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Jasmone 1 - ChemistforChrist

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Synthesis of JASMONE<br />

O<br />

O<br />

DIBAL-H<br />

For a lactone the best<br />

reducing agent to a lactole<br />

is DIBAL-H, for an ester this<br />

might not work so well!<br />

H<br />

OH<br />

O<br />

Thermodynamically more<br />

stable due to being<br />

a trisubstituted double<br />

bond.<br />

O<br />

under the acidc<br />

conditions:<br />

isomerization<br />

MeLi<br />

Me<br />

<strong>Jasmone</strong><br />

O<br />

O<br />

OH<br />

More resources available at<br />

www.chemistforchrist.de<br />

OH<br />

O<br />

CrO 3, verd. H 2SO 4, Aceton<br />

Jones-Reagent<br />

Me CrO3, verd. H2SO4, Aceton<br />

OH<br />

Jones-Reagent<br />

CrO 3<br />

HO<br />

tertiary alcohol,<br />

H 2O is eliminated<br />

Wittig<br />

tertiary alcohol cannot be oxidized!<br />

The textbook Clayden, Greeves, Warren and Wothers (Oxford University press, p. 951, 2004 reprint) proposes<br />

a different mechanism:<br />

OH (VI)<br />

OH<br />

O<br />

R Li<br />

R<br />

O<br />

CrO3 R (VI)<br />

CrO<br />

orange<br />

3<br />

OH<br />

R CrO 3<br />

H<br />

O OH<br />

Cr<br />

O O<br />

R<br />

H<br />

[3,3]<br />

HO OH<br />

Cr<br />

O O<br />

H<br />

H R<br />

Me<br />

O<br />

(III)<br />

CrO3 R<br />

+ Cr(IV)<br />

green<br />

PPh 3<br />

OH<br />

Me<br />

Me<br />

gives Cr(III) and Cr(VI)<br />

by disproportionation.<br />

The first step is the formation of a chromate ester but this intermediate has no proton to lose, so it transfers<br />

the chromate to the other end of the allylic system where there is a proton. The chromate transfer can be<br />

drawn as a [3,3] sigmatropic rearrangement.<br />

Very general method:<br />

R' MgBr<br />

O<br />

R<br />

HO<br />

R'<br />

R'<br />

R H<br />

R<br />

OH<br />

Ox<br />

R'<br />

R<br />

O

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