Q77 of 77 Page 193

Predict the reagent or the product in the following reaction sequence.


When p-Nitrotoluene undergoes reduction, it forms p-methyl benzyl amine. The reagents for ‘1’ are Sn + HCl or Fe + HCl which cause the reduction due to release of hydrogen. Even hydrogen gas passed over finely divided nickel, platinum or palladium in the presence of steam works to drive the reaction.


N-Methylaniline undergoes acetylation in the presence of a strong base such as pyridine to form N-methylacetanilide. The next step is nitration, with the reagents given which are HNO3 and H2SO4. Due the presence of the acetyl group, nitration leads to the formation of 2-Nitro-4-methylacetanilide. Hence, ‘2’ is 2-Nitro-4-methylacetanilide.


This molecule undergoes a reaction with reagent ‘3’ to give 2-nitro-4-methylaniline. This means ‘2’ undergoes hydrolysis to give 2-nitro-4-methylaniline. ‘3’ is H2O/H+.


2-nitro-4-methylaniline further reacts with NaNO2/HCl which forms a diazonium salt, hence ‘4’ is a diazonium salt 2-nitro-4-diazonium chloride.


After reaction with reagent ‘5’, the product formed does not have the N2+Cl- group at 4C. This means the diazonium chloride has been displaced by H. Hence the reagent ‘5’ is a mild reducing agent like hypophosphorous acid H3PO2 (phosphinic acid) or ethanol. The reagents themselves get oxidised to phosphorous acid and ethanal, respectively.



More from this chapter

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73

Assertion : Aromatic 1° amines can be prepared by Gabriel Phthalimide Synthesis.

Reason : Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.


Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.


 74

Assertion : Acetanilide is less basic than aniline.

Reason : Acetylation of aniline results in decrease of electron density on nitrogen.


Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.


 75

A hydrocarbon ‘A’, (C4H8) on reaction with HCl gives a compound ‘B’, (C4H9Cl), which on reaction with 1 mol of NH3 gives compound ‘C’, (C4H11N). On reacting with NaNO2 and HCl followed by treatment with water, compound ‘C’ yields an optically active alcohol, ‘D’. Ozonolysis of ‘A’ gives 2 mols of acetaldehyde. Identify compounds ‘A’ to ‘D’. Explain the reactions involved.

 76

A colourless substance ‘A’ (C6H7N) is sparingly soluble in water and gives a water soluble compound ‘B’ on treating with mineral acid. On reacting with CHCl3 and alcoholic potash ‘A’ produces an obnoxious smell due to the formation of compound ‘C’. Reaction of ‘A’ with benzenesulphonyl chloride gives compound ‘D’ which is soluble in alkali. With NaNO2 and HCl, ‘A’ forms compound ‘E’ which reacts with phenol in alkaline medium to give an orange dye ‘F’. Identify compounds ‘A’ to ‘F’.