(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?

Step by step solution

01

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.

02

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.

03

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.

04

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.

05

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.

06

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)

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