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Assume you have a bottle of enantiomerically pure \((R)-2\)-butanol and any other reagents you might need. Show the reactions you would use to make enantiomerically pure \((S)-2\)-iodobutane. Show the reactions you would use to make enantiomerically pure \((R)-2\)-iodobutane.

Short Answer

Expert verified
Oxidize (R)-2-butanol to 2-butanone. Reduce with (S)-CBS to get (S)-2-butanol or (R)-CBS to get (R)-2-butanol. Substitute with HI to get (S)-2-iodobutane or (R)-2-iodobutane.

Step by step solution

01

Understand the Starting Material

The starting material is (R)-2-butanol, an enantiomerically pure alcohol.
02

Convert (R)-2-butanol to 2-butanone

Oxidize (R)-2-butanol to 2-butanone (butan-2-one) using an oxidizing agent like PCC (Pyridinium chlorochromate).\[ (R)-2-butanol \rightarrow \text{2-butanone} \text{(PCC)} \text{ oxidizing agent.} \]
03

Reduce 2-Butanone to (S)-2-butanol or (R)-2-butanol

To obtain (S)-2-butanol, use a chiral reducing agent such as (S)-CBS catalyst. For (R)-2-butanol, use (R)-CBS catalyst.\[ \text{2-butanone} \rightarrow (S)-2-butanol \text{ (S)-CBS catalyst} \]\[ \text{or} \]\[ \text{2-butanone} \rightarrow (R)-2-butanol \text{ (R)-CBS catalyst} \]
04

Convert (R)-2-butanol or (S)-2-butanol to (R)-2-iodobutane or (S)-2-iodobutane

To convert the alcohols to iodobutane, perform a substitution reaction with HI (Hydroiodic acid). For (S)-2-iodobutane, use (S)-2-butanol and for (R)-2-iodobutane, use (R)-2-butanol.\[(S)-2-butanol + HI \rightarrow (S)-2-iodobutane \]\[(R)-2-butanol + HI \rightarrow (R)-2-iodobutane\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chirality
Chirality is a property of a molecule that makes it non-superimposable on its mirror image, just like your left and right hands. Molecules with chirality often contain a carbon atom attached to four different groups, known as a chiral center. For example, in (R)-2-butanol, the carbon atom connected to the hydroxyl (OH) group is a chiral center because it is bonded to four different groups. Chiral molecules are important in chemistry, particularly in pharmaceuticals, where one enantiomer (one of the two mirror image forms) can be therapeutically active, while the other might be inactive or harmful. This makes understanding and controlling chirality crucial in synthetic chemistry.
Oxidation
Oxidation involves the loss of electrons or increase in oxidation state by a molecule, atom, or ion. In organic chemistry, oxidation often refers to the addition of oxygen or the removal of hydrogen. For example, in the exercise, (R)-2-butanol is oxidized to 2-butanone (butan-2-one) using an oxidizing agent such as pyridinium chlorochromate (PCC). The reaction removes two hydrogen atoms from the alcohol group and converts it into a ketone. Oxidizing agents like PCC are selected based on their ability to selectively oxidize primary alcohols to aldehydes and secondary alcohols to ketones without further oxidation to carboxylic acids.
Reduction
Reduction is the process of gaining electrons or a decrease in oxidation state by a molecule, atom, or ion. In organic chemistry, reduction typically involves adding hydrogen or removing oxygen. In synthesizing chiral compounds, a chiral reducing agent can be used to ensure that the reduction produces a specific enantiomer. For example, in the provided problem, 2-butanone is reduced to either (S)-2-butanol or (R)-2-butanol using chiral reducing agents like (S)-CBS or (R)-CBS catalysts. The choice of catalyst determines the stereochemistry of the resulting alcohol, ensuring the production of one enantiomer over the other.
Substitution Reaction
A substitution reaction involves replacing one functional group in a molecule with another. In the context of this problem, (R)-2-butanol or (S)-2-butanol undergoes a substitution reaction with hydroiodic acid (HI) to replace the hydroxyl (OH) group with an iodine (I) atom, forming (R)-2-iodobutane or (S)-2-iodobutane. This type of reaction is known as a nucleophilic substitution, where the iodide ion (I-) from HI acts as a nucleophile, attacking the electrophilic carbon that is bonded to the OH group, resulting in the formation of iodobutane.
Chiral Catalysts
Chiral catalysts are used in asymmetric synthesis to favor the formation of a particular enantiomer. These catalysts have a chiral component that influences the stereochemistry of the reaction, ensuring the production of one enantiomer over the other. In the given exercise, chiral catalysts like (S)-CBS and (R)-CBS help to reduce 2-butanone selectively to (S)-2-butanol and (R)-2-butanol, respectively. This selective reduction is crucial in synthesizing enantiomerically pure compounds, which are often necessary in pharmaceuticals and other applications where the specific 3D arrangement of the atoms can affect the compound's activity and efficacy.

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Most popular questions from this chapter

Use the following \(\mathrm{p} K_{\mathrm{a}}\) data to decide whether sodium amide would be effective at producing vinyl anions \(\left(\mathrm{R}_{2} \mathrm{C}=\overline{\mathrm{C}} \mathrm{H}\right) \cdot \mathrm{p} K_{\mathrm{a}}\) : ammonia, 38 ; vinyl hydrogen, \(\sim 46\). Write the acid-base reaction for this process.

Predict the product, if any, of the following reactions: (a) \(\mathrm{HO}^{-}+\mathrm{CH}_{3} \mathrm{Br}\) (b) \(\mathrm{HO}^{-}+\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{H}_{2} \mathrm{O}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) (d) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{Br} \rightarrow\)

\(7.11\) Explain the following change in rate for the \(\mathrm{S}_{\mathrm{N}} 2\) reaction: $$ \mathrm{R}-\mathrm{O}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{I} \rightarrow \mathrm{R}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{I}^{-} $$ Rate for \(\mathrm{R}=\mathrm{CH}_{3}\) is much faster than that for \(\mathrm{R}=\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}\)

The calculated heat of formation for \((E)-\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}_{2}^{+}\) is \(227 \mathrm{kcal} / \mathrm{mol}(949 \mathrm{~kJ} / \mathrm{mol})\) and for \((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{2}^{+}\)is \(221 \mathrm{kcal} / \mathrm{mol}(926 \mathrm{~kJ} / \mathrm{mol})\). Draw the two cations showing the three-dimensional perspective and explain why one cation is more stable than the other.

The \(\mathrm{S}_{\mathrm{N}} 2\) reaction that occurs between an alkyl halide and a nucleophile is a very important process in organic chemistry. For this reason, it is necessary to recognize the nucleophilicity of various reagents. In the following pairs of compounds, indicate the more nucleophilic reagent. Explain your reasoning. (a) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N} \quad\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{P}\) (b) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N} \quad\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{~N}^{-}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SNa} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OK}\) There is much detail in this chapter, but that is not usually a problem for most students. Good nucleophiles can be distinguished from good bases, good leaving groups can be identified, and so on. What is hard is to learn to adjust reaction conditions (reagents, temperature, and solvent) so as to achieve the desired selectivity to favor the product of just one of these four reactions. That's hard in practice as well as in answering a question on an exam or problem set. To a certain extent, we all have to learn to live with this problem!

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