$3^{o}$ alcohol $+$ Lucas reagent $\rightarrow$ Immediate turbidity.
$2^{o}$ alcohol $+$ Lucas reagent $\rightarrow$ Turbidity after $5$ mins.
$1^{o}$ alcohol $+$ Lucas reagent $\rightarrow$ No reaction. Thus, the required alcohol is $2 -$ methylpropan $- 2 -$ ol, i.e.
$\mathop {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,C{H_3}} \\
| \\
{{H_3}C - C - C{H_3}}
\end{array}} \\
| \\
{\,\,\,\,\,OH}
\end{array}}\limits_{({3^o}\,\,alcohol)} $
${C_3}{H_7}OH\xrightarrow{{conc\,{H_2}S{O_4}}}X\xrightarrow{{B{r_2}}}Y\xrightarrow[{alkolic\,KOH}]{{high\,level}}Z$
જ્યારે $KMnO_4$ સાથે રિફ્લેક્સ કરે
$(A)$ નું બંધારણ શું હશે ?
મૂળ સંયોજનમાં પણ આ ગુણધર્મો દર્શાવવામાં આવ્યા હતા
$\mathop {{C_9}{H_{12}}{O_2}}\limits_{(A)} \,\xrightarrow{{Na}}\,{H_2}$ મુક્ત
$\mathop {{C_9}{H_{12}}{O_2}}\limits_{(A)} \,\xrightarrow{{Br}}$ કોઈ ઝડપી પ્રક્રિયા નથી
$\mathop {{C_9}{H_{12}}{O_2}}\limits_{(A)} \,\xrightarrow[{cool}]{{Cr{O_3}\,/\,{H^ + }}}{C_9}{H_8}{O_3}$