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Hi everyone,
I'm not sure if I'm missing something, or if the Princeton Review CBT diagnostic 4150 is distinctly in error.
Problem 18 on the Biological Sciences section asks:
"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?"
My reasoning:
2-propanone is acetone, the simplest methyl ketone. There is a methyl group on each side, and either of these methyl groups, if deprotonated by a base, will give an equivalent product. 2x the chance of reaction = good. Also, there is good resonance stabilization of the anion after deprotonation, and the anion is primary even without resonance stabilization. Thus, the anion generated by the 2-propanone can attack the aldehyde (propanal) and make the aldol condensation product. In addition, the aldehyde is less sterically hindered, and thus more open to nucleophilic attack by the 2-propanone carbanion.
Princeton's solution says:
This is the product of an aldol reaction, resulting from the enolate of propanal adding to the ketone (propanone).
Why would propanal get deprotonated, when there is only one site, and it's secondary? 2-propanone has two equivalent primary sites that can be deprotonated, and the carbanion can then have nucleophilic attack on the wide-open aldehyde that's completely sterically unhindered to attack. Even if we made the anion on the propanal, it'd have to attack the 2-propanone, which isn't as sterically accessible.
Am I just dumb? Thanks! =)
I'm not sure if I'm missing something, or if the Princeton Review CBT diagnostic 4150 is distinctly in error.
Problem 18 on the Biological Sciences section asks:
"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?"
My reasoning:
2-propanone is acetone, the simplest methyl ketone. There is a methyl group on each side, and either of these methyl groups, if deprotonated by a base, will give an equivalent product. 2x the chance of reaction = good. Also, there is good resonance stabilization of the anion after deprotonation, and the anion is primary even without resonance stabilization. Thus, the anion generated by the 2-propanone can attack the aldehyde (propanal) and make the aldol condensation product. In addition, the aldehyde is less sterically hindered, and thus more open to nucleophilic attack by the 2-propanone carbanion.
Princeton's solution says:
This is the product of an aldol reaction, resulting from the enolate of propanal adding to the ketone (propanone).
Why would propanal get deprotonated, when there is only one site, and it's secondary? 2-propanone has two equivalent primary sites that can be deprotonated, and the carbanion can then have nucleophilic attack on the wide-open aldehyde that's completely sterically unhindered to attack. Even if we made the anion on the propanal, it'd have to attack the 2-propanone, which isn't as sterically accessible.
Am I just dumb? Thanks! =)
