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  • A hydrocarbon 'A',( {C}_{4}{H}_{8}) on reaction with {HC}_{1} gives a compound 'B'( {C}_{4}{H}_{9} {Cl}), which on reaction with 1 mol of {NH}_{3} gives compound 'C',( {C}_{4} {H}_{11} {N} ) . On reacting with {NaNO}_{2} and {HC}_{1}

A hydrocarbon 'A', \left ( \text {C}_{4}\text {H}_{8} \right ) on reaction with \text {HCl} gives a compound 'B' \left ( \text {C}_{4}\text {H}_{9}\text {Cl} \right ), which on reaction with 1 mol of \text {NH}_{3} gives compound 'C',\left ( \text {C}_{4}\text {H}_{11}\text {N} \right ) . On reacting with \text {NaNO}_{2} and \text {HCl} followed by treatment with water, compound 'C' yields optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 moles of acetaldehyde. Identify compounds' A' to 'D'. Explain the reactions involved.

Answers (1)

(i) Addition of \text {HCl} to compound 'A' shows that compound 'A' is alkene. Compound 'B' is \text {C}_{4}\text {H}_{9}\text {Cl}

(ii) Compound 'B' reacts with \text {NH}_{2}, it forms amine 'C'.

\\\text {C}_{4}\text {H}_{8}\overset{\text {HCl}}{\rightarrow}\text {C}_{4}\text {H}_{9}\text {Cl}\overset{\text {NH}_{3}}{\rightarrow}\text {C}_{4}\text {H}_{11}\text {N}\; \text {Or}\; \text {C}_{4}\text {H}_{9}\text {NH}_{2}\\\text {(A)}\; \; \; \; \; \; \; \; \; \; \; \; \text {(B)}\; \; \; \; \; \; \; \; \; \; \; \; \; \text {(C)}

(iii) 'C' gives diazonium salt with \text {NaNO}_{2}/\text {HCl}, which yields an optically active alcohol. So, 'C' is aliphatic amine.

(iv) 'A' on Ozonolysis produces 2 moles of \text {CH}_{3}\text {CHO}. So, 'A' is \text {CH}_{3}-\text {CH}=\text {CH}-\text {CH}_{3}\left ( \text {But-2-ene} \right )

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