Official DAT Destroyer Q&A Thread

[densaugeo]

---

Members don't see this ad.

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!

 

[gangazi]

dat destroyer 2016

orgo 188) I have question about acidity... so LDA deprotonates the most acidic proton... I was wondering e- delocalizing effects of Ph... In this case Ph delocalizes the middle H, making it more stable. However, in aromatic electrophilic substituion, Ph was EDG rather than EWG b/c it would not delocalize so that it wouldnt break the aromaticity... How are two concepts different in this case?

orgo #249) says a compound is 70% optically pure. R .. and I remember ANOTHER question which says that sample is 10g and the compound is 20% excess S. The calculation seem to be the same...
Does the "20% excess S" mean "20% optically pure S?"


orgo #264) When NaHCO3 is added, why doesnt phenol also found in the aqueous layer? It seems like HCO3- is acting as base to bring out carboxylic acid...

orgo #290) What is the name of the mechanism that does elimination of N2 in the last step?

orgo #283) Why does it go rearrangement? and how to recognize this on the exam?

Thank you orgoman

 

[dentalstudent2021]

249. yes calculation will be similar. since the product will have 70% R and then the rest of the 30% will be R+S (15% each). therefore the product will have 85% R and 15% S.
264. this is purely memorization. NaHCO3 or NaOH will remove carboxylic acids. NaOH will remove phenols and HCl will remove amines.

 

[gangazi]

249. yes calculation will be similar. since the product will have 70% R and then the rest of the 30% will be R+S (15% each). therefore the product will have 85% R and 15% S.
264. this is purely memorization. NaHCO3 or NaOH will remove carboxylic acids. NaOH will remove phenols and HCl will remove amines.

thank you! i will add that to my flash cards!

 

[gangazi]

249. yes calculation will be similar. since the product will have 70% R and then the rest of the 30% will be R+S (15% each). therefore the product will have 85% R and 15% S.
264. this is purely memorization. NaHCO3 or NaOH will remove carboxylic acids. NaOH will remove phenols and HCl will remove amines.

do you know where i could get more info on this though? separating techniques...

 

[dentalstudent2021]

do you know where i could get more info on this though? separating techniques...

those are the 3 you need to know for acid base extraction for the DAT. all techniques are on Chads organic handout from his coursesaver site.

 

[DAT Destroyer]

dat destroyer 2016

orgo 188) I have question about acidity... so LDA deprotonates the most acidic proton... I was wondering e- delocalizing effects of Ph... In this case Ph delocalizes the middle H, making it more stable. However, in aromatic electrophilic substituion, Ph was EDG rather than EWG b/c it would not delocalize so that it wouldnt break the aromaticity... How are two concepts different in this case?

orgo #249) says a compound is 70% optically pure. R .. and I remember ANOTHER question which says that sample is 10g and the compound is 20% excess S. The calculation seem to be the same...
Does the "20% excess S" mean "20% optically pure S?"


orgo #264) When NaHCO3 is added, why doesnt phenol also found in the aqueous layer? It seems like HCO3- is acting as base to bring out carboxylic acid...

orgo #290) What is the name of the mechanism that does elimination of N2 in the last step?

orgo #283) Why does it go rearrangement? and how to recognize this on the exam?

Thank you orgoman

Orgo 188

When LDA removes a proton, we form a carbanion. This negative charge can be spread out due to the removal of electron density by the phenyl group. In this sense, it is electron withdrawing and aids in anion stabilization. In Aromatic reactions in which we see Electrophilic Substitutions, a phenyl group can donate some electron density to aid in stabilization. As you can see, the difference lies in the INTERMEDIATE....carbanion vs. carbocation.

Dr. Romano

 

[farzi2492]

Math Destroyer Test 3 Q 8: Why does theta equal 210 and 330 instead of 240 and 330 ?

 

[DAT Destroyer]

dat destroyer 2016

orgo 188) I have question about acidity... so LDA deprotonates the most acidic proton... I was wondering e- delocalizing effects of Ph... In this case Ph delocalizes the middle H, making it more stable. However, in aromatic electrophilic substituion, Ph was EDG rather than EWG b/c it would not delocalize so that it wouldnt break the aromaticity... How are two concepts different in this case?

orgo #249) says a compound is 70% optically pure. R .. and I remember ANOTHER question which says that sample is 10g and the compound is 20% excess S. The calculation seem to be the same...
Does the "20% excess S" mean "20% optically pure S?"


orgo #264) When NaHCO3 is added, why doesnt phenol also found in the aqueous layer? It seems like HCO3- is acting as base to bring out carboxylic acid...

orgo #290) What is the name of the mechanism that does elimination of N2 in the last step?

orgo #283) Why does it go rearrangement? and how to recognize this on the exam?

Thank you orgoman

249. I am not sure what question you are referring to exactly.....but 70% excess S means that the other 30% is racemic,,,,,,,thus we have 85% S and 15% R. If we had 10g of a compound,,,,,we have 1.5 g of R isomer.

 

[DAT Destroyer]

Math Destroyer Test 3 Q 8: Why does theta equal 210 and 330 instead of 240 and 330 ?

farzi2492
my apologies - it is actually question 5 test 3

Test 3 question 5

Sin(240) = - sin(240-180) = - sin (60) = - sqrt(3)/2 = -0.866
Sin(210) = - sin (210-180) = - sin(30) = - 0.5

 

[DAT Destroyer]

dat destroyer 2016

orgo 188) I have question about acidity... so LDA deprotonates the most acidic proton... I was wondering e- delocalizing effects of Ph... In this case Ph delocalizes the middle H, making it more stable. However, in aromatic electrophilic substituion, Ph was EDG rather than EWG b/c it would not delocalize so that it wouldnt break the aromaticity... How are two concepts different in this case?

orgo #249) says a compound is 70% optically pure. R .. and I remember ANOTHER question which says that sample is 10g and the compound is 20% excess S. The calculation seem to be the same...
Does the "20% excess S" mean "20% optically pure S?"


orgo #264) When NaHCO3 is added, why doesnt phenol also found in the aqueous layer? It seems like HCO3- is acting as base to bring out carboxylic acid...

orgo #290) What is the name of the mechanism that does elimination of N2 in the last step?

orgo #283) Why does it go rearrangement? and how to recognize this on the exam?

Thank you orgoman

264.

NaHCO3 will not react well with a phenol. The phenol is NOT acidic enough and will result in an UNFAVORABLE equilibrium. For an exercise, draw out the reaction.....Pka of phenol is 10......the Pka of the H2C03 formed is about 6 ,, hence reaction favors the phenol. Bottom line....when doing an extraction......for acids use KOH, NaOH, or NaHC03....for phenols , use NaOH or KOH.

Hope this helps

Dr. Romano

 

[Ashish]

DAT Destroyer Gen Chem #341. Can anyone please explain how the Iodine is being balanced here? I don't understand why 2e- to balance the charge in the second half of the reaction. Thank you.

 

[DAT Destroyer]

DAT Destroyer Gen Chem #341. Can anyone please explain how the Iodine is being balanced here? I don't understand why 2e- to balance the charge in the second half of the reaction. Thank you.

When balancing a Redox reaction, one must balance BOTH mass and charge. First we put a 3 to the left side to balance the masses of the Iodine. Now....the left side has a minus 3 charge and the right side has a minus 1 charge . A nice little trick I teach my students is to go to the side that is MORE POSITIVE.....this will be the side that you need to add the electrons to. Thus, the more positive side ( In this case, the less negative ), is on the right side since it is negative one. Thus, we add 2 electrons to this. Now what do you see ? Hopefully you see that both sides have Minus 3 charges and Iodines are balanced.

Hope this helps.

Dr. Romano

 

[dentalstudent2021]

how are these 2 questions different types of question ? I calculated both the same way, but question 2 is not done this way..

1. Probability of rain is 30%. Probability of Martha's flight delay is 25%. Probability of both these events happening is 6%. What is the probability of either rain or delay.
- for this I do understand do add both probabilities and subtract probabilities of both events.
2. 60% of the gecko population is spotted. 57% is green. 17% are both spotted and green. What is the probability of choosing either green or spotted gecko's ?
- I did the same steps as the above questions and obviously got it wrong.

I don't get how they are different questions. Thanks.

 

[DAT Destroyer]

how are these 2 questions different types of question ? I calculated both the same way, but question 2 is not done this way..

1. Probability of rain is 30%. Probability of Martha's flight delay is 25%. Probability of both these events happening is 6%. What is the probability of either rain or delay.
- for this I do understand do add both probabilities and subtract probabilities of both events.
2. 60% of the gecko population is spotted. 57% is green. 17% are both spotted and green. What is the probability of choosing either green or spotted gecko's ?
- I did the same steps as the above questions and obviously got it wrong.

I don't get how they are different questions. Thanks.

These are not the same problem. Read the question carefully. In the first it states: what is the probability that either it rains or Martha's flight will be delayed. In the second one the question States: what is the probability that it belongs to either Green or Spotted BUT NOT BOTH. For the last one use a Venn Diagram and exclude the intersection. See solution in the back of the book

Hope this helps!

 

[Watergrizzly]

Gchem #106

I just want to make sure I'm crystal clear on this topic and #106 best illustrates my slight confusion. I understand that ONLY temp can affect the K value of a reaction. However, Knew vs Koriginal represents two different reactions correct? This leads to the modification of an already known K value? However, the Kvalue isn't technically changed? So if I were to increase the temp of an endothermic reaction it would increase the K value due to the greater amount of resulting products at equilibrium? Thank you in advance!

 

[dentalstudent2021]

Gchem #106

I just want to make sure I'm crystal clear on this topic and #106 best illustrates my slight confusion. I understand that ONLY temp can affect the K value of a reaction. However, Knew vs Koriginal represents two different reactions correct? This leads to the modification of an already known K value? However, the Kvalue isn't technically changed? So if I were to increase the temp of an endothermic reaction it would increase the K value due to the greater amount of resulting products at equilibrium? Thank you in advance!

could you post the question ?

 

[gangazi]

Gchem #106

I just want to make sure I'm crystal clear on this topic and #106 best illustrates my slight confusion. I understand that ONLY temp can affect the K value of a reaction. However, Knew vs Koriginal represents two different reactions correct? This leads to the modification of an already known K value? However, the Kvalue isn't technically changed? So if I were to increase the temp of an endothermic reaction it would increase the K value due to the greater amount of resulting products at equilibrium? Thank you in advance!

When you add heat to the endothermic rxn., the reaction shifts to right... And Knew in your case would be Q value

 

[DAT Destroyer]

Gchem #106

I just want to make sure I'm crystal clear on this topic and #106 best illustrates my slight confusion. I understand that ONLY temp can affect the K value of a reaction. However, Knew vs Koriginal represents two different reactions correct? This leads to the modification of an already known K value? However, the Kvalue isn't technically changed? So if I were to increase the temp of an endothermic reaction it would increase the K value due to the greater amount of resulting products at equilibrium? Thank you in advance!

I love your reasoning. You have an excellent grasp of the topic.When finding a Keq,, this value applies for a reaction exactly as written. If you change the phases of any component such as to , double all of the stoichiometric terms, then it's a new reaction and thus a new Keq.

Hope this helps.

Dr. Romano

 

[Watergrizzly]

I love your reasoning. You have an excellent grasp of the topic.When finding a Keq,, this value applies for a reaction exactly as written. If you change the phases of any component such as to , double all of the stoichiometric terms, then it's a new reaction and thus a new Keq.

Hope this helps.

Dr. Romano

Perfect! Thank you so much! Definitely a welcome reassurance.

 

[Ashish]

Organic Chem # 161. What is the reasoning for the switch in atoms while converting from Newman Projection to Fisher's projection? When I draw out the Fisher's projection by mentally visualizing the Newman projection from right side, both appear identical (i.e Chlorine away from the plane of paper and H towards me which would translate that the molecules are identical?

Thanks!

 

[DAT Destroyer]

Organic Chem # 161. What is the reasoning for the switch in atoms while converting from Newman Projection to Fisher's projection? When I draw out the Fisher's projection by mentally visualizing the Newman projection from right side, both appear identical (i.e Chlorine away from the plane of paper and H towards me which would translate that the molecules are identical?

Thanks!

I will assume that you understand how to do R/S from a Fischer projection. Now when we convert it to a Newman......we MUST do a switch. If viewing from the from the front as I highlighted on the problem, the round " ball" carbon has groups attached to it. These groups are horizontally positioned....yes ? A horizontal group means it is coming TOWARDS you......this needs to be viewed AWAY from you by the CIP rules. Thus we simply make the switch. Doing the switch is indeed easier than explaining it without the use of a model. Hope this helps.....If not......The David Klein books has nice graphics that might make it clearer.

Dr. Romano

 

[FutureDent20]

2016 O-chem #70:
So I get why SOCl2, pyridine works best for this reaction. But I thought HCl with primary alcohol undergoes SN2 reaction? or is this an exception when we have a extensive B-branching we get rearranged product?

 

[DAT Destroyer]

2016 O-chem #70:
So I get why SOCl2, pyridine works best for this reaction. But I thought HCl with primary alcohol undergoes SN2 reaction? or is this an exception when we have a extensive B-branching we get rearranged product?

That is correct......SOCl2 works nicely on primary and secondary alcohols with no rearrangement. Using HCl or HBr can cause rearrangements. It can be an organic chemists worst nightmare. LOL....yes,,,,,beta branching can be the kiss of death. Books teach that a primary alcohol does not rearrange...... That is not always true. If beta branching is present, we protonate and SIMULTANEOUSLY do the shift to form a carbocation. The Carey and Klein text book are one of the few texts to show this. You are good.....Yes....primary alcohols do SN2 most of the time. Hope you enjoyed this exception !!! In Organic Chemistry....like dentistry. and medicine , .exceptions . abound !!!!

Dr. Romano

 

[FutureDent20]

That is correct......SOCl2 works nicely on primary and secondary alcohols with no rearrangement. Using HCl or HBr can cause rearrangements. It can be an organic chemists worst nightmare. LOL....yes,,,,,beta branching can be the kiss of death. Books teach that a primary alcohol does not rearrange...... That is not always true. If beta branching is present, we protonate and SIMULTANEOUSLY do the shift to form a carbocation. The Carey and Klein text book are one of the few texts to show this. You are good.....Yes....primary alcohols do SN2 most of the time. Hope you enjoyed this exception !!! In Organic Chemistry....like dentistry. and medicine , .exceptions . abound !!!!

Dr. Romano

Again and again, Thank you so much for your time and great explanations!! oo Trust me Dr. Romano, my exception list is getting longer and longer, BUT it's a good thing, I am getting more and more confident about this section (Thanks to DAT destroyer and organic odyssey).

 

[workofshart]

#75 of the QR problems... I'm having trouble understanding the concept that sees ln|sin / cos| = ln|sin| - ln|cos| ?

I get the first part regarding tan but I'm lost at the answer.

 

[gangazi]

DAT Destroyer Biology 2016 #72 explanation for d... "ions diffuse across membranes down their concentration gradients" -- I thought the ions needed help of carrier protein in order to diffuse across membrane

 

[gangazi]

#75 of the QR problems... I'm having trouble understanding the concept that sees ln|sin / cos| = ln|sin| - ln|cos| ?

I get the first part regarding tan but I'm lost at the answer.

http://www.purplemath.com/modules/logrules.htm
basically Log (anything / something) = log (anything) - log (something)

 

[gangazi]

2016 DAT BIO #99) I thought both aldosterone and vasopressin affect water reabsorption... So I thought both decrease in aldosterone and a decrease in vasopressin would result in the increased urine production. Am I missing something? Thank you orgoman!

 

[nextgendental]

@ gangazi I'm pretty sure vasopressin (ADH) is inhibited by alcohol. Which results in less reabsorption of water and increased urine. Aldosterone is involved in regulating salt. So when salt is reabsorbed, the salt pulls the water with it. So aldosterone is unrelated in this case. Pretty sure.


Sent from my iPhone using SDN mobile

 

[DAT Destroyer]

#75 of the QR problems... I'm having trouble understanding the concept that sees ln|sin / cos| = ln|sin| - ln|cos| ?

I get the first part regarding tan but I'm lost at the answer.

Need to know properties of Log/ Ln
Ln(AB)= ln(A)+ln(B)
Ln(A/B)= ln(A) - ln(B)
That's why ln[sinx/cosx] = ln sinx - ln cosx

 

[DAT Destroyer]

DAT Destroyer Biology 2016 #72 explanation for d... "ions diffuse across membranes down their concentration gradients" -- I thought the ions needed help of carrier protein in order to diffuse across membrane

The explanation for question #72 is correct, ions diffuse down their concentration gradient because they move through the membrane channels. Many ions are small enough to move via channels, not carriers. Bigger molecules like hormones need a carrier protein or a receptor to bind to.

 

[DAT Destroyer]

2016 DAT BIO #99) I thought both aldosterone and vasopressin affect water reabsorption... So I thought both decrease in aldosterone and a decrease in vasopressin would result in the increased urine production. Am I missing something? Thank you orgoman!

For question# 99, Alcohol inhibits Vasopressin, not Aldosterone. That is what will result in increase urine output.
You are correct in your knowledge about both hormones, however it Aldosterone does not apply to this question.

Hope this helps.

 

[FutureDent20]

2016 O-chem #114:
I am trying to make more sense out of this question: For B what does it mean when we have a lone pair like this? does it mean one of the carbons have this lone pair and the other carbon is Sp3 therefore non-aromatic?

 

[gangazi]

@ gangazi I'm pretty sure vasopressin (ADH) is inhibited by alcohol. Which results in less reabsorption of water and increased urine. Aldosterone is involved in regulating salt. So when salt is reabsorbed, the salt pulls the water with it. So aldosterone is unrelated in this case. Pretty sure.


Sent from my iPhone using SDN mobile

thanks !

 

[DAT Destroyer]

2016 O-chem #114:
I am trying to make more sense out of this question: For B what does it mean when we have a lone pair like this? does it mean one of the carbons have this lone pair and the other carbon is Sp3 therefore non-aromatic?

A lone pair is a p-orbital which will allow electron delocalization. An sp3 Carbon, as you pointed to, represents a carbon that has no p orbital. If no available p-orbital, electron delocalization is not possible. Thus, a molecule will NOT display any aromatic behavior.

Hope this helps.

Dr. Romano

 

[Ashish]

General Chemistry #180: From Bronsted Lowry definition of Base (H+ acceptor), wouldn't ClO4- readily accept H+ than HClO4? I totally understand that weak bases are conjugates of strong acids but the question asks when put in water and am I thinking along the line as to which molecule readily behaves as a base when the amphoteric H2O behaves as an acid. Since HClO4 is such a strong acid, wouldn't it tend to react the weakest as a base?
I feel like I am way overthinking this but just wanted to clarify for the peace of mind. Thanks.

 

[dentalstudent2021]

General Chemistry #180: From Bronsted Lowry definition of Base (H+ acceptor), wouldn't ClO4- readily accept H+ than HClO4? I totally understand that weak bases are conjugates of strong acids but the question asks when put in water and am I thinking along the line as to which molecule readily behaves as a base when the amphoteric H2O behaves as an acid. Since HClO4 is such a strong acid, wouldn't it tend to react the weakest as a base?
I feel like I am way overthinking this but just wanted to clarify for the peace of mind. Thanks.

This one got me as well. the reason is that, it's a SA (strong electrolyte) which means that it would just donate its H+ rather than accepting another one. on the other hand ClO4- (the answer) is the CB of a SA thus a much weaker base compared to the other options

 

[DAT Destroyer]

General Chemistry #180: From Bronsted Lowry definition of Base (H+ acceptor), wouldn't ClO4- readily accept H+ than HClO4? I totally understand that weak bases are conjugates of strong acids but the question asks when put in water and am I thinking along the line as to which molecule readily behaves as a base when the amphoteric H2O behaves as an acid. Since HClO4 is such a strong acid, wouldn't it tend to react the weakest as a base?
I feel like I am way overthinking this but just wanted to clarify for the peace of mind. Thanks.

Do not overthink this question. HClO4 is an acid,,,,,and NEVER a base. ClO4- is indeed a base, and since it comes from one of the strongest acids to be seen.. and would be a hopelessly weak base !!!

Dr. Romano

 

[gangazi]

For question# 99, Alcohol inhibits Vasopressin, not Aldosterone. That is what will result in increase urine output.
You are correct in your knowledge about both hormones, however it Aldosterone does not apply to this question.

Hope this helps.

okay thank you! I hope your post pops up into my head when they give me both answer choices!!

 

[Ashish]

GC #232: When Ammonium does hydrolysis to form NH3, is n't saying that the Ammonium ion is not soluble in water? Simply going by the fact that salts of Ammonium are generally soluble and thus are spectator ions. Thanks!

 

[DAT Destroyer]

GC #232: When Ammonium does hydrolysis to form NH3, is n't saying that the Ammonium ion is not soluble in water? Simply going by the fact that salts of Ammonium are generally soluble and thus are spectator ions. Thanks!

Ammonia is remarkably soluble in water. Ammonia reacts with water to form the ammonium ion. Without this reaction, life would be impossible on Earth !! Compounds containing Group 1 metals and ammonium compounds are ALL water soluble. Review any standard General Chem text for further clarity.

Hope this helps

Dr. Romano

 

[rak173]

2016 dat destroyer OC #58: It says in the answer explanation that structure I is chiral; the Cl is attached to CH2. How come this structure is chiral? Thanks.

 

[DAT Destroyer]

2016 dat destroyer OC #58: It says in the answer explanation that structure I is chiral; the Cl is attached to CH2. How come this structure is chiral? Thanks.

You were bagged !!! Look closer.... you missed the trick. I is indeed chiral......look at the second carbon , this is where the chirality was generated. Just because the Chlorine lands at a certain position does NOT automatically make the exact position chiral. Chlorination at C1 made C2 chiral. No...I and 4 are not the same.....simply name them and you will clearly see they are different. This is an IMPORTANT problem.

Hope this helps

Dr. Romano

 

[FutureDent20]

2016 O-chem destroyer #191:
For answer choice B, I don't understand why we do the second switch? I thought the point was to put the H on dash therefore we need to replace it with CH3. Now why CH3 and Br are switching places?

 

[DAT Destroyer]

2016 O-chem destroyer #191:
For answer choice B, I don't understand why we do the second switch? I thought the point was to put the H on dash therefore we need to replace it with CH3. Now why CH3 and Br are switching places?

When a single " switch " is made, you create the ENANTIOMER........thus one additional " switch " is needed to get back to the equivalent structure !!!

Hope this helps

Dr. Romano

 

[rdub]

I have a question in Math Destroyer 2016...Test1 #24

There is no explanation for the answer. Can someone explain how to do this?

 

[dentalstudent2021]

I have a question in Math Destroyer 2016...Test1 #24

There is no explanation for the answer. Can someone explain how to do this?

could you write out the question here ?

 

[Likkriue]

A graph for any absolute value expression will always have a "V" like shape because it will take into account both the negative and positive value for x-4. And as the shaded portion is UNDER the line, this means Y should be less than it. Im no math wiz but im sure a pre-calc textbook would offer a better explanation.

 

[dentalstudent2021]

answer E ? basically testing you on the absolute value graph. a y =|x| graph would be dead on the (0,0) point. so here its shifted to the RIGHT, so it would be y = |x-4| ....now its asking about the shaded area which is under our graph...so you would simply write y < or = |x-4|

 

[FutureDent20]

When a single " switch " is made, you create the ENANTIOMER........thus one additional " switch " is needed to get back to the equivalent structure !!!

Hope this helps

Dr. Romano

Thank you for your response Dr. Romano. This does make sense but I am afraid I might make a mistake. What If I think about it as fisher projection? meaning simply turn it into fisher projection and then assign R/S as follow:
Is this a correct way or I am just making things more complicated for myself?