Question

In the presence of a trace of peroxide or under the influence of ultraviolet light, 1-octene reacts: (a) with \(\mathrm{CHCI}_{3}\) to form \(1,1,1\) -trichlorononane; (b) with \(\mathrm{CHBr}_{3}\) to form \(1,1,3\) -tribromononane; (c) with \(\mathrm{CBrCl}_{3}\) to form \(1,1,1\) -trichloro-3-bromononane; (d) with \(\mathrm{H}-\mathrm{S}-\mathrm{CH}_{2} \mathrm{COOH}\) (thioglycolic acid) to yield \(\mathrm{n}-\mathrm{C}_{8} \mathrm{H}_{17}-\mathrm{S}-\mathrm{CH}_{2} \mathrm{COOH}\)

Short answer

In this exercise, we analyzed the reactions of 1-octene with various reagents, resulting in the formation of different compounds. The short answers are: Reaction (a): \(C_{8}H_{16} + 3CHCl_{3} \rightarrow C_{9}H_{18}Cl_{3}\) Reaction (b): \(C_{8}H_{16} + 3CHBr_{3} \rightarrow C_{9}H_{18}Br_{3}\) Reaction (c): \(C_{8}H_{16} + CBrCl_{3} \rightarrow C_{9}H_{17}Cl_{3}Br\) Reaction (d): \(C_{8}H_{16} + HSCH_{2}COOH \rightarrow C_{8}H_{17}SCH_{2}COOH\)

Step-by-step solution

Reaction (a)

1-octene will react with CHCl₃ (chloroform) to form 1,1,1-trichlorononane. The first step is to write down the balanced chemical equation for the reaction: \[C_{8}H_{16} + 3CHCl_{3} \rightarrow C_{9}H_{18}Cl_{3}\]

Reaction (b)

1-octene reacts with CHBr₃ (bromoform) to form 1,1,3-tribromononane. The balanced chemical equation for the reaction is as follows: \[C_{8}H_{16} + 3CHBr_{3} \rightarrow C_{9}H_{18}Br_{3}\]

Reaction (c)

1-octene reacts with CBrCl₃ to form 1,1,1-trichloro-3-bromononane. The balanced chemical equation for this reaction is: \[C_{8}H_{16} + CBrCl_{3} \rightarrow C_{9}H_{17}Cl_{3}Br\]

Reaction (d)

1-octene reacts with thioglycolic acid (H-S-CH₂COOH) to form n-C₈H₁₇-S-CH₂COOH. The balanced chemical equation for the reaction is given below: \[C_{8}H_{16} + HSCH_{2}COOH \rightarrow C_{8}H_{17}SCH_{2}COOH\]

Mechanism explanation (optional)

The mechanism of the reactions can vary depending on the nature of the reagents and reaction conditions. However, some reactions can be assumed to proceed via free radical mechanisms. In such cases, the formation of a free radical occurs on the double bond of 1-octene, which then reacts with the reagents to form the products. It is important to note that understanding the detailed mechanism of each reaction requires further examination and might not be compulsory at the high school level.

Key concepts

Chemical Reaction Equations

In organic chemistry, the ability to write balanced chemical reaction equations is fundamental. It represents the transformation from reactants to products, indicating the proportion of molecules or moles involved. For example, when 1-octene reacts with chloroform (CHCl₃), the equation is
\[\begin{equation}C_{8}H_{16} + 3CHCl_{3} \rightarrow C_{9}H_{18}Cl_{3}\right.\right.\right.\right.\right.\right.\end{equation}\]Here, coefficients before chemical formulas reflect the number of molecules required to balance the reaction. The equation must follow the law of conservation of mass, ensuring that the number of atoms for each element is equal on both sides. Writing such equations demands attention to both molecular formulas and how atoms rearrange during reactions. By mastering this, students will find it easier to predict the outcomes of reactions and understand the underlying stoichiometry.

Free Radical Reactions

Free radical reactions are a pivotal concept in organic chemistry, particularly when discussing reactions under specific conditions such as the presence of peroxides or ultraviolet light. A free radical is an atom or molecule with an unpaired electron, making it highly reactive. In the case of 1-octene, a free radical is generated at the alkene's double bond. This reactive intermediate can then attack other molecules, like CHCl₃ or CHBr₃, leading to halogenation products.

The general steps involves the homolytic cleavage of a bond—generating two free radicals—and subsequent reactions where these radicals seek to pair their unpaired electrons, ultimately forming new covalent bonds. Understanding these reaction mechanisms helps explain the formation of various products from seemingly similar conditions. Remember, the detailed mechanism could require a more in-depth analysis of the reaction pathway, involving initiation, propagation, and termination steps.

Halogenation of Alkenes

Halogenation of alkenes involves the addition of halogen atoms to the carbon-carbon double bond of an alkene. This reaction is often facilitated by free radicals when the reaction system is exposed to initiators like peroxides or UV light. For example, when 1-octene undergoes halogenation, different products can be formed depending on the halogenated reagent.

• With chloroform (CHCl₃), it results in 1,1,1-trichlorononane,
• With bromoform (CHBr₃), it leads to 1,1,3-tribromononane,
• And with a mixed halogenated compound (CBrCl₃), we get 1,1,1-trichloro-3-bromononane.

The product's structure reflects the addition pattern of halogens, showcasing the versatility of alkene chemistry. A proper understanding of such reactions enables students to predict the type of halogenation products formed, which is essential for synthesizing specific organic compounds in both laboratory and industrial settings.