa \text{ (D)}\)
3) \(C_6H_5CHO \xrightarrow{Zn-Hg/HCl} C_6H_5CH_3 + H_2O\) (Toluene)
In simple words: Acetaldehyde undergoes aldol condensation, while benzaldehyde undergoes a Cannizzaro reaction and Clemmensen reduction to form toluene.

Exam Tip: Aliphatic aldehydes with \(\alpha\)-hydrogens undergo aldol condensation; aromatic aldehydes without \(\alpha\)-hydrogens undergo Cannizzaro reaction.

Question 9. An organic compound A (\(C_8H_6\)) on treatment with \(H_2SO_4\) and \(HgSO_4\) gives B which also can be obtained from the reaction b/w benzene and an acid chloride in presence of \(AlCl_3\) (anhydrous). B on treatment with \(I_2/NaOH\) gives C and a yellow compound D. Identify from A to D. Write the sequence of reactions.
Answer: A = Phenylacetylene (\(C_6H_5C\equiv CH\))
B = Acetophenone (\(C_6H_5COCH_3\))
C = Sodium benzoate (\(C_6H_5COONa\))
D = Iodoform (\(CHI_3\), yellow compound)
Reactions:
1) \(C_6H_5-C\equiv CH + H_2O \xrightarrow{H_2SO_4/HgSO_4} C_6H_5COCH_3 \text{ (B)}\)
2) \(C_6H_6 + CH_3COCl \xrightarrow{\text{anhydrous }AlCl_3} C_6H_5COCH_3 \text{ (B)} + HCl\)
3) \(C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa \text{ (C)} + CHI_3 \text{ (D, yellow ppt)}\)
In simple words: Hydration of phenylacetylene or Friedel-Crafts acylation of benzene both yield acetophenone, which reacts with iodine and sodium hydroxide to form iodoform.

Exam Tip: Hydration of alkynes follows Markovnikov's rule to yield ketones (except acetylene which gives acetaldehyde).

Question 10. Two moles of an organic compound A on treatment with strong base gives compounds B and C. Compound B on dehydrogenation with Cu gives A while on acidification gives a carboxylic acid D with MF \(CH_2O_2\). Identify A to D and write the sequence of reactions.
Answer: A = Formaldehyde (\(HCHO\))
B = Methanol (\(CH_3OH\))
C = Sodium formate (\(HCOONa\))
D = Formic acid (\(HCOOH\))
Reactions:
1) Compound (D) with MF \(CH_2O_2\) is \(HCOOH\).
2) Compound (C) is \(HCOONa \xrightarrow{H^+} HCOOH \text{ (D)}\)
3) Compound (A) is \(HCHO\). \(2HCHO \xrightarrow{\text{Conc. }NaOH} CH_3OH \text{ (B)} + HCOONa \text{ (C)}\)
4) \(CH_3OH \xrightarrow{Cu/573K} HCHO \text{ (A)} + H_2\)
In simple words: Formaldehyde undergoes Cannizzaro reaction with a strong base to form methanol and sodium formate, and methanol can be dehydrogenated back into formaldehyde.

Exam Tip: Cannizzaro reaction of formaldehyde produces one molecule of alcohol and one molecule of salt of carboxylic acid.

Question 11. An organic compound A (\(C_8H_{16}O_2\)) was hydrolysed with dil \(H_2SO_4\) to yield a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid gives compound B. C on dehydration gives D which reacts with HBr to give optically active compound E. Identify A to E with the help of equations.
Answer: A = Butyl butyrate / Propyl pentanoate ester derivative (such as \(CH_3CH_2CH_2COOC_4H_9\))
B = Butyric acid / Carboxylic acid (\(CH_3CH_2CH_2COOH\))
C = Butanol (\(CH_3CH_2CH_2CH_2OH\))
D = But-1-ene (\(CH_3CH_2CH=CH_2\))
E = 2-bromobutane (\(CH_3CH_2CH(Br)CH_3\), optically active due to chiral carbon)
Reactions:
1) \(CH_3CH_2CH_2COOC_4H_9 \xrightarrow{H_3O^+} CH_3CH_2CH_2COOH \text{ (B)} + C_4H_9OH \text{ (C)}\)
2) \(CH_3CH_2CH_2CH_2OH \xrightarrow{[O]} CH_3CH_2CH_2COOH \text{ (B)}\)
3) \(CH_3CH_2CH_2CH_2OH \xrightarrow{Cu/573K, -H_2O} CH_3CH_2CH=CH_2 \text{ (D)}\)
4) \(CH_3CH_2CH=CH_2 + HBr \rightarrow CH_3CH_2-CH(Br)-CH_3 \text{ (E)}\)
In simple words: An ester hydrolyzes into an acid and alcohol, which dehydrates into an alkene and adds HBr to form a chiral, optically active bromoalkane.

Exam Tip: Presence of a chiral carbon (attached to four different groups) makes compound E optically active.