Question

The product obtained during the following reaction is $$ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br} \stackrel{\mathrm{ABr}_{3} \longrightarrow}{\longrightarrow} $$

Short answer

The product obtained during the reaction with ABr3 is 1-iodopropane (\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I}\)).

Step-by-step solution

Identify the reaction type and the reagent

The reaction provided is an example of a halogen-displacement reaction, where the halogen atom present in the reagent ABr3 (where A is a halogen element) will replace the halogen atom present in the given compound. The given compound is CH3-CH2-CH2-Br, which contains a bromine atom.

Determine the halogen atom in the reagent

In the given reagent ABr3, we know that A is a halogen element and B is bromine. We have to consider any halogen higher in the periodic table for a halogen-displacement reaction. Therefore, we can replace the bromine atom in the given compound with a halogen higher in the periodic table than bromine, so our reaction will look like this: $$ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br} \stackrel{\mathrm{ABr}_{3} \longrightarrow}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{A} $$

Provide a possible higher halogen element

Given the halogen elements in the periodic table in the following order: Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I) and Astatine (At). We need a higher halogen than Bromine; thus, we can consider Iodine (I) as the halogen from the reagent ABr3 (which becomes IBr3 in this case). $$ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br} \stackrel{\mathrm{IBr}_{3} \longrightarrow}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I} $$

Determine the final product

Now that we know the reagent and the given compound, we can find out the final product after the halogen-displacement occurs. So, the final product of the reaction will be an alkyl iodide, which is: $$ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I} $$ The product obtained during the following reaction is CH3-CH2-CH2-I (1-iodopropane).

Key concepts

Alkyl Halide Formation

Alkyl halides are essential compounds in organic chemistry and play a vital role in various reactions. The formation of alkyl halides involves the introduction of a halogen atom (such as fluorine, chlorine, bromine, or iodine) into an alkane. This transformation is fundamental in creating reactive intermediates for further organic reactions. Alkyl halides are typically formed through halogenation reactions, where a hydrogen atom in an alkane is replaced with a halogen atom. These reactions can occur under different conditions:

• In the presence of light or heat, as with free radical halogenations.
• Using halogenation agents like CBR₄ or PBr₃ .

In the exercise given, Br in propane is replaced with I utilizing a halogen-displacement reaction, forming an alkyl iodide. This process highlights the significance of alkyl halide formation in chemistry.

Periodic Table of Halogens

Halogens belong to Group 17 of the periodic table and possess unique properties that make them intriguing for chemists. They include Fluorine ( F ), Chlorine ( Cl ), Bromine ( Br ), Iodine ( I ), and Astatine ( At ). Halogens are known for their high reactivity and tendency to form compounds with other elements. In a halogen displacement reaction, a more reactive halogen will replace a less reactive one in a compound. The periodic table hierarchy is an essential factor:

• F is the most electronegative and reactive.
• I is less reactive than F, Cl, and Br but more than At .

The displacement reaction in our exercise involves using I to replace Br , as I is higher in the periodic table.

Organic Reaction Mechanism

Understanding organic reaction mechanisms is crucial for identifying the steps involved in chemical changes. In the context of halogen displacement reactions, the mechanism involves: 1. Breaking the bond between the alkyl group and the more electronegative halogen ( Br, in this case). 2. Introducing the new halogen ( I ) to the alkyl group. This change is part of a substitution reaction mechanism, where one atom or group is replaced by another. Substitution reactions can be either:

• SN1: Occurs with a unimolecular rate-determining step, often producing a carbocation intermediate.
• SN2: Involves a bimolecular collision step without intermediate formation, typically leading to inversion of configuration.

In the example of 1-iodopropane formation from 1-bromopropane using IBr₃, the reaction is more akin to an SN1 mechanism where the initial bond-breaking facilitates the incoming halogen's attachment. The knowledge of such mechanisms aids in predicting products and reaction pathways.