When ethyl iodide and propyl iodide react with Na in the presence of ether, they form .

The correct option is B Three alkanes

Let's recall Wurtz Reaction.

Alkylhalide reacts with metallic sodium in the presence of dry ether to form symmetrical alkanes containing double the number of C atoms present in the alkyl halide.

$R–X+2Na+X–R\underset{ether}{\overset{Dry}{\to }}R–R+2NaX$

$C{H}_3–Br+2Na+Br–C{H}_3⟶C{H}_3–C{H}_3+2NaBr$

Now,

$C_2{H}_{5}l+2Na+C_3{H}_{7}l\underset{ether}{\overset{Dry}{\to }}{C}_2{H}_5-C_3{H}_7+2Nal$

$Pentane$

$C_2{H}_{5}l+2Na+C_2{H}_{5}l\underset{ether}{\overset{Dry}{\to }}{C}_2{H}_5–C_2{H}_5+2Nal$

$Butane$

$C_3{H}_{7}l+2Na+C_3{H}_{7}l\underset{Ether}{\overset{Dry}{\to }}{C}_3{H}_7–C_3{H}_7+2Nal$

$Hexane$