Official DAT Destroyer Q&A Thread

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Hi guys, since I'm currently going through the DAT Destroyer and I'm sure many of you guys are as well I felt this thread would be helpful. I usually have various questions while going through Destroyer regarding why a certain answer choice is correct or other times I am in need of a more detailed explanation.

Instead of making multiple threads each time we have questions, I thought we could just post them here and anyone can chime in with an explanation or further clarification.

Please don't post entire questions as that would be a violation of copyright. You can ask specific questions regarding a particular problem in the Destroyers.

Hopefully others find this a useful thread. Thanks!

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Dr. Romano (@orgoman22 )

For the OC, 2017 Q318, after the initial base the work up is with H3O+. Because there was no mechanism shown, I am assuming the mechanism goes something like this:

1) Deprot one of the alpha H
2) Nuc attack on one of the C=O by the enolate (This attack will cause the C=O to go to C-O-)
3) The electrons come back down causing the ester group to leave

This is what is intuitive to me as a probable mechanism for this question. But my question is with the H3O+ workup, wouldn't this cause the other ester group to leave to leave, seeing that the reaction is favorable.... meaning from the nucleophilic acyl sub reactions (ester to acid via H3O+).

Thank you for your help!
 
I also have a question about OC #318, 2017 DAT destroyer,

What if we removed the OC2H5 on the right side, and the H to the right of the C=O, on the left side? Wouldn't there still be another H to deprotonate to form the double bond?

Thank you for the help Dr. Romano!

Edit: @dl2501 , the link for the video I found is below. Hopefully it can help answer your question.

 
Last edited:
Hi Nancy and Dr. Romano,

I have a question on ketone vs. aldehyde.

So, when we compare the acidity between ketone and aldehyde... Since Ketone has 2 methyl groups and 1 carbonyl group, which makes it MORE DEACTIVATING than the aldehyde, Ketone is MORE BASIC than aldehyde, correct?

Then, why is ketone LESS REACTIVE than aldehyde in nucleophilic substitution when ketone is MORE BASIC/ MORE DEACTIVATED? I thought relatively basic/deactivated compounds are usually unstable, which makes them more reactive.

How do I interchangeably use these 2 concepts of REACTIVITY and ACIDITY?
What am I missing here?

Thank you!
Lets first start with acidity. The aldehyde is generally a bit more acidic. It is easier for a base to remove the outer proton to yield our needed enolate anion. This is seen nicely in the Haloform reaction. Now lets talk about reactivity. Aldehydes are more reactive than ketones for two reasons. 1. Steric 2. Electronic. Sterically it is easier to attack the aldehyde since an R group and a Hydrogen is presented versus two R groups in the ketone. Secondly is electronic which is a bit more tougher to understand. Draw out a ketone. Take dimethyl ketone as your example. Now draw a resonance structure by moving the electrons from the carbonyl and moving them out. This will give the resonance form in which the Oxygen has a negative charge and the Carbon is positive. Recall that R groups like methyl DONATE electron density,hence we are essentially stabilizing the resonance hybrid using 2 R groups. An aldehyde is done is the same way, but has only one R group, and a hydrogen. Thus there is less resonance stabilization and therefore would be more reactive. Hopefully you can see the synergy of steric and electronic effects at work. Therefore , we can say that as a nice general rule, aldehydes are more reactive than ketones. The pKa of acetaldyde is about 17, while that of dimethylketone is about 18. A factor of 10 when comparing the acidities.

Hope this helps.

Dr. Romano
 
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Dr. Romano (@orgoman22 )

For the OC, 2017 Q318, after the initial base the work up is with H3O+. Because there was no mechanism shown, I am assuming the mechanism goes something like this:

1) Deprot one of the alpha H
2) Nuc attack on one of the C=O by the enolate (This attack will cause the C=O to go to C-O-)
3) The electrons come back down causing the ester group to leave

This is what is intuitive to me as a probable mechanism for this question. But my question is with the H3O+ workup, wouldn't this cause the other ester group to leave to leave, seeing that the reaction is favorable.... meaning from the nucleophilic acyl sub reactions (ester to acid via H3O+).

Thank you for your help!
When assigning group priority, you go by atomic number. Cl would beat an O since Cl has an atomic number of 17, and O is 8. A Cl would beat a COOH group.....simply look and compare the Carbon and the Chlorine. Also.....when doing a Fischer, if group #4 is on the horizontal, the answer is the reverse of what you think !!!
When assigning group priority, you go by atomic number. Cl would beat an O since Cl has an atomic number of 17, and O is 8. A Cl would beat a COOH group.....simply look and compare the Carbon and the Chlorine. Also.....when doing a Fischer, if group #4 is on the horizontal, the answer is the reverse of what you think !!! Review these rules...known as the Cahn Ingold Prelog rules from a standard text written by a PhD chemist. I would recommend the easiest read,,,,,David Klein.

Hope this helps.

Dr. Romano
 
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Question #256 in Organic Chemistry,
I see that the bulky base has removed the less substituted proton. The substrate is secondary. When a bulky base is employed, does it always remove the less substituted proton? I thought it only did that if the substrate was tertiary? Help me!
Let's go over the rules of elimination. If the substrate is a primary halide, Elimination ...E2 product is made when a large sterically hindered base such as LDA or t-butoxide is used. If the halide is secondary or tertiary, and we use LDA or t-butoxide, the major E2 product will be the LESS substituted alkene as shown in problem #256 that you reference. If the base was changed to either OH-, CH30-, C2H50- for example, you would get the E2 product but the more substituted alkene. We call this the Zaitsev product. If you have the text by Klein, Carey, or Solomon's you can see more examples.

Hope this helps.

Keep going, do ALL the work in the Destroyer, you will be happy on test day

Dr. Jim Romano
 
When assigning group priority, you go by atomic number. Cl would beat an O since Cl has an atomic number of 17, and O is 8. A Cl would beat a COOH group.....simply look and compare the Carbon and the Chlorine. Also.....when doing a Fischer, if group #4 is on the horizontal, the answer is the reverse of what you think !!!
When assigning group priority, you go by atomic number. Cl would beat an O since Cl has an atomic number of 17, and O is 8. A Cl would beat a COOH group.....simply look and compare the Carbon and the Chlorine. Also.....when doing a Fischer, if group #4 is on the horizontal, the answer is the reverse of what you think !!! Review these rules...known as the Cahn Ingold Prelog rules from a standard text written by a PhD chemist. I would recommend the easiest read,,,,,David Klein.

Hope this helps.

Dr. Romano


I am still a bit confused...because after the dieckmann (the ring formation), there is still an ester group. If you were to treat an ester group with H3O+ it would form a carboxylic acid. I am just not seeing why a carboxylic acid wouldn't form in this situation.

It is kind of parallel to the Malonic ester synthesis, where after the gringard attack, there is workup with H3O+ to remove the OCh2Ch3, to form carboxylic acids, one of which is removed via beta-decarboxation.

Thank you!!
 
I also have a question about OC #318, 2017 DAT destroyer,

What if we removed the OC2H5 on the right side, and the H to the right of the C=O, on the left side? Wouldn't there still be another H to deprotonate to form the double bond?

Thank you for the help Dr. Romano!

Edit: @dl2501 , the link for the video I found is below. Hopefully it can help answer your question.




Hey!

Yeah the video definitely helped with he mechanism part of it, and which was what I initially assumed. I am still confused with the H3O workup part because I assumed with H3O you would remove an ester (which you see in NucAcyl Reactions). And it seems the mechanism is similar to Malonic Ester, which after H3O there is formation of CarbAcid, one of which is lost via B-Decarb.

Also, would you know why for 264 after deprotonation, there isn't a nuc_attack on one of the carbonyls?

Thanks for the video btw!

@orgoman22
 
Let's go over the rules of elimination. If the substrate is a primary halide, Elimination ...E2 product is made when a large sterically hindered base such as LDA or t-butoxide is used. If the halide is secondary or tertiary, and we use LDA or t-butoxide, the major E2 product will be the LESS substituted alkene as shown in problem #256 that you reference. If the base was changed to either OH-, CH30-, C2H50- for example, you would get the E2 product but the more substituted alkene. We call this the Zaitsev product. If you have the text by Klein, Carey, or Solomon's you can see more examples.

Hope this helps.

Keep going, do ALL the work in the Destroyer, you will be happy on test day

Dr. Jim Romano
Oh okay! So tertiary and secondary give the less substituted alkene! Thanks!
 
I am still a bit confused...because after the dieckmann (the ring formation), there is still an ester group. If you were to treat an ester group with H3O+ it would form a carboxylic acid. I am just not seeing why a carboxylic acid wouldn't form in this situation.

It is kind of parallel to the Malonic ester synthesis, where after the gringard attack, there is workup with H3O+ to remove the OCh2Ch3, to form carboxylic acids, one of which is removed via beta-decarboxation.

Thank you!!
This is a great question. This reaction was the famous Dieckmann.Condensation. The use of base was our first step to generate the enolate. Then we attack as we do in a Claisen condensation...and then set up our leaving group. Recall, this reaction is nothing more than an intramolecular Claisen. The acid used in work-up is dilute and in very small amount and would not hydrolyze the ester. You have a great eye, and organic chemistry teachers love students like you. I think you can see why care must be taken in this final step. Harsh acidic conditions would indeed result in an ester hydrolysis.

Keep up the great work.

Hope this helps.

Dr. Romano
 
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This is a great question. This reaction was the famous Dieckmann.Condensation. The use of base was our first step to generate the enolate. Then we attack as we do in a Claisen condensation...and then set up our leaving group. Recall, this reaction is nothing more than an intramolecular Claisen. The acid used in work-up is dilute and in very small amount and would not hydrolyze the ester. You have a great eye, and organic chemistry teachers love students like you. I think you can see why care must be taken in this final step. Harsh acidic conditions would indeed result in an ester hydrolysis.

Keep up the great work.

Hope this helps.

Dr. Romano


That makes so much more sense, thank you so much for that Dr. Romano (@orgoman22 ).

And just a quick question on OC264, is there a reason why after the initial deprotonation of the super acidic hydrogen, there is not a further reaction (perhaps a self-Aldol)? Additionally, T-Butyl is huge, so can I assume that the deprotonation of the least hindered acidic H (on the end methyl) can also be viable?

Thank you again!
 
When assigning group priority, you go by atomic number. Cl would beat an O since Cl has an atomic number of 17, and O is 8. A Cl would beat a COOH group.....simply look and compare the Carbon and the Chlorine. Also.....when doing a Fischer, if group #4 is on the horizontal, the answer is the reverse of what you think !!!
When assigning group priority, you go by atomic number. Cl would beat an O since Cl has an atomic number of 17, and O is 8. A Cl would beat a COOH group.....simply look and compare the Carbon and the Chlorine. Also.....when doing a Fischer, if group #4 is on the horizontal, the answer is the reverse of what you think !!! Review these rules...known as the Cahn Ingold Prelog rules from a standard text written by a PhD chemist. I would recommend the easiest read,,,,,David Klein.

Hope this helps.

Dr. Romano
Thank you Dr. Romano!
 
Hey!

Yeah the video definitely helped with he mechanism part of it, and which was what I initially assumed. I am still confused with the H3O workup part because I assumed with H3O you would remove an ester (which you see in NucAcyl Reactions). And it seems the mechanism is similar to Malonic Ester, which after H3O there is formation of CarbAcid, one of which is lost via B-Decarb.

Also, would you know why for 264 after deprotonation, there isn't a nuc_attack on one of the carbonyls?

Thanks for the video btw!

@orgoman22
Sorry, was away. I also see that Dr. Romano answered the question. Good Luck!
 
Hello Dr. Romano,

referring to OChem #318 DAT Destroyer, '17

So in the picture below, the product on the left was formed and was one of the answers (B), but could the product on the right have been formed, but just isn't one of the answer choices? If we removed the och2ch3 on the right, could we have gotten the product i drew on the right?

Thank you for your help!
IMG_0279.JPG
 
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Changing the amount of heat energy usually causes a temperature change. However, DURING the phase change, the temperature stays the same even though the heat energy changes. This energy is going into changing the phase and not into raising the temperature. That's why water doesn't get hotter while it boils. The temperature remains constant until the phase change is complete. If a solid is being melted,,,,,during this transition, temperature and kinetic energy are constant, but bonds are being broken. This means potential energy is increasing.

Hope this helps.

Dr. Jim Romano
.

I agree with everything you said, but I was just curious about the kinetic energy. That is, if a reaction is exothermic, like when you go from a gas to a liquid, I would expect the K.E. to decrease because the particles are moving less (and that a decrease in temp means that there is a lower average Kinetic Energy). But the explanation to 282GC is such that if there is a reaction that is exothermic, then there will be an increase in K.E. But how is it if you are going from gas to liquid, that there is a decrease in Kinetic Energy? I mean I definitely agree with you about the P.E., that is gases have higher P.E. than liquids... But the K.E. explanation just seems counterintuitive.

Thank you in advance.
 
Hello Dr. Romano. I had a question about #25 on the Math Destroyer Practice Test 5

So I understand how side a can be solved by tan (0) = a/2, but I am not seeing how b is cot (0) b/2. Maybe because b is in a different triangle that I am not seeing how the opposite side would be 2 for that angle.
 
I agree with everything you said, but I was just curious about the kinetic energy. That is, if a reaction is exothermic, like when you go from a gas to a liquid, I would expect the K.E. to decrease because the particles are moving less (and that a decrease in temp means that there is a lower average Kinetic Energy). But the explanation to 282GC is such that if there is a reaction that is exothermic, then there will be an increase in K.E. But how is it if you are going from gas to liquid, that there is a decrease in Kinetic Energy? I mean I definitely agree with you about the P.E., that is gases have higher P.E. than liquids... But the K.E. explanation just seems counterintuitive.

I mean the only way that I see it intuitively is if you are talking about the

Thank you in advance. .

Hello Dr. Romano. I had a question about #25 on the Math Destroyer Practice Test 5

So I understand how side a can be solved by tan (0) = a/2, but I am not seeing how b is cot (0) b/2. Maybe because b is in a different triangle that I am not seeing how the opposite side would be 2 for that angle.


@Dallas Dentist

I think I can help you answer this question.

For the bigger triangle, Tan = a/2 (you then rearrange to find a because that is one part the length of L) or a= 2 x tan => 2tan

For the smaller triangle, the Tan would be 2/b or Tan=2/b. Now just rearrange to find b (which is the other length of L), and when you rearrange to get b you get b=2/tan. Now you can either do the following step two ways.... 1) automatically see that because tan is on the bottom and thus automatically know that the inverse of it (to get it to the numerator) is just cot or 2) separate the function to get b = (2/1) x (1/tan) . This second way just makes it easier to see that 1/tan is ACTUALLY just cot.... so you get b= (2/1) x (1/tan) [or simply cot] -----> . b= 2 x cot => 2cot

And then you rearrange and simply so you have a= 2tan and b= 2cot => a + b = L => 2tan + 2cot => simplify => 2(tan+cot).

@orgoman22
 
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Hello Dr. Romano. I had a question about #25 on the Math Destroyer Practice Test 5

So I understand how side a can be solved by tan (0) = a/2, but I am not seeing how b is cot (0) b/2. Maybe because b is in a different triangle that I am not seeing how the opposite side would be 2 for that angle.
For side b: tan (theta) = 2/b. Solve for b to get: b = 2/tan(theta) = 2 cot (theta)
Remember: 1/tan = cot

Hope this helps..
 
Hello Dr Romano,

For the DAT Destroyer Bio, do the questions get increasingly more difficult towards the end, or is it a random mix? Just wondering if I should pick and choose what questions I answer in a sitting!
 
Changing the amount of heat energy usually causes a temperature change. However, DURING the phase change, the temperature stays the same even though the heat energy changes. This energy is going into changing the phase and not into raising the temperature. That's why water doesn't get hotter while it boils. The temperature remains constant until the phase change is complete. If a solid is being melted,,,,,during this transition, temperature and kinetic energy are constant, but bonds are being broken. This means potential energy is increasing.

Hope this helps.

Dr. Jim Romano
.

Dr. Romano,

I agree with everything you said, but I was just curious about the kinetic energy. That is, if a reaction is exothermic, like when you go from a gas to a liquid, I would expect the K.E. to decrease because the particles are moving less (and that a decrease in temp means that there is a lower average Kinetic Energy). But the explanation to 282GC is such that if there is a reaction that is exothermic, then there will be anincrease in K.E. But how is it if you are going from gas to liquid, that there is a decrease in Kinetic Energy? I mean I definitely agree with you about the P.E., that is gases have higher P.E. than liquids... But the K.E. explanation just seems counterintuitive.

Also another question for OC:

Regarding, OC264, is there a reason why after the initial deprotonation of the super acidic hydrogen, there is not a further reaction (perhaps a self-Aldol)? Additionally, T-Butyl is huge, so can I assume that the deprotonation of the least hindered acidic H (on the end methyl) can also be viable?

Thank you in advance.
 
Hello Dr Romano,

Slighty confused about a biology topic.. On the DAT Destroyer bio #50 (2017) solution says Neutrophils make up 60% of the white blood cell population (thus majority). Question 54 then says that monocytes make up the majority of white blood cells. However, Monocytes and Neutrophils are structurally different, so im slightly confused what is making up the majority of WBC, Neutrophils or Monocytes? I understand they originate from the same precursor lineage
 
Hello Dr Romano,

Slighty confused about a biology topic.. On the DAT Destroyer bio #50 (2017) solution says Neutrophils make up 60% of the white blood cell population (thus majority). Question 54 then says that monocytes make up the majority of white blood cells. However, Monocytes and Neutrophils are structurally different, so im slightly confused what is making up the majority of WBC, Neutrophils or Monocytes? I understand they originate from the same precursor lineage
Question asks which is false. It is false that monocytes make up most of WBC. Neutrophils do.

There is a mnemonic for WBC: Never Let Monkeys Eat Bananas. Neutrophils, Lympho, Mono, Eosinophils, Basophils.
 
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Hello Dr Romano,

Slighty confused about a biology topic.. On the DAT Destroyer bio #50 (2017) solution says Neutrophils make up 60% of the white blood cell population (thus majority). Question 54 then says that monocytes make up the majority of white blood cells. However, Monocytes and Neutrophils are structurally different, so im slightly confused what is making up the majority of WBC, Neutrophils or Monocytes? I understand they originate from the same precursor lineage

I think I can help answer this one.

A quick google search can help clear this answer up (and I am in no means mocking you), but after struggling over and over in terms of memorizing this, the trend is this: 1) Neutrophils, 2) Lymphocytes, 3) Monocytes, 4) Eosinophils, and 5) Basophils. The mnemonic that I use to help me is: Never Let Monkeys Eat Bananas.

In terms of the question, 54 asks which of the following is FALSE? And thus the answer is E, which is false.

@orgoman22 Can you confirm?
 
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I think I can help answer this one.

A quick google search can help clear this answer up (and I am in no means mocking you), but after struggling over and over in terms of memorizing this, the trend is this: 1) Neutrophils, 2) Lymphocytes, 3) Monocytes, 4) Eosinophils, and 5) Basophils. The mnemonic that I use to help me is: Never Let Monkeys Eat Bananas.

In terms of the question, 54 asks which of the following is FALSE? And thus the answer is E, which is false.

@orgoman22 Can you confirm?
Great mnemonic ! thanks!
 
Dr. Romano,
Can you explain #110 QR (2017 dat destroyer)? This seems like a relatively simple problem but I don't exactly understand what the question is asking..

Thanks in advance!
 
Dr. Romano,

I agree with everything you said, but I was just curious about the kinetic energy. That is, if a reaction is exothermic, like when you go from a gas to a liquid, I would expect the K.E. to decrease because the particles are moving less (and that a decrease in temp means that there is a lower average Kinetic Energy). But the explanation to 282GC is such that if there is a reaction that is exothermic, then there will be anincrease in K.E. But how is it if you are going from gas to liquid, that there is a decrease in Kinetic Energy? I mean I definitely agree with you about the P.E., that is gases have higher P.E. than liquids... But the K.E. explanation just seems counterintuitive.

Also another question for OC:

Regarding, OC264, is there a reason why after the initial deprotonation of the super acidic hydrogen, there is not a further reaction (perhaps a self-Aldol)? Additionally, T-Butyl is huge, so can I assume that the deprotonation of the least hindered acidic H (on the end methyl) can also be viable?

Thank you in advance..

If a reaction evolves heat, we say it is exothermic. The surroundings get warmer. If a reaction is endothermic, heat is absorbed and the surroundings gets colder. Don't confuse average kinetic energy up with kinetic energy. During a phase change, average kinetic energy is constant. If a reaction gives off heat......simply draw an energy diagram. You will clearly see that the potential energy of the products are less than the reactants....and vice versa for an endothermic reaction. During an exothermic process, as heat is released kinetic energy increases......For OC 264... No this molecule is a damn monster,,,,,Aldol condensations work only with smaller ketones. Keep it simple......you are over thinking ......all I asked for was essentially the intermediate.....clearly shown in choice B.

Hope this helps.

Dr. Romano
 
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Dr. Romano,
Can you explain #110 QR (2017 dat destroyer)? This seems like a relatively simple problem but I don't exactly understand what the question is asking..

Thanks in advance!
First reduce the fractions : 50/2 25. 65/13 =5. 40/2 = 20
now: 25 - 5 + 20 = 40. Then 40 = (p/100)*80
reduce the right hand side to get: 40 = 4p/5. Cross multiply and solve for p. p=50
 
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Dr. Romano (@orgoman22 ),

This post is referring to the Math Destroyer Questions. I am having a bit trouble understanding the T9Q28 (regarding the gecko population). I can see that it is the same set up as T1Q27 (regarding Green Village population), but how do these two problems differ from that of T3Q15?

For T1Q27 and T9Q28, they are both venn diagram-like problems, that is for these you take the percentages and subtract the "both" from each and then add it up. But I don't see how T3Q15 is different (there is also another problem that I can seem to find regarding weather percentages that has the same set up as T3Q15), and for these two you add up the first two percentages and subtract the both.

Thank you so much!
 
Dr. Romano (@orgoman22 ),

This post is referring to the Math Destroyer Questions. I am having a bit trouble understanding the T9Q28 (regarding the gecko population). I can see that it is the same set up as T1Q27 (regarding Green Village population), but how do these two problems differ from that of T3Q15?

For T1Q27 and T9Q28, they are both venn diagram-like problems, that is for these you take the percentages and subtract the "both" from each and then add it up. But I don't see how T3Q15 is different (there is also another problem that I can seem to find regarding weather percentages that has the same set up as T3Q15), and for these two you add up the first two percentages and subtract the both.

Thank you so much!
T1Q27 and T3Q15 are very similar. You can use Venn diagram. T9Q28 is different because of the added condition ( BUT NOT BOTH). As the solution indicates you exclude 17%.
cleardot.gif
 
Dr. Romano,

#208 GC (2017)

I understand how the molar solubility with the common ion is 0.16M.. However, I got a molar solubility of ~0.02M without the common ion..
Logically, I thought the molar solubility with the common ion would decrease due to the reaction shifting towards the reactant side (it would be less soluble). But this question contradicts my understanding of this topic. Can you explain why the molar solubility of PbCl2 increases here with the common ion?

Thanks!
 
Dr. Romano,

#208 GC (2017)

I understand how the molar solubility with the common ion is 0.16M.. However, I got a molar solubility of ~0.02M without the common ion..
Logically, I thought the molar solubility with the common ion would decrease due to the reaction shifting towards the reactant side (it would be less soluble). But this question contradicts my understanding of this topic. Can you explain why the molar solubility of PbCl2 increases here with the common ion?

Thanks!
When doing this problem, we changed the KCl to be 0.1M.......You will then see it will be fine. We have corrected this. In water it would be calculated to be 0.0159M.....the mechanics of the problem is fine, but simply use 0.1M for KCl.....this will give you a molar solubility of 0.00163M. Indeed, the common ion effect is operating rendering our salt less solubility.

Hope this helps.

Dr. Romano
 
T1Q27 and T3Q15 are very similar. You can use Venn diagram. T9Q28 is different because of the added condition ( BUT NOT BOTH). As the solution indicates you exclude 17%.
cleardot.gif
Hello Dr Romano (@orgoman22 )

I am actually also stuck on Q.13 T3 of the math destroyer. When I apply the venn diagram is doesnt come out with the correct answer. What I have done is is 14% population is blue eye and 8% both, im calculating 14-8 = 6% as the population on only blue eyes. I also do the same with the left hand population (25-8) giving me 17% just left handed. So then the population of either blue eye or left handed is 17% + 6% ...these are my calcuations which seem to be wrong, please could you explain this? Thank you!
 
Hello Dr Romano (@orgoman22 )

I am actually also stuck on Q.13 T3 of the math destroyer. When I apply the venn diagram is doesnt come out with the correct answer. What I have done is is 14% population is blue eye and 8% both, im calculating 14-8 = 6% as the population on only blue eyes. I also do the same with the left hand population (25-8) giving me 17% just left handed. So then the population of either blue eye or left handed is 17% + 6% ...these are my calcuations which seem to be wrong, please could you explain this? Thank you!
Since the question is asking for either left handed or blue eyes the answer will be: 6+8+17=31%. Your answer would be correct if the question asked: blue eyes or left handed BUT NOT BOTH.
 
Hi Dr.Romano,

This is for the 2017 DAT Destroyer:

For OChem #54: I am confused how this reaction works, is it that Pyr (pyridine?) acts a base that removes a proton from the carboxylic acid, and then the carboxylate anion attacks the carbonyl of the acid chloride with Cl leaving to produce an anhydride? Isn't Pyr a weak base and carboxylate a bad nucleophile?

For Ochem #57: The last step is SN2 with OCH3 replacing the leaving group, but couldn't E2 also occur since we have a strong base and good leaving group?
 
Hi Dr.Romano,

This is for the 2017 DAT Destroyer:

For OChem #54: I am confused how this reaction works, is it that Pyr (pyridine?) acts a base that removes a proton from the carboxylic acid, and then the carboxylate anion attacks the carbonyl of the acid chloride with Cl leaving to produce an anhydride? Isn't Pyr a weak base and carboxylate a bad nucleophile?

For Ochem #57: The last step is SN2 with OCH3 replacing the leaving group, but couldn't E2 also occur since we have a strong base and good leaving group?

#54
Here is part of the mechanism.....the reaction is called a nucleophilic acyl substitution reaction. The first step is attack by the double bonded O onto the carbon of the acyl halide........the Pyr is used in the final step to simply remove off the proton, it is acting as a base .

For # 57.....Yes indeed, an E2 would be a competing reaction. since a strong base and a great nucelophile , namely CH30- is employed. It turns out that SN2 usually predominates over E2 when Tosylate is the leaving group. Since this problem only has substitution products, ignore the competing E2 product. This is an important problem. The DAT exam likes questions like this... where one must have to think in sequence. Keep up the great work, you have some fine questions !

Hope this helps.

Dr. Jim Romano
 

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Dr. Romano (@orgoman22 ),

I have a quick question regarding the stability of alkenes. That is, if you were to compare a primary alkene to a secondary alkene, I would say that the secondary alkene is more stable than a primary alkene. But What if you compared a primary alkene to a Z-secondary alkene? I know that the E-secondary alkene will be more stable than the Z. But because a Z-alkene can have steric hinderance (on the same side), could it be possible that the primary alkene be more stable because it lacks any sort of sterics?

Thank you!
 
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DAT Destroyer (QR) #38 Help

Number 38 on the QR section of the DAT Destroyer 2017 asks to "solve for x in v(x+5) = x-1." The solution explains how to square both sides to end up with results x = -1 or 4 and concludes that only 4 is the answer. It seems like "-1" works as well since the results of square roots can be positive or negative. Can someone explain why only "4" is the answer? Thanks
 
Please could someone explain this very confusing concept to me. So for electrochemistry, in galvanic cells, oxidation occurs at the anode, whilst reduction occurs at the cathode. At the anode, therefore if we have a piece of Zinc metal it will LOSE its electrons/deposit them at the anode and Zn2+ goes into solution. Those electrons are conducted throughout the circuit to the cathode when a Cu2+ ion for example will gain that electron forming copper. So a question on the DAT bootcamp gen chem exam, says that depositing an electron is a gain of electrons that occurs at the cathode. I am so confused by this, I thought electrons are deposited at the anode!!

I would be so grateful if someone to explain this to me!! (@orgoman22 ? :) )
 
Please could someone explain this very confusing concept to me. So for electrochemistry, in galvanic cells, oxidation occurs at the anode, whilst reduction occurs at the cathode. At the anode, therefore if we have a piece of Zinc metal it will LOSE its electrons/deposit them at the anode and Zn2+ goes into solution. Those electrons are conducted throughout the circuit to the cathode when a Cu2+ ion for example will gain that electron forming copper. So a question on the DAT bootcamp gen chem exam, says that depositing an electron is a gain of electrons that occurs at the cathode. I am so confused by this, I thought electrons are deposited at the anode!!

I would be so grateful if someone to explain this to me!! (@orgoman22 ? :) )
Lets consider the following example and use it as a prototype. Zn(s) + Cu++-------> Cu(s) + Zn++. Recall that oxidation occurs on the surface of the solid anode, while reduction occurs on the solid surface of the cathode. Here we see Zinc being oxidized, since it goes from 0 to 2,,,,,,thus it is the anode, Since it is the anode,,,,we lose electrons........these electrons will now be used at the cathode, The cathode will gain electrons. At the cathode,,,,,which is the other solid,,,,namely copper,,,,,,the Copper ions use these electrons to do the reduction reaction. Thus at the anode ..we have....Zn(s)------>Zn++ + 2 electrons.........then at the cathode, we have Cu++ + 2 electrons -------->Cu(s). Thus I hope you now see it ......We are losing mass at the anode,,,,and gaining mass, namely metallic copper at the cathode. Do yourself a big favor, pick up your college textbook and look at the drawn figures of the anode and cathode set-up. You should also be able to calculatesuch things as cell potentials, free energy, and identify the oxidants and reductants from such reactions.

I hope this helps

Dr. Romano
 
Lets consider the following example and use it as a prototype. Zn(s) + Cu++-------> Cu(s) + Zn++. Recall that oxidation occurs on the surface of the solid anode, while reduction occurs on the solid surface of the cathode. Here we see Zinc being oxidized, since it goes from 0 to 2,,,,,,thus it is the anode, Since it is the anode,,,,we lose electrons........these electrons will now be used at the cathode, The cathode will gain electrons. At the cathode,,,,,which is the other solid,,,,namely copper,,,,,,the Copper ions use these electrons to do the reduction reaction. Thus at the anode ..we have....Zn(s)------>Zn++ + 2 electrons.........then at the cathode, we have Cu++ + 2 electrons -------->Cu(s). Thus I hope you now see it ......We are losing mass at the anode,,,,and gaining mass, namely metallic copper at the cathode. Do yourself a big favor, pick up your college textbook and look at the drawn figures of the anode and cathode set-up. You should also be able to calculatesuch things as cell potentials, free energy, and identify the oxidants and reductants from such reactions.

I hope this helps

Dr. Romano
Thank you, that was very helpful!! My main confusion was with the wording of electrons being 'deposited' and i had interpreted that as electrons being deposited, as in lost. However, it seems what the question meant is that electrons are being deposited (given to) the cathode - for reduction to take place.
 
Dr. Romano (@orgoman22 ),

I have a quick question regarding the stability of alkenes. That is, if you were to compare a primary alkene to a secondary alkene, I would say that the secondary alkene is more stable than a primary alkene. But What if you compared a primary alkene to a Z-secondary alkene? I know that the E-secondary alkene will be more stable than the Z. But because a Z-alkene can have steric hinderance (on the same side), could it be possible that the primary alkene be more stable because it lacks any sort of sterics?

Thank you!
You asked a very interesting question. As a nice general rule , the cis isomer, or Z isomer where applicable, would be more stable. I actually went to the Graham Solomons text for you to give you the experimental data. The 1-butene isomer had a heat of hydrogenation of -30.3 kcal, while the cis -2-butene isomer was -26.6 kcal. The more heat given off, would indicate the less stable isomer. This is a good general rule. If the 2 R groups were very very large, the primary alkene would predominate. Thus, as you can see, it is not always 100% predictable. For your purposes, however, the disubstituted, cis or Z isomer generally will be more stable. Thanks for such a wonderful and new question never asked here before.

Hope this helps.

Dr. Romano
 
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You asked a very interesting question. As a nice general rule , the cis isomer, or Z isomer where applicable, would be more stable. I actually went to the Graham Solomons text for you to give you the experimental data. The 1-butene isomer had a heat of hydrogenation of -30.3 kcal, while the cis -2-butene isomer was -26.6 kcal. The more heat given off, would indicate the less stable isomer. This is a good general rule. If the 2 R groups were very very large, the primary alkene would predominate. Thus, as you can see, it is not always 100% predictable. For your purposes, however, the disubstituted, cis or Z isomer generally will be more stable. Thanks for such a wonderful and new question never asked here before.

Hope this helps.

Dr. Romano


I guess it was not a very "quick question," sorry! It was one of the questions in destroyer that made me think, and I know it is extra, but I am just trying to prepare for the worst storm.

Thank you so much for your help.

-David
 
I guess it was not a very "quick question," sorry! It was one of the questions in destroyer that made me think, and I know it is extra, but I am just trying to prepare for the worst storm.

Thank you so much for your help.

-David
Never hesitate to ask questions...I will respond asap when not working with my classroom students.

Prepare for the storm, the DAT Beast is a very sneaky fellow.
 
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So on Gen Chem Destroyer, pg 166, #105, i understand that delta H of formation is delta H products - delta H reactants. I understand that we reverse the second reaction and not the first because we need CuO in the product. However, when we reverse the second rxn, why do we not reverse the delta H, making it -10 instead of 10? In that scenario it would be -140 - (-10) which equals -130. But the answer for this problem is -150. Please help me understand why we dont reverse the second delta H! Thanks!
 
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