Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: dmc3 on May 19, 2009, 03:22:39 PM
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I have a question on why it is important to use anhydrous ethanol when running a reaction in EtOK/EtOH.
Also is 0.22 grams of dibenzyl ketone the correct answer for 1 to 1 molar ratio of benzil and dibenzyl ketone if I synthesized 0.22 g of benzil?
Benzil = 210.23 g/mol
Dibenzyl ketone = 210.27 g/mol
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I have a question on why it is important to use anhydrous ethanol when running a reaction in EtOK/EtOH.
Depends on the reaction. What're your thoughts?
Also is 0.22 grams of dibenzyl ketone the correct answer for 1 to 1 molar ratio of benzil and dibenzyl ketone if I synthesized 0.22 g of benzil?
Benzil = 210.23 g/mol
Dibenzyl ketone = 210.27 g/mol
Assuming that's your limiting reagent, yes
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anhydrous EtOH needs to be used for one simple reason. Making EtOK involves dissolving metallic Na in EtOH. Imagine what happens when you have water in the reaction mixture. At best, it would dissolve very rapidly, but more likely it will catch fire andyou'llbe in trouble.
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anhydrous EtOH needs to be used for one simple reason. Making EtOK involves dissolving metallic Na in EtOH. Imagine what happens when you have water in the reaction mixture. At best, it would dissolve very rapidly, but more likely it will catch fire andyou'llbe in trouble.
Transmutation Na :rarrow: K ? :) Though I don't think you need as strong a acid as the metal to form the salt...
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Thought of potassium, wrote sodium. Same rule applies, but potassium is even more dangerous in contact with water.
You can use KOH, sure. But still presence of water shifts the reaction left. Moreover, EtOK formed from KOH is present in the reaction mixture in small amount because water is formed. In case of pure metal, it' no longer the case.