Question

The organic product obtained in the following reaction is $$ \left(\mathrm{CH}_{3}-\mathrm{CH}_{2}\right)_{3} \mathrm{~B}+\mathrm{Cl}_{2} \longrightarrow $$ (A) \(\left(\mathrm{CCl}_{3}-\mathrm{CH}_{2}\right)_{3} \mathrm{~B}\) (B) \(\left(\mathrm{CH}_{3}-\mathrm{CCl}_{2}\right)_{3} \mathrm{~B}\) (C) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Cl}\) (D) \(\mathrm{CCl}_{3}-\mathrm{CCl}_{3}\)

Short answer

The short answer is: (C) \(\text{CH}_{3}-\text{CH}_{2}-\text{Cl}\).

Step-by-step solution

Identify the reactants and possible products

The reactants are given as the trialkylborane compound \((\text{CH}_{3}-\text{CH}_{2})_{3} \text{B}\) and the halogen gas \(\text{Cl}_{2}\). We need to find out how these reactants interact to form the product. The possible products, each corresponding to one of the given answer choices, are: (A) \(\left(\text{CCl}_{3}-\text{CH}_{2}\right)_{3} \text{B}\) (B) \(\left(\text{CH}_{3}-\text{CCl}_{2}\right)_{3} \text{B}\) (C) \(\text{CH}_{3}-\text{CH}_{2}-\text{Cl}\) (D) \(\text{CCl}_{3}-\text{CCl}_{3}\)

Show the reaction mechanism between trialkylborane and chlorine gas

The B-Cl bond formed via reaction between the boron atom in trialkylborane compound and the chlorine molecule during the reaction is highly polarized. Because boron is electron deficient and has an empty p-orbital, it will have a positive partial charge, while the chlorine atom will have a negative partial charge. This polarized bond allows for the nucleophilic attack by a lone pair of electrons from a chlorine atom in a second chlorine molecule (\(\text{Cl}_{2}\)). The reaction will proceed via single electron transfer (SET) mechanism.

Identify the most stable product

The reaction will continue via a single electron transfer (SET) mechanism with chlorine's lone pair of electrons forming a bond with boron via its vacant p-orbital, and cleaving one of the B-C bonds. This leads to the formation of a new C-Cl bond and creates a chloride ion. The most stable product in this case will be (C) \(\text{CH}_{3}-\text{CH}_{2}-\text{Cl}\), which is a primary alkyl halide formed. The final answer is (C) \(\text{CH}_{3}-\text{CH}_{2}-\text{Cl}\).

Key concepts

Organic Reaction Mechanism

Organic reaction mechanisms describe the step-by-step processes by which organic reactions occur. They detail how reactants transform into products, highlighting the movement of electrons during the reaction. One common feature of organic reaction mechanisms is the consideration of electron flow using arrows to indicate the movement.

• Reactions often involve changes at specific sites in the molecules where bonds are made and broken.
• These processes include the breaking of covalent bonds between atoms and the formation of new ones.
• The stability of intermediates plays a significant role in determining the reaction pathway and the final products.

In the exercise, the trialkylborane compound \((\text{CH}_3-\text{CH}_2)_3 \text{B}\) reacts with chlorine gas (\(\text{Cl}_2\)).
The reaction undergoes a mechanism where the boron atom's empty p-orbital accepts an electron pair from chlorine, leading to bond formation with cleavage of a B-C bond. Understanding reaction mechanisms is crucial as they enable chemists to predict the outcome and determine conditions that favor specific products.

Single Electron Transfer Mechanism

The single electron transfer (SET) mechanism involves the movement of one electron from one atom or molecule to another. Unlike typical covalent reactions where pairs of electrons move, SET reactions are characterized by one-at-a-time electron shifts.

• This mechanism often occurs in reactions involving radicals.
• Radicals are species with unpaired electrons, making them very reactive.
• SET is prevalent in organic reactions involving electron-rich and electron-poor species, like the interaction between trialkylboranes and halogens.

In the given exercise, the SET mechanism explains how the chlorine molecule can interact with the boron atom in trialkylborane.
Chlorine, a diatomic molecule, splits, and one of its atoms donates a single electron to the electron-deficient boron. This initiation through SET sets off a series of reactions leading to the formation of the final product, \(\text{CH}_3-\text{CH}_2-\text{Cl}\). Understanding SET can help clarify why certain reactions proceed in one direction, emphasizing the role of electron motion rather than just chemical bonds.

Boron Chemistry

Boron chemistry is a fascinating subfield of chemistry that demonstrates unique properties due to boron's electron deficiency. Unlike other elements in Group 13, boron forms compounds that are often characterized by incomplete octets, making it an electron-deficient element.

• Boron tends to form covalent compounds, but due to its trivalent nature, the compounds often seek additional electrons.
• Boron compounds are involved in various applications, from material science to medicine.
• In organic chemistry, trialkylboranes are known for their ability to participate in reactions with diverse compounds, including halogens.

In the given reaction, the trialkylborane compound \((\text{CH}_3-\text{CH}_2)_3 \text{B}\) exhibits its classic chemistry.
Boron, with its electron-deficient empty p-orbital, engages in a reaction with chlorine, forming new bonds and displaying the reactive nature.
This enables boron to serve as a versatile component in developing organic transformations. Its ability to form stable bonds while being electron-deficient opens a range of chemical possibilities, making boron chemistry essential to many modern technologies.