${C_2}{H_2}\xrightarrow{{{O_3}}}X\xrightarrow{{Zn/C{H_3}COOH}}Y$
$I. \,CH_3 - H$
$II.\, CH_3CH_2-H$
$III.\, CH_2 = CH - CH_2-H$
$IV.\, C_6H_5$









$C{{H}_{3}}CH\,=\,\,C{{H}_{2}}\,+\,\,{{H}_{2}}O\,\,+\,\,[O]\,\,\underset{Acid}{\mathop{\xrightarrow{KMn{{O}_{4}}}}}\,$$\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2}} \\
{\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,OH\,\,\,\,\,\,\,OH}
\end{array}$ $\xrightarrow{{[O]}}{\mkern 1mu} {\mkern 1mu} X{\mkern 1mu} {\mkern 1mu} + {\mkern 1mu} {\mkern 1mu} HCOOH$
$X$ શું છે ?
$C{{H}_{3}}\,-\,\,\overset{\overset{C{{H}_{3}}}{\mathop{|}}\,}{\mathop{C}}\,H\,\,-\,\,C{{H}_{3}}(excess)\,\,+\,\,B{{r}_{2}}\,\xrightarrow{hv}$
$\,\,\,\,C{{H}_{3}}\,-\,\,\overset{\overset{C{{H}_{3}}}{\mathop{|}}\,}{\mathop{\underset{\underset{Br}{\mathop{|}}\,}{\mathop{C}}\,}}\,\,\,-\,\,C{{H}_{3}}\,+\,\,C{{H}_{3}}\,-\,\,CH\,\,-\,\,C{{H}_{2}}\,-\,\,Br$

$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{{H_3}C - C - CH = C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{H_2}O/{H^ \oplus }}}{\mkern 1mu} \mathop A\limits_{Major\,product} \, + \,\mathop B\limits_{Minor\,product} $
(image) $\xrightarrow[{KOH}]{{KMn{O_4}}}\,B\,\xrightarrow[{FeC{L_3}}]{{B{r_2}}}\,C\,\xrightarrow[{{H^ + }}]{{{C_2}{H_5}OH}}\,D$
$D$ શું હશે?
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\mathop C\limits_7 {H_3} - \mathop C\limits_6 - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 H - \mathop C\limits_2 \equiv \mathop C\limits_1 H} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
જે નીચેના ક્રમમાં છે.
પ્રક્રિયામાં $A$(મુખ્યત્વે) ...... હશે.
$\mathop C\limits_6 {H_3} - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 {H_2} - \mathop C\limits_2 \equiv \mathop C\limits_1 H$
કાર્બન્સ $1, 3$ અને $5$ ના સંકરણની સ્થિતિ નીચેના ક્રમમાં છે, તો સાચો ક્રમ શોધો.
$CH_3CH_2-C \equiv CH+ HCl \rightarrow B \xrightarrow{{HI}} C$
$C{H_3}C \equiv \,C\,C{H_2}C{H_3}\mathop {\xrightarrow[{(2)\;Hydrolysis}]{}}\limits^{(1)\;\;{O_3}} $ .......
$C{H_3}COOH\xrightarrow{{LiAl{H_4}}}A\mathop {\xrightarrow{{{H^ + }}}}\limits_{443\,K} B\xrightarrow{{B{r_2}}}C\mathop {\xrightarrow{{alc.}}}\limits_{KOH} D$
$C{H_2} = C{H_2}\xrightarrow[{oxid}]{{Hypochloro}}$ $M\xrightarrow{R}\begin{array}{*{20}{c}}
{C{H_2} - OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2} - OH}
\end{array}$
$C{H_2} = CH - C{H_3} + HBr \to C{H_3}CHBr - C{H_3}$
$reagent$ શું હશે?
$(i)$ $C{H_3} - C \equiv C - C{H_3}$
$(ii)$ $C{H_3} - C{H_2} - C{H_2} - C{H_3}$
$(iii)$ $C{H_3} - C{H_2} - C \equiv CH$
$(iv)$ $C{H_3} - CH = C{H_2}$