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Topic: synthesis  (Read 3666 times)

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Offline kgothatso

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synthesis
« on: July 22, 2010, 09:11:06 AM »
i am having a problem understanding amd appling the rules of mechanism and synthesis theoretically

Offline OC pro

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Re: synthesis
« Reply #1 on: July 22, 2010, 12:09:52 PM »
And how do you think we can help? Have you tried to read through some textbooks yet?

Offline Jorriss

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Re: synthesis
« Reply #2 on: July 22, 2010, 08:10:49 PM »
This is a difficult question to answer, because even though as broad as your question was, it's a fair question.

Say, you had the question, for example,
'Starting from 1-hexene, 1-butyne, bromoethane, iodomethane and any reagent needed
(you do not need to use all of these compounds), synthesize

hexanal'

Where do you start? Do you start from your reactants? Do you look at the product? What initially goes through your mind?

Offline Wald_ron

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Re: synthesis
« Reply #3 on: July 22, 2010, 08:54:55 PM »
When I think about basic mechanisms I always think about electron movements and stability.

For example CH3-CH3>(CH3)2CH=CH(CH3)2
Stability is determined by Pka.

What Pka really means is if an ion is formed , will the ion be stable. If it is stable , it will be formed easily and thus have a low transition state. The less stable the conjugate base the more energy it will take to form.

A conjugate base of CH3-CH3 is CH(+)3-CH2(-) which will absolutely never ever form on it's own and would require ourageous amounts of heat and pressure to even form.

So CH3-CH3 + HBr--> no reaction
Why , CH3-CH3 will never ever allow an opening for HBr to get in

Now.

The conjugate base of (CH3)2CH=CH(CH3)2 has a higher probability of forming.
(CH3)2C(-)H -C(+)H(CH3)2

How do we know this using the Basic laws of mechanisms?
1) The CH3 groups on the double bonds stabilize carbocations (Carbons with a positive charge on them)  This effect increases with more electronegative atoms on the double bond going right across the periodic table. Thus O> N> C. 

So (CH3)2CH=CH(CH3)2  + HBr -->(CH3)2CH(+)CHBr(CH3)2 --> (CH3)2CH2CHBr(CH3)2

So lets pretend

(CH3)2CH=CH(CH3)2 + HI -->  what would happen OP?
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Offline orgopete

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Re: synthesis
« Reply #4 on: July 23, 2010, 02:14:43 PM »
I don't know if this question has a straight forward answer. I have found it surprising that some professors teach organic chemistry and especially emphasize "retrosynthetic analysis" with a minimum amount of mechanisms. In speaking to some of those students they use a table to identify functional group transformations and to match starting materials and products. For the original poster, I can understand confusion if you happen to be in a similar class.

Because this pedagogy emphasizes strategy over mechanism, it shouldn't be surprising that students might have difficulty in predicting the product of a reaction to which they are trying to use the reverse process. It is like trying to solve long division problems while being unable to multiply.

I had always been quite successful in solving retrosynthesis problems despite my lack of exposure to an educational emphasis upon them. My education was heavily mechanistic. I found this mechanistic approach has served me very very well in a long and productive industrial career. Consequently, in teaching, I tried to avoid asking students to use or suggest a retrosynthesis of any reaction they could not write the mechanism for. In my usage, that means that unless you can predict the products of a reaction, then I did not expect you to predict the starting materials for that same reaction. Although that is not exactly the question being asked, I do think what I have described does (or should) apply to the vast majority of those practicing organic synthesis today.

If you wished to be as dogmatic as I, you can take any bond and analyze it as the product of radical couplings, an electrocyclic reaction, or the two polar opposite ionic couplings. You can then plug and chug your way through all of the reagents that meet those requirements. Obviously, you need to know the mechanisms in order to know which reagents can be radicals, anions, cations, and their equivalents or precursors.
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