Question

A compound of formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) is found to give only a single substance, \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Br}_{3}\), on further substitution. What is the structure of the \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) isomer and of its substitution product?

Short answer

1,2-dibromopropene; substitution product is 1,2,3-tribromopropane.

Step-by-step solution

Analyze the Molecular Formula

The compound has the molecular formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\). This indicates there are 3 carbon atoms, 6 hydrogen atoms, and 2 bromine atoms. The degree of unsaturation is calculated as (2C + 2) - H + N - X/2 = (2*3 + 2) - 6 + 0 - 2/2 = 1. Therefore, the molecule must contain either one double bond or a ring.

Consider Possible Structures

Given the degree of unsaturation, the compound could be a dibrominated alkene with a double bond. Since only one single product \(\mathrm{C}_{3}\mathrm{H}_{5}\mathrm{Br}_{3}\) is formed upon further substitution, the other hydrogen atoms must be equivalent before substitution, implying symmetry in the original structure.

Propose the Isomer Structure

The simplest symmetrical dibromoalkene fitting this description is 1,2-dibromopropene. In this structure, the double bond is between the first and second carbon atoms, making all the hydrogens on the third carbon atom equivalent, allowing them to be substituted by bromine atoms.

Analyze the Substitution Product

In further substitution to form \(\mathrm{C}_{3}\mathrm{H}_{5}\mathrm{Br}_{3}\), a bromine atom replaces one of the hydrogens from the third carbon of 1,2-dibromopropene. This can only occur once due to symmetry, yielding 1,2,3-tribromopropane.

Confirm the Structures

1,2-dibromopropene is symmetrical and allows for the uniform substitution to form only one tribrominated product. Thus, the structures are consistent with the given reaction conditions and products.

Key concepts

Degree of Unsaturation

The degree of unsaturation in a molecular formula is a powerful tool for understanding a compound's structure. It indicates the total number of pi bonds and rings within a molecule. In our exercise, we dealt with the compound \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\). By applying the formula for degree of unsaturation:\[\text{Degree of Unsaturation} = \frac{2C + 2 - H - X + N}{2}\]we determined a value of 1, which suggests the presence of one double bond, a ring, or a combination equivalent to one of these features. Understanding this calculation helps identify potential structural features before considering specific atom arrangements. Multiple degrees of unsaturation could mean various combinations of rings and multiple bonds, adding complexity to the structural determination, but in this simple case, it points clearly towards the presence of one double bond.

Symmetrical Structure

A symmetrical structure means that certain parts of the molecule are mirror images or rotationally identical, implying uniformity and equivalence in chemical environments. In our case, the compound \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) must feature symmetry because it reacts to form only one type of tribrominated product. This indicates that the hydrogen atoms available for bromine substitution are equivalent due to symmetry. In 1,2-dibromopropene, this symmetry arises because the double bond between the first and second carbon atoms ensures that the hydrogens bound to the third carbon are all in the same environment. This is a crucial consideration, as a lack of symmetry would lead to multiple substitution products rather than a single one. This feature greatly simplifies chemical reactions and analysis.

Alkene

Alkenes are hydrocarbons containing a carbon-carbon double bond. They are unsaturated, which means they can undergo addition reactions. In the case of \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\), identifying it as a dibrominated alkene is essential.When determining potential structures, recognizing the presence of an alkene helps narrow down configurations. The double bond not only affects the degree of unsaturation but also dictates how the molecule can react chemically. For alkenes, the stability of configuration (cis or trans) and the position of any substituents like bromines are critical. They influence the overall reactivity and the feasibility of further reactions.Understanding that our compound is an alkene explains the core reactivity and forms a basis for predicting how the molecule will interact under different chemical conditions.

Substitution Reaction

Substitution reactions involve replacing an atom or group in a molecule with another atom or group. In organic chemistry, this often involves halogens like bromine. For the compound \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\), further bromine substitution is a key focus.When 1,2-dibromopropene undergoes substitution, a hydrogen atom on the third carbon, due to the symmetrical environment of the molecule, is replaced by a bromine atom. The reaction forms a single new compound, \(\mathrm{C}_{3}\mathrm{H}_{5}\mathrm{Br}_{3}\), due to the availability and equivalence of the hydrogen atoms originally part of the structure.This uniformity in substitution illustrates the importance of symmetry again. In symmetrical compounds, substitution can create products that appear consistent, reducing complexity in structural outcomes and enhancing predictability in chemical reactions. Substitution reactions are a foundation in synthetic chemistry, underpinning the creation of derivatives and altered compounds.