Q: Q7. How will you convert:

The conversions are carried out through a series of reactions as detailed below:

Q: Q8. How will you convert:\n(i) Ethanoic acid into methanamine\n(ii) Hexanenitrile into 1-aminopentane\n(iii) Methanol to ethanoic acid\n(iv) Ethanamine into methanamine\n(v) Ethanoic a

The conversions are carried out through a series of reactions as follows:

Q: Q9. Write the reactions of (i) an aromatic and (ii) an aliphatic primary amine with nitrous acid.

Nitrous acid (HNO₂) is unstable and is prepared *in situ* by mixing sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at low temperatures (0-5°C).\n\n**(i) Reaction of an aliphatic primary amine with nitrous acid:**\nAliphatic primary amines react with nitrous acid to form unstable aliphatic diazonium salts, which immediately decompose to form alcohols, nitrogen gas, and hydrochloric acid. This reaction is often used to distinguish primary aliphatic amines from other amines, as it produces a brisk effervescence of nitrogen gas.\n\n**General Reaction:**\n`R-NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → [R-N₂⁺Cl⁻] (unstable) → R-OH + N₂↑ + HCl`\n\n**Example (Ethylamine):**\n`CH₃CH₂NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → [CH₃CH₂N₂⁺Cl⁻] → CH₃CH₂OH + N₂↑ + HCl`\n\n**(ii) Reaction of an aromatic primary amine with nitrous acid:**\nAromatic primary amines react with nitrous acid at low temperatures (0-5°C) to form stable arenediazonium salts. This reaction is known as diazotisation and is a very important reaction in organic synthesis, as the diazonium salts can be converted into a variety of other aromatic compounds.\n\n**General Reaction (Diazotisation):**\n`Ar-NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → Ar-N₂⁺Cl⁻ + 2H₂O`\n\n**Example (Aniline):**\n`C₆H₅NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → C₆H₅N₂⁺Cl⁻ + 2H₂O`\n(Benzene diazonium chloride)

Question 1

Q1. Classify the following amines as primary, secondary or tertiary:
(a) CH3—CH2—NH2
(b) (CH3)2NH
(c) CH3—NH—CH(CH3)2
(d) (CH3)3N
(e) C6H5—NH2
(f) C6H5—N(CH3)2
(g) CH3—CH2—N(CH3

Accepted Answer

Amines are classified as primary (1°), secondary (2°), or tertiary (3°) based on the number of alkyl or aryl groups directly attached to the nitrogen atom.

(a) CH3—CH2—NH2: **Primary (1°)**
    (The nitrogen atom is attached to one ethyl group and two hydrogen atoms.)

(b) (CH3)2NH: **Secondary (2°)**
    (The nitrogen atom is attached to two methyl groups and one hydrogen atom.)

(c) CH3—NH—CH(CH3)2: **Secondary (2°)**
    (The nitrogen atom is attached to one methyl group, one isopropyl group, and one hydrogen atom.)

(d) (CH3)3N: **Tertiary (3°)**
    (The nitrogen atom is attached to three methyl groups.)

(e) C6H5—NH2: **Primary (1°)**
    (The nitrogen atom is attached to one phenyl group and two hydrogen atoms.)

(f) C6H5—N(CH3)2: **Tertiary (3°)**
    (The nitrogen atom is attached to one phenyl group and two methyl groups.)

(g) CH3—CH2—N(CH3)—CH2—CH3: **Tertiary (3°)**
    (The nitrogen atom is attached to two ethyl groups and one methyl group.)

Question 2

Q2. Give the IUPAC names of the following compounds and classify them as primary, secondary or tertiary amines.
(a) CH3CH(CH3)CH2NH2
(b) CH3(CH2)2NH2
(c) CH3NHCH(CH3)2
(d) (CH3)3CN

Accepted Answer

The IUPAC naming of amines follows specific rules:
1.  **Primary amines:** The longest carbon chain containing the -NH2 group is identified. The 'e' of the corresponding alkane is replaced by 'amine'. The position of the amino group and any substituents are indicated by numbers.
2.  **Secondary and Tertiary amines:** These are named as N-substituted derivatives of a primary amine. The largest alkyl group attached to nitrogen is chosen as the parent amine, and the other alkyl groups are treated as N-substituents.
3.  **Aromatic amines:** Aniline is the parent compound. Substituents on the ring are numbered, and substituents on the nitrogen are indicated by 'N-'.

Classification is based on the number of alkyl/aryl groups attached directly to the nitrogen atom (1° for one, 2° for two, 3° for three).

Here are the IUPAC names and classifications for the given compounds:

(a) **CH3CH(CH3)CH2NH2**
    *   **Structure:** (CH3)2CH-CH2-NH2
    *   **IUPAC Name:** 2-Methylpropan-1-amine
    *   **Classification:** Primary (1°) amine (Nitrogen is attached to one alkyl group, -CH2CH(CH3)2).

(b) **CH3(CH2)2NH2**
    *   **Structure:** CH3-CH2-CH2-NH2
    *   **IUPAC Name:** Propan-1-amine
    *   **Classification:** Primary (1°) amine (Nitrogen is attached to one alkyl group, -CH2CH2CH3).

(c) **CH3NHCH(CH3)2**
    *   **Structure:** (CH3)2CH-NH-CH3
    *   **IUPAC Name:** N-Methylpropan-2-amine
    *   **Classification:** Secondary (2°) amine (Nitrogen is attached to one methyl and one isopropyl group).

(d) **(CH3)3CNH2**
    *   **Structure:** (CH3)3C-NH2
    *   **IUPAC Name:** 2-Methylpropan-2-amine
    *   **Classification:** Primary (1°) amine (Nitrogen is attached to one alkyl group, -C(CH3)3).

(e) **C6H5NHCH3**
    *   **Structure:** C6H5-NH-CH3
    *   **IUPAC Name:** N-Methylaniline (or N-Methylbenzenamine)
    *   **Classification:** Secondary (2°) amine (Nitrogen is attached to one phenyl and one methyl group).

(f) **(CH3CH2)2NCH3**
    *   **Structure:** (CH3CH2)2N-CH3
    *   **IUPAC Name:** N,N-Diethylethanamine (or N,N-Diethylmethanamine if methyl is parent, but ethyl is larger so N,N-Diethylethanamine is incorrect. It should be N,N-Diethylethanamine if the parent is ethylamine, but here the parent is methanamine with two ethyl groups attached to N. So N,N-Diethylmethanamine is correct.) Let's re-evaluate. The largest alkyl group is ethyl. So it should be N-Ethyl-N-methylethanamine. No, the parent chain is determined by the longest carbon chain attached to N. Here, we have two ethyl groups and one methyl group. If we take ethyl as the parent, it's N-Ethyl-N-methylethanamine. If we take methyl as the parent, it's N,N-Diethylethanamine. IUPAC prefers the largest group as the parent. So, N-Ethyl-N-methylethanamine. Let's recheck the common convention. For tertiary amines, if all groups are different, the largest is parent. If two are same, and one is different, the two same groups are N,N-di-substituents on the parent of the different group. So, N,N-Diethylethanamine is incorrect. It should be N-Ethyl-N-methylethanamine. Wait, if the parent is ethanamine, then the other ethyl and methyl are substituents. So N-Ethyl-N-methylethanamine. This is correct.     *   **IUPAC Name:** N-Ethyl-N-methylethanamine (The parent amine is ethanamine, with an ethyl and a methyl group on the nitrogen.)     *   **Classification:** Tertiary (3°) amine (Nitrogen is attached to two ethyl groups and one methyl group).

(g) **m-BrC6H4NH2**
    *   **Structure:** A benzene ring with an -NH2 group and a -Br group at meta positions to each other.
    *   **IUPAC Name:** 3-Bromobenzenamine (or m-Bromoaniline)
    *   **Classification:** Primary (1°) amine (Nitrogen is attached to one aryl group, C6H4Br).

Question 3

Q3. Write structures and IUPAC names of different isomeric amines corresponding to the molecular formula C4H11N.

Accepted Answer

The molecular formula C4H11N indicates that these are saturated amines. We need to find all possible structural isomers, including primary, secondary, and tertiary amines.

**1. Primary Amines (1°):** (R-NH2)
   These will have one alkyl group attached to the nitrogen atom.

*   **Butan-1-amine**
       *   Structure: CH3—CH2—CH2—CH2—NH2
       *   IUPAC Name: Butan-1-amine

*   **Butan-2-amine**
       *   Structure: CH3—CH2—CH(NH2)—CH3
       *   IUPAC Name: Butan-2-amine

*   **2-Methylpropan-1-amine**
       *   Structure: (CH3)2CH—CH2—NH2
       *   IUPAC Name: 2-Methylpropan-1-amine

*   **2-Methylpropan-2-amine**
       *   Structure: (CH3)3C—NH2
       *   IUPAC Name: 2-Methylpropan-2-amine

**2. Secondary Amines (2°):** (R—NH—R')
   These will have two alkyl groups attached to the nitrogen atom.

*   **N-Methylpropan-1-amine**
       *   Structure: CH3—CH2—CH2—NH—CH3
       *   IUPAC Name: N-Methylpropan-1-amine

*   **N-Methylpropan-2-amine**
       *   Structure: (CH3)2CH—NH—CH3
       *   IUPAC Name: N-Methylpropan-2-amine

*   **N-Ethylethanamine**
       *   Structure: CH3—CH2—NH—CH2—CH3
       *   IUPAC Name: N-Ethylethanamine

**3. Tertiary Amines (3°):** (R—N(R')—R'')
   These will have three alkyl groups attached to the nitrogen atom.

*   **N,N-Dimethylethanamine**
       *   Structure: CH3—CH2—N(CH3)—CH3
       *   IUPAC Name: N,N-Dimethylethanamine

In total, there are **8** isomeric amines for the molecular formula C4H11N.

Question 4

Q4. Arrange the following in increasing order of their basic strength:
(a) C₂H₅NH₂, C₆H₅NH₂, NH₃, (C₂H₅)₂NH
(b) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(c) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N,

Accepted Answer

The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom for donation. This availability is influenced by: 1. **Inductive effect (+I effect):** Alkyl groups are electron-donating, increasing electron density on nitrogen and thus increasing basicity. More alkyl groups generally mean higher basicity. 2. **Resonance effect:** Aromatic amines (like aniline) have the lone pair on nitrogen delocalized into the benzene ring, making it less available for protonation, thus decreasing basicity. 3. **Steric hindrance:** In tertiary amines, bulky alkyl groups can hinder the approach of a proton, reducing basicity, especially in aqueous solution. 4. **Solvation effect:** In aqueous solution, the stability of the conjugate acid (protonated amine) due to hydrogen bonding with water molecules plays a crucial role. Primary amines form more H-bonds than secondary, and secondary more than tertiary. This effect often counteracts the inductive effect, leading to an order of basicity in aqueous solution that is not simply 3° > 2° > 1°. Let's apply these principles to each series: **(a) C₂H₅NH₂, C₆H₅NH₂, NH₃, (C₂H₅)₂NH** * C₆H₅NH₂ (Aniline) is the least basic due to resonance stabilization of the lone pair with the benzene ring. * NH₃ is less basic than aliphatic amines due to the absence of +I effect from alkyl groups. * C₂H₅NH₂ (primary amine) has one ethyl group (+I effect). * (C₂H₅)₂NH (secondary amine) has two ethyl groups (+I effect) and better solvation than tertiary, making it generally more basic than primary in aqueous solution. * **Order of basic strength:** C₆H₅NH₂ < NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH **(b) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂** * C₆H₅NH₂ is the least basic due to resonance. * Among ethyl amines in aqueous solution, the order is typically secondary > primary > tertiary due to a balance of +I effect and solvation. * **Order of basic strength:** C₆H₅NH₂ < (C₂H₅)₃N < C₂H₅NH₂ < (C₂H₅)₂NH **(c) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, NH₃** * NH₃ is less basic than methyl amines. * Among methyl amines in aqueous solution, the order is typically secondary > primary > tertiary due to a balance of +I effect and solvation. * **Order of basic strength:** (CH₃)₃N < NH₃ < CH₃NH₂ < (CH₃)₂NH **(d) C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₆H₅)₂NH, CH₃NH₂** * (C₆H₅)₂NH (diphenylamine) is the least basic because the lone pair on nitrogen is delocalized into *two* benzene rings, making it highly unavailable. * C₆H₅NH₂ (aniline) is less basic than aliphatic amines due to one benzene ring's resonance effect. * C₆H₅N(CH₃)₂ (N,N-dimethylaniline) is more basic than aniline because the two methyl groups exert a +I effect, pushing electron density towards nitrogen, partially counteracting the resonance effect. However, it's still less basic than aliphatic amines. * CH₃NH₂ (methylamine) is an aliphatic primary amine, highly basic due to the +I effect of the methyl group and good solvation. * **Order of basic strength:** (C₆H₅)₂NH < C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂

Question 5

Q5. Discuss the following:
(a) Gabriel phthalimide synthesis
(b) Hoffmann bromamide degradation reaction

Accepted Answer

### (a) Gabriel Phthalimide Synthesis

**Introduction:**
Gabriel phthalimide synthesis is a method used for the preparation of primary amines. It is particularly useful for synthesizing pure primary aliphatic amines, as it avoids the formation of secondary and tertiary amines, which often occurs during the direct ammonolysis of alkyl halides. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

**Reaction:**
The synthesis involves three main steps:
1.  **Formation of potassium phthalimide:** Phthalimide reacts with an alcoholic solution of potassium hydroxide to form potassium phthalimide.
    *   **Reaction:** Phthalimide + KOH → Potassium Phthalimide + H₂O
2.  **N-alkylation:** Potassium phthalimide, being a strong nucleophile, reacts with an alkyl halide (preferably primary alkyl halide) to form N-alkylphthalimide.
    *   **Reaction:** Potassium Phthalimide + R-X → N-Alkylphthalimide + KX
3.  **Hydrolysis (or hydrazinolysis):** The N-alkylphthalimide is then hydrolyzed with aqueous acid or base, or more commonly, treated with hydrazine (hydrazinolysis) to yield the primary amine and phthalic acid (or phthalate salt) or phthalhydrazide, respectively.
    *   **Reaction (Hydrolysis):** N-Alkylphthalimide + 2H₂O (acid/base) → R-NH₂ + Phthalic acid/salt
    *   **Reaction (Hydrazinolysis):** N-Alkylphthalimide + N₂H₄ → R-NH₂ + Phthalhydrazide

**Mechanism (Simplified):**
1.  Phthalimide, being acidic (due to the electron-withdrawing carbonyl groups), loses a proton to KOH, forming the phthalimide anion (a strong nucleophile).
2.  The phthalimide anion attacks the alkyl halide (R-X) in an S_N2 reaction, displacing the halide and forming N-alkylphthalimide.
3.  Hydrolysis or hydrazinolysis then cleaves the N-C bonds, releasing the primary amine.

**Advantages:**
*   Produces pure primary amines, free from secondary and tertiary amines.
*   Useful for preparing primary aliphatic amines.

**Limitations:**
*   Aromatic primary amines cannot be prepared because aryl halides do not undergo S_N2 reactions with the phthalimide anion.
*   Tertiary alkyl halides are not suitable due to elimination reactions.

**NEET Relevance:** High. This is a very important named reaction. Questions often involve identifying the product, reagents, or the limitation regarding aromatic amines.

### (b) Hoffmann Bromamide Degradation Reaction

**Introduction:**
Hoffmann bromamide degradation (also known as Hoffmann rearrangement or Hoffmann hypobromite degradation) is an organic reaction used for the preparation of primary amines by the degradation of an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. The most significant feature of this reaction is that the amine formed contains one carbon atom less than the parent amide.

**Reaction:**
An amide (RCONH₂) reacts with bromine (Br₂) and a strong base (NaOH or KOH) to form a primary amine (RNH₂), sodium carbonate, sodium bromide, and water.

*   **General Reaction:** RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O

**Mechanism (Simplified):**
1.  **Deprotonation of amide:** The amide reacts with the base to form an N-anion.
2.  **Bromination:** The N-anion reacts with bromine to form N-bromoamide.
3.  **Second deprotonation:** The N-bromoamide is deprotonated by the base to form an N-bromoamide anion.
4.  **Rearrangement:** The N-bromoamide anion undergoes a rearrangement where the alkyl or aryl group (R) migrates from the carbonyl carbon to the nitrogen atom, simultaneously with the expulsion of the bromide ion. This forms an isocyanate (R-N=C=O).
5.  **Hydrolysis of isocyanate:** The isocyanate is then hydrolyzed by water (in the presence of base) to form a carbamic acid, which quickly decarboxylates to yield the primary amine and carbon dioxide (which reacts with NaOH to form Na₂CO₃).

**Key Features:**
*   **Carbon Shortening:** The product amine has one carbon atom less than the starting amide. This is a crucial characteristic.
*   **Stereospecificity:** If the R group is chiral, its configuration is retained during migration.
*   **Preparation of Primary Amines:** Exclusively yields primary amines.

**Advantages:**
*   A good method for descending a homologous series (reducing the carbon chain by one).
*   Provides pure primary amines.

**Limitations:**
*   Only primary amines can be prepared.
*   Requires careful control of reaction conditions.

**NEET Relevance:** High. This is a very important named reaction. Questions frequently test the carbon-shortening aspect, the reagents used, and the type of amine formed.

Question 6

Q6. Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide degradation reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Ace

Accepted Answer

Here are short notes on the specified reactions and processes:

**(i) Carbylamine reaction (Isocyanide Test)**
This reaction is a characteristic test for primary amines (both aliphatic and aromatic). When a primary amine is heated with chloroform (CHCl₃) and an alcoholic solution of potassium hydroxide (KOH), it forms an isocyanide (carbylamine), which has a highly unpleasant or foul smell. Secondary and tertiary amines do not give this reaction.

**General Reaction:**
R-NH₂ + CHCl₃ + 3KOH (alc.) → R-NC + 3KCl + 3H₂O
(Primary Amine) (Chloroform) (Potassium hydroxide) (Alkyl/Aryl Isocyanide)

**Example:**
CH₃NH₂ + CHCl₃ + 3KOH (alc.) → CH₃NC + 3KCl + 3H₂O
(Methanamine) (Methyl Isocyanide)

**(ii) Diazotisation**
Diazotisation is the process of converting a primary aromatic amine into an arenediazonium salt. This reaction is carried out by treating the primary aromatic amine with nitrous acid (HNO₂) at a low temperature (0-5°C or 273-278 K). Nitrous acid is usually generated *in situ* by reacting sodium nitrite (NaNO₂) with a mineral acid like hydrochloric acid (HCl).

**General Reaction:**
Ar-NH₂ + NaNO₂ + 2HCl → [Ar-N₂⁺Cl⁻] + NaCl + 2H₂O
(Primary Aromatic Amine) (Sodium Nitrite) (Hydrochloric Acid) (Arenediazonium Chloride)

**Example:**
C₆H₅NH₂ + NaNO₂ + 2HCl → [C₆H₅-N₂⁺Cl⁻] + NaCl + 2H₂O
(Aniline) (Benzenediazonium Chloride)

Aliphatic diazonium salts are highly unstable and decompose rapidly, even at low temperatures.

**(iii) Hofmann’s bromamide degradation reaction**
This reaction is a method for the preparation of primary amines by degrading an amide. When an amide is treated with bromine (Br₂) in an aqueous or alcoholic solution of sodium hydroxide (NaOH), it yields a primary amine that contains one carbon atom less than the original amide. This reaction is useful for 'stepping down' a homologous series.

**General Reaction:**
R-CO-NH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(Amide) (Bromine) (Sodium Hydroxide) (Primary Amine)

**Example:**
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(Acetamide) (Methanamine)

**(iv) Coupling reaction**
Coupling reactions are characteristic reactions of arenediazonium salts, particularly with electron-rich aromatic compounds like phenols and aromatic amines. In these reactions, the diazonium ion acts as a weak electrophile and attacks the activated aromatic ring, typically at the para-position, to form brightly coloured azo compounds (azo dyes). The nitrogen atoms of the diazo group (-N=N-) link the two aromatic rings.

**Example with Phenol (in alkaline medium):**
[C₆H₅-N₂⁺Cl⁻] + C₆H₅OH → C₆H₅-N=N-C₆H₄-OH (p-hydroxyazobenzene, orange dye)
(Benzenediazonium Chloride) (Phenol)

**Example with Aniline (in weakly acidic medium):**
[C₆H₅-N₂⁺Cl⁻] + C₆H₅NH₂ → C₆H₅-N=N-C₆H₄-NH₂ (p-aminoazobenzene, yellow dye)
(Benzenediazonium Chloride) (Aniline)

**(v) Ammonolysis**
Ammonolysis refers to the cleavage of a carbon-halogen bond by ammonia. Alkyl halides react with an alcoholic solution of ammonia to form primary amines. However, the primary amine formed is still nucleophilic and can further react with the alkyl halide to form secondary, tertiary amines, and eventually quaternary ammonium salts. This makes it a less suitable method for preparing pure primary amines due to the formation of a mixture of products.

**Reaction Sequence:**
1. R-X + NH₃ → R-NH₃⁺X⁻ (Substituted ammonium salt)
2. R-NH₃⁺X⁻ + NaOH → R-NH₂ + NaX + H₂O (Primary amine)

**Further reactions leading to a mixture:**
R-NH₂ + R-X → R₂NH + HX (Secondary amine)
R₂NH + R-X → R₃N + HX (Tertiary amine)
R₃N + R-X → R₄N⁺X⁻ (Quaternary ammonium salt)

**(vi) Acetylation**
Acetylation is the process of introducing an acetyl group (CH₃CO-) into a molecule. Amines react with acetylating agents like acetyl chloride (CH₃COCl) or acetic anhydride ((CH₃CO)₂O) to form N-substituted amides. This reaction involves the replacement of a hydrogen atom on the nitrogen with an acetyl group. Acetylation is often used to protect the amino group or to reduce its basicity for further reactions like electrophilic substitution in aromatic amines.

**General Reaction with Acetyl Chloride:**
R-NH₂ + CH₃COCl → R-NH-CO-CH₃ + HCl
(Amine) (Acetyl Chloride) (N-Alkylacetamide)

**General Reaction with Acetic Anhydride:**
R-NH₂ + (CH₃CO)₂O → R-NH-CO-CH₃ + CH₃COOH
(Amine) (Acetic Anhydride) (N-Alkylacetamide)

**(vii) Gabriel phthalimide synthesis**
Gabriel phthalimide synthesis is an excellent method for the preparation of pure primary aliphatic amines. It avoids the formation of secondary and tertiary amines, which is a common problem in ammonolysis of alkyl halides. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

**Reaction Sequence:**
1. **Formation of Potassium Phthalimide:** Phthalimide reacts with ethanolic potassium hydroxide (KOH) to form potassium phthalimide.
   Phthalimide + KOH → Potassium Phthalimide + H₂O

2. **Alkylation:** Potassium phthalimide undergoes an SN₂ reaction with an alkyl halide (R-X) to form N-alkylphthalimide.
   Potassium Phthalimide + R-X → N-Alkylphthalimide + KX

3. **Hydrolysis/Hydrazinolysis:** The N-alkylphthalimide is then hydrolysed with aqueous acid or base, or treated with hydrazine (N₂H₄), to yield the primary amine.
   N-Alkylphthalimide + 2H₂O/H⁺ or OH⁻ → R-NH₂ + Phthalic acid
   OR
   N-Alkylphthalimide + N₂H₄ → R-NH₂ + Phthalhydrazide

**Example:**
1. Phthalimide + KOH → Potassium Phthalimide
2. Potassium Phthalimide + CH₃CH₂Br → N-Ethylphthalimide + KBr
3. N-Ethylphthalimide + N₂H₄ → CH₃CH₂NH₂ + Phthalhydrazide
(Ethanamine)

Question 7

Q7. How will you convert:

Accepted Answer

The conversions are carried out through a series of reactions as detailed below:

Question 8

Q8. How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic a

Accepted Answer

The conversions are carried out through a series of reactions as follows:

Question 9

Q9. Write the reactions of (i) an aromatic and (ii) an aliphatic primary amine with nitrous acid.

Accepted Answer

Nitrous acid (HNO₂) is unstable and is prepared *in situ* by mixing sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at low temperatures (0-5°C).

**(i) Reaction of an aliphatic primary amine with nitrous acid:**
Aliphatic primary amines react with nitrous acid to form unstable aliphatic diazonium salts, which immediately decompose to form alcohols, nitrogen gas, and hydrochloric acid. This reaction is often used to distinguish primary aliphatic amines from other amines, as it produces a brisk effervescence of nitrogen gas.

**General Reaction:**
`R-NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → [R-N₂⁺Cl⁻] (unstable) → R-OH + N₂↑ + HCl`

**Example (Ethylamine):**
`CH₃CH₂NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → [CH₃CH₂N₂⁺Cl⁻] → CH₃CH₂OH + N₂↑ + HCl`

**(ii) Reaction of an aromatic primary amine with nitrous acid:**
Aromatic primary amines react with nitrous acid at low temperatures (0-5°C) to form stable arenediazonium salts. This reaction is known as diazotisation and is a very important reaction in organic synthesis, as the diazonium salts can be converted into a variety of other aromatic compounds.

**General Reaction (Diazotisation):**
`Ar-NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → Ar-N₂⁺Cl⁻ + 2H₂O`

**Example (Aniline):**
`C₆H₅NH₂ + HNO₂ (NaNO₂/HCl, 0-5°C) → C₆H₅N₂⁺Cl⁻ + 2H₂O`
(Benzene diazonium chloride)

Chapter 13: Amines