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The Wittig Reaction: Substitution of =O by =C

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<strong>The</strong> <strong>Wittig</strong> <strong>Reaction</strong>: <strong>Substitution</strong> <strong>of</strong> <strong>=O</strong> <strong>by</strong> <strong>=C</strong><br />

Ph<br />

Ph P CH 2<br />

Ph<br />

(carbon nucleophile)<br />

Ph<br />

Ph<br />

P CH 2<br />

Ph<br />

a phosphonium ylide<br />

Ph =<br />

Ph 3 P<br />

O<br />

[Ad N ]<br />

O<br />

a betaine<br />

This net substitution proceeds <strong>by</strong><br />

an addition-elimination sequence.<br />

CH 2<br />

+ Ph3 P CH 2<br />

+<br />

[A N ]<br />

This elimination<br />

involves breakdown <strong>of</strong><br />

the 4-membered ring<br />

O<br />

Ph 3 P<br />

an oxaphosphetane<br />

Ph 3 P<strong>=O</strong><br />

Phosphonium ylides (powerful nucleophiles)<br />

add to aldehyde or ketone carbonyl groups via<br />

normal [Ad N ]. This tetrahedral intermediate, a<br />

betaine, (BAY-tah-een) has no leaving group,<br />

yet it’s not stable. Because phosphorous has a<br />

high affinity for oxygen, the oxyanion attacks<br />

the positively charged P via [A N ] to make a 4-<br />

membered ring (called a phosphaoxetane).<br />

This strained phosphaoxetane reacts further to<br />

split into Ph 3 P<strong>=O</strong> and an alkene. This<br />

elimination is a little funky, but you should be<br />

able to draw a reasonable set <strong>of</strong> curved arrows<br />

(and identify the frontier orbitals).


Formation <strong>of</strong> Ylides<br />

Ph 3 P + CH 3 I<br />

triphenylphosphine<br />

[S N 2]<br />

H<br />

Ph 3 P C H<br />

H<br />

methyltriphenylphosphonium<br />

iodide<br />

[pt]<br />

I<br />

Ph 3 P CH 2 +<br />

+ LiI<br />

CH 3 CH 2 CH 2 CH 3<br />

butane<br />

CH 3 CH 2 CH 2 CH 2 Li<br />

butyllithium<br />

(strong base)<br />

http://www.chemtube3d.com/

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