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Topic: Acetone reaction scheme  (Read 24712 times)

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Offline Rutherford

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Re: Acetone reaction scheme
« Reply #60 on: February 14, 2013, 10:54:55 AM »
Not the reduction, but the hydrolysis with H2O (below LiAlH4).

Offline Dan

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Re: Acetone reaction scheme
« Reply #61 on: February 14, 2013, 11:04:37 AM »
You need an acid catalyst to hydrolyse a ketal, water alone won't do it.
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Offline Rutherford

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Re: Acetone reaction scheme
« Reply #62 on: February 14, 2013, 12:12:14 PM »
Oh, wait I wrote step S :rarrow: T which is the one with HCl, so HCl should decompose it.

Offline Dan

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Re: Acetone reaction scheme
« Reply #63 on: February 14, 2013, 12:32:01 PM »
Please post clearly what you think each structure is and label them with the letters as in the schemes so as to make them unambiguous.
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Offline Rutherford

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Re: Acetone reaction scheme
« Reply #64 on: February 14, 2013, 12:44:17 PM »
I intended to do that at the end. Here it is.
« Last Edit: February 14, 2013, 01:03:52 PM by Raderford »

Offline discodermolide

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Re: Acetone reaction scheme
« Reply #65 on: February 14, 2013, 01:13:14 PM »
An interesting way of α-bromination of acetone with concomitant ketallisation. (see page 1 of thread)
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Offline Rutherford

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Re: Acetone reaction scheme
« Reply #66 on: February 14, 2013, 01:29:18 PM »
I got a problem with N and O isomers (the ketals in the first scheme). The tip was: N and O are isomers; ωH = 3.8%; ωC = 22.9%.
The formula of the ketal is C5H10O2Br2 (the left compound in the picture). Its molar mass is 262 g/mol while the molar mass according to the elemental analysis should be 263 g/mol. There is a difference here  ???.

Offline Dan

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Re: Acetone reaction scheme
« Reply #67 on: February 14, 2013, 01:41:46 PM »
It's just a rounding error. For 262 g/mol you get...

For the 10H: (10 x 1)/262 = 0.038 (3.8%)
For the 12C: (5 x 12)/262 = 0.229 (22.9%)
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Offline Rutherford

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Re: Acetone reaction scheme
« Reply #68 on: February 14, 2013, 01:44:49 PM »
Okay, I suppose that the scheme I posted is right.

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