Question

when ethyle bromide and methyle bromide is subjected to wurtz reaction the products are ?

Answer 1

when ethyle bromide and methyle bromide is subjected to wurtz reaction the products are ?

Solution:

Wurtz reaction is unspecific, so if there are several active compounds in mixture, they combine in all possible ways:

$$\mathrm{C_2H_5Br} + \mathrm{Na} \rightarrow \mathrm{C_2H_5} \bullet + \mathrm{NaBr}$$ $$\mathrm{CH_3Br} + \mathrm{Na} \rightarrow \mathrm{CH_3} \bullet + \mathrm{NaBr}$$ $$2\mathrm{C_2H_5} \bullet \rightarrow \mathrm{C_2H_5} \cdot \mathrm{C_2H_5}$$ $$\mathrm{C_2H_5} \bullet + \mathrm{CH_3} \bullet \rightarrow \mathrm{C_2H_5} \cdot \mathrm{CH_3}$$ $$2\mathrm{CH_3} \bullet \rightarrow \mathrm{CH_3} \cdot \mathrm{CH_3}$$ $$\mathrm{C_2H_5Br} + \mathrm{CH_3Br} + \mathrm{Na} \rightarrow \mathrm{C_2H_5} \cdot \mathrm{C_2H_5} + \mathrm{C_2H_5} \cdot \mathrm{CH_3} + \mathrm{CH_3} \cdot \mathrm{CH_3} + \mathrm{NaBr}$$

So in this mixture as products we have ethane $(\mathrm{CH_3} \cdot \mathrm{CH_3})$, propane $(\mathrm{C_2H_5} \cdot \mathrm{CH_3})$ and butane $(\mathrm{C_2H_5} \cdot \mathrm{C_2H_5})$.