Skip to main content
Ch.4 - The Study of Chemical Reactions
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
Chapter 4, Problem 49a

The chlorination of pentane gives a mixture of three monochlorinated products.
a. Draw their structures.

Verified step by step guidance
1
Step 1: Understand the problem. The chlorination of pentane involves replacing one hydrogen atom in the pentane molecule with a chlorine atom. This reaction typically occurs via a free radical mechanism, and the products depend on the different types of hydrogen atoms present in the molecule.
Step 2: Analyze the structure of pentane. Pentane (C5H12) is a straight-chain alkane with five carbon atoms. Each carbon atom has a specific number of hydrogens attached, depending on its position in the chain.
Step 3: Identify the unique positions for chlorination. In pentane, there are three types of hydrogens based on the carbon they are attached to: primary hydrogens (on the terminal carbons), secondary hydrogens (on the carbons adjacent to the terminal carbons), and tertiary hydrogens (if present, but pentane does not have tertiary hydrogens).
Step 4: Draw the monochlorinated products. Replace one hydrogen atom at each unique position with a chlorine atom. This will result in three distinct products: 1-chloropentane (chlorine attached to the first carbon), 2-chloropentane (chlorine attached to the second carbon), and 3-chloropentane (chlorine attached to the third carbon).
Step 5: Verify the structures. Ensure that each product has the correct molecular formula (C5H11Cl) and that the chlorine atom is placed at the correct position on the pentane chain. These are the three monochlorinated products resulting from the chlorination of pentane.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Alkanes

Alkanes are saturated hydrocarbons consisting only of carbon and hydrogen atoms, connected by single bonds. They follow the general formula CnH2n+2, where 'n' is the number of carbon atoms. Pentane, with five carbon atoms, is an example of an alkane, and its structure is crucial for understanding the chlorination process.
Recommended video:
Guided course
02:47
Functionalizing Alkanes

Free Radical Halogenation

Free radical halogenation is a reaction mechanism where alkanes react with halogens (like chlorine) to form alkyl halides. This process involves three steps: initiation, propagation, and termination. During chlorination, chlorine radicals abstract hydrogen atoms from pentane, leading to the formation of various monochlorinated products.
Recommended video:
Guided course
12:11
Radical Chain Reaction Mechanism.

Isomerism

Isomerism refers to the existence of compounds with the same molecular formula but different structural arrangements. In the case of chlorinated pentane, the three monochlorinated products arise from the substitution of hydrogen atoms at different positions on the pentane chain, resulting in distinct structural isomers that must be drawn to illustrate their unique configurations.
Recommended video:
Guided course
06:47
Monosaccharides - D and L Isomerism
Related Practice
Textbook Question

The chlorination of pentane gives a mixture of three monochlorinated products.

b. Predict the ratios in which these monochlorination products will be formed, remembering that a chlorine atom abstracts a secondary hydrogen about 4.5 times as fast as it abstracts a primary hydrogen.

7
views
Textbook Question

When exactly 1 mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a ­chlorination reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane, as well as ­unreacted methane.

a. Explain how a mixture is formed from this stoichiometric mixture of reactants, and propose mechanisms for the ­formation of these compounds from chloromethane.

1
views
Textbook Question

a. Draw the structure of the transition state for the second propagation step in the chlorination of methane.

Show whether the transition state is product-like or reactant-like and which of the two partial bonds is stronger.

2
views
Textbook Question

Peroxides are often added to free-radical reactions as initiators because the oxygen–oxygen bond cleaves homolytically rather easily. For example, the bond-dissociation enthalpy of the O―O bond in hydrogen peroxide (H―O―O―H) is only 213 kJ/mol (51 kcal/mol). Give a mechanism for the hydrogen peroxide-initiated reaction of cyclopentane with chlorine. The BDE for HO―Cl is 210 kJ/mol (50 kcal/mol).

1
views
Textbook Question

For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed.

(f)

1
views
Textbook Question

When exactly 1 mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a ­chlorination reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane, as well as ­unreacted methane.

b. How would you run this reaction to get a good conversion of methane to CH3Cl? Of methane to CCl4?

3
views