Ncert Exercises & Solutions

Q.75 A hydrocarbon 'A' (C₄H₈) on reaction with HCI gives a compound 'B,' (C₄H₉Cl), which on reaction with 1 mol of NH₃ gives compound 'C,' (C₄H₁₁N). On reacting with NaNO₂ and HCI followed by treatment with water, compound 'C' yields optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 mols of acetaldehyde. Identify compounds 'A' to 'D'. Explain the reactions involved.

(i) Addition of HCI to compound "A' shows that compound 'A' is an alkene. Compound B ' is C₄H₉Cl.

(ii) Compound E ' reacts with NH2, it forms amine 'C'.

(iii) C' gives diazonium salt with NaNO₂/ HCI, which yields optically active alcohol. So, 'C' is an aliphatic amine.

(iv) A on ozonolysis produces 2 moles of CH₃CHO. So. 'A is CH₃ —CH= CH—CH₃ (But-2-ene).

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