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Topic: Williamson Ether Synthesis  (Read 1581 times)

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Offline neobenzene

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Williamson Ether Synthesis
« on: February 08, 2013, 11:59:25 PM »
Hi...I've been recently taught that in the Williamson ether synthesis, the halide to be reacted, should be a primary halide. This is because if the halide is secondary or tertiary, due to greater sterric crowding, elimination dominates over substitution, and an alkene is formed.

Now, what if the halide I'm using is benzyl chloride....benzene ring with a -CH2Cl attached to it. I'm  not sure how to draw it using the SMILES engine.

Anyways, my question is, if we use benzyl chloride, and any sodium alkoxide (RONa), what will the product be? Will it be an ether, since the halide is primary, or will it be an alkene, since the halide has a bulky benzene ring?

Thanks!

Offline discodermolide

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Re: Williamson Ether Synthesis
« Reply #1 on: February 09, 2013, 02:10:28 AM »
Benzyl chloride cannot form an alkene. So you will form the ether.
ClCC1=CC=CC=C1
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Offline neobenzene

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Re: Williamson Ether Synthesis
« Reply #2 on: February 09, 2013, 03:22:45 AM »
Why can't it form an alkene?

Offline discodermolide

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Re: Williamson Ether Synthesis
« Reply #3 on: February 09, 2013, 03:36:23 AM »
Because there is no proton to initiate the elimination.
You cannot break the aromaticity.
You will not get
C=C1=CC=CC=C1
or
C=C1C=CC=CC1
or
C=C1C=CCC=C1

The first molecule does not make any sense.
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Offline neobenzene

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Re: Williamson Ether Synthesis
« Reply #4 on: February 09, 2013, 05:20:43 AM »
Thanks!!

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