November 21, 2024, 11:28:15 AM
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CH3CHBrCHBrCH3 + NaOH/C2H5OH + reflux-->?

CH2=CHCH=CH2
2 (66.7%)
CH3C--CCH3
1 (33.3%)

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Topic: Elimination for CH3CHBrCHBrCH3  (Read 6380 times)

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hongsk

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Elimination for CH3CHBrCHBrCH3
« on: April 17, 2006, 02:40:32 AM »
CH3CHBrCHBrCH3 + NaOH/C2H5OH + reflux-->?

CH2=CHCH=CH2 or CH3C--CCH3 as the major product?

'--' =triple bond

Which is more stable?

Thank you very much.
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Offline FeLiXe

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Re: Elimination for CH3CHBrCHBrCH3
« Reply #1 on: April 17, 2006, 05:47:37 PM »
I am pretty sure the conjugated diene is more stable.

In my book it says you get the alkyne with a strong base. I don't think NaOH goes as a strong base.
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hongsk

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Re: Elimination for CH3CHBrCHBrCH3
« Reply #2 on: April 18, 2006, 04:33:06 AM »
Quote from: FeLiXe on April 17, 2006, 05:47:37 PM
I am pretty sure the conjugated diene is more stable.

In my book it says you get the alkyne with a strong base. I don't think NaOH goes as a strong base.

but in the view of mechanism, CH3CHBrCHBr2CH3 will first change to CH3CBr=CHCH3 as the major product, so it seems that             CH3C--CCH3 will be the final major product
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