Question

Write structures for all the structural isomers of compounds with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\) in which Cl and \(\mathrm{Br}\) are bonded to a double-bonded carbon.

Short answer

Answer: The two structural isomers are: 1. ClC=C-CH2-CH2Br 2. BrC=C-CH2-CH2Cl

Step-by-step solution

Determine the degrees of unsaturation (DOU)

The degrees of unsaturation will give us an idea of how many double bonds or rings are required within the compound structure. We can calculate the DOU using the following formula: \(DOU = \frac{( 2 \times n_{C} + 2 - n_{H} + n_{X})}{2}\) Where \(n_{C}\) is the number of carbons, \(n_{H}\) is the number of hydrogens, and \(n_{X}\) is the number of halogens in the formula. In our case, \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\) we have: \(DOU = \frac{( 2 \times 4 + 2 - 6 + 2)}{2} = \frac{6}{2} = 3\)

Identify the number of carbons and halogens

We have 4 carbon atoms, each of which can form four bonds. Additionally, we have two halogens in the molecular formula - one Cl and one Br.

List carbon skeleton possibilities

Since we are dealing with a 4-carbon molecule, we can have three main possibilities for the base carbon skeleton: - Linear 4-carbon structure: C-C-C-C - 3-carbon chain with a 1-carbon branch: C-C-C branching off to another C - Ring structure of 4 carbons: C-C-C-C in a cyclic form

Fulfill DOU requirements with double bonds or rings

We need 3 degrees of unsaturation, so for each carbon skeleton possibility, we need to add double bonds or rings to fulfill this requirement. Note that Cl and Br need to bond to a double-bonded carbon. For the linear 4-carbon structure: - Two double bonds: C=C-C-C (one possible location for Cl and Br) - One double bond and a ring: Not possible, as the Cl and Br cannot be bonded to a double-bonded carbon in this structure. For the 3-carbon chain structure with a 1-carbon branch: - Not possible, as we cannot generate 3 degrees of unsaturation while also ensuring Cl and Br are bonded to a double-bonded carbon. For the ring structure of 4 carbons: - One double bond in the cyclic structure: Not possible, as it only generates 2 degrees of unsaturation. Thus, the only possible structure is C=C-C-C with Cl and Br attached to the double bonded Cs.

Draw the structural isomers of \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\)

With the given information that Cl and Br are bonded to double bonded carbons and the only valid carbon skeleton being the linear 4 carbon structure with two double bonds, we can draw the structural isomers: 1. \(\mathrm{ClC=C-CH}_{2}\mathrm{-CH}_{2}\mathrm{Br}\) 2. \(\mathrm{BrC=C-CH}_{2}\mathrm{-CH}_{2}\mathrm{Cl}\) These are the two structural isomers of \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\) where Cl and Br are bonded to a double-bonded carbon.