Answer :
2CH3COCl + Cd(CH3)2 -------------------> 2CH3COCH3 + CdCl2
The reaction of acetyl chloride and dimethyl cadmium is an alkylation reaction.
We know that, the formula of acetyl chloride is [tex]CH_3COCl[/tex] and chemical formula for dimethyl cadmium is [tex](CH_3)_2CdCl[/tex].
The reaction is given below:
[tex]\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{Cl}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CdCl} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}+\mathrm{CdCl}_{2}[/tex]
Here dimethyl cadmium acts as alkylating reagent that provide methyl group which can replace the [tex]Cl^-[/tex] group from acetyl chloride. Hence, acetyl chloride acts as alkylating reagent.
The final product formed is called as 2-butanone as it is a ketone with >C=O (carbonyl group) at 2nd position.