Alcohols and Ethers - Result Question 50
55. 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 $MeMgBr, 0.42 g$ of $X$ gives $224 mL$ of $CH _4$ 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.
$(1992,5 M)$
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Solution:
- Compound ’ $X$ ’ $\underrightarrow{\text{Lucas reagent}}$ No reaction at room temperature.
$C _5 H _8 O \xrightarrow[AgNO _3]{\text { Ammoniacal }}$ ppt, $\quad X \xrightarrow[CH _3 MgBr]{\text { Excess of }} CH _4 ;$
$X \xrightarrow[\text { HI excess }]{H _2 / Pt} n$-pentane
Above information suggests that $X$ has a terminal triple bond and it contains a primary $-OH$ group.
$ \begin{aligned} \Rightarrow H-C \equiv C-CH_2-CH_2-CH_2 OH \xrightarrow[]{Ag\left(NH_3\right)_2^{+}} \\ & ~ \\ Ag-C≡C-CH₂CH₂CH₂OH \end{aligned} $
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