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Chapter 11: Q35P (page 578)

A student wanted to use the Williamson ether synthesis to make (R)-2-ethoxybutane. He remembered that the Williamson synthesis involves an SN2 displacement, which takes place with inversion of configuration. He ordered a bottle of (S)-butan-2-ol for his chiral starting material. He also remembered that the SN2goes best on primary halides and tosylates, so he made ethyl tosylate and sodium (S)-but-2-oxide. After warming these reagents together, he obtained an excellent yield of 2-ethoxybutane.

(a) What enantiomer of 2-ethoxybutane did he obtain? Explain how this enantiomer results from the SN2 reaction of ethyl tosylate with sodium (S)-but-2-oxide.

(b) What would have been the best synthesis of (R)-2-ethoxybutane?

(c) How can this student convert the rest of his bottle of (S)-butan-2-ol to (R)-2-ethoxybutane?

Short Answer

Expert verified

While using Williamson ether synthesis to make R-O-R , choose the less hindered alkyl group to serve as the alkyl halide (R'-X), because it will make a better SN2 substrate. Choose the more hindered alkyl group to form the alkoxide (R-O-), because it is less sensitive to steric hindrance in the reaction.





Step by step solution

01

Williamson ether synthesis

While using Williamson ether synthesis to make R-O-R , choose the less hindered alkyl group to serve as the alkyl halide (R'-X), because it will make a better SN2 substrate. Choose the more hindered alkyl group to form the alkoxide (R-O-), because it is less sensitive to steric hindrance in the reaction.

02

Reaction and explanation

(a) The enantiomer of (S)-2-ethoxybutane is,

Enantiomer of (S)-2-ethoxy butane

When Sodium (S)-but-2-oxide reacted with ethyl tosylate forms (S)-2-ethoxybutane is formed.

Formation of (S)-2-ethoxybutane

(b) First (R)-butan-2-ol is taken and make an anion, then react with ethyl tosylate form (R)-2-ethoxy butane.

(c) (S)-butan-2-ol is treated with /pyridine form (S)-2-butyl tosylate then it is treated with at low temperature to form (R)-2-ethoxybutane.

Formation of (R)-2-ethoxybutane

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Most popular questions from this chapter

Under normal circumstances, tertiary alcohols are not oxidized. However, when the tertiary alcohol is allylic, it can undergo a migration of the double bond (called an allylic shift) and subsequent oxidation of the alcohol. A particularly effective reagent for this reaction is Bobbitt鈥檚 reagent, similar to TEMPO used in many oxidations. (M. Shibuya et al., J. Org. Chem., 2008, 73, 4750.)

Show the expected product when each of these 3掳 allylic alcohols is oxidized by Bobbitt鈥檚 reagent.

(a)

(b)

(c)

(d)

Predict the products of the reactions of the following compounds with:

(1) chromic acid or excess sodium hypochlorite with acetic acid.

(2)PCC or NaOCl (1 equivalent) with TEMPO.

(a)cyclohexanol (b)1-methylcyclohexanol

(c)cyclopentylmethanol (d)cyclohexanone

(e)cyclohexane (f)1-phenylpropan-1-ol

(g)hexan-1-ol (h)acetaldehyde, CH3CHO

Both cis- and trans-2-methylcyclohexanol undergo dehydration in warm sulfuric acid to give 1-methylcyclohexene as the major alkene product. These alcohols can also be converted to alkenes by tosylation usingand pyridine, followed by elimination using KOC(CH3)3as a strong base. Under these basic conditions, the tosylate of cis-2-methylcyclohexanol eliminates to give mostly 1-methylcyclohexene, but the tosylate of trans-2-methylcyclohexanol eliminates to give only 3-methylcyclohexene. Explain how this stereochemical difference in reactants controls a regiochemical difference in the products of the basic elimination, but not in the acid-catalyzed elimination.

Predict the products of the following reactions.

(a) cyclohexylmethanol + TsCl / pyridine (b) product of (a) + LiAlH4

(c) 1- methylcyclohexanol + H2SO4, heat (d) product of (c) + H2, Pt

Neopentyl alcohol, (CH3)3CCH2OH, reacts with concentrated HBr to give 2-bromo-2-methylbutane, a rearranged product. Propose a mechanism for the formation of this product.
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