Question

The reaction of an alkene X with bromine produces a compound Y, which has $22.22\%$C, $3.71\%$H and $74.07\%$Br. The ozonolysis of alkene X gives only one product. The alkene X is?[Given: atomic mass of $C=12;H=1;Br=80$].

• A. Ethylene
• B. $1$-butene
• C. $2$-butene
• D. $3$-hexene

Answer 1

The correct option is C $2$-butene

The compound Y, has $22.22\%$C, $3.71\%$H and $74.07\%$Br.

100 g of compound Y will have 22.22 g C, 3.71 g H and 74.07 g Br.

The number of moles of $C=\frac{22.22g}{12g/mol}=1.85mol$

The number of moles of $H=\frac{3.71g}{1g/mol}=3.71mol$

The number of moles of $Br=\frac{74.07g}{80g/mol}=0.926mol$

The mole ratio $C:H:Br=1.85:3.71:0.926=2:4:1$

The empirical formula of compound Y is $C_2{H}_{4}Br$ which represents dibromobutane with molecular formula $C_4{H}_{8}B{r}_2$

Dibromobutane is obtained by bromination of butene which can be 1-butene or 2-butene.

The ozonolysis of alkene X gives only one product. Hence, alkene X is 2-butene.

$C{H}_3-CH=CH-C{H}_3\stackrel{\text{ozonolysis}}{\to }2C{H}_3-CHO$

$\underset{\text{X}}{C{H}_3-CH=CH-C{H}_3}+B{r}_2\stackrel{\text{bromination}}{\to }\underset{\text{Y}}{C{H}_3-\underset{\underset{\text{Br}}{|}}{CH}-\underset{\underset{\text{Br}}{|}}{CH}-C{H}_3}$