- A${(C{H_3})_2}CH - C{H_2}CH = C{H_2}$
- ✓${(C{H_3})_2}CH - CH = CH - C{H_3}$
- C${(C{H_3})_2}CH - C{H_2}CH = CH - C{H_3}$
- D${(C{H_3})_2}C = CHC{H_2}C{H_3}$
$\left( CH _3\right)_2 CH - CH _2 CH = CH _2$ is named as $4$-methyl pent-$1$-ene.
$\left( CH _3\right)_2 CH - CH _2- CH = CH - CH _3$ is named as $5$-methyl hex-$2$-ene.
$\left( CH _3\right)_2 C = CH - CH _2- CH _3$ is named as $4$-methyl pent-$3$-ene.
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hence $A$ $\xrightarrow{{conc.\,{H_2}S{O_4}}}$ $B,$ $B$ is :
$\gamma_{1} A +\gamma_{2} B \rightarrow \gamma_{3} C +\gamma_{4} D$
Concentration of $C$ changes from $10\, mmol$ appearance of $D$ is $1.5$ times the rate of disappearance of $B$ which is twice the rate of disappearance $A$. The rate of appearance of $D$ has been experimentally determined to be $9 \,m\,mol$ $dm ^{-3} s ^{-1}$. Therefore the rate of reaction is $......\,m\,mol\, dm ^{-3} \,s ^{-1}$. (Nearest Integer)
(Molar mass of $Ca(OH)_2, Na_2SO_4$ and $CaSO_4$ are $74, 143$ and $136\, g\, mol^{-1}$ respectively; $K_{sp}$ of $Ca(OH)_2$ is $5.5 \times 10^{-6}$)