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Topic: reactions: substitutions, ketones and iodoform  (Read 5826 times)

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ajbarr

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reactions: substitutions, ketones and iodoform
« on: July 16, 2004, 01:04:19 AM »
I have three problems that I am struggling with:

CH3CH2CH2Br + CH3C(with three CH3's)attached to an 0-Na+
Is it really as simple as the Br and the Na "leaving" and the rest of the atoms attaching?

CH3C=OCH3 (a ketone) + H2
my friend thinks it is propane but I think it is propanol but am not sure where the second H goes.

thirdly, CH3C=OCH3 ( a ketone) + 3 NaOI gives me ? and then ? reacted with NaOH gives me ?
I think this might be an Iodoform reaction but am really lost here.

Thanks for any help. :)

Offline AWK

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Re:reactions: substitutions, ketones and iodoform
« Reply #1 on: July 16, 2004, 03:00:06 AM »
CH3CH2CH2Br + CH3C(with three CH3's)attached to an 0-Na+
This is a simple method of ether synthesis

CH3CH2CH2Br + (CH3)3ONa = CH3CH2CH2-O-C(CH3)3
Depending on reaction conditions a small amount of an elimination product CH2CH=CH2 can be expected.

CH3C=OCH3 (a ketone) + H2
(CH3)2C=O + H2 = CH3-CHOH-CH3 2-propanol (or isopropyl alcohol)


CH3C=OCH3 ( a ketone) + 3 NaOI
CH3-CO-CH3 + 3 NaOI --> CH3-CO-CI3
CH3-CO-CI3 + NaOH = CH3COONa + CHI3
« Last Edit: July 16, 2004, 03:00:49 AM by AWK »
AWK

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