Official DAT Destroyer Q&A Thread

[densaugeo]

---

Members don't see this ad.

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!

 

[DAT Destroyer]

Thank you for your response Dr. Romano. This does make sense but I am afraid I might make a mistake. What If I think about it as fisher projection? meaning simply turn it into fisher projection and then assign R/S as follow:
Is this a correct way or I am just making things more complicated for myself?

Here is a simple way to do this, as I teach my students....1. Number the Group Priorities 2. If Group #4 , the lowest priority group, is on the horizontal plane, simply reverse the answer ! For example, in the above picture the group priorities appeared to be clockwise....HOWEVER since #4 was horizontal it was reversed to an S. If Group #4 was vertical.....no need to reverse your answer......you are fine. The DAT exam is heavy on Fischer notation.....If you can do a Fischer with one, two, and THREE chiral centers you will be fine. I have examples of this in the DESTROYER book.....know them !

Hope this helps

Dr. Romano

 

[Ashish]

Chem # 354: Why is XeF4 non polar? It because of its 3D orientation that causes the 2 lone pair of electrons to cancel each other? Thanks!

 

[DAT Destroyer]

Chem # 354: Why is XeF4 non polar? It because of its 3D orientation that causes the 2 lone pair of electrons to cancel each other? Thanks!

This molecule has a square planar shape and symmetrical. Therefore dipoles will cancel out and make it nonpolar.
Hope this helps.

 

[GoSpursGo]

This is a friendly reminder that while discussion is great, please refrain from posting copywritten materials.

 

[FutureDent20]

Here is a simple way to do this, as I teach my students....1. Number the Group Priorities 2. If Group #4 , the lowest priority group, is on the horizontal plane, simply reverse the answer ! For example, in the above picture the group priorities appeared to be clockwise....HOWEVER since #4 was horizontal it was reversed to an S. If Group #4 was vertical.....no need to reverse your answer......you are fine. The DAT exam is heavy on Fischer notation.....If you can do a Fischer with one, two, and THREE chiral centers you will be fine. I have examples of this in the DESTROYER book.....know them !

Hope this helps

Dr. Romano

Your method makes TOTAL sense now! THANK YOU Dr. Romano!!!!!!

 

[rdub]

Gen Chem question.... In 2016 DAT Destroyer, the solution for #44 says that when the temperature is the same, molecules of gas have the same kinetic energy. Then on #68, it says that the following statement is false: "All molecules of an ideal gas have the same kinetic energy at constant temperature".

This seems contradicting.

 

[Likkriue]

Gen Chem question.... In 2016 DAT Destroyer, the solution for #44 says that when the temperature is the same, molecules of gas have the same kinetic energy. Then on #68, it says that the following statement is false: "All molecules of an ideal gas have the same kinetic energy at constant temperature".

This seems contradicting.

You are missing the word "average", all molecules have the same AVERAGE kinetic energy at constant temperature. They do not have the same kinetic energy.

Chad goes over this in his video, you should watch it again to get a better idea.

 

[FutureDent20]

2016 DAT destroyer #269:
So relating PE and KE to exothermic vs. Endothermic. Can you see if I have a correct understanding please.
So, in exothermic reaction heat is given off to the surrounding (increase in temp), hence increase in KE of surrounding (since increase in temp = increase in KE). furthermore, since PE is the energy that has the potential to be converted into other forms of energy (i.e. KE), therefore, PE of surrounding decreases since the energy is converted into KE. Now in terms of system, can we say since system lost heat to the surrounding (decrease in temp), KE of the system decreases and PE of the system increases?
Is this flawed? cause I though since we can talk about ∆S of the system and surrounding therefore we can relate it to PE and KE of the system and surrounding as well.

 

[DAT Destroyer]

2016 DAT destroyer #269:
So relating PE and KE to exothermic vs. Endothermic. Can you see if I have a correct understanding please.
So, in exothermic reaction heat is given off to the surrounding (increase in temp), hence increase in KE of surrounding (since increase in temp = increase in KE). furthermore, since PE is the energy that has the potential to be converted into other forms of energy (i.e. KE), therefore, PE of surrounding decreases since the energy is converted into KE. Now in terms of system, can we say since system lost heat to the surrounding (decrease in temp), KE of the system decreases and PE of the system increases?

Is this flawed? cause I though since we can talk about ∆S of the system and surrounding therefore we can relate it to PE and KE of the system and surrounding as well.

Here is the bottom line....lets keep it simple......In an EXOTHERMIC reaction......heat is released and the temperature of the surrounds INCREASE.....things heat up. Now draw a graph of the exothermic reaction. You will notice that the POTENTIAL Energy of the products are LESS than the reactants. Now.....In an ENDOTHERMIC reaction, it is the opposite. Heat is absorbed, hence the surrounding are COLDER......and the graph will show the Potential Energy of the products are higher than the reactants. Understanding this will make you ready for DAT Beast !!!!

Hope this helps

Dr. Romano

 

[FutureDent20]

Here is the bottom line....lets keep it simple......In an EXOTHERMIC reaction......heat is released and the temperature of the surrounds INCREASE.....things heat up. Now draw a graph of the exothermic reaction. You will notice that the POTENTIAL Energy of the products are LESS than the reactants. Now.....In an ENDOTHERMIC reaction, it is the opposite. Heat is absorbed, hence the surrounding are COLDER......and the graph will show the Potential Energy of the products are higher than the reactants. Understanding this will make you ready for DAT Beast !!!!

Hope this helps

Dr. Romano

That I do understand completely thanks to Destroyer questions and your explanation above. Thank you so much Dr. Romano!

 

[DAT Destroyer]

That I do understand completely thanks to Destroyer questions and your explanation above. Thank you so much Dr. Romano!

Always happy to help, if you have more questions feel free to post them and I do my best to answer asap

 

[FutureDent20]

Always happy to help, if you have more questions feel free to post them and I do my best to answer asap

I truly appreciate it! I actually have another question please,
2016 OC destroyer #301:
During the second step (anhydride hydrolysis), treating it with H3O+ I understand how we form the dicarboxy acid, what i don't understand is how come the acid doesn't react with the alkene?

 

[DAT Destroyer]

I truly appreciate it! I actually have another question please,
2016 OC destroyer #301:
During the second step (anhydride hydrolysis), treating it with H3O+ I understand how we form the dicarboxy acid, what i don't understand is how come the acid doesn't react with the alkene?

Great call.....In real life it can....and MIGHT be a side product !!!!! However, we can control the amount and concentration of acid to prevent reaction with the alkene functionality. None of the choices suggest this, this simply use the acid to break the anhydride linkage. However, I am glad to see that you are cognizant of the fact that acid could react with alkenes to produce an alcohol.

Hope this helps

Dr. Romano

 

[rdub]

Question on Math Destroyer 2016: Test 2 #38

I can get to the point where A=100*sin(20)cos(20), but I don't understand how to use the double angle formulas to turn this into 50*sin(40). Can someone walk me through this? I know its probably really easy I just can't seem to figure it out.

 

[DAT Destroyer]

Question on Math Destroyer 2016: Test 2 #38

I can get to the point where A=100*sin(20)cos(20), but I don't understand how to use the double angle formulas to turn this into 50*sin(40). Can someone walk me through this? I know its probably really easy I just can't seem to figure it out.

Sin(2A) = 2sin(A)cos(A).
This is the identity used to solve the problem.
Let's work backward:
100*sin(20)cos(20) = 50*2sin(20)cos(20) = 50*sin(2*20)cos(2*20) = 50*sin(40)cos(40)

Hope this helps!

 

[rdub]

Sin(2A) = 2sin(A)cos(A).
This is the identity used to solve the problem.
Let's work backward:
100*sin(20)cos(20) = 50*2sin(20)cos(20) = 50*sin(2*20)cos(2*20) = 50*sin(40)cos(40)

Hope this helps!

This is helpful, but I still don't see where 50*sin(40) comes from as mentioned in the solutions.

 

[DAT Destroyer]

This is helpful, but I still don't see where 50*sin(40) comes from as mentioned in the solutions.

Sorry I made a mistake in the last one 🙁
100*sin(20)cos(20) = 50*2sin(20)cos(20) = 50*sin(2*20) = 50 sin(40)

 

[Ashish]

Hey guys,

I just ran into some contradicting concepts and can't figure out which one is wrong. In Biochemical pathways like respiration and photosynthesis, we are told that energy is generated or given away when ATP breaks down into ADP, AMP and so forth (delta H < 0, exergonic). But when doing Bond enthalpy problems in general chem, the example states that delta H is positive when bond breaks (endergonic process).

Any explanation to this is greatly appreciated.

Thanks

 

[rdub]

Hey guys,

I just ran into some contradicting concepts and can't figure out which one is wrong. In Biochemical pathways like respiration and photosynthesis, we are told that energy is generated or given away when ATP breaks down into ADP, AMP and so forth (delta H < 0, exergonic). But when doing Bond enthalpy problems in general chem, the example states that delta H is positive when bond breaks (endergonic process).

Any explanation to this is greatly appreciated.

Thanks

My understanding is that the phosphate bonds in ATP are actually somewhat weak, so they require a minimal amount of energy to break (you are correct in assuming it requires energy to break the bond). Energy is released overall because the products of hydrolysis (ADP and Pi) are much lower in energy than the ATP itself.

So although it does require energy to break off the phosphate, a much larger amount of energy is released due to the more stable products.

 

[DAT Destroyer]

Hi, I have a quick question in the organic Odyssey... I attached the pic of my question but for reference its page 74, chap 4, number 30 , 2016 version.

My question is, regarding number 30. The answer is E.... I understand A but not why C is included.... I do not know how to tell when a fisher projection like C is optically active? My understanding is ( correct me if I am wrong) that all enantiomers are optically active. But for some fisher projections I see that sometimes you get an R,S and sometimes like in this case you have an S,S configueration for the fisher projection but regardless of the configeration result, I cant seem to use that to deduce anything because how can I tell if its an enantiomer (to tell if its optically active) if there's no other exact image related to it to compare it too to see if it checks out as an enantiomer or diasteromer or same compound etc




C is optically because it has chiral carbons and lacks symmetry. If a molecule has 1 chiral carbon it will have an enantiomer. In choice C, we have 2 chiral carbons,,,,,,thus 4 stereoisomers are possible......RR RS SR SS.......RR and SS represent enatiomeric pairs, but RR SR and SS , RS represent diastereomeric pairs. It would be DAT SUICIDE if you dont understand this for the exam . My advice.....Get the Klein book......an older edition is fine. The chapter on stereochemistry has many problems that will give you the tools needed to work thru the rest of DESTROYER problems. Hope this helps.

 

[rdub]

DAT Destroyer 2016 - Gen Chem #285

In the solutions, you immediately divide 8 moles of N2 and 8 moles of H2 in half. Why are these divided by two?

 

[Mrhyde]

Hi, I have a quick question in the organic Odyssey... I attached the pic of my question but for reference its page 74, chap 4, number 30 , 2016 version.

My question is, regarding number 30. The answer is E.... I understand A but not why C is included.... I do not know how to tell when a fisher projection like C is optically active? My understanding is ( correct me if I am wrong) that all enantiomers are optically active. But for some fisher projections I see that sometimes you get an R,S and sometimes like in this case you have an S,S configueration for the fisher projection but regardless of the configeration result, I cant seem to use that to deduce anything because how can I tell if its an enantiomer (to tell if its optically active) if there's no other exact image related to it to compare it too to see if it checks out as an enantiomer or diasteromer or same compound etc




C is optically because it has chiral carbons and lacks symmetry. If a molecule has 1 chiral carbon it will have an enantiomer. In choice C, we have 2 chiral carbons,,,,,,thus 4 stereoisomers are possible......RR RS SR SS.......RR and SS represent enatiomeric pairs, but RR SR and SS , RS represent diastereomeric pairs. It would be DAT SUICIDE if you dont understand this for the exam . My advice.....Get the Klein book......an older edition is fine. The chapter on stereochemistry has many problems that will give you the tools needed to work thru the rest of DESTROYER problems. Hope this helps.

Thank you, I actually have the Klein book, I just blanked out on that one question but I read up and cleared it thank you.

Also on the Odyssey 2016 - page 102 number 7 .... I just wanted to make sure that I am right :

I know that SN1 or SN2 reactions CANT happen on an sp2 hybridized carbon atom - we would get "No reaction"

- So does that mean E1 AND E2 CAN happen on an sp2 hybridized carbon atom - or is it only E2 like in the case of this question ?

 

[dentalstudent2021]

.

 

[dentalstudent2021]

DAT Destroyer 2016 - Gen Chem #285

In the solutions, you immediately divide 8 moles of N2 and 8 moles of H2 in half. Why are these divided by two?

aren't those the # of Liters ? could you post the question ? GO eagles!!

 

[rdub]

aren't those the # of Liters ? could you post the question ? GO eagles!!

Yes you are right! I looked at it again and realized it was just taking the moles/L to get the molarity. Not sure how I missed that but thanks for the help.

 

[rdub]

2016 Ochem #90
How is this possible? The product circled in red is what is listed in the solutions. The explanation says a rearrangement takes place, but I can't figure out how this is possible in the mechanism. The product circled in black is what I am getting.

 

[DAT Destroyer]

2016 Ochem #90
How is this possible? The product circled in red is what is listed in the solutions. The explanation says a rearrangement takes place, but I can't figure out how this is possible in the mechanism. The product circled in black is what I am getting.

As soon as the Chlorine atom is bound to the Aluminum as you have drawn , the Chlorine-Aluminum complex leaves SIMULTANEOUSLY as a Hydride shift occurs.....this gives the isopropyl carbocation which becomes the electrophilic specie captured by the benzene ring. Any orgo text will show this reaction if you need to see it fully worked out.

Hope this helps.

Dr. Romano

 

[SmilezAllDay]

Destroyer 2016 .... G-Chem # 229

Just a quick question, I understand how to set up the problem with mCAT = mCAT. I don't understand the change in temperature tho. For the metal the answer key states 100-40 C (initial - final) but for the water its 40 - 30 C (final - initial). It didn't matter for the answers because they were all positive but how do I know to subtract initial - final temp or vice versa (if I do final - initial for both, the answer comes out negative)? Also, I'm not so sure specific heat can be negative, its the amount of heat needed too raise the temp by 1 degree Celsius.

 

[rdub]

Destroyer 2016 .... G-Chem # 229

Just a quick question, I understand how to set up the problem with mCAT = mCAT. I don't understand the change in temperature tho. For the metal the answer key states 100-40 C (initial - final) but for the water its 40 - 30 C (final - initial). It didn't matter for the answers because they were all positive but how do I know to subtract initial - final temp or vice versa (if I do final - initial for both, the answer comes out negative)? Also, I'm not so sure specific heat can be negative, its the amount of heat needed too raise the temp by 1 degree Celsius.

The water started at 30 degrees, and after the metal was put in the water was 40 degrees. This means the temperature of the water had a net gain of 10 degrees.
Just remember for the q=mcat equation you are looking for the change in temp (delta T). The metal was cooled down from 100 to 40 so delta T is (100-40). The water was heated from 30 to 40 (hence 40-30). So whether or not you subtract 40 from 30 or 30 from 40, just take the absolute value for your delta T. Hope that makes sense. If not, I'm sure Dr. Romano will be around soon.

 

[DAT Destroyer]

The water started at 30 degrees, and after the metal was put in the water was 40 degrees. This means the temperature of the water had a net gain of 10 degrees.
Just remember for the q=mcat equation you are looking for the change in temp (delta T). The metal was cooled down from 100 to 40 so delta T is (100-40). The water was heated from 30 to 40 (hence 40-30). So whether or not you subtract 40 from 30 or 30 from 40, just take the absolute value for your delta T. Hope that makes sense. If not, I'm sure Dr. Romano will be around soon.

Looks good,,,,,,specific heat must be a positive value.

Dr. Romano

 

[rak173]

Destroyer 2016 .... G-Chem # 229

Just a quick question, I understand how to set up the problem with mCAT = mCAT. I don't understand the change in temperature tho. For the metal the answer key states 100-40 C (initial - final) but for the water its 40 - 30 C (final - initial). It didn't matter for the answers because they were all positive but how do I know to subtract initial - final temp or vice versa (if I do final - initial for both, the answer comes out negative)? Also, I'm not so sure specific heat can be negative, its the amount of heat needed too raise the temp by 1 degree Celsius.

I use this formula -q (of metal) = q (of water). It's -ve for metal since it's losing heat and +ve for water since it's gaining heat. And for the temperature, I always use Tf - Ti.

 

[rdub]

2016 OChem #137. This problem is asking which compound reacts fastest in an SN1 reaction.
Comparing the two intermediates below, obviously the one on the right is a much more stable carbocation. My logic tells me that since the one on the right is stabalized via resonance, it wouldn't react as fast. The one on the left is so unstable, wouldn't it be more electrophilic, and thus react faster?

The solutions say that the more stable the carbocation, the faster it reacts. Just trying to make sense of this.

 

[DAT Destroyer]

2016 OChem #137. This problem is asking which compound reacts fastest in an SN1 reaction.
Comparing the two intermediates below, obviously the one on the right is a much more stable carbocation. My logic tells me that since the one on the right is stabalized via resonance, it wouldn't react as fast. The one on the left is so unstable, wouldn't it be more electrophilic, and thus react faster?

The solutions say that the more stable the carbocation, the faster it reacts. Just trying to make sense of this.

You are missing a critical point, An unstable carbocation forms very slowly. The more stable carbocation will form quickly and therefore will be attacked by the nucleophile.

Hope this helps

Dr. Romano

 

[rak173]

Hello Dr. Romano,

#255 Gen chem 2016: It's about how pH changes affect the indication (protonate it or deprotonate it). I don;t quiet understand this. When a pH is larger than the indicator's pka, how does that deprotonate the indicator and vice versa when the indicator is protonated by a lower pH? How does this work?

Thanks.

 

[DAT Destroyer]

Hello Dr. Romano,

#255 Gen chem 2016: It's about how pH changes affect the indication (protonate it or deprotonate it). I don;t quiet understand this. When a pH is larger than the indicator's pka, how does that deprotonate the indicator and vice versa when the indicator is protonated by a lower pH? How does this work?

Thanks.

If you are at a pH value below that of the pK of the indicator, you will be in the protonated form. When you are at a pH above the pK of the indicator, you will be in the deprotonated form.

These are VERY important questions, make sure you understand them.

Dr. Romano

 

[Ashish]

GC Destroyer 198.
In This question, I do not have any troubles with the calculation part itself. Instead, I am having troubles understanding what is going on conceptually. Please correct me if I am wrong. My understanding was that common ion decreased the molar solubility but here, it is seen that the molar solubility actually increased when a common ion was placed in the solution. When I calculated the molar solubility of Cl- in just Pbcl2, the molar solubility is ~2.6X 10^-3 mols. The solubility of pbcl2 in Kcl is 1.6x10^-1 as the answer gives us which is greater than molar solubility of pbcl2. AM I missing an important point here>

 

[farzoo101]

GC question 31 2014

Answer is that AgBr is most soluble in NH3. From prior Chad's solubility rules i thought Ag+ complexes were not soluble. Any explanation would be appreciated.

 

[rdub]

2016 Gen Chem #7
This is a very simple problem, but I had a thought I wanted to run by you. The question simply states "monoprotic acid" but does not say if it is a strong acid or a weak acid. Wouldn't it be possible, if it were a monoprotic acid such as HF, to require twice the moles of acid in order to neutralize the base? Why should we assume that the monoprotic acid completely dissociates in a 1:1 ratio with the base?

 

[Ashish]

2016 Gen Chem #7
This is a very simple problem, but I had a thought I wanted to run by you. The question simply states "monoprotic acid" but does not say if it is a strong acid or a weak acid. Wouldn't it be possible, if it were a monoprotic acid such as HF, to require twice the moles of acid in order to neutralize the base? Why should we assume that the monoprotic acid completely dissociates in a 1:1 ratio with the base?

Chad explains this in his Titration video. No matter the strength of acids or base, it takes the same amount of mole of acid to neutralize a base and vice versa. Neutral means pH =7 . 1 mole of monoprotic acid dissociates to give 1 mol of H+ and same with the base in the question ( 1 mol of OH-). 1 mol of H+ and 1 mol of OH- leaves no OH- or H+ in the solution and thus the pH depends on H2O which is 7. Hope I made some sense.

 

[Ashish]

GC question 31 2014

Answer is that AgBr is most soluble in NH3. From prior Chad's solubility rules i thought Ag+ complexes were not soluble. Any explanation would be appreciated.

Beware! This is asking the solubility of AgBr in NH3 not the solubility of AgBr (NH3) in water or a different solvent that you are thinking of.

 

[nextgendental]

GC question 31 2014

Answer is that AgBr is most soluble in NH3. From prior Chad's solubility rules i thought Ag+ complexes were not soluble. Any explanation would be appreciated.

Chads solubility rules state that Most Ag, Pb, and Hg2 salts are insoluble IN WATER.


Sent from my iPhone using SDN mobile

 

[DAT Destroyer]

2016 Gen Chem #7
This is a very simple problem, but I had a thought I wanted to run by you. The question simply states "monoprotic acid" but does not say if it is a strong acid or a weak acid. Wouldn't it be possible, if it were a monoprotic acid such as HF, to require twice the moles of acid in order to neutralize the base? Why should we assume that the monoprotic acid completely dissociates in a 1:1 ratio with the base?

You must make a simple assumption which is to realize that since neutralization occurred on a mono basic specie the acid must have been monoprotic and in equal amounts. Remember, neutralization means Equal moles.

Hope this helps

Dr. Romano

 

[Dentoni]

Hi Dr. Romano and Dr. Nancy, I have a question about electron configuration. Why is it that V (Z=23) is: [Ar] 4s2 3d3..... I thought it should be like: [Ar] 3d5... because it's more stable form of the configuration.. Any insight/advice is really appreciated.

 

[nextgendental]

Hi Dr. Romano and Dr. Nancy, I have a question about electron configuration. Why is it that V (Z=23) is: [Ar] 4s2 3d3..... I thought it should be like: [Ar] 3d5... because it's more stable form of the configuration.. Any insight/advice is really appreciated.

The rule your referring is all shells must be full or half full. That rule only applies to Mo, Cr, Cu, Ag and Au because they will be more stable with its s and p sub shells will be full or half full. You just took both electrons from the s and put it in the p. Which would mean that s has no electrons and p is half full which doesn't fit the rule.


Sent from my iPhone using SDN mobile

 

[DAT Destroyer]

Hi Dr. Romano and Dr. Nancy, I have a question about electron configuration. Why is it that V (Z=23) is: [Ar] 4s2 3d3..... I thought it should be like: [Ar] 3d5... because it's more stable form of the configuration.. Any insight/advice is really appreciated.

Unfortunately, you must take a leap of faith and trust the quantum chemist calculation on energy levels.

For most atoms, the order of electron filling is 1s 2s 2p 3s 3p 4s 3d 4p.

Notice that 4s comes in at a lower energy value than the 3d level. This is important to know for the DAT.

Also be familiar with 3 anomalies: Cu, Cr, and Mo.

Hope this helps

Dr. Romano

 

[gangazi]

Destroyer orgo #78
Why do we need 3 equivalents of NaNH3/NH3? Doesn't it go E2 two times? Just wondering if I am missing something
Thank you Dr. Romano

 

[rdub]

This is a reaction from Road Map 2. I am confused as to why the nitrogen attacks the carbon. Drawing a resonance structure clearly shows a partial negative on the carbon that I circled in red, and a partial positive on the nitrogen. Why would a partially positive nitrogen act as the nucleophile over a partially negative carbon?

 

[nextgendental]

This is a reaction from Road Map 2. I am confused as to why the nitrogen attacks the carbon. Drawing a resonance structure clearly shows a partial negative on the carbon that I circled in red, and a partial positive on the nitrogen. Why would a partially positive nitrogen act as the nucleophile over a partially negative carbon?

That nitrogen has a lone pair so it will attack the carbonyl. The partial negative charge is due to resonance. Only double bonds and lone pairs can attack. Since there is no true double bond due to the resonance there's no way for it to attack. I'm pretty sure that's the case but we'll wait for the orgoman.


Sent from my iPhone using SDN mobile

 

[Ashish]

This is a reaction from Road Map 2. I am confused as to why the nitrogen attacks the carbon. Drawing a resonance structure clearly shows a partial negative on the carbon that I circled in red, and a partial positive on the nitrogen. Why would a partially positive nitrogen act as the nucleophile over a partially negative carbon?

sp2 hybridized carbon don't take part in Sn1, Sn2, E1 or E2. An amide is formed above. Think about why you can't displace the bromine in bromo beneze by introducing it to a strong nucleophile like CN-. In the above reaction, Sp2 hybridized carbon cannot attack the carbonyl carbon because it simply does not behave as a nucleophile. This can also be explained from the orbital perspective. sp2 hybridized carbon has negative charge stabilized due to greater 's' orbital like characteristic which is less in sp3 hybridized carbon (hence loosely held electrons pair that the nucleophile can donate to the electrophile)....kinda analogus as to why an alkyne is more acidic than a alkene and alkane!!! Hope this made some sense.

 

[rdub]

@nextgendental there are plenty of reactions where a benzene acts as a nucleophile despite having no true double bonds.
@Ashish I am not talking about a substitution or elimination reaction. I am thinking of it as the ring attacking the carbon and forming a tetrahedral intermediate, and then kicking off Cl. There are many reactions where this occurs with a benzene ring.

NH2 is a strong EDG and so I figured it would cause the benzene to add the group ortho/para. Still not sure what I'm missing.