Question

Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.

Answer 1

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the I.D.H (hydrogen deficiency index):

$I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}$

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

$I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1$

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).

See figure 1

I hope it helps!