Official DAT Destroyer Q&A Thread

[densaugeo]

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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[uhds]

For 27. Consider the Kw @25 degrees Celsius =[H+][OH-] both of which are 1*10-7. Notice that this value (for the H+ concentration) is FAR GREATER than the given [h+] of the HCl solution. Thus, this is the value that you would use to find the pH. It is BECAUSE the concentration of the solution is too dilute.
For 30 - Not quite. Here, we are looking at an isoelectronic species (with the exception of F-). All this means, is that the species with the greater negative charge has the larger radius.

Thank you so much for your help!

 

[1229ko]

Dr. Romano,
Math Destroyer test 2 #18.. why isn't the answer 1/square root 5??

 

[Pearl E. White]

Dr. Romano,
Math Destroyer test 2 #18.. why isn't the answer 1/square root 5??

It is! But you need to rationalize the denominator! 1/sqrt5 *sqrt5/sqrt5 = sqrt5/5. I hope that helps!

 

[Pearl E. White]

Hello Dr Romano,

for Q2 on the Math Destroyer (2017) test 5... If Ted is mowing the lawn in 2 hours and thats equaly to 2/5 of the lawn, then eliot mows 3/5 of the lawn in 4hr 48mins, why arent we adding the two times together (i.e 2 hours + 2hr 48 mins) im really struggling to understand this even after reading the solutions!

You should add the times together. The answer is 4 hours and 48 minutes. I'm not sure how you came up with the other numbers though. Do you mind explaining more?

 

[deleted768895]

You should add the times together. The answer is 4 hours and 48 minutes. I'm not sure how you came up with the other numbers though. Do you mind explaining more?

Well it says Ted does 2/5 of the lawn in 2 hrs, leaving 3/5 of the lawn to be done by elliot who can do 1Lawn/8hrs...so that mean he does 3/5 of the lawn in 4hrs and 48mins. So surely together they do 4hrs 48 + 2hrs? but thats wrong

 

[1229ko]

It is! But you need to rationalize the denominator! 1/sqrt5 *sqrt5/sqrt5 = sqrt5/5. I hope that helps!

Thank you!

 

[Pearl E. White]

Well it says Ted does 2/5 of the lawn in 2 hrs, leaving 3/5 of the lawn to be done by elliot who can do 1Lawn/8hrs...so that mean he does 3/5 of the lawn in 4hrs and 48mins. So surely together they do 4hrs 48 + 2hrs? but thats wrong

Umm...hmm...I am not sure how those numbers came about. I'll explain how I did it and hopefully it will help. They are working together for 2 hours. So, the first step is to find the combined rate and apply that for 2 hours.
1/5 + 1/8 = 1/x
8/40 + 5/40 =13 lawns/40 hours is the combined rate. So multiply by 2 (because they are working together at this rate for 2 hours)
13/40 * 2 = 13/20 lawn is done. That leaves 7/20 of a lawn for Eliot to do by himself. His working rate is given at 1 lawn/8 hours.
7/20 lawn * 8 hours/1 lawn = 14/5 hours = 2 and 4/5 hours. From here, we need to add the additional 2 hours where they were working together which gives us a total of 4 and 4/5 hours of work AKA 4 hours and 48 minutes.

 

[deleted768895]

Umm...hmm...I am not sure how those numbers came about. I'll explain how I did it and hopefully it will help. They are working together for 2 hours. So, the first step is to find the combined rate and apply that for 2 hours.
1/5 + 1/8 = 1/x
8/40 + 5/40 =13 lawns/40 hours is the combined rate. So multiply by 2 (because they are working together at this rate for 2 hours)
13/40 * 2 = 13/20 lawn is done. That leaves 7/20 of a lawn for Eliot to do by himself. His working rate is given at 1 lawn/8 hours.
7/20 lawn * 8 hours/1 lawn = 14/5 hours = 2 and 4/5 hours. From here, we need to add the additional 2 hours where they were working together which gives us a total of 4 and 4/5 hours of work AKA 4 hours and 48 minutes.

i totally understand how you've done it! thank you so much i actually didn't click that they were only working together for 2 hours, I was kinda confused by the wording of the solution.

 

[Wererew]

Dr. Romano,

I am confusing myself with the HNMR values and pka/acidity values. Are they correlated at all, what am I missing here? (Coming from #126 and #130)

Thank you

 

[Pearl E. White]

Dr. Romano,

I am confusing myself with the HNMR values and pka/acidity values. Are they correlated at all, what am I missing here? (Coming from #126 and #130)

Thank you

I would approach these 2 questions separately. DM me if you would like an explanation on attacking these questions.

 

[deleted768895]

Hello Dr Romano

Q311 DAT Destroyer Organic Chem (2017). How do we know whether heating a carboxylic acid leads to CO2 formation, or that it loses an H20 to form a cyclic ring (when its a dicarboxylic acid).? Would heating a dicarboxylic acid not be a decarboxylation?

 

[ColumbiaOrtho]

Hello, I just have one question about ossification process. On problem #99 from Destroyer asks which would be true about the stress being placed on the bones after a marathon runner ran 100 miles in a race. The answer choice A was "Osteoclastic activity increases, and calcium phosphate is removed from the bone". The answer choice B was "Osteoblastic activity increases, and calcium phosphate is released into the matrix". The correct answer was B, but I do not see why answer choice A is wrong. I thought during the ossification, the bone first gets resorbed by osteoclasts, which break down the bone and increases calcium concentration in the blood, and then gets deposited by osteoblasts. Wouldn't putting a consistent force induce an osteoclastic activity first?

 

[Wererew]

OC #140,

Would the order from least to highest frequency be

B<A<E<D<C?

Thank you!

 

[uhds]

Hello,

On 2015 Math Destroyer Test 1 #10:
The area of the garden plus the walkway is π(x+6)^2=π(x^2+12x+36). The area of the garden alone is πx^2, so you subtract that from π(x^2+12x+36) and you get π(12x+36)=12π(x+3), how did you go from π(12x+36) to 12π(x+3)? And why did only x^2 get subtracted and not πx^2?

 

[DAT Destroyer]

Hello Dr Romano

Q311 DAT Destroyer Organic Chem (2017). How do we know whether heating a carboxylic acid leads to CO2 formation, or that it loses an H20 to form a cyclic ring (when its a dicarboxylic acid).? Would heating a dicarboxylic acid not be a decarboxylation?

Here is a good rule of thumb that my professor gave me.......A dicarboxylic acid will lose water to form the cyclic anhydride if it can for a 5 or 6 membered ring. If you can't form a 5 or 6 membered ring, results are often poor. Also,,,,,if you see a beta keto acid, it would decarboxylate. If you understand these 3 facts, you will set for the DAT exam.

Hope this helps.

Dr. Romano

 

[deleted768895]

Here is a good rule of thumb that my professor gave me.......A dicarboxylic acid will lose water to form the cyclic anhydride if it can for a 5 or 6 membered ring. If you can't form a 5 or 6 membered ring, results are often poor. Also,,,,,if you see a beta keto acid, it would decarboxylate. If you understand these 3 facts, you will set for the DAT exam.

Hope this helps.

Dr. Romano

Thank you!! I've made a mental note of this!

 

[DAT Destroyer]

Hello,

On 2015 Math Destroyer Test 1 #10:
The area of the garden plus the walkway is π(x+6)^2=π(x^2+12x+36). The area of the garden alone is πx^2, so you subtract that from π(x^2+12x+36) and you get π(12x+36)=12π(x+3), how did you go from π(12x+36) to 12π(x+3)? And why did only x^2 get subtracted and not πx^2?

To go from π(12x+36) to 12π(x+3) you factor out 12.
πx^2 gets cancelled

 

[DAT Destroyer]

OC #140,

Would the order from least to highest frequency be

B<A<E<D<C?

Thank you!

E would have the lowest,,,,,we would see 1460 for the methylene groups. B would still have sp2 CH stretch around 2900.......Thus,,,,,,E<B<A<D<C.......Focus on the C-H stretches,,,,,,,,alkynes would be around 3300,,,,,alkenes around 3030....a bit less for B. alkanes come in on sp3 carbons around 2950. For the DAT exam, stick to the main functional groups,.

Dr. Romano

 

[uhds]

To go from π(12x+36) to 12π(x+3) you factor out 12.
πx^2 gets cancelled

So the equation is π(x^2+12x+36)-πx^2 to get π(12x+36)? Sorry, I have trouble following this.

 

[DAT Destroyer]

So the equation is π(x^2+12x+36)-πx^2 to get π(12x+36)? Sorry, I have trouble following this.

You have π(x^2+12x+36) - πx^2. Distribute π to get:
πx^2+12xπ+36π-πx^2 = 12xπ+36π = 12π(x+3)

 

[uhds]

You have π(x^2+12x+36) - πx^2. Distribute π to get:
πx^2+12xπ+36π-πx^2 = 12xπ+36π = 12π(x+3)

Got it, thank you so much!

 

[deleted768895]

Hello Dr Romano,

on the DAT Destroyer 2017 book, it mentions that you now offer online tutoring. Where would i be able to access this?
Thanks

 

[deleted768895]

Hello Dr Romano,

I have a general Chemistry question... So with anion/cation atomic radii, i know the general trend is...the more negative the ion the latger the radius, the more positive the ion the small the radius. So what is the trend with isoelectronic series when they have the same # of electrons? Am I correct in thinking it is the proton number that factors the size of the radius? that is, the ion with the largest atomic number will be the smallest?
Thanks!

 

[Pearl E. White]

Hello Dr Romano,

I have a general Chemistry question... So with anion/cation atomic radii, i know the general trend is...the more negative the ion the latger the radius, the more positive the ion the small the radius. So what is the trend with isoelectronic series when they have the same # of electrons? Am I correct in thinking it is the proton number that factors the size of the radius? that is, the ion with the largest atomic number will be the smallest?
Thanks!

In an isoelectronic series, the atom with the most negative charge will have the larger radius.

 

[DAT Destroyer]

Hello Dr Romano,

I have a general Chemistry question... So with anion/cation atomic radii, i know the general trend is...the more negative the ion the latger the radius, the more positive the ion the small the radius. So what is the trend with isoelectronic series when they have the same # of electrons? Am I correct in thinking it is the proton number that factors the size of the radius? that is, the ion with the largest atomic number will be the smallest?
Thanks!

For isoelctronic species such as F-, Ne, you have to calculate what is known as the effective nuclear charge. I have shown this in problem # 364 in our 2017 book. The greater the nuclear charge,,,,the smaller the specie. Great question. If done correctly, F- will be be +7 and Ne would be +8...thus Ne would be smaller. I looked up the literature values.....F- was 119pm, while Ne was 69pm.

Hope this helps.

Dr. Romano

 

[deleted768895]

For isoelctronic species such as F-, Ne, you have to calculate what is known as the effective nuclear charge. I have shown this in problem # 364 in our 2017 book. The greater the nuclear charge,,,,the smaller the specie. Great question. If done correctly, F- will be be +7 and Ne would be +8...thus Ne would be smaller. I looked up the literature values.....F- was 119pm, while Ne was 69pm.

Hope this helps.

Dr. Romano

That makes a lot of sense, thanks!! But in the converse senario, when you just have a bunch of ions with different numbers of electrons and different charges, the general rule is more negative = larger ion, more positive = small ion?

 

[uhds]

Hello,

I have a question about problem 33 in the 2015 DAT Destroyer, when it says rate=K[Br2]^3[H2O]^3 is this another type of formula that is not rate law?

Edit: I forgot to add this was for the general chemistry section.

 

[DAT Destroyer]

That makes a lot of sense, thanks!! But in the converse senario, when you just have a bunch of ions with different numbers of electrons and different charges, the general rule is more negative = larger ion, more positive = small ion?

That is correct......You got it !!!! Keep up the great work.

 

[dl2501]

Dr. Romano,

For OC106 regarding chromatography, I understand the answer because I remember the order from several sources, but why exactly is an ester less polar than an aldehyde or ketone? Without memorizing the order, I would suppose that ester be more polar due to the presence of two oxygen, which can be in resonance.

Also, I was wondering for the same question, if B would elute before E because B only has one OH group.... Would that make sense?

Thank you.

 

[DAT Destroyer]

Dr. Romano,

For OC106 regarding chromatography, I understand the answer because I remember the order from several sources, but why exactly is an ester less polar than an aldehyde or ketone? Without memorizing the order, I would suppose that ester be more polar due to the presence of two oxygen, which can be in resonance.

Also, I was wondering for the same question, if B would elute before E because B only has one OH group.... Would that make sense?

Thank you.

The ester is able to assume a conformation that allows partial cancellation of its dipole,,,,,This is not something that is obvious, so you must rely on experimental evidence. Choice E is more polar than B...thus we would expect a slower elution rate. Thus, yes, you are correct that B would indeed elute ahead of E.

I hope this helps.

Dr. Romano

 

[dl2501]

The ester is able to assume a conformation that allows partial cancellation of its dipole,,,,,This is not something that is obvious, so you must rely on experimental evidence. Choice E is more polar than B...thus we would expect a slower elution rate. Thus, yes, you are correct that B would indeed elute ahead of E.

I hope this helps.

Dr. Romano




View attachment 222479

Thank you so much, I totally missed the positive positive as well.

 

[dl2501]

Dr Romano,

I am having a bit of a hard time understanding the GC197 2017. NaCl is ionic and it will fully dissociate if it is put in aqueous solution into Na+ and Cl-. If the question is asking which is the correct order of NaCl from lowest to highest concentration, would it not imply that because NaCl does not fully dissociate in non-polar solvents like hexane, that it, hexane, would have the highest concentration?

Thank you for your help.

 

[deleted768895]

Dr Romano,

I am having a bit of a hard time understanding the GC197 2017. NaCl is ionic and it will fully dissociate if it is put in aqueous solution into Na+ and Cl-. If the question is asking which is the correct order of NaCl from lowest to highest concentration, would it not imply that because NaCl does not fully dissociate in non-polar solvents like hexane, that it, hexane, would have the highest concentration?

Thank you for your help.

I think the question is just asking which order NaCl would be most soluble in

 

[deleted768895]

Hello Dr Romano,

For question 316 DAT Destroyer Organic Chem (2017) How can you tell from the molecular formula that its a fatty acid?

 

[dl2501]

Hello Dr Romano,

For question 316 DAT Destroyer Organic Chem (2017) How can you tell from the molecular formula that its a fatty acid?

Yeah, I think that makes sense. I was just getting a but tripped up about the wording, which seemed weird to me.

But to quickly answer your question, a terpene only has C, steroids (and correct me if I am wrong) only has C's (cholesterol - a derivative of steroid, has OH), anhydride always have at least 3Os (2 carbonyl with O in the middle), and proteins have at least 1 N in it (refer to the backbone amino and COOH group. Fatty acids are carboxylic acids at the heart of it (and when it is with a glycerol, I am pretty sure they are considered esters) (waxes are fatty acid-like but they are esters with OH (this may be a bit mroe info). So just by definition you can tell that because fatty acids are COOH derivatives, it can pertain to the formula. But just to double check, you can apply the DOU formula ((2C+2-H-X)/2) (X=halogen NOT O, do not count O in this formula (and N count as -1)). So (2(14)+2-28)/2 =1. There is 1 degree of unsaturation which you get in the carbonyl of a carboxylic acid. And just to add on, and totally unnecessary at this point, if you did this first, you can rule out anhydride and steroids, which have 2 and 4 DOU, respectively.

Hope that helps.

 

[deleted768895]

Yeah, I think that makes sense. I was just getting a but tripped up about the wording, which seemed weird to me.

But to quickly answer your question, a terpene only has C, steroids (and correct me if I am wrong) only has C's (cholesterol - a derivative of steroid, has OH), anhydride always have at least 3Os (2 carbonyl with O in the middle), and proteins have at least 1 N in it (refer to the backbone amino and COOH group. Fatty acids are carboxylic acids at the heart of it (and when it is with a glycerol, I am pretty sure they are considered esters) (waxes are fatty acid-like but they are esters with OH (this may be a bit mroe info). So just by definition you can tell that because fatty acids are COOH derivatives, it can pertain to the formula. But just to double check, you can apply the DOU formula ((2C+2-H-X)/2) (X=halogen NOT O, do not count O in this formula (and N count as -1)). So (2(14)+2-28)/2 =1. There is 1 degree of unsaturation which you get in the carbonyl of a carboxylic acid. And just to add on, and totally unnecessary at this point, if you did this first, you can rule out anhydride and steroids, which have 2 and 4 DOU, respectively.

Hope that helps.

Thank you

 

[1229ko]

Dr. Romano,
Destroyer 2017 #92. I understand that due to the steric hindrance of LDA it's "easier/faster" for it to deprotonate the least crowded alpha hydrogen... But after you depronate the alpha hydrogen and form the enolate, I thought it was in equilibrium and the more stable enolate will predominate.. And with this logic I'm getting answer A. But if the question indicated that LDA was in slight excess I would get answer B because the less stable enolate anion cannot equilibriate with a more stable one. Is my reasoning completely wrong?
Thanks in advance!

 

[niagra_falls]

#67 in the bio section of 2016 version: Microvilli and villi associated with the small intestine are involved in:
the answer i put down was Increasing the absorption surface and breaking food into smaller particles
but the correct answer is only increasing the absorption surface.
why is this? I thought that the small intestine also continued to break down food and get as much nutrients out of the ingested food before it goes into the large intestine.

 

[DAT Destroyer]

#67 in the bio section of 2016 version: Microvilli and villi associated with the small intestine are involved in:
the answer i put down was Increasing the absorption surface and breaking food into smaller particles
but the correct answer is only increasing the absorption surface.
why is this? I thought that the small intestine also continued to break down food and get as much nutrients out of the ingested food before it goes into the large intestine.

They do increase surface area, but enzymes break down particles, not microvilli. Microvilli and villi help the absorption of the digested material.
Breakdown is done by proteolytic enzymes, such as Trypsin and Chimotrypsin.

Hope this helps.

 

[uhds]

Hello,

In the 2015 Math Destroyer Practice Test 1: The shaded region shown is bound by lines y=x+4 and y=-2x+10. I'm having trouble with the signs in the answer. Why is the sign y ≤ x +4 and y ≥ -2x+10, not y ≥ x +4?

Thank you.

 

[DAT Destroyer]

Dr. Romano,
Destroyer 2017 #92. I understand that due to the steric hindrance of LDA it's "easier/faster" for it to deprotonate the least crowded alpha hydrogen... But after you depronate the alpha hydrogen and form the enolate, I thought it was in equilibrium and the more stable enolate will predominate.. And with this logic I'm getting answer A. But if the question indicated that LDA was in slight excess I would get answer B because the less stable enolate anion cannot equilibriate with a more stable one. Is my reasoning completely wrong?
Thanks in advance!

When using LDA....think Speed which is KINETICS, not thermodynamics. Given enough time , an equilibrium could indeed be established, but that is way overthinking this problem. Hope this helps.

Dr. Romano

 

[DAT Destroyer]

Hello Dr Romano,

For question 316 DAT Destroyer Organic Chem (2017) How can you tell from the molecular formula that its a fatty acid?

Lets keep life simple.....Lipids are a very diverse family, but a fatty acid is simple to spot......Consider Palmitic acid CH3(CH2)14COOH......Look at the enormous amount of carbons and hydrogens with only two oxygens..... If you see this , it is a slam dunk to be most likely a fatty acid !!! Fatty acids can be saturated or unsaturated, but all have many carbons and many hydrogens. See any organic textbook and look at a few !!!!

Hope this helps.

Dr. Romano

 

[Nasim89]

Hello Dr Romano,
Could you please explain how to rank in order of increasing pKa?
organic odyssey 2015, Chapter 2, #25
A.ch3ch2so3h B.ch3ch2so2h C.ch3ch2cooh D.ch3ch2ch2oh
I am kind of no sure why A and B are more acidic than C.
Thank you so much!

 

[dl2501]

Dr. Romano (@orgoman22 ),

Q1) I think I understand the concept behind GC350; that is, because you asked for pH of 3.3 and gave us a pKa of HF for 3.3, then you do the MV=MV, and then divide by two because what you ultimately want is the half-eq volume. But how would this question differ/how would you approach a problem like this if the pKa of HF was not given? Or perhaps if the pKa was not the half-equivalence?

Q2) For GC352, I agree that the true answer is E, but I am unsure about the explanation. It says that the Sr++ is smaller than S-, since it forms a positive ion. I was taught that, in the case of an isoelectronic series, where there is an equivalent number of electrons, the anion would be larger than the cation. So if you were to compare Na+ to F-, Na+ would be smaller because there is a bigger effective nuclear charge and thus, pull the electrons closer vs. F- which would have less charge per electron. But in the case of Strontium cation it would be Kr, and the sulfur the anion would be Ar. Kr and Ar are separated by an entire row, and usually if you go down the periodic table, the atomic radius increases, thus I would assume that although Sr++ is a cation, and that S- is an anion, that Sr++ would still be bigger because it is an entire period bigger.

Q3) In terms of counting valence electrons for transition metals, what would be the best approach? Transition metals such as Sc would have 2 valence since 4s is the highest principle number, but elements like Cr,Mo, Cu, Ag would only have 1 valence... Would this be correct to say?

Thank you for your help.

 

[themuffinman11]

For 2017 Destroyer Ochem: #303: when we want to add Chlorine to the benzene via electrophilic aromatic substitution, shouldn't the solvent be AlCl3 or FeCl3 instead of water?

 

[DAT Destroyer]

For 2017 Destroyer Ochem: #303: when we want to add Chlorine to the benzene via electrophilic aromatic substitution, shouldn't the solvent be AlCl3 or FeCl3 instead of water?

I get many questions on this. Since the reduction done in the first step gave the NH2 group which is SUPER SUPER activating, a catalyst such as FeCl3 or AlCl3 is not even needed. I have successfully done this reaction with just water !!!! Usually, if you halogenate a benzene ring, we use the Lewis acid catalyst, but if an NH2 group is present, there is no need.

Hope this helps.

Dr. Jim Romano

 

[DAT Destroyer]

Hello Dr Romano,
Could you please explain how to rank in order of increasing pKa?
organic odyssey 2015, Chapter 2, #25
A.ch3ch2so3h B.ch3ch2so2h C.ch3ch2cooh D.ch3ch2ch2oh
I am kind of no sure why A and B are more acidic than C.
Thank you so much!

I must drill.....no pun intended,,,,,,3 words into my students brains when I lecture on acidity.......EXAMINE THE ANION. The anion that is the most stable will be that acid with the smallest pKa, and consequently the most acidic. The compound A is a sulfonic acid, and would have more resonance forms and greater RESONANCE stabilization than B, which is a sulfenic acid. Next....comes choice c which is called a carboxy acid.......the number of resonance forms is less than in A or B, hence would not be as acidic. Finally, an alcohol which has a pKa of 16 is the least acidic. No resonance stabilization occurs in this alkoxide anion, hence will be the weakest acid. If this is still baffling to you, the David Klein text does a nice job in presenting the needed nuts and bolts understanding. This is a vital topic that you need to master.

Hope this helps.

Dr. Romano

 

[DAT Destroyer]

Hello,

In the 2015 Math Destroyer Practice Test 1: The shaded region shown is bound by lines y=x+4 and y=-2x+10. I'm having trouble with the signs in the answer. Why is the sign y ≤ x +4 and y ≥ -2x+10, not y ≥ x +4?

Thank you.

y<= x+4 because the shaded region is below y=x+4

 

[uhds]

y<= x+4 because the shaded region is below y=x+4

Thank you for your help.

 

[deleted768895]

Hello,

I was wondering if someone could help clarify this question I came across on Qvault bio test..
The question asks
"Which of the following changes would decrease the activity of an enzyme?"
a) Increasing temperature moderately
b) decreasing substrate concentration
c) Increasing enzyme concentration
d) maintaining the surrounding environment at enzymes optimal pH
e) none of the above

Ok so I went for (e) none of the above. The correct answer was (b) decreasing substrate concentration. At first this seemed like the obvious answer choice, but when I thought more into in, what my interpretation of enzyme activity is, its Vmax, and interpret that as the '#of active sites available for reaction', thus I don't see why decreasing the substrate concentration would reduce enzyme activity, I think that reducing substrate concentration merely reduces the rate of reaction. I think i've made a very basic question more complex..