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Topic: Synthesis of 2-methyl-1-phenylbutan-2-ol  (Read 3690 times)

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Offline Violagirl

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Synthesis of 2-methyl-1-phenylbutan-2-ol
« on: October 24, 2011, 10:33:44 PM »
For a problem in my book for O Chem II lecture, I need to find a way to synthesize 2-methyl-1-phenylbutan-2-ol. I wanted to know if what I found was correct if someone would be able to double-check it, it'd be greatly appreciated!

Starting with benzene, I used CH3CH2CH(CH3)COCl with AlCl3, then used Zn(Hg), HCl to get rid of the double bond to O, then used Cl2 with hv to get the Cl onto the beta carbon with where the 2 methyl group is. Then used NaOH to get Cl to leave as a leaving group and replaced it with the OH. Is this right at all??

Offline Honclbrif

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Re: Synthesis of 2-methyl-1-phenylbutan-2-ol
« Reply #1 on: October 24, 2011, 11:15:41 PM »
The easiest answer to any synthesis question is "just buy it"* But that tends to earn you the title of "smartass" from your proff without any credit to go with it, or "lazy cash waster" from your PI if you're a grad student.

But seriously, you have a tendency to greatly over-think organic chem problems. Some tips:

1) If its after the first week of organic chem, you will never, ever, ever see radical halogenation with UV light again (I know, AIBN-initiated reactions or the calculation of rate laws are different stories) but if you find yourself writing "Br2 + UV", you're starting to write fiction.

2) When in doubt: carbonyl chemistry. I don't know if its the most versatile functional group known, but its waaaaaaaay up there.

*Aldrich cat. no. PH010917. Yes its CPR, but you can buy it if you have $US 100 lying around.
Individual results may vary

Offline orgopete

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Re: Synthesis of 2-methyl-1-phenylbutan-2-ol
« Reply #2 on: October 25, 2011, 08:12:32 AM »
It appears as though the question must also state that benzene was a starting material, but I don't know that. Given the steps suggested, are they an indication of the chemistry being discussed in class?

Otherwise, from a practical point of view, I'd probably follow Honclbrif's advice. Since there is an alcohol, I'd consider it could be the product of an addition to a C=O. There are several ways that reaction could take place. Would any of those choices match the chemistry being discussed in class?
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