Official DAT Destroyer Q&A Thread

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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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I apologize if this has been asked before but I couldn't find it. The question is in regards to angiosperms. 2014 edition question 15 in Bio. Why is cross pollination, like double fertilization, not specific to angiosperms as well? I thought cross pollination could only be done by flowering plants.

Flowering plants can also undergo self pollination, not only cross pollination
 
Question 350 of the 2016 Destroyer:

I tried to solve this question using P1V1 = P2V2, giving me an answer of 41.9 torr. However, I read the explanation and it doesn't really make sense to me. I might be lacking some knowledge here, but why are we not using the equation?

Thank you for your time
The question is asking you for the % of O2. The equation that you reference will not help. This equation deals with initial pressures and volumes, as well as well as final conditions. This is not what is being asked. Half of 760 ( 1 atm ) is 380 torr, thus you need to know the % of the gas here. This is given in the info chart.

Hope this helps.

Dr. Jim Romano
DAT Destroyer
 
BIO #15 in 2016 DAT Destroyer.

if this question applied to calcium, would inhibition cause a higher concentration of Ca2+ outside of the cell?
 
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BIO #15 in 2016 DAT Destroyer.

if this question applied to calcium, would inhibition cause a higher concentration of Ca2+ outside of the cell?
That depends on what you are blocking. Calcium is mostly pumped from inside to the outside of the cell. So, if Calcium pump is blocked, more calcium will accumulate inside the cell. Calcium channels bring calcium into the cell, so if they are blocked then calcium ion will accumulate outside of the cell.

Hope this helps.
 
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#232 in the 2016 chem destroyer

Would Br- also be considered a product of hydrolysis since putting NH4Br in water separates them
 
#232 in the 2016 chem destroyer

Would Br- also be considered a product of hydrolysis since putting NH4Br in water separates them
No. Br- is the anion of a STRONG ACID, namely HBr .It would be a spectator ion. Remember.....anions of strong acids such as Br-, Cl-. I-, NO3-, ClO4- don not do hydrolysis since they are so weak. Cations of strong bases such as Na+. Li+, K+, Rb+. Cs+, Ca++, Ba++. Sr++ also are weak and do not participate in hydrolysis reactions. This is an important concept that 99% of students never really understand. I hope this helps. Your question was an excellent one.

Keep up the great work.

Dr. Jim Romano, orgoman22
 
No. Br- is the anion of a STRONG ACID, namely HBr .It would be a spectator ion. Remember.....anions of strong acids such as Br-, Cl-. I-, NO3-, ClO4- don not do hydrolysis since they are so weak. Cations of strong bases such as Na+. Li+, K+, Rb+. Cs+, Ca++, Ba++. Sr++ also are weak and do not participate in hydrolysis reactions. This is an important concept that 99% of students never really understand. I hope this helps. Your question was an excellent one.

Keep up the great work.

Dr. Jim Romano, orgoman22

I'm afraid I'm still part of that 99%. From my understanding, hydrolysis means to break apart a molecule using water, and the products of hydrolysis is anything that has arisen from this break up. So for example, if X dissociates in water (X --> Y + Z), then Y and Z are the products of hydrolysis. Furthermore, I think that anions of strong acids and cations of strong bases do not participate in hydrolysis reactions simply because they are 1) too weak like you mentioned 2) atoms that exist alone can't be further broken down using water alone 3) the molecules made up of two or more compounds such as ClO4- can't undergo hydrolysis because they are polar and stabilized by intermolecular forces (dipole-dipole interactions with h2o in solution, dispersion forces, etc) and splitting them up would require an unfavourable reaction route that is high in energy

What would be wrong with my thinking here?
 
DAT Destroyer (2016) Bio #291

The answer is E, which I'm confused about. The question is referring to a red blood cell, and I know A is correct, but how does it swell without the presence of a cell wall like in plants?
 
DAT Destroyer (2016) Bio #291

The answer is E, which I'm confused about. The question is referring to a red blood cell, and I know A is correct, but how does it swell without the presence of a cell wall like in plants?
Your thinking is correct but you are not considering everything. If you are taking into account how much water was given, then both answers apply. First, red blood cell has to swell. Then, if the water continues coming in will burst. Think of it as a balloon filling up with air. At first it swells, but if you keep adding air into it, it will eventually burst.

Answer is correct.

Hope this helps.
 
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Your thinking is correct but you are not considering everything. If you are taking into account how much water was given, then both answers apply. First, red blood cell has to swell. Then, if the water continues coming in will burst. Think of it as a balloon filling up with air. At first it swells, but if you keep adding air into it, it will eventually burst.

Answer is correct.

Hope this helps.
hey sorry to bug you but you missed my question! thanks for your time
 
hey sorry to bug you but you missed my question! thanks for your time
Consider anions such as I- or Br-. These are stable. They are large atoms, which can stabilize their negative charges quite well. Thus, if you approach it from this perspective, it should be reasonably easy to see. Anion stabilization is the main reason. Anions of strong acids are STABLE as well.........and from a thermodynamic point of view, they would lose stability if they did react. The free energy would be Large and Positive. Likewise, cations of strong bases such as Li+ , K+, and Na would give enormously large Free energy values if they did react with a species such as OH-. This does not have to do with bonding, as you suggest. but purely thermodynamic. I hope this helps. If not, the Physical Chemistry text by my friend Raymond Chang could give more details .

Dr. Romano
 
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I'm a little confused about spontaneity and reversibility. I've read that spontaneous reactions are irreversible. However, isn't this not true since you can input energy to make it go the other direction?
 
I'm a little confused about spontaneity and reversibility. I've read that spontaneous reactions are irreversible. However, isn't this not true since you can input energy to make it go the other direction?
A spontaneous reaction, means it can go forward and has a negative Delta G.
If you are putting in energy. The reaction will have a positive Delta G. Therefore, reaction is no longer considered spontaneous.

Hope this helps.
 
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General question regarding esters:

In esterification, we typically add H+ to deprotonate the carbonyl oxygen first, and in Claisen condensations, we add H3O+ at the end to stop deprotonating the alpha hydrogen. But wouldn't the H+ also turn the ester into a carboxylic acid?
 
General question regarding esters:

In esterification, we typically add H+ to deprotonate the carbonyl oxygen first, and in Claisen condensations, we add H3O+ at the end to stop deprotonating the alpha hydrogen. But wouldn't the H+ also turn the ester into a carboxylic acid?
The first step of the Fischer esterification is protonation of the carbonyl group. This makes the carbonyl group more reactive towards the nucleophilic attack by the alcohol. In the Claisen reaction, we are dealing with a very different situation. We start in a basic medium. Sodium ethoxide or sodium methoxide first forms an ester enolate intermediate . This enolate attacks a second molecule. No acid is used up to this point. An anion is generated in the final mechanistic step......and acid simply stabilizes the final product to the neutral form for eventual isolation. Two very different mechanisms are being used. The first uses acid as a protonating source, the other as what we call a quench. Hope this helps.
 
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The first step of the Fischer esterification is protonation of the carbonyl group. This makes the carbonyl group more reactive towards the nucleophilic attack by the alcohol. In the Claisen reaction, we are dealing with a very different situation. We start in a basic medium. Sodium ethoxide or sodium methoxide first forms an ester enolate intermediate . This enolate attacks a second molecule. No acid is used up to this point. An anion is generated in the final mechanistic step......and acid simply stabilizes the final product to the neutral form for eventual isolation. Two very different mechanisms are being used. The first uses acid as a protonating source, the other as what we call a quench. Hope this helps.
Thank you Dr. Romano!
 
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DAT Destroyer (2016) Gen Chem #79

For #79, I solved the problem the wrong way and got the correct answer lol. In the solutions, it says that Aluminum is the limiting reagent, which i don't understand. Would my logic be wrong in this situation?
Photo on 2-28-17 at 12.09 AM #3.jpg
 
2016 Math destroyer, Test 10 #20
how do you know that the 2 triangles are similar and their sides are proportional?
thanks
 
Freezing point of a .5m Ca[C2H3O2] solution is -2.8 degrees celcius. What would be the freezing point of .5m NaC2H3O2 solution assuming identical experimental conditions?

Dr. Romano, (or anyone who is proficient in this topic), Could you break this down for me a little bit? Even after reading the solution- I am not able to mentally approach this problem. I.E. what is the first step etc. How would i solve this? Thank you in advance :)
 
Dr. Romano,

Quick question on the boiling points of toluene (111C) and diethyl ether (35C) (Organic Chemistry, #41, 2017 version): it seems to me that diethyl ether has dipole-dipole interactions while toluene only has London Dispersion. Is it because of toluene's bigger size that it has a higher boiling point, despite the weaker intermolecular forces?

Thank you.
 
Freezing point of a .5m Ca[C2H3O2] solution is -2.8 degrees celcius. What would be the freezing point of .5m NaC2H3O2 solution assuming identical experimental conditions?

Dr. Romano, (or anyone who is proficient in this topic), Could you break this down for me a little bit? Even after reading the solution- I am not able to mentally approach this problem. I.E. what is the first step etc. How would i solve this? Thank you in advance :)
The concept is this......the more particles of solute that is dissolved in the solvent will cause the freezing point to lower, and the boiling point to be raised. Calcium acetate gives a total of 3 particles.....one Ca++ and 2 Acetate ions for a total of 3 particles . Sodium acetate only gives 2 particles...Na+ and acetate. Thus sodium acetate will lower the freezing point but NOT AS MUCH since there are fewer particles dissociated. So we are looking for a value not quite as low as -2.8 degrees. Hope this helps.
 
Hello, Dr Romano,

On organic chemistry odyssey, chapter 14, question 27(2015 version I believe), I was not sure why N-ethylpropanamide would hydrolyze slower than ethylacetate. I looked up wiki and stated that the acetate ion is the least reactive one. I also answered amide would be the slowest but now I am confused by this. Could you clarify this?
wiki: https://en.wikipedia.org/wiki/Nucleophilic_acyl_substitution

Another question, chapter 14 question 44, saponification is based promoted ester hydrolysis. Are other ester/amide hydrolysis also base promoted? and acid promoted hydrolysis? If not, what would be the examples of base catalyzed reaction and acid catalyzed reaction? I am a bit confused by the wording.

Lastly, on chapter 18 question 50, I was confused by your solution because I thought the positive amino acids move to the cathode and the negative amino acids move to the anode. How do you determine which amino acids are negative given the PI value?

Thank you so so much!
 
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Hi Dr. Ramano,

2 questions regarding the solubility of ionic solutes:

1) Ag3PO4 is considered neutral, is that correct?
Because Ag+ is acidic and PO4^3- is basic, so the overall specie could be approximately neutral.

2) If we add H+ to a solution with Ag3PO4, will it increase its solubility?
I think it will (but want to double check with you), because in DAT Destroyer, you mentioned that only anions of weak acids will participate in acid-base reactions. So PO4^3- will react with water to form OH-, and adding H+ will result in an increased PO4^3- solubility.
 
Dr. Romano,

Quick question on the boiling points of toluene (111C) and diethyl ether (35C) (Organic Chemistry, #41, 2017 version): it seems to me that diethyl ether has dipole-dipole interactions while toluene only has London Dispersion. Is it because of toluene's bigger size that it has a higher boiling point, despite the weaker intermolecular forces?

Thank you.
Indeed so. This is a great example when Van Der Waals interactions greatly trump dipole-dipole interactions. Recall, ethers are very volatile compounds and have very low boiling points. This suggest that there will be weak intermolecular attractions. Hope this helps.
 
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Indeed so. This is a great example when Van Der Waals interactions greatly trump dipole-dipole interactions. Recall, ethers are very volatile compounds and have very low boiling points. This suggest that there will be weak intermolecular attractions. Hope this helps.
Thank you for clarifying this for me. It makes a lot of sense now. And thank you so much for promptly answering our questions.
 
Hey Dr. Romano, You never got to my questoin about the Al being the limiting agent for Gen Chem #79 on the 2016 Dat Destroyer. I got Cr2O3 as the limiting reagent.

Thank you for always responding even though you have a busy schedule.
 
Dr. Romano,

I had a conceptual question regarding DAT destroyer 2016 version Organic Chemistry question 12.

For this question, I completely understand why the answer is B, but I am a bit confused as to why we would not see dehydration occur to form a stable diene compound. When solving the problem, I assumed that this would be a minor product and hence, is not the answer in this case, but was wondering what would inhibit the alcohol group from leaving in presence of H3O. I was thinking that the Grignard or water could act as the base and cause E1 to form Diene or even react as SN1, but in almost all problems I see I never see dehydration as part of the option in this sort of question. Could you clarify why we might not see a dehydration in this case, but do witness dehydration in cases such as in Claisen-Schmidt condensation.

Thank you in advance for your help and time!!
 
Dr. R

Organic chemistry #212

i understand that we are looking at a simple self claison reaction.
however, i have a question about the second step....if you followed the claison reaction with acid (specificallly H3O+) wouldn't that cause the ester to be changed to a COOH. the answer that was in the book had kept it as an ester. I assume that you wanted us to understand that the acid was placed there to reprotinate the final product since deprotination would occur between the carbonyls. but isn't reprotination usually done with water instead of acid to specifically avoid changing the final ester to carboxylic acid?


ps. thank you so much for writing the most challenging thing in the world. i cry tears of joy and misery at the same time. i didnt know that was possible.
 

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Dr. R

Organic chemistry #212

i understand that we are looking at a simple self claison reaction.
however, i have a question about the second step....if you followed the claison reaction with acid (specificallly H3O+) wouldn't that cause the ester to be changed to a COOH. the answer that was in the book had kept it as an ester. I assume that you wanted us to understand that the acid was placed there to reprotinate the final product since deprotination would occur between the carbonyls. but isn't reprotination usually done with water instead of acid to specifically avoid changing the final ester to carboxylic acid?


ps. thank you so much for writing the most challenging thing in the world. i cry tears of joy and misery at the same time. i didnt know that was possible.
You need to examine the mechanism of the Claisen carefully. The final step acid is indeed employed as you stated, however it is used for a purpose, The purpose is to protonate the anion in the final step. This acid then becomes water, and will not react with the final product, which is our beta keto ester.

Hope this helps

Dr. Romano
 
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You need to examine the mechanism of the Claisen carefully. The final step acid is indeed employed as you stated, however it is used for a purpose, The purpose is to protonate the anion in the final step. This acid then becomes water, and will not react with the final product, which is our beta keto ester.

Hope this helps

Dr. Romano

thank you so much for the clarification! you are greatly appreciated. :)
 
Hello, Dr Romano,

On organic chemistry odyssey, chapter 14, question 27(2015 version I believe), I was not sure why N-ethylpropanamide would hydrolyze slower than ethylacetate. I looked up wiki and stated that the acetate ion is the least reactive one. I also answered amide would be the slowest but now I am confused by this. Could you clarify this?
wiki: https://en.wikipedia.org/wiki/Nucleophilic_acyl_substitution

Another question, chapter 14 question 44, saponification is based promoted ester hydrolysis. Are other ester/amide hydrolysis also base promoted? and acid promoted hydrolysis? If not, what would be the examples of base catalyzed reaction and acid catalyzed reaction? I am a bit confused by the wording.

Lastly, on chapter 18 question 50, I was confused by your solution because I thought the positive amino acids move to the cathode and the negative amino acids move to the anode. How do you determine which amino acids are negative given the PI value?

Thank you so so much!


Dr Romano,

I know there are a lot of questions to be answered but don't forget about this one!!

Thank you so much! and I really appreciate this :)
 
Dr Romano,

I know there are a lot of questions to be answered but don't forget about this one!!

Thank you so much! and I really appreciate this :)

Question:

On organic chemistry odyssey, chapter 14, question 27(2015 version I believe), I was not sure why N-ethylpropanamide would hydrolyze slower than ethylacetate. I looked up wiki and stated that the acetate ion is the least reactive one. I also answered amide would be the slowest but now I am confused by this. Could you clarify this?

Response:
Due to their electronic similarity to esters, amides can also undergo hydrolysis, but they are less susceptible to nucleophilic attack such as that involved in hydrolysis reactions, mainly due to their electronic stabilization. Resonance is a dominating factor in amide lack of reactivity. The acetate ion that you reference does not hydrolyze in acid, and you are comparing two different things. Thus, the bottom line is this.....Amides and esters both can hydrolyze in acid, but amides are much slower and more vigorous conditions are employed. For example, it took me 3 hours , 100 degrees C, with 70% Sulfuric acid to hydrolyze an aromatic amide !!!! If you want more on this, the David Klein text can help. ....Acetate is indeed UNREACTIVE in nuclephilic acyl substitution reactions as Wiki stated. Hydrolysis in acid or base is a DIFFERENT mechanism. Hope that is cleared up now.


Dr. Romano
 
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Destroyer version '17

Gen Chem #73. Why is D false? I thought all ideal gases had the same kinetic energy at constant temperatures. Isn't that kind of how Graham law of effusion/diffusion was derived? Set the kinetic energies of 2 molecules equal to each other and then get the ratios of molarity and rates. The explanation says because ideal gases can have different speeds. Well that's true if they had different weights too.
Thanks in advance for any possible clearing of this confusion.
 
Destroyer version '17

Gen Chem #73. Why is D false? I thought all ideal gases had the same kinetic energy at constant temperatures. Isn't that kind of how Graham law of effusion/diffusion was derived? Set the kinetic energies of 2 molecules equal to each other and then get the ratios of molarity and rates. The explanation says because ideal gases can have different speeds. Well that's true if they had different weights too.
Thanks in advance for any possible clearing of this confusion.
If you are at a given temperature, molecules have different kinetic energies,,,,,but they have an AVERAGE kinetic energy. Here is how I explain this to my students.....Think of it as a group of children in a class. They clearly have different energies. Some are sleepy, some are alert, some...who knows ? Clearly they are not all at the same energy. However.....I could, however, give you an average rating of say 7 on a scale of 1 to 10. Hopefully you see,,,,,at a given temperature,,,,AVERAGE kinetic energy can be stated, but individually the molecules are not the same.

Hope this helps.

Dr. Romano
 
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If you are at a given temperature, molecules have different kinetic energies,,,,,but they have an AVERAGE kinetic energy. Here is how I explain this to my students.....Think of it as a group of children in a class. They clearly have different energies. Some are sleepy, some are alert, some...who knows ? Clearly they are not all at the same energy. However.....I could, however, give you an average rating of say 7 on a scale of 1 to 10. Hopefully you see,,,,,at a given temperature,,,,AVERAGE kinetic energy can be stated, but individually the molecules are not the same.

Hope this helps.

Dr. Romano
Yes, this definitely clears things up a bit. Thank you for your response. I see I need to focus on the word "average".
 
i posted a question but figured it out. please disregard! sorry
 
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Hey Dr. Romano,

(2016) DAT Destroyer Bio #524

1.) would the overall charge of lysine by 0 when it's at pH 11.5?
2.)at pH, 7, would lysine's overall charge be +1?
 
Hey Dr. Romano,

(2016) DAT Destroyer Bio #524

1.) would the overall charge of lysine by 0 when it's at pH 11.5?
2.)at pH, 7, would lysine's overall charge be +1?
You are correct about pH 7. However at pH 11.5 Lysine would have a -1 charge, if you consider both NH3 groups deprotonated.

Hope this helps.
 
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Dr. Romano,

Sorry for bugging you like this, but it seems you have missed my question again! I understand how busy you must be, so thank you in advance for your input and help! To save time, I will repost my question here! Looking forward to hearing from you soon, thank you so much for your time and help as always!!

I had a conceptual question regarding DAT destroyer 2016 version Organic Chemistry question 12.

For this question, I completely understand why the answer is B, but I am a bit confused as to why we would not see dehydration occur to form a stable diene compound. When solving the problem, I assumed that this would be a minor product and hence, is not the answer in this case, but was wondering what would inhibit the alcohol group from leaving in presence of H3O. I was thinking that the Grignard or water could act as the base and cause E1 to form Diene or even react as SN1, but in almost all problems I see I never see dehydration as part of the option in this sort of question. Could you clarify why we might not see a dehydration in this case, but do witness dehydration in cases such as in Claisen-Schmidt condensation.

Thank you in advance for your help and time!!
 
Dr. Romano,

Sorry for bugging you like this, but it seems you have missed my question again! I understand how busy you must be, so thank you in advance for your input and help! To save time, I will repost my question here! Looking forward to hearing from you soon, thank you so much for your time and help as always!!

I had a conceptual question regarding DAT destroyer 2016 version Organic Chemistry question 12.

For this question, I completely understand why the answer is B, but I am a bit confused as to why we would not see dehydration occur to form a stable diene compound. When solving the problem, I assumed that this would be a minor product and hence, is not the answer in this case, but was wondering what would inhibit the alcohol group from leaving in presence of H3O. I was thinking that the Grignard or water could act as the base and cause E1 to form Diene or even react as SN1, but in almost all problems I see I never see dehydration as part of the option in this sort of question. Could you clarify why we might not see a dehydration in this case, but do witness dehydration in cases such as in Claisen-Schmidt condensation.

Thank you in advance for your help and time!!

In the first step, we oxidized the molecule to a ketone. After the Grignard reaction was done, we did what is called a quench. Very dilute acid...a small amount in an ice bath !!! This would be enough to form the product, but NOT enough to protonate the product and do the E1 as you suggested. It is done cold in an ice bath if you recall from lab. Clearly insufficient to make the product reaction further. I am glad to see such a pedantic eye and perspicacious analysis of product formation !!!

Dr. Romano

Hope this helps.
 
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Dr. Romano,

Cna you please explain Gen chem #107 in the 2016 destroyer. I am confused as to how NA2SO4 as an i of 3 and how we can even use this property to differentiate freezing points. I thought freezing point depression formula found a change in temperature, not the overall actual freezing point? Hopefully that just made sense, haha.
 
On #113 Gen Chem 2016 destroyer, in the solutions, can you please explain how FeCl3 and AlCl3 are acidic? I was trying to understand it as explained through the solution but Im not quite getting it. Thank You!
 
On #113 Gen Chem 2016 destroyer, in the solutions, can you please explain how FeCl3 and AlCl3 are acidic? I was trying to understand it as explained through the solution but Im not quite getting it. Thank You!
Do a little trick......which is shown in a later problem. Write out AlCl3......underneath it write out HOH ( water ).....and criss cross the two...we obtain Aluminum hydroxide and Hydrochloric acid.......Aluminum hydroxide is a weak base, Hydrochloric acid is a strong acid. These are the two " parent molecules " that reacted to give the salts you have referenced. The stronger component determines the pH.

I hope this helps.....If not......I wrote a clear solution to this in another problem.

Dr. Romano
 
In the first step, we oxidized the molecule to a ketone. After the Grignard reaction was done, we did what is called a quench. Very dilute acid...a small amount in an ice bath !!! This would be enough to form the product, but NOT enough to protonate the product and do the E1 as you suggested. It is done cold in an ice bath if you recall from lab. Clearly insufficient to make the product reaction further. I am glad to see such a pedantic eye and perspicacious analysis of product formation !!!

Dr. Romano

Hope this helps.

Thank you so much for the clarification!! Most people get pretty annoyed with such specific questions like mine, but thank you for taking the time to answer them, they really help solidifying my understanding.
 
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Dr Romano,

Is it true that when counting the hybridization of an atom, the radical around it does NOT count, but when considering the molecular shape, a radical counts as an electron cloud?

Say for NO2, the ONO bond angle in NO2 is slightly over 120. But does the central N atom have an sp hybridization instead of sp2?

Thank you for your time in advance!
 
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.
 
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