budullewraagh
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Registered: 1-8-2004
Location: new york
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synthesis of meta chloro propylbenzene
this week's exam the last question asked for a synthesis of meta chloro propylbenzene.
in notes, halides were considered to be weak as electron donating and electron withdrawing groups, but basically mostly neutral between the two.
my proposed synthesis was the chlorination of benzene with AlCl3 as a catalyst, then meta addition of propane by using propyl chloride and AlCl3.
the answer key says to add propyl chloride with AlCl3 and then to add chlorine, reduce the C=O with NaBH4 (i would have just done a wolff-kishner)
then eliminate with H2SO4 and hydrogenate with H2 on a Pd catalyst. other routes would involve using TsCl on the -OH and eliminating with a strong
base, then hydrogenating with H2 on a Pd catalyst.
i recognize how the "correct" answer would yield more chlorine in the meta position, but even if each of those 5 steps have a 75% yield, the
overall yield is only 23.7%. i believe that competition between ortho/para and meta addition of propyl chloride would result in at least 25% meta
product, especially considering the fact that Cl is the second most electronegative halogen.
does anyone agree?
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docberto
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Registered: 29-5-2005
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First of all, there is a difference between propyl chloride and propionyl chloride. Friedel-Crafts alkylation with propyl chloride results in a
isopropyl group. The secondary carbocation is much more stable than the primary.
Halogens on an aromatic ring are ortho/para directors. I don't think meta will happen much in EAS.
Third, eliminating the alkyl leaving group with a strong base would likely substitute in for the halo group on the aromatic.
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