Nitro Compounds
Class 12 Chemistry Notes • NEB Syllabus 2083
Syllabus, 15 solved multiple choice questions, the full textbook exercise with solutions and diagrams, handwritten and typed PDF notes and frequently asked questions on nitroalkanes and nitrobenzene.
1. Syllabus
Unit 15: Nitro Compounds
These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Practise the multiple choice questions first, then work through the solved numerical and subjective questions, and finish with the PDF notes for quick revision.
2. Multiple Choice Questions
Question 1. Which of the following is called an ambident group?
- a) Amino
- b) Nitro
- c) Aldehyde
- d) Alcohol
Correct Answer: b) Nitro
Reason: The nitro (nitrite) group can bond through either of two atoms, nitrogen or oxygen. Silver nitrite therefore gives the nitroalkane \(\mathrm{R-NO_2}\) (through N) and the alkyl nitrite \(\mathrm{R-O-N=O}\) (through O). A group with two possible points of attachment is ambident. Amino, aldehyde and alcohol groups have one point of attachment.
Question 2. What type of isomerism is shown by the following structures? \(\mathrm{CH_3CH_2CH_2CH_2NO_2}\) (1-nitrobutane) and \(\mathrm{CH_3CH_2CH(NO_2)CH_3}\) (2-nitrobutane)
- a) Chain isomerism
- b) Position isomerism
- c) Functional isomerism
- d) Metamerism
Correct Answer: b) Position isomerism
Reason: Both have the molecular formula \(\mathrm{C_4H_9NO_2}\) and the same four-carbon chain, but the -NO₂ group is on carbon 1 in one and on carbon 2 in the other. Same chain, different position of the functional group: position isomerism.
Question 3. What would be the product when ethane is heated with concentrated nitric acid at about 430 °C?
- a) Ethyl nitrite is obtained
- b) Ethanoic acid is obtained
- c) Ethanol is obtained
- d) Nitroethane is obtained
Correct Answer: d) Nitroethane is obtained
Reason: Nitration of an alkane in the vapour phase at 400 to 430 °C replaces one hydrogen by the -NO₂ group: \(\mathrm{C_2H_6+HNO_3\xrightarrow{430^\circ C}C_2H_5NO_2+H_2O}\).
Question 4. What would be the product Z in the following reaction? \(\mathrm{CH_3CH_2I+AgNO_2}\) (alcoholic, heat) gives Z + AgI
- a) Ethane
- b) Ethyne
- c) Nitromethane
- d) Nitroethane
Correct Answer: d) Nitroethane
Reason: Silver nitrite is covalent, so the lone pair on nitrogen attacks the alkyl carbon and the nitroalkane is the main product: \(\mathrm{CH_3CH_2I+AgNO_2\xrightarrow{alc.,\ \Delta}CH_3CH_2NO_2+AgI}\). The alkyl nitrite is formed only as the minor product.
Question 5. Study the following table and choose the correct alternative.
| Reactants | Condition / reagent | Major products |
|---|---|---|
| \(\mathrm{CH_3CH_2NO_2}\) | Sn/HCl | P |
| Q | \(\mathrm{AgNO_2}\) / heat | \(\mathrm{CH_3NO_2}\) |
| \(\mathrm{CH_3CH_2NO_2}\) | R | \(\mathrm{CH_3CH_2NHOH}\) |
- a) P – N-ethylethanamine, Q – Ethyl iodide, R – Zn and NH₄Cl
- b) P – Ethanamine, Q – Ethyl iodide, R – Zn and NH₄Cl
- c) P – N-ethylethanamine, Q – Methyl iodide, R – Zn
- d) P – Ethanamine, Q – Methyl iodide, R – Zn and NH₄Cl
Correct Answer: d) P – Ethanamine, Q – Methyl iodide, R – Zn and NH₄Cl
Reason: P: \(\mathrm{CH_3CH_2NO_2+6[H]\xrightarrow{Sn/HCl}CH_3CH_2NH_2+2H_2O}\) (primary amine, ethanamine). Q: \(\mathrm{CH_3I+AgNO_2\longrightarrow CH_3NO_2+AgI}\), so Q is methyl iodide. R: \(\mathrm{CH_3CH_2NO_2+4[H]\xrightarrow{Zn/NH_4Cl}CH_3CH_2NHOH+H_2O}\), so R is zinc with ammonium chloride (neutral medium).
Question 6. Which of the following is a secondary nitroalkane?
- a) Nitropropane
- b) Nitroethane
- c) Nitrobutane
- d) 2-Nitropropane
Correct Answer: d) 2-Nitropropane
Reason: In 2-nitropropane, \(\mathrm{CH_3CH(NO_2)CH_3}\), the carbon carrying the -NO₂ group is joined to two other carbon atoms, so it is a secondary nitroalkane. Nitroethane and 1-nitropropane or 1-nitrobutane (the -NO₂ on the end carbon) are primary.
Question 7. Which of the following is used to convert nitroethane into ethanamine?
- a) Li
- b) Sn
- c) LiAlH₄ / heat
- d) Sn / heat
Correct Answer: c) LiAlH₄ / heat
Reason: Among the given options only the metal hydride \(\mathrm{LiAlH_4}\) (a strong reducing agent) reduces \(\mathrm{-NO_2}\) to \(\mathrm{-NH_2}\): \(\mathrm{CH_3CH_2NO_2\xrightarrow{LiAlH_4,\ \Delta}CH_3CH_2NH_2}\). Tin works only with HCl (Sn/HCl); tin or lithium alone is not a reducing system for this reaction.
Question 8. What could be the product X in the following reaction? \(\mathrm{CH_3CH_2NO_2}\) (LiAlH₄ / heat) gives X
- a) N-ethyl hydroxylamine
- b) Ethanamine
- c) N-methyl hydroxylamine
- d) Propanamine
Correct Answer: b) Ethanamine
Reason: Lithium aluminium hydride reduces the nitro group of nitroalkanes completely to the primary amine with the same carbon skeleton: \(\mathrm{CH_3CH_2NO_2+6[H]\longrightarrow CH_3CH_2NH_2+2H_2O}\). The hydroxylamine forms only in neutral medium (Zn/NH₄Cl).
Question 9. What is the process of addition of a nitro group into a benzene ring called?
- a) Nucleophilic addition
- b) Electrophilic addition
- c) Nucleophilic substitution
- d) Electrophilic substitution
Correct Answer: d) Electrophilic substitution
Reason: The electrophile is the nitronium ion \(\mathrm{NO_2^+}\) (from conc. \(\mathrm{HNO_3}\) and conc. \(\mathrm{H_2SO_4}\)). It attacks the π electrons of the ring and replaces one hydrogen, so the aromatic ring is retained: electrophilic substitution, not addition.
Question 10. What would be the product P in the following reaction sequence? M (conc. HNO₃ / conc. H₂SO₄) gives N; N (LiAlH₄ / heat) gives O; O reacts with benzene diazonium chloride to give P.
Figure: Coupling of benzene diazonium chloride with aniline gives p-aminoazobenzene (the –N=N– group enters para to –NH₂)
- a) Aniline
- b) Phenol
- c) p-Hydroxy diazobenzene
- d) p-Amino diazobenzene
Correct Answer: d) p-Amino diazobenzene
Reason: M is benzene, N is nitrobenzene and O is aniline (the textbook reduction of nitrobenzene to the amine). Aniline couples with benzene diazonium chloride at the para position of the ring, giving p-aminoazobenzene (p-amino diazobenzene): \(\mathrm{C_6H_5N_2Cl+C_6H_5NH_2\longrightarrow H_2NC_6H_4N=NC_6H_5+HCl}\). The coupling partner must be an activated ring such as an amine or phenol; with phenol the product would be p-hydroxy diazobenzene.
Question 11. What are the required conditions for sulphonation of nitrobenzene?
- a) H₂SO₄
- b) H₂SO₄ and heat
- c) Conc. H₂SO₄
- d) Conc. H₂SO₄ and heat
Correct Answer: d) Conc. H₂SO₄ and heat
Reason: The deactivated ring of nitrobenzene needs a strong sulphonating agent and heating: \(\mathrm{C_6H_5NO_2+H_2SO_4\xrightarrow{conc.,\ \Delta}m\text{-}O_2NC_6H_4SO_3H+H_2O}\). The product is m-nitrobenzenesulphonic acid.
Question 12. Which of the following is correct about the information given in the table?
| Reactants | Condition | Major products |
|---|---|---|
| \(\mathrm{C_6H_5NO_2}\) | Sn/HCl | P |
| \(\mathrm{C_6H_5NO_2}\) | Zn/NaOH/Δ | Q |
| \(\mathrm{C_6H_5NO_2}\) | R | \(\mathrm{C_6H_5NHOH}\) |
- a) P – Aniline, Q – Azoxybenzene, R – Zn/NH₄Cl
- b) P – Hydroxylamine, Q – Azoxybenzene, R – Zn/NH₄Cl
- c) P – Aniline, Q – Hydrazobenzene, R – Zn/NH₄Cl
- d) P – Hydroxylamine, Q – Hydrazoxybenzene, R – Zn/NH₄Cl
Correct Answer: c) P – Aniline, Q – Hydrazobenzene, R – Zn/NH₄Cl
Reason: P: \(\mathrm{C_6H_5NO_2+6[H]\xrightarrow{Sn/HCl}C_6H_5NH_2+2H_2O}\) (aniline). Q: zinc with NaOH reduces two molecules to hydrazobenzene, \(\mathrm{2C_6H_5NO_2+10[H]\xrightarrow{Zn/NaOH}C_6H_5NHNHC_6H_5+4H_2O}\). R: \(\mathrm{C_6H_5NO_2+4[H]\xrightarrow{Zn/NH_4Cl}C_6H_5NHOH+H_2O}\) (N-phenylhydroxylamine in neutral medium).
Question 13. What would be the compounds M, N and O in the following reaction sequence? M (conc. HNO₃) gives N; N (Sn/HCl) gives O. M is an aromatic hydrocarbon of molecular mass 78 amu.
- a) M – Ethyne, N – Nitrobenzene, O – Toluene
- b) M – Benzene, N – Nitrobenzene, O – Chlorobenzene
- c) M – Benzene, N – Aniline, O – Toluene
- d) M – Benzene, N – Nitrobenzene, O – Aniline
Correct Answer: d) M – Benzene, N – Nitrobenzene, O – Aniline
Reason: The aromatic hydrocarbon of molecular mass 78 is benzene, \(\mathrm{C_6H_6}\) (6 × 12 + 6 × 1 = 78). \(\mathrm{C_6H_6\xrightarrow{HNO_3/H_2SO_4}C_6H_5NO_2\xrightarrow{Sn/HCl}C_6H_5NH_2}\). Nitration gives nitrobenzene and Sn/HCl reduces it to aniline.
Question 14. What is the nature of the –NO₂ group during an electrophilic substitution reaction?
- a) o-director
- b) m-director
- c) p-director
- d) o and p director
Correct Answer: b) m-director
Reason: The -NO₂ group withdraws electrons by the -I and -R effects, which leaves the ortho and para positions with the most positive character. The electrophile therefore attacks the meta position, where the electron density is highest. The group is a meta director and a ring deactivator.
Question 15. What is the compound Z in the following reaction? Nitrobenzene (Br₂ / FeBr₃) gives Z
Figure: Product Z, m-bromonitrobenzene
- a) o-Bromonitrobenzene
- b) p-Bromonitrobenzene
- c) m-Bromonitrobenzene
- d) Nitrobenzene
Correct Answer: c) m-Bromonitrobenzene
Reason: \(\mathrm{C_6H_5NO_2+Br_2\xrightarrow{FeBr_3}m\text{-}BrC_6H_4NO_2+HBr}\). The -NO₂ group is a meta director, so bromine enters position 3.
3. Numerical and Subjective Questions
Short Answer Questions
Complete the table
| Points to be compared | Nitration | Sulphonation | Halogenation |
|---|---|---|---|
| Reactants | Nitrobenzene and nitric acid | Nitrobenzene and sulphuric acid | Nitrobenzene and bromine (or chlorine) |
| Conditions | Conc. \(\mathrm{HNO_3}\) and conc. \(\mathrm{H_2SO_4}\), 398 K (fuming \(\mathrm{HNO_3}\) for a third nitro group) | Conc. \(\mathrm{H_2SO_4}\) and heat | Anhydrous \(\mathrm{FeBr_3}\) (or \(\mathrm{FeCl_3}\)) |
| Products | m-Dinitrobenzene (1,3,5-trinitrobenzene with fuming acid) | m-Nitrobenzenesulphonic acid | m-Bromonitrobenzene (m-chloronitrobenzene) |
All three are electrophilic substitution reactions. The -NO₂ group is electron-withdrawing, so the ring is deactivated and the incoming group enters the meta position.
Figure: The –NO₂ group is a meta director, so every incoming group enters at position 3
Uses
Identify P, Q, R and S
P. Nitration of benzene with conc. \(\mathrm{HNO_3}\) / conc. \(\mathrm{H_2SO_4}\) gives nitrobenzene. Q. Sn/HCl reduces the nitro group to the amino group: aniline. R. Aniline (a primary amine) with chloroform and KOH gives phenyl isocyanide (carbylamine reaction). S. Catalytic hydrogenation of the isocyanide carbon gives the secondary amine, N-methylaniline.
| Name | Formula | |
|---|---|---|
| P | Nitrobenzene | \(\mathrm{C_6H_5NO_2}\) |
| Q | Aniline | \(\mathrm{C_6H_5NH_2}\) |
| R | Phenyl isocyanide | \(\mathrm{C_6H_5NC}\) |
| S | N-Methylaniline | \(\mathrm{C_6H_5NHCH_3}\) |
Figure: Reaction sequence from benzene to P, Q, R and S
Identify A, B, C, D and E
A. Aniline with \(\mathrm{NaNO_2}\) and HCl at 0 to 5 °C gives benzene diazonium chloride. B. Warming the diazonium salt with water replaces \(\mathrm{-N_2^+}\) by -OH: phenol. C. Distilling phenol with zinc dust removes the -OH group and gives benzene. D. Nitration of benzene gives nitrobenzene. E. Sn/HCl reduces nitrobenzene back to aniline.
| Name | Formula | |
|---|---|---|
| A | Benzene diazonium chloride | \(\mathrm{C_6H_5N_2^+Cl^-}\) |
| B | Phenol | \(\mathrm{C_6H_5OH}\) |
| C | Benzene | \(\mathrm{C_6H_6}\) |
| D | Nitrobenzene | \(\mathrm{C_6H_5NO_2}\) |
| E | Aniline | \(\mathrm{C_6H_5NH_2}\) |
Figure (a): Aniline to A, B and C
Figure (b): C to D and E. The sequence returns to aniline
Convert nitrobenzene into phenol and treat it with ferric chloride
Nitrobenzene is reduced to aniline, aniline is diazotised at 0 to 5 °C, and the diazonium salt is hydrolysed by warm water to phenol. Phenol gives a violet colour with neutral ferric chloride solution (a test for phenol).
Figure: Conversion of nitrobenzene into phenol; phenol then gives a violet colour with neutral FeCl₃
Figure: Phenol gives a violet colour with neutral ferric chloride solution
Complete the table
| Reactants | Conditions | Products |
|---|---|---|
| Benzene | Conc. \(\mathrm{HNO_3}\) / conc. \(\mathrm{H_2SO_4}\), 60 °C | Nitrobenzene |
| Nitrobenzene | \(\mathrm{NaNO_2}\) / dil. HCl at low temperature | No reaction (this reagent diazotises primary aromatic amines, not nitro compounds) |
| Nitrobenzene | Zn / \(\mathrm{CH_3OH}\) (alkaline) / heat | Azobenzene, \(\mathrm{C_6H_5N=NC_6H_5}\) |
| Nitrobenzene | Zn / \(\mathrm{NH_4Cl}\) | N-Phenylhydroxylamine |
| Nitrobenzene | Fuming \(\mathrm{HNO_3}\) / conc. \(\mathrm{H_2SO_4}\), heat | 1,3,5-Trinitrobenzene |
Reactions.
Figure: Reduction of nitrobenzene in alkaline medium. The product depends on the reducing agent
Reduction of nitroethane in neutral medium
In neutral medium (zinc with ammonium chloride, heated) the nitro group is reduced only as far as the hydroxylamine, so nitroethane gives N-ethylhydroxylamine.
Differences
| Catalytic reduction | Electrolytic reduction |
|---|---|
| Hydrogen gas with Ni, Pt or Pd is used. | Reduction is done at the cathode of an electrolytic cell in strongly acidic medium. |
| The product is aniline. | The product is N-phenylhydroxylamine, which rearranges to p-aminophenol. |
| \(\mathrm{C_6H_5NO_2+3H_2\xrightarrow{Ni}C_6H_5NH_2+2H_2O}\) | \(\mathrm{C_6H_5NO_2+4[H]\longrightarrow C_6H_5NHOH+H_2O}\), then rearrangement to \(\mathrm{p\text{-}H_2NC_6H_4OH}\) |
4. View PDF Notes
i. Handwritten Notes
ii. Typed PDF Notes
5. Frequently Asked Questions
What are nitro compounds? What is the functional group of nitro compounds? (Nitro compounds class 12 notes)
How are nitroalkanes named? Give examples and their isomerism.
How are nitroalkanes prepared?
What are the physical properties of nitroalkanes?
How are nitroalkanes reduced? What are the products in acidic, neutral and catalytic reduction?
How is nitrobenzene prepared? Write the mechanism and the physical properties of nitrobenzene.
What are the products of reduction of nitrobenzene in acidic, neutral, alkaline, electrolytic and catalytic conditions?
Why is the nitro group a meta director? Explain the electrophilic substitution of nitrobenzene.
What are the uses of nitro compounds?
Where can I get Nitro compounds class 12 notes pdf, notes pdf and handwritten notes? (Class 12 Chemistry Book PDF Nepal)
Which book is useful for Unit 15 Nitro Compounds? Is there a Pioneer Chemistry class 12 pdf or a CDC book?
What topics are included in Unit 15 Nitro Compounds of NEB Class 12 Chemistry?
Where can I get Class 12 Chemistry notes of NEB (handwritten, PDF, Nepal)?
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