Aniline reacts with acid chlorides and acid anhydries in the presence of pyridine to form anilides. In ammine, there are three bond pairs whereas one lone pair. Aniline reacts with alkyl halides to form a mixture of secondary amines,tertiary amines and quaternary salts. Note: Partial delocalisation of the electron is the delocalisation of the pi-electron and the effect is also known as resonance due to which the molecule becomes more stable. Therefore, option A is the correct answer. Here, the complex formed is known as anilinium chloride. The balanced chemical reaction between aniline and hydrochloric acid is: Acids, metal powders, flammable liquids, chlorates, nitrates, sulfur, finely divided organic or combustible materials. Due to which the formation of the complex the aniline will dissolve in the solution of hydrochloric acid. In the protonation process, the negative anion from the acid comes and attaches to the positive charge that is present on the amine. Now, the amine is able to undergo the protonation by an acid that is hydrochloric acid. When aniline dissolves in the hydrochloric acid then the lone pair that is present on the amine group will undergo partial delocalisation. Now, the hydrochloric acid as the name tells has acidic nature so when it will react with aniline the acid-base reaction will take place. The molecule is formed when the amine group attaches on the benzene ring. Aniline is an organic compound which has basic nature. In the given question, we have to choose the correct product formed by the reaction between aniline and hydrochloric acid. ^ each ion peak M, M+1, M+2, M+3 is equally represented and given the reaction conditions pulegone itself does not fragment in absence of amine.Hint: For this problem, we have to write the complete reaction of the aniline with hydrochloric acid and after the product is formed we have to study its nature and properties to choose the correct option.'On the Mechanism of the Skraup−Doebner−Von Miller Quinoline Synthesis'.

Hexacyclic Compounds: Pyridine, Quinoline, and Isoquinoline'. Because α-amino protons are not available in this model compound the reaction is not taken to the fully fledged quinoline. elimination of one aniline molecule through 8 and rearomatization leads to final product 9. In the next step 5 reacts with a second aniline molecule in a nucleophilic conjugate addition to imine 6 and subsequent electrophilic addition and proton transfer to leads to 7. This intermediate then fragments to the imine 4a and the saturated cyclohexanone 4b in a non-reversible reaction and both fragments recombine in a condensation reaction to the conjugated imine 5. The first step is a nucleophilic conjugate addition of the amine with the enol to the amine ketone 3 in a reversible reaction. In this study 4-isopropylaniline 1 is reacted with a mixture (50:50)of ordinary pulegone and the 13C-enriched isomer 2 and the reaction mechanism is outlined in scheme 2 with the labeled carbon identified with a red dot. A 2006 study proposes a fragmentation-recombination mechanism based on carbon isotope scrambling experiments. The reaction mechanism for this reaction and the related Skraup synthesis is a matter of debate. The reaction is catalyzed by Lewis acids such as tin tetrachloride and scandium(III) triflate and Brønsted acids such as p-toluenesulfonic acid, perchloric acid, amberlite and iodine. When the α,β-unsaturated carbonyl compound is prepared in situ from two carbonyl compounds (via an Aldol condensation), the reaction is known as the Beyer method for quinolines. This reaction is also known as the Skraup-Doebner-Von Miller quinoline synthesis, and is named after the Czech chemist Zdenko Hans Skraup (1850–1910), and the Germans Oscar Döbner (Doebner) (1850–1907) and Wilhelm von Miller (1848–1899).

Aniline reacts with perchloric acid