Question

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

Short answer

There are 6 structural isomers of compounds with molecular formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\) that have chlorine and bromine bonded to a double-bonded carbon atom.

Step-by-step solution

Identify the Base Structure

Let's start with the arrangement of the carbon atoms. A molecular formula of \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\) implies that there's a double bond, as deduced by the degree of unsaturation from the formula: Degree of Unsaturation = \(\frac{2 \times n + 2 - X}{2}\), where n is the number of carbon atoms, and X is the number of hydrogen, halogen atoms. Degree of Unsaturation = \(\frac{2 \times 4 + 2 - 6 - 1 - 1}{2} = 1\) Since there's only one degree of unsaturation, we will form only one double bond in the molecules.

Determine Position of Double Bond

Now, we'll determine the possible positions for the double bond across the four carbon atoms. There are three possibilities for the double bond, which are: 1. Carbon 1 and Carbon 2 2. Carbon 2 and Carbon 3 3. Carbon 3 and Carbon 4

Attach Chlorine and Bromine

Next, we'll attach the chlorine and bromine atoms to the double-bonded carbon atoms as per the given condition. It can be done in two ways: 1. Chlorine attached to first double-bonded carbon and bromine attached to the second double-bonded carbon 2. Bromine attached to first double-bonded carbon and chlorine attached to the second double-bonded carbon

Combine Double Bonds and Halogen Positions

For the three double bond possibilities, let's have each possible arrangement of chlorine and bromine: 1. Double bond between Carbon 1 and Carbon 2 - Chlorine attached to Carbon 1, Bromine attached to Carbon 2 - Bromine attached to Carbon 1, Chlorine attached to Carbon 2 2. Double bond between Carbon 2 and Carbon 3 - Chlorine attached to Carbon 2, Bromine attached to Carbon 3 - Bromine attached to Carbon 2, Chlorine attached to Carbon 3 3. Double bond between Carbon 3 and Carbon 4 - Chlorine attached to Carbon 3, Bromine attached to Carbon 4 - Bromine attached to Carbon 3, Chlorine attached to Carbon 4

Draw Structures

Now that we have all possible combinations, let's draw them: 1.1. \(\ce{C(Cl)=C(Br)-CH2-CH3}\) 1.2. \(\ce{C(Br)=C(Cl)-CH2-CH3}\) 2.1. \(\ce{CH2-C(Cl)=C(Br)-CH3}\) 2.2. \(\ce{CH2-C(Br)=C(Cl)-CH3}\) 3.1. \(\ce{CH3-CH2-C(Cl)=C(Br)}\) 3.2. \(\ce{CH3-CH2-C(Br)=C(Cl)}\) These six structures represent all the structural isomers of compounds with molecular formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\) that have chlorine and bromine bonded to a double-bonded carbon atom.

Key concepts

Molecular Formula

The molecular formula is like a recipe for a compound. It tells us the number and types of atoms in a molecule. For the exercise, the molecular formula is \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\). Let's break this down:

• Four carbon atoms (\(\mathrm{C}_{4}\))
• Six hydrogen atoms (\(\mathrm{H}_{6}\))
• One chlorine atom (\(\mathrm{Cl}\))
• One bromine atom (\(\mathrm{Br}\))

This formula implies the structure of the compound but doesn't specify how the atoms are connected. In organic chemistry, knowing the connectivity is crucial because different connections can form different molecules. This is where identifying structural isomers comes in handy as it explores all possible unique arrangements of the given atoms.

Degree of Unsaturation

The degree of unsaturation indicates the number of rings or double bonds in a molecule. It helps us understand how many multiple bonds or ring structures a molecule contains. For a molecular formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\), we calculate as follows:

\[\text{Degree of Unsaturation} = \frac{2n + 2 - X}{2}\]where \(n\) is the number of carbon atoms, and \(X\) is the total number of hydrogen, chlorine, and bromine atoms. Plugging in the values:

\[\text{Degree of Unsaturation} = \frac{2 \times 4 + 2 - 6 - 2}{2} = 1\]This result means that the molecule must have one double bond since there is one degree of unsaturation. This handy calculation allows chemists to predict the types of bonds in a compound just from the molecular formula itself.

Organic Chemistry

Organic Chemistry is the study of carbon compounds, usually containing carbon-carbon bonds. It is a vast field of chemistry that is central to understanding the structure and properties of organic molecules. In this exercise, we explore structural isomers, which are compounds with the same molecular formula but different arrangements of atoms.

By altering the connectivity of the atoms, you can form isomers with unique properties. For instance, the placement of chlorine and bromine in \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{ClBr}\) impacts how the molecule behaves in reactions. Each isomer can show different physical and chemical properties, highlighting the intricate beauty and complexity within organic chemistry.

Double Bond Positioning

Double bond positioning is crucial in determining molecular structure and properties. A double bond involves the sharing of four electrons between two atoms, and its location can significantly alter an organic molecule's behavior. In the given exercise, the formula allows us three primary locations for the double bond:

• Between Carbon 1 and Carbon 2
• Between Carbon 2 and Carbon 3
• Between Carbon 3 and Carbon 4

These different positions create different structural isomers. Additionally, as specified in the exercise, chlorine, and bromine must be attached to a double-bonded carbon. This requirement restricts the halogens' potential bonding positions based on where the double bond is situated. Alterations in double bond positioning can greatly affect isomer properties and identification.