Question

What happens when:
(i) bromoethane is treated with zinc and hydrochloric acid.
(ii) hydrogen is passed into 2-bromopropane in the presence of palladium.

Answer 1

The reaction between zinc and hydrochloricacid is Zn + HCl = H2 + ZnCl2. It is a single replacement reaction where zinc metal displaces the hydrogen to form hydrogen gas and zinc chloride, a salt. Zinc reacts quickly with the acid to form bubbles of hydrogen.

Answer 2

"Generally, when an "alkyl halide" is treated with "zinc" and "hydrochloric acid", it is "reduced" to the "respective alkane".

The "nascent hydrogen" formed by the "reaction" between "zinc" and "HCl acid" reduces "ethyl chloride to ethane".

(i) Ethane is formed.

$\begin{matrix} { CH }_{ 3 }{ CH }_{ 2 }-Br \\ Bromoethane \end{matrix}\quad +\quad 2\left[ H \right] \quad \xrightarrow [ Zn/HCl ]{ \left( Reduction \right)  } \quad { CH }_{ 3 }{ CH }_{ 3 }\quad +\quad HBr$

(ii) Propane is obtained.

$\quad \begin{matrix} { CH }_{ 3 }- & CH & -Br \\ \quad  & | & \quad  \\ \quad  & { CH }_{ 3 } & \quad  \end{matrix}\quad +\quad { H }_{ 2 }\quad \xrightarrow [ Pd ]{ \left( Reduction \right)  } \quad \begin{matrix} { CH }_{ 3 }- & { CH }_{ 2 } \\ \quad  & | \\ \quad  & { CH }_{ 3 } \end{matrix}\left( Propane \right) \quad +\quad Br$

When 2-bromopropane is reduced in presence palladium to form propane."