Question

Preparation and reactions of alcohols Alcohols can by prepared from alkyl halides, epoxides, alkenes, etc. When treated with acid, \(\mathrm{C}-\mathrm{OH}\) bond is cleaved heterolytically to form a carbocation which could then undergo substitution, elimination or rearrangement depending on the conditions and stability of the carbocation. When \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) is refluxed with metallic sodium in dry ethers the major product obtained is (a) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\)

Short answer

Answer: The major product is CH₂=CH-CH=CH₂ (b).

Step-by-step solution

Identifying the type of reaction

The given reaction is a reaction between an alcohol and metallic sodium in a dry ether. This is called a sodium metal reduction reaction. It is a form of reduction reaction where the metallic sodium reduces the alcohol.

Determine the reaction mechanism

In a sodium metal reduction reaction, the hydrogen atom of the \(\mathrm{OH}\) group on the alcohol undergoes abstraction by the sodium metal. The sodium metal donates an electron to the hydrogen atom, effectively releasing it as a hydride ion (\(\mathrm{H}^{-}\)). Concurrently, the sodium replaces the hydrogen on the \(\mathrm{OH}\) group, forming a sodium alkoxide intermediate. The hydride ion then attacks the carbon atom bonded to the halogen atom (\(\mathrm{Br}\)), resulting in the formation of a double bond and the expulsion of the halogen ion (\(\mathrm{Br}^-\)).

Apply the mechanism to the given alcohol

To find the major product, apply the sodium metal reduction mechanism to the given alcohol: \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\). First, the sodium metal abstracts the hydrogen from the \(\mathrm{OH}\) group: \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{Na} \rightarrow \mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa} + \mathrm{H}^{-}\) Next, the hydride ion attacks the carbon atom bonded to the bromine atom, causing the double bond formation and the expulsion of the bromide ion: \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa} + \mathrm{H}^{-} \rightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} + \mathrm{NaBr}\)

Identify the major product

The major product formed after refluxing the given alcohol with metallic sodium in dry ethers is: (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\)

Key concepts

Sodium Metal Reduction

Sodium metal reduction is a fascinating type of reaction involving the interaction between alcohols and sodium metal. When alcohol is treated with metallic sodium, it acts as a reducing agent. This means sodium facilitates the removal of hydrogen from the hydroxyl (\( \text{OH} \)) group present in alcohol. Think of sodium as taking on a transforming role, where it donates an electron to the hydrogen of the OH group, effectively creating a hydride ion (\( \text{H}^- \)). The reaction typically proceeds in a dry environment, such as dry ether, to ensure that moisture does not interfere with the process. This interaction results in the formation of an intermediate compound called sodium alkoxide, which is crucial to the next steps in the reaction mechanism.

Alkoxide Formation

The formation of an alkoxide is a pivotal step in sodium metal reduction reactions. Here, when sodium interacts with the hydrogen atom of the hydroxyl (\( \text{OH} \)) group, it forms sodium hydroxide and releases hydrogen gas (\( \text{H}_2 \)). This creates a new chemical species known as sodium alkoxide ([R-O-Na]). This compound plays a significant role as it is much more reactive than the initial alcohol.
The sodium alkoxide is characterized by the metal atom directly bonded to the oxygen, replacing the hydrogen from the hydroxyl group. For example, when sodium reacts with an alcohol derived from BrCH₂CH₂CH₂CH₂OH, the resulting compound is BrCH₂CH₂CH₂CH₂ONa. This transition from alcohol to alkoxide is vital for subsequent chemical interactions in the reaction process.

Carbocation Mechanism

In organic chemistry, carbocations represent a class of intermediates that play a vital role in many mechanisms. They are positively charged carbon atoms, usually formed during reactions that involve heterolytic bond cleavage, such as in alcohol reactions. Here, the carbon atom forming the carbocation can undergo a variety of reactions.
In the specific context of sodium metal reduction, after the formation of sodium alkoxide, a hydride ion produced can further influence the reaction. It can initiate the elimination of a halogen, such as bromine, from an adjacent carbon. The resulting structure can stabilize by forming multiple bonds, commonly leading to alkyne or alkene structures. This process highlights the complexity and significance of carbocations in organic transformations.

Heterolytic Cleavage

Heterolytic cleavage is an essential concept in understanding reactions involving alcohols and other organic compounds. It refers to the breaking of a chemical bond in which both electrons from the bond pair are transferred to one of the atoms, resulting in charged intermediates. This is different from homolytic cleavage, where the electrons are equally divided.
In the context of the step-by-step solution provided, heterolytic cleavage becomes relevant when considering the conversion of the alkoxide into the final product. This step involves the breaking of the carbon-bromine bond, facilitated by the action of the hydride ion, resulting in the ejection of a bromide ion (\(\text{Br}^-\)). This step further leads to the formation of a carbon-carbon double bond characteristic of the final structure, allowing complex organic transformations to occur efficiently.