
Methyl iodide and Zn/Cu couple in alcohol produces:
A. Ethane
B. Methane
C. Both A and B
D. none of the above
Answer
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Hint: To determine the answer to this question we should know the function of the Zn/Cu couple in alcohol. The function of the Zn/Cu couple in alcohol is reduction. The Zn/Cu couple in alcohol is a reducing agent. We will find the product formed by the reduction of methyl iodide.
Complete solution:
The Zn/Cu couple in alcohol is used for the addition of hydrogen, so they cause reduction and are known as reducing agents.
The Zn/Cu couple in alcohol is used for the reduction of alkyl halide.
The Zn/Cu couple in alcohol causes the removal of halide from an alkyl halide and adds hydrogen to give hydrocarbon.
The Zn/Cu couple in alcohol causes the removal of iodine from methyl halide and adds hydrogen to give methane.
The reaction is shown as follows:
${\text{C}}{{\text{H}}_{\text{3}}} - {\text{I}}\,{\text{ + }}\,{{\text{H}}_2}\,\mathop {\mathop \to \limits_{{\text{alcohol}}} }\limits^{{\text{Zn/Cu}}} \,{\text{C}}{{\text{H}}_{\text{3}}} - {\text{H}}\,{\text{ + }}\,{\text{HI}}$
So, the methyl iodide and Zn/Cu couple in alcohol produce methane.
Therefore, option (B) methane is correct.
Note:The addition of hydrogen is known as reduction whereas the removal of hydrogen is known as oxidation. The addition of electrons is also known as reduction. The metals generally used for the reduction of unsaturated hydrocarbons to saturated hydrocarbons are platinum, palladium and nickel. In place of Zn/Cu couples in alcohol, the zinc in presence of acid can also be used for the reduction of alkyl halide to form alkane. The compounds having carbon and hydrogen only are known as hydrocarbons. The saturated hydrocarbons are known as alkane and unsaturated hydrocarbons are known as alkene and alkyne.
Complete solution:
The Zn/Cu couple in alcohol is used for the addition of hydrogen, so they cause reduction and are known as reducing agents.
The Zn/Cu couple in alcohol is used for the reduction of alkyl halide.
The Zn/Cu couple in alcohol causes the removal of halide from an alkyl halide and adds hydrogen to give hydrocarbon.
The Zn/Cu couple in alcohol causes the removal of iodine from methyl halide and adds hydrogen to give methane.
The reaction is shown as follows:
${\text{C}}{{\text{H}}_{\text{3}}} - {\text{I}}\,{\text{ + }}\,{{\text{H}}_2}\,\mathop {\mathop \to \limits_{{\text{alcohol}}} }\limits^{{\text{Zn/Cu}}} \,{\text{C}}{{\text{H}}_{\text{3}}} - {\text{H}}\,{\text{ + }}\,{\text{HI}}$
So, the methyl iodide and Zn/Cu couple in alcohol produce methane.
Therefore, option (B) methane is correct.
Note:The addition of hydrogen is known as reduction whereas the removal of hydrogen is known as oxidation. The addition of electrons is also known as reduction. The metals generally used for the reduction of unsaturated hydrocarbons to saturated hydrocarbons are platinum, palladium and nickel. In place of Zn/Cu couples in alcohol, the zinc in presence of acid can also be used for the reduction of alkyl halide to form alkane. The compounds having carbon and hydrogen only are known as hydrocarbons. The saturated hydrocarbons are known as alkane and unsaturated hydrocarbons are known as alkene and alkyne.
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