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!

 

[Denevistas]

If you look at the right side of the equation, the charge is 6+, because
(2Cr3+)

On the left side, the charge is 12+. So to equal out the charges, you need to place 6 electrons on the left. This will contribute to 6- charge and you get 6+ on the left side. The charges must be balanced.

Hope this helps.
.

🙂
Thanks Dr. Jim
Hope the next question isnt such elementary!

 

[DAT Destroyer]

.
🙂
Thanks Dr. Jim
Hope the next question isnt such elementary!

Always happy to help!

 

[plusalpha]

The question I have is about a really fundamental concept, but I don't seem to have nailed it down, so I'd like to know the answer once and for all.
I have trouble distinguishing when to use the alkoxide as a base and when to use it as a nucleophile... I remember Chad say in his video that an alkoxide with a tertiary carbon goes E2, while with a primary carbon it goes SN2. Is this correct? Also, will the solvent play a role in differentiating E2 and SN2?

 

[plusalpha]

For OC question #267 in the 2015 Destroyer, I don't understand that second step where two Cl's are added ortho to the NH2 group on the aromatic ring. Is it an identical step like Cl2/AlCl3?

 

[DAT Destroyer]

For OC question #267 in the 2015 Destroyer, I don't understand that second step where two Cl's are added ortho to the NH2 group on the aromatic ring. Is it an identical step like Cl2/AlCl3?

The NH2 group is EXTREMELY activating....... and will add both groups ortho to it. This group is one of the most reactive of all groups. In real life, if we wanted to only do it once, we would have to protect it with acetyl chloride. Thus if you had aniline and brominated it, we would get the bromines in the ortho and para positions !

Hope this helps.

Dr. Romano

 

[Frank22]

The NH2 group is EXTREMELY activating....... and will add both groups ortho to it. This group is one of the most reactive of all groups. In real life, if we wanted to only do it once, we would have to protect it with acetyl chloride. Thus if you had aniline and brominated it, we would get the bromines in the ortho and para positions !

Hope this helps.

Dr. Romano

Hey doc, what would be more activating; -NR2 or -OR groups? Happy Halloween by the way 🙂!

By the way, I've been reading a lot about the use of C60 fullerene as a blocker of HIV-1 protease; organic chemistry is such a fascinating field.

 

[DAT Destroyer]

Hey doc, what would be more activating; -NR2 or -OR groups? Happy Halloween by the way 🙂!

By the way, I've been reading a lot about the use of C60 fullerene as a blocker of HIV-1 protease; organic chemistry is such a fascinating field.

Happy Halloween back to you!

An NR2 group would trump an OR group. If you would love to see a wonderful ranking of groups, see the text by Francis Carey.

 

[plusalpha]

The NH2 group is EXTREMELY activating....... and will add both groups ortho to it. This group is one of the most reactive of all groups. In real life, if we wanted to only do it once, we would have to protect it with acetyl chloride. Thus if you had aniline and brominated it, we would get the bromines in the ortho and para positions !

Hope this helps.

Dr. Romano

Thank you for your prompt reply Dr. Romano.
For the answer to GC question #2, why are we using double negatives to represent the rate of the reaction? Doesn't the negative sign in the equation (-delta[x]/delta t) already signify that the concentration of X (a reactant) is decreasing? When we do 0.5-0.6, I'm thinking that the negative sign that results from the calculation is in fact changing the original rate equation back to positive.

 

[Izzo10]

Hi, I have a question about Gen Chem, Q: 145. For some reason I'm having trouble wrapping my head around exactly what this change in solubility will look like. For instance, this is how I went about trying to solve it:

Question asks how solubility will be affected, so I put each compound in H2O:
CuCl(s) + H2O(l) --> CuOH(s?) + HCl(aq)
Now I add in some acid and see what happens:
CuOH(s) + HCl(aq) + H+(aq) --> H2O + Cu2+(aq) + HCl(aq)

So from here I start getting confused as to what this might mean in terms of the solubility of CuCl(s). Can you please break this down very simply step by step so I can understand what everything means. Because it appears there is a severe gap in my understanding. Thank you!

 

[DAT Destroyer]

Thank you for your prompt reply Dr. Romano.
For the answer to GC question #2, why are we using double negatives to represent the rate of the reaction? Doesn't the negative sign in the equation (-delta[x]/delta t) already signify that the concentration of X (a reactant) is decreasing? When we do 0.5-0.6, I'm thinking that the negative sign that results from the calculation is in fact changing the original rate equation back to positive.

Yes.....the rate of a reaction is a POSITIVE quantity that tells us how the concentration of a reactant or product changes with time. The negative sign simply means that a reactant is decreasing,,,,,and thus will allow you to arrive at a positive rate. You indeed have the right idea.


Hope this helps.

Dr. Romano

 

[DAT Destroyer]

Hi, I have a question about Gen Chem, Q: 145. For some reason I'm having trouble wrapping my head around exactly what this change in solubility will look like. For instance, this is how I went about trying to solve it:

Question asks how solubility will be affected, so I put each compound in H2O:
CuCl(s) + H2O(l) --> CuOH(s?) + HCl(aq)
Now I add in some acid and see what happens:
CuOH(s) + HCl(aq) + H+(aq) --> H2O + Cu2+(aq) + HCl(aq)

So from here I start getting confused as to what this might mean in terms of the solubility of CuCl(s). Can you please break this down very simply step by step so I can understand what everything means. Because it appears there is a severe gap in my understanding. Thank you!

Your way is long, and inefficient. The concept is very important to understand. The solid would be soluble if and only if an acid -base reaction can occur. Focus on the anions.... Br-....Cl-.....I-.......these anions are all weak.....since they come from strong acids. Essentially, they do ZERO.....they are spectators. Thus our rule is that anions of strong acids and cations of strong bases are spectators....a rule that few students ever consider. The compound with a Fluorine would be affected by pH because the F- is an anion of a relatively weak acid. Thus,,,,,,F- concentration would dramatically change, hence the salt solubility.

Hope this helps.

Dr. Romano

 

[Izzo10]

This helps immensely. Thank you!

 

[DAT Destroyer]

This helps immensely. Thank you!

Always happy to help.

Dr. Romano

 

[plusalpha]

In the solution to GC question #335 about Beers law, shouldn't the extinction coefficient constant (e) and cell path length (b) be grouped together in a parentheses? This may sound like a petty objection, but if the operation 0.800/(11,000)(1) is calculated as is, the 0.800 is divided by 11,000 and then MULTIPLIED by 1. In this case, the 1 doesn't change the answer, but if the cell path length were a different value, the answer definitely would have been different. In other words, the answer technically should be c=A/(eb)...? Wanna hear what other people think about this. Thanks!

 

[Lnguyen7]

In the solution to GC question #335 about Beers law, shouldn't the extinction coefficient constant (e) and cell path length (b) be grouped together in a parentheses? This may sound like a petty objection, but if the operation 0.800/(11,000)(1) is calculated as is, the 0.800 is divided by 11,000 and then MULTIPLIED by 1. In this case, the 1 doesn't change the answer, but if the cell path length were a different value, the answer definitely would have been different. In other words, the answer technically should be c=A/(eb)...? Wanna hear what other people think about this. Thanks!

A = ebc, so c = A/(eb). You are right.

 

[plusalpha]

Is there a "special" way to recognize whether ketones/aldehyes will undergo Aldol condensation when a reagent that can act as both a nucleophile or a base? For example, in OC question number 68, dilute NaOH is added to propanal. I predicted that the OH- will act as a nucleophile and attack the carbonyl carbon. I know mechanism of the Aldol condensation, I just have trouble easily recognizing that it will be an Aldol reaction..

 

[DAT Destroyer]

Is there a "special" way to recognize whether ketones/aldehyes will undergo Aldol condensation when a reagent that can act as both a nucleophile or a base? For example, in OC question number 68, dilute NaOH is added to propanal. I predicted that the OH- will act as a nucleophile and attack the carbonyl carbon. I know mechanism of the Aldol condensation, I just have trouble easily recognizing that it will be an Aldol reaction..

You have a GREAT question.....Once upon a time.....many moons ago, I asked my professor the exact same thing !!!!! The Aldol condensation occurs if you have an aldehyde or simple ketone reacting with dilute base such as NaOH or acid. Now.....not any aldehyde or ketone reacts, but ones that have an ALPHA hydrogen. In base, the first step is the formation of the enolate......in acid, we see the formation of the enol. The second step is the nucelophilic attack onto the second molecule of reactant. Now,,,,,,you asked,,,,,why cant the base attack the carbonyl and open it up . LOL......Most authors NEVER address this perspicacious question.....the answer is simple....IT DOES !!!!! However, the product is a diol, and the reaction quickly reverts back to what you started with. The other alternative is abstraction of the alpha proton, namely the ALDOL. This reaction is favored much much more than the former process. I am glad to see you are alert. A great scientist , doctor, or dentist must always be alert to other possible outcomes.

Thanks for a great question.

I hope this helps.

Dr. Romano

 

[plusalpha]

You have a GREAT question.....Once upon a time.....many moons ago, I asked my professor the exact same thing !!!!! The Aldol condensation occurs if you have an aldehyde or simple ketone reacting with dilute base such as NaOH or acid. Now.....not any aldehyde or ketone reacts, but ones that have an ALPHA hydrogen. In base, the first step is the formation of the enolate......in acid, we see the formation of the enol. The second step is the nucelophilic attack onto the second molecule of reactant. Now,,,,,,you asked,,,,,why cant the base attack the carbonyl and open it up . LOL......Most authors NEVER address this perspicacious question.....the answer is simple....IT DOES !!!!! However, the product is a diol, and the reaction quickly reverts back to what you started with. The other alternative is abstraction of the alpha proton, namely the ALDOL. This reaction is favored much much more than the former process. I am glad to see you are alert. A great scientist , doctor, or dentist must always be alert to other possible outcomes.

Thanks for a great question.

I hope this helps.

Dr. Romano

And thank YOU sir for your relentless replies with answers to our questions!

 

[DAT Destroyer]

And thank YOU sir for your relentless replies with answers to our questions!

Always glad to help, keep up the good work champ!

 

[plusalpha]

For question #26 of Math Destroyer 2015 practice test 5, is the fact that the bottom triangle is a 30-60-90 triangle an example of an assumption that we have to make on test day? From the information given in the question and the diagram alone, we can't say that the angle that LOOKS like a right angle is actually a 90 degree angle. If anyone could clarify what we can/cannot assume in the QR section diagrams, that would be great!

 

[DAT Destroyer]

For question #26 of Math Destroyer 2015 practice test 5, is the fact that the bottom triangle is a 30-60-90 triangle an example of an assumption that we have to make on test day? From the information given in the question and the diagram alone, we can't say that the angle that LOOKS like a right angle is actually a 90 degree angle. If anyone could clarify what we can/cannot assume in the QR section diagrams, that would be great!

To solve this problem, we have to assume that a bottom triangle is right triangle ( 30-60-90), otherwise you can't solve the problem. A good place to start is look at the answers. Since they all involve trig functions, it's a hint that the triangle is right .

Hope this helps!

 

[plusalpha]

To solve this problem, we have to assume that a bottom triangle is right triangle ( 30-60-90), otherwise you can't solve the problem. A good place to start is look at the answers. Since they all involve trig functions, it's a hint that the triangle is right .

Hope this helps!

Yup it helped a lot, thanks Dr. Romano!

In GC question 270 in Destroyer 2015, it asks for what temperature will a fixed amount of gas occupy the specified volume. Because the question did not specify the unit of temperature, I thought that answer choice B would also have been a correct choice. The Torr units cancel out and so do the Liter units, and we're left with only degrees celsius in our units, which looks perfectly legitimate to me. So my question is, in the gen chem section if the question does not ask for the specific temperature unit, do we assume that it has to be always in Kelvin and not degrees celsius?

 

[DAT Destroyer]

Yup it helped a lot, thanks Dr. Romano!

In GC question 270 in Destroyer 2015, it asks for what temperature will a fixed amount of gas occupy the specified volume. Because the question did not specify the unit of temperature, I thought that answer choice B would also have been a correct choice. The Torr units cancel out and so do the Liter units, and we're left with only degrees celsius in our units, which looks perfectly legitimate to me. So my question is, in the gen chem section if the question does not ask for the specific temperature unit, do we assume that it has to be always in Kelvin and not degrees celsius?

Good thoughtful question. Celsius seems to be getting a bad name these days, especially in General Chemistry calculations. When you use any of these gas laws, ALWAYS work in Kelvin. If a gas was at 0 degree Celsius, which is perfectly reasonable, you cannot plug a zero into any equation and get a result . However, after you get your answer in Kelvin, check the answer choices….There are many tricky and evil exam makers out there ( not mentioning any names !!! ). The final answer can always be changed from your Kelvin answer to Celsius or Fahrenheit.



Hope this helps.



Dr. Romano

 

[dentalstudent2021]

Math destroyer Test 5 #24

The sum of two numbers if 123. if the larger number is divided by the smaller, the quotient is 5 and the remainder is 3. what is the larger number ?

--don't understand this equation L = 5s+3. L being the larger number and s being the smaller number.
Thanks.

 

[DAT Destroyer]

Math destroyer Test 5 #24

The sum of two numbers if 123. if the larger number is divided by the smaller, the quotient is 5 and the remainder is 3. what is the larger number ?

--don't understand this equation L = 5s+3. L being the larger number and s being the smaller number.
Thanks.

When the larger number is divided by the smaller number, the quotient is 5 and the remainder is 3. This translates to: L/S = 5 R 3 which means L = 5S + 3. Take for example when you divide 17 by 3, you get 5 and the remainder is 2. This is the same as saying 17 = 5*3 + 2

Hope this helps..

 

[NYDent22]

Hey
#62 orgo 2015

Shouldn't the N be negative since it is coming to replace the electrons that the iodine took with it??
Ty

 

[DAT Destroyer]

Hey
#62 orgo 2015

Shouldn't the N be negative since it is coming to replace the electrons that the iodine took with it??
Ty

No. NH3 is ammonia. It is neutral. Whenever Nitrogen has 3 bonds and one lone pair, it will have 0 formal charge. Nitrogen with 2 bonds and 2 lone pairs have negative charge, and Nitrogen with 4 bonds and no lone pairs have + charge.

This is an Sn2 reaction. Ammonia is a good nucleophile. Nucleophile does not always have to carry a negative charge. It must be able to donate electrons.

Hope this helps.

 

[NYDent22]

Thank you!! It helped a lot!

 

[kittycat090198]

Hi!

For #273 of the GC section, the explanation in the back says "Thus, 2x=1" I guess I'm just confused as to what you're setting equal to each other to solve for this? It might be because I'm overtired but I keep looking at it over and over and nothing is jumping out to me.

Thanks!

 

[DAT Destroyer]

Hi!

For #273 of the GC section, the explanation in the back says "Thus, 2x=1" I guess I'm just confused as to what you're setting equal to each other to solve for this? It might be because I'm overtired but I keep looking at it over and over and nothing is jumping out to me.

Thanks!

This is an equilibrium problem. You are forming 2 moles of Ammonia in a 2 liter mixture.
So the molarity of Ammonia is 2/2 = 1.

If you look at the balanced reaction, you will see that you the ratio of Nitrogen to ammonia is 1:2 respectively. Since the coefficient of ammonia is 2, therefore it is 2x. The problem tells you that equilibrium concentration of ammonia is 2 moles. So 2 moles/ 2 liters = 1.
That is where you get 2x=1.

Hope this helps.

 

[kittycat090198]

Hi -

For destroyer 2015, OC # 31, it's asking for the compound that is most basic.
In the explanation in the back it says that electron withdrawing groups decrease the basicity of a compound, and while I understand that, it says that choices B, C, and E all have EWG.

According to this, http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch12/subseff.gif choice C's substituent is mildly activating and therefore an EDG, so shouldn't this increase basicity? Same goes for B's :NH2 group, considered by this list to be highly activating and therefore also an EDG.


Thanks!

 

[DAT Destroyer]

Hi -

For destroyer 2015, OC # 31, it's asking for the compound that is most basic.
In the explanation in the back it says that electron withdrawing groups decrease the basicity of a compound, and while I understand that, it says that choices B, C, and E all have EWG.

According to this, http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch12/subseff.gif choice C's substituent is mildly activating and therefore an EDG, so shouldn't this increase basicity? Same goes for B's :NH2 group, considered by this list to be highly activating and therefore also an EDG.


Thanks!

Hello kittycat090198,

Looking at Activating and Deactivating groups is NOT a good way to do this. Focus your attention on the Nitrogen. Aromatic amines are not very good bases,,,,,since the benzene ring itself removes electron density OFF THE NITROGEN, and makes it less available to capture a proton. This eliminates B and E. Amides have resonance....thus are horrible bases.....thus C is gone. A and B battle it out . B has electrons in an sp2 orbital..... A has electrons in an sp3 orbital. The electrons are more loosely available in the sp3 compound, thus can be used to capture a proton. A is correct.

Hope this helps.

Dr. Romano

 

[kittycat090198]

Hello kittycat090198,

Looking at Activating and Deactivating groups is NOT a good way to do this. Focus your attention on the Nitrogen. Aromatic amines are not very good bases,,,,,since the benzene ring itself removes electron density OFF THE NITROGEN, and makes it less available to capture a proton. This eliminates B and E. Amides have resonance....thus are horrible bases.....thus C is gone. A and B battle it out . B has electrons in an sp2 orbital..... A has electrons in an sp3 orbital. The electrons are more loosely available in the sp3 compound, thus can be used to capture a proton. A is correct.

Hope this helps.

Dr. Romano

Awesome thanks! I was just confused as to why the EWG/EDG thing was mentioned in the solutions as I was using the logic you just explained to choose my answer!

 

[DAT Destroyer]

Awesome thanks! I was just confused as to why the EWG/EDG thing was mentioned in the solutions as I was using the logic you just explained to choose my answer!

Glad I could help! Keep up the great work.

Dr. Romano

 

[plusalpha]

Gen. Chemistry # 238 in the 2015 destroyer:

Are rate constants denoted by the capital K or lowercase k? The rate constants in the question are shown as capital K's so I'm confused.

 

[Dr. Dan]

They MIGHT be interchangeable (as it could just be a notational oversight), but I've only ever seen them as lowercase k. Most often, we see capital K used for Kelvin units and equilibrium constants, but the people who can best answer your question are the ones who wrote the book (Dr. Romano and Nancy).
Edit: Just checked my copy of the book, and it says that K1 and K2 are rate constants, so the use of a large K is should simply be a matter of choice in notation.

 

[fittycent]

Rate constants deal with kinetics. Equilibrium constants deal with thermodynamics.

Rate constants are lowercase k. Equilibrium constants are capital K.

The one time you'd see these presented at the same time would be... the one instance in which kinetics and thermodynamics intersect (at least in intro gchem material):

Keq = kforward/kbackward

capital K is equilibrium constant (thermo)
lower-case k is rate constant (kinetics)

 

[omarski]

Hey! I got a question in regards to problems 86 in the organic chemistry section. It's asking which group will deactivate the benzene toward reaction with electrophiles?

I'm a little confused with the answer provided. It says " o/p directors activate the ring, however NO (nitric oxide?) and Halogens are o/p directors, but deactivate the ring. All meta directors deactivate the ring."
For the correct answer c, do I have the right idea? I'm thinking that there's a positive charge on the nitrogen and it must be electron withdrawing. I mean it's going to pull electrons towards the positive charge , and since it clearly doesn't have any to donate, there is no way it'd be an electron donating group, right?

As for answer d, why does it not have electron withdrawing properties? Wouldn't the oxygens have some pull due to their electronegativity in comparison to the carbon atom they are bonded with? My train of thought is this, if something like NO is electron withdrawing, it makes sense to me since Oxygen is more electronegative than Nitrogen, thus pulling electrons in to further stabilize the benzene ring. I think the same way when looking at a carbonyl and justifying why it is also electron withdrawing. The oxygen will make the carbon it is attached to a little positive, attracting electrons in, due to oxygen's electronegativity 'hogging' on the electrons from the carbon. Is this train of thought wrong? I can't seem to figure out why answer D isn't electron withdrawing as well.
edit: I've been thinking about this question for the last day. What I noticed is that oxygen is more electronegative than carbon, but probably not enough to make a big enough deal? Things like carbonyls have that extra bond to help distribute the charge a little differently due to resonance?

Also for # 88 in organic chemistry as well, how much more of a major product is the para position over the ortho position? I thought para is preferred, but ortho has twice as many spots, so it comes out pretty evenly. But I'm trying to think of whether or not the size of what is attached is playing a role for where the Cl attaches?

 

[plusalpha]

I think this question was asked before by someone else, but I did not see a definitive answer, so I'll ask again.

334 Bio destroyer: Is the term "allele frequency" denoted by q^2 or just q? The solution solves the question by using q^2 as the allele frequency, but I thought q=frequency of the ALLELE and q^2 = frequency of the homozygous GENOTYPE. Can someone clear this up for me? Thanks

 

[DAT Destroyer]

Hey! I got a question in regards to problems 86 in the organic chemistry section. It's asking which group will deactivate the benzene toward reaction with electrophiles?

I'm a little confused with the answer provided. It says " o/p directors activate the ring, however NO (nitric oxide?) and Halogens are o/p directors, but deactivate the ring. All meta directors deactivate the ring."
For the correct answer c, do I have the right idea? I'm thinking that there's a positive charge on the nitrogen and it must be electron withdrawing. I mean it's going to pull electrons towards the positive charge , and since it clearly doesn't have any to donate, there is no way it'd be an electron donating group, right?

As for answer d, why does it not have electron withdrawing properties? Wouldn't the oxygens have some pull due to their electronegativity in comparison to the carbon atom they are bonded with? My train of thought is this, if something like NO is electron withdrawing, it makes sense to me since Oxygen is more electronegative than Nitrogen, thus pulling electrons in to further stabilize the benzene ring. I think the same way when looking at a carbonyl and justifying why it is also electron withdrawing. The oxygen will make the carbon it is attached to a little positive, attracting electrons in, due to oxygen's electronegativity 'hogging' on the electrons from the carbon. Is this train of thought wrong? I can't seem to figure out why answer D isn't electron withdrawing as well.
edit: I've been thinking about this question for the last day. What I noticed is that oxygen is more electronegative than carbon, but probably not enough to make a big enough deal? Things like carbonyls have that extra bond to help distribute the charge a little differently due to resonance?

Also for # 88 in organic chemistry as well, how much more of a major product is the para position over the ortho position? I thought para is preferred, but ortho has twice as many spots, so it comes out pretty evenly. But I'm trying to think of whether or not the size of what is attached is playing a role for where the Cl attaches?

I like your train of thought. Let us consider the NO group, also known as the nitroso group. When on a benzene ring it acts in the same way as a halogen. It does indeed withdraw electrons and thus raises the energy needed for the transition state, thus the reaction is slowed..HOWEVER, in the same fashion as a halogen, the group donates electrons by the resonance effect. This additional resonance form can be seen in the ortho and para pathway only. Two effects are operating : induction and resonance. The reaction is slowed by the inductive effect but the ortho and para pathway give an additional resonance form, which means the energy barrier will be lower to achieve. Hopefully you can see that the NO group can be thought of as a halogen sitting on the benzene ring. Normally the para isomer is the major product, due to steric reasons. However if there is a methyl group on the benzene ring the ortho isomer is often favored because there are two ortho positions and only one para position. As the group gets bigger, steric factors would outweigh statistical factors. For example, if toluene was nitrated we get 60 percent ortho and 40 percent para. For the DAT exam they would almost always want the para isomer.

Hope this helps.

Dr. Jim Romano

 

[omarski]

Thank you so much! Thinking of the nitroso group as a halogen made this concept sink in.
I always forget induction factoring into the equation and not just resonance. I need to repeat "CARDIO" to myself a million times I guess, ha.

edit: I've got another question! #140 on organic chemistry. In regards to nucleophile strength. I know and understand the answer through this understanding. My train of thought for this question was this: S is larger than O. So it is less electronegative. Because the electrons are not held in as tightly (due to electronegativity being less in S), S is less 'selfish' and allows for the electrons to be used for nucleophilic attack more readily.

Here is the part that confuses me: S is larger than O, less electronegative than O, thus is more polarized.

What is meant by more polarized? It has more distribution of charge? Because it is larger and thus less electronegative, it doesn't 'want' electrons as badly, so how is it more polarized? I'm assuming my definition of polarized is being confused with polar? For polar molecules, the more uneven the distribution of charge between atoms, the more polar. Causes of uneven distribution of charge are a result of large differences in electronegativity between atoms of molecules. Because S is less electronegative than O, the uneven distribution of charge would be less in comparison. Thus making the the molecule with S less polar overall in comparison to the molecule with O.

 

[DAT Destroyer]

Thank you so much! Thinking of the nitroso group as a halogen made this concept sink in.
I always forget induction factoring into the equation and not just resonance. I need to repeat "CARDIO" to myself a million times I guess, ha.

edit: I've got another question! #140 on organic chemistry. In regards to nucleophile strength. I know and understand the answer through this understanding. My train of thought for this question was this: S is larger than O. So it is less electronegative. Because the electrons are not held in as tightly (due to electronegativity being less in S), S is less 'selfish' and allows for the electrons to be used for nucleophilic attack more readily.

Here is the part that confuses me: S is larger than O, less electronegative than O, thus is more polarized.

What is meant by more polarized? It has more distribution of charge? Because it is larger and thus less electronegative, it doesn't 'want' electrons as badly, so how is it more polarized? I'm assuming my definition of polarized is being confused with polar? For polar molecules, the more uneven the distribution of charge between atoms, the more polar. Causes of uneven distribution of charge are a result of large differences in electronegativity between atoms of molecules. Because S is less electronegative than O, the uneven distribution of charge would be less in comparison. Thus making the the molecule with S less polar overall in comparison to the molecule with O.

You seem to have the concept of polar compounds well understood. The term polarized and polar are often used in different context. A molecule that is polarizable such as I- or HS- have very loosely held electrons and can be used for attacking purposes. This is very useful especially in reactions such as SN2 which are favored by strong nucleophiles.

Hope this helps..

Dr. Romano

 

[aj30]

Dr. Romano,
Mr. Ali here from class, back with questions.
Gen Chem # 213 shows a graph of "Energy" vs " Reaction Coordinate" and you state in choice c) that "the reaction is exothermic at -15kJ per mol" (and this is correct, you say in the answer).

My problem with this is that given that ∆G and ∆H are totally separate (though often related) measures the graph can't be showing both ∆G quantities (like activation energy) and a ∆H quantity like exo vs endothermic. So if the term "Energy" on the Y axis refers to ∆G, we can see activation energy, but not whether it's exothermic. All we can tell is that it has a -∆G, not whether it has a + or - ∆H, which may or may not be true, depending on other factors (∆G=∆H - T∆S).

I looked in my college textbooks and all of them, from McMurry ch. 6, to Brown Lemay ch. 14, to Campbell (Bio) ch. 8 all seem to discuss these "reaction progress" charts purely in terms of ∆G, with no mention of ∆H (Campell even replaces the term "Energy" on the Y axis with "free energy").

Am I missing something here in terms of ∆G-∆H relation or is the above correct?

Thanks

 

[streacy1]

Hey All,

I am working on some DAT Destroyer (2015) Orgo problems, and am really stumped on #92. I see how the correct answer is correct, but in the solutions it makes it seem like NaI would not do E2 with a secondary halide in acetone. I thought it would do a mix. Anyone think they can shed some light on this? Thanks!

 

[DAT Destroyer]

Hey All,

I am working on some DAT Destroyer (2015) Orgo problems, and am really stumped on #92. I see how the correct answer is correct, but in the solutions it makes it seem like NaI would not do E2 with a secondary halide in acetone. I thought it would do a mix. Anyone think they can shed some light on this? Thanks!

Not a chance . The E2 reaction almost always occurs on a secondary or tertiary halide when a strong base such as NH2-, methoxide, ethoxide, t-butoxide are employed. I- is a great nucleophile , but a poor base. This reaction is the famous Finkelstein reaction and is the prototype reaction of the SN2 reaction mechanism.

Hope this helps.

Dr. Jim Romano

 

[omarski]

Hey All,

I am working on some DAT Destroyer (2015) Orgo problems, and am really stumped on #92. I see how the correct answer is correct, but in the solutions it makes it seem like NaI would not do E2 with a secondary halide in acetone. I thought it would do a mix. Anyone think they can shed some light on this? Thanks!

I think E2 and SN2 oftentimes both occur simultaneously. However, the favored product is highly dependent on whether the nucleophile is a strong base or not.
NaI is a strong nucleophile but a weak base. Also it is in an aprotic solvent. This automatically limits the reaction to be either E2 or SN2 as SN1 and E1 require carbocation stabilization through polar protic solvents. So now that we know it can only be SN2 or E2, it comes down to two different things. What type of carbon is the leaving group on, and what type of nucleophile is being used? Because it is a secondary carbon, it has no limitations for SN2 or E2. If it was tertiary, SN2 wouldn't be possible due to backside attack being blocked. So now the final thing it can come down to is what nucleophile/base is being used. Since NaI is a strong nucleophile but weak base, the majority of product should be SN2 rather than E2.
Also remember this: Most strong nucleophiles have a negative charge. Because of the bond type between Na+ + I- being ionic, the Iodine should be viewed as having a negative charge.

Dr. Romano,
Mr. Ali here from class, back with questions.
Gen Chem # 213 shows a graph of "Energy" vs " Reaction Coordinate" and you state in choice c) that "the reaction is exothermic at -15kJ per mol" (and this is correct, you say in the answer).

My problem with this is that given that ∆G and ∆H are totally separate (though often related) measures the graph can't be showing both ∆G quantities (like activation energy) and a ∆H quantity like exo vs endothermic. So if the term "Energy" on the Y axis refers to ∆G, we can see activation energy, but not whether it's exothermic. All we can tell is that it has a -∆G, not whether it has a + or - ∆H, which may or may not be true, depending on other factors (∆G=∆H - T∆S).

I looked in my college textbooks and all of them, from McMurry ch. 6, to Brown Lemay ch. 14, to Campbell (Bio) ch. 8 all seem to discuss these "reaction progress" charts purely in terms of ∆G, with no mention of ∆H (Campell even replaces the term "Energy" on the Y axis with "free energy").

Am I missing something here in terms of ∆G-∆H relation or is the above correct?

Thanks

Summary: You're mixing two different concepts that explain different things, together.
I can see where your confusion is arising. I highly recommend if you have an account for course-saver to watch Chad's video on the topic. I'll try and explain the fundamental concept of this topic myself. Activation energy is shown in these graphs. It shows the "barrier of entry" or amount of energy required for the reaction to proceed in a specific direction (either forward or reverse depending on what you're looking for). So here is where ∆H comes into play. The difference between the starting energy and the ending energy (basically heat being released or absorbed) is what ∆H is indicating.
You're looking at two separate concepts. The Reaction Coordinate Diagrams are strictly discussing kinetics. Kinetics main purpose is to solve how fast a reaction can occur; something thermodynamics isn't capable of doing. Because of this, the "pathway" of a reaction is super important in kinetics. This is where the idea of reaction coordinate diagrams comes into play. When solving problems about reaction coordinate diagrams I recommend reading these 3 suggestions below:
1. Look at the energy levels of the reactants and the energy level of the products
2. If heat is being released (exothermic) the products will be at a lower energy level than the reactants
3. Similarly, if heat is being absorbed (endothermic) the products will be at a higher energy level than the reactants

The formula ∆G=∆H - T∆S is used to figure out whether or not a reaction will be spontaneous based on a few different variables. Again, I highly recommend watching Chad's videos if you have time, as he explains this topic extremely well. The formula can also help you solve for an unknown variable if given the others. This formula is a thermodynamics topic. What I mean by this is, thermodynamics is there to explain whether a reaction may spontaneously happen or not. What it fails to do is show the pathway taken (how long it would take).
ex: Chad gives a really good example of this. He says, he was supposed to meet his friend in Argentina, but instead of going straight to Argentina, he looped around the world twice then got to Argentina. Thermodynamics in this sense, would just tell you that he got to Argentina, eventually. Kinetics would tell you, he took a really bad route to Argentina.

So what I'm trying to get at is this, you're mixing two different fundamental topics. I hope my explanation made a little bit of sense, and that it is correct haha. Let me know if you need more explanations.

 

[streacy1]

Not a chance . The E2 reaction almost always occurs on a secondary or tertiary halide when a strong base such as NH2-, methoxide, ethoxide, t-butoxide are employed. I- is a great nucleophile , but a poor base. This reaction is the famous Finkelstein reaction and is the prototype reaction of the SN2 reaction mechanism.

Hope this helps.

Dr. Jim Romano

Thank you!!

 

[DAT Destroyer]

Thank you!!

Welcome! Keep up the good work!

 

[aj30]

I think E2 and SN2 oftentimes both occur simultaneously. However, the favored product is highly dependent on whether the nucleophile is a strong base or not.
NaI is a strong nucleophile but a weak base. Also it is in an aprotic solvent. This automatically limits the reaction to be either E2 or SN2 as SN1 and E1 require carbocation stabilization through polar protic solvents. So now that we know it can only be SN2 or E2, it comes down to two different things. What type of carbon is the leaving group on, and what type of nucleophile is being used? Because it is a secondary carbon, it has no limitations for SN2 or E2. If it was tertiary, SN2 wouldn't be possible due to backside attack being blocked. So now the final thing it can come down to is what nucleophile/base is being used. Since NaI is a strong nucleophile but weak base, the majority of product should be SN2 rather than E2.
Also remember this: Most strong nucleophiles have a negative charge. Because of the bond type between Na+ + I- being ionic, the Iodine should be viewed as having a negative charge.

Summary: You're mixing two different concepts that explain different things, together.
I can see where your confusion is arising. I highly recommend if you have an account for course-saver to watch Chad's video on the topic. I'll try and explain the fundamental concept of this topic myself. Activation energy is shown in these graphs. It shows the "barrier of entry" or amount of energy required for the reaction to proceed in a specific direction (either forward or reverse depending on what you're looking for). So here is where ∆H comes into play. The difference between the starting energy and the ending energy (basically heat being released or absorbed) is what ∆H is indicating.
You're looking at two separate concepts. The Reaction Coordinate Diagrams are strictly discussing kinetics. Kinetics main purpose is to solve how fast a reaction can occur; something thermodynamics isn't capable of doing. Because of this, the "pathway" of a reaction is super important in kinetics. This is where the idea of reaction coordinate diagrams comes into play. When solving problems about reaction coordinate diagrams I recommend reading these 3 suggestions below:
1. Look at the energy levels of the reactants and the energy level of the products
2. If heat is being released (exothermic) the products will be at a lower energy level than the reactants
3. Similarly, if heat is being absorbed (endothermic) the products will be at a higher energy level than the reactants

The formula ∆G=∆H - T∆S is used to figure out whether or not a reaction will be spontaneous based on a few different variables. Again, I highly recommend watching Chad's videos if you have time, as he explains this topic extremely well. The formula can also help you solve for an unknown variable if given the others. This formula is a thermodynamics topic. What I mean by this is, thermodynamics is there to explain whether a reaction may spontaneously happen or not. What it fails to do is show the pathway taken (how long it would take).
ex: Chad gives a really good example of this. He says, he was supposed to meet his friend in Argentina, but instead of going straight to Argentina, he looped around the world twice then got to Argentina. Thermodynamics in this sense, would just tell you that he got to Argentina, eventually. Kinetics would tell you, he took a really bad route to Argentina.

So what I'm trying to get at is this, you're mixing two different fundamental topics. I hope my explanation made a little bit of sense, and that it is correct haha. Let me know if you need more explanations.

Thanks for the response.
Your answer is not perfectly clear but the gist of it-that I'm confusing the 2 topics-is incorrect. My whole point was exactly that-that ∆G and ∆H are totally unrelated and so can't both be represented in the same curve. I don't see how you answer that.

 

[omarski]

Hey question in regards to Organic Chemistry section #226
"The following has a rotation of:"
I assigned R and S and knew the C2H5 would be achiral, so the compound would be meso, meaning the overall rotation is 0. But my question is this, if the two NO2 groups were not cancelling each other's rotation, the answer would be d. cannot tell , correct? From what I understand R and S does not determine + or - rotation and that this can only be experimentally tested and varies from molecule to molecule. In this case, because a meso compound's definition, the rotation by definition, must be zero.

Also question in regards to Organic Chemistry section #222

Because the Bromine is a primary halide, wouldn't the base only have one real choice for which Hydrogen to remove while the leaving group leaves to form the alkene?

For example, if the Bromine was a secondary halide, there would be two options for where the alkene would form. Both the zaitsev and anti-zaitsev product are possible, but the determinant would be which base is used. If the anti-zaitsev product formed, then we could assume a bulky base that is sterically hindered was used, which would remove a few of the answer choices.

But in this case, because it is a primary halide and there is only one real product that can form due to the location of the Bromine, wouldn't that enable any decently strong base to react and form an alkene?

The solution says : "If a primary halide reacts, we get the E2 product, but the less substituted alkene!"
I guess my question is this: Because of the location of Bromine, isn't the E2 alkene product limited to forming in only one location, the less substituted alkene, regardless?

edit:
Stumbled upon another question for #234 in Organic Chemistry section.
If the question were to ask which had the highest boiling point, the answer would be c correct? since it has both hydrogen bonding and Ionic bonding.
c>a>d>e>b was the order I assigned from highest boiling point to lowest.