Question

A hydrocarbon ‘A’, \[\ce{(C4H8)}\] on reaction with \[\ce{HCl}\] gives a compound ‘B’, \[\ce{(C4H9Cl)}\], which on reaction with 1 mol of \[\ce{NH3}\] gives compound ‘C’, \[\ce{(C4H11N)}\]. On reacting with \[\ce{NaNO2}\] and \[\ce{HCl}\] 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.

Answer 1

\[\ce{(A) ->[Ozonolysis] 2CH3CHO}\]

\[\ce{\underset{(A)}{C4H8} ->[HCl] \underset{(B)}{C4H9Cl}}\]

Addition of \[\ce{HCl}\] has occurred on ‘A’. This implies ‘A’ is an alkene.

\[\ce{\underset{(B)}{C4H9Cl} -> \underset{(C)}{C4H11N}}\]

\[\ce{Cl}\] in compound ‘B’ is substituted by \[\ce{NH2}\] to give ‘C’.

\[\ce{(C) ->[NaNO2/HCl][H2O] (D)}\]

‘C’ gives a diazonium salt with \[\ce{NaNO2/HCl}\] that liberates \[\ce{N2}\] to give optically active alcohol. This means that ‘C’ is an aliphatic amine. Number of carbon atoms in amine is same as in compound ‘A’.

Since products of ozonolysis of compound ‘A’ are \[\ce{CH3 - CH = O}\] and \[\ce{O = CH - CH3}\]. The compound ‘A’ is \[\ce{CH3 - CH = CH - CH3}\]

On the basis of structure of ‘A’ reactions can be explained as follows:

\[\begin{array}{cc}
\phantom{}\ce{CH3 - CH = CH - CH3 ->[HCl] CH3CH2 - CH - CH3}\phantom{.}\\
\phantom{..........}|\phantom{...........................}|\phantom{......}\\
\phantom{......}\ce{\underset{(A)}{Cl}}\phantom{.........................}\ce{\underset{(B)}{Cl}}\phantom{.}\\
\end{array}\]

\[\begin{array}{cc}
\phantom{.}\ce{CH3 - CH2 - CH - CH3 ->[NH3] CH3 - CH2 - CH - CH3}\phantom{..}\\
\phantom{...............}|\phantom{..............................}|\phantom{...........}\\
\phantom{........}\ce{\underset{(B)}{Cl}}\phantom{............................}\ce{\underset{(C)}{NH2}}\phantom{..}\\
\end{array}\]

\[\begin{array}{cc}
\phantom{............................................}\ce{CH2CH3}\\
\phantom{.....................................}|\\
\phantom{.}\ce{CH3 - CH2 - CH - CH3 ->[NaNO2/HCl][H2O] CH3 - C - H}\phantom{..}\\
\phantom{...............}|\phantom{...................................}|\phantom{..............}\\
\phantom{..........}\ce{\underset{(C)}{NH2}}\phantom{...........................}\ce{\underset{(D)}{\underset{(Optically active)}{OH}}}\phantom{..}\\
\end{array}\]