Question

$0.34$ gram of a hydrocarbon when treated with methyl magnesium iodide gives $112$mL of $C{H}_4$ at STP. Possible structure of the hydrocarbon is:

• A. $(C{H}_3{)}_{2}CH-C\equiv C-H$
• B. $C{H}_3-(C{H}_2{)}_3-C\equiv CH$
• C. $(C{H}_3{)}_{3}C-C\equiv C-H$
• D. both (B) and (C).

Answer 1

The correct option is A $(C{H}_3{)}_{2}CH-C\equiv C-H$

Since, $112$ $mL$ of $C{H}_4=\left(\frac{112}{22400}\right)=0.005$ $mole$

$\because 1$ $mole$ hydrocarbon $\stackrel{C{H}_{3}MgBr}{⟶}1$ $mole$ $C{H}_4$

$\therefore 0.005$ $mole=$ mole of hydrocarbon

$\therefore M_o=\left(\frac{0.34}{0.005}\right)=68$ $gmo{l}^{-1}$

$\therefore (C{H}_3{)}_{2}CH-C\equiv CH$