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Topic: Oxidation of alcohol and Elimination  (Read 9982 times)

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Offline Dolphinsiu

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Oxidation of alcohol and Elimination
« on: April 07, 2007, 11:39:52 AM »
In my organic chem quiz,

I leave blank on these two questions during quiz as I really don't know how to do.

1.Why tertiary alcohol cannot be oxidized by potassium dichromate?(10%)
2.Why anti-coplanar configuration of halogen alkane favours elimination?(10%)

However, I still want to know the actual reasons. Do all of you know how to think out of this question? Thank you!

Offline joemok

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Re: Oxidation of alcohol and Elimination
« Reply #1 on: April 07, 2007, 01:26:33 PM »
My language may not be good enough, but i have some ideas about question 1.

Alcohol can be oxidized to ketone, or aldehyde which is further oxidized to carboxylic acid
 ----- in these process, the C--OH becomes C =O
 ----- for tertiary alcohol, the carbon that the -OH group bonded to hass no H connected so elimination of the H does not takes place.
 ----- generally, the alkyl group will not be eliminated. In which case C would have 5 bonds that is not possible.

Offline Yggdrasil

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Re: Oxidation of alcohol and Elimination
« Reply #2 on: April 07, 2007, 01:46:53 PM »
2.Why anti-coplanar configuration of halogen alkane favours elimination?(10%)

In an E2 reaction, think of the transition state.  For the hydrogen and halogen that leave, both leave behind two sp3 orbitals which become p-orbitals.  What geometry must these p-orbitals have in order for them to form a double bond?  Of the two geometries that satisfy these constraints, which is more sterically favorable?

Offline Ψ×Ψ

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Re: Oxidation of alcohol and Elimination
« Reply #3 on: April 07, 2007, 05:56:31 PM »
1.Why tertiary alcohol cannot be oxidized by potassium dichromate?(10%)

If the alcohol is tertiary, there are three carbons attached to the alcohol carbon.  What are you going to do, rip them off?  They aren't going anywhere.  (If you oxidize to the ketone, you have a double bond from C to O and then three more single bonds from C to other C's.  Five bonds is not a happy number for carbon.)

Offline Dolphinsiu

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Re: Oxidation of alcohol and Elimination
« Reply #4 on: April 08, 2007, 03:59:03 AM »
Thank you!

I only don't know question 2.

For staggered geometry, anti-coplanar is more sterically favorable. Why?
« Last Edit: April 08, 2007, 04:47:22 AM by Dolphinsiu »

Offline Yggdrasil

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Re: Oxidation of alcohol and Elimination
« Reply #5 on: April 08, 2007, 02:27:15 PM »
In what two situations could you get coplanar hydrogens?

Offline Dolphinsiu

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Re: Oxidation of alcohol and Elimination
« Reply #6 on: April 09, 2007, 05:18:59 AM »
I only think of Eclipsed configuration! Why have two situations?

Offline english

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Re: Oxidation of alcohol and Elimination
« Reply #7 on: April 09, 2007, 06:26:56 AM »
Try not to memorize why 3° alcohols cannot undergo oxidation with Na2Cr2O7 (or K2Cr2O7).

Draw mechanisms!

Offline Dolphinsiu

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Re: Oxidation of alcohol and Elimination
« Reply #8 on: April 09, 2007, 06:35:57 AM »
The mechanism? how to draw really?

Offline english

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Re: Oxidation of alcohol and Elimination
« Reply #9 on: April 09, 2007, 02:25:33 PM »
The mechanism? how to draw really?

Chromic acid is produced from dichromate in strong acid. 

Chromic acid acts as a "docking site" for your molecule of alcohol to latch onto, via an SN2 reaction. 

Go from there.  I can tell you that there will be an elimination-like step somewhere.  Try to intuitively go about moving things around, and figure out how to get that double bond, so that the O-Cr bond breaks, and your =O forms.


Offline Dolphinsiu

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Re: Oxidation of alcohol and Elimination
« Reply #10 on: April 27, 2007, 01:23:33 PM »
Do anyone know how to do Q.2?

My Org Chem Quiz get 0 marks on this question when I answer it ''it is due to less steric effect''.

Why I cannot get any marks?

Do anyone have idea on answering Q.2? Thank you!


Offline english

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Re: Oxidation of alcohol and Elimination
« Reply #11 on: April 27, 2007, 02:01:04 PM »
Do anyone know how to do Q.2?

My Org Chem Quiz get 0 marks on this question when I answer it ''it is due to less steric effect''.

Why I cannot get any marks?

Do anyone have idea on answering Q.2? Thank you!



I'm assuming you're referring to anti-periplanar arrangement?  Different usage...

Anyway, you must have this arrangement simply because of electron-electron repulsions that can occur between your base and your halide ion if they were on the same side of the C—C bond. 

There is also an MO explanation as well.


I'm assuming you got marked wrong because you were not specific.  Simply saying "sterics" doesn't clarify the situation.

Offline Dolphinsiu

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Re: Oxidation of alcohol and Elimination
« Reply #12 on: April 30, 2007, 05:22:31 AM »
Do you know what is high order elimination? Which one belongs to? E1(Unimolecular elimination), E2(bimolecular elimination) or Eco1(Conjugated base elimination)?

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