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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Hydrocarbons
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$CH_3CH_2CH_2CH_3\quad\underrightarrow {Cl_2/h\nu} \quad A+B$(Monochlorination products).The approximate ratio of percentage yields of $A$ and $B$ formed in the reaction is

$\begin{array}{1 1}(a)\;50 : 50&(b)\;72 : 28\\(c)\;45 : 55&(d)\;60 : 40\end{array}$
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$CH_3CH_2CH_2CH_3\quad\underrightarrow{Cl_2/h\nu} \quad CH_3-CH_2-CH_2-CH_2Cl(A)$(Relative amount)
$\Rightarrow $No of equivalent hydrogen $\times $ reactivity
$\Rightarrow 6\times 1=6$
$CH_3CH_2CH_2CH_3\quad\underrightarrow{Cl_2/h\nu} \quad CH_3-CH-Cl-CH_2-CH_3(B)$(Relative amount)
$\Rightarrow $No of equivalent hydrogen $\times $ reactivity
$\Rightarrow 4\times 3.8=15.2$
Total amount$=6+15.2=21.2$
$\%$ of A=$\large\frac{6}{21.2}$$\times 100=28.3$
$\%$ of B=$\large\frac{15.2}{21.2}$$\times 100=71.7$
Hence (b) is the correct option.
answered Mar 20, 2014 by sreemathi.v
edited Jul 14 by meena.p

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