The product formed in the first step of the reaction of $CH_3-CH_2-CH(Br)-CH_2-CH(Br)-CH_3$ with excess $Mg / Et_2O$ $(Et = C_2H_5)$ is
- A$CH_3-CH_2-CH(MgBr)-CH_2-CH(MgBr)-CH_3$
- B$CH_3-CH_2-CH(CH_3-CH_2-CH-CH_2-CH_3)-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$
- C$A$ cyclopropane derivative
- D$MgBr-CH(CH_2CH_3)-CH_2-CH(MgBr)-CH_3$
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Statement-$I$ : Kolbe's electrolysis of sodium propionate gives $n$-hexane as product.
Statement-$II$ : In Kolbe's process,$CO_2$ is liberated at the anode and $H_2$ is liberated at the cathode.
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Statement-$I$ : Kolbe's electrolysis of sodium propionate gives $n$-hexane as product.
Statement-$II$ : In Kolbe's process,$CO_2$ is liberated at the anode and $H_2$ is liberated at the cathode.
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