Question

Using your reaction roadmap as a guide, show how to convert butane into 2 -butyne. Show all reagents and all molecules synthesized along the way.

Short answer

Answer: The step-by-step procedure to convert butane into 2-butyne involves the following steps: 1. Halogenation of butane with bromine (Br2) in the presence of light (hν) to obtain 2-bromobutane. 2. Substitution of bromine in 2-bromobutane with an azide group (N3) using sodium azide (NaN3) to form 2-azidobutane. 3. Reduction of the azide group in 2-azidobutane to an amine group (NH2) using lithium aluminum hydride (LiAlH4), resulting in 2-aminobutane. 4. Conversion of 2-aminobutane to 2-chlorobutane by treating it with thionyl chloride (SOCl2). 5. Double dehydrohalogenation of 2-chlorobutane using a strong base like potassium tert-butoxide (KOtBu) to eliminate the halogen and a hydrogen atom, forming 2-butyne.

Step-by-step solution

Halogenation of Butane

To convert butane into 2-butyne, first, we need to add a halogen (Br in this case) to the second carbon atom of the butane molecule. This can be achieved by treating butane with bromine (Br2) in the presence of light (hν) to obtain 2-bromobutane. The reaction is as follows: $$ C_4H_{10} + Br_2 \xrightarrow{h\nu} C_4H_9Br + HBr $$

Substitution with NaN3

Next, we need to replace the bromine atom in 2-bromobutane with an azide group (N3). This can be done using sodium azide (NaN3) in a substitution reaction. The NaN3 reacts with 2-bromobutane to form 2-azidobutane and sodium bromide. The reaction is as follows: $$ C_4H_9Br + NaN_3 \rightarrow C_4H_9N_3 + NaBr $$

Reduction of Azide to Amine

Now, we need to reduce the azide group in 2-azidobutane to an amine group (NH2). This can be achieved by treating 2-azidobutane with lithium aluminum hydride (LiAlH4) in a reduction reaction. The product obtained is 2-aminobutane. $$ C_4H_9N_3 + LiAlH_4 \rightarrow C_4H_{11}N + LiAlH_3N $$

Double Dehydrohalogenation

To convert 2-aminobutane into 2-butyne, we first need to convert it into 2-chlorobutane by treating it with thionyl chloride (SOCl2). The reaction is as follows: $$ C_4H_{11}N + SOCl_2 \rightarrow C_4H_9Cl + HCl + SO_2 $$ To achieve the alkyne (2-butyne) formation, we need to undergo a double dehydrohalogenation. This can be done by treating 2-chlorobutane with two equivalents of a strong base, like potassium tert-butoxide (KOtBu), to eliminate the halogen accompanied by a removal of a hydrogen atom. First, dehydrohalogenation forms 1-butyne: $$ C_4H_9Cl + KOtBu \rightarrow C_4H_8 + KCl + HOtBu $$ Next, a second dehydrohalogenation forms 2-butyne: $$ C_4H_8 + KOtBu \rightarrow C_4H_6 + KCl + HOtBu $$ So the final product 2-butyne has been achieved by the stepwise conversion of butane.

Key concepts

Halogenation Reaction

The synthesis of 2-butyne starts with a halogenation reaction, where a halogen is added to an organic compound. When butane is treated with bromine in the presence of ultraviolet light, a bromine atom attaches to the second carbon of the butane molecule to form 2-bromobutane. During this process, one of the bromine atoms from the Br₂ molecule replaces a hydrogen atom on the butane, resulting in the formation of hydrobromic acid as a byproduct. Halogenation is a type of substitution reaction, crucial in transforming hydrocarbons to more reactive intermediates for further chemical modifications.

Halogens like chlorine and bromine are commonly used due to their reactivity. This reaction typically requires the presence of light or another radical initiator to propagate the chain reaction needed to replace hydrogen with the halogen on the organic molecule.

Importance in Organic Synthesis

In organic synthesis, halogenation is a vital step. Not only does it allow for the introduction of functional groups into hydrocarbon chains, but the resultant haloalkanes are also typically more reactive and serve as precursors for other functional groups through subsequent reactions.

Nucleophilic Substitution

Following halogenation, the synthesis requires a nucleophilic substitution reaction. This reaction involves replacing a leaving group with a nucleophile. In the conversion to 2-butyne, the leaving group is a bromine atom, and the nucleophile is azide (N₃^-). When sodium azide reacts with 2-bromobutane, the nucleophile replaces the bromine to form 2-azidobutane.

Nucleophilic substitutions can proceed through two primary mechanisms: SN1 and SN2. The mechanism used depends on the structure of the alkyl halide and the reaction conditions. Particularly in this example, the reaction likely proceeds via an SN2 mechanism, where the displacement takes place in a single, concerted step, because the substrate is a primary halide which favors the SN2 pathway over the SN1.

Key Features of SN2 Reactions

• Backside attack by the nucleophile on the carbon bearing the leaving group.
• Bimolecular, with the rate dependent on the concentration of both the alkyl halide and the nucleophile.
• Inversion of configuration at the carbon center where substitution occurs, making it a valuable reaction in synthesizing enantiomerically pure compounds.

Reduction Reaction

Reduction reactions are fundamental in organic chemistry and involve the gain of electrons by a molecule, often by the addition of hydrogen. In our pathway to synthesize 2-butyne, the reduction step involves the conversion of the azide group in 2-azidobutane to an amine group to form 2-aminobutane.

This transformation is carried out using lithium aluminum hydride (LiAlH₄), a powerful reducing agent that can convert various functional groups, like azides, to amines. Reduction reactions like this are crucial to manipulate functional groups and build molecular complexity in synthesis.

Role of LiAlH₄ in Reduction

LiAlH₄ is particularly useful due to its ability to donate a hydride ion (H^-) and thus act as a source of electrons for the reduction. Such reductions are sensitive to the surrounding environment and usually require anhydrous (water-free) conditions to prevent reaction with water, which can decompose the reagent.

Dehydrohalogenation

The final steps in synthesizing 2-butyne involve dehydrohalogenation, which is the removal of a hydrogen and a halogen (a hydrogen halide) from an alkyl halide to form an alkene or alkyne. In this case, 2-chlorobutane is treated with a strong base, potassium tert-butoxide, to form first 1-butyne and eventually 2-butyne after subsequent treatment.

Dehydrohalogenation is an example of an elimination reaction, as opposed to the substitution reactions discussed earlier. A strong base is used to abstract a proton adjacent to the carbon with the halogen, leading to the formation of a double or triple bond in the carbon skeleton.

Aspects of Dehydrohalogenation

• Requires a strong base, such as potassium tert-butoxide.
• Produces alkenes or alkynes depending on the number of dehydrohalogenation steps.
• Can lead to isomerization, where the position of the double or triple bond might shift to form a more stable molecule.

Dehydrohalogenation is critical for controlling the location of double or triple bonds in organic molecules, which in turn determines their reactivity and properties.