Answer :
When CH₃Br is treated with sodium, two different types of aluminum compounds are formed.
When CH₃Br reacts with sodium, it undergoes a nucleophilic substitution reaction where the bromine atom is replaced by a methyl group. The resulting product is CH₃AlBr₂. Additionally, a side reaction occurs where the sodium displaces the methyl group from CH₃Br, forming CH₃Na. Subsequently, CH₃Na reacts with AlBr₃ to produce another aluminum compound, CH₃₂AlBr.
Overall, the reaction yields two distinct aluminum compounds: CH₃AlBr₂ and CH₃₂AlBr. The formation of these compounds can be explained by the reaction mechanism involving the nucleophilic substitution of CH₃Br by sodium, leading to the generation of methyl aluminum bromide and dimethyl aluminum bromide as the final products.
Complete quwestion:
How many different types of aluminum compounds are formed when CH₃Br is treated with sodium?
