Do You Know Amines?

Explanation
Compound A is an amide. The treatment of A with NH3 results in the formation of an amine, compound B. Heating compound B leads to the formation of an isocyanate, compound C. Compound C, when treated with Br2 in the presence of KOH, undergoes a Hofmann degradation reaction, producing ethylamine. Therefore, compound A is an amide.

Explanation
Gabriel's synthesis is a method used to prepare primary amines. It involves the reaction of phthalimide with an alkyl halide followed by hydrolysis and decarboxylation. However, N-Methylbenzylamine cannot be prepared by Gabriel's synthesis because it is a secondary amine, not a primary amine. In Gabriel's synthesis, only primary amines can be obtained.

Explanation
NaOH-Br2 is the reagent that yields methylamine when treated with acetamide. This is because NaOH (sodium hydroxide) is a strong base that can deprotonate the amide group in acetamide, forming the corresponding carboxylate ion. Br2 (bromine) is an electrophilic reagent that can react with the carboxylate ion, replacing the oxygen with a bromine atom. This intermediate can then undergo a rearrangement reaction to form methylamine.

Explanation
Based on the given information, compound (A) is treated with nitrous acid to yield an alcohol and N2 gas. This suggests that (A) contains an amine group (NH2), which reacts with nitrous acid to form a diazonium salt. (A) is also warmed with CHCl3 and caustic potash to give (C), which is then reduced to isopropylmethylamine. This indicates that (A) contains an alkyl group that can be converted to an amine through reduction. Therefore, the structure of (A) is likely an alkyl amine with the formula C3H9N.

Explanation
The most basic compound is the one that can easily accept a proton (H+) to form a stable conjugate acid. In general, amines are more basic than alcohols and ethers. Among the given compounds, ammonia (NH3) is the most basic because it has a lone pair of electrons on the nitrogen atom, which can readily accept a proton to form NH4+. The other compounds, water (H2O) and methanol (CH3OH), also have lone pairs of electrons, but they are less basic compared to ammonia.

Explanation
The basicity of a compound is determined by its ability to donate a lone pair of electrons. In this case, CH3CH2NH2 (I) is an amine compound, which has a lone pair of electrons on the nitrogen atom. This lone pair can easily be donated, making it a strong base. C6H5CONH2 (III) is an amide compound, which also has a lone pair of electrons on the nitrogen atom. However, the presence of the phenyl group (C6H5) attached to the nitrogen atom reduces the availability of the lone pair, making it a weaker base compared to I. CH3CONH2 (II) is also an amide compound, but it does not have any electron-donating groups attached to the nitrogen atom, making it the weakest base among the three compounds. Therefore, the correct order of basicity is I > III > II.

Explanation
Primary amines can be directly oxidized to the corresponding nitro compound by potassium permanganate. This is because primary amines have a hydrogen atom attached to the nitrogen atom, which can be easily oxidized by potassium permanganate. Secondary amines and tertiary amines do not have this hydrogen atom and therefore cannot be directly oxidized to nitro compounds by potassium permanganate.

Explanation
When the optically active compound (X) is treated with NaNO2/HCl, it undergoes a reaction called the Sandmeyer reaction. In this reaction, the compound is converted into a diazonium salt. The diazonium salt can then undergo various reactions depending on the conditions. In this case, it is converted into a 2° alcohol. However, the conversion of the diazonium salt to the alcohol involves the loss of the chiral center, resulting in a racemic mixture of 2° alcohols. This is because the reaction proceeds through a carbocation intermediate, which can undergo both front-side and back-side attacks, leading to the formation of both R and S configurations of the alcohol.

Explanation
Sulphanilic acid exists as zwitterions. Zwitterions are molecules that have both positive and negative charges, but overall they have a neutral charge. In the case of sulphanilic acid, it contains an amino group (-NH2) and a sulfonic acid group (-SO3H), which can both donate and accept protons. This allows the molecule to exist in a zwitterionic form, with the amino group being positively charged and the sulfonic acid group being negatively charged. This property makes sulphanilic acid useful in various chemical reactions and applications.

Explanation
Aryldiazonium salts can undergo various reactions, including diazotization, Sandmeyer reaction, and coupling reactions. However, one reaction that is not observed with aryldiazonium salts is the Cannizzaro reaction. The Cannizzaro reaction is a disproportionation reaction of aldehydes or ketones in the presence of a strong base, resulting in the formation of an alcohol and a carboxylic acid. Since aryldiazonium salts do not contain any carbonyl group, they cannot undergo the Cannizzaro reaction.

Explanation
Nitrous acid (HNO2) is used in the diazotization of arylamine as it acts as a source of electrophilic nitrosonium ion (NO+). The nitrosonium ion is a highly reactive electrophile that can react with the arylamine to form a diazonium salt. This reaction is an important step in the synthesis of various organic compounds, such as azo dyes. The other options listed (suppressing hydrolysis of phenol and neutralizing the base liberated) are not the primary roles of nitrous acid in this context.

Explanation
When aniline is treated with bromine water, bromination occurs at the para position of the benzene ring, resulting in p-bromoaniline. When p-bromoaniline is treated with an aqueous solution of sodium nitrite in the presence of dilute HCl, a diazonium salt is formed. This diazonium salt can be converted into a tetrafluoroborate salt. Heating the tetrafluoroborate salt leads to the elimination of nitrogen gas and the formation of a fluorobenzene derivative. In this case, the bromine substituents on the benzene ring will be replaced by fluorine atoms, resulting in 2, 4, 6-tribromofluorobenzene as the final product.