Question

Using your roadmap as a guide, show how to convert 2 -methylbutane into racemic 3 -bromo-2-methyl-2-butanol. Show all reagents and all molecules synthesized along the way.

Short answer

Question: Show the conversion of 2-methylbutane into racemic 3-bromo-2-methyl-2-butanol. Answer: The conversion of 2-methylbutane into racemic 3-bromo-2-methyl-2-butanol involves the following steps: 1. Acid-catalyzed dehydration with H2SO4 to form 2-methyl-2-butene. 2. Halogenation with Br2 to form 3-bromo-2-methyl-2-butene. 3. Hydroboration-oxidation using BH3 and H2O2/NaOH to yield the racemic mixture of 3-bromo-2-methyl-2-butanol.

Step-by-step solution

Formation of 2-methyl-2-butene

Start with the given compound, 2-methylbutane. First, we need to convert this into an alkene. To do this, we can use a strong acid like H2SO4 to carry out an acid-catalyzed dehydration reaction: 2-methylbutane + H2SO4 -> 2-methyl-2-butene + H2O This reaction removes a water molecule from the starting material, forming the alkene 2-methyl-2-butene.

Halogenation to form 3-bromo-2-methyl-2-butene

Next, we need to introduce a bromine atom to the alkene to form 3-bromo-2-methyl-2-butene. To do this, we will use a halogenation reaction, which involves adding bromine across the double bond using a reagent like Br2: 2-methyl-2-butene + Br2 -> 3-bromo-2-methyl-2-butene This reaction forms the product 3-bromo-2-methyl-2-butene, where the Br atom has been added to the 3rd carbon of the chain.

Formation of the racemic mixture of 3-bromo-2-methyl-2-butanol

Finally, we need to convert the 3-bromo-2-methyl-2-butene into the racemic 3-bromo-2-methyl-2-butanol. To achieve this, we perform hydroboration-oxidation using reagents like BH3 and H2O2 in the presence of a base, such as NaOH: 3-bromo-2-methyl-2-butene + BH3 -> intermediate alkylborane intermediate alkylborane + H2O2 / NaOH -> racemic 3-bromo-2-methyl-2-butanol This two-step process first forms an intermediate alkylborane by adding BH3 to the alkene. The second step involves oxidation with hydrogen peroxide (H2O2) and a base like NaOH to yield the racemic mixture of 3-bromo-2-methyl-2-butanol. The racemic mixture indicates that equal amounts of both enantiomers are produced.

Key concepts

Acid-Catalyzed Dehydration

Acid-catalyzed dehydration is a chemical reaction where an alcohol loses a water molecule, forming an alkene in the presence of an acid catalyst like sulfuric acid (H2SO4). This type of reaction is often used in organic chemistry to create carbon-carbon double bonds.

In the context of converting 2-methylbutane to an alkene, the dehydration involves the removal of a hydrogen atom and a hydroxyl group (OH) from adjacent carbon atoms within the alcohol molecule. This process typically proceeds through a carbocation intermediate and can be influenced by factors such as the structure of the alcohol and the strength of the acid.

The mechanism of this reaction includes the following key steps:

• Protonation of the alcohol to form an alkyloxonium ion.
• Loss of a water molecule from the alkyloxonium ion to form a carbocation.
• Elimination of a hydrogen atom (deprotonation) from the carbocation to yield the alkene.

Halogenation Reaction

Halogenation is a reaction where a halogen molecule, such as bromine (Br2), is added to an unsaturated compound like an alkene. This reaction is particularly important for introducing halogen atoms into an organic molecule.

During the halogenation of 2-methyl-2-butene, a bromine molecule is added across the carbon-carbon double bond, resulting in the formation of 3-bromo-2-methyl-2-butene. This type of reaction proceeds without rearrangement, meaning that the halogen atom(s) will add to the most stable carbon cations formed during the reaction.

Key steps in the halogenation mechanism include:

• The pi electrons of the double bond attack a bromine molecule, leading to the formation of a cyclic bromonium ion intermediate.
• Nucleophilic attack on the bromonium ion by a bromide ion (Br-) results in the formation of the vicinal dibromide.

Hydroboration-Oxidation

Hydroboration-oxidation is a two-step organic reaction process that is utilized to convert an alkene into an alcohol. Firstly, hydroboration involves the addition of boron and hydrogen atoms across the carbon-carbon double bond. Secondly, oxidation replaces the boron with a hydroxyl group (OH).

In the synthesis of racemic 3-bromo-2-methyl-2-butanol, the alkene undergoes hydroboration to form an alkylborane intermediate. This intermediate is then oxidatively treated to form an alcohol. Notably, hydroboration adds atoms to the less substituted carbon because of its anti-Markovnikov selectivity.

Key aspects of the reaction include:

• The alkene reacts with diborane (BH3) to form an alkylborane intermediate.
• An oxidation step follows, where the alkylborane reacts with hydrogen peroxide (H2O2) and sodium hydroxide (NaOH), resulting in the alcohol.

Racemic Mixture

A racemic mixture refers to a fifty-fifty mix of two enantiomers of a chiral molecule. Enantiomers are mirror images of each other but cannot be superimposed onto one another, similar to a person's left and right hands.

Enantiomers have the same physical and chemical properties except for their behavior towards plane-polarized light and reactions in a chiral environment. One enantiomer rotates plane-polarized light to the right (dextrorotatory), and the other to the left (levorotatory).

In the synthesis of 3-bromo-2-methyl-2-butanol, the hydroboration-oxidation step results in a racemic mixture because the reaction does not distinguish between the formation of the two enantiomers. This leads to the equal production of both forms, which is critical for substances that have chiral centers and need to be used in environments such as drug synthesis where enantiomeric purity can dictate the therapeutic effect.