Question

(a) In a study of chlorination of propane, four products \((A, B,\), \(C\) and D) of formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{C}_{12}\) were isolated. What are their structures? (b) Each was chlorinated further, and the number of trichloro products \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{CI}_{3}\right)\) obtained from each was determined by gas chromatography. A gave one trichloro product; \(B\) gave two? and \(C\) and \(D\) each gave three. What is the structure of \(\mathrm{A} ?\) Of \(\mathrm{B}\) ? of \(\mathrm{C}\) and \(\mathrm{D}\) ? (c) By another synthetic method, compound C was obtained in optically active form. Now what is the structure of \(\mathrm{C}\) ? Of \(\mathrm{D} ?\) (d) When optically active \(\mathrm{C}\) was chlorinated, one of the trichloropropanes (E) obtained was optically active, and the other two were optically inactive. What is the structure of \(E\) ? Of the other two?

Short answer

The structures of the products are: A = 1,1-dichloropropane, B = 1,3-dichloropropane, C = 2,2-dichloropropane (optically active), D = 1,2-dichloropropane, E = 2,2,2-trichloropropane (optically active), and the other two trichloropropanes are 1,2,2-trichloropropane and 1,2,3-trichloropropane (both optically inactive).

Step-by-step solution

Identify possible structures for C3H6Cl2 products

To find possible structures for products with the formula C3H6Cl2, we should consider all the ways in which two chlorine atoms can be added to a propane molecule (C3H8). There are four possible structures for the C3H6Cl2 compounds: 1,1-dichloropropane, 1,2-dichloropropane, 1,3-dichloropropane, and 2,2-dichloropropane.

Match the structures to the products by trichloro product count

Now that we have the four possible structures, we can assign each one to A, B, C, or D based on the number of trichloro products it forms. Product A gives one trichloro product (C3H5Cl3), product B gives two, and products C and D each give three. For compound A to only give 1 trichloro product, the two chlorine atoms must be on the same carbon. Therefore, A = 1,1-dichloropropane. For compound B to give 2 trichloro products, we can analyze the remaining structures and deduce that B = 1,3-dichloropropane. For compounds C and D to give 3 trichloro products, they can be either 1,2-dichloropropane or 2,2-dichloropropane. For now, we cannot confirm their exact structures yet.

Determine the structure of compound C

Product C was obtained in optically active form. This implies that C has a chiral carbon. Only 2,2-dichloropropane has a chiral carbon (the central one). Hence, product C = 2,2-dichloropropane.

Determine the structure of compound D

Since we've already identified the structures of products A, B, and C, the structure of product D must be the remaining one: D = 1,2-dichloropropane.

Identify the optically active trichloropropane (E)

Optically active C (2,2-dichloropropane) gives an optically active trichloropropane E. The only chiral carbon in 2,2-dichloropropane is the central carbon (carbon 2). To preserve this chirality, the chlorine must be added to the same carbon (carbon 2). The structure of E is thus 2,2,2-trichloropropane.

Identify the other two trichloropropane products

The other two trichloropropane products must be formed by adding the chlorine atom to carbons 1 or 3. Their structures are 1,2,2-trichloropropane and 1,2,3-trichloropropane. These two products will be optically inactive. In summary, the structures of the products are as follows: - A = 1,1-dichloropropane - B = 1,3-dichloropropane - C = 2,2-dichloropropane (optically active) - D = 1,2-dichloropropane - E = 2,2,2-trichloropropane (optically active) - Other two trichloropropanes = 1,2,2-trichloropropane and 1,2,3-trichloropropane (optically inactive)

Key concepts

Chiral Molecules

Chirality is a property of a molecule that arises from the presence of an asymmetric carbon atom. This asymmetric carbon has four different substituents, making the molecule non-superimposable on its mirror image. Such molecules are known as chiral molecules. They often appear as pairs known as enantiomers, which are mirror images of each other.

In our case, 2,2-dichloropropane is chiral, as it possesses a central carbon attached to four different groups. This setup is essential in understanding products of reactions like chlorination, where the arrangement of atoms is crucial.

Recognizing and identifying chirality in molecules can greatly influence how they react and interact with other molecules. Chiral molecules have significant roles in fields such as medicinal chemistry, where one enantiomer of a drug might be therapeutic while its mirror image could be harmful.

Optical Activity

Optical activity relates to a molecule's ability to rotate the plane of polarized light. This property is a direct consequence of chirality and is a defining characteristic of chiral molecules. Understanding optical activity is vital in determining the purity and mixture of enantiomers.

When compound C, which is 2,2-dichloropropane, obtained in optically active form, this indicates the presence of an excess of one enantiomer over the other. This unbalanced distribution results in a measurable rotation of polarized light.

In distinguishing between optical isomers, gas chromatography and polarimetry are often utilized. These methods help determine the specific angle of rotation and quantify the specific enantiomers present.

Gas Chromatography

Gas chromatography (GC) is a method used to separate and analyze compounds that can be vaporized without decomposition. It is especially useful in determining the different products of chemical reactions, like the chlorination of propane in this scenario.

In the exercise, GC was used to analyze the trichloro products of the different dichloropropane isomers. The method separates compounds based on their volatility, allowing for precise recognition of the number and type of products formed from further chlorination.

By using gas chromatography, we observe how products differ and identify the structures based on how long they take to travel through the chromatography column. Each compound leaves a distinctive "fingerprint" in the chromatogram, making it easier to match its structure.

Chemical Structures

Chemical structures depict how atoms in a molecule are bonded and arranged in three-dimensional space. Understanding these structures is crucial for predicting the chemical behavior and properties of a molecule.

In the exercise, the chlorination of propane and the resulting dichloropropane products showcase how bonds are formed and broken. For example, 1,1-dichloropropane has chlorine atoms on the same carbon, while 1,2-dichloropropane has them on adjacent carbons, influencing the number and type of products formed during further chlorination.

Being able to predict structures like those from chlorinated hydrocarbons highlights the importance of knowing chemical structures. It assists in interpreting reaction mechanisms and anticipating changes during chemical transformations, which is fundamental in chemistry and various practical applications.