
- ✓$a > c > b > d $
- B$a > b > c > d$
- C$d > b > c > a$
- D$a > d > b > c$

$3^{\circ}>2^{\circ}>1^{\circ}$
The order of leaving group in $E 2$ elimination reaction is shown below :
$R - I > R - Br > R - Cl$
Therefore, the correct order for reaction with alcoholic $KOH$ is as follows :
$a > c > b > d$
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Product $(B)$ is
List $-I$
$(A){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_6} + CO + HCl\,\xrightarrow{{AlC{l_3}}}\,{C_6}{H_5}CHO$
$(B){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_5} + C{H_3}$ $\xrightarrow{{Cr{O_2}C{l_2}}}\,{C_6}{H_5}CHO$
$(C){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_5}C{H_2}Br\, + \,C{H_3}Br$ $\xrightarrow{{Na/ether}}\,{C_6}{H_5}C{H_2}C{H_3}$
$(D){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_6}\, + \,{(C{H_3})_2}C = C{H_2}$ $\xrightarrow{{{H_2}S{O_4}}}\,{C_6}{H_5}C{(C{H_3})_3}$
List $-II$
$(a)$ Friedel-Crafts reaction
$(b)$ Wurtz-Fitting reaction
$(c)$ Gattermann-Koch Aldehyde synthesis
$(d)$ Etard’s reaction
