100+ Essential Organic Chemistry Reasoning Questions: CBSE Class 12 Board Exam

This post contains 100+ organic chemistry reasoning questions which are important for the CBSE Class 12 Board Exam.

1. Why is sulphuric acid is not used during reaction of alcohols with KI?

Ans: Sulphuric acid converts the KI into HI and then oxidizes into I₂.

2. Arrange each set of compounds in order of increasing boiling point.

(i) Bromoethane , Bromoform, Chloromethane, Dibromomethane.

(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Ans: As atomic mass increases boiling point increases. While branching compounds has spherical shape, so has less contact as compared to the straight chain compounds, hence less boiling point for branching chain alkyl halides.

(i) Chloromethane(CH₃Cl) < Bromoethane(CH₃-CH₂-Br) < Dibromomethane(CH₂Br₂) < Bromoform (CHBr₃)
(ii) Isopropyl chloride (CH₃CH(Cl)CH₃) < 1-Chloropropane (CH₃CH₂CH₂Cl) < 1-Chlorobutane (CH₃CH₂CH₂CH₂Cl)

3. Which of the following undergo S_N² reaction faster?

Ans: (i) (b) is more faster to undergo S_N² reaction because it is primary alkyl halide.
(ii) (c) is more faster to undergo S_N² reaction because I is better leaving group than Cl due to its large size than Cl.

4. Predict the order of reactivity of the following compounds in S_N¹ & S_N² reactions.

(i) The four isomeric bromo butanes.

(ii) C₆H₅CH₂Br, C₆H₅CH₂(C₆H₅)Br, C₆H₅CH₂(CH₃)Br, C₆H₅CH₂(CH₃)(C₆H₅)Br.

Ans: (i) CH₃CH₂CH₂CH₂Br < CH₃CH(CH₃)CH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr order for S_N¹ mechanism.

CH₃CH₂CH₂CH₂Br > CH₃CH(CH₃)CH₂Br > CH₃CH₂CH(Br)CH₃ > (CH₃)₃CBr order for S_N² mechanism.

(ii) C₆H₅CH₂Br < C₆H₅CH(CH₃)Br < C₆H₅CH(C₆H₅)Br < C₆H₅C(CH₃)(C₆H₅)Br order for S_N¹ mechanism.

C₆H₅CH₂Br > C₆H₅CH(CH₃)Br > C₆H₅CH(C₆H₅)Br > C₆H₅C(CH₃)(C₆H₅)Br order for S_N² mechanism.

5. Although chlorine is an electron withdrawing group yet it is ortho and para directing in electrophilic aromatic substitution reactions. Why?

Ans: Cl is electron withdrawing by inductive effect, and is electron releasing by resonance. It increases negative charge density at ortho and para positions as resonance effect is more dominating as compared to inductive effect.

6. Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N² mechanism? Explain your answer.

a. CH₃CH₂CH₂CH₂Br & CH₃CH₂CH(Br)CH₃

b. CH₃CH₂CH(Br)CH₃ & (CH₃)₃CBr

c. CH₃CH(CH₃)CH₂CH₂Br & CH₃CH₂CH(CH₃)CH₂Br

Ans: a. CH₃CH₂CH₂CH₂Br as primary alkyl halide is more reactive than secondary alkyl halide.

b. CH₃CH₂CH(Br)CH₃ as secondary alkyl halide is more reactive than tertiary alkyl halide.

c. CH₃CH(CH₃)CH₂CH₂Br as methyl group is far from Br it will less stabilize the carbocation formed and facilitate the S_N² mechanism.

7. Which of the following compound go faster S_N¹ reaction?

Ans: 1. First compound is more stable because tertiary is more reactive than secondary alkyl halide for S_N¹ mechanism.

2. First compound is more stable because secondary is more reactive than primary for S_N¹ mechanism.

8. Which one of the following has highest boiling point: CH₂Cl₂, CHCl₃, CCl₄.

Ans: Net Dipole moment of CH₂Cl₂ & CCl₄ is zero due to cancellation of dipole moment of bonds due to its symmetric structure. Whereas net dipole moment of CHCl₃ is not zero, as its structure is not symmetric.

9. Which compound in each of the following pairs react faster in S_N² reaction with OH?

(i) CH₃Br or CH₃I (ii) (CH₃)₃CCl or CH₃Cl.

Ans: CH₃I is faster reactive by S_N² mechanism because C-I bond strength is less than C-Br bond. So, CH₃Cl reacts faster than (CH₃)₃CCl by S_N² mechanism because 1⁰ alkyl halide more reactive by S_N² mechanism.

10. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.

Ans: KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs. The attack takes place mainly through carbon atom not through nitrogen atom since C—C is more stable than C—N bond. However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

11. Why does NO₂ group show its effect only at ortho and para positions and not at meta positions?

Ans: The presence of NO₂ group at ortho and para positions withdraws the electron density form the benzene ring and thus facilitates the attack of the nucleophile on halo arene. The carbanaion thus formed is stabilized through resonance. The negative charge appeared at ortho and para positions with respect to the halogen substituent is stabilized by NO₂ group while in case of meta nitro benzene, none of the resonating structures bear the negative charge on carbon atom bearing the NO₂ group. Therefore, the presence of nitro group at meta position does not stabilize the negative charge and no effect on reactivity is observed by the presence of NO₂ group at meta position.

12. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. Why?

Ans: It is due to electron with drawing effect of phenyl group which decrease dipole moment of chloro benzene.

13. Alkyl halides, though polar, are immiscible with water. Explain.

Ans: Due to absence of hydrogen bond in alkyl halides with water.

14. Grignard reagent should be prepared under anhydrous conditions. Give reason.

Ans: Grignard reagent highly reactive and on reaction with water forms alkanes.

RMgX + H₂O → RH + Mg(X)OH.

15. Vinyl chloride is less reactive towards nucleophilic substitution. Explain.

Ans: Due to resonance, between C—Cl bond in vinyl chloride show partial double bond character.

16. Chloroform stored in dark colored bottles by completely filling it it. Explain.

Ans: Chloroform react with air in presence of sun light it forms poisonous phosogene.

2 CHCl₃ + O₂ → 2 COCl₂+ 2 HCl

17. Out of C₆H₅CH₂Cl & C₆H₅CH(Cl)C₆H₅, which is more easily hydrolyzed by KOH?

Ans: The second compound is more resonance stabilized, hence easily forms carbocation and react faster by S_N¹ mechanism.

18. The treatment of alkyl halides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH alkenes are major products. Explain.

Ans: Substitution of alkyl halide gives alcohols and elimination gives alkenes and there is a competition between substitution and elimination. In the presence of low polarity solvent like alcohol favours elimination where as in the presence of high polarity solvent like water substitution favoured.

19. Bond angle of ethers is slightly greater than tetrahedral bond angle. Explain.

Ans: Due to repulsion of lone pairs of electrons on the oxygen atom and the repulsion between bulky alkyl groups in ethers bond angles is slightly greater than tetrahedral bond angle.

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20. Arrange the following sets of compounds in order of their increasing boiling points:

a. Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.

b. Pentan-1-ol, n-butane, pentanal, ethoxyethane.

Ans: a. Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.

b. n-Butane, ethoxyethane, pentanal and pentan-1-ol.

21. Acidity of alcohols is 1⁰>2⁰>3⁰, explain?

Ans: An electron-releasing group (–CH₃, –C₂H₅) increases electron density on oxygen tending to decrease the polarity of O-H bond. This decreases the acid strength. For this reason, the acid strength of alcohols decreases in the given order.

22. Arrange the following compounds in increasing order of their acid strength: Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.

Ans: Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3,5-dinitrophenol, 2,4, 6-trinitrophenol.

23. For the reaction of alcohols with carboxylic acid or anhydrides to form esters, conc. H₂SO₄ is added, while for the reaction of alcohol with acid chloride, base like pyridine is used. Explain.

Ans: R-OH + RCOOH/RCOOCOR → RCOOR + H₂O

R-OH + RCOCl → RCOOR + HCl

Sulphuric acid removes the water and prevent backward reaction, base like pyridine removes the acid HCl and prevent backward reaction.

24. The relative dehydration of alcohols is Tertiary>Secondar>Primary. Give reason.

Ans: Dehydration is carried out by carbocation mechanism, and stability of carbocation is Tertiary>Secondary>Primary.

Read more: Unique Guide to Organic Conversions (Cheat Sheet included)

25. What is intermediate in Riemer –Tiemann reaction?

Ans: :CHCl₂

26. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Ans: Due to hydrogen bond.

27. Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.

Ans: Methoxy group on benzene ring donates the electrons and activate the ring and creates negative charge on the ortho and para positions, hence electrophile is attracted at these positions.

28. Arrange the following compounds in the increasing order of their boiling points: CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, H₅C₂-O-C₂H₅, CH₃CH₂CH₂CH₂CH₃.

Ans: CH₃CH₂CH₂CH₂CH₃ < H₅C₂-O-C₂H₅ < CH₃CH₂CH₂CHO < CH₃CH₂CH₂CH₂OH.

Alcohols with hydrogen bond has more boiling point. Ethers have less dipole dipole moment as compare with aldehydes and alkanes least with weak Vander Waals forces.

29. Aldehydes and ketones are soluble in water. Explain.

Ans: They soluble in water due to formation of hydrogen bond between carbonyl compounds and water.

30. Toluene with CrO₃ and acetic anhydride forms benzaldehyde. Explain.

Ans: Toluene forms benzylidene diacetate intermediate with CrO₃ and acetic anhydride.

31. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.

Ans: The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.

32. Carbonyl carbon of carboxylic acid is less electronegative than aldehydes and ketones. Give reason.

Ans: The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible resonance structure shown below:

33. Carboxylic acids are having higher boiling points than aldehydes, ketones and even alcohols of comparable molecular masses. Explain.

Ans: Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses. This is due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding. The hydrogen bonds are not broken completely even in the vapour phase. In fact, most carboxylic acids exist as dimer in the vapour phase or in the aprotic solvents.

34. Carboxylic acids are stronger acids than phenol though both possess resonance stabilization of respective anions. Explain.

Ans: The higher acidity of carboxylic acids as compared to phenols can be understood similarly. The conjugate base of carboxylic acid, a carboxylate ion, is stabilised by two equivalent resonance structures in which the negative charge is at the more electronegative oxygen atom. The conjugate base of phenol, a phenoxide ion, has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom.

Therefore, resonance in phenoxide ion is not as important as it is in carboxylate ion. Further, the negative charge is delocalised over two electronegative oxygen atoms in carboxylate ion whereas it is less effectively delocalised over one oxygen atom and less electronegative carbon atoms in phenoxide ion . Thus, the carboxylate ion is more stabilised than phenoxide ion, so carboxylic acids are more acidic than phenols.

35. Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6- Trimethylcyclohexanone does not.

(ii) There are two –NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Ans: i) Due to steric effect and electron releasing effect.

ii) The other of the two NH₂ groups is involved in resonance with C=O group.

iii) To prevent backward reaction

36. Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. Explain.

Ans: Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (S_N²) of alkyl halides by the anion formed by the phthalimide. In this reaction, Potassium salt of phthalimide is formed. It reacts readily with alkyl halide to form the corresponding alkyl derivative.

But aryl halide does not react with potassium phthalimide. Because C-X bond in haloarene (alkyl halide) is difficult to be cleaved due to partial double bond character, moreover, aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide. So, aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.

37. Boiling point of primary, secondary and tertiary amines is different. Give reason.

Ans: The intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as follows: Primary > Secondary > Tertiary

38. Generally aniline direct bromination gives tri-substituted product to get mono-substituted product, what is to be done?

Ans: This can be done by protecting the -NH₂ group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.

39. Account for the following:

(i) pK_b of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Ans: i) More pK_b less basic, less basicity of aniline is due to involvement of lone pair in resonance.

ii) In case of aniline, its aromatic nature makes it insoluble in water.

iii) Due to electron releasing nature of methyl group (+I effect) , methyl amine is more basic than water. Thus, it produces OH^– ions by accepting H⁺ ions from water.

CH₃-NH₂ + H-OH → CH₃NH₃⁺ + OH^–

Ferric chloride dissociates into water to form Fe³⁺ & Cl^– ions. OH^– ions reacts with Fe³⁺ ions to form precipitate of hydrated ferric oxide.

2Fe³⁺ + 6OH^– → Fe₂O₃.3H₂O

iv) In the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing.

v) Aniline does not undergo Friedel-Crafts reaction because being a Lewis base it forms salt with anhydrous AlCl₃, which is a Lewis acid.

vi) Because diazonium salts of aromatic amines are stabilized by resonance.

vii) Because by Gabriel phthalimide synthesis, we get pure primary amine unlike the ammonolysis where we get the mixture of products.

40. Give plausible explanation for each of the following:

i. Why are amines less acidic than alcohols of comparable molecular masses?

ii. Why do primary amines have higher boiling point than tertiary amines?

iii. Why are aliphatic amines stronger bases than aromatic amines?

Ans: i. Nitrogen is less electronegative than oxygen, which makes less polarization of N—H bond as compare with O—H bond.

ii. There are two hydrogen atom present in primary amine, one hydrogen atom in secondary amine and no hydrogen atom in tertiary amine. So, hydrogen bonding is present in primary amine upto greater extent while in tertiary amine there is no hydrogen bonding. Hence primary amines have higher boiling point than tertiary amines.

iii. In aromatic amines lone pair present on nitrogen atom involved in resonance, hence unable to attack on H⁺ ion to show its basic character.

41. Preparation of ethers by dehydration of alcohols is not suitable for the using of secondary and tertiary alcohols. Give reason.

Ans: Dehydration of secondary and tertiary alcohols to give corresponding ethers is successful as elimination competes over substitution and as a consequence, alkenes are easily formed.

42. The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

C₂H₅ONa + (CH₃)₃-Cl → C₂H₅-O-C(CH₃)₃

i. What would be the major product of this reaction?

ii. Write a suitable reaction for the preparation of t-butyl ethyl ether.

Ans: i. The major product of the given reaction is 2-methylprop-1-ene. It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution reaction.

ii. (CH₃)₃-ONa + C₂H₅Cl → C₂H₅-O-C(CH₃)₃

43. Which of the following is appropriate for the preparation of anisole?

a) Bromo Benzene & Sodium Methoxide b) Phenoxide & Methyl bromide.

Ans: Set b) is appropriate because nucleophilic substitution on aromatic ring is difficult due to partial double bond character due to the resonance of lone pair of bromine with benzene ring.

44. The bond angle in alcohol is slightly less than the tetrahedral bond angle. Why is it so?

Ans: It is due to the repulsion between the unpaired electron pairs of oxygen.

45. Carbon-oxygen bond length in phenol is less than the alcohol?

Ans: It is due to the partial double bond character between C and O in phenol.

46. Boiling point of the alcohol is more than the alkenes. Comment.

Ans: Due to intermolecular hydrogen in alcohols it has high boiling point.

47. Branched alcohols are having low boiling point compare with normal straight chain alcohols? Explain.

OR

Boiling point of t-butyl alcohol is less than the n-butyl alcohol. Explain.

Ans: As branching increases surface area decreases and Vander Waal force decreases, hence boiling point decreases. And also in branched alcohols due to electron releasing effect strength of hydrogen decreases.

48. Arrange the following compounds in increasing order of their boiling points.

CH₃CH₂OH, CH₃CHO, CH₃OCH₃, CH₃CH₂CH₃

Ans: Molecular mass of these compounds are similar; ethanol is having intermolecular H-bond hence more boiling point. Among acetaldehyde and ether, acetaldehyde has strong dipole attractions than the ethers. Propane has weak Vander Waal forces. Hence the boiling point order is CH₃CH₂OH > CH₃CHO > CH₃OCH₃ > CH₃CH₂CH₃.

49. Arrange the following in increasing order of their reactivity towards nucleophilic addition reaction.

HCHO, CH₃CHO, CH₃COCH₃

Ans: CH₃COCH₃ < CH₃CHO < HCHO

The order is due to Electron releasing and Steric effect.

50. Ethers have low boiling point compare with corresponding alcohols. Give reasons.

Ans: Alcohols can form intermolecular H-bond, where as ethers do not. Hence, alcohols are having higher boiling point than ethers.

51. Alcohols are bronsted bases or proton acceptors. Explain.

Ans: It is due to presence of unshared electron pair over oxygen atom which makes alcohols proton acceptors. Proton acceptors are known as Bronsted bases.

52. Alcohols are weak acids than water. Explain.

Ans: In both the cases, due to polar O-H bond, behave as acids. But, in alcohols due to presence of alkyl groups, which are electron releasing, hence O-H bond is less polarized, shows weak acidic character.

53. Arrange the acidity order for 1⁰, 2⁰ & 3⁰ isomeric alcohols (C₄H₁₀O)?

Ans: Electron releasing group (alkyl group) destabilizes the anion and hence, decrease the acidity. For simplest case of alcohols, primary alcohols are more acidic than secondary alcohols which are more acidic than tertiary alcohols.

The acidity order is: Primary alcohol > Secondary alcohol > Tertiary alcohol.

54. Phenol is stronger acid than alcohol. Explain.

Or

Phenol is weak proton acceptor than alcohol. Explain.

Ans: Acidity of the substance depends on stability of anion after losing H⁺ ion. Since phenoxide ion undergoes resonance stabilization compare to alkoxide, phenol is more acidic than alcohol.

55. Arrange the following compounds in increasing order of their acidic strength: Propan-1-ol, 4-Methyl phenol, Phenol, 3,5-dinitrophenol, 2,4,6 Trinitrophenol.

Ans: Propan-1-ol < 4-Methoxy phenol < Phenol < 3,5-dinitrophenol < 2, 4, 6-Trinitrophenol.

Phenol is more acidic than alcohol due to resonance stability of phenoxide ion. Electron releasing group, i.e., methoxy (OCH₃) decreases acidity. Electron withdrawing group, i.e., nitro group (NO₂) increases acidity. More number of electron withdrawing groups, more will be the acidity.

56. Ortho-nitrophenol and para-nitrophenol can be separated by distillation. Explain.

Or

Boiling point of para-nitrophenol is more than the ortho-nitrophenol. Explain.

Ans: Ortho-nitrophenol is a steam volatile compound due to presence of intramolecular hydrogen bonding whereas para-nitrophenol is having intermolecular hydrogen bond. Intramolecular hydrogen bond decreases boiling point whereas intermolecular hydrogen bond increases boiling point.

57. The usual halogenation of benzene takes place of in presence of Lewis acid catalyst like AlCl₃, where as phenol can directly react with bromine. Explain?

Ans: In case of halogenation of benzene, Br₂ is polarized by lewis acid FeBr₃, but in case of phenol, the polarization of Br₂ takes place by phenol due to highly activating effect of -OH group present on the benzene ring.

58. Primary alkyl halide reacts with sodium alkoxide and forms ethers in good yield but t-alkyl halides yield fewer amounts of ethers. Explain.

Ans: Since tertiary carbocation is more stable than primary carbocation and it will form alkenes instead of ether.

59. Explain the fact that in aryl alkyl ethers (i) alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho & para position.

Ans: Alkoxy group is electron releasing to the benzene ring (+M effect) and by resonance negative charge obtained at ortho & para positions only. Hence electrophile attracts at ortho & para positions. And due to +M effect ring became activated.

60. t- butyl methyl ether reacts with HI forms methanol & t-butyl iodide. Explain.

Ans: Since tertiary carbocation is more stable and reaction is followed by carbocation mechanism, so it gives t-butyl iodide and methanol.

61. The commercial ethanol is mixed with copper sulphate & pyridine. Explain.

Ans: Commercial ethanol is mixed with CuSO₄ & pyridine to make it unfit for drinking. It is known as denaturation of alcohol.

62. Explain why are alcohols comparatively more soluble in water than the corresponding hydrocarbons?

Ans: It is due to Hydrogen Bond formation with water.

63. Explain how does the -OH group to a carbon of benzene ring activates it towards electrophilic substitution?

Ans: -OH group donates its non bonded electrons to resonance with benzene and creates negative charge at ortho and para-positions, hence ring is activated towards electrophilic substitution reaction.

64. Unlike phenols, alcohols are easily protonated. Give reason.

Ans: Alcohols act as proton acceptors or Bronsted bases. It is due to presence of unshared electron pair over oxygen. In case of phenol lone pair is involving in resonance. Hence, it can not be protanated easily.

65. Reaction of alcohols with hydrogen halides (HX) is of the following order: 3⁰ alcohols > 2⁰ alcohols > 1⁰ alcohols. Explain.

Ans: This reaction takes place through carbocation mechanism since the stability of carbocation is 3 2 1. The reactivity order is also same.

66. Anisole react with HI gives phenol & methyl iodide but not iodo benzene and methanol. Explain.

Ans: Nucleophilic substitution on aromatic ring is difficult due to partial double bond character between oxygen and carbon of benzene ring.

67. Name the intermediate or electrophile formed during Reimer-Tiemann reaction?

Ans: Dichloro carbon :CCl₂.

68. Phenol react with bromine in the presence of carbon disulphide gives mono bromophenol whereas with water gives tribromophenol. Explain.

Ans: CS₂ is a non-polar solvent whereas water is a polar solvent. In the presence of polar solvent phenol polarizes and gives mole electrophilic substitution.

69. Ethers made of similar kind of polar bonds (symmetric) also possess some net dipole moment. Explain.

Ans: Due to its bent structure.

70. While separating a mixture of ortho & para nitro phenols by steam distillation, name the isomer which is steam volatile. Give reason.

Ans: Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bond.

71. Explain why ortho-nitrophenol is more acidic than ortho-methoxyphenol?

Ans: Nitro group is an electron withdrawing group which increases acidity where as methoxy group is an electron releasing group which decreases acidity. So, ortho-nitrophenol is more acidic than orth-methoxyphenol.

72. Phenol is more active towards electrophilic substitution than benzene. Explain.

Ans: OH group present on the phenol is electron releasing and activates the benzene ring through resonance.

73. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reasons.

Ans: Secondary and tertiary alcohols will give alkenes easily by dehydration.

74. Explain why is propanol higher boiling than butane?

Ans: Due to intermolecular hydrogen bond present in alcohols, they possess high boiling point.

75. Explain why are alcohols comparatively more soluble in water than the corresponding hydrocarbons?

Ans: Due to intermolecular hydrogen bonding of alcohol molecule with water molecule. While in case of hydrocarbons, hydrogen bonding is not possible.

76. Aldehydes and ketones possess more dipole moment than ethers. Explain.

Ans: Due to presence of double bond more polarity which increases dipole moment.

77. Arrange the following compounds in increasing order of their boiling points: CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃.

Ans: CH₃CH₂OH > CH₃CHO > CH₃OCH₃ > CH₃CH₂CH₃.

78. Why it is necessary to use sulphuric acid in nitration of benzene?

Ans: Sulphuric acid helps in the formation of NO₂⁺ electrophile, which is easy to substitute on benzene.

79. Acetic acid is halogenated in the presence of red P and Cl₂ but formic acid cannot be halogenated in the same way. Give reason.

Ans: There is no alpha hydrogen in formic acid.

80. Carboxylic acid are having more boiling point than alcohols of same molecular mass. Give reason.

Ans: In carboxylic acids, strength of H-bonding is stronger than alcohols and forms dimmer in carboxylic acids which increases boiling point.

81. Carboxylic acid can not give characteristic reactions of aldehydes and ketones though having carbonyl group CO. Why?

Ans: Carbonyl group in carboxylic acid is resonance stabilized as carboxylate ion . Hence it can not give nucleophilic addition reactions given by carbonyl group.

82. Formaldehyde gives Cannizaro’s reaction but acetaldehyde cannot. Give reason.

Ans: Cannizaro’s reaction is given by any aldehydes without alpha hydrogen, hence acetaldehyde cannot give this reaction.

83. Why do aldehydes and ketones have high dipole moment?

Ans: Due to presence of polarisable C=O bond in it.

84. Phenol is acidic but do not react with sodium bicarbonate solution. Give reason.

Ans: Phenol is a weak acid, hence it cannot react with weak base like sodium bicarbonate.

85. Boiling point of glycol is higher than alcohol of same molecular mass. Give reason.

Ans: As the number of –OH group increases strength of H-bond increases and hence boiling point also increases.

86. Nitration of phenol is easier than benzene. Explain.

Ans: Due to presence of OH group in phenol which is electron releasing group, the ring gets activated and thus, electrophilic substitution is easier in case of phenol.

87. Why methyl amine has lower boiling point than methanol?

Ans: Amines possess weak hydrogen bond than alcohols, hence methanol has higher boiling point than methyl amine.

88. Why is it difficult to prepare pure amines by ammonalsis of alkyl halides?

Ans: Reaction of alkyl halides with ammonia gives mixture of amines because secondary and tertiary amines also behave as nucleophilic and undergoes substitution with alkyl halides.

89. Electrophilic substitution in case of aromatic amines takes place more readily than benzene. Explain.

Ans: NH₂ group present on the benzene shows electron releasing effect and it activates the benzene ring. Hence, electrophilic substitution takes place more readily in case of aromatic amines than benzene.

90. Why does silver chloride dissolve in methyl amine solution?

Ans: Methyl amine forms a complex compound with silver chloride and that is soluble in water.

2CH₃NH₂ + AgCl → [Ag(CH₃NH₂)₂]Cl

91. Tertiary amines do not undergo acylation. Explain.

Ans: As these are not having hydrogen on nitrogen, they cannot give acylation reaction with acid chlorides.

92. Why is diazotization of aniline carried in ice cold solution?

Ans: Diazonium salts are stable at cold conditions (0-5 ⁰C) only.

93. Why does amides less basic than amines?

Ans: Lone pair present on nitrogen atom in amides is involved in resonance, so lone pair are not available for protonation, hence, it is less basic than amines.

94. Why is ethylamine more basic than aniline?

Ans: In ethyl amine, lone pair is not involved in resonance, whereas in aniline, lone pair on nitrogen is involved in resonance. Hence, ethyl amine is more basic than aniline.

95. Acid catalyzed dehydration of tert-butanol is faster than n-butanol, why?

Ans: Dehydration is taking place through carbocation mechanism as tertiary carbocation is more stable formed in case of tert-butanol. Hence, acid catalysed dehydration of tert- butanol is faster than n-butanol.

96. Sodium bisulphate is used for purification of aldehydes and ketones, why?

Ans: Aldehydes and ketones react with a saturated solution of NaHSO₃ to form white crystalline precipitates. These products are known as “aldehyde bisulphite” and “ketone bisulphite”. The precipitates are filtered. The precipitates can be decomposed by aqueous HCl or NaOH to regenerate the compounds in pure form.

Check out: What is Substitution Reaction and its types?

97. Hydrazones of acetaldehyde are not prepared in highly acidic medium. Give reason.

Ans: Hydrazones are prepared from hydrazines and it is basic in nature. In highly acidic medium, it forms salt.

98. Why are amides amphoteric in nature?

Ans: Lone pair present in N of amides in involving in resonance and less available for donating, hence behave as weak bases. And during resonance it forms +ve charge on nitrogen which has tendency to lose proton and behave as acid. So, amides are amphoteric in nature.

99. Highly branched carboxylic acids are less acidic than unbranched carboxylic acids .Why?

Ans: As branching increases, electron releasing effect increases, which decreases acidity due to the decrease in polarity of O-H bond.

100. Aniline dissolves in aqueous HCl, why?

Ans: Aniline is basic in nature, hence, it is soluble in acids like HCl.

101. A weakly basic solution favours coupling of benzene diazonium chloride with phenol. Explain.

Ans: As the basic condition favours the formation of phenoxide ion, thus increasing the reactivity of the ring.

102. Why are aryl diazonium ion more stable than alkyl diazonium ion?

Ans: Aryl diazonium ion is stabilized by resonance but in alkyl diazonium ion, no resonance for stabilization.

103. Ethers possess a dipole moment even if the alkyl radicals in the molecule are identical. Explain.

Ans: Due to sp³ hybridisation, ethers posses bent structure, hence net dipole moment is not zero and will not cancel.

104. Why di-tert butyl ether cannot be prepared by Williamson’s synthesis?

Ans: As tertiary carbocation is more stable, it has less tendency to undergo nucleophilic substitution (S_N² reaction occur in Williamson synthesis). Hence it cannot be prepared.

105. Treatment of C₆H₅CHO with HCN gives a mixture of two isomers, which cannot be separated even by very careful fractional distillation. Why?

Ans: This reaction gives racemic mixture of two optical isomers which are having almost all same boiling point and similar chemical properties. Hence they cannot be separated by fractional distillation.

Read more: Discover the Quirky World of 51 Organic Chemistry Reactions!

106. Electrophilic substitution on nitrobenzene is difficult compare with benzene. Explain.

Ans: Nitro group present on the benzene is electron withdrawing and deactivates the ring for substitution. Hence it is difficult.

107. The presence of electron withdrawing groups on benzene facilitates nucleophilic substitution. Explain.

Ans: Due to electron withdrawing effect it will stabilize the carbanion intermediate. Trinitrochlorobenzene+KOH Trinitrophenol

108. Tertiary amines are having low boiling point compare with primary and secondary amines. Explain.

Ans: Due to absence of hydrogen bond in tertiary amines, they possess low boiling point as compared with primary and secondary amines.

109. In solutions, basic strength of amines is secondary>tertiary>primary. Explain.

Ans: Basic strength of amines depends on ammonium cation in solutions. Ammonium stability not only depends on electron releasing effect but also H-bonding, steric factor. Hence, the above order of basic strength is correct.

110. In gas phase, basic strength of amines is tertiary>secondary>primary. Explain.

Ans: In gas phase, there is no solvation effect. It only depends on electron releasing effect. Hence, the above order is correct for basic strength in gas phase.

111. Acylation on amines and phenols takes place directly, whereas acylation on benzene requires AlCl₃ catalyst. Give reasons.

Ans: Reaction of amines and phenol with acid chlorides is nucleophilic substitution, which takes place directly due to presence of lone pair on both oxygen and nitrogen. Whereas acid chloride on reaction with benzene in electrophilic substitution which requires a catalyst like AlCl₃ to form an electrophile.

112. Acid chlorides give pungent smell in air. Explain.

Ans: Acid chlorides undergo hydrolysis with atmospheric moisture and gives HCl fumes, which are pungent in smell.

113. Nitration of benzene gives substantial amount of meta product, although –NH₂ is ortho and para-directing group. Why?

Ans: Nitration takes place in the presence of H⁺ ions which protanates on amine and forms which behaves as electron withdrawing group,vhence it forms some amount of meta product.

114. Before nitration, aniline is acylated. Explain.

Ans: To protect -NH₂ group from oxidation and to prevent the formation of meta product.

115. Reactivity order of carboxylic acid derivatives are Acid chlorides > Anhydrides > Esters > Amides. Explain the reason.

Ans: The order of reactivity of carboxylic acid derivatives is Acid Chlorides > Acid Anhydrides > Esters > Amides. Acid chlorides are most reactive and acid amides are least reactive. This is due to the following basicity order of the leaving group:
Cl⁻< RCOO⁻< RO⁻< NH⁻

116. Explain why cyanides are soluble in water whereas isocyanides are insoluble?

Ans: Cyanides can form hydrogen bond with water whereas isocyanides cannot.

117. Preparation of alkyl halides from alcohols by using SOCl₂ is preferable. Explain.

Ans: When SOCl₂ is used, side products are gases and forms pure alkyl halides.

R-OH + SOCl₂ → R-Cl + SO₂+ HCl

Check out: Important Organic Chemistry Questions for CBSE Board Exam 2024

118. Alkyl halides react with KNO2 give alkyl nitrites whereas AgNO₂ gives nitro alkanes. Explain.

Ans: KNO₂ is an ionic compound and forms NO₂^– which is an ambidentate nucleophile. It can form bond with nitrogen and oxygen. Since C-O bond is stable, it forms alkyl nitrites (R-CH₂-O-N=O). Whereas AgNO₂ is covalent compound, oxygen is not free, bonding takes through nitrogen and forms nitroalkanes (R-CH₂-NO₂).

119. During preparation of esters by reaction of carboxylic acids and alcohols, the ester formed is distilled out as soon as it is formed. Explain.

Ans: Formation of ester is a reversible reaction. To prevent backward reaction,it is distilled out immediately.

120. Electrophilic substitution on benzoic acid takes place at meta position. Give reason.

Ans: Since -COOH group is electron withdrawing group, it acts as meta directing group in resonance, as it creates positive charge on ortho and para position. And electrophile is positive and feel repulsion at ortho and para position. Hence it goes to the meta position.

Some Important Reasoning Questions in Organic Chemistry Asked in CBSE Board

1. Electrophilic substitution in case of amines takes place more readily than benzene. Explain.
2. CH₃CONH₂ is a weaker base than CH₃CH₂NH₂. Explain.
3. Nitro compounds have higher boiling points than hydrocarbons having almost same molecular mass. Explain.
4. o-nitrophenol is more acidic than o-methoxyphenol. Explain.
5. Glycerol is used in cosmetics. Why?
6. Explain the observed Kb order: Et₂NH> Et₃N>EtNH₂, in aqueous solution.
7. Aldehydes are more reactive than ketones towards nucleophilic reaction. Explain.
8. Electrophilic substitution in benzoic acid takes place at meta position. Explain.
9. Why are primary amines higher boiling than tertiary amines?
10. Alkylamines are stronger base than arylamine. Explain.
11. Acetanilide is preferred as good solvent in organic chemistry. Give reason.
12. Carboxylic acid do not gives characteristic reactions of carbonyl group. Why?
13. Treatment of benzaldehyde with HCN gives a mixture of two isomers which can be separated even by fractional distillation. Give reason.
14. NaHSO₃ is used for purification of aldehydes and ketone. Why?
15. How is the basic strength of aromatic amine affected by the presence of an electron releasing group on the benzene ring?
16. How an –OH group attached to a carbon in the benzene ring activates benzene towards electrophilic substitution?
17. Nitrobenzene does not undergo Friedel Crafts reaction. Give reason.
18. Methylamine in water reacts with FeCl₃ to precipitate Fe(OH)₃. Explain.
19. Methylamine is stronger base than ammonia. Give reason.
20. Why are primary amine higher boiling than tertiary amine?
21. Carboxylic acids have much higher acidity than phenol. Explain.
22. In aqueous solution, secondary amines are more basic than tertiary amines. Explain.
23. Amines are more basic than comparable alcohol. Explain.
24. Aromatic amines are less basic than ammonia and aliphatic amines. Explain.
25. Aldehydes are more reactive than ketone towards nucleophilic reaction. Explain.
26. Chloroacetic acid is stronger than acetic acid. Explain.
27. Even in mild condition, aniline on bromination gives 2,4,6-tribromoaniline. Explain.
28. Diazonium ion acts as electrophile. Explain.
29. Nucleophilic substitution of p-nitrochlorobenzene is easier than that of chlorobenzene. Explain.
30. Amines are more basic than comparable alcohol. Explain.
31. It is difficult to prepare pure amines by ammonolysis of R-X. Explain.
32. Electrophilic substitution in case of aromatic amines takes place more readily than in benzene. Explain.
33. In contrast to arenas, aliphatic hydrocarbons do not undergo nitration. Explain.
34. Ethers possess a net dipole moment even if the alkyl radicals in the molecule is identical. Explain.
35. Sodium bisulphate is used for purification of aldehydes and ketones. Explain.
36. Most aromatic acids are solid but the aliphatic acids are liquid. Explain.
37. Aniline is a weaker base than cyclohexyl amine. Explain.
38. Benzoic acid is stronger than acetic acid. Explain.

Do comment down, if you have your own list of questions for addition in this list.

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