Amines

Step 1: Effect of acetyl chloride on aniline. Acetylation of aniline with acetyl chloride decreases the electron density on the benzene ring, making it less reactive in electrophilic substitution reactions.
Step 2: Conclusion. Thus, acetyl chloride decreases the activation of the benzene ring in aniline.

Step 1: Bond angles in methane and trimethylamine. The C–N–C bond angle in trimethylamine is very similar to the H–C–H bond angle in methane due to the sp hybridization of both carbon and nitrogen atoms.
Step 2: Conclusion. Thus, the C–N–C bond angle in trimethylamine does not differ from the H–C–H bond angle in methane.

Step 1: Understand the reaction of acylazides. Acylazides ( ) undergo a nucleophilic substitution reaction with water or other nucleophiles, producing amides ( ).
Step 2: Conclusion. Thus, acylazide in acidic or alkaline medium reacts to form R–NH .

Step 1: Basic strength of alkyl amines. The basic strength of alkyl amines decreases as the number of alkyl groups increases because alkyl groups are electron-donating and increase the electron density on nitrogen. More alkyl groups make the nitrogen less available to accept protons.
Step 2: Conclusion. Thus, the basic strength order is RNH >R NH>R N.

Step 1: Understanding base strengths.
The basicity of amines is influenced by the availability of the lone pair of electrons on nitrogen. In alkylamines, as the number of alkyl groups increases, the electron-donating ability increases, making the amine more basic.

Step 2: Conclusion.
Thus, the correct order of base strengths is .

Final Answer:

Step 1: Reaction of diazonium salts.
When diazonium salts are heated or boiled in water, they undergo hydrolysis, leading to the formation of phenol.

Step 2: Conclusion.
Thus, the correct product formed is phenol , so the correct answer is option (C).

Final Answer:

Step 1: Understanding carbamylation.
The carbamylation reaction involves the reaction of a primary amine with isocyanates to form a carbamate (urethane) derivative. This reaction typically occurs with primary amines.

Step 2: Conclusion.
Thus, the correct answer is option (A).

Final Answer:

Step 1: Write the compound formula.
Acetamide is:

Step 2: Acidic hydrolysis of amides.
Amides on hydrolysis with acid produce carboxylic acid and ammonium salt.

Step 3: Identify the organic product.
The main organic product is acetic acid (CH COOH).
Final Answer:

Step 1: Meaning of diazotization.
Diazotization is the reaction of primary aromatic amines with nitrous acid (NaNO /HCl at 0–5°C) to form diazonium salts.
Step 2: Check each option.
(A) Aniline (C H NH ) is primary aromatic amine undergoes diazotization.
(C) Toluidine derivative still primary aromatic amine undergoes diazotization.
(D) Nitroaniline derivative still aromatic primary amine (though slower) undergoes diazotization.
(B) Benzylamine (C H CH NH ) is primary aliphatic amine (amino group not directly on aromatic ring).
Aliphatic diazonium salts are unstable and decompose immediately, so it is considered that it does not undergo diazotization.
Final Answer:

Step 1: Recall difference between and .
- gives alkyl cyanides ( ) because attacks through carbon.
- gives alkyl isocyanides ( ) because attack occurs through nitrogen.
Step 2: Apply to given reaction.

So product is ethyl isocyanide.
Step 3: Check which statement is true.
In , ethyl group is attached to nitrogen:

Thus statement (III) is true.
Final Answer:

Step 1: Recall Gattermann reaction.
Gattermann reaction replaces diazonium group with halogen using copper powder and HX.
Step 2: For chlorine substitution.
When is introduced, reagent used is:

Step 3: Identify correct pair.
Thus:

Final Answer:

Step 1: Identify required conversion.
This is conversion of an amide to an amine with one carbon less.
Step 2: Name of reaction.
This is Hofmann bromamide reaction (Hofmann rearrangement).
Step 3: Reagent used.
Hofmann rearrangement uses in presence of strong base which forms in situ.
Thus reagent is .
Final Answer:

Step 1: Identify reduction product based on conditions.
Nitrobenzene reduction can give different products depending on medium.
In alkaline medium, partial reduction and coupling leads to azo compounds.
Step 2: Alkaline reduction produces azobenzene.
Using in presence of alcohol (methanol), nitrobenzene reduces and couples to form:
which is azobenzene.
Step 3: Why other reagents do not form azo.
gives aniline.
is mild acidic, gives phenylhydroxylamine/aniline.
Final Answer:

Step 1: Basicity depends on availability of lone pair.
More available lone pair stronger base.
Step 2: Compare aromatic amines.
In aniline and its derivatives, lone pair on nitrogen delocalizes into benzene ring by resonance, reducing basicity.
Step 3: Triphenylamine case.
In , lone pair is delocalized into three phenyl rings.
So resonance withdrawal is maximum and lone pair becomes least available for protonation.
Hence it is least basic.
Final Answer:

Step 1: Reaction type (Nucleophilic substitution).
p-nitrobromobenzene has a strong electron withdrawing group at para position.
This activates benzene ring towards nucleophilic substitution (SNAr).
Step 2: Role of sodium ethoxide.
Sodium ethoxide acts as nucleophile.
It replaces bromine on aromatic ring.
Step 3: Product formed.
Replacement of Br by gives p-nitrophenetole.
Final Answer:

Step 1: Understand the reaction.
The self-condensation of two moles of ethyl acetate in the presence of sodium ethoxide leads to the formation of acetoacetic ester. This reaction is known as the Claisen condensation.
Step 2: Conclusion.
Thus, the product of the self-condensation is acetoacetic ester.
Final Answer:

Step 1: Understand basicity.
The basicity of an amine is determined by the availability of the lone pair on nitrogen for protonation. Substituents that donate electron density (like methoxy) make the amine more basic, while electron-withdrawing groups (like nitro) decrease basicity.
Step 2: Conclusion.
p-Methoxyaniline, with its electron-donating methoxy group, will be the most basic among the options.
Final Answer:

Step 1: Reaction with and NaOH.
The reaction of phthalimide with and NaOH leads to the formation of anthranilic acid through a nucleophilic substitution.
Step 2: Explanation.
Thus, phthalimide reacts with bromine and NaOH to produce anthranilic acid.
Final Answer:

Step 1: Understanding the reactions.
The dye test is a method to distinguish between different amines based on their reactivity. -toluidine and benzyl amine exhibit different behaviors during the test, which allows their differentiation.
Step 2: Conclusion.
Hence, the dye test is the correct way to distinguish between -toluidine and benzyl amine.
Final Answer:

In aniline, p-methoxyaniline and p-methyl aniline, the lone pair of electrons on the Natom is delocalised on the benzene ring while in benzylamine it is delocalised, and more available for donation. Hence benzylamine is most basic among the given.

Among the given options, the ion is the most basic, as it can readily accept protons due to the lone pair on nitrogen.

Step 2: Conclusion.
The most basic compound is , corresponding to option (b).

Step 1: Understanding the Claisen condensation mechanism.
In a Claisen condensation, a strong base removes a proton from the -carbon of an ester. The resulting carbanion acts as a nucleophile and forms a new bond with another ester molecule.

Step 2: Conclusion.
All the statements provided are correct, corresponding to option (4).

Step 1: Definition of -amino acids.
In -amino acids, the amino group is attached to the carbon atom that is adjacent to the carboxyl group. Both structures given in options (1) and (2) describe -amino acids.

Step 2: Conclusion.
The correct answer is both (1) and (2), corresponding to option (3).

Step 1: Reagent action.
Anisole reacts with these reagents to form methylation of the aromatic ring with a methyl group from .

Step 2: Conclusion.
The major product formed is , corresponding to option (4).

Step 1: Understanding the reaction.
In the reduction of benzene diazonium chloride with hypophosphorous acid and water, red phosphorus ( ) is commonly used as the catalyst. This reduces the diazonium ion to produce benzene.

Step 2: Conclusion.
The catalyst used in this reaction is red phosphorus, corresponding to option (2).

Step 1: LiAlH4 Reduction.
Lithium aluminium hydride (LiAlH4) is a strong reducing agent and can reduce nitriles (R-C≡N) to primary amines (R-NH2).
Step 2: Conclusion.
The correct answer is (A), .

Step 1: Identify the reaction type.
The azo compound shown is likely formed by a diazotization reaction involving ethylamine and CH CH .
Step 2: Conclusion.
Thus, the starting reagents needed are CH CH and ethylamine.
Final Answer:

Step 1: Analyze the reaction sequence.
Compound A must be nitrobenzene, as its reduction and subsequent treatment with CHCl and alcoholic KOH lead to the formation of N-methyl aniline.
Step 2: Conclusion.
Thus, the compound A is nitrobenzene.
Final Answer:

Step 1: The reaction of aniline with nitrous acid leads to diazotization to form a diazonium salt.
Step 2: On further reaction with HNO₃, a substitution reaction occurs, producing a compound with a nitro group attached to the phenyl ring.

Final Answer:

Step 1: Nitrous acid reacts with primary amines to form alcohols and release gas. The reaction suggests a primary amine group.
Step 2: The structure of the compound formed on reduction suggests an amine group with two alkyl groups attached. This is consistent with isopropylmethylamine.

Final Answer:

Step 1: The reaction of nitrosamines with concentrated forms secondary amines and is classified as a nitroso reaction.
Step 2: The reaction follows the mechanism of the Liebermann nitroso reaction, which involves the formation of secondary amines.

Final Answer:

This sequence of reactions converts propanoic acid to salicylic acid by first converting it to an acyl chloride using thionyl chloride (SOCl2), then reacting with ammonia (NH3), followed by bromination and hydrolysis.
Final Answer:

A zwitterion is a molecule with both positive and negative charges on different atoms. Compound (1) exists in this state due to the amine and carboxyl groups.
Final Answer:

Step 1: Reduction of nitrobenzene.
Aniline can be obtained by reducing nitrobenzene with zinc in hydrochloric acid.

Step 2: Final product.
Thus, the correct reaction is , which produces aniline as the main product.