Amines

Answer (c) Any primary amine

This reaction is known as Sandmeyer?s reaction.

The substance which can donate a pair of| electrons is called Lewis base.

Amines contain a lone pair of electron on| nitrogen atom, so behave as Lewis base.

When ethyl bromide reacts with alcoholic KCN, propane nitrile is obtained as main product.

Due to delocalisation of lone pair of electrons on the nitrogen atom in aniline, it is weaker basic than benzaldehyde.

Hofmann bromamide reaction : this method is used for the preparation of any amine as well as aniline. During the reaction amide is treated with and alkali to get primary amine. This reaction is used to descend the series in which carbon atom is removed as carbonate ion reaction is given as

Compound is benzyl amine It reacts with at to form benzyl alcohol and gas.



Benzyl amine show s following five isomers.

Hofmann bromamide reaction is

This reaction is given by primary amide only. Option (b) is a secondary amide, so it will not undergo. Hofmann bromamide reaction.

is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.



Isonitriles on reduction with lithium aluminium hydride produce secondary amines

In the preparation of primary amines from phthalimide, is not used during the process. In this reaction Phthalimide, base, are used

Presence of nitro group (EWG) increases the acidic strength, and picric acid is trinitrophenol.

or is called as Hinsberg?? reagent. It is used for the distinction of and amine

Aniline (C₆H₅NH₂) and N-methylaniline (C₆H₅NHCH₃) are both aromatic amines, but they react differently with certain reagents due to the difference in the substituent attached to the nitrogen. One of the distinguishing tests for these compounds is the Hinsberg test, which uses chloroform (CHCl₃) and alcoholic potassium hydroxide (KOH). In this test: - Aniline reacts with chloroform and alcoholic potassium hydroxide to form a soluble product (the product will be a sulfonamide). - N-Methylaniline, however, does not react in the same way and does not form the sulfonamide under these conditions. Thus, the suitable reagents to differentiate Aniline from N-methylaniline are chloroform and alcoholic potassium hydroxide.

- (I) Benzenesulphonyl Chloride: This compound reacts with amines to form sulphonamide derivatives. It is used in the Hinsberg test, so it corresponds to (b) Hinsberg reagent.
- (II) Sulphanilic acid: This compound has an amino group and a sulfonic acid group, and it can exist as a zwitterion under certain conditions. Therefore, it corresponds to (a) Zwitterion.
- (III) Alkyl Diazonium salts: These salts are used in the azo dye synthesis and can react to form colored dyes. Thus, it corresponds to (c) Dyes. - (IV) Ary1 Diazonium salts: Ary1 diazonium salts can undergo reduction reactions to form alcohols, making them relevant to (d) Conversion to alcohols.
Thus, the correct matching is 1-a, II-b, III-c, IV-d. The correct answer is Option (3).

Step 1: Understanding the reaction sequence.
This is a multi-step synthesis starting from benzene. We need to identify the product at each step to determine the final product 'D'.
Step 2: Identifying compounds A, B, and C.

• Benzene A: The reaction of benzene with a nitrating mixture (Conc. HNO + Conc. H SO ) is electrophilic aromatic substitution (nitration). The product A is Nitrobenzene (C H NO ).
  
• A B: The reaction of nitrobenzene with Fe/HCl is a standard method for the reduction of the nitro group to an amino group. The product B is Aniline (C H NH ).
  
• B C: The reaction of aniline with NaNO /HCl at low temperature (273K or 0 C) is diazotization. The product C is Benzenediazonium chloride (C H N Cl ).
  

Step 3: Identifying the final reaction and product D.
The benzenediazonium chloride (C) is treated with ethanol (C H OH). In this reaction, ethanol acts as a reducing agent, and the diazonium group is replaced by a hydrogen atom, yielding benzene. Ethanol itself gets oxidized.
The reaction is:
Ethanol (an alcohol) is oxidized to Ethanal (an aldehyde). Therefore, the organic compound 'D' is ethanal.
Step 4: Final Answer:
The organic compound 'D' formed in the reaction is Ethanal (CH CHO).

Step 1: Understanding the Basicity of Alkylamines:
The basicity of an amine in the aqueous phase depends on the availability of the lone pair of electrons on the nitrogen atom for donation to a proton (H ) and the stability of the resulting conjugate acid (alkylammonium ion) in water.
Step 2: Analyzing the Factors Affecting Basicity in Aqueous Phase:
Several factors collectively determine the basic strength of alkylamines in water:
(A) Inductive Effect (+I Effect): Alkyl groups (like -CH , -C H ) are electron-donating. They push electron density towards the nitrogen atom, increasing the availability of the lone pair and thus increasing basicity. Based on this effect alone, the order of basicity would be: Tertiary (3 )>Secondary (2 )>Primary (1 )>Ammonia. This factor is crucial.
(B) Solvation Effect (Hydration): After an amine accepts a proton, it forms a positively charged alkylammonium ion (RNH , R NH , R NH ). This ion is stabilized by hydrogen bonding with surrounding water molecules. The more hydrogen atoms attached to the nitrogen in the cation, the greater the extent of hydrogen bonding and the more stable the ion. - Primary amine conjugate acid (RNH ) has 3 H-atoms for H-bonding. - Secondary amine conjugate acid (R NH ) has 2 H-atoms. - Tertiary amine conjugate acid (R NH ) has 1 H-atom. Thus, the order of stabilization by solvation is 1 >2 >3 . This effect enhances the basicity of primary and secondary amines relative to tertiary amines. This is a very important factor in the aqueous phase.
(C) Steric Hindrance: Bulky alkyl groups around the nitrogen atom can physically block the lone pair, making it difficult for a proton to approach (steric hindrance to protonation). They also hinder the solvation of the conjugate acid. This effect is most pronounced in tertiary amines and tends to decrease their basicity. This factor is also significant.
(D) Hyperconjugation Effect: Hyperconjugation involves the delocalization of -electrons into an adjacent empty or partially filled p-orbital or a -orbital. It is primarily used to explain the stability of carbocations, alkenes, and free radicals. While it might have a very minor, indirect influence, it is not considered a primary or deciding factor in determining the basic strength of amines. The basicity is dominated by the interplay of the inductive effect, solvation, and steric hindrance.
Step 3: Conclusion:
The inductive effect, solvation effect, and steric hindrance are the three major factors that decide the basic strength of alkylamines in the aqueous phase. The hyperconjugation effect is not a direct or significant factor in this context.
Step 4: Final Answer:
The basic strength of alkylamines in the aqueous phase is not decided by the hyperconjugation effect.

Step 1: Understanding the Reaction Conditions:
The question describes the nitration of aniline, which is an electrophilic aromatic substitution. The reaction is carried out in a strong acidic medium (typically a mixture of concentrated nitric acid and concentrated sulfuric acid).
Step 2: Analyzing the Directing Influence of the Amino Group:
The amino group (-NH ) is a powerful activating group due to the lone pair of electrons on the nitrogen, which can be delocalized into the benzene ring (+R effect). This increases the electron density at the ortho and para positions, making the -NH group an ortho, para-director. Therefore, options (A) and (C) are incorrect.
Step 3: Considering the Effect of the Strong Acidic Medium:
Aniline is a weak base due to the lone pair on the nitrogen atom. In a strongly acidic medium, the -NH group gets protonated to form the anilinium ion (-NH ). This protonation happens to a large extent.
Step 4: Analyzing the Directing Influence of the Anilinium Ion:
The anilinium ion (-NH ) has a positive charge on the nitrogen atom adjacent to the ring. This group strongly withdraws electron density from the benzene ring through the inductive effect (-I effect). It is a strongly deactivating group. Deactivating groups direct incoming electrophiles to the meta position.
Because a significant portion of the aniline exists as the anilinium ion in the reaction mixture, electrophilic attack by the nitronium ion (NO ) occurs at the meta position of the anilinium ion, leading to the formation of a substantial amount of m-nitroaniline (about 47%). The ortho and para products are also formed (about 2% and 51% respectively) from the small amount of unprotonated aniline that is still present.
Step 5: Evaluating the Options:
- (A) and (C) are incorrect as the -NH group is o,p-directing.
- (B) correctly states that aniline is present as the anilinium ion, which is the reason for meta-direction.
- (D) is incorrect; o, p, and m-nitroaniline are isomers and have the same molar mass.
Step 6: Final Answer:
The formation of a significant amount of m-nitroaniline is due to the presence of the meta-directing anilinium ion in the strong acidic medium.