another ochem
Started by allstardentist
the answer is C. My guess is because of steric hindrance. Anyone have thoughts on this problem?
I also agree that the answer is C. 1st it is an aldehyde and 2nd there is no steric strain. Answer A and B are Carbo acids so they are there ruled out.... C,D and E are all primary alcohol however, D is attached to 2R sub. on the Carbon and 3R sub. on the Carbon for E. Hope this clears it up!allstardentist said:
jackbauer! said:
C is an aldehyde, D is a secondary alcohol, and E is a tertiary alcohol. You name alcohols based on the carbon substitution. So if it's an alcohol attached to a tertiary carbon, it will be a tertiary alcohol. Amines on the other hand are named based on their own attachments, not the carbon that they may be attached to. So a Nitrogen with two attached R groups will be a secondary amine, three attached R groups, tertiary amine.
http://en.wikipedia.org/wiki/Alcohol#Primary.2C_secondary.2C_and_tertiary_alcohols
http://en.wikipedia.org/wiki/Amine
FYI, B is a Ketone, not a carboxylic acid. Tertiary alcohols can be oxidized, but resist it.
Best of luck!
Here is a site of the order of organic oxidation
D is still the right answer🙂
http://www.csun.edu/~hcchm007/333oloc.pdf
D is still the right answer🙂
http://www.csun.edu/~hcchm007/333oloc.pdf
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The real answer is C not D according to kaplan.
I don't think you get it jack. All-star has the answers and it is C because of the steric hinderance exhibited by D, otherwise D would be your answer if there was only 1R groups connected instead of 2.
allstardentist said:
i just finished ochem last semester. i can tell you that the answer is primary alcholo. whenever you stuck in any porblem that ask about oxidiation ask yourself which one can make for O.it can be aldehyde b/c still have to oxidize to oxygen till become carboxylic acid to have the maxium o. the the answer is D.
RING12 said:
D is not a primary alcohol if you look at it carefully.
D is definitely not a primary, but a secondary.
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i'm taking ochem this summer and my professor said it was a bogus poorly worded question...because the "most easily oxidized" doesn't have meaning in Organic chemistry without crisp detailed assumptions regarding solvents, nucleophiles, electrophiles, environment, etc.
He said he could drum up experimental conditions that would make every answer the "easiest to oxidize".
I secondarily did a fair search of online material looking for "easiest to oxidize" etc. I found no clear cut statement which would of course support my professors position. So I wouldn't sweat it too much.
He said he could drum up experimental conditions that would make every answer the "easiest to oxidize".
I secondarily did a fair search of online material looking for "easiest to oxidize" etc. I found no clear cut statement which would of course support my professors position. So I wouldn't sweat it too much.
First of all, not every question has a "clear cut" answer however there is a "major" principle reflected in the answer. I believe the question is eluding to the oxidation levels of the functional groups. The reasoning for my answer is the fact that the aldehyde, when oxidized will be a carboxylic acid
(Note, oxidation is defined as a reaction that introduces a more electronegative element to or removes a molecular H from the system.)
On the other hand, the secondary alcohol in D can only be oxidized to a ketone.
Looking at the oxidation levels between both fxnal groups, the aldehyde is "easily" oxidized to become a more oxidative group than the ketone.
(Note, oxidation is defined as a reaction that introduces a more electronegative element to or removes a molecular H from the system.)
On the other hand, the secondary alcohol in D can only be oxidized to a ketone.
Looking at the oxidation levels between both fxnal groups, the aldehyde is "easily" oxidized to become a more oxidative group than the ketone.
