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: Convert 2-methylbutane into racemic 3-bromo-2-methyl-2-butanol and show all reagents and molecules synthesized along the way. Answer: 1. Perform dehydrohalogenation of 2-methylbutane to form 2-methyl-2-butene using HCl, KOH, and ethanol. 2. Brominate 2-methyl-2-butene to form 2,3-dibromo-2-methylbutane with Br2. 3. Carry out a substitution reaction using water and NaOH to replace one bromine atom in 2,3-dibromo-2-methylbutane, forming racemic 3-bromo-2-methyl-2-butanol and NaBr.

Step-by-step solution

Dehydrohalogenation of 2-methylbutane to form 2-methyl-2-butene

To accomplish this, we first react 2-methylbutane with HCl to form 2-chloro-2-methylbutane, followed by dehydrohalogenation using a strong base. Reagents: HCl, KOH, and ethanol Reaction steps: 1. 2-methylbutane + HCl → 2-chloro-2-methylbutane 2. 2-chloro-2-methylbutane + KOH (in ethanol) → 2-methyl-2-butene + KCl

Bromination of 2-methyl-2-butene to form 2,3-dibromo-2-methylbutane

The next step is bromination of the double bond in 2-methyl-2-butene. We will use Br2 as a reagent for this purpose. Reagent: Br2 Reaction step: 1. 2-methyl-2-butene + Br2 → 2,3-dibromo-2-methylbutane

Substitution reaction to form racemic 3-bromo-2-methyl-2-butanol

In the final step, we perform a substitution reaction using water to replace one of the bromine atoms on the 2,3-dibromo-2-methylbutane. This will result in the formation of racemic 3-bromo-2-methyl-2-butanol. Reagents: H2O, NaOH Reaction steps: 1. 2,3-dibromo-2-methylbutane + H2O (in the presence of NaOH) → racemic 3-bromo-2-methyl-2-butanol + NaBr Now, we have successfully converted 2-methylbutane into racemic 3-bromo-2-methyl-2-butanol and showed all reagents and molecules synthesized along the way.

Key concepts

Dehydrohalogenation

Dehydrohalogenation is a crucial organic chemistry reaction that involves the removal of a hydrogen halide from a molecule, leading to the formation of an alkene. This reaction typically involves two key steps:

• Addition of a halogen (such as chlorine or bromine) to an alkane to form a haloalkane.
• Use of a strong base, like potassium hydroxide (KOH), to remove a hydrogen atom and a halogen atom, resulting in the formation of a double bond.

In the case of 2-methylbutane, the journey begins with the formation of 2-chloro-2-methylbutane when it reacts with hydrochloric acid (HCl).
This prepares the compound for the dehydrohalogenation step, where KOH in ethanol facilitates the removal of HCl, transforming the molecule into 2-methyl-2-butene, an alkene.
This process is beneficial for converting saturated compounds into more reactive unsaturated molecules that can undergo further chemical reactions like bromination.

Bromination

Bromination is a type of halogenation specifically involving the addition of bromine to a molecule. It is particularly valuable in organic synthesis due to its ability to add bromine across double bonds, effectively saturating them.
In the context of transforming 2-methyl-2-butene, bromination plays a key role. When 2-methyl-2-butene is exposed to bromine ( Br_2 ), an addition reaction occurs.
This results in the formation of 2,3-dibromo-2-methylbutane.

• The electron-rich double bond in 2-methyl-2-butene makes it particularly susceptible to attack by the bromine molecules.
• The product is a dibromo compound, which can be precisely manipulated in subsequent chemical reactions such as substitution.

Bromination is a versatile reaction that is not only used to add bromine across double bonds but can also add bromine to other unsaturated compounds, aiding in the synthesis of more complex molecules.

Substitution Reaction

A substitution reaction involves replacing one atom or group in a molecule with another. These reactions are fundamental in organic synthesis. In the transformation from 2,3-dibromo-2-methylbutane to racemic 3-bromo-2-methyl-2-butanol, a substitution reaction is utilized.

• Using water (H2O) in the presence of a strong base like sodium hydroxide (NaOH), one of the bromine atoms on the dibromo compound is replaced by a hydroxyl group (OH).
• This results in the formation of an alcohol, namely 3-bromo-2-methyl-2-butanol.

The term 'racemic' indicates that the product contains an equal mixture of two enantiomers, which are mirror-image forms of the molecule.
This aspect is important in organic chemistry, especially when considering the physiological activity of different enantiomers in biological systems.
The substitution process is essential as it allows chemists to modify and create specific functional groups, leading to the development of desired compounds.