How do you draw and name all isomers of C_3H_4Cl_2?

Sep 22, 2016

You first draw the parent compound propene ${C}_{3} {H}_{6}$

Explanation:

There is a double bond in it: $C {H}_{2} = C H - C {H}_{3}$

Now replace two of the $H$'s with $C l$. You can do this in 5 ways. (remember the double bond always has the lowest number)

$C C {l}_{2} = C H - C {H}_{3}$ 1,1-dichloropropene

$C H C l = C C l - C {H}_{3}$ 1,2-dichloropropene

$C H C l = C H - C {H}_{2} C l$ 1,3-dichloropropene

$C {H}_{2} = C C l - C {H}_{2} C l$ 2,3-dichloropropene

$C {H}_{2} = C H - C H C {l}_{2}$ 3,3-dichloropropene

Then there's another parent compound ${C}_{3} {H}_{6}$ called cyclopropane , where the $C$'s form a triangle of $C {H}_{2}$'s

Here there are only two possibilities to replace two $H$'s:

Both on the same $C$ 1,1-dichlorocyclopropane

On two neighbouring $C$'s 1.2-dichlorocyclopropane