The Law Posted December 10, 2009 Report Share Posted December 10, 2009 Question from TPR test 5 biology: Based on the information in the passage, which of the following is the most likely product of the reaction between propanal and 2-propanone, in the presence of base? Relevant info in passage: In general, the addition reactions of both the acid-catalyzed and the base-catalyzed condensations are readily reversible. The equilibrium constant for the addition phase is generally unfavorable for acyclic ketones and favorable for aldehydes and cyclic compounds. On the other hand, the equilibrium constant for the dehydration phase is generally favorable since the resultant product is stabilized by resonance. Therefore, even under conditions which are unfavorable for addition, the reaction may be driven to completion by the favorability of dehydration, which is not easily reversible. I got it wrong because I chose the aldehyde to attack the ketone. I guess, now in hindsight, that was dumb because the ketone has 2 alpha hydrogens that are both stabilized by resonance and no other carbons to induct.... but the reason I picked the aldehyde as the nuleophile was it says addition phase is generally unfavorable for acyclic ketones and favorable for aldehydes and cyclic compounds. Can someone help me find out what I'm missing here? Thanks! Link to comment Share on other sites More sharing options...
janny_jan Posted December 11, 2009 Report Share Posted December 11, 2009 The alpha hydrogen on aldehydes have a pka of 17 and ketones have a pka of 19. Therefore, propanal is a better "acid" and would be deprotonated. I think that the correct answer choice to that question is wrong because it has one too many carbons. I think i remember this one.... the diagnostic is in error because their answer has 7 carbons Link to comment Share on other sites More sharing options...
The Law Posted December 11, 2009 Author Report Share Posted December 11, 2009 The alpha hydrogen on aldehydes have a pka of 17 and ketones have a pka of 19. Therefore, propanal is a better "acid" and would be deprotonated. I think that the correct answer choice to that question is wrong because it has one too many carbons. I think i remember this one.... the diagnostic is in error because their answer has 7 carbons LOL... that is what I was having a problem with... so I'm right then? The aldehyde is what gets deprotonated. It's an easy reaction that's why I got frustrated with this one. Link to comment Share on other sites More sharing options...
The Law Posted December 11, 2009 Author Report Share Posted December 11, 2009 Link to comment Share on other sites More sharing options...
avenir001 Posted December 11, 2009 Report Share Posted December 11, 2009 u're right that the aldehyde is deprotonated and attacks the ketone, but none of their answers looks right to me. the only one that comes close is A, but as jan said it has too many carbons. the correct answer would look like this: (CH3)2--CH(OH)--CH(CH3)--CHO Link to comment Share on other sites More sharing options...
The Law Posted December 11, 2009 Author Report Share Posted December 11, 2009 u're right that the aldehyde is deprotonated and attacks the ketone, but none of their answers looks right to me. the only one that comes close is A, but as jan said it has too many carbons. the correct answer would look like this: (CH3)2--CH(OH)--CH(CH3)--CHO SWEET! That means I maybe got 13 on that practice. SUH WEET. haha Link to comment Share on other sites More sharing options...
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