Official DAT Destroyer Q&A Thread

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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!
 
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florida21 said:
Does an ester or an ether have acidic hydrogens? That Destroyer says that an ether doesn't have an acidic hydrogen however, wouldn't the carbon next to the oxygen lose an H and this would be stabilized by resonance the oxygen?
Click to expand...


Ethers have nothing to stabilize the anion !!!!!! Ethers are not acidic at all. Esters with alpha protons are somewhat acidic.....pKa around 25.

We exploit this in reactions such as the Claisen Condensation.

Thankfully,,,,,ethers are not acidic,,,,,we run reactions like Grignards in them,,,,,and they make excellent solvents.

Hope this Helps

Dr. Romano
 
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DetermineDent said:
On 2015 Version of DAT Destroyer:
GChem #12
Could we have used deltaT=-iKfm for this problem?
Cause when I used that, I got -4.2 for deltaT.

Does the explanation in the solution on using "i" always apply for boiling pt. and melting pt.?
Click to expand...


Anytime you have a boiling point or melting point problem , use the i term. If we have a nonelectrolyte such as a protein or sugar...No big deal... i = 1. In problem 12.....No need for the formula. We need to think fast......Both are electrolytes.....yes ? Therefore the first compound gives 3 particles,,,,, the second compound 2 particles,,,,,,

Now....ask your self a question..... " What is the Concept " ? The more particles the LOWER the freezing point and the higher the boiling point. With this in mind.....clearly the answer is the one with the lesser particles will freeze at a temperature not quite as low. -1.9 seems like a logical choice.

THIS IS A HUGE DAT-style problem.


Hope this helps.

Dr. Romano
 
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orgoman22 said:
As a RULE.....NEVER use the white powder LiAlH4 with water.....it makes an explosive cocktail !!!! You must do it in steps.....thus we write it as 1. LiAlH4, Et20 ..then 2. Water or H30+. This is just a fine point.....

For the DAT just make sure you know that NaBH4 reduces aldehydes, ketones, and acyl halides,,,,,,,,while the POWERFUL and MIGHTY LiAlH4 reduces aldehydes, ketones, acyl halides, esters, nitriles and most importantly carboxy acids !

Hope this helps.

Dr. Romano
Click to expand...

Thank you for the clarification and I understand your overall point! I may be getting bogged down with the details but I'd rather know more than less.
 
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When practicing QR questions in the destroyer book, is it best to do it with no calculator at all? Or have it in hand for questions that I think would take too long by hand?
 
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510586 said:
Can someone explain why we round 29.25 up for number 51? QR part of destroyer
Click to expand...
1. You can't have a fraction of a hit, it has to be an integer.
2. 29.25 has to be round up because if she got 29 hits, she would get an average of 0.324. The question asks for an average of 0.325
 
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so does average imply at least .325? Because I think 29 is closer than 30. I understand rounding up for questions like need 5 cups of milk to get 3 cakes or something so 4.5 wouldn't cut it, but average seemed less strict.
 
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Hi,
For OCHEM #21 in the 2013 Destroyer, the solution states that B,C and E are electron withdrawing groups. I though NH2 is donating group. And NH2 is also on A which is the correct answer. I am confused about why B and C are withdrawing groups because on the table in my book it says that those are activating donating groups.
 
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syedj94 said:
Hi,
For OCHEM #21 in the 2013 Destroyer, the solution states that B,C and E are electron withdrawing groups. I though NH2 is donating group. And NH2 is also on A which is the correct answer. I am confused about why B and C are withdrawing groups because on the table in my book it says that those are activating donating groups.
Click to expand...
If this is the question where A is a pentane with an NH2 on it, then the reasoning was because of resonance alone. I think someone about a question similar to this.
 
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Can someone explain number 48 for me for biology 2015? It said that the gene had over 4000 base pairs, but there were 7 introns. So why is the number of nucleotides so much smaller than 4000? I thought it was 4000 because she started out with more than 4000 and lost 7 so it would still be near 4000.
 
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For number 55 of biology, why can't the answer be cold blooded? Isn't cold blooded another word for poikilotherm?
 
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510586 said:
Can someone explain number 48 for me for biology 2015? It said that the gene had over 4000 base pairs, but there were 7 introns. So why is the number of nucleotides so much smaller than 4000? I thought it was 4000 because she started out with more than 4000 and lost 7 so it would still be near 4000.
Click to expand...
A mature mRNA does not contain introns (non coding region). Initially she has 4000 base pairs (not more than 4000), after the mRNA is processed, the intron is cut out and so you expect the number of base pairs is less than 4000.
 
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Lnguyen7 said:
A mature mRNA does not contain introns (non coding region). Initially she has 4000 base pairs (not more than 4000), after the mRNA is processed, the intron is cut out and so you expect the number of base pairs is less than 4000.
Click to expand...
It said she has over 4000 base pairs in the question. I know the introns would cause a smaller number, but how do we know the magnitude?
 
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510586 said:
It said she has over 4000 base pairs in the question. I know the introns would cause a smaller number, but how do we know the magnitude?
Click to expand...
Ah, I just read the question again, she indeed has over 4000 pairs. Anyway, I don't think you can know the magnitude here without some kinds of experiment. If there was, I'm interested to know! Those numbers in the choices are just arbitrary in my opinion. 1 nucleotide is highly unlikely too 🙂
 
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Lnguyen7 said:
Ah, I just read the question again, she indeed has over 4000 pairs. Anyway, I don't think you can know the magnitude here without some kinds of experiment. If there was, I'm interested to know! Those numbers in the choices are just arbitrary in my opinion. 1 nucleotide is highly unlikely too 🙂
Click to expand...
Yea it's ok I guess the main point is just that introns are removed lol
 
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DetermineDent said:
#101 in Biology, 2015 ed.
What does it mean by "linear chromosome"?

I guess I'm confused cause DNA is a double helix..
Click to expand...

Mammalian chromosomes have a linear shape. Bacterial chromosomes are circular. Chromosomes consist of DNA, which is double stranded. Double stranded DNA is packed into the linear chromosome through a serious of folding processes.

Hope this helps
 
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510586 said:
What is the difference between poikilotherm and cold blooded?
Click to expand...

Cold blooded, is another terminology used to describe Ectotherms.

Ectotherms are species that can generate very little internal heat. Therefore, they are living in an environment that has relatively constant temperature and maintain their body temperature similar to the temperature of the environment. If, the temperature varies, ectotherms usually seek another source of heat.

Some ectotherms that live in an environment with a constant temperature, eventually start maintaining that constant internal (body) temperature. They become homeotherms.

Poikilotherms, on the other hand, are species that can survive in the environment with a wide range of temperatures, and can vary their body temperatures according to the environment.

Hope this helps.
 
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orgoman22 said:
Cold blooded, is another terminology used to describe Ectotherms.

Ectotherms are species that can generate very little internal heat. Therefore, they are living in an environment that has relatively constant temperature and maintain their body temperature similar to the temperature of the environment. If, the temperature varies, ectotherms usually seek another source of heat.

Some ectotherms that live in an environment with a constant temperature, eventually start maintaining that constant internal (body) temperature. They become homeotherms.

Poikilotherms, on the other hand, are species that can survive in the environment with a wide range of temperatures, and can vary their body temperatures according to the environment.

Hope this helps.
Click to expand...
Thanks!
 
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For number 240 on OC destroyer 2015, why does the LiAlH4 also not reduce it? Doesn't LiAlH4/H3O+ reduce it? Is Et2O different somehow?
 
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510586 said:
For number 240 on OC destroyer 2015, why does the LiAlH4 also not reduce it? Doesn't LiAlH4/H3O+ reduce it? Is Et2O different somehow?
Click to expand...

This is a very very VERY important concept,,,,,,,Choices a and b represent the Wolfe-Kishner and Clemmensen reductions,,,,,,both convert the carbonyl into the CH2 group.....called methylene...... LiAlH4 also reduces carbonyl compounds but produces the SECONDARY ALCOHOL !!!!! Choice c also gives the secondary alcohol.

Hope this helps

Dr. Romano
 
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orgoman22 said:
This is a very very VERY important concept,,,,,,,Choices a and b represent the Wolfe-Kishner and Clemmensen reductions,,,,,,both convert the carbonyl into the CH2 group.....called methylene...... LiAlH4 also reduces carbonyl compounds but produces the SECONDARY ALCOHOL !!!!! Choice c also gives the secondary alcohol.

Hope this helps

Dr. Romano
Click to expand...
That's odd I must have learned it wrong then. They treated a benzyllic carbonyl as something you could reduce all the way to h's with lialh4. Thanks
 
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Do amides have a higher boiling point than alcohol because they have resonance as well? Or only the ability to form dimers? Also how is their H bonding better than alcohols? Finally, are esters also flat or only amides? Thanks
 
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DetermineDent said:
#180 on organic chemistry..
Is that the most acidic proton because of resonance?
Click to expand...

Think about it like this- what hydrogens can you remove? You can remove a hydrogen and leave a primary carboanion. Or you can remove one from the rings with double bonds and ruin aromaticity. Or you can remove a hydrogen and make a secondary carboanion. A secondary carboanion is more stable since you're not messing with aromaticity. And yes, in regards to resonance, I believe those electrons might shift up into an P orbital leaving the carbon SP2 (not 100% sure, but I think that's correct). Either way you have resonance and a secondary carboanion.
 
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DetermineDent said:
#180 on organic chemistry..
Is that the most acidic proton because of resonance?
Click to expand...

This is a slam dunk. The proton between the two benzene rings is VERY acidic,,,,,,the anion is flanked by two electron withdrawing benzene rings which can offer resonance stabilization.

Dr. Romano
 
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510586 said:
Do amides have a higher boiling point than alcohol because they have resonance as well? Or only the ability to form dimers? Also how is their H bonding better than alcohols? Finally, are esters also flat or only amides? Thanks
Click to expand...

Amides have the highest boiling and melting points because they are partially ionic !!!!!! The amide is composed of resonance contributors which allow it to have partial CHARGES.......hydrogen bonding also can occur, but this is secondary. Dimers form for carboxy acids....and along with intermolecular hydrogen bonding......have boiling and melting points HIGHER than alcohols but less than amides. Part of the ester is in the SAME plane.....the sp2 atoms AND the atoms DIRECTLY bonded to them.

Hope this helps

Dr. Romano
 
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can anyone explain why esters are less water soluble than aldehydes/ketones? An ester has 2 oxygens that can act as a H bond acceptor while an aldehyde/ketone only has 1 oxygen.
 
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#158 on gen chem. This may be a silly question but I'm not understanding it..
For deltaG=-nFE

Why did we multiply 3*2 for n instead of adding them?
 
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florida21 said:
can anyone explain why esters are less water soluble than aldehydes/ketones? An ester has 2 oxygens that can act as a H bond acceptor while an aldehyde/ketone only has 1 oxygen.
Click to expand...

florida21

Esters can form hydrogen bonds with water,,,,,,so can aldehydes and ketones.Small esters have boiling points similar to those of aldehydes and ketones with the same number of carbon atoms. If the ester , aldehyde or ketone exceeds 4 carbons solubility in water drops precipitously.

Hope this helps.

Dr. Romano
 
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For common ion effect, does the extra compound in the solution have to be soluble to have an effect? Ex: AgCl(s) <=> Ag+(aq.) + Cl-(aq.) (Ksp = 1.8 x 10-10) but with .5 M PbCl2. Would the Cl2 count as 2? Would it be factored in?
 
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510586 said:
For common ion effect, does the extra compound in the solution have to be soluble to have an effect? Ex: AgCl(s) <=> Ag+(aq.) + Cl-(aq.) (Ksp = 1.8 x 10-10) but with .5 M PbCl2. Would the Cl2 count as 2? Would it be factored in?
Click to expand...

AgCl has a small Ksp.....thus is not very soluble.....only a small amount of Chloride ions are produced. Lead Chloride is MUCH MUCH more soluble.....It has a Ksp of 2 exp -5....thus to answer your question......YES,,,,,the additional ion has to be reasonably soluble to produce the additional ion. Here we get 2 x 0.5 moles of chloride ions since lead chloride yields 1 mole of Pb and 2 moles of chloride . Notice how the addition of the common ion SHIFTS the equilibrium to the LEFT......this produces more AgCl, thereby DECREASING its solubility.

Hope this helps.

Dr. Romano
 
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orgoman22 said:
AgCl has a small Ksp.....thus is not very soluble.....only a small amount of Chloride ions are produced. Lead Chloride is MUCH MUCH more soluble.....It has a Ksp of 2 exp -5....thus to answer your question......YES,,,,,the additional ion has to be reasonably soluble to produce the additional ion. Here we get 2 x 0.5 moles of chloride ions since lead chloride yields 1 mole of Pb and 2 moles of chloride . Notice how the addition of the common ion SHIFTS the equilibrium to the LEFT......this produces more AgCl, thereby DECREASING its solubility.

Hope this helps.

Dr. Romano
Click to expand...
Sorry I was trying to find something very very insoluble as the extra compound. So if it was straight up insoluble, it wouldn't change the solubility of the one dissolving, correect?
 
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510586 said:
Sorry I was trying to find something very very insoluble as the extra compound. So if it was straight up insoluble, it wouldn't change the solubility of the one dissolving, correect?
Click to expand...

Yes that is correct!

For the DAT.....I assure you.....the added compound would indeed be soluble....or soluble within reason !!!!
 
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Does resonance not cause a higher boiling point? I understand the dimer part but doesn't resonance stabilize as well? For acetic acid
 
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510586 said:
Does resonance not cause a higher boiling point? I understand the dimer part but doesn't resonance stabilize as well? For acetic acid
Click to expand...


Boiling point is governed by intermolecular attractions. Acetic acid has the ability to hydrogen bond with other acetic acid molecules,,,,,in ADDITION....the presence of two hydrogen bonding functional groups in this compound raises the boiling point even further. We form dimeric structures. Resonance plays no role here. It does stabilize the anion, but has little to do with the high boiling point. Any organic text will show you a wonderful picture of a carboxy acid dimer.

Hope this helps.

Dr. Romano
 
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orgoman22 said:
Boiling point is governed by intermolecular attractions. Acetic acid has the ability to hydrogen bond with other acetic acid molecules,,,,,in ADDITION....the presence of two hydrogen bonding functional groups in this compound raises the boiling point even further. We form dimeric structures. Resonance plays no role here. It does stabilize the anion, but has little to do with the high boiling point. Any organic text will show you a wonderful picture of a carboxy acid dimer.

Hope this helps.

Dr. Romano
Click to expand...
It does thanks
 
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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.
 
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