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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[DAT Destroyer]

Destroyer Version 2017,
Chemistry #274, I was wondering why the answer is D. The explanation says 4 moles of ions, which would be answer choice B, right? If not, I'm confused as to why we need the 6 *10^23 number. I thought 6 *10^23 was for when we wanted molecules, or atoms; not moles. Thank you in advance.

The explanation is correct it was a typo on the letter choice, and you are indeed correct the choice is B. It has been corrected.

Thanks..Nancy

 

[whatisit350]

The explanation is correct it was a typo on the letter choice, and you are indeed correct the choice is B. It has been corrected.

Thanks..Nancy

Oh ok, great. Thank you for clearing that up. And, I'm sorry but I have another question.

Just a few questions down, Question #291, wouldn't X be able to represent both OH- and H3O+? If I set X to equal OH- and get the pH that way, then the answer is B, instead of C.

Thank you again for taking the time to answer all of our questions. I know we all truly appreciate that.

 

[boogahead10]

<2016 Math Destroyer Test 17 #14>
Dear Dr. Romano,

Hopefully this is the main board you mentioned!

So, I was able to get by understanding all the other problems from Math Destroyer, until I reached this question.
It is, as I mentioned before, 2016 Math Destroyer Test 17 #14 that starts with "in a set of data consisting of 1000 numbers, the mean is 15 and the standard deviation is 2..."

I was able to solve the new mean, but I wasn't able to solve for the new std dv.
The answer key multiples 0.75 x 2 = 1.5
Where is this 2 from? What is the formula to solve for the new std dv without the hassle of figuring out the variance, etc.

Thank you in advance!

 

[DAT Destroyer]

Oh ok, great. Thank you for clearing that up. And, I'm sorry but I have another question.

Just a few questions down, Question #291, wouldn't X be able to represent both OH- and H3O+? If I set X to equal OH- and get the pH that way, then the answer is B, instead of C.

Thank you again for taking the time to answer all of our questions. I know we all truly appreciate that.

Please do not forget that pH is a (-log of H+ concentration). You have the value of H+ in this problem, so there is no need to use OH and pOH. Also you cannot assume that ph+pOH =14 in here, because Kw is 5*10^-14, not 1*10^-14.
So either way you solve it, answer would be the same.
Hope this helps.

 

[DAT Destroyer]

<2016 Math Destroyer Test 17 #14>
Dear Dr. Romano,

Hopefully this is the main board you mentioned!

So, I was able to get by understanding all the other problems from Math Destroyer, until I reached this question.
It is, as I mentioned before, 2016 Math Destroyer Test 17 #14 that starts with "in a set of data consisting of 1000 numbers, the mean is 15 and the standard deviation is 2..."

I was able to solve the new mean, but I wasn't able to solve for the new std dv.
The answer key multiples 0.75 x 2 = 1.5
Where is this 2 from? What is the formula to solve for the new std dv without the hassle of figuring out the variance, etc.

Thank you in advance!

2 is the original standard deviation. Adding the same number to each number on the set doesn't change the standard deviation, but multiplying each number in the set by the same number does.
The new standard deviation is : 2*0.75=1.5

 

[whatisit350]

Please do not forget that pH is a (-log of H+ concentration). You have the value of H+ in this problem, so there is no need to use OH and pOH. Also you cannot assume that ph+pOH =14 in here, because Kw is 5*10^-14, not 1*10^-14.
So either way you solve it, answer would be the same.
Hope this helps.

Ah ok, yes this makes sense. Thank you for explaining that.

 

[whatisit350]

Destroyer version 2017.
General Chemistry #364. I was just wondering if you could please explain why the valence shell electrons of Co was only 2. I thought it would be 9; 2 from the 4S, and 7 from the 3D.

 

[boogahead10]

<Math Destroyer 2016 Test 12 #22>
Dear Dr. Romano,

I'm going through test 12, and I am having a hard time with a lot of these problems again...
I'm having a hard time understanding the solution to #22 especially.
I even wrote out the numbers, but I still don't understand how to approach this.
It says, "we have an additional factor of 5 each in 25 and 50, yielding a total of 12 factors of 5."
Aren't numbers 25 and 50 included in the total number 10 that were divisible by 5?
Why must we add 25 and 50 again and obtain 12?
Could you please help me?

Thank you in advance!

 

[DAT Destroyer]

Destroyer version 2017.
General Chemistry #364. I was just wondering if you could please explain why the valence shell electrons of Co was only 2. I thought it would be 9; 2 from the 4S, and 7 from the 3D.

I have answered this question before, and it is a GREAT one ! The organic chemist would indeed say 9, as a nice approximation when dealing with Organo-mettallic mechanisms and studies. However, the General Chemist and Physical Chemist use the more traditional method. Go to the highest principle quantum number. In Cobalt it is n=4. This represents the FOURTH principle energy level, hence is considered the valence shell. In General Chemistry, this is your best approach. Thus , if we form the Co+2 cation, we will be removing electrons from the 4s sublevel. For more details, the text by Raymond Chang does an excellent job explaining this with further examples.

Here is an example for you....(d) Fe: 1s22s22p63s23p64s23d6. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital Fe2+: 1s22s22p63s23p64s23d6 . I hope you can see...we removed electrons from the VALENCE shell,,,,here it is 2.

Hope this helps

Dr. Romano

 

[DAT Destroyer]

<Math Destroyer 2016 Test 12 #22>
Dear Dr. Romano,

I'm going through test 12, and I am having a hard time with a lot of these problems again...
I'm having a hard time understanding the solution to #22 especially.
I even wrote out the numbers, but I still don't understand how to approach this.
It says, "we have an additional factor of 5 each in 25 and 50, yielding a total of 12 factors of 5."
Aren't numbers 25 and 50 included in the total number 10 that were divisible by 5?
Why must we add 25 and 50 again and obtain 12?
Could you please help me?

Thank you in advance!

The multiples of 5 are: 5 10 15 20 25 30 35 40 45 50. So far we have 10 numbers.
However each multiple of 25 taken together with some even number will produce 2 multiples of 10.( 25*4=100) There are 2 multiples of 25; 25 and 50 which makes the number of zeros 12.
Hope this helps

 

[whatisit350]

I have answered this question before, and it is a GREAT one ! The organic chemist would indeed say 9, as a nice approximation when dealing with Organo-mettallic mechanisms and studies. However, the General Chemist and Physical Chemist use the more traditional method. Go to the highest principle quantum number. In Cobalt it is n=4. This represents the FOURTH principle energy level, hence is considered the valence shell. In General Chemistry, this is your best approach. Thus , if we form the Co+2 cation, we will be removing electrons from the 4s sublevel. For more details, the text by Raymond Chang does an excellent job explaining this with further examples.

Here is an example for you....(d) Fe: 1s22s22p63s23p64s23d6. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital Fe2+: 1s22s22p63s23p64s23d6 . I hope you can see...we removed electrons from the VALENCE shell,,,,here it is 2.

Hope this helps

Dr. Romano

Hmm ok, yes it makes sense when you think about it like that. But I think I need to dig further on this, and will probably check out the resource you suggested. Thank you for the direction.

 

[socal6]

ORGO (2017) #92

Hi Dr. Romano,

If you are presented with more than 1 α-H in a given molecule, why is it that tert-butoxide is always able to remove the more acidic α-H, whereas LDA is only able to remove the more accessible/less hindered α-H. I always assumed the rules would be the same for any large bulky bases.

Thank you!

 

[DAT Destroyer]

ORGO (2017) #92

Hi Dr. Romano,

If you are presented with more than 1 α-H in a given molecule, why is it that tert-butoxide is always able to remove the more acidic α-H, whereas LDA is only able to remove the more accessible/less hindered α-H. I always assumed the rules would be the same for any large bulky bases.

Thank you!

The more acidic alpha hydrogen is ALSO the more accessible. The hydrogen on the outside which is also the easiest to remove will have the smaller pKa. Recall, the haloform reaction. The outer hydrogen was abstracted in preference to the inner hydrogen. Although the inner hydrogen would result in the more stable anion, kinetics dominates thermodynamics here. In problem, 92, LDA or t-butoxide would have both removed the outer hydrogen. We call this intermediate the kinetic intermediate. This allows us to attack the halide in an SN2 fashion to finish the job.

I hope this helps.

Dr. Romano

 

[socal6]

The more acidic alpha hydrogen is ALSO the more accessible. The hydrogen on the outside which is also the easiest to remove will have the smaller pKa. Recall, the haloform reaction. The outer hydrogen was abstracted in preference to the inner hydrogen. Although the inner hydrogen would result in the more stable anion, kinetics dominates thermodynamics here. In problem, 92, LDA or t-butoxide would have both removed the outer hydrogen. We call this intermediate the kinetic intermediate. This allows us to attack the halide in an SN2 fashion to finish the job.

I hope this helps.

Dr. Romano

Yes it does! Thank you so much!

 

[DAT Destroyer]

Yes it does! Thank you so much!

Always happy to help, feel free to post more questions.

Dr. Romano

 

[socal6]

Always happy to help, feel free to post more questions.

Dr. Romano

ORGO (2017) #264

Sorry for the double post, but I found the question I was talking about! Could you please explain why in this scenario tert-butoxide abstracts the double α-H as opposed to the more accessible α-H on the methyl attached to the carbonyl? Thank you so much!

 

[boogahead10]

<Gen Chem Destroyer 2016 #187>
I deduced these from reading the problem: Delta H>0 and Delta S >0.
But, shouldn't Delta G >0 because the temperature decreased...?

Thank you!!

 

[DAT Destroyer]

ORGO (2017) #264

Sorry for the double post, but I found the question I was talking about! Could you please explain why in this scenario tert-butoxide abstracts the double α-H as opposed to the more accessible α-H on the methyl attached to the carbonyl? Thank you so much!

Great question. Normally, t-butoxide would remove the most accessible hydrogen as you so perspicaciously pointed out. However, look closely. A double alpha proton lurks. This proton is VERY acidic, with a pKa of about 7. Removal of this proton, which is much more acidic than the outer proton.....pKa of about 17...also would yield a resonance stabilized anion. As I tell all my Advanced Organic Chemistry students.....NEVER bet against resonance. Betting against resonance will make you a poor man. Thus, having a lower pKa, and yielding a resonance stabilized enolate anion more than compensates for the steric effect that you reference. For further reading, I like the text book by David Klein, PhD.

I hope this helps.

Dr. Jim Romano

 

[socal6]

Great question. Normally, t-butoxide would remove the most accessible hydrogen as you so perspicaciously pointed out. However, look closely. A double alpha proton lurks. This proton is VERY acidic, with a pKa of about 7. Removal of this proton, which is much more acidic than the outer proton.....pKa of about 17...also would yield a resonance stabilized anion. As I tell all my Advanced Organic Chemistry students.....NEVER bet against resonance. Betting against resonance will make you a poor man. Thus, having a lower pKa, and yielding a resonance stabilized enolate anion more than compensates for the steric effect that you reference. For further reading, I like the text book by David Klein, PhD.

I hope this helps.

Dr. Jim Romano

Thanks again Dr. Romano! I actually get it now!

 

[iphone2020]

General chemistry question number 78. Why the answer is not B? The co2 is no polar but we have oxygen in our structure so would not that count! The oxygen can make bond with hydrogen in water.


Sent from my iPhone using SDN mobile

 

[bellab17]

Hi,

I was wondering if you could please clarify regarding the following questions:

-Question 8 for ochem---- could one say that the reason e) is correct bc the structure on the left has better quality resonance structures due to the two carbonyls so therefore its more stable, less reactive and a weaker base?
For choice d) is this answer wrong bc oxygen is an electronegative atom and therefore makes the compound less basic compared to the carbon atom in the ring in the other structure?

-q13 for ochem....how do you know how many cooh to put based on side chainoxidation?

-q10 for ochem...what concept was tested here? I looked back in my notes and couldnt figure out how they came up with those answers? Why are the br placed in those positions?

- q19 for ochem....i dont understand how to best solve this problem based on cnh2n plus 2?

-q37 for ochem....can one say a) is correct and the most basic bc it doesnt have any resonance but the other choices do. So bc a) has no resonance its less stable, more reactive so most basic/stronger base?
Thanks so much!

 

[DAT Destroyer]

Hi,

I was wondering if you could please clarify regarding the following questions:

-Question 8 for ochem---- could one say that the reason e) is correct bc the structure on the left has better quality resonance structures due to the two carbonyls so therefore its more stable, less reactive and a weaker base?
For choice d) is this answer wrong bc oxygen is an electronegative atom and therefore makes the compound less basic compared to the carbon atom in the ring in the other structure?

-q13 for ochem....how do you know how many cooh to put based on side chainoxidation?

-q10 for ochem...what concept was tested here? I looked back in my notes and couldnt figure out how they came up with those answers? Why are the br placed in those positions?

- q19 for ochem....i dont understand how to best solve this problem based on cnh2n plus 2?

-q37 for ochem....can one say a) is correct and the most basic bc it doesnt have any resonance but the other choices do. So bc a) has no resonance its less stable, more reactive so most basic/stronger base?
Thanks so much!

-Question 8 for ochem---- could one say that the reason e) is correct bc the structure on the left has better quality resonance structures due to the two carbonyls so therefore its more stable, less reactive and a weaker base?
For choice d) is this answer wrong bc oxygen is an electronegative atom and therefore makes the compound less basic compared to the carbon atom in the ring in the other structure?


Lets start with choice E......Their is more resonance in the second structure, hence the electron density is removed off the N....hence it is less basic. In Choice D, yes.....O is more electronegative, thus will remove some electron density off the N making it less basic. Good job, you have the idea !!!! Hope this helps.

 

[bellab17]

Yes thank you. When you say electron density what are you referring to in regards to basicity? I guess that is what is confusing me??? I just thought, the more resonance stuctures and quality resonance structures equals more stable so less reactive and weaker base. How do i apply the reasoning behind electron density here?

Thx!

-Question 8 for ochem---- could one say that the reason e) is correct bc the structure on the left has better quality resonance structures due to the two carbonyls so therefore its more stable, less reactive and a weaker base?
For choice d) is this answer wrong bc oxygen is an electronegative atom and therefore makes the compound less basic compared to the carbon atom in the ring in the other structure?


Lets start with choice E......Their is more resonance in the second structure, hence the electron density is removed off the N....hence it is less basic. In Choice D, yes.....O is more electronegative, thus will remove some electron density off the N making it less basic. Good job, you have the idea !!!! Hope this helps.

s

 

[DAT Destroyer]

Yes....an amine is a abase, thus has the electron density that will enable it to capture a proton. If the amine has resonance, this electron density is delocalized, and render it a less effective base. . Hope this helps.

 

[DAT Destroyer]

Hi,

I was wondering if you could please clarify regarding the following questions:

-Question 8 for ochem---- could one say that the reason e) is correct bc the structure on the left has better quality resonance structures due to the two carbonyls so therefore its more stable, less reactive and a weaker base?
For choice d) is this answer wrong bc oxygen is an electronegative atom and therefore makes the compound less basic compared to the carbon atom in the ring in the other structure?

-q13 for ochem....how do you know how many cooh to put based on side chainoxidation?

-q10 for ochem...what concept was tested here? I looked back in my notes and couldnt figure out how they came up with those answers? Why are the br placed in those positions?

- q19 for ochem....i dont understand how to best solve this problem based on cnh2n plus 2?

-q37 for ochem....can one say a) is correct and the most basic bc it doesnt have any resonance but the other choices do. So bc a) has no resonance its less stable, more reactive so most basic/stronger base?
Thanks so much!

Orgo #13 2017 Destroyer

When doing a side chain oxidation on an aromatic ring, we oxidize all R groups that have a benzylic hydrogen to COOH. Selectivity does not work.....thus always assume all groups that can oxidize will. If we had t-butyl-ethylbenzene for example what do you get ? Well.....t-butyl lacks benzylic hydrogens, so only oxidize the ethyl group. We get 4-t-butylbenzoic acid. This above example is a bit tricky because of the ring adjacent to the benzene . Any Organic text will show you more examples for this very important and fundamental reaction. Hope this helps.

 

[bellab17]

Okay, thanks so much!

Orgo #13 2017 Destroyer

When doing a side chain oxidation on an aromatic ring, we oxidize all R groups that have a benzylic hydrogen to COOH. Selectivity does not work.....thus always assume all groups that can oxidize will. If we had t-butyl-ethylbenzene for example what do you get ? Well.....t-butyl lacks benzylic hydrogens, so only oxidize the ethyl group. We get 4-t-butylbenzoic acid. This above example is a bit tricky because of the ring adjacent to the benzene . Any Organic text will show you more examples for this very important and fundamental reaction. Hope this helps.
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[DAT Destroyer]

Hi,

I was wondering if you could please clarify regarding the following questions:

-Question 8 for ochem---- could one say that the reason e) is correct bc the structure on the left has better quality resonance structures due to the two carbonyls so therefore its more stable, less reactive and a weaker base?
For choice d) is this answer wrong bc oxygen is an electronegative atom and therefore makes the compound less basic compared to the carbon atom in the ring in the other structure?

-q13 for ochem....how do you know how many cooh to put based on side chainoxidation?

-q10 for ochem...what concept was tested here? I looked back in my notes and couldnt figure out how they came up with those answers? Why are the br placed in those positions?

- q19 for ochem....i dont understand how to best solve this problem based on cnh2n plus 2?

-q37 for ochem....can one say a) is correct and the most basic bc it doesnt have any resonance but the other choices do. So bc a) has no resonance its less stable, more reactive so most basic/stronger base?
Thanks so much!

Question 10
The concept is from Chapter 2,,,,,writing isomers and recognizing structures that are the same and isomeric. This is a very important problem. If you need a review on this, the text by David Klein is easy and very well written. Hope this helps.

 

[DAT Destroyer]

Hi,

I was wondering if you could please clarify regarding the following questions:

-Question 8 for ochem---- could one say that the reason e) is correct bc the structure on the left has better quality resonance structures due to the two carbonyls so therefore its more stable, less reactive and a weaker base?
For choice d) is this answer wrong bc oxygen is an electronegative atom and therefore makes the compound less basic compared to the carbon atom in the ring in the other structure?

-q13 for ochem....how do you know how many cooh to put based on side chainoxidation?

-q10 for ochem...what concept was tested here? I looked back in my notes and couldnt figure out how they came up with those answers? Why are the br placed in those positions?

- q19 for ochem....i dont understand how to best solve this problem based on cnh2n plus 2?

-q37 for ochem....can one say a) is correct and the most basic bc it doesnt have any resonance but the other choices do. So bc a) has no resonance its less stable, more reactive so most basic/stronger base?
Thanks so much!

q19 for ochem....i dont understand how to best solve this problem based on cnh2n plus 2?

You first need to consult a text and learn how to calculate the degree of unsaturation. If you see the Hs are double +2 the amount of Carbons, we have ZERO degrees of unsaturation . For example ....C5H12......or C7H16O. Both have zero degres of unsaturation. This means the compound has only single bonds,,,,,,no rings,,,no double bonds, no triple bonds. A degree of 1 means a compound has 1 double bond or 1 ring. 2 degrees mean 2 double bonds, or 2 rings, or a double and a ring...or a triple bond. In this problem, we had zero degrees, Notice oxygen does not count , but for every N, we remove a hydrogen. If the degree was zero....only single bonds result,,,,,thus this compound cannot be an amide, since it has a double bond ! The Klein book does a decent job in explaining this, but any orgo book will give you more examples.

Hope this helps!

Dr. Jim Romano

 

[bellab17]

Yes, sorry, i realized i was not reading the problem correctly at first, but eventually saw what i did wrong in my first approach. Thank you again for taking out the time to explain, always very helpful!

q19 for ochem....i dont understand how to best solve this problem based on cnh2n plus 2?

You first need to consult a text and learn how to calculate the degree of unsaturation. If you see the Hs are double +2 the amount of Carbons, we have ZERO degrees of unsaturation . For example ....C5H12......or C7H16O. Both have zero degres of unsaturation. This means the compound has only single bonds,,,,,,no rings,,,no double bonds, no triple bonds. A degree of 1 means a compound has 1 double bond or 1 ring. 2 degrees mean 2 double bonds, or 2 rings, or a double and a ring...or a triple bond. In this problem, we had zero degrees, Notice oxygen does not count , but for every N, we remove a hydrogen. If the degree was zero....only single bonds result,,,,,thus this compound cannot be an amide, since it has a double bond ! The Klein book does a decent job in explaining this, but any orgo book will give you more examples.

Hope this helps!

Dr. Jim Romano

 

[wizkidz]

Dr.Romano- Question #63 Orgo Destroyer 2016-Is there a methodically way to do this or do we have to be able to visualize this and draw everything out?

 

[wizkidz]

Also a question on #66 in Orgo 2016-I understand the formation of the enamine after the first reaction, but dont understand how the reaction proceeds from here. I know the enamine rxn is reversible with H30+ but unsure how this works here. Thank you!

 

[charcot bouchard]

The question in question (no pun intended) asks:

The Ksp of PbCl2 is 1.6 x 10^-5 at 25 degrees C. What is the solubility of PbCl2 in 0.01M KCl?

First, let's find the molar solubility of PbCl2 with no common ion effect:

Ksp = 1.6 x 10^-5 = x(2x)^2
1.6 x 10^-5 = 4x^3
x = 0.0158

Now, if we solve the question asked using 0.01M Cl- as the common ion:

Ksp = 1.6 x 10^-5 = x(0.01)^2
1.6 x 10^-5 = 0.0001x
x = 0.16

0.16 is the correct answer given in DAT Destroyer. My question is: why is the molar solubility of PbCl2 so much higher when you start with Cl- in solution? Is that not supposed to deter PbCl2 from dissolving? Doesn't this problem contradict what you would expect if you understand the common ion effect?

 

[DAT Destroyer]

The question in question (no pun intended) asks:

The Ksp of PbCl2 is 1.6 x 10^-5 at 25 degrees C. What is the solubility of PbCl2 in 0.01M KCl?

First, let's find the molar solubility of PbCl2 with no common ion effect:

Ksp = 1.6 x 10^-5 = x(2x)^2
1.6 x 10^-5 = 4x^3
x = 0.0158

Now, if we solve the question asked using 0.01M Cl- as the common ion:

Ksp = 1.6 x 10^-5 = x(0.01)^2
1.6 x 10^-5 = 0.001x
x = 0.16

0.16 is the correct answer given in DAT Destroyer. My question is: why is the molar solubility of PbCl2 so much higher when you start with Cl- in solution? Is that not supposed to deter PbCl2 from dissolving? Doesn't this problem contradict what you would expect if you understand the common ion effect?

I re-did the calculation. You are indeed correct referencing the common ion effect. The molar solubility must DECREASE. Use 0.1M as the Chloride concentration. In water, as you said we get .016M......using the new value of 0.1 M chloride ion we get 0.0016M....which now makes physical sense. I apologize for the typo.

Problem was done fine.....a number was just slightly off.
Only the very best of students would find this. I thank you for pointing this out.

Keep up the good work

Hope this helps.

Dr. Romano

 

[charcot bouchard]

I re-did the calculation. You are indeed correct referencing the common ion effect. The molar solubility must DECREASE. Use 0.1M as the Chloride concentration. In water, as you said we get .016M......using the new value of 0.1 M chloride ion we get 0.0016M....which now makes physical sense. I apologize for the typo.

Problem was done fine.....a number was just slightly off.
Only the very best of students would find this. I thank you for pointing this out.

Keep up the good work

Hope this helps.

Dr. Romano

Thank you Dr. Romano, that does help. One follow-up question for you:

Why does the common ion effect fail when we start with a 0.01M Cl- concentration, as in the typo problem? It would make sense to me if the molar solubility with 0.01M Cl in the starting solution was very close to the solubility in water, given that 0.01M is such a small concentration; however, the molar solubility increases by a factor of 10. Can you help me understand this phenomenon please?

Thank you,

Parker Green

 

[DAT Destroyer]

Thank you Dr. Romano, that does help. One follow-up question for you:

Why does the common ion effect fail when we start with a 0.01M Cl- concentration, as in the typo problem? It would make sense to me if the molar solubility with 0.01M Cl in the starting solution was very close to the solubility in water, given that 0.01M is such a small concentration; however, the molar solubility increases by a factor of 10. Can you help me understand this phenomenon please?

Thank you,

Parker Green

When we did the approximation we assumed the Cl- donation from the original salt was NEGLIGIBLE as compared to the amount we were adding. At 0.01M, we are too small in concentration to omit the amount from the Lead Chloride dissociation. ..i.e. the 2x term. When we neglect a term in any equilibrium expression, we must always be certain that the amount is very small. The Brady text has a great example of this very concept. If you cant find the book most standard texts such as Zumdahl, Chang, or Slowinski do a great job. Hope this helps.

 

[charcot bouchard]

When we did the approximation we assumed the Cl- donation from the original salt was NEGLIGIBLE as compared to the amount we were adding. At 0.01M, we are too small in concentration to omit the amount from the Lead Chloride dissociation. ..i.e. the 2x term. When we neglect a term in any equilibrium expression, we must always be certain that the amount is very small. The Brady text has a great example of this very concept. If you cant find the book most standard texts such as Zumdahl, Chang, or Slowinski do a great job. Hope this helps.

Thank you, that makes sense.

Came across another Gen Chem question today. It is #267 in the 2017 Destroyer. It asks which Indicator you should use to, "determine the pH of an NH4Cl solution?"

The answer key says that the NH4+ will react with water to produce H3O+ ions. It then says: "Since H3O+ are produced, our solution would be acidic with pH of about 5 or 6." How do we know the pH will be around 5 or 6? The question does not give a concentration or amount of NH4Cl at any point.

Thank you,
Parker Green

 

[DAT Destroyer]

Thank you, that makes sense.

Came across another Gen Chem question today. It is #267 in the 2017 Destroyer. It asks which Indicator you should use to, "determine the pH of an NH4Cl solution?"

The answer key says that the NH4+ will react with water to produce H3O+ ions. It then says: "Since H3O+ are produced, our solution would be acidic with pH of about 5 or 6." How do we know the pH will be around 5 or 6? The question does not give a concentration or amount of NH4Cl at any point.

Thank you,
Parker Green

This question did not ask you to calculate the pH value exactly. However, Ammonium ion is a weak acid. You should know it will most likely be in a pH range 4-6. Choice C is therefore correct.

Hope this helps.

 

[charcot bouchard]

This question did not ask you to calculate the pH value exactly. However, Ammonium ion is a weak acid. You should know it will most likely be in a pH range 4-6. Choice C is therefore correct.

Hope this helps.

Ok, ok, I should have seen that. That makes sense. Thank you.

I keep thinking each question I ask you will be my last, but then I find another! I appreciate your help. Here is one: From the 2017 Math Destroyer #4 reads: The endpoints of the diagonal of a square are located at (2,4) and (6,2). Find the area of the square.

If I am understanding this correctly, we are given two points of the corner of a square, the top-left corner and the bottom-right, then asked to find the area of the square. However, this square appears to have a height and length that are not equal to each other. Would this not be a rectangle? If this were a square this problem should be much easier to solve than the method given in the answer key. You should just be able to square the difference between either the x coordinates or the y coordinates to find the area of a square, correct?

Thank you,

Parker Green

 

[DAT Destroyer]

Ok, ok, I should have seen that. That makes sense. Thank you.

I keep thinking each question I ask you will be my last, but then I find another! I appreciate your help. Here is one: From the 2017 Math Destroyer #4 reads: The endpoints of the diagonal of a square are located at (2,4) and (6,2). Find the area of the square.

If I am understanding this correctly, we are given two points of the corner of a square, the top-left corner and the bottom-right, then asked to find the area of the square. However, this square appears to have a height and length that are not equal to each other. Would this not be a rectangle? If this were a square this problem should be much easier to solve than the method given in the answer key. You should just be able to square the difference between either the x coordinates or the y coordinates to find the area of a square, correct?

Thank you,

Parker Green

It is a square. A tilted one.
See the attached picture.
It's a rough sketch

 

[IWCBTYJ]

Dr. Romano,

As always, thank you for your time! I had a question that I have been struggling to understand fully and was hoping if you could help me. For question 79 on 2016 DAT destroyer, you said that CN is not a good leaving group and thus the reaction does not occur which makes perfect sense. However, I also know that CN- will do a 1,4 addition to alpha beta unsaturated compounds. The reason for the latter statement, I believe, is that CN- is "relatively" a good leaving group compared to CH3 and other groups that add 1,2 and thus able to act as a leaving group from the initially form 1,2 product to form a more stable 1,4 product. With this said, I guess I'm a bit confused as to how using my logic for why CN- is a 1,4 addition group to alpha beta unsaturated carbonyls, we wouldn't see a product forming in question 79!

Thank you so much for your help! your time and assistance is much appreciated!

 

[DAT Destroyer]

Dr. Romano,

As always, thank you for your time! I had a question that I have been struggling to understand fully and was hoping if you could help me. For question 79 on 2016 DAT destroyer, you said that CN is not a good leaving group and thus the reaction does not occur which makes perfect sense. However, I also know that CN- will do a 1,4 addition to alpha beta unsaturated compounds. The reason for the latter statement, I believe, is that CN- is "relatively" a good leaving group compared to CH3 and other groups that add 1,2 and thus able to act as a leaving group from the initially form 1,2 product to form a more stable 1,4 product. With this said, I guess I'm a bit confused as to how using my logic for why CN- is a 1,4 addition group to alpha beta unsaturated carbonyls, we wouldn't see a product forming in question 79!

Thank you so much for your help! your time and assistance is much appreciated!

Great question. The reaction is easy to do, but undergrad books do not go into an explanation .
CN- is a powerful nucleophile because it has a high energy HOMO......this is an orbital that will do the " attack ". The HOMO must attack a LUMO........if the orbitals are compatible in size and energy, a reaction can occur. In 1,4- unsaturated compounds, the LUMO is at position 4..residing on the beta carbon......thus the reaction readily occurs. Alcohols, thiols, Iodine, Phosphorus nucleophiles, cuprates, amines also will attack position number 4. We sometimes call these species " soft "......meaning that they are under orbital control. Grignards , Alkyl lithium reagents, and hydrides are called " hard " meaning they react with little orbital guidance, but are driven solely by CHARGE. In other words, they attack the carbonyl group, NOT the #4 or beta position. As far as leaving group ability, the CN- is a POOR leaving group in reactions such as SN1 or SN2...or most elimination reactions. The CN- as a leaving group is not well stabilized. However, in some reactions under BASIC conditions such as the Benzoin condensation, we can see the CN- group leave, however this is beyond the scope of the DAT. Let me summarize.......CN- or Cuprates go to beta carbon.......orbital controlled. Grignards , hydrides like LiAlH4,or alkyl lithium reagents go to carbonyl to form the alcohol since they are charged controlled. In most reactions, CN- is a horrendous leaving group especially Sn1 or Sn2 processes.

I hope this helps.

Dr. Romano

 

[IWCBTYJ]

Great question. The reaction is easy to do, but undergrad books do not go into an explanation .
CN- is a powerful nucleophile because it has a high energy HOMO......this is an orbital that will do the " attack ". The HOMO must attack a LUMO........if the orbitals are compatible in size and energy, a reaction can occur. In 1,4- unsaturated compounds, the LUMO is at position 4..residing on the beta carbon......thus the reaction readily occurs. Alcohols, thiols, Iodine, Phosphorus nucleophiles, cuprates, amines also will attack position number 4. We sometimes call these species " soft "......meaning that they are under orbital control. Grignards , Alkyl lithium reagents, and hydrides are called " hard " meaning they react with little orbital guidance, but are driven solely by CHARGE. In other words, they attack the carbonyl group, NOT the #4 or beta position. As far as leaving group ability, the CN- is a POOR leaving group in reactions such as SN1 or SN2...or most elimination reactions. The CN- as a leaving group is not well stabilized. However, in some reactions under BASIC conditions such as the Benzoin condensation, we can see the CN- group leave, however this is beyond the scope of the DAT. Let me summarize.......CN- or Cuprates go to beta carbon.......orbital controlled. Grignards , hydrides like LiAlH4,or alkyl lithium reagents go to carbonyl to form the alcohol since they are charged controlled. In most reactions, CN- is a horrendous leaving group especially Sn1 or Sn2 processes.

I hope this helps.

Dr. Romano

This clarifies much! Thank you so much!

 

[GDragon_1988]

Dear Dr. Romano,

I created a question that was similar to #115 Gen Chem (2016) and was wondering if you could tell me if i got them correct.

what would be produced if chloric acid, HClO3 was mixed with sulfuric acid, H2SO4?

HSO4-, ClO2+, H2O

would my products be right in this situation?

also, if i was to mix Nitrous acid with sulfuric acid, would i get HSO4-, NO++, H2O as products?

Thanks

 

[GDragon_1988]

Dr. Romano, #186 Orgo DAT Destroyer (2016).

I narrowed the answer choices to A and D, but I'm still confused. Couldn't I turn the sawhorse into a newman projection? I viewed choice D as anti-conformation if i turned it into a newman.

 

[wizkidz]

Dr Romano- i had a question on Orgo Destroyer 2016 #259. Why does the acyl halide add to the amine portion rather than the ring?

 

[Faefly]

Hi I have some questions about dat destroyer
For example, is general chemistry a set of tests. What I mean, does every 30 questions represent random questions from all of the chapters in gen chem or are the questions organized by topics? like every 20 questions correspond to a certain topic.
Same question goes for the qr section in the dat destroyer.
Also, how many questions are there per section? Bio, genchem, ochem, and qr.
Finally, how many questions are recommended to solve each day?

 

[music314]

Dr. Romano, #186 Orgo DAT Destroyer (2016).

I narrowed the answer choices to A and D, but I'm still confused. Couldn't I turn the sawhorse into a newman projection? I viewed choice D as anti-conformation if i turned it into a newman.

For this question, you are correct that choice D would be correct if it had been shown as a Newman Projection. However, it is shown as a sawhorse, and that's why the answer is choice A. But the truth still stands that a staggered anti conformation is higher in stability than a staggered gauche conformation. The point is that the question wanted you to recognize that a sawhorse is not the same as a Newman Projection. I hope that helps.

 

[GDragon_1988]

For this question, you are correct that choice D would be correct if it had been shown as a Newman Projection. However, it is shown as a sawhorse, and that's why the answer is choice A. But the truth still stands that a staggered anti conformation is higher in stability than a staggered gauche conformation. The point is that the question wanted you to recognize that a sawhorse is not the same as a Newman Projection. I hope that helps.

i see what you're saying. I guess I overanalyzed the problem. thank you

 

[DAT Destroyer]

Dr Romano- i had a question on Orgo Destroyer 2016 #259. Why does the acyl halide add to the amine portion rather than the ring?

This in NOT a Friedel Craft reaction........The Freidel Craft reaction will FAIL if a meta director is present or an NH2 group. In the case of the NH2 group, this highly reactive functionality will react with the ALCl3 catalyst and highly deactivate the molecule. This reaction has no AlCl3 catalyst, and clearly is not a Friedel Craft reaction. The NH2 group simply reacts with the acyl halide in a nucleophilic acyl substitution reaction. This is a great way to deactivate the Amino group when doing a synthetic methodology.

Hope this helps.

Dr. Jim Romano

 

[GDragon_1988]

Hey Dr. Romano------ Gen Chem #139 (2016 DAT Destroyer)

if the question said there was still a decrease from 25C but an increase in Kw instead, would it mean that the reaction would be exothermic in that situation?