Question

An allylide on hydrolysis gives allylene. The alkaline earth metal cation of allylides dissolves in dry ether in the presence of alkyl halide to form Grignard reagent. The allylide is-
A. $M{g_2}{C_3}$
B. $C{e_2}{C_3}$
C. $Mn{C_2}$
D. $MgB{r_2}$

Answer 1

Hint: Through chemistry a carbide typically represents a carbon compound and a metal compound. The method for creating carbide coatings on a metal component is in metallurgy, carbidizing or carburizing.

Complete step-by-step answer:
Since we know that allylides belong to the species of ionic carbides.
The formation of Grignard reagent is only possible in the presence of Magnesium.
The general equation for the same is-
$ \Rightarrow R - X + Mg\xrightarrow[{ether}]{{dry}}RMgX$
Where, R-X is from the alkyl halide chain in which R belongs to the alkyl group and X belongs to the halogen group.
We reacted this alkyl halide with Magnesium in order to form a Grignard reagent in presence of dry ether. The resultant product was the Grignard reagent.
Since we used Mg (alkaline metal) in the formation of Grignard reagent, it is quite clear through the question as well that the allylide used to form Grignard reagent in the presence of dry ether will be of magnesium. So, either option A or option D should be correct.
Since option D is not an ionic carbide, the only option left is option A, which is the correct answer.

Note: Any group of reagents based on the interactions of magnesium and organic halide, typically in the vicinity of an ether, and having a general formula RMgX, where R is an organic group and X would be a halogen: used during the Grignard reaction. The application of an excess of Grignart 's reagent to an ester or lactone creates a tertiary alcohol in which two alkyl groups are equivalent and an additional Grignard to a nitrile produces a non-symmetric ketone by means of an intermediate metalloimine.