Amines

Step 1: Understanding the Reaction
The reaction is the acylation (specifically, acetylation) of 3-aminobenzamide with acetic anhydride ((CH CO) O). We need to determine which of the two nitrogen atoms in the molecule will react to form the major product.
Step 2: Comparing the Nucleophilicity of the Nitrogen Atoms
The starting material, 3-aminobenzamide, has two nitrogen atoms, each with a lone pair of electrons:
1. The aromatic amine nitrogen (-NH ): The lone pair on this nitrogen is delocalized into the benzene ring through resonance. This reduces its availability and nucleophilicity compared to a simple alkyl amine, but it is still quite nucleophilic.
2. The amide nitrogen (-CONH ): The lone pair on this nitrogen is strongly delocalized onto the adjacent carbonyl oxygen atom. This resonance is very significant, making the amide nitrogen much less basic and significantly less nucleophilic than the aromatic amine nitrogen.
Step 3: Predicting the Site of Acylation
Acetic anhydride is an electrophile. The reaction is a nucleophilic acyl substitution where the nucleophile is one of the nitrogen atoms. Since the aromatic amine nitrogen is significantly more nucleophilic than the amide nitrogen, it will preferentially attack the electrophilic carbonyl carbon of the acetic anhydride.
The reaction is:

The amino group (-NH ) on the ring is converted to an acetamido group (-NHCOCH ).
Step 4: Final Answer
The major product 'P' is 3-(acetylamino)benzamide, where acylation has occurred at the more nucleophilic aromatic amino group. This corresponds to the structure given in option (A).

To determine the compounds ‘A’ and ‘B’, we need to understand the reactions shown.

The given reaction involves the conversion of an alkyl halide to different compounds using different reagents: AgCN and NaCN. These reagents differ in their behavior due to the difference in the nature of the cyanide ion attachment to silver and sodium.

Reaction with AgCN:

• Silver cyanide (AgCN) has a covalent cyanide ion. This nature leads to the formation of isocyanide when reacting with alkyl halides.
• Thus, when the alkyl chloride reacts with AgCN, the major product formed is an isocyanide, i.e., R-NC (alkyl isocyanide).

Reaction with NaCN:

• Sodium cyanide (NaCN), on the other hand, has an ionic cyanide ion, which can easily donate the cyanide ion.
• When the alkyl chloride reacts with NaCN, the major product formed is a nitrile, i.e., R-CN (alkyl cyanide).

Based on the reactions, the compounds can be identified as follows:

Fig. Correct structures of ‘A’ and ‘B’.

Therefore, for the given question, the correct compounds ‘A’ and ‘B’ are alkyl isocyanide and alkyl cyanide, respectively.

The correct answer is (C) : (B) and (D) are correct statements.

Hence, H₂SO₄acts as an acid.

The correct answer is (C) :
The most appropriate option is (C) as one group is far enough from –COOH group

To determine the product ‘B’, we need to analyze the given reaction sequence step-by-step:

The first step involves the reduction of the cyano group (−CN) using diisobutylaluminium hydride (DIBAL-H) followed by hydrolysis.

This transforms the −CN group into an aldehyde group.

Thus, compound 'A' is formed by converting the nitrile to an aldehyde.

In the second step, compound ‘A’ reacts with acetaldehyde (CH₃CHO) under aldol condensation conditions (presence of dilute NaOH and heat).

Aldol condensation occurs when enolate ions, formed from aldehydes or ketones in basic conditions, react with another aldehyde or ketone. Here, the enolate form of acetaldehyde will add to the aldehyde formed from the initial reduction.

This reaction leads to the formation of a β-hydroxy ketone or aldehyde, which can further undergo dehydration to form an α,β-unsaturated carbonyl compound.

Therefore, the major product ‘B’ is the unsaturated carbonyl compound formed through the aldol condensation reaction.

The correct option representing product ‘B’ is:

The given reaction involves two steps. Let's analyze each step:

Step 1: Acid Chloride Reaction with Ammonia

The first reaction involves compound 'A' (C₆H₅Cl₂O) reacting with ammonia (NH₃). This step typically forms an amide. Since compound 'A' is described as having the formula C₆H₅Cl₂O, it can be inferred to be an acyl chloride. The general reaction with ammonia forms an amide:

1. Formation of an amide should clarify the intermediate product

Therefore, the intermediate compound formed is C₆H₅CINH₂O, matching the description given as C₆H₅ClNO.

Step 2: Hoffmann Bromamide Degradation

In the presence of Br₂ and NaOH, the amide undergoes Hoffmann Bromamide Degradation. This reaction removes a carbonyl group, ultimately forming an amine while decreasing the carbon count by one:

The final product has an amino group (NH₂) in the para position relative to the chloro group, and a methyl group (CH₃). This corresponds to 4-Chloro-3-methylaniline.

Following these steps, you can verify that the correct compound 'A' aligns with the initial reactant that leads to this product.

Correct structure of compound 'A' (Acyl Chloride)

Thus, the correct answer is the third option, as it matches with the starting acyl chloride structure required for this reaction sequence.

The correct answer is (C) :

The correct answer is (C) : On treatment with secondary amine, it leads to a product, that is soluble in alkali

Due to formation of anilinium ion in acidic medium meta product is also obtained in significant amount

Aniline does not undergo Friedel Craft reaction because the reagent being electron deficient acts as a Lewis acid.

So, the correct option is (C): (A) is true, but (R) is false.

The correct answer is (B) : alcohol

The correct answer is (D) :
is the strongest base among the given compounds due to the maximum +I effect and the lone pair of N is not in dynamic state so it can be donated easily.

The Hinsberg reagent is a chemical used to distinguish primary, secondary, and tertiary amines. It typically consists of benzene sulfonyl chloride. When primary amines react with the Hinsberg reagent, they form a sulfonamide that is soluble in alkali. Secondary amines form a compound that is insoluble in alkali, while tertiary amines do not react with the Hinsberg reagent.

In the question provided, we have four images as options for identifying the Hinsberg reagent. Let's examine each option:

Correct option: Benzene sulfonyl chloride (Hinsberg reagent)

From the above analysis, the correct option is the benzene sulfonyl chloride, which is represented by the first image:

• The first image corresponds to the chemical structure of benzene sulfonyl chloride.
• The other images represent different chemical structures that are not used as the Hinsberg reagent.

Therefore, the correct answer is the first image, which depicts the Hinsberg reagent.

The correct answer is (D) : Positively charged nitrogen at benzene ring.

Lewis acids are electron deficient species and nitrogen atom of aniline has a lone pair of electron. When Lewis acid reacts with aniline, then a salt forms in which nitrogen acquires a positive reaction and further reaction is not possible.

To solve this problem, we need to identify the organic compound 'A' based on the given properties and reactions it undergoes. Let's analyze the details step-by-step:

• Compound Properties: The compound 'A' contains nitrogen and chlorine and is acidic in nature, as it turns litmus red.
• Molecular Weight: The given molecular weight of 'A' is around 131±2.
• Reaction with NaOH: When treated with aqueous NaOH, a liquid separates that contains nitrogen but not chlorine. This suggests the formation of a compound that has nitrogen but has removed the chlorine, likely as a gaseous product like HCl.
• Diazotization: The liquid obtained after treatment with NaOH, when reacted with nitrous acid and phenol, gives an orange precipitate. This indicates the presence of a primary aromatic amine, as diazotization followed by coupling with phenol yields azo dyes.

Based on these observations, the compound 'A' is likely to be an aromatic amine with a hydrochloride form, as aromatic amines can undergo diazotization and azo coupling reactions.

Let's consider the options provided and their structures:

The compound depicted in the image corresponds to a hydrochloride salt of an aromatic amine. It fits all the conditions:

• Dissolves in water to form an acidic solution (due to HCl present).
• Has a molecular weight close to 131 g/mol.
• Reacts with NaOH to release HCl and form a free amine.
• Undergoes diazotization and coupling with phenol to form an azo dye, indicating it is a primary aromatic amine.

In aqueous medium, the basic strength of amines depends on:
1. Electron density on the nitrogen atom (due to inductive effect of methyl groups).
2. Solvation effect, which stabilizes the conjugate acid after accepting a proton.

- Dimethylamine (Me NH): The best combination of inductive effect and solvation, making it the most basic.
- Methylamine (MeNH ): Slightly less basic due to less inductive effect but better solvation.
- Trimethylamine (Me N): Weaker basicity because bulky groups hinder solvation.
- Ammonia (NH ): Lowest basicity due to the absence of inductive effect.

Thus, the basic strength order is Me NH MeNH Me N NH . Basicity in aqueous medium is influenced by both inductive effects and solvation. Dimethylamine balances these effects best, while trimethylamine suffers from steric hindrance.

(I) CH₃-COOH + Br₂ (Red P) Hell Volhard Zelinsky Reaction (Q)

• This reaction involves the bromination of carboxylic acids at the alpha carbon position.
• It requires the presence of phosphorus (P).

(II) CH₃-C=O-CH₃ + NaOI Iodoform Reaction (R)

• The iodoform reaction identifies compounds with a methyl ketone group (-CO-CH₃).
• It produces a yellow precipitate of iodoform (CHI₃).

(III) Benzene + CO + HCl + AlCl₃ Gattermann Koch Reaction (P)

• This reaction involves the formylation of benzene to form benzaldehyde.
• It requires carbon monoxide (CO), hydrogen chloride (HCl), and AlCl₃ as catalysts.

Correct Matches:

• (I) → Q
• (II) → R
• (III) → P

The correct answer is 3.

Analysis of Statements on Methyl Orange:

Statement I:

Methyl orange is not a weak acid; it is an acid-base indicator that exists in different structural forms depending on the pH of the solution:

• In acidic solutions, methyl orange exists in its protonated form.
• In basic solutions, it undergoes deprotonation to form a conjugate base.

Conclusion: Since methyl orange is not classified as a weak acid, Statement I is incorrect.

Statement II:

The color intensity of methyl orange depends on its structural form:

• The quinonoid form of methyl orange is more intensely or deeply colored and is responsible for the orange/red coloration in acidic media.
• The benzenoid form is less intensely colored and contributes to the yellow coloration in basic media.

Since the statement incorrectly claims that the benzenoid form is more intensely colored, Statement II is also incorrect.

Final Conclusion:

Since both Statement I and Statement II are incorrect, the correct answer is option (1).

Statement I: Pure aniline and other arylamines are indeed colourless liquids.
Statement II: Aniline becomes coloured on storage, not due to reduction, but due to oxidation by air and light, forming brown-coloured quinonoid compounds. Hence, Statement II is incorrect.

The correct answer is (A) : A-III, B-IV, C-II, D-I

Amines are organic compounds that contain a nitrogen atom bonded to one or more alkyl or aryl groups. They are weak bases and react with acids to form salts.
The strength of a base is measured by its pKb value, which is the negative logarithm of its base dissociation constant (Kb). The higher the pKb value, the weaker the base.
List I contains the names of different amines, while List II contains their respective pKb values.
To match the amines with their correct pKb values, we need to compare their structures and the effects of the different alkyl or aryl groups attached to the nitrogen atom on their basicity.

The given reaction sequence is;

Step 1: Diazotization. The first step involves the reaction of the given compound (2-hydroxycyclohexanone) with sodium nitrite (NaNO2) and hydrochloric acid (HCl). This is a diazotization reaction, which converts the amino group (-NH2) of an aromatic amine to a diazonium salt (-N⁺ 2Cl⁻). This reaction doesn’t happen with the carbonyl group. Since the starting compound shown is not an amino compound, it needs to have NH2 for diazotization reaction to happen. Hence it would be more logical if the question used a starting compound of 2-aminocyclohexanol rather than 2-hydroxycyclohexanone. Let’s assume that the starting material is

Then the diazonium salt is formed

Step 2: Reduction of Diazonium salt. The second step involves the reaction of the diazonium salt (A) with ammonium hydrosulfide (NH4SH) in water. This re duces the diazonium salt to the corresponding amine, with nitrogen gas as a byproduct.
So the final product B is 2-aminocyclohexanol.

To solve this problem, let's analyze each statement individually and provide the correct reasoning behind why both statements are true.

• Statement I: Aniline reacts with concentrated followed by heating at 453–473 K to form p-aminobenzene sulphonic acid, which gives a blood red color in the 'Lassaigne's test'.

Explanation:

• In concentrated , aniline undergoes sulfonation to form p-aminobenzene sulphonic acid (also known as sulphanilic acid) when heated in the range of 453–473 K.
• This compound, upon Lassaigne's test for sulphur, forms sodium thiocyanate, which reacts with ferric chloride to give a blood-red color.

This establishes that Statement I is true.

• Statement II: In Friedel-Craft's alkylation and acylation reactions, aniline forms a salt with the catalyst. Due to this, the nitrogen of aniline acquires a positive charge and acts as a deactivating group.

Explanation:

• Aniline is a base and reacts with Lewis acids such as to form a salt. The nitrogen atom of aniline gets protonated, acquiring a positive charge.
• This positively charged nitrogen significantly reduces the electron density of the aromatic ring, making it less reactive in the Friedel-Craft's reactions, hence acting as a deactivating group.

This confirms that Statement II is also true.

Since both statements are true, the correct option is: Both Statement I and Statement II are true.