nabilesmail

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Dec 25, 2009
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I'm a bit confused on this problem mostly because I don't know where the 3rd carbon lies. Usually I go by the most oxidized carbon as C #1. Can someone help me with this, I guess it is more of a naming question
 

MedPR

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I'm a bit confused on this problem mostly because I don't know where the 3rd carbon lies. Usually I go by the most oxidized carbon as C #1. Can someone help me with this, I guess it is more of a naming question
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dnovikov

10+ Year Member
Jun 6, 2008
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lidocaine ^^^

3-hydroxylidocaine i think? ^^^^

Question goes as follows: To convert lidocaine into 3-hydroxylidocaine, what must be added to it?
A. Strong acid in water
B. Strong base in water
C. An oxidizing agent
D. A reducing agent
 
Mar 13, 2012
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I believe the answer is C, oxidizing agent.

You don't really need to know nomenclature to figure this question out. If you look at the picture, the only difference between the two molecules is a hydroxyl group on the benzene ring. Even if you're not given the pictures of the structures, from the names of the compound you see that "hydroxy" has been added. Thus it is an oxidation reaction, and that's all you need to know.

Sidenote: Carbon numbering in this question would be a waste of time. But if you care to touch up on it, looking at the structure, the hydroxyl group is on Carbon 3, and the amine/carbon substituent is on carbon 1 of the ring. The MCAT would not expect you to know a structure off of a root like "lidocaine". You can probably logically assume that 3-hydroxy lidocaine differs from lidocaine by a single hydroxyl group and since there is a benzene ring present, it is on the third carbon.
 
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