Samacheer Kalvi Class 12 Chemistry Solutions Chapter 13 Organic Nitrogen Compounds

Part - I Text Book Evaluation

I. Choose the Correct Answer

 

Question 1. Which of the following reagent can be used to convert nitrobenzene to aniline
(a) Sn/HCI
(b) ZnHg/NaOH
(c) LiAlH4
(d) All of these
Answer: (a) Sn/HCI
In simple words: To change nitrobenzene into aniline, you need to use a special chemical called Sn/HCI. It helps remove oxygen and add hydrogen, making it an amine.

🎯 Exam Tip: Remember common reducing agents like Sn/HCl, Fe/HCl, and H2/Ni for converting nitro compounds to amines, as this is a frequent reaction in organic chemistry.

 

Question 2. The method by which aniline cannot be prepared is
(a) degradation of benzamide with Br2/NaOH
(b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.
(c) Hydrolysis of phenylcyanide with acidic solution.
(d) reduction of nitrobenzene by Sn/HCI
Answer: (b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.
In simple words: Aniline cannot be made using the Gabriel phthalimide synthesis method when chlorobenzene is used, because aryl halides (like chlorobenzene) do not easily react with the phthalimide salt to form N-aryl phthalimides. This is an important exception to remember for preparing amines.

🎯 Exam Tip: Understand the limitations of specific synthesis methods, especially Gabriel phthalimide synthesis, which is not suitable for preparing aromatic primary amines directly.

 

Question 3. Which one of the following will not undergo Hofmann bromamide reaction
(a) CH3CONHCH3
(b) CH3CH2CONH2
(c) CH3CONH2
(d) C6H5CONH2
Answer: (a) CH3CONHCH3
In simple words: The Hofmann bromamide reaction only works with primary amides (amides with -NH2). CH3CONHCH3 is a secondary amide because it has an alkyl group attached to the nitrogen instead of another hydrogen, so it will not react.

🎯 Exam Tip: Hofmann bromamide degradation is specific to primary amides, converting them into primary amines with one less carbon atom. Always check the amide structure before applying this reaction.

 

Question. Assertion: Acetamide on reaction with KOH and bromine gives acetic acid
Reason: Bromine catalyses hydrolysis of acetamide.
(a) if both assertion and reason are true and reason is the correct explanation of assertion
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer: (d) both assertion and reason are false.
In simple words: The assertion is wrong because acetamide in this reaction makes methylamine, not acetic acid. The reason is also wrong because bromine acts as an oxidant, not a catalyst for hydrolysis.

🎯 Exam Tip: Carefully recall the products of Hofmann bromamide degradation, which always yields an amine with one carbon less than the starting amide, not a carboxylic acid.

 

Question 5. \( \text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{aq NaOH}} \text{A} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{B} \xrightarrow{\text{NH}_3} \text{C} \xrightarrow{\text{Br}_2/\text{NaOH}} \text{D} \)
(a) bromomethane
(b) \( \alpha \)-Bromo sodium acetate
(c) methanamine
(d) acetamide
Answer: (c) methanamine
Solution:
\( \text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{aq NaOH}} \text{CH}_3\text{CH}_2\text{-OH} \xrightarrow{\text{KMnO}_4} \text{CH}_3\text{-COOH} \xrightarrow{\text{NH}_3} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{-NH}_2 \)
In simple words: This chemical process changes bromoethane into ethanol, then to acetic acid, then to acetamide, and finally into methanamine. Each step changes the functional group in a specific way.

🎯 Exam Tip: When analyzing a reaction sequence, identify the reagents and the type of reaction (substitution, oxidation, amidation, degradation) at each step to determine the intermediate and final products.

 

Question 6. Which one of the following nitro compounds does not react with nitrous acid?
(a) CH3-CH2-CH2-N02
(b) (CH3)2 CH-CH2 NO2
(c) (CH3)3CNO2
(d) CH3-C(CH3)(CH-NO2)
Answer: (c) (CH3)3CNO2
Solution:
3° Nitroalkane
In simple words: Tertiary nitroalkanes, like (CH3)3CNO2, do not react with nitrous acid. This is because they lack a hydrogen atom on the carbon that holds the nitro group, which is needed for the reaction to happen.

🎯 Exam Tip: Remember that the reaction of nitro compounds with nitrous acid depends on the presence of \( \alpha \)-hydrogens; tertiary nitroalkanes lack these, so they do not react.

 

Question 7. Aniline + benzoylchloride \( \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{-NH-COC}_6\text{H}_5 \) this reaction is known as
(a) Friedel - Crafts reaction
(b) HVZ reaction
(c) Schotten – Baumann reaction
(d) None of these
Answer: (c) Schotten – Baumann reaction
In simple words: When aniline (an amine) reacts with benzoyl chloride, it forms an N-phenylbenzamide. This specific type of acylation reaction is called the Schotten-Baumann reaction.

🎯 Exam Tip: The Schotten-Baumann reaction is a key method for benzoylation of amines or phenols, typically carried out in a two-phase system (aqueous alkali and an organic solvent).

 

Question 8. The product formed by the reaction of an aldehyde with a primary amine is
(a) carboxylic acid
(b) aromatic acid
(c) Schiff's base
(d) ketone
Answer: (c) Schiff's base
In simple words: When an aldehyde and a primary amine react, they lose a water molecule and form a new compound called a Schiff's base, which contains a carbon-nitrogen double bond. This reaction is also known as imine formation.

🎯 Exam Tip: Schiff's bases are important intermediates in organic synthesis and can be identified by their \( \text{C=N} \) (imine) functional group, formed by the condensation of aldehydes/ketones with primary amines.

 

Question 9. Which of the following reaction is not correct
(a) \( \text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{CH}_2\text{OH}+\text{N}_2 \)
(b) \( \text{(CH}_3\text{)}_2\text{N-} \xrightarrow{\text{NaNO}_2/\text{HC}} \text{(CH}_3\text{)}_2\text{N-} \text{N}=\text{NCl} \)
(c) \( \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{NH}_2 \)
(d) None of these
Answer: (b) \( \text{(CH}_3\text{)}_2\text{N-} \xrightarrow{\text{NaNO}_2/\text{HC}} \text{(CH}_3\text{)}_2\text{N-} \text{N}=\text{NCl} \)
Solution:
p – nitrosation takes places, the product is \( \text{(CH}_3\text{)}_2\text{N-} \text{NO} \)
In simple words: The reaction shown in option (b) is wrong. When a secondary aromatic amine (like N,N-dimethylaniline) reacts with nitrous acid, it undergoes p-nitrosation to form an N-nitrosoamine, not a diazonium salt.

🎯 Exam Tip: Differentiate between the reactions of primary, secondary, and tertiary amines with nitrous acid; secondary amines form N-nitrosoamines, which are yellow oily compounds, while primary aromatic amines form diazonium salts.

 

Question 10. When aniline reacts with acetic anhydride the product formed is
(a) o – amirioacetophenone
(b) m – aminoacetophenone
(c) p – aminoacetophenone
(d) acetanilide
Answer: (d) acetanilide
Solution:
\( \text{C}_6\text{H}_5\text{-NH}_2 + \text{CH}_3\text{-C(=O)-O-C(=O)-CH}_3 \xrightarrow{} \text{C}_6\text{H}_5\text{-NH-C(=O)CH}_3 \)
acetanilide
In simple words: When aniline reacts with acetic anhydride, it undergoes a process called acetylation. This means an acetyl group from the acetic anhydride attaches to the nitrogen of aniline, forming acetanilide.

🎯 Exam Tip: Acetylation is a common method to protect the amino group in aniline, making it less reactive towards electrophilic substitution reactions, and acetanilide is the resulting amide.

 

Question 11. The order of basic strength for methyl substituted amines in aqueous solution is
(a) N(CH3)3> N(CH3)2 H > N(CH3)H2> NH3
(b) N(CH3)H,> N(CH3)2H > N(CH3)3 > NH3
(c) NH3> N(CH3)H2> N(CH3)2H > N(CH3)3
(d) N(CH3)2 H > N(CH3)H2> N(CH3)3 > NH3
Answer: (d) N(CH3)2 H > N(CH3)H2> N(CH3)3 > NH3
In simple words: In water, secondary methyl amines are the strongest bases, followed by primary methyl amines, then tertiary methyl amines, and finally ammonia. This order is due to a combination of electron-donating effect and hydrogen bonding with water.

🎯 Exam Tip: The basic strength of amines in aqueous solution is influenced by inductive effect, solvation effect (hydrogen bonding with water), and steric hindrance. For methyl-substituted amines, the order is usually 2° > 1° > 3° > NH3 due to a balance of these factors.

 

Question 12. `NO2` ring with `Br` -> `NO2` ring with `Br` and `N=N-Cl`. 'A' is
(a) H3PO2 and H2O
(b) H+/H2O
(c) HgSO4 / H2SO4
(d) Cu2Cl2
Answer: (a) H3PO2 and H2O
In simple words: This reaction shows how a diazonium salt, which has the `N=N-Cl` group, can be changed back into just an aromatic ring with a bromine. The chemicals H3PO2 and H2O are used to remove the diazonium group and replace it with hydrogen.

🎯 Exam Tip: \( \text{H}_3\text{PO}_2 \) (hypophosphorous acid) and ethanol are common reagents used for the reduction of diazonium salts to arenes (replacing the diazonium group with hydrogen).

 

Question 13. \( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{A} \xrightarrow{\text{NaNO}_2/\text{HCl}, \text{273K}} \text{B} \xrightarrow{\text{H}_2\text{O}, \text{283K}} \text{C} \) 'C' is
(a) C6H5-OH
(b) C6H5-CH2OH
(c) C6H5-CHO
(d) C6H5NH2
Answer: (a) C6H5-OH
Solution:
\( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{H}_2\text{O}} \text{C}_6\text{H}_5\text{-OH} + \text{N}_2 + \text{HCl} \)
In simple words: Nitrobenzene is first reduced to aniline (A). Then, aniline reacts with nitrous acid to form benzenediazonium chloride (B). Finally, benzenediazonium chloride reacts with water to form phenol (C), releasing nitrogen gas.

🎯 Exam Tip: This sequence demonstrates the preparation of phenol from nitrobenzene, involving reduction to an amine, diazotization to a diazonium salt, and then hydrolysis of the diazonium salt.

 

Question 14. Nitrobenzene on reaction with Con HNO3 / H2SO4 at 80-100°C forms which one of the following products?
(a) 1, 4 – dinitrobenzene
(b) 2, 4, 6 – tirnitrobenzene
(c) 1, 2 – dinitrobenzene
(d) 1, 3 – dinitrobenzene
Answer: (d) 1, 3 -dinitrobenzene
Solution:
NO2 attached to benzene ring + HNO3/H2SO4 at 80-100°C forms 1,3-dinitrobenzene.
In simple words: When nitrobenzene reacts with strong nitric and sulfuric acids at high temperatures, the nitro group already on the ring directs the new nitro group to attach at the meta (1,3) position, forming 1,3-dinitrobenzene.

🎯 Exam Tip: The nitro group is a strong meta-directing and deactivating group in electrophilic aromatic substitution reactions, which means new substituents will attach at the meta position.

 

Question 15. \( \text{C}_5\text{H}_{13}\text{N} \) reacts with \( \text{HNO}_2 \) to give an optically active compound – The compound is
(a) pentan-1 -amine
(b) pentan-2-amine
(c) N, N – dimethyipropan -2-amine
(d) N-methylbutan-2-amine
Answer: (d) N-methylbutan-2-amine
In simple words: N-methylbutan-2-amine is the specific amine that, when reacted with nitrous acid, produces an alcohol that is optically active. This happens because the reaction creates a chiral center.

🎯 Exam Tip: Primary amines react with nitrous acid to form alcohols, which may be optically active if a chiral center is formed. Analyze the structure of the resulting alcohol to confirm chirality.

 

Question 16. Secondary nitro alkanes react with nitrous acid to form
(a) red solution
(b) blue solution
(c) green solution
(d) yellow solution
Answer: (b) blue solution
In simple words: When a secondary nitro alkane mixes with nitrous acid, it creates a "pseudonitrol" compound. This compound then reacts with an alkali to turn into a blue solution, which is a key test to identify secondary nitro alkanes.

🎯 Exam Tip: Remember the color tests for distinguishing primary, secondary, and tertiary nitroalkanes with nitrous acid: primary gives red, secondary gives blue, and tertiary gives no color change (no reaction).

 

Question 17. Which of the following amines does not undergo acetylation?
(a) t-butylamine
(b) ethylamine
(c) diethylamine
(d) triethylamine
Answer: (d) triethylamine (3°amine)
In simple words: Triethylamine is a tertiary amine. It has three alkyl groups attached to the nitrogen atom and no hydrogen atoms on the nitrogen. Acetylation requires a hydrogen atom on the nitrogen to be replaced by an acetyl group, so tertiary amines cannot undergo this reaction.

🎯 Exam Tip: Acetylation occurs when an acetyl group ( \( \text{CH}_3\text{CO-} \) ) replaces a hydrogen atom on a nitrogen or oxygen. Tertiary amines lack the necessary hydrogen on the nitrogen, hence they cannot be acetylated.

 

Question 18. Which one of the following is most basic?
(a) 2, 4 – dichloroaniline
(b) 2, 4 – dimethylaniline
(c) 2, 4 – dinitroaniline
(d) 2, 4 – dibromoaniline
Answer: (b) 2,4-dimethylaniline
Solution: CH3 is a+I group, all other – I group +T group increase the electron density on NH2 and hence increase the basis nature.
In simple words: 2,4-dimethylaniline is the most basic among the options because the two methyl (\( \text{CH}_3 \)) groups are electron-donating, which increases the electron density on the nitrogen atom of the amine and makes it more willing to accept a proton. Other groups like chloro, nitro, and bromo are electron-withdrawing, making the amine less basic.

🎯 Exam Tip: The basicity of anilines is strongly influenced by the electron-donating or electron-withdrawing nature of substituents on the benzene ring. Electron-donating groups increase basicity, while electron-withdrawing groups decrease it.

 

Question 19. When `R-N=O` (where R is a generic alkyl group) is reduced with Sn/HCI the pair of compounds formed are
(a) Ethanol, hydroxylamine hydrochloride
(b) Ethanol, ammonium hydroxide
(c) Ethanol, NH2OH
(d) C3H5NH2, H2O
Answer: (a) Ethanol, hydroxylamine hydrochloride
In simple words: When an alkyl nitrite (like ethyl nitrite, which would be \( \text{CH}_3\text{CH}_2\text{-O-N=O} \)) is reduced with Sn/HCl, it breaks down. It forms an alcohol (like ethanol) and hydroxylamine hydrochloride, which is \( \text{NH}_2\text{OH.HCl} \).

🎯 Exam Tip: Remember that the reduction of alkyl nitrites with strong reducing agents like Sn/HCl cleaves the N-O bond, yielding the corresponding alcohol and hydroxylamine. This is a characteristic reaction to identify nitrites.

 

Question 20. IUPAC name for the amine \( \text{CH}_3\text{-N}(\text{CH}_3)\text{-C}(\text{CH}_3)(\text{C}_2\text{H}_5)\text{-CH}_2\text{-CH}_3 \) is
(a) 3 – Bimethy lamino – 3 – methyl pentane
(b) 3(N,N – Triethyl) – 3 – amino pentane
(c) 3-N,N – trimethyl pentanamine
(d) 3 – (N,N – Dimethyl amino) – 3-methyl pentane
Answer: (d) 3 – (N,N – Dimethyl amino) -3- methyl pentane
In simple words: To name this amine, find the longest carbon chain with the nitrogen substituent, which is a pentane. The nitrogen has two methyl groups (N,N-dimethylamino) and is attached to the third carbon, which also has a methyl group. So it is 3-(N,N-dimethylamino)-3-methylpentane.

🎯 Exam Tip: For complex amines, identify the longest carbon chain containing the nitrogen or the carbon to which the nitrogen is attached. Then, systematically name the substituents on both the nitrogen and the carbon chain according to IUPAC rules, prioritizing the amine group.

 

Question 21. `Benzene-C≡N` with `OCH3` at para position `+ CH3MgBr` \( \xrightarrow{\text{H}_3\text{O}^+} \) P product. 'P' in the above reaction is
(a) `Benzene-CH2OH` with `OCH3` at para position
(b) `Benzene-C(=O)CH3` with `OCH3` at para position
(c) `Benzene-CHO` with `OCH3` at para position
(d) `Benzene-COOH` with `OCH3` at para position
Answer: (b) `Benzene-C(=O)CH3` with `OCH3` at para position
In simple words: When a nitrile, like 4-methoxybenzonitrile, reacts with a Grignard reagent (`CH3MgBr`), and then water, it forms a ketone. The `C≡N` group becomes a `C(=O)CH3` group. This is a useful way to make ketones.

🎯 Exam Tip: Remember that nitriles react with Grignard reagents followed by hydrolysis to yield ketones, where the alkyl group from the Grignard reagent becomes part of the ketone. This reaction is highly regioselective.

 

Question 22. Ammonium salt of benzoic acid is heated strongly with P2O5 and the product so formed is reduced and then treated with NaNO2/ HCl at low temperature. The final compound formed is
(a) Benzene diazonium chloride
(b) Benzyl alcohol
(c) Phenol
(d) Nitrosobenzene
Answer: (b) Benzyl alcohol
Solution:
\( \text{C}_6\text{H}_5\text{COONH}_4 \xrightarrow{\text{P}_2\text{O}_5, \Delta} \text{C}_6\text{H}_5\text{-C}\equiv\text{N} \xrightarrow{\text{LiAlH}_4} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{C}_6\text{H}_5\text{CH}_2\text{OH} \)
In simple words: Starting with ammonium benzoate, heating with \( \text{P}_2\text{O}_5 \) converts it to benzonitrile. Reducing benzonitrile with \( \text{LiAlH}_4 \) gives benzylamine. Finally, treating benzylamine with nitrous acid forms benzyl alcohol.

🎯 Exam Tip: This reaction sequence illustrates how a carboxylic acid derivative can be converted into an alcohol, often through an amine intermediate using diazotization, which replaces the amino group with hydroxyl.

 

Question 23. Identify X in the sequence given below.
`Chlorobenzene-NH2` \( \xrightarrow{\text{CHCl}_3/\text{KOH}} \) (Y) \( \xrightarrow{\text{HCl/(300K)}} \) X + methanoic acid
(a) `Chlorobenzene-NH2`
(b) `Chlorobenzene-C≡N`
(c) `Chlorobenzene-N≡C`
(d) `Chlorobenzene-NH-CH3`
Answer: (c) `Chlorobenzene-N≡C`
In simple words: The first step is a carbylamine reaction, where the primary amine (chlorobenzene-NH2) reacts with chloroform and KOH to form an isocyanide (Y). Then, hydrolysis of this isocyanide (Y) with HCl yields a primary amine (X) and methanoic acid. Therefore, X is the isocyanide before hydrolysis.

🎯 Exam Tip: The carbylamine reaction (also known as isocyanide test) is a distinctive test for primary amines, converting them into foul-smelling isocyanides. Isocyanides can then be hydrolyzed to primary amines and formic acid.

 

Question 24. Among the following, the reaction that proceeds through an electrophilic substitution, is:
(a) `Benzene-N2Cl` \( \xrightarrow{\text{Cu}_2\text{Cl}_2} \) `Benzene-Cl` + N2
(b) `Benzene` \( \xrightarrow{\text{Cl}_2/\text{AlCl}_3} \) `Benzene-Cl` + HCl
(c) `Benzene-CH2OH` \( \xrightarrow{\text{UV light}/\text{Cl}_2} \) `Benzene-CH2Cl` and `Benzene-CCl3`
(d) `Benzene-CH2OH` \( \xrightarrow{\text{Heat}/\text{Delta}} \) `Benzene-CH2Cl` + H2O
Answer: (b) `Benzene` \( \xrightarrow{\text{Cl}_2/\text{AlCl}_3} \) `Benzene-Cl` + HCl
Explanation:
a) Nucleophilic substitution
b) Electrophilic substitution
In simple words: Electrophilic substitution is a reaction where an atom or group on an aromatic ring is replaced by an "electron-loving" species. In option (b), chlorine with \( \text{AlCl}_3 \) acts as an electrophile, replacing a hydrogen on the benzene ring to form chlorobenzene, which is a classic example of electrophilic aromatic substitution.

🎯 Exam Tip: Recognize electrophilic aromatic substitution reactions, which typically involve a Lewis acid catalyst (like \( \text{AlCl}_3 \)) to generate an electrophile, which then attacks the electron-rich aromatic ring.

 

Question 25. The major product of the following reaction
`Benzene-COOH` with `COOH` at ortho position + NH3 \( \xrightarrow{\text{strong heating}} \)
(a) `Benzene-COOH` with `CONH2` at ortho position
(b) `Phthalimide (cyclic amide)`
(c) `Benzene-COOH` with `NH2` at ortho position
(d) `Phthalic acid with two NH2` groups
Answer: (b) `Phthalimide (cyclic amide)`
Solution:
`Phthalic acid` \( \xrightarrow{\text{+ NH}_3} \) `Ammonium phthalate` \( \xrightarrow{\text{Heating}} \) `Phthalamide` \( \xrightarrow{\text{Heating}} \) `Phthalimide`
In simple words: When phthalic acid is heated strongly with ammonia, it first forms an ammonium salt, then a diamide, and finally cyclizes to form phthalimide. Phthalimide is a cyclic amide.

🎯 Exam Tip: Cyclic anhydrides and dicarboxylic acids readily react with ammonia and heat to form cyclic imides, such as phthalimide, which is a stable and important organic intermediate.

 

II. Short Answer Questions

 

Question 1. Write down the possible isomers of the C4H9NO2 and give their IUPAC names
Answer:

In simple words: Isomers are different chemicals that have the same number and types of atoms but are arranged differently. For C4H9NO2, there can be various ways to attach the nitro group or form nitrites, leading to different structural forms, each with its unique name.

🎯 Exam Tip: When writing isomers, consider chain isomerism, positional isomerism for the nitro group, and functional isomerism (nitroalkanes vs. alkyl nitrites). Systematically draw all possible structures to avoid missing any.

 

Question 2. There are two isomers with the formula CH3 NO2. How will you distinguish between them?
Answer:
The formula \( \text{CH}_3\text{NO}_2 \) primarily refers to nitromethane, which exists in two tautomeric forms: the nitro form and the aci-form (also called nitronic acid). These forms can be distinguished by their chemical properties:

• Primary and secondary nitroalkanes with a hydrogen atom on the alpha-carbon exhibit tautomerism, meaning they can switch between the nitro form and the aci form.


• Tertiary nitroalkanes do not show tautomerism because they lack an alpha-hydrogen atom, which is essential for this change.


• Nitromethane (a primary nitroalkane) specifically exists in two tautomeric forms: the nitro form and the aci-form. The aci-form is more acidic.

Here's how to distinguish them based on a general understanding of nitroalkane isomers and their properties (assuming the question might imply a broader context or typo in formula):

In simple words: The two forms of nitromethane (nitro and aci) act differently. The aci-form is more acidic and dissolves quickly in NaOH, while the nitro form is less acidic and dissolves slowly. The aci-form also gives a red-brown color with iron chloride and conducts electricity better.

🎯 Exam Tip: Tautomerism in nitroalkanes involves the migration of a hydrogen atom and a double bond. The acidic nature of the aci-form is key for distinguishing it, often tested using basic solutions or ferric chloride.

 

Question 3. What happens when
(i) 2 - Nitropropane boiled with HCl
(ii). Nitrobenzene electrolytic reduction in strongly acidic medium.
(iii). Oxidation of tert - butylamine with KMnO4
(iv). Oxidation of acetone oxime with trifluoromethoxy acetic acid.
Answer:
(i) When 2-nitropropane is boiled with HCl, it forms acetone, nitrous oxide, and water. This reaction breaks down the nitro compound into a ketone.
\[ \begin{array}{l} \mathrm{CH}_{3}-\underset{\mid}{\mathrm{C}}-\mathrm{NO}_{2} \quad \xrightarrow{\mathrm{HCl} / \mathrm{H}_{2} \mathrm{O}} \mathrm{CH}_{3}-\underset{\|}{\mathrm{C}}=\mathrm{O}+\mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{CH}_{3} \quad \text { (Acetone) (Nitrous oxide)} \end{array} \]
(ii) The electrolytic reduction of nitrobenzene in a strongly acidic medium produces p-aminophenol. This is a special type of reduction where the product depends on the pH of the solution.
\[ \mathrm{NO}_{2} \quad \xrightarrow{\text { Electrolytic reduction }} \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+2 \mathrm{H}_{2} \mathrm{O} \quad \text { (Aniline) } \quad \xrightarrow{4[\mathrm{H}]} \quad \mathrm{NHOH} \quad \text { (Phenylhydroxylamine) } \quad \xrightarrow{+[\mathrm{H}]} \quad \mathrm{NH}_{2} \quad \text { (p-aminophenol)} \]
(iii) The oxidation of tert-butylamine with potassium permanganate (KMnO4) gives 2-methyl-2-nitropropane. This reaction converts the amine group into a nitro group.
\[ \begin{array}{l} \mathrm{CH}_{3}-\underset{\mid}{\mathrm{C}}-\mathrm{NH}_{2} \quad \xrightarrow{\mathrm{KMnO}_{4} / 3[\mathrm{O}]} \quad \mathrm{CH}_{3}-\underset{\mid}{\mathrm{C}}-\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{CH}_{3} \quad \text { (t-butylamine) } \quad \mathrm{CH}_{3} \quad \text { (2-methyl-2-nitropropane)} \end{array} \]
(iv) When acetone oxime is oxidized with trifluoromethoxy acetic acid (CF3COOOH), it forms 2-nitropropane. This is a common method for synthesizing nitro compounds from oximes.
\[ \begin{array}{l} \mathrm{CH}_{3}-\underset{\mid}{\mathrm{C}}=\mathrm{NOH} \quad \xrightarrow{\mathrm{CF}_{3} \mathrm{COOOH}} \quad \mathrm{CH}_{3}-\underset{\mid}{\mathrm{C}}-\mathrm{NO}_{2} \\ \mathrm{CH}_{3} \quad \text { (Acetone oxime) } \quad \mathrm{CH}_{3} \quad \text { (2-nitropropane)} \end{array} \]
In simple words: When these compounds react, they undergo changes like breaking bonds, adding atoms, or changing functional groups. For example, nitropropane turns into acetone, and nitrobenzene can become p-aminophenol.

🎯 Exam Tip: Remember that reaction products can vary significantly with changes in reaction conditions like pH, temperature, and specific reagents used.

 

Question 5. Identify compounds A, B, and C in the following sequence of reactions.
(i) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2} \xrightarrow{\mathrm{Fe} / \mathrm{HCl}} \mathrm{A} \xrightarrow{\mathrm{HNO}_{2}} \mathrm{B} \xrightarrow{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}} \mathrm{C} \)
(ii) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}_{2} \mathrm{Cl} \xrightarrow{\mathrm{CuCN}} \mathrm{A} \xrightarrow{\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}} \mathrm{B} \xrightarrow{\mathrm{NH}_{3}} \mathrm{C} \)
(iii) \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \xrightarrow{\mathrm{NaCN}} \mathrm{A} \xrightarrow{\mathrm{OH}^{-}} \mathrm{B} \xrightarrow{\mathrm{NaOH}+\mathrm{Br}_{2}} \mathrm{C} \)
(iv) \( \mathrm{CH}_{3} \mathrm{NH}_{2} \xrightarrow{\mathrm{CH}_{3} \mathrm{Br}} \mathrm{A} \xrightarrow{\mathrm{CH}_{3} \mathrm{COCl}} \mathrm{B} \xrightarrow{\mathrm{B}_{2} \mathrm{H}_{6}} \mathrm{C} \)
(v) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \xrightarrow{(\mathrm{CH}_{3} \mathrm{CO})_{2} \mathrm{O}} \mathrm{A} \xrightarrow{\mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}} \mathrm{B} \xrightarrow{\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}} \mathrm{C} \)
(vi) Identify B in the following reaction:
\[ \mathrm{N}_{2} \mathrm{Cl}^{-} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{OH}, \mathrm{pH} (9-10)} \mathrm{B} \]
(vii) Identify A, B, C in the following reaction:
\[ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NC} \xrightarrow{\mathrm{HgO}} \mathrm{A} \xrightarrow{\mathrm{H}_{2} \mathrm{O}} \mathrm{B} \xrightarrow{\mathrm{NaNO}_{2} / \mathrm{HCl}, \mathrm{H}_{2} \mathrm{O}} \mathrm{C} \]
Answer:
(i) Here's how the compounds transform:
\[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2} \xrightarrow{\mathrm{Fe} / \mathrm{HCl}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \quad \text{(A) Aniline} \quad \xrightarrow{\mathrm{HNO}_{2}} \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}_{2} \mathrm{Cl} \quad \text{(B) Benzenediazonium Chloride} \quad \xrightarrow{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}} \quad \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{N}=\mathrm{N}-\mathrm{OH} \quad \text{(C) p-hydroxyazobenzene} \]
(ii) Here's how the compounds transform:
\[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}_{2} \mathrm{Cl} \xrightarrow{\mathrm{CuCN}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CN} \quad \text{(A) Phenylcyanide} \quad \xrightarrow{\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}} \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} \quad \text{(B) Benzoic acid} \quad \xrightarrow{\mathrm{NH}_{3}} \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONH}_{4} \quad \text{(C) Ammonium benzoate} \]
(iii) Here's how the compounds transform:
\[ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \xrightarrow{\mathrm{NaCN}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN} \quad \text{(A) Ethylcyanide} \quad \xrightarrow{\mathrm{OH}^{-}, \text {Partial hydrolysis}} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2} \quad \text{(B) Propanamide} \quad \xrightarrow{\mathrm{Br}_{2} / \mathrm{NaOH}, \text{Hoffmann's degradation}} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} \quad \text{(C) Ethylamine} \]
(iv) Here's how the compounds transform:
\[ \mathrm{CH}_{3} \mathrm{NH}_{2} \quad \text{(Methylamine)} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{Br}} \quad \mathrm{CH}_{3} \mathrm{NHCH}_{3} \quad \text{(A) N-Methyl methanamine} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{COCl}} \quad \mathrm{CH}_{3} \mathrm{CON}(\mathrm{CH}_{3})_{2} \quad \text{(B) N, N-dimethylacetamide} \quad \xrightarrow{\mathrm{B}_{2} \mathrm{H}_{6}} \quad \mathrm{CH}_{3}-\underset{\mid}{\mathrm{C}}-\mathrm{N}(\mathrm{CH}_{3})_{2} \quad \text{(C) 1-hydroxy-N, N-dimethylethanamine)} \]
(v) Here's how the compounds transform:
\[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \quad \text{(Aniline)} \quad \xrightarrow{(\mathrm{CH}_{3} \mathrm{CO})_{2} \mathrm{O}, \text{Pyridine}} \quad \mathrm{NHCOCH}_{3} \quad \text{(A) Acetanilide} \quad \xrightarrow{\mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}, 288 \mathrm{~K}} \quad \mathrm{NHCOCH}_{3} \quad \text{(B) p-nitroacetanilide} \quad \xrightarrow{\mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}} \quad \mathrm{NH}_{2} \quad \text{(p-nitroaniline)} \]
(vi) Here's how the compounds transform:
\[ \mathrm{N}_{2} \mathrm{Cl}^{-} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{OH}, \mathrm{pH} (9-10)} \quad \mathrm{N}(\mathrm{CH}_{3})_{2}-\mathrm{N}=\mathrm{N}-\mathrm{N}(\mathrm{CH}_{3})_{2} \quad \text{(A) p-N,N dimethyl amino azobenzene} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{OH}, \mathrm{pH} (4-5)} \quad \mathrm{N}=\mathrm{N}-\mathrm{CH}_{2}-\mathrm{C}_{6} \mathrm{H}_{5} \quad \text{(B) 2-Phenylazo-4-methylphenol} \quad \xrightarrow{\mathrm{pH} (4-5)} \quad \mathrm{NH}_{2} \quad \text{(C) p-aminoazo benzene)} \]
(vii) Here's how the compounds transform:
\[ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NC} \xrightarrow{\mathrm{HgO}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{N=C=O} \quad \text{(A) Ethylisocyanate} \quad \xrightarrow{\mathrm{H}_{2} \mathrm{O}} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} \quad \text{(B) Ethylamine} \quad \xrightarrow{\mathrm{NaNO}_{2} / \mathrm{HCl}, \mathrm{H}_{2} \mathrm{O}} \quad \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \quad \text{(C) Ethylalcohol} \]
In simple words: These are different chemical reactions where one compound changes into others. By following the steps with various reagents and conditions, we can see how the starting material is transformed into different products, labeled A, B, and C.

🎯 Exam Tip: Pay close attention to the specific reagents and conditions (like temperature or solvent) over the reaction arrows, as they dictate the type of transformation and the final product.

 

Question 6. Write short notes on the following
• Hofmann's bromide reaction
• Ammonolysis
• Gabriel phthalimide synthesis
• Schotten - Baumann reaction
• Carbylamine reaction
• Mustard oil reaction
• Coupling reaction
• Diazotisation
• Gomberg reaction
Answer:
I. Hofmann's bromamide reaction:
In this reaction, an acid amide is changed into an amine. The amine produced has one less carbon atom than the original amide. It uses bromine (\( \mathrm{Br}_{2} \)) and potassium hydroxide (\( \mathrm{KOH} \)). This is a useful way to shorten a carbon chain.
\[ \mathrm{CH}_{3}-\underset{\|}{\mathrm{C}}-\mathrm{NH}_{2} \quad \xrightarrow{\mathrm{Br}_{2} / \mathrm{KOH}} \quad \mathrm{CH}_{3}-\mathrm{NH}_{2}+\mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{KBr}+\mathrm{H}_{2} \mathrm{O} \]
II. Ammonolysis:
Ammonolysis happens when alkyl halides are heated with alcoholic ammonia in a sealed tube. This process creates a mix of primary (1°), secondary (2°), and tertiary (3°) amines, along with quaternary ammonium salts. This reaction uses ammonia to replace a halogen atom.
\[ \mathrm{CH}_{3}-\mathrm{Br} \quad \xrightarrow{\mathrm{NH}_{3}} \quad \mathrm{CH}_{3} \mathrm{NH}_{2} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{Br}} \quad \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{Br}} \quad \left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} \quad \xrightarrow{\mathrm{CH}_{3} \mathrm{Br}} \quad \left(\mathrm{CH}_{3}\right)_{4} \mathrm{N}^{+} \mathrm{Br}^{-} \]
III. Gabriel phthalimide synthesis:
In this method, phthalimide is first treated with alcoholic potassium hydroxide (\( \mathrm{KOH} \)) to make potassium phthalimide. Then, this is heated with an alkyl halide, and finally, alkaline hydrolysis gives a primary amine. This method is good for making pure primary amines because it avoids forming secondary or tertiary amines. Aniline, an aromatic primary amine, cannot be made this way.
\[ \begin{array}{l} \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{CO})_{2} \mathrm{NH} \quad \xrightarrow{\text { alcoholic } \mathrm{KOH}} \quad \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{CO})_{2} \mathrm{NK} \quad \xrightarrow{\mathrm{R}-\mathrm{X}} \quad \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{CO})_{2} \mathrm{NR} \quad \xrightarrow{\text { aqueous } \mathrm{KOH}} \quad \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2}+\mathrm{RNH}_{2} \\ \text { (Phthalimide)} \quad \text { (Potassium phthalimide)} \quad \text { (N-alkyl phthalimide)} \quad \text { (Phthalic acid)} \quad \text { (1° amine)} \end{array} \]
IV. Schotten - Baumann reaction:
This reaction is used to prepare N-alkyl benzamides or N-phenylbenzamides. It involves benzoylation, where an amine reacts with benzoyl chloride in the presence of sodium hydroxide (\( \mathrm{NaOH} \)). This is a helpful way to add a benzoyl group to an amine.
\[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{C}_{6} \mathrm{H}_{5}-\underset{\|}{\mathrm{C}}-\mathrm{Cl} \quad \xrightarrow{\text { Pyridine}} \quad \mathrm{C}_{6} \mathrm{H}_{5}-\underset{\|}{\mathrm{C}}-\mathrm{NHC}_{6} \mathrm{H}_{5}+\mathrm{HCl} \\ \text { (Aniline) (Benzoyl chloride)} \quad \text { (N-phenylbenzamide)} \]
V. Carbylamine reaction:
Primary amines react with chloroform (\( \mathrm{CHCl}_{3} \)) and alcoholic potassium hydroxide (\( \mathrm{KOH} \)) to produce isocyanides, also known as carbylamines. These compounds have a very unpleasant smell. This reaction is used as a test to identify primary amines, as secondary and tertiary amines do not show this reaction.
\[ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{CHCl}_{3}+3 \mathrm{KOH} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_{2} \mathrm{O} \]
VI. Mustard oil reaction:
When primary amines react with carbon disulfide (\( \mathrm{CS}_{2} \)), they form N-alkyl dithiocarbamic acid. This acid, when treated with mercuric chloride (\( \mathrm{HgCl}_{2} \)), gives an alkyl isothiocyanate. These isothiocyanates have a smell similar to mustard oil, which gives the reaction its name.
\[ \mathrm{CH}_{3}-\mathrm{N}-\mathrm{H}+\mathrm{C}=\mathrm{S} \rightarrow \mathrm{CH}_{3}-\mathrm{NH}-\underset{\|}{\mathrm{C}}-\mathrm{SH} \quad \xrightarrow{\mathrm{HgCl}_{2}} \quad \mathrm{CH}_{3}-\mathrm{N}=\mathrm{C}=\mathrm{S}+\mathrm{HgS}+2 \mathrm{HCl} \\ \mathrm{H} \quad \text { (N-Methyldithiocarbamic acid)} \quad \text { (Methyl isothiocyanate (Mustard oil smell))} \]
VII. Coupling reaction:
Benzenediazonium chloride reacts with electron-rich aromatic compounds, like phenol or aniline, to form brightly colored azo compounds. This reaction happens through electrophilic substitution, usually at the para position. These colored compounds are often used as dyes.
\[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{Cl}+\mathrm{H}-\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{OH} \quad \xrightarrow{\mathrm{pH}(9-10), 273-278 \mathrm{~K}} \quad \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{N}=\mathrm{N}-\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{OH} \quad \text {(p-hydroxyazobenzene (orange dye))} \]
VIII. Diazotisation:
This is the process where aniline reacts with nitrous acid (\( \mathrm{NaNO}_{2} / \mathrm{HCl} \)) at a low temperature (273-278K) to produce benzene diazonium chloride. This is a very important reaction for making other aromatic compounds.
\[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{NaNO}_{2}+2 \mathrm{HCl} \quad \xrightarrow{273-278 \mathrm{~K}} \quad \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{N}=\mathrm{N}-\mathrm{Cl}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \]
IX. Gomberg reaction:
In the Gomberg reaction, benzenediazonium chloride reacts with benzene in the presence of sodium hydroxide (\( \mathrm{NaOH} \)) to form biphenyl. This is a way to create a carbon-carbon bond between two aromatic rings.
\[ \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{N}=\mathrm{N}-\mathrm{Cl}+\mathrm{H}-\mathrm{C}_{6} \mathrm{H}_{5} \quad \xrightarrow{\mathrm{NaOH}} \quad \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{C}_{6} \mathrm{H}_{5}+\mathrm{N}_{2}+\mathrm{HCl} \]
In simple words: These are different named chemical reactions that help us make new organic compounds or identify specific groups. Each reaction has a special set of ingredients and conditions that lead to a particular product.

🎯 Exam Tip: When writing about named reactions, always mention the reactants, reagents, key conditions (like temperature), and the main product to score full marks.

 

Question 7. How will you distinguish between Primary, Secondary, and tertiary aliphatic amines
Answer:

In simple words: We can tell primary, secondary, and tertiary amines apart by how they react with different chemicals. Some reactions, like with chloroform, only happen with primary amines, while others give different products or no reaction at all with secondary or tertiary amines.

🎯 Exam Tip: Focus on unique reactions for each amine type, such as the carbylamine test for primary amines, which produces a distinct, unpleasant smell.

 

Question 8. Account for the following
• Aniline does not undergo Friedel - Crafts reaction
• Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
• \( \mathrm{pK}_{\mathrm{b}} \) of aniline is more than that of methylamine
• Gabriel phthalimide synthesis is preferred for synthesising-primary amines
Answer:
(i) Aniline does not undergo Friedel - Crafts reaction: Aniline is basic, meaning it has a lone pair of electrons on its nitrogen atom. When aniline reacts with a Lewis acid like anhydrous \( \mathrm{AlCl}_{3} \) (which is used in Friedel-Crafts reactions), it forms an adduct. This adduct prevents further electrophilic substitution reactions from happening. The \( \mathrm{AlCl}_{3} \) acts as a proton acceptor, making the amino group unavailable for other reactions.
(ii) Diazonium salts of aromatic amines are more stable than those of aliphatic amines: Aromatic diazonium salts are more stable because the positive charge on the diazonium group (\( -\mathrm{N}_{2}^{+} \)) is spread out (delocalized) over the benzene ring through resonance. This spreading of charge makes the ion more stable. Aliphatic diazonium salts do not have this resonance stabilization, so they are very unstable and break down quickly.
\[ \mathrm{N} = \mathrm{N}: \quad \mathrm{N}^{+} = \mathrm{N}: \quad \mathrm{N} = \mathrm{N} \]
(iii) \( \mathrm{pK}_{\mathrm{b}} \) of aniline is more than that of methylamine: Aniline is less basic than methylamine. In aniline, the lone pair of electrons on the nitrogen atom is directly attached to the benzene ring. This lone pair gets pulled into the benzene ring through resonance, making it less available to accept a proton. Methylamine, on the other hand, has an electron-donating alkyl group (\( \mathrm{CH}_{3} \)) that pushes electrons towards the nitrogen, making its lone pair more available for protonation. Therefore, aniline's \( \mathrm{pK}_{\mathrm{b}} \) value is higher, meaning it's a weaker base.
(iv) Gabriel phthalimide synthesis is preferred for synthesizing primary amines: This method is highly favored because it only produces primary amines. Secondary and tertiary amines are not formed in this synthesis, which means you get a very pure primary amine without needing extensive purification. This makes it a clean and efficient way to prepare primary amines.
(v) Ethylamine is soluble in water, whereas aniline is not: Ethylamine can form strong intermolecular hydrogen bonds with water molecules, making it soluble. Aniline, however, has a large hydrophobic phenyl (\( \mathrm{C}_{6} \mathrm{H}_{5} \)) group. This large non-polar part reduces its ability to form hydrogen bonds with water significantly, making aniline largely insoluble in water. The size of the non-polar part really affects solubility.
(vi) Amines are more basic than amides: In amides, the carbonyl group (\( \mathrm{C}=\mathrm{O} \)) is highly electronegative and pulls electrons away from the nitrogen atom. This makes the lone pair of electrons on the amide nitrogen less available to accept a proton, reducing its basicity. In amines, alkyl groups are typically electron-donating, pushing electrons towards the nitrogen and making its lone pair more readily available to accept a proton. This electron-donating effect increases the basicity of amines.
(vii) Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline: During the nitration of aniline, strong acidic conditions are used. In acidic solution, the amino group (\( -\mathrm{NH}_{2} \)) gets protonated to form the anilinium ion (\( -\mathrm{NH}_{3}^{+} \)). The anilinium ion is a meta-directing and deactivating group. Therefore, a significant amount of m-nitroaniline is formed, alongside the expected ortho and para products.
In simple words: Aniline acts differently in reactions compared to simpler amines due to its structure. Its electron cloud interacts with the benzene ring, making its nitrogen less available for certain reactions and causing it to behave unusually in others, like nitration.

🎯 Exam Tip: For "account for" questions, always explain the underlying chemical principle or structural feature (like resonance or inductive effects) that causes the observed behavior. This shows a deep understanding of the concepts.

 

Question 9. Arrange the following
(i) In increasing order of solubility in water \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
(ii) In increasing order of basic strength
a) aniline, p - toludine and p - nitroaniline
b) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{P}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} \)
Answer:
(i) Increasing order of solubility in water: \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} < (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} < \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
Aromatic amines like aniline are less soluble than aliphatic amines. Among aliphatic amines, solubility decreases as the size of the alkyl group increases because the hydrophobic part becomes larger. Solubility is generally inversely proportional to molar mass for similar compounds.
Solubility \( \propto \frac{1}{\text{Molar mass}} \)
(ii) Increasing order of basic strength:
a) \( \text{p - nitroaniline} < \text{aniline} < \text{p - toludine} \)
The p-nitro group is electron-withdrawing (-I effect), which decreases the basic strength. The p-methyl group is electron-releasing (+I effect), which increases the basic strength. So, nitroaniline is the weakest, and p-toludine is the strongest.
b) \( \text{P-Cl-C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \)
The p-chloro group has an electron-withdrawing effect, making it less basic than aniline. N-methylaniline (\( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} \)) is more basic than aniline due to the electron-donating methyl group.
(iii) Decreasing order of basic strength in gas phase: \( (\mathrm{C}_{2} \mathrm{H}_{5})_{3} \mathrm{N} > (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} > \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} > \mathrm{NH}_{3} \)
In the gas phase, there is no solvation effect to consider. The basic strength mainly depends on the +I effect (electron-donating effect) of the alkyl groups. The more alkyl groups present, the stronger the +I effect, and thus the stronger the base.
(iv) Increasing order of boiling point: \( (\mathrm{CH}_{3})_{2} \mathrm{NH} < \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \)
Amines generally have lower boiling points compared to alcohols of similar molar mass. This is because amines have weaker hydrogen bonds than alcohols. Among amines, primary amines form stronger hydrogen bonds than secondary amines, and tertiary amines form the weakest (or no) hydrogen bonds.
(v) Decreasing order of \( \mathrm{pK}_{\mathrm{b}} \) values: \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} < \mathrm{CH}_{3} \mathrm{NH}_{2} < (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} \)
A lower \( \mathrm{pK}_{\mathrm{b}} \) value means a stronger base. Ethylamine is generally more basic than methylamine in aqueous solution due to a better balance of inductive and solvation effects. Aniline derivatives are less basic due to resonance.
(vi) Increasing order of basic strength: \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}(\mathrm{CH}_{3})_{2} < \mathrm{CH}_{3} \mathrm{NH}_{2} < (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} \)
Aromatic amines are weaker bases than aliphatic amines. Among aliphatic amines, the order can vary based on the alkyl group. Secondary amines are often more basic than primary or tertiary amines in aqueous solution.
(vii) Decreasing order of basic strength: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} > \mathrm{CH}_{3} \mathrm{NH}_{2} > \mathrm{NH}_{3} > \mathrm{O}_{2} \mathrm{N}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} \)
Ethylamine is generally more basic than methylamine, which is more basic than ammonia. Nitroaniline is much weaker due to the strong electron-withdrawing nitro group.
In simple words: The order of how easily amines dissolve in water or how strong they act as a base depends on their specific structure, like how many carbon chains are attached and if there's a benzene ring. Electron-donating groups make them stronger bases, while electron-withdrawing groups make them weaker.

🎯 Exam Tip: When arranging amines by basicity in aqueous solution, remember the interplay of inductive effect, steric hindrance, and solvation effects. For boiling points, prioritize hydrogen bonding capability.

 

Question 10. How will you prepare propan - 1 - amine from
(i) butane nitrile
(ii) propanamide
(iii) 1-nitropropane
Answer:
(i) From butane nitrile: Butane nitrile can be converted into propan-1-amine through two steps: first, partial hydrolysis to form propanamide, followed by Hoffmann's degradation. This sequence effectively reduces the carbon chain length by one carbon atom.
\[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CN} \quad \xrightarrow{\mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{OH}^{-}, \text{Partial Hydrolysis}} \quad \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\underset{\|}{\mathrm{C}}-\mathrm{NH}_{2} \quad \xrightarrow{\mathrm{Br}_{2} / \mathrm{KOH}, \text{Hoffmann's degradation}} \quad \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} \\ \text { (Butane nitrile)} \quad \text { (Butanamide)} \quad \text { (Propan-1-amine)} \]
(ii) From propanamide: Propanamide can be directly reduced to propan-1-amine using lithium aluminum hydride (\( \mathrm{LiAlH}_{4} \)) followed by hydrolysis. This reaction is a powerful method for converting amides into amines without changing the carbon chain length.
\[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CONH}_{2} \quad \xrightarrow{\text{(i) } \mathrm{LiAlH}_{4}, \text{(ii) } \mathrm{H}_{2} \mathrm{O}} \quad \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} \\ \text { (Propanamide)} \quad \text { (Propan-1-amine)} \]
(iii) From 1-nitropropane: 1-nitropropane can be reduced to propan-1-amine using tin (\( \mathrm{Sn} \)) and hydrochloric acid (\( \mathrm{HCl} \)), which provides the necessary reducing hydrogen atoms. This is a common method for converting nitro compounds into amines.
\[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NO}_{2} \quad \xrightarrow{\mathrm{Sn} / \mathrm{HCl}, 6[\mathrm{H}]} \quad \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \\ \text { (1-nitropropane)} \quad \text { (Propan-1-amine)} \]
In simple words: To make propan-1-amine, you can start from different compounds. For example, butane nitrile needs two steps (one to shorten it, one to change the group), propanamide can be directly changed with a strong reducing agent, and 1-nitropropane simply needs a reduction.

🎯 Exam Tip: Remember specific reagents for reduction (like \( \mathrm{LiAlH}_{4} \) for amides) and for Hoffmann's degradation ( \( \mathrm{Br}_{2} / \mathrm{KOH} \)) as they are key to these transformations.

 

Question 11. Identify A,B,C and D \( \mathrm{CH}_{3}-\mathrm{NO}_{2} \xrightarrow{\mathrm{LiA}/\mathrm{H}_{4}} \mathrm{A} \xrightarrow{2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}} \mathrm{B} \xrightarrow{\mathrm{H}_{2} \mathrm{SO}_{4}} \mathrm{C} \)
Answer:
Here's the reaction sequence with the identified compounds:
\[ \mathrm{CH}_{3}-\mathrm{NO}_{2} \xrightarrow{\mathrm{LiAlH}_{4}} \mathrm{CH}_{3}-\mathrm{NH}_{2} \quad \text{(A)} \quad \xrightarrow{2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}} \quad \left(\mathrm{CH}_{3}-\mathrm{CH}_{2}\right)_{2}-\mathrm{N}-\mathrm{CH}_{3} \quad \text{(B)} \quad \xrightarrow{\mathrm{H}_{2} \mathrm{SO}_{4}} \quad \left(\mathrm{CH}_{3}-\mathrm{CH}_{2}\right)_{2}-\mathrm{N}^{+}(\mathrm{CH}_{3}) \mathrm{HSO}_{4}^{-} \quad \text{(C)} \]

In simple words: Starting with nitromethane, a series of reactions leads to different compounds. First, it's reduced to a primary amine (A), then reacts with ethyl bromide to form a tertiary amine (B), which then forms a salt (C) with sulfuric acid.

🎯 Exam Tip: When identifying compounds in a reaction sequence, carefully consider the type of reagent and its characteristic reaction to predict each intermediate and product correctly.

 

Question 12. How will you convert diethylamine into
(i) N,N - dimethylacetamide
(ii) N - nitrosodiethylamine
Answer:
(i) To convert diethylamine into N,N - dimethylacetamide: Diethylamine can react with acetyl chloride to form N,N-diethylacetamide. This is an acylation reaction where an acetyl group is added to the amine.
\[ \mathrm{CH}_{3} \mathrm{COCl}+(\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} \rightarrow \mathrm{CH}_{3} \mathrm{CON}(\mathrm{C}_{2} \mathrm{H}_{5})_{2}+\mathrm{HCl} \\ \text { (Acetylchloride) (Diethylamine)} \quad \text { (N,N-diethylacetamide)} \]
(ii) To convert diethylamine into N - nitrosodiethylamine: Diethylamine reacts with nitrous acid (\( \mathrm{HON=O} \)), which is typically generated in situ from sodium nitrite (\( \mathrm{NaNO}_{2} \)) and hydrochloric acid (\( \mathrm{HCl} \)), to form N-nitrosodiethylamine. This is a common test for secondary amines.
\[ (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}+\mathrm{HON}=\mathrm{O} \quad \xrightarrow{\mathrm{NaNO}_{2} / \mathrm{HCl}} \quad (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{N}-\mathrm{N}=\mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \text { (Nitrous acid)} \quad \text { (N-nitrosodiethylamine)} \]
In simple words: We can change diethylamine into other compounds by reacting it with different chemicals. For instance, to get N,N-diethylacetamide, we use acetyl chloride, and to make N-nitrosodiethylamine, we use nitrous acid.

🎯 Exam Tip: For conversions, always identify the functional group changes needed and select the appropriate reagents for each step. Remember that nitrous acid reactions are often used to distinguish between primary, secondary, and tertiary amines.

 

Question 13. Identify A, B and C
Answer:

The reaction starts with pentan-1,5-dioic acid. Reacting it with \( \text{SOCl}_2 \) gives glycocyl chloride (A). When glycocyl chloride (A) reacts with \( \text{NH}_3 \), it forms glutamide (B). Finally, reducing glutamide (B) with \( \text{LiAlH}_4 \) yields pentan-1,5-diamine (C). Each step transforms the compound, leading to the final product.
In simple words: First, pentan-1,5-dioic acid turns into glycocyl chloride (A). Then, glycocyl chloride (A) changes to glutamide (B). Finally, glutamide (B) is converted into pentan-1,5-diamine (C).

🎯 Exam Tip: Remember to identify the functional group changes at each step, especially the conversion from carboxylic acid to acyl chloride, then to amide, and finally to amine, as these are common organic reactions.

 

Question 14. Identify A,B,C and D
Answer:

When aniline reacts with benzaldehyde, it forms a Schiff's base (A). This reaction is a condensation reaction, where water is removed. Nitrating aniline (B) with \( \text{Con HNO}_3/\text{H}_2\text{SO}_4 \) can lead to different nitro-substituted products, depending on conditions and substituent effects. The reaction of the Schiff's base with nitrating mixture (B) leads to the formation of p-Nitro benzal aniline (C) and o-Nitro benzal aniline (D). These are isomers that vary in the position of the nitro group on the benzene ring.
In simple words: Aniline and benzaldehyde combine to make a Schiff's base (A). When this base reacts with a nitrating mixture, it forms two main products: p-Nitro benzal aniline (C) and o-Nitro benzal aniline (D), which are similar but have their nitro group in different spots.

🎯 Exam Tip: Remember that aniline reacts with aldehydes to form Schiff's bases, and nitration reactions introduce a nitro group to the aromatic ring, often producing ortho and para isomers. Always consider the directing effects of existing substituents.

 

Question 15. Complete the following reaction
Answer:

The reaction starts with phthalimide reacting with potassium hydroxide (KOH) and ammonia, forming N-isopropylphthalimide (B). This intermediate then reacts with isopropyl bromide (C) and undergoes hydrolysis in acidic conditions \( \text{H}_2\text{O/H}^+\), leading to the final product propanoic acid (D). This process involves several steps of substitution and hydrolysis, demonstrating how a cyclic amide can be transformed into a carboxylic acid through a series of chemical reactions.
In simple words: Phthalimide reacts with KOH and ammonia to become N-isopropylphthalimide (B). This then reacts with isopropyl bromide (C) and water with acid to finally make propanoic acid (D).

🎯 Exam Tip: Pay close attention to the reagents and reaction conditions, especially for hydrolysis, as they determine the final product. Identifying intermediates correctly is key to understanding the full reaction pathway.

 

Question 16. Predict A,B, C and D for the following reaction
Answer:

The given reaction sequence starts with phthalimide (A) reacting with ammonia and heat, then with KOH. This forms an intermediate, which then reacts with isopropyl bromide (B) to form N-isopropyl-1-phthalimide (C). Finally, hydrolysis of (C) with \( \text{H}_2\text{O/H}^+ \) yields phthalic acid (D). This series of steps is a variation of the Gabriel phthalimide synthesis, where an alkyl halide is introduced and then hydrolyzed to form an acid.
In simple words: Phthalimide (A) reacts with ammonia and then KOH. This forms an intermediate that combines with isopropyl bromide (B) to make N-isopropyl-1-phthalimide (C). Adding water and acid to (C) gives phthalic acid (D).

🎯 Exam Tip: The Gabriel phthalimide synthesis is a common method for preparing primary amines, but its modifications can lead to different products like acids or N-substituted phthalimides depending on the final hydrolysis conditions.

 

Question 17. A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of \( \text{CO}_2 \). (B) on heating with liquid ammonia followed by treating with \( \text{Br}_2 \)/\(\text{KOH} \) gives (C) which on treating with \( \text{NaNO}_2 \) and \( \text{HCl} \) at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.
Answer: The monobasic acid (D) has a molecular mass of 74. We can determine its formula using this information. For a monobasic acid, the general formula is \( \text{C}_{\text{n}}\text{H}_{2\text{n}+1}\text{COOH} \). The molar mass is calculated as: \( 12 \times n + (2n+1) \times 1 + 12 + 2 \times 16 + 1 \times 1 = 74 \). This simplifies to: \( 14n + 46 = 74 \). Subtracting 46 from both sides gives: \( 14n = 28 \). Dividing by 14 gives: \( n = 2 \). So, compound (D) is \( \text{C}_2\text{H}_5\text{COOH} \), which is propanoic acid (\( \text{CH}_3\text{CH}_2\text{COOH} \)). This simple calculation helps us identify the final product precisely.

Based on the steps, the compounds are:

In simple words: We find out that compound (D) is propanoic acid by calculating its formula from its molecular weight. Then, working backward through the reactions, we identify (A) as 1,1-dibromopropane, (B) as butanoic acid, and (C) as 1-aminopropane.

🎯 Exam Tip: When dealing with complex synthesis problems, always work backward from the final product's known properties, like molecular mass, to deduce its structure. Then, systematically follow the reaction steps to identify intermediate compounds.

 

Question 18. Identify A to E in the following frequency of reactions
Answer:

The reaction sequence begins with toluene reacting with \( \text{CH}_3\text{Cl} \)/\(\text{AlCl}_3 \), undergoing Friedel-Crafts alkylation. Nitration of toluene with \( \text{HNO}_3 \)/\(\text{H}_2\text{SO}_4 \) yields o-nitro toluene (A) and p-nitro toluene. Then, o-nitro toluene (A) is reduced by \( \text{Sn/HCl} \) to o-amino toluene (B and C). Next, o-amino toluene (C) undergoes diazotization with \( \text{NaNO}_2 \)/\(\text{HCl} \) at low temperature (\( 0^\circ\text{C} \)) to form o-methyl benzene diazonium chloride (D). Finally, the diazonium salt (D) reacts with \( \text{CuCN} \) (Sandmeyer reaction) to give o-cyano toluene (E). This demonstrates a classic series of aromatic transformations.
In simple words: First, toluene is turned into o-nitro toluene (A). Then, o-nitro toluene becomes o-amino toluene (B and C). Next, o-amino toluene changes to o-methyl benzene diazonium chloride (D). Lastly, this diazonium compound forms o-cyano toluene (E).

🎯 Exam Tip: Remember common named reactions like nitration, reduction of nitro groups, diazotization, and the Sandmeyer reaction, as they are fundamental in organic synthesis. Pay attention to the conditions for each step, especially temperature for diazotization.

 

III. Evaluate Yourself

 

Question 1. Write all possible isomers for the following compounds.
(i) \( \text{C}_2\text{H}_5\text{NO}_2 \)
(ii) \( \text{C}_3\text{H}_7-\text{NO}_2 \)
Answer:

Isomers are different compounds that have the same molecular formula but different arrangements of atoms. For \( \text{C}_2\text{H}_5\text{NO}_2 \), the two main functional isomers are nitroethane (where \( \text{NO}_2 \) is directly attached to carbon) and ethyl nitrite (where \( \text{NO}_2 \) is attached through an oxygen atom, \( \text{-O-N=O} \)). For \( \text{C}_3\text{H}_7\text{NO}_2 \), there are position isomers for the nitro group (1-nitropropane and 2-nitropropane) and functional isomers as nitrites (propyl nitrite and isopropyl nitrite). Understanding these arrangements is key to isomerism.
In simple words: \( \text{C}_2\text{H}_5\text{NO}_2 \) can be nitroethane or ethyl nitrite. \( \text{C}_3\text{H}_7\text{NO}_2 \) can be 1-nitropropane, 2-nitropropane, propyl nitrite, or isopropyl nitrite. These are different ways to arrange the same atoms.

🎯 Exam Tip: When writing isomers, always consider both structural (chain and position) and functional group isomerism. For nitro compounds, look for both nitroalkanes (R-NO2) and alkyl nitrites (R-O-N=O) as distinct functional isomers.

 

Question 2. Find out the product of the following reactions.
(i) \( \text{CH}_3\text{CH}(\text{Cl})\text{COOH} \) \( \xrightarrow{\text{(i)NaNO}_2} \) ? [X] \( \xrightarrow{\text{(ii)}\text{H}_2\text{O}/\Delta} \)
(ii) \( \text{CH}_3-\text{CH}_2-\text{Br} + \text{NaNO}_2 \) \( \xrightarrow{\text{alcohol } \Delta} \) [Y]
Answer:
(i) \( \text{CH}_3\text{CH}(\text{Cl})\text{COOH} \) \( \xrightarrow{\text{i) NaNO}_2 \text{ ii) H}_2\text{O}} \)
\( \implies \text{CH}_3-\text{CH}_2-\text{NO}_2 + \text{CO}_2 + \text{NaCl} \)
(x) Nitro ethane

(ii) \( \text{CH}_3-\text{CH}_2-\text{Br} + \text{NaNO}_2 \) \( \xrightarrow{\text{alcohol } \Delta} \)
\( \implies \text{CH}_3-\text{CH}_2-\text{NO}_2 + \text{NaBr} \)
Nitro ethane

In the first reaction, 2-chloropropanoic acid reacts with sodium nitrite and then undergoes hydrolysis with heating. The chlorine atom is replaced by a nitro group, and a decarboxylation reaction happens, where carbon dioxide is released. This results in the formation of nitroethane. In the second reaction, ethyl bromide reacts with sodium nitrite in an alcoholic medium with heat. This is a nucleophilic substitution reaction where the bromine atom is replaced by a nitro group, forming nitroethane. Both reactions lead to the same product, nitroethane, through different pathways.
In simple words: The first reaction turns a chloro-acid into nitroethane by replacing chlorine and removing carbon dioxide. The second reaction turns ethyl bromide into nitroethane by replacing bromine. Both end up making nitroethane.

🎯 Exam Tip: Pay attention to the role of \( \text{NaNO}_2 \) in these reactions; it can act as a source of nitrite ion for nucleophilic substitution, often forming nitroalkanes or alkyl nitrites depending on the conditions and substrate. Decarboxylation is also a key step to recognize when \( \text{CO}_2 \) is liberated.

 

Question 3. Predict the major product that would be obtained on nitration of the following compounds?
(i) Toluene with a carboxylic acid group
(ii) Toluene with a nitro group
(iii) Toluene with a dinitro group
Answer:

When nitrating toluene-4-carboxylic acid with a mixture of concentrated nitric acid and sulfuric acid, the major product is 2-nitro-4-methylbenzoic acid. The methyl group is an ortho/para director and activating, while the carboxylic acid group is a meta director and deactivating. However, the position is already fixed in the starting material. Nitration of nitrobenzene, which already has a deactivating nitro group, leads predominantly to 1,3-dinitrobenzene, as the existing nitro group directs the new nitro group to the meta position. For 1,3,5-trinitrobenzene, further nitration is highly unlikely because the benzene ring is severely deactivated by the three strong electron-withdrawing nitro groups, making it very resistant to electrophilic attack. This shows how existing groups on a benzene ring influence where new groups attach.
In simple words: When you add a nitro group to toluene-4-carboxylic acid, you mostly get 2-nitro-4-methylbenzoic acid. If you add it to nitrobenzene, you mostly get 1,3-dinitrobenzene. But if you try to add it to 1,3,5-trinitrobenzene, it usually won't react because the ring is already too "full" with nitro groups.

🎯 Exam Tip: Always analyze the directing and activating/deactivating effects of existing substituents on an aromatic ring when predicting nitration products. Strongly deactivating groups like \( \text{NO}_2 \) make further electrophilic substitution difficult or impossible.

 

Question 4. Draw the structure of the following compounds
(i) Neopentylamine
(ii) Tert – butylamine
(iii) \( \alpha \)-amino propionaldehyde
(iv) Tribenzylamine
(v) N – ethyl – N – methylhexan – 3- amine
Answer:

These are the structural formulas for the given amine compounds. Neopentylamine and tert-butylamine are primary amines with specific branched alkyl groups. Alpha-amino propionaldehyde is an amino aldehyde. Tribenzylamine is a tertiary amine with three benzyl groups attached to nitrogen, making it quite bulky. N-ethyl-N-methylhexan-3-amine is a tertiary amine with different alkyl substituents on the nitrogen atom and a hexane backbone. Each structure reflects the name precisely.
In simple words: Here are the chemical drawings for each name: Neopentylamine is a specific kind of primary amine, like tert-butylamine. \( \alpha \)-amino propionaldehyde has both an amino group and an aldehyde group. Tribenzylamine has three benzyl groups on the nitrogen. N-ethyl-N-methylhexan-3-amine is a complex amine with ethyl and methyl groups on the nitrogen and a hexane chain.

🎯 Exam Tip: When drawing structures from IUPAC names, always identify the parent chain and the functional group (amine in this case) first. Then, attach substituents in the correct positions, ensuring valency rules (e.g., nitrogen forms three bonds) are followed. For common names like "neopentyl" or "tert-butyl", recall their specific branched structures.

 

Question 5. Give the correct IUPAC names for the following amines
Answer:

The IUPAC naming system provides clear and unambiguous names for chemical compounds based on their structure. For amines, the longest carbon chain containing the amine group is identified as the parent chain, and the position of the amino group is indicated. Substituents on the nitrogen atom are denoted with an 'N-' prefix. For cyclic amines or those with other functional groups like phenols, the amine group's position relative to other groups determines the name. These names help accurately describe each compound.
In simple words: The names describe where the amine group is on the carbon chain and if any other groups are attached to the nitrogen atom. For example, pentan-2-amine means a five-carbon chain with the amine on the second carbon. Cyclohexanamine is a ring with an amine.

🎯 Exam Tip: To correctly name amines, always identify the longest carbon chain containing the nitrogen, number it to give the amine group the lowest possible number, and use 'N-' for any substituents directly on the nitrogen atom. Remember common names like 'aniline' for benzenamine.

 

12th Chemistry Guide Organic Nitrogen Compounds Additional Questions and Answers

 

I. Choose the correct answer

 

Question 1. Nitro ethane and ethyl nitrite are
(a) chain isomers
(b) position isomers
(c) functional isomers
(d) tautomers
Answer: (c) functional isomers
In simple words: Nitroethane and ethyl nitrite have the same chemical formula but different arrangements of atoms and different functional groups. This makes them functional isomers.

🎯 Exam Tip: Functional isomers have the same molecular formula but different functional groups, leading to distinct chemical properties.

 

Question 2. 1-nitrobutane and 2-nitrobutane are
(a) chain isomers
(b) Position isomers
(c) functional isomers
(d) tautomers
Answer: (b) Position isomers
In simple words: These two compounds have the same carbon chain and the same functional group, but the nitro group is attached to a different carbon atom in each. That's why they are position isomers.

🎯 Exam Tip: Position isomers have the same carbon skeleton and functional groups, but the position of the functional group or substituent differs.

 

Question 3. 1-nitrobutane and 2- methyl-1-nitro propane are
(a) chain isomers
(b) Position isomers
(c) funtional isomers
(d) tautomers
Answer: (a) chain isomers
In simple words: These two compounds have the same molecular formula and functional group but differ in the arrangement of the carbon atoms in their main chain. This makes them chain isomers.

🎯 Exam Tip: Chain isomers have different arrangements of carbon atoms in the main chain or skeleton, leading to different branching patterns.

 

Question 4. \( \text{CH}_3-\text{N}=\text{O} \) and \( \text{CH}_2=\text{NOH} \) are
(a) chain isomers
(b) position isomers
(c) functional isomers
(d) tautomers
Answer: (d) tautomers
In simple words: These two compounds can easily change back and forth between each other by moving a hydrogen atom and a double bond. This special type of isomerism is called tautomerism.

🎯 Exam Tip: Tautomerism involves the rapid interconversion of two isomers that differ in the position of a hydrogen atom and a double bond, typically found in compounds with \( \alpha \)-hydrogens adjacent to a carbonyl or nitro group.

 

Question 5. Which among the following is a tertiai nitroalkane?
(a) 2-nitro propane
(b) 2-methyl-.1-nitro propane
(c) 2-methyl – 2-nitropropane
(d) nitro ethane
Answer: (c) 2-methyl – 2-nitropropane
In simple words: A tertiary nitroalkane has the nitro group attached to a carbon atom that is bonded to three other carbon atoms. In 2-methyl-2-nitropropane, the carbon holding the nitro group is indeed bonded to three other carbon atoms.

🎯 Exam Tip: To identify a tertiary nitroalkane, locate the carbon atom directly bonded to the nitro group. If this carbon atom is also bonded to three other carbon atoms (and no hydrogens), it's a tertiary nitroalkane.

 

Question 6. Which among the following does not sho tautomerism?
(a) 2-nitro propane
(b) 2-methyl-1-nitropropane
(c) 2-methyl-2-nitropropane
(d) nitro ethane
Answer: (c) 2-methyl-2-nitropropane
In simple words: Tautomerism needs a hydrogen atom on the carbon next to the nitro group. 2-methyl-2-nitropropane does not have such a hydrogen, so it cannot undergo tautomerism.

🎯 Exam Tip: Tautomerism in nitroalkanes requires the presence of an \( \alpha \)-hydrogen (a hydrogen atom on the carbon atom adjacent to the nitro group). Tertiary nitroalkanes lack \( \alpha \)-hydrogens and therefore do not exhibit tautomerism.

 

Question 7. Which nitro alkane does not dissolve in NaOH?
(a) 2-nitro propane
(b) 2-methyl – 1 – nitro propane
(c) 2 – methyl-2-nitropropane
(d) nitroethane
Answer: (c) 2 – methyl-2-nitropropane
In simple words: Nitroalkanes need to have a hydrogen atom on the carbon next to the nitro group to react with NaOH. 2-methyl-2-nitropropane doesn't have this special hydrogen, so it can't dissolve in NaOH.

🎯 Exam Tip: Primary and secondary nitroalkanes have \( \alpha \)-hydrogens, which makes them acidic enough to react with strong bases like NaOH and dissolve. Tertiary nitroalkanes lack \( \alpha \)-hydrogens and are therefore not acidic and do not react with NaOH.

 

Question 8. Which among the following does not react with Nitrous acid.
(a) 2 – nitro propane
(b) 2-methyl-1-nitropropane
(c) 2 – methyl – 2 – nitro propane
(d) nitro ethane
Answer: (c) 2 – methyl – 2 – nitro propane
In simple words: Nitroalkanes need a hydrogen atom next to the nitro group to react with nitrous acid. Since 2-methyl-2-nitropropane does not have this hydrogen, it cannot react with nitrous acid.

🎯 Exam Tip: The reaction of nitroalkanes with nitrous acid depends on the number of \( \alpha \)-hydrogens. Primary nitroalkanes give nitrolic acids, secondary nitroalkanes give pseudonitrols, while tertiary nitroalkanes (lacking \( \alpha \)-hydrogens) do not react with nitrous acid.

 

Question 9. The correct order of acidic nature of nitro alkanes is
(a) nitro propane> nitroethane > nitro methane
(b) nitro propane \( < \) nitroethane \( < \) nitromethane
(c) nitro methane \( < \) nitro ethane \( < \) nitro propane
(d) nitro methane \( > \) nitro propane
Answer: (b) nitro propane \( < \) nitroethane \( < \) nitromethane
In simple words: The acidity of nitroalkanes decreases as the number of alkyl groups attached to the \( \alpha \)-carbon increases. So, nitromethane is the most acidic, followed by nitroethane, and then nitropropane, because nitromethane has fewer electron-donating alkyl groups that destabilize the conjugate base.

🎯 Exam Tip: The acidity of nitroalkanes is due to the electron-withdrawing effect of the nitro group and the ability to form a resonance-stabilized carbanion (aci-form). Electron-donating alkyl groups attached to the \( \alpha \)-carbon destabilize this carbanion, reducing acidity. Thus, more alkyl groups mean less acidity.

 

Question 10. \( \text{CH}_3-\text{CH}_2-\text{Br} + \text{KNO}_2 \) \( \xrightarrow{\text{Ethanol } / \Delta} \) \( \text{CH}_3-\text{CH}_2-\text{NO}_2 + \text{KBr} \) The above reaction follows the mechanism
(a) E1
(b) E2
(c) \( \text{SN}^1 \)
(d) \( \text{SN}^2 \)
Answer: (d) \( \text{SN}^2 \)
In simple words: This reaction is a nucleophilic substitution where a new group replaces an old one, happening in one step. Since ethyl bromide is a primary alkyl halide and potassium nitrite is a good nucleophile, it favors the \( \text{SN}^2 \) pathway.

🎯 Exam Tip: Primary alkyl halides typically undergo \( \text{SN}^2 \) reactions with strong nucleophiles, especially in polar aprotic or slightly polar protic solvents like ethanol. The reaction is concerted, meaning bond breaking and bond forming occur simultaneously.

 

Question 11. Acetaldoxime is converted into nitroethane by reaction with
(a) \( \text{HNO}_2 \)
(b) \( \text{CF}_3\text{COOOH} \)
(c) \( \text{KMnO}_4 \)
(d) \( \text{HNO}_3 \)/\(\text{H}_2\text{SO}_4 \)
Answer: (b) \( \text{CF}_3\text{COOOH} \)
In simple words: To change acetaldoxime into nitroethane, you use trifluoroperacetic acid (\( \text{CF}_3\text{COOOH} \)). This reagent helps add an oxygen atom in the right place, turning the oxime into a nitro group.

🎯 Exam Tip: Peroxy acids (like trifluoroperacetic acid) are commonly used as oxidizing agents to convert aldoximes into nitroalkanes. This is a useful synthetic transformation in organic chemistry.

 

Question 12. Oil of mirbane is
(a) nitro ethane
(b) nitro propane
(c) nitro benzene
(d) nitro aniline
Answer: (c) nitrobenzene
In simple words: "Oil of mirbane" is the common name for nitrobenzene. It has an almond-like smell and was historically used as a cheaper substitute for almond oil in soaps and perfumes.

🎯 Exam Tip: Remember common names for important organic compounds, as they often appear in questions. Nitrobenzene's characteristic almond-like odor is a distinguishing feature.

 

Question 13. Direct nitration of nitro benzene gives
(a) o-dinitro benzene
(b) m-dinitro benzene
(c) p-dinitro benzene
(d) All of the options
Answer: (b) m-dinitro benzene
In simple words: When you directly add another nitro group to nitrobenzene, the existing nitro group tells the new one to go to the meta position. So, the main product you get is m-dinitrobenzene.

🎯 Exam Tip: The nitro group (\( \text{-NO}_2 \)) is a strong meta-directing and deactivating group in electrophilic aromatic substitution reactions. Therefore, direct nitration of nitrobenzene yields m-dinitrobenzene as the major product.

 

Question 14. P – diamino benzene can be converted to p – dinitro benzene by
(a) Caros acid
(b) persuiphuric acid
(c) peroxy trifluoro acetic acid
(d) All of the options
Answer: (d) All of the options
In simple words: You can turn p-diamino benzene into p-dinitro benzene using any of these strong oxidizing agents: Caro's acid, persulfuric acid, or peroxy trifluoroacetic acid. They all help replace the amino groups with nitro groups.

🎯 Exam Tip: The conversion of amino groups to nitro groups (or other oxidizing processes) typically requires strong oxidizing agents. Caro's acid (\( \text{H}_2\text{SO}_5 \)), persulfuric acid (\( \text{H}_2\text{S}_2\text{O}_8 \)), and peroxy trifluoroacetic acid (\( \text{CF}_3\text{COOOH} \)) are all powerful oxidants capable of this transformation.

 

Question 15. Nef carbonyl synthesis given by
(a) \( \text{C}_6\text{H}_5\text{CHO} \)
(b) \( \text{CH}_3-\text{C}(\text{CH}_3)(\text{NO}_2) \)
(c) \( \text{C}_6\text{H}_5\text{CH}_2\text{NO}_2 \)
(d) All of the options
Answer: (a) \( \text{C}_6\text{H}_5\text{CHO} \)
In simple words: The Nef carbonyl synthesis converts a primary or secondary nitroalkane (after forming its salt) into an aldehyde or ketone. So, \( \text{C}_6\text{H}_5\text{CHO} \) (benzaldehyde) is an example of a product formed by this synthesis.

🎯 Exam Tip: The Nef reaction is a synthetic procedure used to convert primary or secondary nitroalkanes (via their nitronate salts) into aldehydes or ketones. Tertiary nitroalkanes do not undergo this reaction because they lack an \( \alpha \)-hydrogen to form the nitronate salt.

 

Question 16. In electrophilic substititution reaction of nitro benzene, the – \( \text{NO}_2 \) group acts as
(a) activating group and m – directing
(b) deactivating group and m-directing
(c) activating group and o,p – directing
(d) deactivating group and o,p -directing
Answer: (b) deactivating group and m-directing
In simple words: In reactions where new groups are added to nitrobenzene, the \( \text{NO}_2 \) group makes the ring less reactive. It also guides any new groups to attach to the meta position, not the ortho or para positions.

🎯 Exam Tip: Electron-withdrawing groups like \( \text{-NO}_2 \), \( \text{-CHO} \), \( \text{-COOH} \), and \( \text{-CN} \) are typically deactivating and meta-directing in electrophilic aromatic substitution reactions. They pull electron density away from the ring, making it less attractive to electrophiles, and concentrate the remaining electron density at the meta positions.

 

Question 17. Identify X in the following sequence of reaction
Answer: (c) CH3CH2CH2COCl
In simple words: To get the final amine `CH3CH2CH2NH2` through Hofmann degradation, you need to start with an amide that has one more carbon atom. This means the amide should be `CH3CH2CH2CONH2`. This amide can be formed by reacting `CH3CH2CH2COCl` (butanoyl chloride) with ammonia. So, X is `CH3CH2CH2COCl`.

🎯 Exam Tip: Remember that Hofmann bromamide degradation shortens the carbon chain by one atom, meaning the starting amide must have one more carbon than the target amine.

 

Question 18. The reducing agent used to reduce alkylcyanides to primary amines is
(a) H2/Ni
(b) LiAlH4
(c) Na/C2H5OH
(d) All of the options
Answer: (d) All of the options
In simple words: Alkyl cyanides can be changed into primary amines using several strong reducing agents. These include hydrogen gas with a nickel catalyst, lithium aluminum hydride, or sodium in ethanol. All these methods add hydrogen atoms to the cyanide group, turning it into a primary amine group.

🎯 Exam Tip: Familiarize yourself with common reducing agents in organic chemistry, as many functional groups can be reduced by multiple reagents, often with different selectivities.

 

Question 19. The reducing agent used in Mendius reaction is
(a) H2/Ni
(b) LiAlH4
(c) Na/C2H5OH
(d) NaBH4
Answer: (c) Na/C2H5OH
In simple words: The Mendius reaction uses sodium in ethanol to change a nitrile into a primary amine. This reaction is a specific type of reduction that adds hydrogen to the nitrile group. This method is good for making primary amines from nitriles.

🎯 Exam Tip: Associate specific named reactions (like Mendius reaction) with their characteristic reagents and transformations to quickly recall the answer.

 

Question 20. Reduction of alkyl iso cyanides give
(a) primary amines
(b) secondary amines
(c) tertiary amines
(d) None of the options
Answer: (b) secondary amines
In simple words: When you reduce alkyl isocyanides, the nitrogen atom gains an extra hydrogen, and the carbon atom also gains hydrogen. This process forms secondary amines, which have two alkyl groups attached to the nitrogen.

🎯 Exam Tip: Distinguish carefully between the reduction products of alkyl nitriles (primary amines) and alkyl isocyanides (secondary amines).

 

Question 21. A \( \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{B} \). A and B are respectively
(a) CH3NH2, CH3CH2NH2
(b) CH3OH, CH2CH2OH
(c) CH3CH2NH2, CH3NH2
(d) CH3CN, CH3CH2CN
Answer: (c) CH3CH2NH2, CH3NH2
In simple words: Acetamide (`CH3CONH2`) can be changed in two ways. When it is reduced with lithium aluminum hydride (`LiAlH4`), it forms ethylamine (`CH3CH2NH2`). When it is treated with bromine and potassium hydroxide (`Br2/KOH`) in the Hofmann degradation, it forms methylamine (`CH3NH2`), losing one carbon atom. So, A is ethylamine and B is methylamine.

🎯 Exam Tip: Remember that LiAlH4 reduces amides to amines with the same number of carbon atoms, while Hofmann degradation reduces amides to amines with one less carbon atom.

 

Question 22. Gabriel phthalimide synthesis is used for the preparation of
(a) aliphatic primary amines
(b) aromatic primary amines
(c) both (a) & (b)
(d) None of the options
Answer: (a) aliphatic primary amines
In simple words: The Gabriel phthalimide synthesis is a special method used to make primary amines where the nitrogen is connected to an alkyl group. It cannot be used to make primary amines where the nitrogen is directly connected to a benzene ring.

🎯 Exam Tip: Understand the scope and limitations of named reactions; Gabriel phthalimide synthesis is highly selective for aliphatic primary amines due to the SN2 mechanism.

 

Question 23. Which among the following can not be prepared by Gabriel phthalimide synthesis?
(a) Methanamine
(b) ethanamine
(c) benzenamine
(d) propananiine
Answer: (c) benzenamine
In simple words: Gabriel phthalimide synthesis cannot make primary amines like benzenamine (also called aniline) where the amine group is directly attached to a benzene ring. This is because aryl halides do not react well with the phthalimide anion, as they do not undergo the necessary SN2 reaction.

🎯 Exam Tip: Aryl halides (like bromobenzene) are unreactive in SN2 reactions with the phthalimide anion, making this synthesis unsuitable for aromatic primary amines.

 

Question 24. Two molecules of propannitrile in the presence of Na/Ether to form 3-imino- 2-methylpentanenitrile. This reaction is known as
(a) Baltz - schiemann reaction
(b) Thorpe nitrile condensation
(c) Gomberg reaction
(d) hotten - Baurnann reaction
Answer: (b) Thorpe nitrile condensation
In simple words: The Thorpe nitrile condensation is a chemical reaction where two molecules of a nitrile combine together. This reaction happens with a strong base like sodium metal in ether to form a new compound called an iminonitrile. This specific reaction forms `3-imino-2-methylpentanenitrile`.

🎯 Exam Tip: Recognize that "condensation" reactions often involve joining two smaller molecules to form a larger one, typically with the loss of a small molecule like water or in this case, direct addition.

 

Question 25. In Hoffmann's ammonolysis, if excess alkyl halide is taken, the final product is
(a) primary amine
(b) secondary amine
(c) tertiary amine
(d) quaternary ammonium salt
Answer: (d) quaternary ammonium salt
In simple words: If you use a lot of alkyl halide in Hofmann's ammonolysis, the reaction won't stop at making just primary, secondary, or tertiary amines. The amines will keep reacting with the extra alkyl halide until they form a salt where the nitrogen atom is bonded to four alkyl groups, which is called a quaternary ammonium salt.

🎯 Exam Tip: Control the stoichiometry of reactants in ammonolysis; excess alkyl halide always favors the formation of quaternary ammonium salts.

 

Question 26. Hoffmann's ammonolysis reaction is an example of
(a) nucleophilic addition
(b) nucleophilic substitution
(c) electrophilic addition
(d) electrophilic substitution
Answer: (b) nucleophilic substitution
In simple words: In Hofmann's ammonolysis, the ammonia molecule acts as a nucleophile, meaning it has a free pair of electrons to attack another molecule. It replaces the halide atom in the alkyl halide. This type of reaction, where one group replaces another, is called nucleophilic substitution.

🎯 Exam Tip: Classify reactions by their fundamental mechanism; the attack by a lone pair (from ammonia) on an electron-deficient carbon (of the alkyl halide) is a hallmark of nucleophilic substitution.

 

Question 27. In Hoffmann's ammonolysis the order of reactivity of alkylhalides with amines is
(a) RI > RBr > RCl
(b) RI < RBr < RCl
(c) RI > RBr < RCl
(d) RI < RBr> RCl
Answer: (a) RI > RBr > RCl
In simple words: In Hofmann's ammonolysis, the alkyl halide reacts faster when the leaving group is better. Iodine is a larger atom and a better leaving group than bromine, and bromine is a better leaving group than chlorine. So, alkyl iodides (`RI`) react the fastest, followed by alkyl bromides (`RBr`), and then alkyl chlorides (`RCl`) react the slowest.

🎯 Exam Tip: Reactivity of alkyl halides in nucleophilic substitution generally depends on the strength of the C-X bond and the leaving group ability, with iodide being the best leaving group.

 

Question 28. CH3Br \( \xrightarrow{\text{NaN}_3} \) A \( \xrightarrow{\text{LiAlH}_4} \) B. A and B are respectively
(a) CH3Na, CH3CN
(b) CH3NH2,CH3N3
(c) CH3N3 CH3NH2
(d) CH3NH2,CH3CN
Answer: (c) CH3N3 CH3NH2
In simple words: First, methyl bromide (`CH3Br`) reacts with sodium azide (`NaN3`) to form methyl azide (`CH3N3`), so A is methyl azide. Next, methyl azide (`CH3N3`) is reduced by lithium aluminum hydride (`LiAlH4`) to produce methylamine (`CH3NH2`), making B methylamine. This is a common way to make primary amines.

🎯 Exam Tip: Remember that azides (`R-N3`) are readily reduced to primary amines (`R-NH2`) by common reducing agents like LiAlH4.

 

Question 29. The correct ascending order of basic strength of alkylamines in aqueous solution is.
(a) R2NH > RNH2 > R3N > NH3
(b) NH3 > R3N > RNH2 > R2NH
(c) NH3 < R3N < RNH2 < R2NH
(d) R2NH < RNH2 < R3N < NH3
Answer: (c) NH3 < R3N < RNH2 < R2NH
In simple words: In general, for methyl substituted amines in water, the order of increasing basic strength is: ammonia (`NH3`), then tertiary amine (`R3N`), then primary amine (`RNH2`), and finally secondary amine (`R2NH`) which is the strongest. This order is influenced by how well the amines dissolve in water, their electron-donating power, and how much space the groups around the nitrogen take up.

🎯 Exam Tip: The basicity order of amines in aqueous solution is complex, influenced by inductive effect, steric hindrance, and solvation effects, often leading to secondary amines being the most basic.

 

Question 30. Among the three types of amines, secondary amine is more basic due to
(a) +1 effect
(b) steric effect
(c) hydration effect
(d) All of the options
Answer: (d) All of the options
In simple words: Secondary amines are often the most basic because of a good balance of different factors. Alkyl groups have an electron-donating (+I) effect, pushing electrons towards the nitrogen and making it more basic. There's also a hydration effect, where the charged ammonium ion is stabilized by water molecules. Steric effects (how much space the groups take up) can sometimes reduce basicity, but in secondary amines, these factors often combine to make them very basic.

🎯 Exam Tip: Remember that the basicity of amines is a combination of inductive effects, solvation, and steric hindrance, making secondary amines often the most basic due to an optimal balance of these factors.

 

Question 31. In case of substituted aniline, the groups which increase the basic strength are
(i) -CH3
(ii) -NO2
(iii) -OCH3
(iv) -Cl
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (b) (i) & (iii)
In simple words: Groups that donate electrons to the benzene ring will increase the basic strength of aniline, making the nitrogen's lone pair more available. Methyl (`-CH3`) and methoxy (`-OCH3`) groups are both electron-donating. Nitro (`-NO2`) and chloro (`-Cl`) groups are electron-withdrawing, which would decrease basicity.

🎯 Exam Tip: Electron-donating groups increase the basicity of amines, while electron-withdrawing groups decrease it, especially when attached to the aromatic ring.

 

Question 32. The correct order of basic strength of amines is
(a) C6 H5 - NH2 > C6 H5 CH2 NH2> NH3 > C6 H5 CH2 NHCH3 > C2 H5 NH2
(b) C6 H5 CH2 NH2 > C6 H5 CH2 NHCH3 > NH3 > C6 H5 NH2 > C2H5NH2
(c) C2 H5 NH2 > C6H5 CH2 NH2 > NH3 > C6 H5 CH2 NHCH3 > C6 H5 NH2
(d) C6 H5CH2 NH CH3 > C6 H5 CH2 NH2 > NH3 > C2 H5 NH2 > C6 H5 NH2
Answer: (c) C2 H5 NH2 > C6H5 CH2 NH2 > NH3 > C6 H5 CH2 NHCH3 > C6 H5 NH2
In simple words: This order shows how easily each amine can accept a proton. Ethylamine (`C2H5NH2`) is generally the strongest base here because its ethyl group donates electrons. Benzylamine (`C6H5CH2NH2`) comes next, being stronger than plain ammonia (`NH3`). Interestingly, N-methylbenzylamine (`C6H5CH2NHCH3`), a secondary aralkylamine, is shown as less basic than ammonia, possibly due to a combination of steric and inductive effects. Aniline (`C6H5NH2`) is the weakest base because its nitrogen's electron pair is shared with the benzene ring, making it less available.

🎯 Exam Tip: Remember that aliphatic amines are generally more basic than ammonia, and ammonia is more basic than aromatic amines, with electron-donating groups increasing basicity and electron-withdrawing groups decreasing it.

 

Question 33. Schotten - Baumann reaction is an example of
(a) nucleophilic substitution
(b) nucleophilic addition
(c) electrophilic substitution
(d) electropholic addition
Answer: (a) nucleophilic substitution
In simple words: The Schotten-Baumann reaction involves an amine reacting with an acyl chloride or anhydride, usually in the presence of a base. The nitrogen atom of the amine uses its lone pair of electrons to attack the carbon of the acyl group, replacing a chloride ion. This type of replacement, where an electron-rich group attacks, is called nucleophilic substitution.

🎯 Exam Tip: Recognize that acylation reactions of amines, like the Schotten-Baumann reaction, are typically nucleophilic acyl substitutions where the amine acts as the nucleophile.

 

Question 34. The product 'D' of the reaction CH3Cl \( \xrightarrow{\text{KCN}} \) A \( \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \) B \( \xrightarrow{\text{NH}_3} \) C \( \xrightarrow{\Delta} \) D
(a) CH3CH2NH2
(b) CH3CN
(c) HCONH2
(d) CH3CONH2
Answer: (d) CH3CONH2
In simple words: This reaction starts with methyl chloride (`CH3Cl`), which reacts with potassium cyanide (`KCN`) to form methyl cyanide (`CH3CN`), compound A. Then, methyl cyanide is hydrolyzed with acid and water to get acetic acid (`CH3COOH`), compound B. Acetic acid then reacts with ammonia (`NH3`) to form ammonium acetate (`CH3COONH4`), compound C. Finally, heating ammonium acetate (`C`) makes it lose water and turn into acetamide (`CH3CONH2`), which is compound D.

🎯 Exam Tip: Remember the common transformations of nitriles: reduction to amines, and hydrolysis to carboxylic acids, which can then be converted to amides.

 

Question 35. The test used to identify a primary amine is
(a) iodoform test
(b) silver mirror test
(c) libermann's nitroso test
(d) carbylamine test
Answer: (d) carbylamine test
In simple words: The carbylamine test is a very specific way to check for primary amines. When a primary amine is heated with chloroform and alcoholic potassium hydroxide, it forms an isocyanide, which has a very bad and strong smell. This distinct smell confirms the presence of a primary amine.

🎯 Exam Tip: The carbylamine test is highly characteristic for primary amines, so a foul-smelling isocyanide indicates their presence.

 

Question 36. The test used to identify a secondary amine is
(a) iodo form test
(b) silver mirror test
(c) Libermann's nitroso test
(d) carbylamine test
Answer: (c) Libermann's nitroso test
In simple words: The Libermann's nitroso test is used to detect secondary amines. When a secondary amine reacts with nitrous acid, it forms an N-nitrosoamine, which is an oily yellow compound. This oily yellow layer, when warmed with phenol and concentrated sulfuric acid, gives a red solution that turns blue or green when made alkaline, indicating a secondary amine.

🎯 Exam Tip: Associate the characteristic oily yellow nitrosoamine product with the Libermann's nitroso test for secondary amines.

 

Question 37. When ethylamine is treated with chloroform and alcoholic KOH an unpleasant smell is evolved due to the formation of
(a) ethylcyanide
(b) ethylisocyanide
(c) ethylisocyanate
(d) ethylcyanate
Answer: (b) ethylisocyanide
In simple words: This is the carbylamine reaction, a test for primary amines. Ethylamine (`CH3CH2NH2`), a primary amine, reacts with chloroform (`CHCl3`) and a base like alcoholic potassium hydroxide (`KOH`). This reaction produces ethylisocyanide (`CH3CH2NC`), which is known for its extremely strong and unpleasant smell. This distinct odor is how primary amines are identified.

🎯 Exam Tip: The formation of a foul-smelling isocyanide is the key indicator of a positive carbylamine test for primary amines.

 

Question 38. C6H5N + 2Cl- + H3PO2 + H2O → C6H6 + H3PO3 + HCl + N2 This reaction proceeds through
(a) SN¹ mechanism
(b) SN² mechanism
(c) free radical mechanism
(d) elimination reaction
Answer: (c) free radical mechanism
In simple words: This reaction shows the reduction of a diazonium salt (`C6H5N2+Cl-`) to benzene (`C6H6`) using hypophosphorous acid (`H3PO2`). This process happens through a free radical mechanism, where single electrons are involved in breaking and forming bonds. This is a special way to remove the diazonium group.

🎯 Exam Tip: Reductions of diazonium salts with hypophosphorous acid are classic examples of free radical mechanisms in aromatic chemistry.

 

Question 39. Benzene diazonium chloride is converted into chloro benzene by
(i) Sandmeyer reaction
(ii) Gatter mann reaction
(iii) Gomberg reaction
(a) (i) & (ii)
(b) (i.) & (iii)
(c) (ii) & (iii)
(d) (i) (ii) & (iii)
Answer: (a) (i) & (ii)
In simple words: Benzene diazonium chloride can be changed into chlorobenzene using either the Sandmeyer reaction or the Gattermann reaction. Both reactions involve using a copper catalyst to substitute the diazonium group with a chlorine atom. The Gomberg reaction, however, is used to make biphenyls.

🎯 Exam Tip: Differentiate between Sandmeyer and Gattermann reactions, which use different copper reagents (CuCl/HCl vs. Cu/HCl) but achieve the same halogenation of diazonium salts.

 

Question 40. \( \text{C}_6\text{H}_5-\text{N}_2\text{Cl} \xrightarrow{\text{SnCl}_2+\text{HCl}} \text{A} \). The product A is
(a)
(b)
(c) \( \text{NH}_2-\text{NH}_2 \)
(d)
Answer: (b)
In simple words: Benzene diazonium chloride is a diazonium salt. When it is treated with a reducing agent like stannous chloride in hydrochloric acid (`SnCl2/HCl`), it undergoes a reduction reaction. This specific reduction converts the diazonium group into a hydrazine group, forming phenylhydrazine (`C6H5NHNH2`). Phenylhydrazine is a key reagent in organic synthesis.

🎯 Exam Tip: Remember that SnCl2/HCl is a common reagent for reducing diazonium salts to phenylhydrazines.

 

Question 41. Hydrogen cyanide and hydrogen isocyanide are
(a) chain isomers
(b) position isomers
(c) tautomers
(d) Geometrical isomers
Answer: (c) tautomers
In simple words: Tautomers are a type of structural isomer that can quickly change into each other through the movement of a hydrogen atom and a change in chemical bonds. While `HCN` (hydrogen cyanide) and `HNC` (hydrogen isocyanide) are generally considered functional isomers due to different connectivity, some contexts refer to their rapid interconversion as a form of tautomerism, particularly in gas phase. They both have the same molecular formula but different arrangements of atoms.

🎯 Exam Tip: Focus on the definition of tautomerism: isomers that interconvert readily via proton migration, often accompanied by changes in bond positions. For HCN/HNC, this is a subtle point, often classified as functional isomerism.

 

Question 42. The IUPAC name of C6H5CN is
(a) phenyl cyanide
(b) benzene carbonitrile
(c) benzonitrile
(d) benzene cyanide
Answer: (b) benzene carbonitrile
In simple words: The compound `C6H5CN` consists of a benzene ring (`C6H5`) attached to a cyanide group (`-CN`). In IUPAC naming, when a cyanide group is directly linked to a benzene ring, the compound is named as a "carbonitrile" derivative of benzene. So, the systematic name is benzene carbonitrile. It is also commonly known as benzonitrile.

🎯 Exam Tip: Learn the systematic IUPAC nomenclature rules for nitriles, especially when they are attached to aromatic rings, where "carbonitrile" is used.

 

Question 43. A \( \xrightarrow{\text{P}_2\text{O}_5, -\text{H}_2\text{O}} \text{CH}_3-\text{CH}_2-\text{CN} \). Reactant A is
(a) CH3CN
(b) CH3NH2
(c) CH3CONH2
(d) CH3CH2CONH2
Answer: (d) CH3CH2CONH2
In simple words: The reaction shows a compound losing water (`-H2O`) in the presence of phosphorus pentoxide (`P2O5`) to form propanenitrile (`CH3CH2CN`). This is a common method for dehydrating amides to nitriles. Therefore, the starting compound A must be propanamide (`CH3CH2CONH2`). The `P2O5` acts as a dehydrating agent.

🎯 Exam Tip: Remember that P2O5 is a strong dehydrating agent often used to convert amides into nitriles.

 

Question 44. Methyl magnesium chloride is converted into ethanenitrile by treating with
(a) cyanogen
(b) cyanogen chloride
(c) methyl cyanide
(d) hydrogen cyanide
Answer: (b) cyanogen chloride
In simple words: Methyl magnesium chloride (`CH3MgCl`) is a Grignard reagent. To convert it into ethanenitrile (`CH3CN`), you need to add a cyanide group (`-CN`). Cyanogen chloride (`Cl-CN`) is the perfect reagent for this, as the methyl group (`CH3-`) from the Grignard reagent will attack the carbon of the cyanogen chloride, kicking out the chlorine and forming `CH3CN`.

🎯 Exam Tip: Grignard reagents are powerful nucleophiles that react with electrophilic carbon sources, like cyanogen chloride, to form new carbon-carbon bonds.

 

Question 45. Which one of the following compound is a strong base?
(a)
(b)
(c)
(d)
Answer: (c)
In simple words: A stronger base means the nitrogen's lone pair of electrons is more available to accept a proton. Electron-donating groups increase this availability. The methoxy (`-OCH3`) group is a strong electron-donating group, especially when in the para position, as it can donate electrons through resonance, increasing the electron density on the nitrogen and making the amine more basic. Other options have less effective electron-donating groups or electron-withdrawing groups (`-NO2`) which would decrease basicity.

🎯 Exam Tip: Remember that electron-donating groups (like -OCH3 and -CH3) increase the basicity of aniline by stabilizing the conjugate acid and making the lone pair more available, while electron-withdrawing groups (like -NO2) decrease it.

 

Question 46. Which among the following will not undergo diazotisation?
(a) m- toluidine
(b) aniline
(c) p - amino phenol
(d) benzylamine
Answer: (d) benzylamine
In simple words: Diazotisation is a reaction that primarily converts primary aromatic amines into diazonium salts. Aniline, m-toluidine, and p-aminophenol are all primary aromatic amines, so they can undergo this reaction. Benzylamine (`C6H5CH2NH2`) is a primary aralkylamine, meaning its amine group is not directly attached to the benzene ring. Primary aliphatic and aralkylamines react with nitrous acid to form very unstable diazonium salts that immediately decompose, so they do not undergo stable diazotisation.

🎯 Exam Tip: The key difference is that stable diazonium salts are formed only from primary aromatic amines, not from aliphatic or aralkyl primary amines.

 

Question 47. An organic compound (A) C3H9N when treated with nitrous acid, gave an alcohol (B) and N2 gas. (A) on warming with CHCl3 and Caustic potash gave (C) which on reduction gave isopropyl methylamine. Organic compound (A) is
(a) CH3CH2NH CH3
(b) CH3 CH2 CH2 NH2
(c) CH3-CH(CH3)-NH2
(d) CH3-N-(CH3)-CH3
Answer: (c) CH3-CH(CH3)-NH2
In simple words: Compound A has the formula `C3H9N`. Since it reacts with nitrous acid to form an alcohol and nitrogen gas, and it gives a positive carbylamine test (reacts with `CHCl3` and `KOH`), it must be a primary amine. The carbylamine product (C) on reduction forms isopropyl methylamine (`(CH3)2CH-NH-CH3`). This means that (C) was isopropyl isocyanide (`(CH3)2CH-NC`), and therefore, the original primary amine (A) was isopropylamine (`(CH3)2CH-NH2`).

🎯 Exam Tip: Use the sequence of reactions (nitrous acid test, carbylamine test, and reduction of isocyanide) as a detective trail to deduce the structure of the unknown amine, paying close attention to carbon chain changes.

 

Question 48. CH3CN \( \xrightarrow{\text{Na+C}_2\text{H}_5\text{OH}} \) A \( \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4} \) B \( \xrightarrow{\text{HNO}_2} \) C. Identify A, B and C respectively.
(a) Ethanamide, Ethanamine, Ethanol
(b) Ethanamine, Ethanol, Ethanoic acid
(c) Ethanol, Ethanal, Ethanoic acid
(d) Ethanamine, Ethanal, Ethanoic acid
Answer: (b) Ethanamine, Ethanol, Ethanoic acid
In simple words: The reaction starts with methyl cyanide (`CH3CN`). When it is reduced by sodium in ethanol (Mendius reaction), it forms ethanamine (`CH3CH2NH2`), which is compound A. Next, ethanamine (A) reacts with nitrous acid (`HNO2`) to form ethanol (`CH3CH2OH`), which is compound B. Finally, ethanol (B) is oxidized by potassium dichromate in sulfuric acid (`K2Cr2O7/H2SO4`) to yield ethanoic acid (`CH3COOH`), compound C.

🎯 Exam Tip: Understand the sequence of functional group transformations: nitriles reduce to amines, primary aliphatic amines react with nitrous acid to form alcohols, and alcohols can be oxidized to carboxylic acids.

 

Question 49. Which of the following nitro compounds when reacted with nitrous acid followed by treatment with alkali produces blue colour?
(a) 2-methyl-2-nitropropane
(b) 2-methyl-1-nitropropane
(c) 2-nitropropane
(d) nitrobezene
Answer: (c) 2-nitropropane
In simple words: This question refers to the Victor Meyer test, which distinguishes between different types of nitroalkanes. Secondary nitroalkanes, like 2-nitropropane, react with nitrous acid to form pseudonitrols. When these pseudonitrols are treated with an alkali, they turn a blue color. Primary nitroalkanes give a red color, while tertiary nitroalkanes give no color change.

🎯 Exam Tip: Remember the Victor Meyer test colors: primary nitroalkanes (red), secondary nitroalkanes (blue), and tertiary nitroalkanes (no color) after reaction with nitrous acid and alkali.

 

II. Assertion and Reason

 

Question 1. Assertion (A): Aniline when exposed to air becomes coloured. Reason (R): Aniline when exposed to air undergoes oxidation.
(a) Both A and R are correct, R explians A
(b) Both A and R are correct, R does not explain A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer: (a) Both A and R are correct, R explians A
In simple words: The assertion is true because aniline changes color when left out in the open. The reason is also true because aniline reacts with oxygen in the air, which is an oxidation process. This oxidation leads to the formation of colored products, explaining why aniline becomes colored.

🎯 Exam Tip: Recognize that many organic compounds, especially amines, can undergo air oxidation to form colored byproducts, a common observation in the lab.

 

Question 2. Assertion (A): Aniline is more basic than ammonia. Reason (R): The lone pair of electron on nitrogen atom in aniline gets delocalised over the benzene ring.
(a) Both A and R are correct, R explains A
(b) Both A and R are correct, R does not explain A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer: (d) A is wrong but R is correct
In simple words: The assertion is false because aniline is actually less basic than ammonia. This is because the lone pair of electrons on the nitrogen atom in aniline is shared with the benzene ring through resonance. This delocalization makes the lone pair less available to accept a proton, which means aniline is a weaker base compared to ammonia. So, the reason is correct, but the assertion is wrong.

🎯 Exam Tip: Understand that resonance stabilization of the lone pair on nitrogen in aromatic amines reduces its availability for protonation, thus decreasing basicity compared to aliphatic amines or ammonia.

 

Question 3. Assertion (A): Nitroalkanes are more acidic than aldehydes, ketones and cyanides. Reason (R): α-H atom of 1° and 2° nitroalkanes, show acidic character because of the electron releasing effect of NO2 group.
(a) Both A and R are correct, R explains A
(b) Both A and R are correct, R does not explain A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer: (c) A is correct but R is wrong
In simple words: The assertion is true; nitroalkanes are indeed more acidic than aldehydes, ketones, and cyanides because the nitro group helps stabilize the negative charge on the alpha carbon when the hydrogen is removed. However, the reason is false because the nitro (`-NO2`) group is a strong electron-withdrawing group, not electron-releasing. Its electron-withdrawing effect, along with resonance, makes the alpha-hydrogens acidic.

🎯 Exam Tip: Always remember that the nitro group (-NO2) is strongly electron-withdrawing, which enhances the acidity of adjacent protons by stabilizing the conjugate base through inductive and resonance effects.

 

Question 4. Assertion (A): Nitro benzene does not undergo Friedel crafts reactions: Reason (R): NO2 group present in the benzene ring is strongly deactivating in nature.
(a) Both A and R are correct, R explains A
(b) Both A and R are correct, R does not explain A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer: (a) Both A and R are correct, R explains A
In simple words: The assertion is correct: nitrobenzene usually does not participate in Friedel-Crafts reactions. The reason given is also correct: the nitro (`-NO2`) group is very good at pulling electrons away from the benzene ring. This makes the ring much less likely to react in electrophilic substitution reactions like Friedel-Crafts. The reason clearly explains why the assertion is true.

🎯 Exam Tip: Recognize that strongly electron-withdrawing groups like -NO2 deactivate the benzene ring to such an extent that Friedel-Crafts reactions (which are electrophilic substitutions) typically do not occur.

III. Pick out the correct statement

 

Question 1.
(i) In amines the nitrogen atom is \( \text{Sp}^3 \) hybridised.
(ii) In amines, the fourth \( \text{Sp}^3 \) hybridised orbital contains a lone pair of electron.
(iii) Due to the presence of lone pair of electron C- N-C bond angle is more than the normal tetrehedral bond angle.
(iv) The C-N-C bond angle of trimethylamine is 1090 which is greater than tetrahedral angle,
(a) (i) & (ii)
(b) (ii)&(iii)
(c) (iii)&(iv)
(d) (I&(iv)
Answer: (a) (i) & (ii)
In simple words: This question asks to find the correct statements about amines. Statements (i) and (ii) are correct because nitrogen in amines is \( \text{sp}^3 \) hybridized and has a lone pair of electrons. Statements (iii) and (iv) are incorrect regarding bond angles.

🎯 Exam Tip: Remember that the bond angle in amines is usually less than the ideal tetrahedral angle (109.5°) due to the lone pair-bond pair repulsion.

 

Question 2.
(i) Tertiary amines form inter molecular hydrogen bond.
(ii) Amines have higher boiling points than alcohols.
(iii) Nitrogen being less electronegative than oxygen, the N-H bond in amines is less polar than O-H bond in alcohols
(iv) Lower aliphatic amines are soluble in water, since they form hydrogen bond with water,
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (c) (iii) & (iv)
In simple words: The correct statements are that the N-H bond in amines is less polar than the O-H bond in alcohols, and smaller aliphatic amines can dissolve in water because they form hydrogen bonds. Tertiary amines do not form hydrogen bonds. Alcohols generally have higher boiling points than amines of similar molecular weight.

🎯 Exam Tip: When comparing properties like boiling point and solubility, always consider hydrogen bonding, polarity, and molecular weight. Tertiary amines lack an N-H bond, preventing them from forming intermolecular hydrogen bonds.

 

Question 3.
(i) In Hofmann's degradation reaction primary amines with one carbon less than the parent amides are formed.
(ii) Gabriel phthalimide synthesis is useful for the preparation of three types of amines.
(iii) Alkyl halides can be converted to primary amines by treating with sodium azide followed by reduction using \( \text{LiAlH}_4 \)
(iv) When phenol reacts with ammonia in presence of anhydrous \( \text{ZnCl}_2 \) gives p - amino phenol.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (i) & (iii)
(d) (i) & (iv)
Answer: (c) (i) & (iii)
In simple words: Hofmann's degradation makes primary amines with one less carbon. Alkyl halides can become primary amines by first reacting with sodium azide, then being reduced. Gabriel phthalimide synthesis is only for primary amines. Phenol reacting with ammonia and \( \text{ZnCl}_2 \) produces aniline, not p-aminophenol.

🎯 Exam Tip: Remember that Gabriel phthalimide synthesis is a very specific method to prepare *pure primary amines* and does not yield secondary or tertiary amines. Hofmann degradation always shortens the carbon chain by one atom.

 

Question 4.
(i) Alkyl cyanides on reduction with \( \text{LiAlH}_4 \) give secondary amines.
(ii) Alkyl cyanides on hydrolysis with alkali or dilute mineral acid give carboxylic acids.
(iii) Nitriles containing a- H atom undergo condensation with esters in presence of sodamide in ether to form ketonitriles.
(iv) Alkycyanides have lower boiling points than analogous acetylenes
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (i) & (iii)
(d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: Alkyl cyanides turn into carboxylic acids when broken down with acid or alkali. Nitriles with an alpha-hydrogen can react with esters using sodamide to form ketonitriles. Alkyl cyanides usually have higher boiling points than similar acetylenes.

🎯 Exam Tip: Note that reduction of alkyl cyanides with \( \text{LiAlH}_4 \) yields *primary amines*, not secondary amines. Hydrolysis reactions are key for converting nitriles into carboxylic acids.

IV. Pick out the incorrect statement

 

Question 1.
(i) Bromobenzene can be converted into nitrobenzene by reacting with \( \text{KNO}_2 \)
(ii) Nitroethane is suspected to cause genetic damage and be harmful to the nervous system.
(iii) 2-methyl – 2-nitro propane reacts with \( \text{HCl} \) undergoing hydrolysis to form 2-methyl – 2 – propanol.
(iv) Ethyl nitrite on reduction with \( \text{Sn} \) / \( \text{HCl} \) gives ethanol.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (i) & (iii)
(d) (iii) & (iv)
Answer: (c) (i) & (iii)
In simple words: Bromobenzene cannot be directly changed into nitrobenzene with \( \text{KNO}_2 \). Also, 2-methyl-2-nitropropane does not change into 2-methyl-2-propanol when broken down with \( \text{HCl} \). Nitroethane is indeed linked to health risks. Ethyl nitrite does produce ethanol upon reduction.

🎯 Exam Tip: Pay attention to the reactivity of aromatic compounds. Halogenated benzenes typically require more complex reactions (like Ullmann reaction or two-step processes) to introduce nitro groups, not a direct \( \text{KNO}_2 \) reaction.

 

Question 2.
(i) Nitroalkanes exhibit functional isomerism with alkyl nitrites.
(ii) Tertiary nitroalkanes exhibit tautomerism due to the absence of a-H atom.
(iii) The electrical conductivity of aciform of nitro alkanes is high.
(iv) When the number of alkyl group attached to the a- carbon of nitroalkane increases, its acidity increases.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (ii) & (iv)
(d) (i) & (iv)
Answer: (c) (ii) & (iv)
In simple words: Tertiary nitroalkanes do not show tautomerism because they lack an alpha-hydrogen atom. Also, as more alkyl groups are added to the alpha-carbon of a nitroalkane, its acidity actually decreases, not increases. Functional isomerism between nitroalkanes and alkyl nitrites is correct, and the aci-form of nitroalkanes does have high electrical conductivity.

🎯 Exam Tip: Remember that tautomerism in nitroalkanes requires an alpha-hydrogen atom, which is absent in tertiary nitroalkanes. Inductive effects from alkyl groups decrease acidity, so more alkyl groups mean less acidic character.

 

Question 3.
(i) Benzoylation of aniline is known as Schotten. Baumann reaction
(ii) The reaction of ethylamine with nitrous acid is known as diazotisation.
(iii) The reaction of primary amine with nitrous acid is known as Libermann's nitroso test.
(iv) The reaction of primary amine with \( \text{CS}_2 \) and \( \text{HgCl}_2 \) is known as Mustard oil reaction.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (ii) & (iv)
(d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: The reaction of ethylamine with nitrous acid is called diazotization. The reaction of a primary amine with nitrous acid is used as Libermann's nitroso test. Benzoylation of amines is the Schotten-Baumann reaction, and the reaction of primary amines with carbon disulfide and mercuric chloride is the Mustard oil reaction. So, statements (ii) and (iii) are incorrect as they misapply the names.

🎯 Exam Tip: It is crucial to distinguish between reactions involving primary, secondary, and tertiary amines, as their reactions with nitrous acid or other reagents can lead to different products and characteristic tests. Benzoylation is Schotten-Baumann, while the primary amine reaction with \( \text{CS}_2 \) is the Mustard oil reaction.

 

Question 4.
(i) Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(ii) Aryl fluorides and iodides can not be prepared by direct halogenation but can be prepared by using benzene diozonium chloride.
(iii) In Sandmeyer reaction and Gattermann reaction the catalyst used are \( \text{Cu} \)/\( \text{HCI} \) and \( \text{Cu}_2\text{Cl}_2 \) / \( \text{HCZ} \) respectively .
(iv) Benzene diazonium chloride when added with boiling water gives benzene.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (c) (iii) & (iv)
In simple words: In Sandmeyer and Gattermann reactions, the catalysts are \( \text{Cu}_2\text{Cl}_2 \)/\( \text{HCl} \) and \( \text{Cu} \)/\( \text{HCl} \), respectively. When benzene diazonium chloride is mixed with boiling water, it forms phenol. Diazo compounds are indeed used as dyes. Aryl fluorides and iodides are not made by direct halogenation but can be made from benzene diazonium chloride.

🎯 Exam Tip: Remember the specific catalysts for Sandmeyer (\( \text{Cu}_2\text{X}_2 \)/\( \text{HX} \)) and Gattermann (\( \text{Cu} \)/\( \text{HX} \)) reactions. Also, the reaction of diazonium salts with water is a key method for preparing phenols.

V. Match the following

 

Question 1.

Answer:
1. \( \text{Zn} \)/\( \text{NH}_4\text{Cl} \) - Phenylhydroxylamine
2. \( \text{SnCl}_2 \)/ \( \text{KOH} \) - Azobenzene
3. \( \text{Sn} \)/\( \text{HCl} \) - Aniline
4. \( \text{Zn} \)/\( \text{NaOH} \) - Hydrazobenzene
5. \( \text{Fe} \)/ steam - Nitrosobenzene
In simple words: This question asks to match different reducing agents with the products they form from nitrobenzene. Each reducing agent works in a special way to change nitrobenzene into a different chemical.

🎯 Exam Tip: Memorize the specific reducing agents and their corresponding products for nitrobenzene, as this is a common topic in organic chemistry examinations. The acidity or basicity of the medium often dictates the outcome.

 

Question 2.

Answer:
1. \( \text{CH}_2=\text{CH}-\text{CH}_2-\text{NH}_2 \) - Prop-2-en-1-amine
2. \( \text{CH}_3-\text{NH}-\text{CH}(\text{CH}_3)_2 \) - N-methylpropan-2-amine
3. \( \text{H}_2\text{N}-(\text{CH}_2)_6-\text{NH}_2 \) - Hexane-1,6-diamine
4. \( \text{CH}_3\text{CH}_2\text{NH}_2 \) - Ethanamine
5. \( \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2 \) - N,N-dimethylaniline
In simple words: This question asks us to correctly name different amines using IUPAC rules. Each amine has a unique structure, and the IUPAC name describes that structure clearly, telling us about the carbon chain, double bonds, and how the amine group is attached.

🎯 Exam Tip: To score full marks, always identify the longest carbon chain containing the amine group, number it to give the amine group the lowest possible number, and correctly name any substituents. For N-substituted amines, indicate the groups attached to the nitrogen atom.

 

Question 3.

Answer:
1. Schotten Baumann reaction - N-phenylbenzamide
2. Diazotisation - Benzene diazonium chloride
3. Carbylamine test - Methyl isocyanide
4. Mustard oil reaction - Methylisothiocyanate
5. Coupling reaction - p-hydroxyazobenzene
In simple words: This question asks to link different chemical reactions with the products they typically form. For example, the Schotten-Baumann reaction produces an N-phenylbenzamide, while the carbylamine test yields an isocyanide, and the coupling reaction creates an azo compound. Each reaction has a distinct result.

🎯 Exam Tip: Understanding the characteristic products for named reactions and functional group tests is crucial. Pay close attention to the starting materials and reagents, as these often determine the reaction's name and outcome.

 

Question 4.

Answer:
1. Chloropicrin - insecticide
2. Ethyl nitrite - diuretic
3. Nitrobenzene - Solvent in perfume industry
4. Nitro ethane - Fuel additive
5. Nitriles - TNT
In simple words: This question asks to match each chemical compound with its correct use. For example, chloropicrin is used as an insecticide, while ethyl nitrite can act as a diuretic. Nitrobenzene is used as a solvent.

🎯 Exam Tip: Knowing the industrial and practical applications of common organic compounds is important. Focus on the main uses of these compounds to easily match them in exams.

VI. Two Mark Questions

 

Question 1. Write a note on acidic nature of nitroalkanes.
Answer:
Primary and secondary nitroalkanes have an alpha-hydrogen atom, making them acidic. This acidity comes from the electron-withdrawing effect of the \( \text{-NO}_2 \) group, which helps stabilize the negative charge after the alpha-hydrogen is removed. Nitroalkanes are more acidic than aldehydes, ketones, esters, and cyanides. Because they are acidic, nitroalkanes can dissolve in sodium hydroxide ( \( \text{NaOH} \)) to form a salt. The aci-nitro form is more acidic than the nitro form. However, if more alkyl groups are attached to the alpha-carbon, the acidity decreases because alkyl groups release electrons, making the alpha-hydrogen harder to remove.
In simple words: Nitroalkanes are acidic because they have a special hydrogen atom next to the nitro group. The nitro group pulls electrons away, making it easier for this hydrogen to leave. But if there are many bulky groups, it becomes less acidic.

🎯 Exam Tip: The key to understanding nitroalkane acidity is the alpha-hydrogen and the electron-withdrawing nature of the nitro group, which stabilizes the conjugate base. Tertiary nitroalkanes lack an alpha-hydrogen and thus are not acidic.

 

Question 2. How is the oil of mirbane prepared?
Answer:
Oil of mirbane, also known as nitrobenzene, is prepared by nitrating benzene. When benzene is heated to 330K (57°C) with a nitrating mixture, which is a combination of concentrated nitric acid ( \( \text{HNO}_3 \)) and concentrated sulfuric acid ( \( \text{H}_2\text{SO}_4 \)), an electrophilic substitution reaction takes place. During this reaction, a nitro group ( \( \text{-NO}_2 \) ) replaces a hydrogen atom on the benzene ring, forming nitrobenzene. This compound has an almond-like smell.

\[ \begin{array}{l} \text{H} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{NO}_2 \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \text{Benzene} \quad \quad \quad \stackrel{\text{Con HNO}_3 / \text{Con H}_2\text{SO}_4}{\text{330K}} \quad \text{Nitrobenzene} \\ \end{array} \]

In simple words: Oil of mirbane is made by mixing benzene with strong nitric acid and strong sulfuric acid, then heating it. This causes a nitro group to attach to the benzene, creating nitrobenzene.

🎯 Exam Tip: Remember the specific reagents (concentrated nitric and sulfuric acids) and temperature (around 330K) for nitration of benzene, which is a classic electrophilic aromatic substitution reaction.

 

Question 3. How is P – dinitrobenzene prepared?
Answer:
Direct nitration of nitrobenzene usually gives m-dinitrobenzene as the major product. To prepare p-dinitrobenzene, an indirect method is used. This method involves converting p-nitroaniline into p-dinitrobenzene. First, p-nitroaniline undergoes diazotization to form a diazonium salt. This diazonium salt is then treated with fluoroboric acid ( \( \text{HBF}_4 \) ) to form a diazonium fluoroborate. Finally, this compound is reacted with a copper salt (like \( \text{Cu} \)/\( \text{NaNO}_2 \) ) to replace the diazonium group with a nitro group at the para position.

\[ \begin{array}{l} \text{NH}_2 \\ \quad \\ \quad \\ \quad \\ \text{p-nitroaniline} \\ \end{array} \quad \stackrel{\text{NaNO}_2/\text{HCl}}{\longrightarrow} \quad \begin{array}{l} \text{N}_2^+\text{Cl}^- \\ \quad \\ \quad \\ \quad \\ \text{p-nitrobenzene diazonium chloride} \\ \end{array} \quad \stackrel{\text{HBF}_4}{\longrightarrow} \quad \begin{array}{l} \text{N}_2^+\text{BF}_4^- \\ \quad \\ \quad \\ \quad \\ \text{p-nitrobenzene diazonium fluoroborate} \\ \end{array} \quad \stackrel{\text{Cu}/\text{NaNO}_2}{\longrightarrow} \quad \begin{array}{l} \text{NO}_2 \\ \quad \\ \quad \\ \quad \\ \text{p-dinitrobenzene} \\ \end{array} \]

In simple words: To make p-dinitrobenzene, we start with p-nitroaniline. First, we change it into a diazonium salt, then into a fluoroborate salt. After that, we use a copper reagent to put a second nitro group in the para position, making the final product. Direct nitration does not produce this.

🎯 Exam Tip: Remember that the strong activating nature of the amino group in aniline makes it difficult to control nitration to get a para product. Therefore, indirect methods involving protection or multi-step synthesis are often required.

 

Question 4. Convert p-diamino benzene into p-dinitro benzene.
Answer:
To convert p-diamino benzene into p-dinitro benzene, the amino groups need to be replaced with nitro groups. This can be achieved using strong oxidizing agents such as Caro's acid ( \( \text{H}_2\text{SO}_5 \) ), persulphuric acid ( \( \text{H}_2\text{S}_2\text{O}_8 \) ), or peroxytrifluoroacetic acid ( \( \text{CF}_3\text{COOOH} \) ). These reagents can directly convert an amino group ( \( \text{-NH}_2 \) ) into a nitro group ( \( \text{-NO}_2 \) ) through an oxidation process.

\[ \begin{array}{l} \quad \text{NH}_2 \\ \quad \quad \\ \quad \quad \\ \quad \quad \\ \text{p-diaminobenzene} \\ \end{array} \quad \stackrel{\text{CF}_3\text{COOOH}}{\longrightarrow} \quad \begin{array}{l} \text{NO}_2 \\ \quad \\ \quad \\ \quad \\ \text{p-dinitrobenzene} \\ \end{array} \]

In simple words: To change p-diamino benzene into p-dinitro benzene, you use a strong chemical that adds oxygen, like Caro's acid. This chemical replaces the amino groups with nitro groups.

🎯 Exam Tip: This conversion is a specific oxidation reaction. Always identify the functional groups being transformed and select appropriate oxidizing agents for replacing amino groups with nitro groups in aromatic systems.

 

Question 5. Write the various reduction stages of nitromethane.
Answer:
Nitromethane can be reduced in several stages depending on the reducing agent used. The reduction typically proceeds as follows:
1. Nitromethane ( \( \text{CH}_3\text{NO}_2 \) ) is first reduced to nitrosomethane ( \( \text{CH}_3\text{N=O} \) ).
2. Nitrosomethane is further reduced to N-methylhydroxylamine ( \( \text{CH}_3\text{NH-OH} \) ).
3. Finally, N-methylhydroxylamine is reduced to methylamine ( \( \text{CH}_3\text{NH}_2 \) ).

\( \text{CH}_3\text{NO}_2 \xrightarrow{+2\text{H}} \text{CH}_3\text{N=O} \xrightarrow{+2\text{H}} \text{CH}_3\text{NH-OH} \xrightarrow{+2\text{H}} \text{CH}_3\text{NH}_2 \)
In simple words: Nitromethane can be changed step-by-step into other compounds by adding hydrogen. It first becomes nitrosomethane, then N-methylhydroxylamine, and finally methylamine.

🎯 Exam Tip: Be aware of the different intermediates formed during the reduction of nitro compounds. The degree of reduction often depends on the strength of the reducing agent and the reaction conditions.

 

Question 6. Write about the reduction of nitro methane in acid medium and neutral medium.
Answer:
The reduction of nitromethane varies depending on the reaction medium:
**Acid Medium:** In an acidic environment (e.g., using \( \text{Sn} \)/\( \text{HCl} \) ), nitromethane is fully reduced to a primary amine, which is methylamine.

\[ \begin{array}{l} \quad \text{CH}_3\text{NO}_2 \quad \quad \stackrel{\text{Sn}/\text{HCl}}{\text{6[H]}} \quad \text{CH}_3\text{NH}_2 \\ \text{Nitromethane} \quad \quad \quad \quad \quad \quad \quad \text{Methylamine} \quad \text{(acid medium)} \\ \end{array} \]

**Neutral Medium:** In a neutral medium (e.g., using \( \text{Zn} \)/\( \text{NH}_4\text{Cl} \) ), nitromethane is reduced to N-methylhydroxylamine.

\[ \begin{array}{l} \quad \text{CH}_3\text{NO}_2 \quad \quad \stackrel{\text{Zn}/\text{NH}_4\text{Cl}}{\text{4[H]}} \quad \text{CH}_3\text{NHOH} \\ \text{Nitromethane} \quad \quad \quad \quad \quad \quad \quad \text{N-methylhydroxylamine} \quad \text{(neutral medium)} \\ \end{array} \]

In simple words: When nitromethane is reduced in acid, it changes into methylamine. But if it is reduced in a neutral liquid, it becomes N-methylhydroxylamine instead. The type of liquid used changes what you get.

🎯 Exam Tip: Always specify the reaction conditions (acidic, neutral, or basic) when discussing the reduction of nitro compounds, as the choice of reducing agent and medium significantly influences the final product.

 

Question 7. What happpens when ethylnitrite is reduced by Sn / Hcl?
Answer:
When ethyl nitrite ( \( \text{CH}_3\text{CH}_2\text{-O-N=O} \) ) is reduced using tin ( \( \text{Sn} \) ) and hydrochloric acid ( \( \text{HCl} \) ), it undergoes a reduction reaction to produce ethanol ( \( \text{CH}_3\text{CH}_2\text{-OH} \) ) and nitrous acid ( \( \text{HNO}_2 \) ).

\[ \text{CH}_3\text{CH}_2\text{-O-N=O} + 6[\text{H}] \xrightarrow{\text{Sn} / \text{HCl}} \text{CH}_3\text{CH}_2\text{-OH} + \text{HNO}_2 \]

In simple words: When ethyl nitrite is treated with tin and hydrochloric acid, it changes into ethanol and nitrous acid.

🎯 Exam Tip: Remember that alkyl nitrites are esters of nitrous acid, and their reduction often leads to the formation of alcohols, especially under strong reducing conditions like \( \text{Sn} \)/\( \text{HCl} \).

 

Question 8. What happens when ethyl nitrite undergoes hydrolysis with acid or base?
Answer:
When ethyl nitrite undergoes hydrolysis, whether with an acid ( \( \text{H}^+ \) ) or a base ( \( \text{OH}^- \) ), it breaks down to produce ethanol ( \( \text{CH}_3\text{CH}_2\text{-OH} \) ) and nitrous acid ( \( \text{HNO}_2 \) ). The nitrous acid, being unstable, can further decompose.

\[ \text{CH}_3\text{CH}_2\text{-O-N=O} + 6[\text{H}] \xrightarrow{\text{OH}^- \text{ (or) H}^+} \text{CH}_3\text{CH}_2\text{-OH} + \text{HNO}_2 \]

In simple words: When ethyl nitrite reacts with either an acid or a base, it breaks apart. This reaction gives you ethanol and nitrous acid.

🎯 Exam Tip: Hydrolysis of esters, including alkyl nitrites, typically involves breaking the ester linkage to yield an alcohol and an acid (or its salt). The presence of acid or base acts as a catalyst for this process.

 

Question 9. What is Nef Carbonyl Synthesis?
Answer:
Nef Carbonyl Synthesis is a chemical reaction used to convert primary and secondary nitroalkanes into aldehydes and ketones, respectively. The reaction involves first treating the nitroalkane with a base to form an aci-nitro salt, which is then hydrolyzed under acidic conditions. Specifically, the aci-nitro salt is reacted with aqueous acid, often sulfuric acid, which leads to the formation of the corresponding carbonyl compound (aldehyde or ketone) and nitrous oxide ( \( \text{N}_2\text{O} \) ).

\[ \text{R}_2\text{CHNO}_2 \quad \stackrel{\text{KOH}}{\longrightarrow} \quad \text{R}_2\text{C=NO}_2^- \text{K}^+ \quad \stackrel{\text{H}_2\text{O}/\text{H}^+}{\longrightarrow} \quad \text{R}_2\text{C=O} + \text{N}_2\text{O} \]\[ \text{CH}_3\text{CH}_2\text{NO}_2 \quad \stackrel{\text{KOH}}{\longrightarrow} \quad \text{CH}_3\text{CH=NO}_2^- \text{K}^+ \quad \stackrel{\text{H}_2\text{O}/\text{H}^+}{\longrightarrow} \quad \text{CH}_3\text{CHO} + \text{N}_2\text{O} \]

In simple words: Nef Carbonyl Synthesis is a way to change special nitro compounds into aldehydes or ketones. You first treat the nitro compound with a base, then use acid and water to get the desired carbonyl compound.

🎯 Exam Tip: Recognize Nef Carbonyl Synthesis as a valuable method for forming C-C bonds and converting nitro groups into carbonyl groups. The key steps involve deprotonation to form an aci-nitro anion, followed by acid-catalyzed hydrolysis.

 

Question 10. Write about the electrolytic reduction of nitro benzene?
Answer:
The electrolytic reduction of nitrobenzene is a process where nitrobenzene is reduced using an electric current. The products formed depend on the acidity of the medium.
In a strongly acidic medium, nitrobenzene is reduced to aniline.
In a weakly acidic or neutral medium, it is reduced to phenylhydroxylamine.
In a weakly basic medium, it forms azoxybenzene, azobenzene, and hydrazobenzene.
In a strong basic medium, it yields azoxybenzene.

\[ \begin{array}{l} \quad \text{NO}_2 \\ \quad \\ \quad \\ \quad \\ \text{Nitrobenzene} \\ \end{array} \quad \stackrel{\text{Electrolytic reduction}}{\longrightarrow} \quad \left\{ \begin{array}{l} \text{NH}_2 \quad \text{(Strongly acidic)} \\ \quad \quad \text{Aniline} \\ \quad \quad \\ \text{NHOH} \quad \text{(Weakly acidic/Neutral)} \\ \quad \quad \text{Phenylhydroxylamine} \\ \quad \quad \\ \text{N=N} \quad \text{(Weakly basic)} \\ \quad \quad \text{Azoxybenzene, Azobenzene, Hydrazobenzene} \\ \end{array} \right. \]

In simple words: Electrolytic reduction changes nitrobenzene using electricity. What you get depends on how acidic or basic the liquid is. It can turn into aniline, phenylhydroxylamine, or other compounds like azoxybenzene.

🎯 Exam Tip: The pH of the solution is a critical factor in determining the product of electrolytic reduction of nitrobenzene. Be sure to differentiate between the products formed in acidic, neutral, and basic media.

 

Question 11. Convert m- dinitrobenzene into m – nitroaniline
Answer:
To convert m-dinitrobenzene into m-nitroaniline, a selective reduction of one of the nitro groups is required. This can be achieved using ammonium sulfide ( \( \text{(NH}_4\text{)}_2\text{S} \) ) as the reducing agent. Ammonium sulfide selectively reduces one nitro group to an amino group without affecting the other.

\[ \begin{array}{l} \quad \text{NO}_2 \\ \quad \quad \\ \quad \quad \\ \quad \quad \\ \text{m-dinitrobenzene} \\ \end{array} \quad \stackrel{+3(\text{NH}_4)_2\text{S}}{\longrightarrow} \quad \begin{array}{l} \quad \text{NO}_2 \\ \quad \quad \\ \quad \quad \\ \quad \quad \\ \text{m-nitroaniline} \\ \end{array} \quad + \quad 6\text{NH}_3 + 2\text{H}_2\text{O} + 3\text{S} \]

In simple words: To change m-dinitrobenzene into m-nitroaniline, you use a special chemical called ammonium sulfide. This chemical only changes one of the nitro groups into an amino group, leaving the other nitro group as it is.

🎯 Exam Tip: Remember that selective reduction is crucial when a molecule has multiple similar functional groups, and you only want to modify one. Ammonium sulfide is a classic reagent for selectively reducing one nitro group in dinitro compounds.

 

Question 12. What happens when
(i) methyl cyanide and
(ii) methyl isocyanide are reduced?
Answer:
(i) When methyl cyanide ( \( \text{CH}_3\text{CN} \) ) is reduced, for example, with sodium amalgam in ethanol ( \( \text{Na(Hg)} \)/\( \text{C}_2\text{H}_5\text{OH} \) ), it yields a primary amine, which is ethanamine (or ethylamine).

\[ \text{CH}_3\text{CN} \xrightarrow{\text{Na(Hg)}/\text{C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{NH}_2 \]

(ii) When methyl isocyanide ( \( \text{CH}_3\text{NC} \) ) is reduced, for example, with sodium amalgam in ethanol ( \( \text{Na(Hg)} \)/\( \text{C}_2\text{H}_5\text{OH} \) ), it produces a secondary amine, which is N-methylmethanamine (or dimethylamine).

\[ \text{CH}_3\text{NC} \xrightarrow{\text{Na(Hg)}/\text{C}_2\text{H}_5\text{OH}} \text{CH}_3\text{NH-CH}_3 \]

In simple words: When methyl cyanide is reduced, it becomes ethanamine, which is a primary amine. But when methyl isocyanide is reduced, it turns into N-methylmethanamine, which is a secondary amine.

🎯 Exam Tip: Distinguish carefully between the reduction products of nitriles (cyanides) and isocyanides. Nitriles typically yield primary amines, while isocyanides yield secondary amines with one of the alkyl groups coming from the isocyanide carbon itself.

 

Question 13. How can you convert acetamide into
(i) ethanamine
(ii) methanamine.
Answer:
(i) To convert acetamide ( \( \text{CH}_3\text{CONH}_2 \) ) into ethanamine ( \( \text{CH}_3\text{CH}_2\text{NH}_2 \) ), reduction is performed. Acetamide can be reduced using a strong reducing agent like lithium aluminium hydride ( \( \text{LiAlH}_4 \) ) followed by hydrolysis.

\[ \text{CH}_3\text{CONH}_2 \quad \xrightarrow{\text{(i) LiAlH}_4 \text{ (ii) H}_2\text{O}} \quad \text{CH}_3\text{CH}_2\text{NH}_2 \]

(ii) To convert acetamide ( \( \text{CH}_3\text{CONH}_2 \) ) into methanamine ( \( \text{CH}_3\text{NH}_2 \) ), Hofmann's bromamide degradation reaction is used. This reaction involves treating acetamide with bromine ( \( \text{Br}_2 \) ) in the presence of a strong base like potassium hydroxide ( \( \text{KOH} \) ). This reaction results in an amine with one carbon atom less than the starting amide.

\[ \text{CH}_3\text{CONH}_2 \quad \xrightarrow{\text{Br}_2 / \text{KOH}} \quad \text{CH}_3\text{NH}_2 + \text{K}_2\text{CO}_3 + \text{KBr} + \text{H}_2\text{O} \]

In simple words: To get ethanamine from acetamide, you reduce it. To get methanamine from acetamide, you use Hofmann's reaction, which also removes one carbon atom.

🎯 Exam Tip: Remember that \( \text{LiAlH}_4 \) reduces amides to amines with the same number of carbon atoms, while Hofmann's bromamide degradation reduces amides to amines with one less carbon atom. These are fundamental reactions for changing the carbon chain length and functional group type.

 

Question 14. Write about Sabatier – Mailhe method of preparing three types of Amines
Answer:
The Sabatier-Mailhe method is a catalytic method for preparing amines by passing the vapor of an alcohol and ammonia over a heated catalyst like alumina ( \( \text{Al}_2\text{O}_3 \) ), tungstic oxide ( \( \text{W}_2\text{O}_5 \) ), or silica at about 400°C. This method can yield a mixture of primary, secondary, and tertiary amines, as well as quaternary ammonium salts, depending on the conditions and the ratio of alcohol to ammonia.

\[ \begin{array}{ll} \text{C}_2\text{H}_5\text{OH} + \text{NH}_3 \quad \xrightarrow{\text{Al}_2\text{O}_3 / 400^\circ\text{C}} & \text{C}_2\text{H}_5\text{NH}_2 \quad \text{(Primary amine)} \\ & \text{(C}_2\text{H}_5\text{)}_2\text{NH} \quad \text{(Secondary amine)} \\ & \text{(C}_2\text{H}_5\text{)}_3\text{N} \quad \text{(Tertiary amine)} \\ \end{array} \]

In simple words: The Sabatier-Mailhe method makes different kinds of amines by heating alcohol vapor and ammonia over a special catalyst. This process can create primary, secondary, and tertiary amines all at once.

🎯 Exam Tip: Understand that the Sabatier-Mailhe method is a high-temperature catalytic process, producing a mixture of amines, unlike some more selective synthesis methods. The catalyst and temperature are crucial for the reaction to occur.

 

Question 15. Write a note on basic strength of aniline.
Answer:
Aniline is less basic than alkylamines and ammonia. This is because the amino ( \( \text{-NH}_2 \) ) group in aniline is directly attached to the benzene ring. The lone pair of electrons on the nitrogen atom in aniline gets delocalized into the benzene ring through resonance. This delocalization makes the lone pair less available for protonation, which is essential for basicity. As a result, aniline is a weaker base.

\[ \begin{array}{ccc} \quad \text{NH}_2 & \quad \text{NH}_2^+ \\ \quad \text{Aniline} \quad & \quad \text{Anilinium ion} \\ \end{array} \]

If electron-releasing groups (like \( \text{-CH}_3 \), \( \text{-OCH}_3 \), \( \text{-NH}_2 \) ) are attached to the benzene ring, they increase the electron density on the nitrogen, thereby increasing the basic strength. Conversely, electron-withdrawing groups (like \( \text{-NO}_2 \), halogens, \( \text{-COOH} \) ) decrease the electron density on nitrogen, which reduces the basic strength of aniline.
In simple words: Aniline is not as strong a base as simple amines because the electron pair on its nitrogen atom spreads out into the benzene ring. This makes the electrons less available to grab a proton, which means it's a weaker base. Groups that push electrons make it stronger, while groups that pull electrons make it weaker.

🎯 Exam Tip: The key concept here is resonance. The delocalization of nitrogen's lone pair into the aromatic ring reduces its availability for donation, directly impacting basicity. Always consider inductive and resonance effects of substituents when comparing basic strengths.

 

Question 16. How is sulphanilic acid prepared?
Answer: Sulphanilic acid is prepared in two steps. First, aniline reacts with concentrated sulfuric acid to form anilinium hydrogen sulphate. This salt is then heated to a high temperature, between \( 453 \) and \( 473 \) K, which causes it to rearrange and form p-aminobenzene sulphonic acid, commonly known as sulphanilic acid. This process demonstrates a key method for adding a sulfonic acid group to an aromatic amine.
In simple words: Sulphanilic acid is made by first mixing aniline with strong sulfuric acid, then heating it up.

🎯 Exam Tip: Remember the temperature range \( (453-473K) \) as it's crucial for the rearrangement to sulphanilic acid. Also, note that the product is a zwitterion due to the presence of both acidic and basic groups.

 

Question 17. How will you convert aliphatic carboxylic acid into aromatic carboxylic acid?
Answer: Aliphatic carboxylic acids can be converted to aromatic carboxylic acids through a series of steps. One method involves first converting benzene diazonium fluoroborate into benzoic acid by heating it with acetic acid. Another general approach is to use reactions that introduce an aromatic ring and then oxidize a side chain to form the carboxylic acid group. This process often involves multi-step organic synthesis.
In simple words: We can change a simple straight-chain acid into a ring-shaped acid. One way is to heat a specific benzene compound with acetic acid to get benzoic acid.

🎯 Exam Tip: Focus on the benzene diazonium fluoroborate to benzoic acid conversion as a direct method, noting the reactants involved (acetic acid and heat).

 

Question 18. Write about the isomerisation of alkylisocyanides.
Answer: Alkyl isocyanides are compounds where the carbon is attached to nitrogen by a triple bond, and then to the alkyl group. When these alkyl isocyanides are strongly heated, typically to about \( 250^\circ C \), they undergo a rearrangement. This process converts them into their more stable isomeric form, which are the alkyl cyanides. This change happens because the C-N triple bond in isocyanides is less stable than the C-C triple bond in cyanides.
\( \text{CH}_3 - \text{N} \equiv \text{C} \xrightarrow{\text{Heat}} \text{CH}_3 - \text{C} \equiv \text{N} \)
In simple words: When alkyl isocyanides are heated, they change into alkyl cyanides. The new form is more stable.

🎯 Exam Tip: Remember that isocyanides (\( \text{R-NC} \)) rearrange to cyanides (\( \text{R-CN} \)) upon heating, indicating that cyanides are the more stable isomers.

VII. Three Mark Questions

 

Question 19. Write the uses of nitroalkanes.
Answer: Nitroalkanes are useful in several ways. They work as good solvents for many organic substances like vinyl polymers, cellulose esters, synthetic rubbers, oils, fats, waxes, and dyes. They are also used in organic synthesis to make other chemicals. Some nitroalkanes, like nitromethane, react with halogens in the presence of an alkali to create trihalogen derivatives such as chloropicrin (\( \text{CCl}_3\text{NO}_2 \)), which is used to sterilize soil. Nitroethane is also used as a fuel additive and as a component in rocket propellants, highlighting their diverse applications.
In simple words: Nitroalkanes are used as good solvents for many things, to make other chemicals, as soil sterilizers, and as fuel.

🎯 Exam Tip: Focus on remembering at least three distinct uses: solvent, organic synthesis intermediate, and fuel/propellant component.

 

Question 20. How is chloropicrin prepared?
Answer: Chloropicrin, also known as trichloronitromethane, is prepared by treating nitromethane (\( \text{CH}_3\text{NO}_2 \)) with chlorine gas (\( \text{Cl}_2 \)) in the presence of sodium hydroxide (\( \text{NaOH} \)). This reaction involves the substitution of all three hydrogen atoms in nitromethane with chlorine atoms. Chloropicrin is a colorless, oily liquid with a pungent odor.
\( \text{CH}_3\text{NO}_2 + 3\text{Cl}_2 \xrightarrow{\text{NaOH}} \text{CCl}_3\text{NO}_2 + 3\text{HCl} \)
In simple words: Chloropicrin is made by reacting nitromethane with chlorine, using sodium hydroxide to help the reaction.

🎯 Exam Tip: The key reactants are nitromethane, chlorine, and a base (NaOH). Remember the product formula \( \text{CCl}_3\text{NO}_2 \).

VII. Three Mark Questions

 

Question 1. How are nitro compounds classified?
Answer: Nitro compounds are organic compounds that contain the nitro (\( \text{-NO}_2 \)) functional group. They are primarily classified into two main types: aliphatic nitro compounds and aromatic nitro compounds.
1. Aliphatic Nitro Compounds: In these compounds, the nitro group is attached to an alkyl chain. They can be further categorized based on the carbon atom to which the nitro group is attached:
(i) Primary (\( 1^\circ \)) Nitroalkanes: The nitro group is attached to a primary carbon atom (e.g., \( \text{CH}_3\text{CH}_2\text{NO}_2 \) - Nitroethane).
(ii) Secondary (\( 2^\circ \)) Nitroalkanes: The nitro group is attached to a secondary carbon atom (e.g., \( \text{CH}_3\text{CH}(\text{NO}_2)\text{CH}_3 \) - 2-nitropropane).
(iii) Tertiary (\( 3^\circ \)) Nitroalkanes: The nitro group is attached to a tertiary carbon atom (e.g., \( \text{(CH}_3\text{)}_3\text{CNO}_2 \) - 2-methyl-2-nitropropane).
2. Aromatic Nitro Compounds: In these compounds, the nitro group is directly attached to an aromatic ring, such as a benzene ring (e.g., Nitrobenzene). This classification helps chemists understand their different chemical properties and reactions.
In simple words: Nitro compounds are sorted into two main groups: those with nitro on a carbon chain (aliphatic) and those with nitro on a ring (aromatic). Aliphatic ones are further split into primary, secondary, and tertiary based on where the nitro group sits.

🎯 Exam Tip: Be sure to distinguish between primary, secondary, and tertiary nitroalkanes based on the substitution of the carbon atom bearing the nitro group.

 

Question 2. Convert (i) acetaldehyde into nitro ethane (ii) acetone into 2-nitropropane
Answer:
(i) Acetaldehyde into Nitroethane:
First, acetaldehyde reacts with hydroxylamine to form acetaldoxime. Then, acetaldoxime is oxidized using a strong oxidizing agent like trifluoroperacetic acid (\( \text{CF}_3\text{COOOH} \)) to convert it into nitroethane. This reaction replaces the oxime group with a nitro group.
\( \text{CH}_3\text{CHO} + \text{NH}_2\text{OH} \longrightarrow \text{CH}_3\text{CH}=\text{NOH} \) (Acetaldehyde to Acetaldoxime)
\( \text{CH}_3\text{CH}=\text{NOH} \xrightarrow{\text{CF}_3\text{COOOH}} \text{CH}_3\text{CH}_2\text{NO}_2 \) (Acetaldoxime to Nitroethane)
(ii) Acetone into 2-nitropropane:
First, acetone reacts with hydroxylamine to form acetonoxime. Then, acetonoxime is oxidized using trifluoroperacetic acid (\( \text{CF}_3\text{COOOH} \)) to convert it into 2-nitropropane. This reaction similarly replaces the oxime group with a nitro group.
\( \text{CH}_3\text{COC}(\text{CH}_3) + \text{NH}_2\text{OH} \longrightarrow \text{CH}_3\text{C}(=\text{NOH})\text{CH}_3 \) (Acetone to Acetonoxime)
\( \text{CH}_3\text{C}(=\text{NOH})\text{CH}_3 \xrightarrow{\text{CF}_3\text{COOOH}} \text{CH}_3\text{CH}(\text{NO}_2)\text{CH}_3 \) (Acetonoxime to 2-nitropropane)
In simple words: To change acetaldehyde to nitroethane, first make it into acetaldoxime, then oxidize it. For acetone to 2-nitropropane, first make it into acetonoxime, then oxidize it the same way.

🎯 Exam Tip: The conversion of both aldehyde/ketone to their respective nitroalkanes involves two key steps: oxime formation (with hydroxylamine) and subsequent oxidation of the oxime (with a peracid like \( \text{CF}_3\text{COOOH} \)).

 

Question 3. Write about the hydrolysis of nitro alkanes
Answer: Hydrolysis of nitroalkanes means breaking them down with water, usually in the presence of an acid. This reaction can be done using concentrated hydrochloric acid (\( \text{HCl} \)) or concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). The outcome depends on the type of nitroalkane:
1. Primary Nitroalkanes: When primary nitroalkanes undergo hydrolysis, they yield carboxylic acids.
\( \text{CH}_3\text{CH}_2\text{NO}_2 \xrightarrow{\text{HCl/H}_2\text{O}} \text{CH}_3\text{COOH} + \text{NH}_2\text{OH} + \text{HCl} \)
2. Secondary Nitroalkanes: Hydrolysis of secondary nitroalkanes produces ketones.
\( \text{(CH}_3\text{)}_2\text{CHNO}_2 \xrightarrow{\text{HCl/H}_2\text{O}} \text{(CH}_3\text{)}_2\text{C}=\text{O} + \text{N}_2\text{O} + \text{H}_2\text{O} \)
3. Tertiary Nitroalkanes: Tertiary nitroalkanes are generally stable and do not undergo hydrolysis under these conditions. This difference in reactivity is due to the structure of the carbon atom attached to the nitro group.
\( \text{(CH}_3\text{)}_3\text{CNO}_2 \xrightarrow{\text{HCl/H}_2\text{O}} \) No reaction
In simple words: Hydrolysis of nitroalkanes means breaking them with water and acid. Primary ones make acids, secondary ones make ketones, and tertiary ones do not react.

🎯 Exam Tip: Remember the products for primary (carboxylic acid) and secondary (ketone) nitroalkane hydrolysis. Also, note that tertiary nitroalkanes are resistant to hydrolysis.

 

Question 4. Write about the classification of amines
Answer: Amines are organic compounds derived from ammonia (\( \text{NH}_3 \)) by replacing one or more hydrogen atoms with alkyl or aryl groups. They are broadly classified into aliphatic and aromatic amines, and further sub-classified based on the number of hydrogen atoms replaced in the ammonia molecule.

This systematic classification helps in understanding the chemical behavior and naming conventions of different amine compounds.
In simple words: Amines are like ammonia but with carbon groups instead of hydrogen. They are classified into aliphatic (chain) and aromatic (ring) types, and then further by how many hydrogens on the nitrogen are replaced (primary, secondary, or tertiary).

🎯 Exam Tip: For each class of amine (primary, secondary, tertiary), remember to give both aliphatic and aromatic examples to demonstrate complete understanding.

 

Question 5. Write notes on structure of amines
Answer: The structure of amines is similar to that of ammonia (\( \text{NH}_3 \)), where the nitrogen atom is trivalent and contains one lone pair of electrons. Here are the key features:
* The nitrogen atom in amines is \( \text{sp}^3 \) hybridized.
* Out of the four \( \text{sp}^3 \) hybridized orbitals, three overlap with hydrogen atoms or carbon atoms of alkyl/aryl groups, forming three sigma bonds.
* The fourth \( \text{sp}^3 \) orbital contains the lone pair of electrons.
* Due to the presence of this lone pair, the geometry of amines is pyramidal, not tetrahedral, even though the nitrogen is \( \text{sp}^3 \) hybridized. The lone pair causes repulsion, pushing the bonding pairs closer.
* The \( \text{C-N-H} \) or \( \text{C-N-C} \) bond angle in amines is less than the ideal tetrahedral angle of \( 109.5^\circ \). For instance, in trimethylamine, the \( \text{C-N-C} \) bond angle is \( 108^\circ \). In contrast, the \( \text{H-N-H} \) bond angle in ammonia is \( 107^\circ \). This variation is influenced by the size of the alkyl groups, with bulkier groups increasing the bond angle slightly due to steric repulsion.
In simple words: Amines are like ammonia, with nitrogen having a lone pair of electrons. The nitrogen atom uses \( \text{sp}^3 \) hybrid orbitals, which gives amines a pyramid shape, making the bond angles smaller than \( 109.5^\circ \).

🎯 Exam Tip: Always mention the \( \text{sp}^3 \) hybridization of nitrogen and the pyramidal geometry due to the lone pair, explaining how it affects bond angles.

 

Question 6. Prepare aniline from (i) chloro benzene (ii) Phenol (iii) nitro benzene
Answer: Aniline can be prepared from chlorobenzene, phenol, and nitrobenzene through different chemical reactions:
(i) From Chlorobenzene: Chlorobenzene reacts with ammonia (\( \text{NH}_3 \)) at high temperatures (\( 200^\circ C \)) and pressure in the presence of a copper(I) oxide (\( \text{Cu}_2\text{O} \)) catalyst to yield aniline.
\( \text{C}_6\text{H}_5\text{Cl} + \text{NH}_3 \xrightarrow{\text{Cu}_2\text{O/200}^\circ \text{C}} \text{C}_6\text{H}_5\text{NH}_2 \)
(ii) From Phenol: Phenol reacts with ammonia (\( \text{NH}_3 \)) at high temperatures (\( 300^\circ C \)) in the presence of anhydrous zinc chloride (\( \text{ZnCl}_2 \)) to form aniline.
\( \text{C}_6\text{H}_5\text{OH} + \text{NH}_3 \xrightarrow{\text{Anhy ZnCl}_2\text{/300}^\circ \text{C}} \text{C}_6\text{H}_5\text{NH}_2 \)
(iii) From Nitrobenzene: Nitrobenzene is reduced to aniline using a reducing agent like tin and hydrochloric acid (\( \text{Sn/HCl} \)). The reaction is a catalytic hydrogenation.
\( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O} \)
In simple words: Aniline can be made from chlorobenzene by reacting it with ammonia and copper oxide, from phenol by reacting it with ammonia and zinc chloride, or from nitrobenzene by reducing it with tin and acid.

🎯 Exam Tip: For each conversion, remember the key starting material, the reagents/catalysts, and the approximate reaction conditions (e.g., high temperature).

 

Question 7. Derive an expression for basic strength of amines.
Answer: Amines are weak bases because the nitrogen atom has a lone pair of electrons it can donate. In an aqueous solution, amines react with water to form an alkylammonium ion and a hydroxide ion. This is an equilibrium reaction:
\( \text{R-NH}_2 + \text{H-OH} \rightleftharpoons \text{R-NH}_3^+ + \text{OH}^- \)
The basicity constant, \( \text{K}_b \), for this equilibrium is given by the expression:
\( \text{K}_b = \frac{[\text{R-NH}_3^+][\text{OH}^-]}{[\text{R-NH}_2]} \)
A larger \( \text{K}_b \) value indicates a stronger base, meaning the amine is better at accepting a proton from water. Conversely, a smaller \( \text{pK}_b \) (where \( \text{pK}_b = -\text{log K}_b \)) corresponds to a stronger base. The equilibrium lies far to the left, which shows that amines are generally weaker bases compared to strong bases like sodium hydroxide (\( \text{NaOH} \)). This constant helps us compare the basic strength of different amines.
In simple words: Amines are weak bases that react with water to make ammonium ions and hydroxide. The basic strength is measured by a constant called \( \text{K}_b \); a higher \( \text{K}_b \) means a stronger base.

🎯 Exam Tip: Clearly write down the equilibrium reaction of an amine with water and the correct expression for \( \text{K}_b \). Remember that higher \( \text{K}_b \) (or lower \( \text{pK}_b \)) means stronger basicity.

 

Question 8. An organic compound (A) on reduction gives compound (B). (B) on treatment with CHCl3 and alcoholic KOH gives (C) . (C) on catalytic reduction gives N-methyl aniline. Identify A, B, C and write its equation.
Answer: Let's identify the compounds step by step:
* Compound (A): When reduced, it gives (B). This is nitrobenzene (\( \text{C}_6\text{H}_5\text{NO}_2 \)).
* Compound (B): Reduction of nitrobenzene gives aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)).
* Compound (C): Aniline reacts with chloroform (\( \text{CHCl}_3 \)) and alcoholic potassium hydroxide (\( \text{KOH} \)) in the carbylamine reaction (isocyanide test) to give phenyl isocyanide (\( \text{C}_6\text{H}_5\text{NC} \)).
* Catalytic reduction of phenyl isocyanide gives N-methylaniline.
Here are the reactions:
1. Reduction of Nitrobenzene (A) to Aniline (B):
\( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Reduction, e.g., [H]}} \text{C}_6\text{H}_5\text{NH}_2 \)
2. Aniline (B) to Phenyl Isocyanide (C) (Carbylamine Reaction):
\( \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc)} \longrightarrow \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
3. Catalytic Reduction of Phenyl Isocyanide (C) to N-methylaniline:
\( \text{C}_6\text{H}_5\text{NC} \xrightarrow{\text{H}_2\text{ catalyst}} \text{C}_6\text{H}_5\text{NHCH}_3 \)

In simple words: (A) is nitrobenzene, which becomes (B) aniline when reduced. Aniline then reacts with chloroform and KOH to make (C) phenyl isocyanide. Finally, (C) is changed into N-methylaniline by catalytic reduction.

🎯 Exam Tip: Recognise the carbylamine reaction (B to C) as a key step, producing an isocyanide, and remember that reduction of isocyanides gives secondary amines.

 

Question 9. Convert aniline into (i) p-bromo aniline (ii) p-nitro aniline
Answer: Aniline can be converted into p-bromoaniline and p-nitroaniline using specific reaction sequences. The amino (\( \text{-NH}_2 \)) group in aniline is a strong activating and ortho-para directing group.
(i) Conversion to p-bromoaniline:
To get only a mono-brominated product at the para position (p-bromoaniline) and avoid multiple brominations, the activating effect of the \( \text{-NH}_2 \) group must be reduced.
1. Acetylation of Aniline: Aniline is first treated with acetic anhydride (\( \text{(CH}_3\text{CO)}_2\text{O} \)) to form acetanilide. The amide group (\( \text{-NHCOCH}_3 \)) is a weaker activating group than \( \text{-NH}_2 \).
\( \text{C}_6\text{H}_5\text{NH}_2 + \text{(CH}_3\text{CO)}_2\text{O} \longrightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{CH}_3\text{COOH} \) (Aniline to Acetanilide)
2. Bromination of Acetanilide: Acetanilide is then brominated with bromine water (\( \text{Br}_2\text{/CH}_3\text{COOH} \)) to give p-bromoacetanilide as the major product.
\( \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{Br}_2 \longrightarrow \text{p-Br-C}_6\text{H}_4\text{NHCOCH}_3 + \text{HBr} \)
3. Hydrolysis of p-bromoacetanilide: The p-bromoacetanilide is hydrolyzed (treated with acid or base) to remove the acetyl group, yielding p-bromoaniline.
\( \text{p-Br-C}_6\text{H}_4\text{NHCOCH}_3 \xrightarrow{\text{H}_3\text{O}^+} \text{p-Br-C}_6\text{H}_4\text{NH}_2 + \text{CH}_3\text{COOH} \)
(ii) Conversion to p-nitroaniline:
Direct nitration of aniline with nitrating mixture (\( \text{HNO}_3 + \text{H}_2\text{SO}_4 \)) leads to the formation of m-nitroaniline along with ortho and para products, due to the oxidation of aniline and protonation of the \( \text{-NH}_2 \) group under acidic conditions. To obtain p-nitroaniline as the major product, the \( \text{-NH}_2 \) group needs to be protected.
1. Acetylation of Aniline: Aniline is first acetylated to form acetanilide to protect the amino group.
\( \text{C}_6\text{H}_5\text{NH}_2 + \text{(CH}_3\text{CO)}_2\text{O} \xrightarrow{\text{Pyridine}} \text{C}_6\text{H}_5\text{NHCOCH}_3 \) (Aniline to Acetanilide)
2. Nitration of Acetanilide: Acetanilide is nitrated with a nitrating mixture (\( \text{HNO}_3/\text{H}_2\text{SO}_4 \)) at low temperature (\( 288 \text{K} \)) to give p-nitroacetanilide as the major product.
\( \text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{HNO}_3\text{/H}_2\text{SO}_4, 288\text{K}} \text{p-NO}_2\text{-C}_6\text{H}_4\text{NHCOCH}_3 \)
3. Hydrolysis of p-nitroacetanilide: The p-nitroacetanilide is then hydrolyzed to remove the acetyl group, yielding p-nitroaniline.
\( \text{p-NO}_2\text{-C}_6\text{H}_4\text{NHCOCH}_3 \xrightarrow{\text{H}^+\text{/H}_2\text{O}} \text{p-NO}_2\text{-C}_6\text{H}_4\text{NH}_2 \)
In simple words: To make p-bromoaniline or p-nitroaniline from aniline, first cover the \( \text{NH}_2 \) group (make it acetanilide). Then, add bromine or a nitro group. Lastly, remove the covering to get the desired product. This stops unwanted reactions at other positions.

🎯 Exam Tip: The key strategy here is "protection of the amino group" by acetylation to control regioselectivity and prevent side reactions during electrophilic substitution (bromination or nitration).

VIII. Five Mark Questions

 

Question 1. Explain structural isomerism exhibited by nitro alkanes
Answer: Nitroalkanes are organic compounds containing a nitro group (\( \text{-NO}_2 \)) attached to a carbon chain. They exhibit various types of structural isomerism, which means they can have the same molecular formula but different arrangements of atoms.
1. Chain Isomerism: This type of isomerism occurs when compounds have the same molecular formula but differ in the arrangement of the carbon skeleton (straight chain vs. branched chain).
* Example: 1-nitrobutane and 2-methyl-1-nitropropane. Both have the formula \( \text{C}_4\text{H}_9\text{NO}_2 \).
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{-NO}_2 \) (1-nitrobutane)
\( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{-NO}_2 \) (2-methyl-1-nitropropane)
2. Position Isomerism: This occurs when compounds have the same carbon skeleton and functional groups, but the functional group is located at a different position on the carbon chain.
* Example: 1-nitrobutane and 2-nitrobutane. Both have the formula \( \text{C}_4\text{H}_9\text{NO}_2 \).
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{-NO}_2 \) (1-nitrobutane)
\( \text{CH}_3\text{CH}_2\text{CH}(\text{NO}_2)\text{CH}_3 \) (2-nitrobutane)
3. Functional Isomerism: This type of isomerism occurs when compounds have the same molecular formula but different functional groups. Nitroalkanes are functional isomers with alkyl nitrites.
* Example: 1-nitrobutane and butyl nitrite. Both have the formula \( \text{C}_4\text{H}_9\text{NO}_2 \).
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{-NO}_2 \) (1-nitrobutane)
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{-O-N=O} \) (butyl nitrite)

In simple words: Nitroalkanes can have different structures even with the same chemical formula. This includes changing the carbon chain, changing where the nitro group is placed, or being a completely different type of compound (like an alkyl nitrite).

🎯 Exam Tip: Provide a clear definition and an example for each type of isomerism (chain, position, functional) when discussing nitroalkanes.

 

Question 2. Explain the reduction reactions of nitro benzene in various medium.
Answer: Nitrobenzene can be reduced to various products depending on the reaction medium and the reducing agent used. Here's an explanation of its reduction in different conditions:
1. In Acidic Medium (e.g., Sn/HCl, Fe/HCl): In a strong acidic medium, nitrobenzene is completely reduced to aniline. This is a common method for preparing aromatic primary amines.
\( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl or Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \) (Aniline)
2. In Neutral Medium (e.g., Zn dust/NH\( _4 \)Cl): In a neutral medium, nitrobenzene is reduced to phenylhydroxylamine.
\( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Zn/NH}_4\text{Cl}} \text{C}_6\text{H}_5\text{NHOH} \) (Phenylhydroxylamine)
3. In Weakly Acidic Medium (e.g., Zn dust/NaOH or Electrolytic reduction at weak acid): When reduction is performed with zinc dust and sodium hydroxide (or by electrolytic reduction in a weakly acidic solution), nitrobenzene forms azoxybenzene.
\( 2\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Zn/NaOH}} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_5 \) (Azoxybenzene)
4. With Specific Reagents (e.g., SnCl\( _2 \)/KOH): Using tin(II) chloride and potassium hydroxide, azoxybenzene can be further reduced to azobenzene.
\( \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_5 \xrightarrow{\text{SnCl}_2\text{/KOH}} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_5 \) (Azobenzene)
5. Strong Alkaline Medium (e.g., Zn/NaOH): In a strong alkaline medium, azobenzene can be further reduced to hydrazobenzene.
\( \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_5 \xrightarrow{\text{Zn/NaOH}} \text{C}_6\text{H}_5\text{-NH-NH-C}_6\text{H}_5 \) (Hydrazobenzene)
These different products highlight the versatility of nitrobenzene in organic synthesis, allowing for the preparation of various nitrogen-containing aromatic compounds.
In simple words: Nitrobenzene can turn into many different things depending on how it's reduced. In acid, it becomes aniline. In neutral conditions, it forms phenylhydroxylamine. In weak alkaline conditions, it can become azoxybenzene, which can be further reduced to azobenzene and then hydrazobenzene in stronger alkali.

🎯 Exam Tip: Pay close attention to the reducing agent and the medium (acidic, neutral, alkaline) as they dictate the final product of nitrobenzene reduction. Remembering at least two specific conditions and their products is helpful.

 

Question 3. An Organic compound (A) – C7H7NO on treatment with Br2 and KOH gives an amine (B), which gives carbylamine test. (B) upon diazotization to give (C). (C) on coupling_with p-cresol to give compound (D). Identify A, B, C and D with necessary reaction.
Answer: Let's break down the reactions to identify compounds A, B, C, and D:
1. Hofmann's Degradation Reaction: Compound (A) (\( \text{C}_7\text{H}_7\text{NO} \)) reacts with \( \text{Br}_2 \) and \( \text{KOH} \) to give an amine (B). This is a Hofmann bromamide degradation reaction, which converts an amide to a primary amine with one less carbon atom. Since B gives the carbylamine test, it must be a primary amine.
\( \text{C}_6\text{H}_5\text{CONH}_2 \text{ (Benzamide, A)} \xrightarrow{\text{Br}_2\text{/KOH}} \text{C}_6\text{H}_5\text{NH}_2 \text{ (Aniline, B)} + \text{KBr} + \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \) Therefore, (A) is Benzamide and (B) is Aniline.
2. Diazotization Reaction: Amine (B) (Aniline) undergoes diazotization to give (C). Diazotization of aniline produces benzene diazonium chloride.
\( \text{C}_6\text{H}_5\text{NH}_2 \text{ (Aniline, B)} + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \text{ (Benzene Diazonium Chloride, C)} + \text{NaCl} + 2\text{H}_2\text{O} \) Therefore, (C) is Benzene Diazonium Chloride.
3. Coupling Reaction: Compound (C) (Benzene Diazonium Chloride) couples with p-cresol to give compound (D). This is an electrophilic substitution reaction where the diazonium ion attacks the electron-rich p-cresol.
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \text{ (C)} + \text{C}_6\text{H}_4(\text{CH}_3)\text{OH} \text{ (p-cresol)} \xrightarrow{\text{pH 9-10}} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_3(\text{CH}_3)\text{OH} \text{ (2-phenylazo-4-methylphenol, D)} \) Therefore, (D) is 2-phenylazo-4-methylphenol.

In simple words: Compound (A) is benzamide, which turns into (B) aniline using bromine and KOH. Aniline then becomes (C) benzene diazonium chloride by reacting with sodium nitrite and HCl. Finally, (C) reacts with p-cresol to form (D) 2-phenylazo-4-methylphenol, a colored dye.

🎯 Exam Tip: This question tests knowledge of Hofmann bromamide degradation, diazotization, and coupling reactions. Identify each reaction type and the functional group changes at each step.

 

Question 4. Write a short note an Libermann's nitroso test ?
Answer: Libermann's nitroso test is a chemical test used to identify secondary amines.
When secondary alkyl or aryl amines react with nitrous acid (\( \text{HNO}_2 \)), they form N-nitrosoamines. Nitrous acid is usually generated in situ from sodium nitrite (\( \text{NaNO}_2 \)) and concentrated hydrochloric acid (\( \text{HCl} \)).
\( \text{R}_2\text{NH} + \text{HON=O} \xrightarrow{\text{Conc HCl}} \text{R}_2\text{N-N=O} + \text{H}_2\text{O} \) (Secondary amine to N-nitrosoamine)
The N-nitrosoamine formed is typically a yellow oily liquid. When this oily liquid is warmed with phenol and concentrated sulfuric acid, it produces a characteristic sequence of colors:
1. It first turns green or blue.
2. Upon dilution with water, it turns red.
3. When made alkaline with sodium hydroxide (\( \text{NaOH} \)), the solution becomes blue or violet.
This unique color change sequence confirms the presence of a secondary amine. Primary and tertiary amines give different or no such color reactions.
In simple words: Libermann's nitroso test helps find secondary amines. When a secondary amine reacts with nitrous acid, it makes a yellow oil. This oil, when mixed with phenol and acid, changes color from green/blue to red when water is added, and then to blue/violet when alkali is added.

🎯 Exam Tip: Remember the specific color changes (green/blue \(\rightarrow \) red \(\rightarrow \) blue/violet) and the reagents (nitrous acid, phenol, conc. \( \text{H}_2\text{SO}_4 \), NaOH) for Libermann's nitroso test, which is specific for secondary amines.

 

Question 5. An Organic compound (A) – CNCI react with methyl magnesium Bromide to give, compound B - (C2H3N). B - upon catalytic reduction to give compound C- (C2H7N). C gives carbylamine test, Identify compound A, B and C and write the reactions.
Answer: Let's identify the compounds step-by-step:
1. Reaction of (A) with Grignard Reagent: Compound (A) (\( \text{CNCl} \)) reacts with methyl magnesium bromide (\( \text{CH}_3\text{MgBr} \)) to give compound (B) with formula \( \text{C}_2\text{H}_3\text{N} \). This is a reaction of cyanogen chloride with a Grignard reagent, leading to the formation of a nitrile.
\( \text{ClCN (Cyanogen chloride, A)} + \text{CH}_3\text{MgBr} \longrightarrow \text{CH}_3\text{CN (Ethanenitrile, B)} + \text{MgBrCl} \) Therefore, (A) is cyanogen chloride and (B) is ethanenitrile (or methyl cyanide).
2. Catalytic Reduction of (B) to (C): Compound (B) (\( \text{C}_2\text{H}_3\text{N} \)) undergoes catalytic reduction to give compound (C) (\( \text{C}_2\text{H}_7\text{N} \)). Reduction of a nitrile yields a primary amine.
\( \text{CH}_3\text{CN (Ethanenitrile, B)} + 2\text{H}_2 \xrightarrow{\text{Ni catalyst}} \text{CH}_3\text{CH}_2\text{NH}_2 \text{ (Ethanamine, C)} \) Therefore, (C) is ethanamine.
3. Carbylamine Test for (C): Compound (C) gives the carbylamine test. The carbylamine test is specific for primary amines. Ethanamine is a primary amine, which confirms the identification of (C).
\( \text{CH}_3\text{CH}_2\text{NH}_2 \text{ (Ethanamine, C)} + \text{CHCl}_3 + 3\text{KOH (alc)} \longrightarrow \text{CH}_3\text{CH}_2\text{NC} \text{ (Ethyl isocyanide)} + 3\text{KCl} + 3\text{H}_2\text{O} \) The ethyl isocyanide formed has a very unpleasant smell, confirming the presence of a primary amine.

In simple words: Compound (A) is cyanogen chloride. It reacts with methyl magnesium bromide to make (B) ethanenitrile. When (B) is reduced, it forms (C) ethanamine, which is a primary amine that passes the carbylamine test.

🎯 Exam Tip: Key points are recognizing the Grignard reaction with cyanogen chloride to form a nitrile, the reduction of nitrile to a primary amine, and the specific carbylamine test for primary amines.

 

Question 6. Explain the action of nitrous acid on three types of amine.
Answer: Nitrous acid (\( \text{HNO}_2 \)), usually generated in situ from \( \text{NaNO}_2 \) and \( \text{HCl} \), reacts differently with primary, secondary, and tertiary amines, making it a useful reagent for distinguishing them.
(a) Primary Amines:
* Aliphatic Primary Amines: React with nitrous acid to form highly unstable aliphatic diazonium salts, which immediately decompose to form alcohols with the liberation of nitrogen gas (\( \text{N}_2 \)) and hydrogen chloride.
\( \text{R-NH}_2 + \text{HNO}_2 \xrightarrow{} [\text{R-N}_2^+\text{Cl}^-] \xrightarrow{\text{H}_2\text{O}} \text{R-OH} + \text{N}_2 + \text{HCl} \)
For example, ethylamine reacts to give ethanol.
* Aromatic Primary Amines: React with nitrous acid at low temperatures (\( 273-278 \text{K} \)) to form stable (for a short time) arenediazonium salts, such as benzene diazonium chloride. These salts are crucial intermediates for many organic syntheses.
\( \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O} \)
(b) Secondary Amines:
Secondary alkyl or aryl amines react with nitrous acid to form N-nitrosoamines, which are yellow oily liquids. This is the basis of Libermann's nitroso test. These compounds are insoluble in water.
\( \text{R}_2\text{NH} + \text{HON=O} \longrightarrow \text{R}_2\text{N-N=O} + \text{H}_2\text{O} \)
(c) Tertiary Amines:
* Aliphatic Tertiary Amines: React with nitrous acid to form trialkylammonium nitrite salts, which are soluble in water.
\( \text{R}_3\text{N} + \text{HNO}_2 \longrightarrow [\text{R}_3\text{NH}]^+\text{NO}_2^- \)
* Aromatic Tertiary Amines: Undergo electrophilic substitution with nitrous acid at the para-position to form p-nitroso-N,N-dimethylaniline (a green-colored compound), which is a p-nitroso derivative.
\( \text{(CH}_3\text{)}_2\text{N-C}_6\text{H}_5 \xrightarrow{\text{HNO}_2\text{/ConHCl}} \text{p-NO-C}_6\text{H}_4\text{N(CH}_3\text{)}_2 \)
This differential reactivity of amines with nitrous acid is a fundamental concept in organic chemistry for identification and synthesis.
In simple words: Nitrous acid reacts differently with different amines. Primary amines make unstable salts that turn into alcohols (aliphatic) or stable diazonium salts (aromatic). Secondary amines make yellow oily N-nitrosoamines. Tertiary amines form soluble salts (aliphatic) or undergo substitution to form nitroso compounds (aromatic).

🎯 Exam Tip: Clearly differentiate between aliphatic and aromatic primary amines with nitrous acid. For secondary amines, mention the N-nitrosoamine product, and for tertiary, note the salt formation or p-nitroso derivative.

 

Question 7. Explian: (i) Sand mayer reaction (ii) Gaftermann reaction (iii) Baltz - schiemann reaction
Answer: These are important named reactions used for replacing the diazonium group in arenediazonium salts with various substituents.
(i) Sandmeyer Reaction:
The Sandmeyer reaction is a synthetic procedure used to synthesize aryl halides (aryl chlorides, aryl bromides) and aryl cyanides from arenediazonium salts. The diazonium salt is treated with a copper(I) salt (e.g., \( \text{Cu}_2\text{Cl}_2 \), \( \text{Cu}_2\text{Br}_2 \), or \( \text{CuCN} \)) dissolved in the corresponding hydrohalic acid (\( \text{HCl} \), \( \text{HBr} \)) or potassium cyanide solution.
\( \text{Ar-N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2\text{/HCl}} \text{Ar-Cl} + \text{N}_2 \)
\( \text{Ar-N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Br}_2\text{/HBr}} \text{Ar-Br} + \text{N}_2 \)
\( \text{Ar-N}_2^+\text{Cl}^- \xrightarrow{\text{CuCN/KCN}} \text{Ar-CN} + \text{N}_2 \)
(ii) Gattermann Reaction:
The Gattermann reaction is similar to the Sandmeyer reaction but uses copper powder (instead of copper(I) salts) with the corresponding hydrohalic acid to replace the diazonium group with a halogen atom. It provides a more convenient method for introducing halogens or cyanide into an aromatic ring.
\( \text{Ar-N}_2^+\text{Cl}^- \xrightarrow{\text{Cu/HCl}} \text{Ar-Cl} + \text{N}_2 \)
\( \text{Ar-N}_2^+\text{Cl}^- \xrightarrow{\text{Cu/HBr}} \text{Ar-Br} + \text{N}_2 \)
(iii) Balz-Schiemann Reaction:
This reaction is used for the synthesis of aryl fluorides from arenediazonium salts. The diazonium salt is first treated with fluoroboric acid (\( \text{HBF}_4 \)) to form an arenediazonium fluoroborate. This salt is then heated, which causes it to decompose and yield the aryl fluoride.
\( \text{Ar-N}_2^+\text{Cl}^- + \text{HBF}_4 \longrightarrow \text{Ar-N}_2^+\text{BF}_4^- + \text{HCl} \)
\( \text{Ar-N}_2^+\text{BF}_4^- \xrightarrow{\text{Heat}} \text{Ar-F} + \text{BF}_3 + \text{N}_2 \)
These reactions are vital for expanding the range of functional groups that can be attached to aromatic rings.
In simple words: These are ways to change a diazonium salt into other compounds. Sandmeyer and Gattermann reactions use copper to add chlorine, bromine, or cyanide. Balz-Schiemann reaction uses fluoroboric acid and heat to add fluorine.

🎯 Exam Tip: Distinguish Sandmeyer from Gattermann by the copper reagent (Cu(I) salts vs. Cu powder). Remember that Balz-Schiemann is specifically for introducing fluorine.

 

Question 8. Convert benzene diazonium chloride into: (i) benzene (ii) iodo benzene (iii) phenol (iv) nitrobenzene
Answer: Benzene diazonium chloride (\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \)) is a versatile intermediate and can be converted into various compounds:
(i) To Benzene:
Benzene diazonium chloride can be converted to benzene by reduction with hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) or ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)). This reaction replaces the diazonium group with hydrogen.
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \xrightarrow{} \text{C}_6\text{H}_6 + \text{H}_3\text{PO}_3 + \text{HCl} + \text{N}_2 \)
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{} \text{C}_6\text{H}_6 + \text{CH}_3\text{CHO} + \text{HCl} + \text{N}_2 \)
(ii) To Iodobenzene:
Benzene diazonium chloride reacts with potassium iodide (\( \text{KI} \)) solution to yield iodobenzene. This is a direct substitution reaction.
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{KI} \longrightarrow \text{C}_6\text{H}_5\text{I} + \text{KCl} + \text{N}_2 \)
(iii) To Phenol:
Benzene diazonium chloride can be converted to phenol by warming it with water or by boiling it with dilute sulfuric acid.
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{H}_2\text{O} \xrightarrow{\text{Warm}} \text{C}_6\text{H}_5\text{OH} + \text{N}_2 + \text{HCl} \)
(iv) To Nitrobenzene:
Converting benzene diazonium chloride to nitrobenzene is generally done through an indirect route. First, the diazonium salt is converted to an aryl fluoride using the Balz-Schiemann reaction, and then the fluoride can be nitrated. A more common method for direct replacement of \( \text{-N}_2^+\text{Cl}^- \) by \( \text{-NO}_2 \) involves using sodium nitrite (\( \text{NaNO}_2 \)) in the presence of copper.
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{NaNO}_2\text{/Cu}} \text{C}_6\text{H}_5\text{NO}_2 + \text{N}_2 + \text{CuCl} \)
In simple words: Benzene diazonium chloride can be changed into: benzene (by reduction), iodobenzene (with potassium iodide), phenol (by warming with water), or nitrobenzene (using sodium nitrite and copper).

🎯 Exam Tip: For each conversion, know the specific reagent required. For benzene, recognize \( \text{H}_3\text{PO}_2 \) or ethanol as reducing agents. For iodobenzene, it's KI, and for phenol, it's water (hydrolysis). The conversion to nitrobenzene involves \( \text{NaNO}_2/\text{Cu} \).

 

Question 9. Explain coupling reactions of benzene diazonium chloride.
Answer: Coupling reactions involve benzene diazonium chloride reacting with electron-rich aromatic compounds such as phenol or aniline. These reactions create brightly colored azo compounds, which are often used as dyes. Typically, coupling takes place at the para position. However, if the para position is already taken, the reaction will happen at the ortho position instead. An electron-donating group at the para position helps the coupling reaction happen more easily. This type of reaction is an electrophilic substitution reaction.
In simple words: Benzene diazonium chloride joins with other special compounds to make colorful azo compounds. This joining usually happens at a specific spot called the para position.

🎯 Exam Tip: Remember that azo compounds are known for their vibrant colors and are widely used in the dye industry, which is a key characteristic to mention.

 

Question 10. Write notes on
(i) Thrope nitrile condensation
(ii) Levine and Hauser acetylation
Answer:
(i) Thrope nitrile condensation: This reaction involves two molecules of an alkyl nitrile, which is a compound with a cyanide group, joining together. It happens when the nitrile has an alpha-hydrogen atom (a hydrogen on the carbon next to the cyanide group) and a base like sodium is present. The process forms a new compound called an iminonitrile. For example, propane nitrile can react with itself to form 3-imino-2-methylpentane nitrile. This is a type of carbon-carbon bond formation.
\( \text{CH}_3\text{CH}_2\text{-CN} + \text{CH}_2\text{-CN} \xrightarrow{\text{Na/Ether}} \text{CH}_3\text{CH}_2\text{-C(=NH)-CH(CH}_3\text{)-CN} \)
In simple words: Two nitrile molecules that have a special hydrogen atom can join together with the help of sodium to make a new, larger nitrile molecule.

🎯 Exam Tip: Remember that Thrope nitrile condensation is a self-condensation reaction, meaning the same type of molecule reacts with itself.

Answer:
(ii) Levine and Hauser acetylation: In this reaction, nitriles that have an alpha-hydrogen atom react with esters. This condensation happens when sodamide is present in ether solvent. The main products formed are ketonitriles. This process is important for creating more complex organic structures by adding a ketone group next to a nitrile. It essentially replaces an ethoxy group with a methyl nitrile equivalent.
In simple words: This reaction helps nitriles (compounds with CN) join with esters (another type of organic compound) to make new compounds called ketonitriles. Sodamide helps this reaction happen.

🎯 Exam Tip: Focus on the key reactants: nitriles with alpha-hydrogens and esters, forming ketonitriles, as this highlights the main function of Levine and Hauser acetylation.

 

Question 11. An aromatic nitro compound (A) on reduction with Sn/HCl gives compound (B) \( \text{C}_6\text{H}_7\text{N} \) which on treatment with Benzoyl chloride in the presence of pyridine to give compound (C). Compound (B) on treatment with \( \text{CH}_3\text{Br} \) to give compound (D) which further reacts with \( \text{NaNO}_2 \)/HCl to give compound (E) with yellow oil liquid. Identify (A) to (E) and write the reactions.
Answer: Let's identify the compounds step-by-step based on the reactions given:
1. Compound (A), nitrobenzene, is reduced by Sn/HCl to form compound (B), aniline.
\( \text{C}_6\text{H}_5\text{NO}_2 \text{ (A)} + \text{6[H]} \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{-NH}_2 \text{ (B)} + \text{2H}_2\text{O} \)
2. Compound (B), aniline, reacts with benzoyl chloride in the presence of pyridine (this is a Schotten-Baumann reaction) to form compound (C), N-phenyl benzamide.
\( \text{C}_6\text{H}_5\text{NH}_2 \text{ (B)} + \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{Pyridine}} \text{C}_6\text{H}_5\text{NH-CO-C}_6\text{H}_5 \text{ (C)} + \text{HCl} \)
3. Compound (B), aniline, also reacts with methyl bromide (\( \text{CH}_3\text{Br} \)) to produce compound (D), N-methylaniline.
\( \text{C}_6\text{H}_5\text{NH}_2 \text{ (B)} + \text{CH}_3\text{Br} \longrightarrow \text{C}_6\text{H}_5\text{NHCH}_3 \text{ (D)} \)
4. Compound (D), N-methylaniline, then reacts with \( \text{NaNO}_2 \)/HCl to give compound (E), N-nitroso methyl phenylamine, which is a yellow oily liquid. This is Libermann's nitroso test for secondary amines.
\( \text{C}_6\text{H}_5\text{NHCH}_3 \text{ (D)} + \text{HNO}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{C}_6\text{H}_5\text{N(NO)CH}_3 \text{ (E)} + \text{H}_2\text{O} \)
Here is a summary of the compounds:

In simple words: We start with a nitro compound (A) and change it into an amine (B). Then, we make different products (C, D, E) from this amine through a series of chemical steps, like adding a benzoyl group or a methyl group, and then reacting with nitrous acid.

🎯 Exam Tip: When solving synthesis problems, identify the functional groups at each step to determine the type of reaction and reagent needed. Pay attention to the conditions (e.g., pyridine, \( \text{NaNO}_2 \)/HCl) as they are crucial for specific transformations.