Official DAT Destroyer Q&A Thread

[densaugeo]

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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!

 

[lazyindy]

do the problems get progressively harder as you go through each section? I did the first 35 questions in GC and got 5 wrong. Is this a good starting point?

 

[DAT Destroyer]

On questions like 194 in destroyer, will we ever be given an acid or base to go with water that has 2x the OH or H+? Would this change the way we calculate ka? Like instead of [H3O+][N3-]/[HN3] it might be [H3O+]^2[X2-] / [H2X] ? Sorry if the question is not clear. I've been using Chad's shortcut H+=(Ka*[A-])^1/2 for these problems to do them faster like he recommended and am wondering when it would not work.

Sure....why not ? You need to be able to find pH and pOH for strong acids and bases which do NOT have pKa or pKb values. The formulas you want to use will not work . For example, calculating the pH of a Calcium Hydroxide solution....or Sulfuric acid solution. .. Before doing the DESTROYER problems, review a source such as a text written by PhD chemist authors such as Brady, Chang, Zumdahl, Slowinski, Masterson or Ebbing. They have great examples.

Dr. Romano

 

[510586]

Sure....why not ? You need to be able to find pH and pOH for strong acids and bases which do NOT have pKa or pKb values. The formulas you want to use will not work . For example, calculating the pH of a Calcium Hydroxide solution....or Sulfuric acid solution. .. Before doing the DESTROYER problems, review a source such as a text written by PhD chemist authors such as Brady, Chang, Zumdahl, Slowinski, Masterson or Ebbing. They have great examples.

Dr. Romano

Oh yea I know how to do the ones without a pKa given. I'm trying to think of a salt that uses the pka stuff but is an exception to that equation. I finished destroyer twice and did Chad quizzes but didn't see any problem with pKa stuff that couldn't be solved with that Chad equation. I think the strong base strong acid thing is a different example (the BaOH2 thing). Augh

 

[510586]

do the problems get progressively harder as you go through each section? I did the first 35 questions in GC and got 5 wrong. Is this a good starting point?

No they are about the same throughout. I actually missed the most in the 1st 40 problems than any other set of 40.

 

[florida21]

Is 1,4 cyclohexanediol nonpolar? I was wondering because the 2 hydroxy groups on the opposite sides should cancel out the dipoles. However, is this molecule still capable of H bonding? We were asked which would elute first in a chromatography column but, I don't know if this molecule would H bond with the polar beads or if it would elute first since the dipole is equal and in opposite directions and technically would cancel out.

 

[DetermineDent]

So I thought RNA viruses with reverse transcriptase are called retroviruses.
But in #299 in bio it says all RNA viruses are called retroviruses..

 

[grivacobae]

I have an Ochem question on DAT Destroyer 2012 #44
"When propanal is treated with dilute sodium hydroxide, what is the major products?
d) 3-hydroxy-2-methylpentanal (this is w/o condensation; loss of H2O)
e) 2-methyl-2-pentenal (this is with condensation; loss of H2O to form alkene) " Answer is d.

I thought Aldol Condensation reaction produces an alkene after the H2O leaves. Even on the explanation it says "... do the Aldol Condensation" but it stops at 3-hydroxy-2-methylpentanal instead of going all the way to 2-methyl-2-pentenal.

Why isn't it going all the way?

 

[DAT Destroyer]

So I thought RNA viruses with reverse transcriptase are called retroviruses.
But in #299 in bio it says all RNA viruses are called retroviruses..

This question presumes the RNA viruses that use Reverse transcriptase make DNA and integrate into the host's genome are retroviruses.

 

[DAT Destroyer]

Is 1,4 cyclohexanediol nonpolar? I was wondering because the 2 hydroxy groups on the opposite sides should cancel out the dipoles. However, is this molecule still capable of H bonding? We were asked which would elute first in a chromatography column but, I don't know if this molecule would H bond with the polar beads or if it would elute first since the dipole is equal and in opposite directions and technically would cancel out.

1,4 cyclohexanediol does indeed have a dipole,,,,,,about 2.0 as a matter of fact for the trans isomer ! There would not be cancellation......You are assuming a flat structure.....recall a cyclohexane will be in a chair conformation. Thus, dipoles certainly do not cancel. The molecule can easily hydrogen bond with itself or another polar molecule.

Hope this helps

Dr. Romano

 

[DetermineDent]

Reviewing second time around now!! Woo!
#2 on orgo

So why isn't the alcohol miscible in water? I thought since they had hydrogen bonding, they would be..

 

[510586]

I have an Ochem question on DAT Destroyer 2012 #44
"When propanal is treated with dilute sodium hydroxide, what is the major products?
d) 3-hydroxy-2-methylpentanal (this is w/o condensation; loss of H2O)
e) 2-methyl-2-pentenal (this is with condensation; loss of H2O to form alkene) " Answer is d.

I thought Aldol Condensation reaction produces an alkene after the H2O leaves. Even on the explanation it says "... do the Aldol Condensation" but it stops at 3-hydroxy-2-methylpentanal instead of going all the way to 2-methyl-2-pentenal.

Why isn't it going all the way?

Heat is required to finish out double bond. If it is just the naoh it makes the one with an oh not alkene. My school also just lumped them together but heat and naoh are separate steps

 

[florida21]

1,4 cyclohexanediol does indeed have a dipole,,,,,,about 2.0 as a matter of fact for the trans isomer ! There would not be cancellation......You are assuming a flat structure.....recall a cyclohexane will be in a chair conformation. Thus, dipoles certainly do not cancel. The molecule can easily hydrogen bond with itself or another polar molecule.

Hope this helps

Dr. Romano

would this thinking work for 1,6 hexanediol, a planar structure or 2,4 heaxanediol?

 

[510586]

Would radical bromination even be possible at this benzene ring (carbon E)?

 

[DAT Destroyer]

Would radical bromination even be possible at this benzene ring (carbon E)? View attachment 193453

Who's work is this?

 

[510586]

Who's work is this?

Organic odyssey. Sorry should have mentioned that.

 

[DAT Destroyer]

Who's work is this?

Give page number will make it easier, will answer a little late today, in lecture til 8pm

 

[510586]

Give page number will make it easier, will answer a little late today, in lecture til 8pm

Chapter 3 number 29 page 46

 

[DAT Destroyer]

Chapter 3 number 29 page 46

Here you go ,had a brief break..

Not a chance in Hell !!!! You have a better chance at winning a fight against Mike Tyson !!!! Removal of the Hydrogen at E would result in the highly UNSTABLE phenyl radical. Do not even think about its abstraction. Removal of hydrogen D results in a very stabilized benzylic radical. Another way to look at it is to say that hydrogen D would have the smallest bond dissociation energy......it is quite easy to remove because the result is a radical of very high stability.


Hope this helps.

 

[510586]

Here you go ,had a brief break..

Not a chance in Hell !!!! You have a better chance at winning a fight against Mike Tyson !!!! Removal of the Hydrogen at E would result in the highly UNSTABLE phenyl radical. Do not even think about its abstraction. Removal of hydrogen D results in a very stabilized benzylic radical. Another way to look at it is to say that hydrogen D would have the smallest bond dissociation energy......it is quite easy to remove because the result is a radical of very high stability.


Hope this helps.

Yea that's what I thought too. I read the question like which was slowest so I picked D because I knew E was impossible. Not sure what the DAT would want me to pick I understood the concept but answered incorrectly 🙁

 

[Retriever22]

For Math Destroyer Test 2 Q14:

How do you get (24x^-2 y^-10)/(3x^-5 y^4) = (8x^5)/(x^2 y^2 y^10) ?

Wouldn't it simplify down to 8x^3 y^-14?

Thanks in advance for your help!

 

[Lnguyen7]

For Math Destroyer Test 2 Q14:

How do you get (24x^-2 y^-10)/(3x^-5 y^4) = (8x^5)/(x^2 y^2 y^10) ?

Wouldn't it simplify down to 8x^3 y^-14?

Thanks in advance for your help!

Did you mean (24x^-2 y^-10)/(3x^-5 y^4) = (8x^5)/(x^2 y^4 y^10) ?

Back to your question. Yes, if you simplify everything, you will get 8x^3 y^-14, but is this one of the choices? If no, then you have to pick the one that is best; in this case (8x^5)/(x^2 y^4 y^10)

 

[ArugulaSalad]

I had a question on 2015 Destroyer Biology, #472. It has a diagram of the lac operon.
The solution says: "F is the repressor (lactose) which is involved in switching the operon off when it binds to the operator." I was under the impression that the repressor was a separate protein that binds to the operator, and that lactose would come and bind to the repressor, inactivating it.
Basically, repressor =/= lactose... right?

 

[510586]

Just to clarify, c2h4 would be trig planar and c2h6 would be tetrahedral correct?

 

[510586]

In nomenclature, do alkenes get higher priority or alkynes

 

[Retriever22]

Did you mean (24x^-2 y^-10)/(3x^-5 y^4) = (8x^5)/(x^2 y^4 y^10) ?

Back to your question. Yes, if you simplify everything, you will get 8x^3 y^-14, but is this one of the choices? If no, then you have to pick the one that is best; in this case (8x^5)/(x^2 y^4 y^10)

Yes sorry about that! And I don't know how I overlooked it before but after looking at the answers again I realize that the answer was just a less simplified form of what I had written down. Thanks for your help!

 

[DAT Destroyer]

Just to clarify, c2h4 would be trig planar and c2h6 would be tetrahedral correct?

Ethene is trigonal planar C2H4.......C2H6 is ethane,,,,and is tetrahedral. Build a model....very easy to confirm this.

Dr. Romano

 

[DAT Destroyer]

In nomenclature, do alkenes get higher priority or alkynes

Believe it or not NEITHER alkene nor alkyne is a priority !!!! If both are in a compound....use the name the gives the OVERALL lowest numbers !!!! However....if a tie exits,,,,then the alkene gets the priority. For example,2 -hexen-4-yne.

Hope this helps.

Dr. Romano

 

[510586]

Believe it or not NEITHER alkene nor alkyne is a priority !!!! If both are in a compound....use the name the gives the OVERALL lowest numbers !!!! However....if a tie exits,,,,then the alkene gets the priority. For example,2 -hexen-4-yne.

Hope this helps.

Dr. Romano

Awesome thanks. Oh wait so alkene gets priority in a tie but normally it's whichever comes first? I thought if it gets priority in a tie it would be the last thing like hexynene

 

[alexis2010]

For Math Destroyer Test 2 Q14:

How do you get (24x^-2 y^-10)/(3x^-5 y^4) = (8x^5)/(x^2 y^2 y^10) ?

Wouldn't it simplify down to 8x^3 y^-14?

Thanks in advance for your help!

(24x^-2y^-10)/(3x^-5y^4) = 8x^3y^-14 = 8x^3/y^14. Since y has a negative exponent we bring it down to make positive.

 

[DAT Destroyer]

Awesome thanks. Oh wait so alkene gets priority in a tie but normally it's whichever comes first? I thought if it gets priority in a tie it would be the last thing like hexynene

"I thought if it gets priority in a tie it would be the last thing like hexynene"

That is not correct, the Klein book shows some really good examples.

 

[510586]

For this question, why wouldn't c also make a broad band at 3300? Don't cooh groups have a broad band from 2500-3500?

 

[Lnguyen7]

For this question, why wouldn't c also make a broad band at 3300? Don't cooh groups have a broad band from 2500-3500? View attachment 193614

The question does not ask anything about the COOH. Not only C, but B and D also give broad band at ~3300. However, you only look for the one that gives a broad band at ~3300 AND is unreactive toward Jones reagent.

 

[510586]

The question does not ask anything about the COOH. Not only C, but B and D also give broad band at ~3300. However, you only look for the one that gives a broad band at ~3300 AND is unreactive toward Jones reagent.

Oh woops thought it said after reacting with. Thanks

 

[510586]

On this question, why is the cut in the middle more stable than the cut on the left? My school notes didn't differentiate.

 

[Lnguyen7]

On this question, why is the cut in the middle more stable than the cut on the left? My school notes didn't differentiate.

Here's what I think. A cut on the left gives you methyl radical while a cut on the right gives you propyl radical. Both carbons are electron deficient, but we know alkyl group is electron donor. Thus a propyl radical is more stable than a methyl radical.

 

[510586]

Oh ok so it's more that we are looking at the radical not the cation?

Here's what I think. A cut on the left gives you methyl radical while a cut on the right gives you propyl radical. Both carbons are electron deficient, but we know alkyl group is electron donor. Thus a propyl radical is more stable than a methyl radical.

 

[Lnguyen7]

Oh ok so it's more that we are looking at the radical not the cation?

I think that we need to look at both. However, the acylium can stabilize by resonance, so the radical is more important.

 

[510586]

I think that we need to look at both. However, the acylium can stabilize by resonance, so the radical is more important.

Got it thanks

 

[ysiraje2]

Hello

This is regarding #99 of the OC section.

It states (summised):

which statements are true?
A. DNA lacks a 3'OH group
B. As the % of Adenine and Thymine increase, so does the melting point of the sample.
C. Sugar and phosphates are outside helix
D. The bases are stacked outside helix
E: More than one of the above
Destroyer Answer: C

My answer: E
My reasoning: C and B are both true...
As the percentage of Adenine and Thymine increases, the melting point of the sample WILL ALSO increase. Just not as much as if Cytosine and Guanine increased...

The answer in the back seems to imply that AT concentrations don't affect Melting Point at all. This question is not asking to compare AT and CG effect on melting point...

 

[510586]

Hello

This is regarding #99 of the OC section.

It states (summised):

which statements are true?
A. DNA lacks a 3'OH group
B. As the % of Adenine and Thymine increase, so does the melting point of the sample.
C. Sugar and phosphates are outside helix
D. The bases are stacked outside helix
E: More than one of the above
Destroyer Answer: C

My answer: E
My reasoning: C and B are both true...
As the percentage of Adenine and Thymine increases, the melting point of the sample WILL ALSO increase. Just not as much as if Cytosine and Guanine increased...

The answer in the back seems to imply that AT concentrations don't affect Melting Point at all. This question is not asking to compare AT and CG effect on melting point...

wouldn't an increase in AT % mean a decrease in GC %? It is % not flat number

 

[DAT Destroyer]

Hello

This is regarding #99 of the OC section.

It states (summised):

which statements are true?
A. DNA lacks a 3'OH group
B. As the % of Adenine and Thymine increase, so does the melting point of the sample.
C. Sugar and phosphates are outside helix
D. The bases are stacked outside helix
E: More than one of the above
Destroyer Answer: C

My answer: E
My reasoning: C and B are both true...
As the percentage of Adenine and Thymine increases, the melting point of the sample WILL ALSO increase. Just not as much as if Cytosine and Guanine increased...

The answer in the back seems to imply that AT concentrations don't affect Melting Point at all. This question is not asking to compare AT and CG effect on melting point...

As the % of Adenine -Guanine increases, this implies the % C-G decreases!!!! More A-T means less hydrogen bonding, hence weaker attractions. Consult the Campbell book for a well-presented chapter on Nucleic Acid Intro for more details.

Hope this helps.

Dr. Romano

 

[shiba_inu]

Hi I have an Ochem question #107 in 2014 edition. I dont understand why the propyne is acting as a Bronstead lowry acid? how do the arrows show that it is donating a proton?

 

[DAT Destroyer]

Hi.

In the 2015 QR Destroyer, on p459, #43 asks: "Given that the large triangle is isosceles, what is its area?"

1. Why do we assume that the base of the triangle labeled 3 is 3, and the base of the triangle labeled 4 is 4?
2. Why do we assume that the height of the isosceles triangle is equal to the base?

Thanks in advance.

To find one side of the largest isosceles triangle we add 3+4 to get 7 which is the base.

Since the triangle is isosceles the height is also 7.

The area of the triangle is therefore: (1/2)*7*7=49/2.

Isosceles means two sides are equal

Hope this helps..

 

[MolarDentist]

Math Destroyer test 1 Q 4
The endpoints of the diagonal of a square are located at (2,4) and (6,2). Find the area of the square.
I understand the solution but we have to assume it's a square. I graphed the endpoint and it's a rectangle. Any help is highly appreciated.

 

[DAT Destroyer]

Math Destroyer test 1 Q 4
The endpoints of the diagonal of a square are located at (2,4) and (6,2). Find the area of the square.
I understand the solution but we have to assume it's a square. I graphed the endpoint and it's a rectangle. Any help is highly appreciated.

It's indeed a square. The picture attached provides a rough sketch. It's a tilted square.

Hope this helps!

 

[MolarDentist]

Ohhhhhh. I didn't look at it that way. Thanks for the quick response.
I'm new to this forum, I just found out about it today from my predent adviser.
I just have 2 questions:
Is there a limit to the number of questions we can post?
If so can someone show me how to search the forum for similar questions I might have.
Thanks

 

[510586]

Can someone please explain number 146 on QR part of DAT destroyer? I'm having trouble following the answer in the back of the book. How does (tanx+cotx)/cscx = sinx/cosx = cosx/sinx?

 

[Lnguyen7]

Can someone please explain number 146 on QR part of DAT destroyer? I'm having trouble following the answer in the back of the book. How does (tanx+cotx)/cscx = sinx/cosx = cosx/sinx?

There is no way (tanx+cotx)/cscx = sinx/cosx = cosx/sinx
But (tanx+cotx)/cscx = (1/cscx)(tanx+cotx) = sinx(sinx/cosx + cosx/sinx) = sin^2(x)/cosx + cosx = 1/cosx = secx

 

[DAT Destroyer]

Can someone please explain number 146 on QR part of DAT destroyer? I'm having trouble following the answer in the back of the book. How does (tanx+cotx)/cscx = sinx/cosx = cosx/sinx?

First split it to get:

Tan(x)/csc (x) + cot(x)/csc(x)

Which is equal to :
[Sin(x)/cos(x)]/[1/sin(x)]+ [cos(x)/sin(x)]/[1/sin (x)] which simplifies to:

Sin^2(x)/cos(x) + cos(x). With the common denominator cos(x) we get:
(Sin^2(x)+cos^2(x))/cos(x) = 1/cos(x) = sec(x)

Hope this helps!

 

[510586]

There is no way (tanx+cotx)/cscx = sinx/cosx = cosx/sinx
But (tanx+cotx)/cscx = (1/cscx)(tanx+cotx) = sinx(sinx/cosx + cosx/sinx) = sin^2(x)/cosx + cosx = 1/c

Yea its what the explanation said

First split it to get:

Tan(x)/csc (x) + cot(x)/csc(x)

Which is equal to :
[Sin(x)/cos(x)]/[1/sin(x)]+ [cos(x)/sin(x)]/[1/sin (x)] which simplifies to:

Sin^2(x)/cos(x) + cos(x). With the common denominator cos(x) we get:
(Sin^2(x)+cos^2(x))/cos(x) = 1/cos(x) = sec(x)

Hope this helps!

This clears it up thanks