The intermediate compound formed in Frankland’s reaction is:
A. \[RZn{I_2}\]
B. \[{R_2}Zn\]
C. \[RZnI\]
D. \[{R_2}ZnI\]

Hint: Frankland reaction is similar to Wurtz reaction. In Frankland reaction, Zn is used in place of Na used in Wurtz reaction.

Complete step by step answer:
Preparation of dialkyl zinc from zinc and alkyl iodide is called Frankland’s reaction. The first synthetic organometallic compound to be formed was diethyl zinc. The step by step mechanism for the reaction is shown below:
               \[R - I + Zn \to R - Zn - I\]
               \[R - Zn - I + Zn \to R - Z{n^ \bullet } + Z{n^ \bullet } - I\]
               \[R - Z{n^ \bullet } + Zn{}^ \bullet - I + R - I \to R - R + Zn{I_2}\]
Clearly, we can see that the intermediate compound formed is R-Zn-I. During the reaction, free radical is formed. The reaction was discovered by Edward Frankland in July 1849, a young chemist in England. This reaction is now regarded as a key event in organic and organometallic chemistry. Symmetric alkanes can be prepared by using this reaction. Dry ether is used as a solvent in this reaction which helps to form free radicals and to prevent Zn from moisture. Haloalkane and Zinc are taken in a ratio of 2:1, which we can see from the mechanism. If they are taken in a ratio of 1:1, RZnI is formed instead of R-R.

Hence, the correct option is (C) RZnI

Additional information: Alkyl zinc has found many applications in organic synthesis because the addition of alkyl zinc to aldehydes in the presence of a chiral catalyst affords secondary alcohols with a high level of enantioselectivity.

Note: A student might consider \[Zn{I_2}\] to be a possible answer as well. But \[Zn{I_2}\] is not an intermediate formed in the reaction but a by-product.