An optically active alcohol (C6H10O) absorbs two moles of hydrogen per mole of A upon catalytic hydrogenation and gives a product B. The compound B is resistant to oxidation by CrO3 and does not show any optical activity. Deduce the structures of A and B. [IIT, 1996]

Question

An optically active alcohol ([latex] C_6H_{10}O [/latex]) absorbs two moles of hydrogen per mole of A upon catalytic hydrogenation and gives a product B. The compound B is resistant to oxidation by [latex] CrO_3 [/latex] and does not show any optical activity. Deduce the structures of A and B.

[IIT, 1996]

Accepted Answer

[latex] \begin{matrix} C _6 H _{10} O +2 H _2 & \overset{ ext {Catalyst}}{\xrightarrow {\hspace{2 cm}} } & C _6 H _{14} O & \overset{CrO_3}{\xrightarrow {\hspace{2 cm}} } & ext {No reaction} \ ext {Compound A} && ext {Compound B} \ ext {Optically active} && ext {optically inactive} \end{matrix} [/latex]

Compound A is unsaturated monohydric alcohol and it has either two C = C bonds or one triple bond.

Compound A is optically active and after hydrogenation optically inactive compound ‘B’ is obtained which is not oxidized with [latex] CrO_3 [/latex] . So in it [latex] -OH [/latex] group is attached with tertiary C-atom i.e., it is tertiary alcohol.

Thus, the possible structure of A is as follows:

[latex] \begin{matrix} && OH \ && | \ CH _3&-&{ }^* C &-& C& \equiv & CH \ && | \ && CH_2 &-&CH_3\end{matrix} [/latex]

[latex] { }^* C ightarrow [/latex] Asymmetric C-atom and due to presence of it, it shows optically activity.

[latex] \begin{matrix} &&OH \ &&| \ CH _3&-&{ }^* C &-& C &\equiv& CH &+&2 H _2 & \overset{ ext{Catalyst}}{\xrightarrow {\hspace{2 cm}} } \ &&|\&& CH_2 &-&CH_3\ && ext {Compound A (optically active)} \end{matrix} [/latex]

[latex] \begin{matrix} &&OH \ &&| \ CH _3&-& C &-& CH _2&-& CH _3 \ && | \ && CH_2 &-&CH_3\ && ext {Compound B} \ && ext{(Optically inactive)} \end{matrix} [/latex]

Question 5.SSE.13: An optically active alcohol (C6H10O) absorbs two moles of hy......

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