Question

1,3 -diiodopropane on reaction with sodium in ether gives (a) \(\mathrm{IH}_{2} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Short answer

Answer: (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}\mathrm{CH}_{2} \mathrm{CH}_{3}\)

Step-by-step solution

Write down the given reactant and conditions

Here you are given 1,3-diiodopropane reacting with sodium in ether. The reactant molecule looks like this: \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I}(\mathrm{1,3-Diiodopropane})\) This molecule reacts with sodium in ether (Na/Et2O) in a Wurtz reaction.

Identify the type of reaction

The given reaction is a Wurtz reaction, which is a coupling reaction of alkyl halides with sodium in ether to form higher alkanes. In this reaction, alkyl halides react with sodium in the presence of dry ether to form a new C-C bond in the resulting molecule.

Mechanism of the Wurtz reaction

In the Wurtz reaction mechanism, two moles of 1,3-diiodopropane reacts with two moles of sodium in dry ether. Sodium, being an alkali metal, loses an electron to form a sodium ion (Na+) and an electron (e-). The electron reacts with the iodine atom in 1,3-diiodopropane, reducing it to iodide (I-) and generating a free radical on the carbon atom. The two carbon-centered radicals react with each other, forming a new C-C bond. The two iodide ions pair up with the two sodium ions to form sodium iodide (2 NaI) as a byproduct. The overall reaction can be represented as follows: 2 \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I}\) + 2 Na \(\xrightarrow{\text{Et}_2\text{O}} \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\) + 2 NaI

Identify the product

Following the mechanism of a Wurtz reaction, the product of the reaction of 1,3-diiodopropane with sodium in ether is: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3} (n-Hexane)\) This matches option (b) in the exercise. Therefore, the correct answer is: (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}\mathrm{CH}_{2} \mathrm{CH}_{3}\)

Key concepts

1,3-diiodopropane

Understanding the structure and properties of 1,3-diiodopropane is crucial when studying reactions like the Wurtz reaction. It is an organic compound with the formula \[\[\begin{align*}\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I}\end{align*}\]\], where the iodine atoms are attached to the first and third carbon atoms.

The presence of these iodine atoms makes 1,3-diiodopropane a reactive species, especially in the presence of sodium metal in an ether solution. In organic synthesis, it's used as an intermediate to form larger molecules through various coupling reactions.

Because iodine is a good leaving group, reactions involving 1,3-diiodopropane can often lead to the formation of new carbon-carbon bonds, substantially modifying the carbon skeleton of the initial molecule. This is exactly what happens in the Wurtz reaction where 1,3-diiodopropane is transformed into a new molecule with an extended carbon chain.

Sodium in Ether

The sodium in ether component of this reaction is a critical aspect for the Wurtz coupling process. Sodium, an alkali metal, performs as a strong reducing agent in this kind of reaction. The role of ether, specifically diethyl ether (\[\[\begin{align*}\text{Et}_2\text{O}\end{align*}\]\]), is equally significant as it provides a non-polar, aprotic environment that stabilizes the reactive intermediate species during the reaction.

When sodium is dissolved in diethyl ether, it tends to donate electrons due to its low ionization energy. These electrons are crucial in creating free radicals from alkyl halides like 1,3-diiodopropane when they come into contact with the alkyl halide, setting the stage for the subsequent coupling reactions.

It is important to handle sodium in ether with care, as the reaction can be quite exothermic and the reagents are quite reactive, posing potential safety hazards if not managed properly.

Alkyl Halides Coupling

The process of alkyl halides coupling is a cornerstone in organic synthesis. It is based on joining two alkyl halide molecules to form a single, longer alkane. In the context of the Wurtz reaction, the alkyl halides involved are typically the same, leading to symmetrical products.

Here's the generalized mechanism: each alkyl halide donates an electron to a sodium atom. This transfer generates two key intermediates: alkyl radicals and sodium halide. The alkyl radicals then couple together to form a new C-C bond, and the sodium halide forms an ionic byproduct.

Key Aspects of Alkyl Halide Reactivity:

• Good leaving group: Halides such as iodine are excellent leaving groups due to their size and polarizability.
• Radical Stability: Formation and subsequent coupling of radicals largely depend on the stability of those radicals.
• Reaction Conditions: The success of the coupling relies on conditions that favor the homolytic cleavage of the C-X bond where X is the halide.

Understanding these factors aids in grasping why alkyl halide coupling is a powerful method for constructing larger organic molecules.

Organic Chemistry Mechanisms

The concept of organic chemistry mechanisms is foundational for comprehending how chemical reactions occur. A reaction mechanism explains the step-by-step sequence of elementary reactions by which overall chemical change occurs.

In the Wurtz reaction, the mechanism reveals how 1,3-diiodopropane undergoes transformations to form n-hexane. Mechanisms in organic chemistry often involve homolytic or heterolytic bond cleavage, electron transfer, radical formation, and recombination.

A proper understanding of the mechanism allows chemists to predict products, identify intermediates, and understand the nuances that dictate reaction conditions. Moreover, mastering mechanisms affords chemists an advantageous position when strategizing synthetic pathways for complex molecules and enhances their predictive capabilities for novel reactions.