An ethereal solution of an alkyl halide preferably the bromide of iodide, is treated with sodium
`2R-X+2Na overset("Ether")toR-R+2NaX`
`2CH_(3)-Br+2Na overset("Ether")toCH_(3)-CH_(3)+2NaBr`
`CH_(3)CH_(2)Br+2Na+CH_(3)Br overset("Ether")toCH_(3)CH_(2)-CH_(3)+2NaBr`
In this reaction, the product has new (C-C) bond with the same type of alkyl halide, the product has symmetery and this helps in deciding the nature of reacting halide. Intermediates are free radicals
Formation of free radical is easiest in
A. `CH_(3)CH_(2)Cl`
B. `CH_(3)CH_(2)Br`
C. `CH_(3)CH_(2)F`
D. `CH_(3)CH_(2)I`
