Haloalkanes and Haloarenes

Mass of organic compound = 0.2 gram

$n_{AgBr}=\frac{0.188}{188}mol$

= 0.001 mol

$n_{Br}=n_{AgBr}=0.001mole$

 Mass of Br = 0.001 * 80g

= 0.08 g

Mass & of Br = $\frac{0.08}{0.2}*100\mathrm{\%}$

= 40 %

Kindly consider the following figure

  $\underset{Chlorocompound}{\underset{?}{A}}\underset{\left(2\right)H_{2}O}{\overset{\left(1\right)O_3}{\to }}Formaldehyde$

$\downarrow$                   

Weight = 1.53 gm

V = 448 ml (STP)

Mole = $\frac{448}{22400}=\frac{weight}{molecularweight}$

$\therefore \frac{1.53}{molecularweight}=\frac{1}{50}$  

$\therefore$ Molecular weight of compound A = 50 * 1.53 = 76.5 gram

Molecular weight 76.5 = C_nH_2n-1 Cl

$\therefore C_n{H}_{2n-1}=41$  

or 12 n + 2n – 1 = 41

Molecular formula = C₃H₅Cl

Number of C- atoms = 3

Yellow precipitate is Agl (s) which only produce by structure IV

Following are the possible isomers of 1, 2-dimethyl cyclopropane

Mass of organic compound = 0.45 gm

Mass of AgBr obtained = 0.36 gm

$\therefore MolesofAgBr=\frac{0.36}{188}$

$\therefore MassofBromine=\frac{0.36}{188}*80=0.1532gm$

$\therefore \mathrm{\%}Brincompound=\frac{0.1532}{0.45}*100=34.04\mathrm{\%}$                

Ethylidene chloride is

1, 1 Dichloro ethane.