Question

Compound $X$ (molecular formula, $C_5{H}_{8}O$ ) does not react appreciably with Lucas reagent at room temperature but gives a precipitate with ammoniacal silver nitrate with excess of $MeMgBr,0.42g$ of $X$ gives $224mL$ of $C{H}_4$ at STP. Treatment of $X$ with $H_2$ in presence of $Pt$ catalyst followed by boiling with excess $HI$, gives $n$-pentane. Suggest structure for $X$ and write the equation involved.

Answer 1

Compound $X\stackrel{\text{Lucas reagent}}{\to }$ No reaction at room temperature .
$C_5{H}_{8}O\underset{AgN{O}_3}{\overset{\text{Ammoniacal}}{\to }}$ ppt, $X\underset{C{H}_{3}MgBr}{\overset{\text{Excess of }}{\to }}C{H}_4$
$X\underset{HI\text{excess}}{\overset{H_2/Pt}{\to }}n-$ pentane
Above information suggest that $X$ has a terminal triple bond and it contain primary $—OH$ group.