Question 5.SSE.8: Compound X (molecular formula, C5H8O) does not react appreci......

Compound ‘X’  \overset{\text { Lucas reagent}} {\xrightarrow {\hspace{2 cm}} }   No reaction at room temperature

(Mol. formula C_5H_8O )

\overset{\text {Ammonical } AgNO_3} {\xrightarrow {\hspace{2 cm}} }   Precinitates \overset{\text {Excess } MeMgBr} {\xrightarrow {\hspace{2 cm}} }  CH_4   \overset{\text {(i )} H_2 / Pt}{\underset{\text {(ii) Boil with excess HI }}{\xrightarrow {\hspace{2 cm}} }}   n-Pentane

Hence in compound X, five C-atoms are present in straight chain. It gives methane with excess GR, so in it acidic hydrogen is present. It gives precipitate with ammonical AgNO_3 . Therefore, it must have acidic hydrogen in the form of alkynic group. It does not give any reaction with Lucas reagent therefore, it has p-alcoholic group.

So the basis of above properties,the possible structure of compound X is given as follows:

HC \equiv C – CH _2- CH _2- CH _2 OH (mol. formula C_5H_8O )

Reaction:

(i) HC \equiv C – CH _2 CH _2- CH _2 OH  \overset{\text {Lucas reagent}}{\underset{\text { at room temp. }}{\xrightarrow {\hspace{2 cm}} }}  \text {No reaction}

(ii) \begin{matrix} HC \equiv C – CH _2 CH _2- CH _2 OH + AgNO _3+ NH _4 OH &\rightarrow & AgC \equiv C &- CH _2 CH _2- CH _2 OH + NH _4 NO _3+ H _2 O \\ &&\text {White ppt.} \end{matrix}

(iii) HC \equiv C – CH _2 CH _2- CH _2 OH +2 MeMgBr \rightarrow BrMg – C \equiv C – CH _2 CH _2- CH _2 O – MgBr +2 CH _4

(iv) \begin{matrix} HC \equiv C – CH _2 CH _2- CH _2 OH +2 H _2 \stackrel{ Pt }{\longrightarrow} CH _3- CH _2- CH _2- CH _2- CH _2 OH & \overset{\text {Boil HI}}{\underset{\left(- I _2,- II _2 O \right)}{\xrightarrow {\hspace{2 cm}} }} & CH _3- CH _2- CH _2- CH _2- CH _3 \\ && \text {n-pentane} \end{matrix}

So the compound ‘X’ is HC \equiv C – CH _2- CH _2 – CH _2 OH