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Topic: why is the optical activity not retained here?  (Read 5360 times)

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Offline a confused chiral girl

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why is the optical activity not retained here?
« on: December 16, 2006, 02:46:14 AM »
I read that this has something to do with alpha/beta Carbons, but these 2 structures are not carbohydrates, and only carbohydrates can have carbons assigned to alpha/beta when they are cyclic. Can someone please explain this to me? The practice question asks by using equations or chemical formulas or words to explain why is the result like that, and I dont have answer key to check.
thank u

ps. why would you use alpha/carbon to explain when there are no anomeric carbons anyway? because anomeric = one C attached to 2 O ..and here, the C is just attached to a double bonded O, which is not 2 O technically.... ???

Offline Yggdrasil

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Re: why is the optical activity not retained here?
« Reply #1 on: December 16, 2006, 03:32:16 AM »
In organic chemistry, alpha, beta, gamma, etc. are usually used to refer to the position of functional groups relative to other functional groups.  Groups which are adjacent to other functional groups are called alpha.  Groups attached to the next carbon over (2 carbons away from the functional group) are called beta, etc.  For example:

In alpha amino acids like glycine (H2N-CH2-COOH), the amino group is on the alpha carbon, the carbon directly adjacent to the carboxylic acid).

In beta-mercaptoethanol (a.k.a. 2-mercaptoethanol, HS-CH2-CH2-OH), the thiol is in the beta position relative to the OH (i.e. there are two carbons between the thiol and the alcohol).

So, given these definitions of alpha and beta, what is special about the alpha hydrogens of a ketone?

Offline a confused chiral girl

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Re: why is the optical activity not retained here?
« Reply #2 on: December 16, 2006, 07:46:46 AM »
is it the first structure, the alpha C is affected, and in the second structure, the alpha Carbon is not affected so optical activity is retained?
and then eventually, racemization will occur.

Please correct me if I am wrong  :-\

Offline Dan

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Re: why is the optical activity not retained here?
« Reply #3 on: December 16, 2006, 07:52:55 AM »
Think about why the alpha position is affected, but the beta position is not.
My research: Google Scholar and Researchgate

Offline Dolphinsiu

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Re: why is the optical activity not retained here?
« Reply #4 on: December 16, 2006, 10:13:09 AM »
Is it related to its intramolecular force(hydrogen bond)?

Offline a confused chiral girl

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Re: why is the optical activity not retained here?
« Reply #5 on: December 16, 2006, 01:36:16 PM »
o, is it because the alpha Carbon is more acidic than the beta carbon, so the beta carbon in the second structure is not affected??

but for optical activity ...you have to look at the acidity?
can someone please explain? thank u in advance

Offline Dan

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Re: why is the optical activity not retained here?
« Reply #6 on: December 16, 2006, 01:40:48 PM »
o, is it because the alpha Carbon is more acidic than the beta carbon, so the beta carbon in the second structure is not affected??

yes

Quote
but for optical activity ...you have to look at the acidity?
can someone please explain? thank u in advance

In this case, the first step of the mechanism for racemisation is deprotonation, so in this case, the acidity will affect racemisation and therefore opical activity.
Can you draw the mechanism for the racemisation of the first compound?
My research: Google Scholar and Researchgate

Offline kryptoniitti

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Re: why is the optical activity not retained here?
« Reply #7 on: December 16, 2006, 02:15:24 PM »
Hint: ketone with base and you get
an enolate in an equilibrium reaction.
Now draw the resonance structures
of the enolates for both molecules.
You see something funny happen to
the other "wedged" methyl group if
you think in terms of equilibrium and
chirality.

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