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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I am using the 2017 Destroyer to prepare for my DAT next week and I had a question on one of the Organic problems. On number 37 the solution says that choice B (Aniline) has an electron withdrawing group, therefore it can be eliminated, but from watching Chad's videos and my studies I was taught that an NH2 group on Benzene should be electron-donating, which should make Benzene more reactive, thus making it more basic. I was wondering if anyone could clear this up for me. Thank you!
You are looking at the WRONG group !!!!! Focus on the NH2 group......The BENZENE ring is withdrawing electron density away from the Amine group, hence rendering it a less effective and therefore a weaker base. Only choice A has electrons totally LOCALIZED on a Nitrogen and in an sp3 orbital...thus A is most basic since these electrons are neither held tightly nor part of a resonance system.

Hope this helps.

Dr. Romano
 
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Dr. Romano,

For OC Question 259 in 2017, I understand the question is asking for the BEST way to do the double inversion, I was wondering if a double inversion would still occur for option C?
 
You are looking at the WRONG group !!!!! Focus on the NH2 group......The BENZENE ring is withdrawing electron density away from the Amine group, hence rendering it a less effective and therefore a weaker base. Only choice A has electrons totally LOCALIZED on a Nitrogen and in an sp3 orbital...thus A is most basic since these electrons are neither held tightly nor part of a resonance system.

Hope this helps.

Dr. Romano
for B, because the Amine is donating (an electron donating group [EDG]) therefore benzene is then withdrawing electron dentistry which in turn becomes a weaker base? So, EDG's activate the ring, and the ring then withdraws electron groups which make the compound a weaker base? But the answer states that electron withdrawing groups make the compound less basic. Do both electron donating and withdrawing crougs make the compound less basic? Thank you for the clarification.
 
Dr. Romano,

For OC Question 259 in 2017, I understand the question is asking for the BEST way to do the double inversion, I was wondering if a double inversion would still occur for option C?
No. HBr would first protonate the molecule, then give the racemic mixture. Only 50% would have a double inversion. Only choice B as shown in the solution completes the job in high yield. This is a very important problem !!!! Hope this helps.
 
for B, because the Amine is donating (an electron donating group [EDG]) therefore benzene is then withdrawing electron dentistry which in turn becomes a weaker base? So, EDG's activate the ring, and the ring then withdraws electron groups which make the compound a weaker base? But the answer states that electron withdrawing groups make the compound less basic. Do both electron donating and withdrawing groups make the compound less basic? Thank you for the clarification.
You seem very confused. An amine is a base. If the electrons are localized on the amine, they can share it easily with a proton and act as a base. If an amine is bonded to a Benzene ring, it is LESS basic. The benzene ring acts as an electron withdrawing group towards the unshared electron pair on the nitrogen atom. Thus...AROMATIC AMINES are weak bases. Forget about activating groups or deactivating groups...that is a DIFFERENT concept. We talk about activating and deactivating groups usually in the context of reaction rates and directing influences....BUT not here. This problem deals with basicity. Any factor that withdraws electron density from the NH2 group......be it resonance, or a benzene ring will render the amine LESS basic. If this still is not clear, you need to consult an organic text written by a PhD chemist. Hope this helps.
 
Hello Dr Romano,

For Q.38 DAT Destroyer 2017 QR, the two possible solutions for X = 4 or -1. The solution says that 4 is the only answer. However, upon substituting x = 4 or x = -1 I still get a correct answer? Please let me know what im doing wrong!
 
Hello Dr Romano,

For Q.38 DAT Destroyer 2017 QR, the two possible solutions for X = 4 or -1. The solution says that 4 is the only answer. However, upon substituting x = 4 or x = -1 I still get a correct answer? Please let me know what im doing wrong!
No. You don't get the correct answer if you plug in x=-1.
You get : 2 = -2
 
@orgoman22

Because the reaction was via hydroboration and it was a syn addition, there was no rearrangements. But will there ever be a case where the starting product will go to answer choice (d) which I thought, initially, was the answer because the ring is more stable (5 vs 4). That is, I do remember earlier on in the book where you mentioned that H2/Pt with a cyclopropene and cyclobutene will cause the ring to open up (linearly). But is there a case where you start with the reactant and then do straight to answer choice d?

Thank you.
 
For the Math Destroyer test #3 (q#15), why don't you subtract 8% from both populations (for a total of popA%+popB%-16)? Basically, why is it popA%+popB%-8? I'm confused, because there are a lot of questions like this on DAT bootcamp and they subtract the shared percentage from both populations for this type of question :/
 
@orgoman22

Because the reaction was via hydroboration and it was a syn addition, there was no rearrangements. But will there ever be a case where the starting product will go to answer choice (d) which I thought, initially, was the answer because the ring is more stable (5 vs 4). That is, I do remember earlier on in the book where you mentioned that H2/Pt with a cyclopropene and cyclobutene will cause the ring to open up (linearly). But is there a case where you start with the reactant and then do straight to answer choice d?

Thank you.
OC 243
Hydroboration is a reaction that employs BH3 and Peroxide- basic workup to give an alcohol with no rearrangements. This is seen in choice A. This reaction is one of our most important reactions in all of Organic Chemistry. It won Herbert Brown the 1979 Nobel prize in Chemistry. I met him 3 times, and he was one of the most kindest people I ever met. This reaction involves intermediates not normally seen in organic chemistry, and consequently no rearrangements, shifts, or ring expansions occur. Choice D can be made by an intramolecular SN2 reaction......For example, if you take 4-iodo-1-buatanol with sodium ethoxide, you would remove the H off the alcohol, and intramolecularly attack C4 to yield the compound you see in D called Tetrahydrofuran or oxacyclopentane .

I hope this helps.

Dr. Romano
 
OC 243
Hydroboration is a reaction that employs BH3 and Peroxide- basic workup to give an alcohol with no rearrangements. This is seen in choice A. This reaction is one of our most important reactions in all of Organic Chemistry. It won Herbert Brown the 1979 Nobel prize in Chemistry. I met him 3 times, and he was one of the most kindest people I ever met. This reaction involves intermediates not normally seen in organic chemistry, and consequently no rearrangements, shifts, or ring expansions occur. Choice D can be made by an intramolecular SN2 reaction......For example, if you take 4-iodo-1-buatanol with sodium ethoxide, you would remove the H off the alcohol, and intramolecularly attack C4 to yield the compound you see in D called Tetrahydrofuran or oxacyclopentane .

I hope this helps.

Dr. Romano

Definitely! Thank you bunches.
 
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OC 269 Destroyer 2017-
I have never seen this reaction to where the carbonyl group adds to the NH2 attached to a benzene ring. Can you please explain this reaction and does it only occur with phenylamine + carbonyls or is it seen with other groups too? I was assuming NH2 would just act as an ortho/para director to the Carbonyl.
 
OC 269 Destroyer 2017-
I have never seen this reaction to where the carbonyl group adds to the NH2 attached to a benzene ring. Can you please explain this reaction and does it only occur with phenylamine + carbonyls or is it seen with other groups too? I was assuming NH2 would just act as an ortho/para director to the Carbonyl.


I had a bit of trouble as well, but I quickly realized after going over my nuc acyl sub reactions that the amine is basically adding to an acid chloride. And an acid chloride is SUPER reactive esp to the amide form (think amide formation). Additionally, and I may be wrong here so do not quote me on it, but to add the acyl group in a benzene reaction, you would need a LA catalyst. The AlCl3 in these reactions (under benzene) is to catalyze (so it takes the Cl). For the question, there was no such catalyst and thus another reason to why it failed to add at the O/P position...

@orgoman22 Can you confirm this?

@Dallas Dentist Let me know if that helps!
 
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I had a bit of trouble as well, but I quickly realized after going over my nuc acyl sub reactions that the amine is basically adding to an acid chloride. And an acid chloride is SUPER reactive esp to the amide form (think amide formation). Additionally, and I may be wrong here so do not quote me on it, but to add the acyl group in a benzene reaction, you would need a LA catalyst. The AlCl3 in these reactions (under benzene) is to catalyze (so it takes the Cl). For the question, there was no such catalyst and thus another reason to why it failed to add at the O/P position...

@orgoman22 Can you confirm this?

@Dallas Dentist Let me know if that helps!
This is a very important reaction. In practice, the NH2 group is very reactive and must be partially deactivated. The NH2 group by itself would not work well with a Friedel Crafts reaction. Note that this reaction is simply a nucleophilic acyl substitution reaction. No Lewis acid catalyst was employed, and is NOT a Friedel Crafts reaction ! By reacting it as shown in the problem,this allowed us to form the amide, which although is an o/p director, it is not nearly as reactive as the amino group.

Hope this helps.

Dr. Romano
 
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Dr. Romano,

I have two questions that came up on OChem back to back.

Q303: I have never seen a Cl2 and water add to the ring. What is this reaction?

Q304: I was under the assumption that this was a wolf-kitchner reaction (NH2NH2), wouldn't this rid the reactant of carbonyls, not turn it into an amine?
 
Dr. Romano,

I have two questions that came up on OChem back to back.

Q303: I have never seen a Cl2 and water add to the ring. What is this reaction?

Q304: I was under the assumption that this was a wolf-kitchner reaction (NH2NH2), wouldn't this rid the reactant of carbonyls, not turn it into an amine?
For 303... In this reaction, we first reduced the nitro group into the SUPER activating group NH2. The next reaction is simply a chlorination. Normally we use FeCl3, or a suitable Lewis acid catalyst, but none is even needed, water will suffice due to the high reactivity of the amine !!! !!! I have done this reaction many times, and water works just as well here since the NH2 is such a super activator.

For 304,,,,,Look closer.....first you did the SN2. then hydrolyzed the ring......we call this the Gabriel reaction. You should instantly recognize this reaction since we get the primary amine with either acid hydrolysis or treatment with hydrazine. The Wolfe Kishner reaction is primarily used for aldehyde and ketone reduction. This compound is an IMIDE......and would not reduce with a Wolfe Kishner. The bottom line is this.....You need to study and recognize the Gabriel synthesis of primary amines !!!

Hope this helps

Dr. Romano
 
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@orgoman22 ,

For one of the Bio Questions, I think it was 99 regarding a marathon runner and the osteoblast activity... I am just a bit confused. Would decreasing calcium reabsorption in bones would lead to an increase in serum calcium levels. This sounds counterintuitive to me, because if you decrease Ca reabsorption in bone by decreasing osteoclast activity, then wouldn't this lead to a lower Ca count in the blood?

Thank you in advance.
 
It is important you complete all the work in the DAT Destroyer, do not follow any study guide that instructs you not to complete the work, we are an independent company and we do not endorse any company that does so . We work with students in our classroom year round and we complete every single problem in the Destroyer.
 
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Transcription start site, is where first RNA nucleotide binds. That is usually downstream (after the promoter site). Promoter site is used to guide RNA polymerase to the correct starting site, promoter is not used to initiate transcription.

Hope this helps.

If the promotor is not used to initiate transcription, then is RNA polymerase responsible for initiating transcription?
 
Hello Dr. Romano,

Questions 4 and 5, 2017 MATH Destroyer. So, after doing the problem, i saw that we are not allow to find the distance with just the simple difference between the points. meaning (2,4) and (6,2) difference between X= 4 and Y=2. Area=4x2 which is wrong. Can you explain why we are not allowed to do this.

After reviewing my answers, i figured out how to do question #4 using the distance formula and getting the right answer. I tried doing the same with question 5, but if I followed the same procedure as question 4, I would have to divide 74 by 2 [ 2s^2=74]. Then giving me s^2=34. but the answer says Area=74. I have attached a picture of how i did both problems following the procedure you wrote in q#4. Can you help me figure problem 5.

Thank you for your help.
 

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Hello Dr. Romano,

Questions 4 and 5, 2017 MATH Destroyer. So, after doing the problem, i saw that we are not allow to find the distance with just the simple difference between the points. meaning (2,4) and (6,2) difference between X= 4 and Y=2. Area=4x2 which is wrong. Can you explain why we are not allowed to do this.

After reviewing my answers, i figured out how to do question #4 using the distance formula and getting the right answer. I tried doing the same with question 5, but if I followed the same procedure as question 4, I would have to divide 74 by 2 [ 2s^2=74]. Then giving me s^2=34. but the answer says Area=74. I have attached a picture of how i did both problems following the procedure you wrote in q#4. Can you help me figure problem 5.

Thank you for your help.
For Question #5 it gave the end points of the side not the diagonal like in question #4
 
Dr. Romano,

For question 349 the actual values are 2.7/5.1 which give a percentage of 52, however you rounded it to 3/5 and got a percentage of 60. Would something like this pop up on the test? Or do they use more precise numbers?
 
Does anyone have a formula sheet for Gen chem? I'm using Mike's videos, and there are quite a few formulas that he doesn't go over and some he ~~does (kinda), but rather explains the concept than explain how to use the formula or even give the formula. (not saying the videos are bad, they're pretty good) Just thought is Dr. Romano, or anyone has a compiled list of gen chem formulas.

Also, the ADA has included new questions on QR. I was hoping someone could pass the along the file Dr. Romano/Nancy uploaded that has some Quant. comparison questions. I couldn't find them in this discussion.

Thanks you for your help!
 
Does anyone have a formula sheet for Gen chem? I'm using Mike's videos, and there are quite a few formulas that he doesn't go over and some he ~~does (kinda), but rather explains the concept than explain how to use the formula or even give the formula. (not saying the videos are bad, they're pretty good) Just thought is Dr. Romano, or anyone has a compiled list of gen chem formulas.

Also, the ADA has included new questions on QR. I was hoping someone could pass the along the file Dr. Romano/Nancy uploaded that has some Quant. comparison questions. I couldn't find them in this discussion.

Thanks you for your help!
 

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Dr. Romano,

For gen chem question 362 : Do we not include stereocenters when counting constitutional isomers?
 
Hello Dr Romano,

Please could you help me with 2 questions... #194 Orgo Chem from the DAT Destroyer 2017, why was the proton that was attacked most acidic? I thought usually LDA attacks an alpha hydrogen when there is a carbonyl carbon, but there wasn't any carbonyls in this example.

Question 2 is from #205 how can you determine correct molecule was meso by working out the R/S configurations?
 
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Hello Dr Romano,

Please could you help me with 2 questions... #194 Orgo Chem from the DAT Destroyer 2017, why was the proton that was attacked most acidic? I thought usually LDA attacks an alpha hydrogen when there is a carbonyl carbon, but there wasn't any carbonyls in this example.

Question 2 is from #205 how can you determine correct molecule was meso by working out the R/S configurations?
#194
No.....you learned incorrectly. LDA is a super powerful base, but a poor nucleophile. It will remove an acidic proton. The proton need not involve a carbonyl group. Here, removal of a proton yields a HIGHLY STABILIZED anion ,,,,,this is what really counts !!!! I hope this helps.
 
Hello Dr Romano,

Please could you help me with 2 questions... #194 Orgo Chem from the DAT Destroyer 2017, why was the proton that was attacked most acidic? I thought usually LDA attacks an alpha hydrogen when there is a carbonyl carbon, but there wasn't any carbonyls in this example.

Question 2 is from #205 how can you determine correct molecule was meso by working out the R/S configurations?
205
If half the molecule is R.....and the other half is S...and..... each side has the SAME groups....it is likely a MESO compound !!!! Half the molecule will rotate polarized light clockwise...half will rotate it counterclockwise thus gives a zero rotation meso compound. Hope this helps.
 
205
If half the molecule is R.....and the other half is S...and..... each side has the SAME groups....it is likely a MESO compound !!!! Half the molecule will rotate polarized light clockwise...half will rotate it counterclockwise thus gives a zero rotation meso compound. Hope this helps.
Thank you for your help Dr Romano :)
 
Dr. Romano (@orgoman22 ),

This is a question regarding GC 282. So I was thought that if you go from a liquid to a gas, it is considered to be endothermic and thus require heat. Now going by this logic, shouldn't this be an increase in the K.E. because temperature is being increased and thus there will be a higher average kinetic energy of the molecules. And the opposite is true for going from gas to liquid. Now the reasoning from 282 is kind of counterintuitive, such that an exothermic reaction releases heat to the surroundings, and so there is an increase in kinetic energy, but a decrease in potential. So if you are going from a gas to a liquid, I agree that the potential energy will decrease because gases have higher potential energies than liquids, but wouldn't there be an overall decrease in kinetic energy? So by the same logic, if a reaction was endothermic, wouldn't you expect a higher average K.E. because it is at a higher temperature?

Thank you in advance for your help!
 
Dr. Romano (@orgoman22 ),

This is a question regarding GC 282. So I was thought that if you go from a liquid to a gas, it is considered to be endothermic and thus require heat. Now going by this logic, shouldn't this be an increase in the K.E. because temperature is being increased and thus there will be a higher average kinetic energy of the molecules. And the opposite is true for going from gas to liquid. Now the reasoning from 282 is kind of counterintuitive, such that an exothermic reaction releases heat to the surroundings, and so there is an increase in kinetic energy, but a decrease in potential. So if you are going from a gas to a liquid, I agree that the potential energy will decrease because gases have higher potential energies than liquids, but wouldn't there be an overall decrease in kinetic energy? So by the same logic, if a reaction was endothermic, wouldn't you expect a higher average K.E. because it is at a higher temperature?

Thank you in advance for your help!
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,

#215 gen chem (dat destroyer 2017):
For this question I got the correct answer but my reasoning is different from the solution you have. I thought a solution of Na2HPO4 would be basic simply because it forms 2NaOH + H3PO4....is this a wrong way to go about answering this question?

Thanks in advance!
 
Dr. Romano,

#215 gen chem (dat destroyer 2017):
For this question I got the correct answer but my reasoning is different from the solution you have. I thought a solution of Na2HPO4 would be basic simply because it forms 2NaOH + H3PO4....is this a wrong way to go about answering this question?

Thanks in advance!
Yes ....Your reasoning has no logic. This is indeed one of the toughest problems in the book. You have to compare the Ka and Kb values. First seeing how it acts if it was an acid,,,,then seeing how it would act as a base. Follow my solution carefully. The DAT will surely be easier, but tackling the tough problems could make the difference between a 21 and a 28 . Hope this helps.
 
Dr. Romano,

#215 gen chem (dat destroyer 2017):
For this question I got the correct answer but my reasoning is different from the solution you have. I thought a solution of Na2HPO4 would be basic simply because it forms 2NaOH + H3PO4....is this a wrong way to go about answering this question?

Thanks in advance!

Keep going and if you have any more questions feel free to post them.

Dr. Romano
 
Is it worrying that I'm only getting about 23/40 on the math destroyer tests? (I'm on test 2) it's very hard! I'm worried for the DAT
 
Hello Dr. Romano, DAT destroyer OCHEM 2017

For problem 178 (looking at answer key) the answer also shows using LiALH4 will reduce the carbonyl (ketone) at the top to an OH. and a carboxylic acid to CH2OH. So, why doesn't it fully convert the ketone at the top to CH2 but instead keeps it an OH?

Edit: i just did #270. so only Wolff-kishner and Clemmensen will fully convert a carbonyl to CH2? LiAlH4 will convert ketones to OH's, and Carboxylic acids to CH2OH? Correct?

Thank you!
 
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Hello Dr. Romano, DAT destroyer OCHEM 2017

For problem 178 (looking at answer key) the answer also shows using LiALH4 will reduce the carbonyl (ketone) at the top to an OH. and a carboxylic acid to CH2OH. So, why doesn't it fully convert the ketone at the top to CH2 but instead keeps it an OH?

Edit: i just did #270. so only Wolff-kishner and Clemmensen will fully convert a carbonyl to CH2? LiAlH4 will convert ketones to OH's, and Carboxylic acids to CH2OH? Correct?

Thank you!
For problem 178. ...There are many reducing agents. Many books are written just dealing with them !!! If you do a Wolfe-Kishner reaction, we take an aldehyde or ketone group and convert it into the CH2 group. The mechanism of both are rather involved and are not needed for the DAT exam. If you use LiAlH4 on and aldehyde or ketone a different mechanism called a hydride transfer is in operation. Aldehydes reduce into primary alcohols, and ketones reduce into secondary alcohols. Also note that nitriles will reduce into primary amines ,esters AND carboxy acids will reduce into primary alcohols. I have shown many examples of this in the DESTROYER book for you to practice. Note that the weaker reducing agent NaBH4 will reduce aldehydes and ketones, but usually does not reduce acids, esters, or nitriles. Keep up the great work.

Hope this helps.
Dr. Jim Romano

"Think 30"
 
Thank you Dr. Romano.

For #306 OChem dat Destroyer 2017,

At carbon 2, H gets 4, Br gets 1 because its highest priority comparing a; C, C, H, and Br. Then, won't COOH get 2nd priority over the R group which has Cl? So, the way I did it in the end it goes clockwise, BUT since H is on the horizontal it converts to S?

isn't O is more electronegative than Cl? (FONCLBrSCH or FONCLBrIdge as some people put it? I've read some people say O>Cl and other Cl>O)
 
Thank you Dr. Romano.

For #306 OChem dat Destroyer 2017,

At carbon 2, H gets 4, Br gets 1 because its highest priority comparing a; C, C, H, and Br. Then, won't COOH get 2nd priority over the R group which has Cl? So, the way I did it in the end it goes clockwise, BUT since H is on the horizontal it converts to S?

isn't O is more electronegative than Cl? (FONCLBrSCH or FONCLBrIdge as some people put it? I've read some people say O>Cl and other Cl>O)
R group with Cl gets higher priority because it has higher atomic mass. In the Fischer projection as drawn with R group getting priority 2, it would be 3S. However, since the H is projecting out of the page we would reverse the configuration to 3R. I hope that is helpful :)
 
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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!
 
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!
 
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