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!

 

[cacajuate]

Alcohols do not get reduced.....only oxidized. Molecules such as aldehydes, ketones, nitriles, acyl halides, alkenes, alkynes, imines, esters, and acids get reduced.

Hope this helps.

Dr. Romano

I should have stated my question better. That particular problem talks about which molecules can first be oxidized and then reduced.

2-pentanol is oxidized by K2Cr2O7 to 2-pentanone which can then be reduced by LiAlH4 or NaBH4 correct? The answer key says that this is not one of the answers.

 

[DAT Destroyer]

I should have stated my question better. That particular problem talks about which molecules can first be oxidized and then reduced.

2-pentanol is oxidized by K2Cr2O7 to 2-pentanone which can then be reduced by LiAlH4 or NaBH4 correct? The answer key says that this is not one of the answers.

No.....the problem does NOT say FIRST oxidized. If it said that.....you would be correct. Re-read this question. You may see it one day again.... 🙂

 

[cacajuate]

😱😱

I did misread the question!

 

[DAT Destroyer]

😱😱

I did misread the question!

Be very careful on the DAT, this is a good lesson that you have learned while studying..Better now than when taking the actual test.

Keep going...

Good luck in your studies..

Take care..Nancy🙂

 

[cacajuate]

Destroyer '15 GC #245

I understand how to get the pH but my understanding was that if the PKa value was below the pH, the molecule would be protonated. With a pH of 9, by my understanding, all the molecules would be protonated which I know isn't right. Where am I going wrong here?

 

[DAT Destroyer]

DAT organic chemistry odyssey chapter 13 question 6,

so in the reaction scheme, i understand that the first reagent will cause a markovnikov addition of OH and H without any carbocation shifts.

the second reagent will convert the secondary alcohol into the ketone.

but what i don't understand is the addition of NH3 to a ketone. I have been told from watching organic chemistry lectures at UC irvine that NH3 when added to an aldehyde or ketone will generate a very unstable "imine" and you typically wouldn't see this reaction. So in theory, wouldn't the correct answer be a ketone?

DAT organic chemistry odyssey chapter 13 question 38,

why is the answer A not D?

DAT organic chemistry odyssey chapter 12 question 10,

why is the group a deactivating group? You clearly see that the sulfur attached to the benzene has a lone pair, so wouldn't this mean that its an electron donor group?

DAT organic chemistry odyssey chapter 12 question 13,

Is the difference between electron pair A and electron pair D purely based on % s character?

DAT organic chemistry odyssey chapter 12 question 36,

why are acetals ortho/para directors if the atom directly bound to the benzene has no lone pairs and has the two oxygen groups directly attached? Is it because the dipoles cancel out and therefore you can treat it like an "alkyl group"?

Also just a general question from the substitution reactions chapters:

If you have a secondary alkyl halide and you react it with a non-bulky but strong base, I know you get a mixture of E2 and SN2. Is E2 favored over Sn2 though?

For example (i am referencing organic chem odyssey problems): chapter 8 question 33 (the Iodine bound to the cyclohexane undergoes E2) and chapter 8 question 9 (the iodine bound to cyclopentane undergoes Sn2 even though there is beta branching from the adjacent methyl group). Just curious on the rationale for the two questions why one is favored over the other

Ch 13 problem #6.... The imine indeed wouldn't be stable.....in the sense of isolating it.....it would be reduced down rapidly. However.... the question ask for the product formed.....Go with the imine.

Ch 13...#38......This reaction illustrates the RACEMIZATION of a ketone in base. Alpha hydrogens are replaced with deuteriums. This is an excellent way to introduce an isotopic label. The key intermediate is the enolate anion......upon reforming the carbonyl group, a D can come in from top or bottom face....thus giving the racemate. The carbonyl group STAYS,,,,,this is a more stable bond than if it was destroyed.

Ch 12 problem #10..... Great question. Wow,,,,You are good. Charged molecule will destabilize the transition state,,,,,more for the ortho and para pathways,,,less for meta,,,,hence the meta pathways dominates. The lone pair would represent d orbital electrons.....this would not contribute to intermediate stabilization. Carbon uses 2p orbitals,,,,S uses 3 p orbitals,,,,,the orbital size not the greatest for overlap. Thus the alkyl sulfonium group is a META director !!!

Chapter 12 problem #36 Acetals are simply Diethers,,,,,,This is a very nice way to transform a meta director such as acetyl into an O/P director !!!!! There is still a Hydrogen that can participate in a hyperconjugative effect,,,,,draw it out......thus this can help stabilize the transition state for ortho and para pathways. Great question....wow.

For the GENERAL QUESTION......There is always a competition set up between SN2 and E2 on secondary halides. I have been most successful using HEAT ...reflux,,,,,,,to drive more E2 product. More heat allows one to do more bond breaking and forming. In a problem such as Problem 33, Chap In Chapter 8., #33......I suppose I should have added heat,,,,to avoid confusion. As a general rule of thumb....A secondary or tertiary halide treated with a strong base such as methoxide, ethoxide, amide,,,,,,,,gives more E2. Although a generalization,,,,a nice rule of thumb. In Chapter 8, problem #9....the minute you see SULFUR,,,,,think SN2 !!! Sulfur is not a good base,,,,but a GREAT nucleophile,,,,,,,and LOVES to do SN2....cyanide as well. Recall, a strong nucleophile favors Sn2,,,,,,

Hope this helps.

Dr. Jim Romano

 

[DAT Destroyer]

Destroyer '15 GC #245

I understand how to get the pH but my understanding was that if the PKa value was below the pH, the molecule would be protonated. With a pH of 9, by my understanding, all the molecules would be protonated which I know isn't right. Where am I going wrong here?

A pH of 9 will favor the DEPROTONATED form of each....since we are beyond the pKa of each. For example,,,,,look at Indicator A...at 9...we are well beyond the pKa of 4.4,,,,thus the deprotonated form is HEAVILY favored and the solution will be blue. This is an important problem.....and likely to be seen one day,,,,soon. 😉

Dr. Romano

 

[PocketRocket]

Destroyer 15 Bio # 206:

I thought that the DNA coding sequence or strand(if that is what they are referring to in the question) will have the same bass sequence as the mRNA transcript. That being said wouldn't the answer therefore be "B" because the mRNA will be GGA?

Thank you very mucho!

 

[StudentDoc1234]

Destroyer 15 Bio # 206:

I thought that the DNA coding sequence or strand(if that is what they are referring to in the question) will have the same bass sequence as the mRNA transcript. That being said wouldn't the answer therefore be "B" because the mRNA will be GGA?

Thank you very mucho!

The mRNA strand will have a compliment sequence to the DNA strand.... so since the DNA strand was GGA the mRNA strand will be CCU. But the question is asking for the tRNA anti-codon, which will compliment the mRNA strand by using GGA.

 

[PocketRocket]

The mRNA strand will have a compliment sequence to the DNA strand.... so since the DNA strand was GGA the mRNA strand will be CCU. But the question is asking for the tRNA anti-codon, which will compliment the mRNA strand by using GGA.

Yes, but DNA is double stranded and we have the coding and the anti-coding strand. The anti-coding strand acts as the template so its the coding strand that the mRNA will resemble. Are we assuming here that the DNA strand they are talking about is the anti-coding strand?

 

[StudentDoc1234]

Yes, but DNA is double stranded and we have the coding and the anti-coding strand. The anti-coding strand acts as the template so its the coding strand that the mRNA will resemble. Are we assuming here that the DNA strand they are talking about is the anti-coding strand?

Oh sorry, I see what you mean... ya based off the answers it looks like we are assuming the template strand. I guess we should always assume they're talking about the template strand since it's the strand RNA polymerase uses to make the mRNA?

 

[tyjacobs]

DAT ochem odyssey chapter 8 question 41

The answer says that SH- is a stronger nucleophile than OH-. However, the reaction is taking place in an aprotic solvent (DMF). So wouldn't that mean that OH- would be a stronger nucleophile in this scenario?

cite: http://chemistry.umeche.maine.edu/CHY251/Ch7-Overhead4.html

 

[DAT Destroyer]

DAT ochem odyssey chapter 8 question 41

The answer says that SH- is a stronger nucleophile than OH-. However, the reaction is taking place in an aprotic solvent (DMF). So wouldn't that mean that OH- would be a stronger nucleophile in this scenario?

cite: http://chemistry.umeche.maine.edu/CHY251/Ch7-Overhead4.html

tyjacobs ,

If you EVER see CN- or anything with Sulfur, immediately entertain the strong possibility of an SN2 reaction. This is exactly what is set up here.....a strong nucleophile,,,,,relatively unhindered halide, polar aprotic solvent......THE PERFECT STORM !!! Sulfur would easily be the best. Also recall, sulfur is large and is less electronegative,,,,,thus will have loosely held electrons that can be used for the backside attack onto the antibonding orbital of the halide.

Hope this helps.

Dr. Romano

 

[Sigg]

Hi everyone,

I was wondering if someone could help me with this DAT Destroyer question:

Gen Chem, question 71.
A student heated Na2SO4 + 10H2O to determine the % of water. Which statement is incorrect?
TRUE -->Heating must be controlled to avoid decomposition of the anhydrous (dry) salt
TRUE -->A sample that is not sufficiently cooled will weigh less
TRUE -->If the % of Na2SO4 was calculated to be lower than expected, a possible reason would be loss of gas from the anhydrous salt
TRUE -->If a salt binds the waters so tightly, it is possible for decomposition to occur
FALSE --> A dessicator is employed to prevent loss of mass due to decomposition during heating

Firstly...I'm wondering where/what topic under general chemistry this falls under? This question kind of concerned me, because I've never intensively dealt with anything like this or studied something in this general topic of hydrates and determining the % of water in my Chemistry classes. I generally understand the facts here and I can memorize them...but is there anything more I should know about this topic??

The second statement in the solutions is explained by saying "it weighs less due to buoyancy because of air current convection" - what exactly does this mean? And does this relate to the 3rd statement of "if the % of Na2SO4 was calculated to be lower than expected, a possible reason would be loss of gas from the anhydrous salt" ?

Also, with the 4th statement "if a salt binds the waters so tightly..." - is there an explanation as to why?

One more thing - the 5th statement is false because dessicators are employed to prevent uptake of moisture, right? And would this uptake of moisture increase the mass?

Thanks! I'm pretty lost about this topic in general so any advice would be helpful!

 

[DAT Destroyer]

Hi everyone,

I was wondering if someone could help me with this DAT Destroyer question:

Gen Chem, question 71.
A student heated Na2SO4 + 10H2O to determine the % of water. Which statement is incorrect?
TRUE -->Heating must be controlled to avoid decomposition of the anhydrous (dry) salt
TRUE -->A sample that is not sufficiently cooled will weigh less
TRUE -->If the % of Na2SO4 was calculated to be lower than expected, a possible reason would be loss of gas from the anhydrous salt
TRUE -->If a salt binds the waters so tightly, it is possible for decomposition to occur
FALSE --> A dessicator is employed to prevent loss of mass due to decomposition during heating

Firstly...I'm wondering where/what topic under general chemistry this falls under? This question kind of concerned me, because I've never intensively dealt with anything like this or studied something in this general topic of hydrates and determining the % of water in my Chemistry classes. I generally understand the facts here and I can memorize them...but is there anything more I should know about this topic??

The second statement in the solutions is explained by saying "it weighs less due to buoyancy because of air current convection" - what exactly does this mean? And does this relate to the 3rd statement of "if the % of Na2SO4 was calculated to be lower than expected, a possible reason would be loss of gas from the anhydrous salt" ?

Also, with the 4th statement "if a salt binds the waters so tightly..." - is there an explanation as to why?

One more thing - the 5th statement is false because dessicators are employed to prevent uptake of moisture, right? And would this uptake of moisture increase the mass?

Thanks! I'm pretty lost about this topic in general so any advice would be helpful!

This question falls under the category of Lab activities. Let's just say that you may see a similar problem one day. Google this lab experiment........It involves removing water from a salt. This was a cornerstone of Freshmen Chem lab.

Dr. Jim Romano

 

[tyjacobs]

dat destroyer organic chem question 60

it is asked how many constitutional isomers exist for the molecule shown. However, if you look at the molecular formula, there is one degree of unsaturation, so technically wouldn't an alkene and a cyclohexane would also fit the bill of being a constitutional isomer? And if you consider the stereoisomers (R,S for cyclohexane and E,Z for the alkene) on top of the ones in the key, there would be more than 8 isomers for the molecule shown

 

[DAT Destroyer]

dat destroyer organic chem question 60

it is asked how many constitutional isomers exist for the molecule shown. However, if you look at the molecular formula, there is one degree of unsaturation, so technically wouldn't an alkene and a cyclohexane would also fit the bill of being a constitutional isomer? And if you consider the stereoisomers (R,S for cyclohexane and E,Z for the alkene) on top of the ones in the key, there would be more than 8 isomers for the molecule shown

I will clarify this question. I want isomers of the below compound........but keeping the cyclopentane ring and the methyl groups. I will reword this very IMPORTANT question so there will be no misunderstanding. Indeed...if you looked at just the molecular formula, many many isomers could be written.

Keep hammering away.

Dr. Romano

 

[jko1016]

Hello Doc,

I had a question on 15' Ochem #31 about basicity.
answer keys says that answer choice b contain electron withdrawing group. Wasn't both NH2 and benzene electron donating group?
I was confused because I understood acidity increase as proton were more "naked" and tried to think that
basicity increase as less naked the proton. -> armomacity donate some e- cloud to nh2 ->higher basicity than cyclopentanamine.

Thanks for the help!

 

[DAT Destroyer]

Hello Doc,

I had a question on 15' Ochem #31 about basicity.
answer keys says that answer choice b contain electron withdrawing group. Wasn't both NH2 and benzene electron donating group?
I was confused because I understood acidity increase as proton were more "naked" and tried to think that
basicity increase as less naked the proton. -> armomacity donate some e- cloud to nh2 ->higher basicity than cyclopentanamine.

Thanks for the help!

Lets review the basics...Amines are bases......they want to capture a proton. They will use their lone pair of electrons to do this Yes ?

OK,,,,,,If a benzene ring or any electron withdrawing group is present...it will decrease the available electron donation by the amine group. Hence,,,,we can say that Aromatic amines such as B or E are not that good as far as basicity is concerned .. Amides are neutral......the electrons are tried up in resonance. This leaves choices A and D. This is tricky...... D has electrons in an sp2 orbital.....

A has electrons in an sp3 orbital. Sp2 orbitals hold electrons tighter......thus will be a poorer base since it would rather not share them the way an sp3 N can. So to recap.....

The lone electron pair on Nitrogen will be less available for protonation and the amine less basic if 1. Electron withdrawing groups are present 2. The lone electron pair is in an sp2 or sp orbital 3. The lone pair is involved in maintaining the aromaticity of a molecule such as in Pyrrole.

Hope this helps.

Dr. Jim Romano

 

[ns718]

Math Destroyer Q23 Test 15.
Why is choice A not the right answer? If we pick 3 marbles they can all be the same color? What am I missing here,?
Thanks

 

[DAT Destroyer]

Math Destroyer Q23 Test 15.
Why is choice A not the right answer? If we pick 3 marbles they can all be the same color? What am I missing here,?
Thanks

The question asks for the number of picks that will GUARANTEE 3 marbles of the same color. Eli has to be extremely lucky to pick 3 marbles of the same color without looking. Like the explanation says look at the worst case scenario. The 7th pick will guarantee 3 marbles of the same color.

Hope this helps!

 

[ns718]

Thank you. It makes sense now.

 

[tyjacobs]

dat destroyer gen chem question 194

The answer says that [H+] will equal [N3-] at equilibrium. However, wouldn't the autoionization of water slightly increase the [H+]?

Wouldn't the ICE box look like this if we wanted to be most accurate (I understand you ignore the 1*10^-7 h3o+ concnetration in general calculations when the acid concentration is relatively high)?

 

[tbogdan89]

In Destroyer 2015 BIO #450. The correct answer is E, and I agree with every part of it except for the word fetus in answer choice A. I thought that the fetus does not actually breath via lungs while in the womb, and it is actually the newborn infant that is allowed to breathe easier due to the surfactant. The answer solution doesnt mention anything about making it easier for the fetus to breathe, if you could clarify I would really appreciate it!

 

[tyjacobs]

DAT biology question 262

It is asked what COULD account for the 2:1 ratio (not asking which is most likely).

Although this is very very unlikely, couldn't mutations account for the ratio?

therefore, wouldn't e be the appropriate answer?

 

[cacajuate]

DAT biology question 262

It is asked what COULD account for the 2:1 ratio (not asking which is most likely).

Although this is very very unlikely, couldn't mutations account for the ratio?

therefore, wouldn't e be the appropriate answer?

Lol solid logic dude. Lethal gene is much more likely therefore it's the appropriate answer.

 

[tooth_hurty]

Destroyer 2015 bio #243

Do we assume that all autosomal recessive disorders skip generations? When solving the problem, it seems as though both autosomal recessive and autosomal dominant could be possible.

 

[skooti]

Dr. Romano,
in the Destroyer 2013 version. GC Q) 194. I don'u understand why only F- hydrolyzes and not Na+ giving us answer B.

Thanks.

 

[DAT Destroyer]

Destroyer 2015 bio #243

Do we assume that all autosomal recessive disorders skip generations? When solving the problem, it seems as though both autosomal recessive and autosomal dominant could be possible.

No. That cannot be assumed. If it is autosomal recessive and both parents are carriers, there is 25% chance of inheriting the disorder in first generation.

 

[Sigg]

Hi Dr. Romano,

2015 Destroyer question #315
the answer states that this statement is false:
"The strontium cation is larger than the sulfur anion since negative ions gain electrons"

In the answers, it states that Sr++ is smaller than S-- since it forms a positive ion. However, looking at the periodic table, Sr is in period 5, and its cation would correspond to the noble gas Krypton, whereas S is in period 3, and its anion would correspond to the noble gas Argon. So wouldn't Sr++ still be larger than S--? I would understand if Ca++ was used, because these ions would be in the same period.

I did a bit of searching on this question, and came across a similar thread, but no consensus was really reached.

http://forums.studentdoctor.net/threads/destroyer-2011-gen-chem-244.821270/

Any insight would be appreciated! Thanks!

 

[BYU4you]

Hi Dr. Romano,

2015 Destroyer question #315
the answer states that this statement is false:
"The strontium cation is larger than the sulfur anion since negative ions gain electrons"

In the answers, it states that Sr++ is smaller than S-- since it forms a positive ion. However, looking at the periodic table, Sr is in period 5, and its cation would correspond to the noble gas Krypton, whereas S is in period 3, and its anion would correspond to the noble gas Argon. So wouldn't Sr++ still be larger than S--? I would understand if Ca++ was used, because these ions would be in the same period.

I did a bit of searching on this question, and came across a similar thread, but no consensus was really reached.

http://forums.studentdoctor.net/threads/destroyer-2011-gen-chem-244.821270/

Any insight would be appreciated! Thanks!

The extra electrons are held out much further than the radius of Sr++ especially since Sr++ is lacking those electrons sticking out.

 

[Sigg]

Hm...okay, but I still can't get over the fact that it's in a totally different period. How would we have known this? This kind of goes against what I thought I knew about about the periodic table.

So is the positively charged ion always going to be larger? Like, what about a comparison between O-- and Sr++...which radius would be bigger? That's now 2 periods different!

 

[510586]

Hi Jack,

Here is a video i found on your tube that might help.



Dr. Romano

Hi Orgoman, in this video, can the nondisjunction of anaphase 2 also be n+1, n-1, n+1, n-1? And is there a nondisjunction where one cell gets all of it in animal cells? I believe that happens in plants but I haven't heard of it for animal cells. Like tetraploidy I think it was.

 

[cacajuate]

Hi Dr. Romano,

2015 Destroyer question #315
the answer states that this statement is false:
"The strontium cation is larger than the sulfur anion since negative ions gain electrons"

In the answers, it states that Sr++ is smaller than S-- since it forms a positive ion. However, looking at the periodic table, Sr is in period 5, and its cation would correspond to the noble gas Krypton, whereas S is in period 3, and its anion would correspond to the noble gas Argon. So wouldn't Sr++ still be larger than S--? I would understand if Ca++ was used, because these ions would be in the same period.

I did a bit of searching on this question, and came across a similar thread, but no consensus was really reached.

http://forums.studentdoctor.net/threads/destroyer-2011-gen-chem-244.821270/

Any insight would be appreciated! Thanks!

yea, i see what you are saying. Usually when comparing radiuses they all need to be in the same isoelectric series. Ca++ would be the better one to compare and the reason why S-- would be the bigger one is because it has more electrons and they would increase the radius.

I don't think Sr++ is a fair comparison.

 

[Sigg]

yea, i see what you are saying. Usually when comparing radiuses they all need to be in the same isoelectric series. Ca++ would be the better one to compare and the reason why S-- would be the bigger one is because it has more electrons and they would increase the radius.

I don't think Sr++ is a fair comparison.

It's question 315 for me in Gen Chem (2015 destroyer). Kind of near the end of the gen chem section, since there are only 335 questions

The full question is this:

Which statement is false?
a) CH3F has London dispersion forces b/w molecules and dipole-dipole attractions
b) CH3COOH has hydrogen bonding as the dominant intermolecular force
c) London dispersion forces exist in all molecular solids and increase as the number of electrons increase
d) SO2 possesses both dipole-dipole attractions and London dispersion forces
e) [the "false" statement] --> The strontium cation is larger than the sulfur anion since negative ions gain electrons

A through D are obviously true...another reason why E might be false is the second half of the statement (which I underlined), which does sound kind of weird...
BUT, in the answers, the Destroyer clearly says that the reason why it's false is because Sr++ is smaller than S-- since it forms a positive ion...which really confuses me

 

[tooth_hurty]

Destroyer bio 2015 #340

Could the cooperative effect of O2 binding to hemoglobin also be referred to as the Bohr effect? Or does the Bohr effect exclusively refer to pH/CO2 concentrations and their effect on O2 affinity? I had initially thought Bohr effect so I wondering if this could technically be an answer as well. Thanks.

 

[StudentDoc1234]

Destroyer bio 2015 #340

Could the cooperative effect of O2 binding to hemoglobin also be referred to as the Bohr effect? Or does the Bohr effect exclusively refer to pH/CO2 concentrations and their effect on O2 affinity? I had initially thought Bohr effect so I wondering if this could technically be an answer as well. Thanks.

Bohr effect is one of the factors that influences whether or not cooperative binding takes place.

 

[StudentDoc1234]

2015 Gchem #212

Could I have just multiplied 1.52 by 22.4 to get the answer for a problem like this that's at STP? The solutions page has a different formula for it using P(MW)=DRT. The difference between my way and the Destroyers way comes down to a difference of just .02

 

[DAT Destroyer]

It's question 315 for me in Gen Chem (2015 destroyer). Kind of near the end of the gen chem section, since there are only 335 questions

The full question is this:

Which statement is false?
a) CH3F has London dispersion forces b/w molecules and dipole-dipole attractions
b) CH3COOH has hydrogen bonding as the dominant intermolecular force
c) London dispersion forces exist in all molecular solids and increase as the number of electrons increase
d) SO2 possesses both dipole-dipole attractions and London dispersion forces
e) [the "false" statement] --> The strontium cation is larger than the sulfur anion since negative ions gain electrons

A through D are obviously true...another reason why E might be false is the second half of the statement (which I underlined), which does sound kind of weird...
BUT, in the answers, the Destroyer clearly says that the reason why it's false is because Sr++ is smaller than S-- since it forms a positive ion...which really confuses me

From the chemical literature values Sr++ has an ionic radius of 132pm, S-- has a value of 170pm, as a general rule, an anion is larger than it's neutral counterpart, while the cation is smaller. Go to a general chemistry text book such as Chang, Zumdahl or Masterson. These books have very nice tables to show you what happens when a neutral atom becomes an ion. Hopefully you will see the more positive an atom becomes the smaller it's radius, and the more negative an atom becomes the larger the radius. If you don't have a GenChem book handy you can find some nice tables on the internet.

Hope this helps

Dr. Romano

 

[DAT Destroyer]

2015 Gchem #212

Could I have just multiplied 1.52 by 22.4 to get the answer for a problem like this that's at STP? The solutions page has a different formula for it using P(MW)=DRT. The difference between my way and the Destroyers way comes down to a difference of just .02

Yes indeed, be careful doing this and make sure you are at STP. I like to show my students the density formula because if you are NOT at STP and multiply by 22.4 things will end poorly🙁

Good job

Dr. Romano

 

[tooth_hurty]

Destroyer 15 Biology #476

Could yeasts also be used as a vector for recombinant DNA cloning since they have plasmids? If so, would fungi also be a correct answer?

 

[DAT Destroyer]

Destroyer 15 Biology #476

Could yeasts also be used as a vector for recombinant DNA cloning since they have plasmids? If so, would fungi also be a correct answer?

Yeast can be uses as vectors. They have an Origin of replication sequences, they have plasmids and they are capable of modifying proteins posttranslationally.

However, not all fungi behave the same as yeast. So, not all can be used as vectors.

Hope this helps..

 

[Ame535]

Hi Dr.Romano,
Please help me understand this.
I understand that Kw is only affected by change in temperature because it is an equilibrium constant. I also know that increasing temperature will favor endothermic reaction, while decreasing it will favor the exothermic reaction. What I don't get is:
1)If we decrease the temp. and Kw has gone smaller this means the reaction does not favor the exothermic reaction, why? So, if temp. has decreased and Kw increased, then we could say that the reaction favors exothermic?
2)What about increasing temp. and Kw decreased does this mean reaction does not favor endothermic?

Thank you

 

[super112]

Help with understanding the explanation in bio section #385. It says surfactant helps reduce the surface tension within the newborn infant's lungs but later it says the lungs are collapsed in newborns. Is there a little bit of room in the lungs where liquid accumulates and the surfactant acts as a mucous to help expel this liquid when the time comes? Thanks

 

[DAT Destroyer]

Hi Dr.Romano,
Please help me understand this.
I understand that Kw is only affected by change in temperature because it is an equilibrium constant. I also know that increasing temperature will favor endothermic reaction, while decreasing it will favor the exothermic reaction. What I don't get is:
1)If we decrease the temp. and Kw has gone smaller this means the reaction does not favor the exothermic reaction, why? So, if temp. has decreased and Kw increased, then we could say that the reaction favors exothermic?
2)What about increasing temp. and Kw decreased does this mean reaction does not favor endothermic?

Thank you

A decrease in temperature will favor an exothermic reaction and therefore the equilibrium constant will increase in value. If the equilibrium was to decrease with a decrease in temperature it means the reaction was endothermic.

If a reaction is exothermic an increase in temperature drives the equilibrium to the left and would decrease the value of the equilibrium constant. See the solution to problem 107 in the 2015 Destroyer. This problem shows a neat little trick to make this very simple I show this to my classroom students and they love it!.

Hope this helps

Dr. Romano

 

[Ame535]

A decrease in temperature will favor an exothermic reaction and therefore the equilibrium constant will increase in value. If the equilibrium was to decrease with a decrease in temperature it means the reaction was endothermic.

If a reaction is exothermic an increase in temperature drives the equilibrium to the left and would decrease the value of the equilibrium constant. See the solution to problem 107 in the 2015 Destroyer. This problem shows a neat little trick to make this very simple I show this to my classroom students and they love it!.

Hope this helps

Dr. Romano

Thank you.

 

[super112]

So on ochem #223, you said the general acidity ranking are aromatic acids>Carboxylic acids>...

Is an exception to this rule phenol vs COOH? Are there other common examples you think DAT might try to slip by us? Thanks

 

[DAT Destroyer]

So on ochem #223, you said the general acidity ranking are aromatic acids>Carboxylic acids>...

Is an exception to this rule phenol vs COOH? Are there other common examples you think DAT might try to slip by us? Thanks

Carboxy acids are much more acid than phenol !!!! How much ? Carboxy acids have pKa values around 5......Phenols around 10.......this is 5 pKa value difference or 100000 times more !!!!!

I highly recommend you know a few pKa values.....I have posted the important ones to know on a previous thread. Also.....make sure you are clear on anion stabilization. A carboxy acid is more acidic than a phenol because its anion is much more stabilized.


Hope this helps.

Dr. Romano

 

[Dentiz]

I have two questions Dr. Romano, one is DAT orgo number 15, it says we can only get two products, cant the first ester react with itself and produce a third product. second question is gen chem # 252, since we have ammonium ion and a phosphate ion, both of which can react as acids and bases respectively, wouldn't that give use OH- and H3o+ ions making it 6 moles of ions. Thank you

@orgoman22

 

[DAT Destroyer]

I have two questions Dr. Romano, one is DAT orgo number 15, it says we can only get two products, cant the first ester react with itself and produce a third product. second question is gen chem # 252, since we have ammonium ion and a phosphate ion, both of which can react as acids and bases respectively, wouldn't that give use OH- and H3o+ ions making it 6 moles of ions. Thank you

@orgoman22

In a problem such as this only consider a 1:1 mole ratio, however in real life a self condensation could be possible, but generally is NOT what we are looking for.

Whenever you do a mixed Aldol or mixed Claissen do not consider the self condensation product.

GenChem 252

Do NOT over think a question such as this. Water gives only a small amount of hydroxide and hydronium ions. Ammonium phosphate disassociates into 3 moles of ammonium and 1 mole of phosphate ions. The best answer is 4 moles.

Hope this helps.

Dr. Romano