The cycloalkane (X) on bromination consumes o | vedclass

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(B) The bromination of a cycloalkane $(X)$ consuming one mole of $Br_2$ implies an addition reaction across a double bond or ring opening,resulting in

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$66$

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(B) The bromination of a cycloalkane $(X)$ consuming one mole of $Br_2$ implies an addition reaction across a double bond or ring opening,resulting in a dibromo derivative $(Y)$.Let the formula of $(Y)$ be $C_n H_m Br_2$.Given the $C:Br$ ratio is $3:1$,we have $\frac{n}{2} = \frac{3}{1}$,which gives $n = 6$.The product $(Y)$ is $C_6 H_{10} Br_2$.The molar mass of $C_6 H_{10} Br_2 = (6 	imes 12) + (10 	imes 1) + (2 	imes 80) = 72 + 10 + 160 = 242 \ g \ mol^{-1}$.The percentage of bromine in $(Y) = \frac{	ext{Mass of } Br}{	ext{Molar mass of } Y} 	imes 100 = \frac{160}{242} 	imes 100 \approx 66.11 \%$.Rounding to the nearest integer,we get $66 \%$.

The cycloalkane $(X)$ on bromination consumes one mole of bromine per mole of $(X)$ and gives product $(Y)$ in which $C:Br$ ratio is $3:1$. The percentage of bromine in the product $(Y)$ is $ . . . . . . \%$ (Nearest integer).
(Given: Molar mass in $g \ mol^{-1}$,$H:1, C:12, O:16, Br:80$)

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The cycloalkane $(X)$ on bromination consumes one mole of bromine per mole of $(X)$ and gives product $(Y)$ in which $C:Br$ ratio is $3:1$. The percentage of bromine in the product $(Y)$ is $ . . . . . . \%$ (Nearest integer).(Given: Molar mass in $g \ mol^{-1}$,$H:1, C:12, O:16, Br:80$)