The number of isomers possible for a dibromo | vedclass

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(B) The molecular weight of the dibromo derivative is $186 \ u$. Given $Br = 80 \ u$,the mass of the alkene part is $186 - (2 	imes 80) = 186 - 160 =

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$3$

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(B) The molecular weight of the dibromo derivative is $186 \ u$. Given $Br = 80 \ u$,the mass of the alkene part is $186 - (2 	imes 80) = 186 - 160 = 26 \ u$. The general formula for an alkene is $C_nH_{2n}$,so $12n + 2n = 26$,which gives $14n = 26$. Since $n$ must be an integer,we re-evaluate: the dibromo derivative of ethene $(C_2H_4)$ is $C_2H_2Br_2$. Its molecular weight is $(2 	imes 12) + (2 	imes 1) + (2 	imes 80) = 24 + 2 + 160 = 186 \ u$. Thus,the compound is $C_2H_2Br_2$. The possible isomers for $C_2H_2Br_2$ are: $1$. $1,1-dibromoethene$ $(CH_2=CBr_2)$ $2$. $cis-1,2-dibromoethene$ $3$. $trans-1,2-dibromoethene$ Therefore,there are $3$ possible isomers.

List-$I$ (Pair of Compounds)	List-$II$ (Type of Isomers)
$A$. $2-$Methylpropene and but$-1-$ene	$I$. Stereoisomers
$B$. $Cis-$but$-2-$ene and $trans-$but$-2-$ene	$II$. Position isomers
$C$. $2-$Butanol and diethyl ether	$III$. Chain isomers
$D$. But$-1-$ene and but$-2-$ene	$IV$. Functional group isomers

The number of isomers possible for a dibromo derivative (Molecular weight $= 186 \ u$) of an alkene is $(Br = 80 \ u)$.

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