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!

 

[DAT Destroyer]

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.

Organic Chemistry section #226

You are correct. The +/- rotation can only be experimentally determined. If one NO2 was on the dash and another was on the wedge, the molecule would be chiral, but we would only be able to determine the rotation experimentally.

Hope this helps.

 

[DAT Destroyer]

Organic Question 222

I want to give you a rule and write it down 5 times. Primary halides LOVE do to the SN2 with most nucleophiles.

There are 2 main exceptions which include strong bases that are big and bulky such as LDA and t-butoxide. These bases do the E2 mechanism. These bases have a very difficult time approaching the molecule from the backside, thus they remove a hydrogen to do the elimination reaction.

Hope this helps

Dr. Romano

 

[DAT Destroyer]

Organic Question 222

I want to give you a rule and write it down 5 times. Primary halides LOVE do to the SN2 with most nucleophiles.

There are 2 main exceptions which include strong bases that are big and bulky such as LDA and t-butoxide. These bases do the E2 mechanism. These bases have a very difficult time approaching the molecule from the backside, thus they remove a hydrogen to do the elimination reaction.

Hope this helps

Dr. Romano

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.

Orgo Question 234

For the amines the quaternary amine is ionic, thus has the highest boiling and melting point. Let us now consider primary, secondary, and tertiary amines. In the gas phase, tertiary is most basic but in solution, the secondary amine is most basic. The explanation goes beyond the scope of the DAT, but involves the stabilization of the protonated amine. We see a balance between hydrogen bonding and electron releases by alkyl groups. The bottom line: In the gas phase, tertiary amines are most basic and in solution, secondary amines are most basic. As far as boiling points go, as long as the number of carbons are the same, the quaternary amine is highest, followed by primary then secondary and tertiary the least since it lacks hydrogen bonding.

Hope this helps.

Dr. Jim Romano

 

[cdy]

I am having a trouble understanding Question 43 in Math section/ DAT Destroyer 2015. Question states, "Given that the large triangle is isosceles, what is its area?" Given is the base of the two small triangles, 3 and 4. How did you get the base of the large triangle which is 4+3=7 and then you considered the height of the large triangle 7 as well?

Thank you.

 

[DAT Destroyer]

I am having a trouble understanding Question 43 in Math section/ DAT Destroyer 2015. Question states, "Given that the large triangle is isosceles, what is its area?" Given is the base of the two small triangles, 3 and 4. How did you get the base of the large triangle which is 4+3=7 and then you considered the height of the large triangle 7 as well?

Thank you.

Because the triangle is isosceles, 2 sides must be equal. Since the base is 7 the height is also 7. The area is then : 7*7/2 = 49/2

Hope this helps...

 

[omarski]

Thank you so much! For the amine question, is this a simple and effective way to look at it? More hydrogens on it = more hydrogens to donate = more acidic? And the exception to this is quarternary since it has ionic bonding, which is a stronger bond type than what all the rest have, thus increasing BP the most?

edit:
Got another question! It might be too late and I'm starting to do something wrong...
It's in regards to General Chemistry # 315

I understand why answer choice E is wrong, because cations are smaller in comparison to their their anion counterpart and the answer is clearly describing opposite of what happens. But something in the answer key is driving me nuts!

The answer says "Sr++ is smaller than S-- since it forms a positive ion."

How is Sr 2+ smaller than S--? I understand the entire concept, anions are gaining electrons, and not enough protons to hold them in tightly; similarly, cations lose electrons, meaning more protons to help hold the electrons tighter. But even when ionizing the Strontium it's still in row 4 while Sulfur when ionized is row 3.

I feel like this answer would hold true if S++ was compared to Se-- since they would have the same number of shells and the Strontium cation would hold the electrons in tighter since more protons available. However, S-- is missing an entire shell of electrons that S++ has, so how can it possibly be bigger?

Am I missing something big here?

 

[cdy]

Because the triangle is isosceles, 2 sides must be equal. Since the base is 7 the height is also 7. The area is then : 7*7/2 = 49/2

Hope this helps...

Yes, thanks!!

 

[omarski]

Hey I have a question in regards to Biology #40

Can endospores actually reproduce through binary fission, everything I read says it enters a dormant stage until the conditions surrounding it are better? Or does being an endospore protect the bacteria in a sense to where it would survive under conditions that it otherwise would not, so it doesn't prevent proliferation, it just delays it?
I can't seem to find anything indicating that they do undergo binary fission though, so I just want to confirm.

 

[DAT Destroyer]

Hey I have a question in regards to Biology #40

Can endospores actually reproduce through binary fission, everything I read says it enters a dormant stage until the conditions surrounding it are better? Or does being an endospore protect the bacteria in a sense to where it would survive under conditions that it otherwise would not, so it doesn't prevent proliferation, it just delays it?
I can't seem to find anything indicating that they do undergo binary fission though, so I just want to confirm.

Certain bacteria form endospores in order to withstand difficult environmental conditions or if they lack certain nutrient(s). Endospore has a hard cover and a copy of the bacterial chromosome. Also water is removed from the endospore.

Endospore, therefore is not designed to reproduce, it is designed to protect the bacterial genome from being degraded by the harsh environmental conditions. Endospores are dormant, but may be viable and perform some daily functions if the conditions are less brutal.


Hope this helps.

 

[NYDent22]

Hey! I have a question regarding the solution for question 31 (Orgo) in 2015 Dat Destroyer.
It says that Choices B,C, & E all have EWG and decrease basicity.
Don't those choices have EDG rather than EWG which therefore increases the basicity?
Thank you

 

[omarski]

Certain bacteria form endospores in order to withstand difficult environmental conditions or if they lack certain nutrient(s). Endospore has a hard cover and a copy of the bacterial chromosome. Also water is removed from the endospore.

Endospore, therefore is not designed to reproduce, it is designed to protect the bacterial genome from being degraded by the harsh environmental conditions. Endospores are dormant, but may be viable and perform some daily functions if the conditions are less brutal.


Hope this helps.

Ok awesome ! That's what I thought happens, just needed to verify that I wasn't misinterpreting what I read. Thanks again 🙂

 

[DAT Destroyer]

Ok awesome ! That's what I thought happens, just needed to verify that I wasn't misinterpreting what I read. Thanks again 🙂

Glad I could help!

Dr. Romano

 

[auoso]

Hi there,

I'm looking for some clarification on Biology #124 (2015), which seems especially relevant to us predents 🙂

124. Which embryonic layer(s) will develop molars, premolars, and canines?

a) Endoderm
b) Ectoderm
c) Mesoderm
d) B and C
e) Placenta

The answer says b) Ectoderm, because they are teeth, but from what I've read, only the enamel of the teeth form from the ectoderm. I assumed then the dentin and other structures come from the mesoderm and thus d) would be the correct answer. This source seems to corroborate. Anybody have thoughts on this? http://discovery.lifemapsc.com/libr...mbryology/chapter-77-development-of-the-teeth

 

[omarski]

Hi there,

I'm looking for some clarification on Biology #124 (2015), which seems especially relevant to us predents 🙂

124. Which embryonic layer(s) will develop molars, premolars, and canines?

a) Endoderm
b) Ectoderm
c) Mesoderm
d) B and C
e) Placenta

The answer says b) Ectoderm, because they are teeth, but from what I've read, only the enamel of the teeth form from the ectoderm. I assumed then the dentin and other structures come from the mesoderm and thus d) would be the correct answer. This source seems to corroborate. Anybody have thoughts on this? http://discovery.lifemapsc.com/libr...mbryology/chapter-77-development-of-the-teeth

It says on wiki and another Student doctor forum post that it's from the ectoderm for mammals.

 

[DAT Destroyer]

Hi there,

I'm looking for some clarification on Biology #124 (2015), which seems especially relevant to us predents 🙂

124. Which embryonic layer(s) will develop molars, premolars, and canines?

a) Endoderm
b) Ectoderm
c) Mesoderm
d) B and C
e) Placenta

The answer says b) Ectoderm, because they are teeth, but from what I've read, only the enamel of the teeth form from the ectoderm. I assumed then the dentin and other structures come from the mesoderm and thus d) would be the correct answer. This source seems to corroborate. Anybody have thoughts on this? http://discovery.lifemapsc.com/libr...mbryology/chapter-77-development-of-the-teeth

The consensus was that teeth develop from ectoderm, now recent evidence suggests that mesoderm is also responsible. Perhaps, it could be true. However, it was viewed as an opinion before, and therefore was not considered as facts by books.

Let's see if more evidence will prove this to be true in the future.

 

[omarski]

This is adding a scenario to Organic Chemistry #227, in case I was presented with different groups to compare.
Hey if I were to compare which leaving group is better , a tosylate group, or a Br/I, who would win? I know the tosylate group is resonance stabilized, but I and Br as acids, are extremely acidic and the more acidic something is, the weaker the conjugate base and the more stable it is. So If given the option to pick between the three I would go I > Br > Tosylate group. Just want to confirm that my train of thought for this is correct. Thanks!
pKa HI = -10
pKa HBr = -9
pKa Sulfonic acid = -3

 

[aj30]

Organic #294 discusses the reaction of a Gringard with a nitrile and says that it will result in a ketone.

My question is, why does the reaction stop there? Shouldn't the Gringard now react with the ketone to make a tertiary alcohol? I know it doesn't specify "2 moles" but that shouldn't make difference. Think about an ester-the Gringard first replaces the -OR group in Nucleophilic Acyl Substitution and then the second Gringard attacks the ketone to make it a tertiary alcohol. There's no need to specify "2 moles"-the reaction always goes to completion unless stated otherwise. It seems to be almost the exact same thing with Gringard + nitrile.
(I know McMurry states the reaction this way too and doesn't discuss the next step but I never knew why.)
Thanks

 

[omarski]

Organic #294 discusses the reaction of a Gringard with a nitrile and says that it will result in a ketone.

My question is, why does the reaction stop there? Shouldn't the Gringard now react with the ketone to make a tertiary alcohol? I know it doesn't specify "2 moles" but that shouldn't make difference. Think about an ester-the Gringard first replaces the -OR group in Nucleophilic Acyl Substitution and then the second Gringard attacks the ketone to make it a tertiary alcohol. There's no need to specify "2 moles"-the reaction always goes to completion unless stated otherwise. It seems to be almost the exact same thing with Gringard + nitrile.
(I know McMurry states the reaction this way too and doesn't discuss the next step but I never knew why.)
Thanks

I think this comes down to a fundamental concept that is really important. Esters have much better leaving groups than both aldehydes and ketones. As a result, nucleophilic substitution can occur (when the leaving group is present), followed by another reaction of nucleophilic addition.
However, aldehydes and ketones both lack a decent leaving group and can only undergo nucleophilic addition, as a result.

 

[aj30]

I think this comes down to a fundamental concept that is really important. Esters have much better leaving groups than both aldehydes and ketones. As a result, nucleophilic substitution can occur (when the leaving group is present), followed by another reaction of nucleophilic addition.
However, aldehydes and ketones both lack a decent leaving group and can only undergo nucleophilic addition, as a result.

The above is certainly true (saying that it's a fundamental concept is even an understatement, it's like Step A of 30 Organic reactions we need to know) but it doesn't answer the question. As I stated in my post, esters undergo two processes on their way to becoming tertiary alcohols with Gringards-
A. They undergo nucleophilic acyl substitution to become ketones
B. Once they are ketones, they undergo nucleophilic addition to become tertiary alcohols.

So, a nitrile first undergoes a reaction to become a ketone (via backwards imine formation mechanism but that's not important). Once it's a ketone, it can now undergo nucleophilic addition like all other ketones, so why doesn't it?

The point of my comparison to ester was to show that we can include 2 different reaction steps, each with different mechanism, in one overall reaction, to say that "an ester + Grignard=ketone". Therefore, why can't we do the same with nitrile?

 

[DAT Destroyer]

This is adding a scenario to Organic Chemistry #227, in case I was presented with different groups to compare.
Hey if I were to compare which leaving group is better , a tosylate group, or a Br/I, who would win? I know the tosylate group is resonance stabilized, but I and Br as acids, are extremely acidic and the more acidic something is, the weaker the conjugate base and the more stable it is. So If given the option to pick between the three I would go I > Br > Tosylate group. Just want to confirm that my train of thought for this is correct. Thanks!
pKa HI = -10
pKa HBr = -9
pKa Sulfonic acid = -3

Organic #294 discusses the reaction of a Gringard with a nitrile and says that it will result in a ketone.

My question is, why does the reaction stop there? Shouldn't the Gringard now react with the ketone to make a tertiary alcohol? I know it doesn't specify "2 moles" but that shouldn't make difference. Think about an ester-the Gringard first replaces the -OR group in Nucleophilic Acyl Substitution and then the second Gringard attacks the ketone to make it a tertiary alcohol. There's no need to specify "2 moles"-the reaction always goes to completion unless stated otherwise. It seems to be almost the exact same thing with Gringard + nitrile.
(I know McMurry states the reaction this way too and doesn't discuss the next step but I never knew why.)
Thanks

During the last step of reacting the nitrile with a Grignard to form a ketone, we have to add water. The addition of water would destroy any Grignard that is present

Hope this helps.

Dr. Romano

 

[DAT Destroyer]

This is adding a scenario to Organic Chemistry #227, in case I was presented with different groups to compare.
Hey if I were to compare which leaving group is better , a tosylate group, or a Br/I, who would win? I know the tosylate group is resonance stabilized, but I and Br as acids, are extremely acidic and the more acidic something is, the weaker the conjugate base and the more stable it is. So If given the option to pick between the three I would go I > Br > Tosylate group. Just want to confirm that my train of thought for this is correct. Thanks!
pKa HI = -10
pKa HBr = -9
pKa Sulfonic acid = -3

Very few groups can beat a tosylate group as a leaving group. This group is well stabilized by resonance and would be beat by few others. [N2 gas could beat it however]. I- is not even close, but an excellent non-resonance stabilized leaving group. Resonance will generally trump other factors such as base strength. I- beats Br- since it is more stabilized.

Hope this helps

Dr. Romano

 

[RosalinaDDS]

DAT Destroyer 2015
Organic Chemistry #2

Why is A and B not miscible in each other?

 

[DAT Destroyer]

DAT Destroyer 2015
Organic Chemistry #2

Why is A and B not miscible in each other?

You misread the answer. It states that in order separate 2 liquids the 2 solvents should not be miscible in each other. That means, if 2 solvents are miscible in each other, you will not be able to separate the 2 liquids.
You need to have a polar and a nonpolar solvent.

Hope this helps.

 

[omarski]

Ok awesome. Thanks again 🙂. Again just trying to fill in the holes!

 

[DAT Destroyer]

Ok awesome. Thanks again 🙂. Again just trying to fill in the holes!

🙂

 

[aj30]

DAT Destroyer 2015 QR # 46 shows a polygon with 5 sides and asks for angle X. Below the figure it's noted that "75° is an acute not obtuse angle". The answer says angle X is 140°.
I understand all the math involved but there seems to be something very wrong with the picture. The angle labeled "75°" is quite a bit larger than a right angle so can't possibly be below 90°. Furthermore, this polygon can't possibly have the proposed angles of 110°, 2 140°'s, and 2 75°''s. That would mean angle X is almost double the 75° angle right near it which is clearly not the case.

Am I missing something or is the picture off? Thanks

 

[DAT Destroyer]

DAT Destroyer 2015 QR # 46 shows a polygon with 5 sides and asks for angle X. Below the figure it's noted that "75° is an acute not obtuse angle". The answer says angle X is 140°.
I understand all the math involved but there seems to be something very wrong with the picture. The angle labeled "75°" is quite a bit larger than a right angle so can't possibly be below 90°. Furthermore, this polygon can't possibly have the proposed angles of 110°, 2 140°'s, and 2 75°''s. That would mean angle X is almost double the 75° angle right near it which is clearly not the case.

Am I missing something or is the picture off? Thanks

The picture on the problem is not drawn to scale, and for the DAT you should never assume that a picture is drawn to scale.

Hope this helps!

 

[omarski]

Hey black horse. The rules for boiling point and melting point are different. Melting points prefer branches for stacking; however, boiling points do not. From what I can deduce about the picture you submitted, the topic is about boiling points not melting points!
Remember that more branching increases melting point but decreases boiling point!

I have a question myself, if you were to compare an alpha hydrogen on a ketone/aldehyde to a hydrogen on an alcohol (OH), would the alcohol's hydrogen be more acidic even though the aldehyde/ketone have resonance stabilization? From what I understand negatively charged carbons are nowhere near as stable as electronegative oxygens. But I was going over a bootcamp problem and it looks like they made resonance seem more important than atom? Also from what I know CARDIO, atom is before resonance. Just need some clarification thanks!

 

[aj30]

Destroyer Gen Chem #43
Choice A) shows the combustion of pentane, C5H12 + O2, to yield CO2 + H2O. Combustion reactions are redox: C is oxidized and O reduced. So far so good.

How do you calculate the original oxidation state of C though? I've never seen an oxidation number of -5.5, for example-oxidation numbers are always whole numbers. So assuming H is +1 (as usual) what is C? Obviously this question pertains not only to combustion of pentane but to other alkanes and possibly other compounds as well.

I figure that because oxidation numbers aren't representations of the real molecule, just a "formality" we assign to it (as Brown LeMay puts it), we can possibly assign half numbers, if we need to. Is this correct?
Thanks

 

[DAT Destroyer]

Destroyer Gen Chem #43
Choice A) shows the combustion of pentane, C5H12 + O2, to yield CO2 + H2O. Combustion reactions are redox: C is oxidized and O reduced. So far so good.

How do you calculate the original oxidation state of C though? I've never seen an oxidation number of -5.5, for example-oxidation numbers are always whole numbers. So assuming H is +1 (as usual) what is C? Obviously this question pertains not only to combustion of pentane but to other alkanes and possibly other compounds as well.

I figure that because oxidation numbers aren't representations of the real molecule, just a "formality" we assign to it (as Brown LeMay puts it), we can possibly assign half numbers, if we need to. Is this correct?
Thanks

n order to calculate the oxidation state of carbon in this molecule, you must look at the structure of Pentane. It is CH3CH2CH2CH2CH3. Therefore some carbons will have different oxidation states. Carbon that is bound to 3 Hydrogens has a -3 oxidation state, while carbon that is bound to 2 Hydrogens has a -2 oxidation state.
When considering oxidation states in Organic molecules, you must remember that Carbon is more electronegative than Hydrogen, therefore H is +1. So if one carbon is bound to 3 H's it has a -3 oxidation state. If carbon is bound to 2 H's it will have -2 oxidation state. When carbon is bound to carbon, the oxidation state does not change because both carbons have the same electronegativity.

Hope this helps.

 

[farzi2492]

QR question 13. how does 2 sq rt. (9) to the third become (3)^3?

 

[DAT Destroyer]

QR question 13. how does 2 sq rt. (9) to the third become (3)^3?

Because sqrt(9) =3 and then you do 3^3 =27

Hope this helps!🙂

 

[farzi2492]

Because sqrt(9) =3 and then you do 3^3 =27

Hope this helps!🙂

just confused as to what happens to the 2 out front otherwise i understand that sqrt of 9 = 3 which then becomes 3^3 = 27

 

[DAT Destroyer]

just confused as to what happens to the 2 out front otherwise i understand that sqrt of 9 = 3 which then becomes 3^3 = 27

 

[farzi2492]

View attachment 199830

that makes more sense!! thanks!!

 

[DAT Destroyer]

that makes more sense!! thanks!!

Glad I could help!🙂

 

[SDDDAT]

Gen Chem #55

Which of the following compounds is most likely to be a molecule?
a)LiCl
b)CaBr2
c)SI4
d)Al2O3
e)CaSO4

The answer says C but I'm not sure why D isn't correct either. From what I understand, a molecule has covalent bonds, which is true for C and D. A, B, and E have ionic bonds so they can't be molecules. Thanks for any help!

 

[DAT Destroyer]

Gen Chem #55

Which of the following compounds is most likely to be a molecule?
a)LiCl
b)CaBr2
c)SI4
d)Al2O3
e)CaSO4

The answer says C but I'm not sure why D isn't correct either. From what I understand, a molecule has covalent bonds, which is true for C and D. A, B, and E have ionic bonds so they can't be molecules. Thanks for any help!

This answer is correct. Choice D "Al3O2" is an ionic molecule. It is metal and nonmetal Aluminum has 3+ charge and Oxygen has 2- charge.

Hope this helps.

 

[SDDDAT]

This answer is correct. Choice D "Al3O2" is an ionic molecule. It is metal and nonmetal Aluminum has 3+ charge and Oxygen has 2- charge.

Hope this helps.

Thanks, didn't realize that Al is a metal... -_-

I have another question, #87 from Gen Chem

Which solution has the lowest freezing point?
a) 0.2 M KCL
b) 1.2 M C6H12O6
c) 1.4 M Na2SO4
d) 2.4 M C2H5OH
e) 1.8 M NaNO3

According to the book the solution is c because the solute will dissolve into 3 particles, causing the greatest freezing point depression. I get why this would occur, but I'm wondering why intermolecular forces wouldn't also play a role here? Since a, c, and e are ionic, wouldn't they have greater freezing points than b and d due to the stronger ionic bonds? If intermolecular forces aren't important here, then when do we consider them?

Thanks for your help!

 

[DAT Destroyer]

Thanks, didn't realize that Al is a metal... -_-

I have another question, #87 from Gen Chem

Which solution has the lowest freezing point?
a) 0.2 M KCL
b) 1.2 M C6H12O6
c) 1.4 M Na2SO4
d) 2.4 M C2H5OH
e) 1.8 M NaNO3

According to the book the solution is c because the solute will dissolve into 3 particles, causing the greatest freezing point depression. I get why this would occur, but I'm wondering why intermolecular forces wouldn't also play a role here? Since a, c, and e are ionic, wouldn't they have greater freezing points than b and d due to the stronger ionic bonds? If intermolecular forces aren't important here, then when do we consider them?

Thanks for your help!

You seem to be confused about Low vs high freezing point.
Higher freezing point means it is harder to freeze, means it freezes at lower temperatures. Higher freezing point means, it is easier to freeze and freezes at higher temperatures.
That is when intermolecular forces and number of particles apply.

Hope this helps.

 

[SDDDAT]

Thank! I think I realized my mistake in thinking that intermolecular forces and number of particles have separate effects on MP.

 

[DAT Destroyer]

Thank! I think I realized my mistake in thinking that intermolecular forces and number of particles have separate effects on MP.

🙂

 

[gambaboy]

Hey there,

Organic Chem #4 from Destroyer 2015,

It asks for which entry is correct when referring to base strength. The answer key said that it was E because that the addition of the O "removes a small amount of electron density away from the Nitrogen", making the Nitrogen a weaker base. I was confused by this because I thought that O-R is an electron-donating group and would donate electrons and make the base stronger. Is this the wrong way of thinking about it?

Another way I got the answer was E, was that I figured that the extra ketone gave the compound more resonance and more stability, which would decrease the strength and reactivity as a base. Could this be a way to approach the question?

Thanks!

 

[DAT Destroyer]

Hey there,

Organic Chem #4 from Destroyer 2015,

It asks for which entry is correct when referring to base strength. The answer key said that it was E because that the addition of the O "removes a small amount of electron density away from the Nitrogen", making the Nitrogen a weaker base. I was confused by this because I thought that O-R is an electron-donating group and would donate electrons and make the base stronger. Is this the wrong way of thinking about it?

Another way I got the answer was E, was that I figured that the extra ketone gave the compound more resonance and more stability, which would decrease the strength and reactivity as a base. Could this be a way to approach the question?

Thanks!

The key concept is to understand that a base needs to accept a proton. Nitrogen is a very good base.....however, if electron density is being taken away from N by resonance perhaps.....or by an OXYGEN by induction , this will remove electron density and render the base weaker. If O is alone, as in the case of a base such as methoxide or ethoxide, then yes it can be quite an effective base. ..... . In choice E.....the choice that has two carbonyls does indeed contribute to stability and removes electron density from the N, thus greatly reducing the electron accepting ability.....i.e. Basicity.

Hope this helps.

Dr. Jim Romano

 

[gambaboy]

The key concept is to understand that a base needs to accept a proton. Nitrogen is a very good base.....however, if electron density is being taken away from N by resonance perhaps.....or by an OXYGEN by induction , this will remove electron density and render the base weaker. If O is alone, as in the case of a base such as methoxide or ethoxide, then yes it can be quite an effective base. ..... . In choice E.....the choice that has two carbonyls does indeed contribute to stability and removes electron density from the N, thus greatly reducing the electron accepting ability.....i.e. Basicity.

Hope this helps.

Dr. Jim Romano

Thanks for the reply.

I'm getting a little confused because under CARDIO rule of base ranking, bases get weaker when they are next to resonance because it removes electron density and weakens the base, as you said.

However, from the definition of electron donating groups (EDG) & electron withdrawing groups, electron donating groups would increase basicity because they have lone pairs of electrons that they can donate, with (-OR) being an electron donating group. Don't these two conflict with each other?

 

[DAT Destroyer]

Thanks for the reply.

I'm getting a little confused because under CARDIO rule of base ranking, bases get weaker when they are next to resonance because it removes electron density and weakens the base, as you said.

However, from the definition of electron donating groups (EDG) & electron withdrawing groups, electron donating groups would increase basicity because they have lone pairs of electrons that they can donate, with (-OR) being an electron donating group. Don't these two conflict with each other?

I prefer to never use silly tricks , but rather understand the concept. This so called CARDIO rules breaks down often and has many exceptions. An electron withdrawing group such as NO2, COOH, Carbonyl, SO3H all increase the acidity of a molecule. These groups will withdraw electron density and STABILIZE the anion....thus render a molecule much less basic, but more acidic. Let us do a concrete example. Consider benzoic acid with an NO2 group and another benzoic acid with a OCH3 group. The nitrobenzoic is more acidic because after the loss of a proton, the ANION is able to be stabilized by the nitro group. This stabilization is reflected in the fact that the negative anion charge is spread out. Any Organic chem text will draw the resonance forms for you to see. The next thing we need to do is to examine the benzoic acid with the OCH3 group. There is some electron withdrawal by induction, but also a MUCH GREATER electron donating effect, by resonance. This will destabilize the anion. Thus....the anion will be less stabilized and hence acidity decreases. Bottom line......Groups like NO2, COOH, CHO, carbonyl, SO3H. CN all increase acidity by electron withdrawal.....groups like CH3, C2H5, R groups, OCH3, OC2H5, N-groups like NH2, OH groups increase basicity by destabilizing the anion.

Hope this helps.

Dr. Jim Romano

 

[aj30]

Organic #54
How do we know the answer is C and not B? B can fit as well as an N-H stretch is very similar to an O-H stretch. It can't be a primary amide because then it should have the double stretch but the secondary amide in B will have only one N-H stretch.
I guess you can say that the N-H stretch shouldn't be so "strong and broad" as in the graph and so that's how we know that peak at 3300 is an alcohol but that's not always the case. In an IR graph of N-methyl acetamide, for example (a primary amide just like here), the N-H stretch is very close to what an O-H stretch would look like.
Thanks

 

[hye340]

for the 2009 dat the organic chem part question 90 the question asks what is the structure of a compound having a molecular formula of c7h12.the answer 1,2 di methyl cyclobutane IS THIS WRONG?cause in the destroyer I never saw anything like that?thanks


http://www.predent.org.vt.edu/admissions/dat-sample-test-2009.pdf

 

[sharkteethdentist]

Gen Chem #238 2016 edition. Why does floride ion after titration reacts with water produce HF and OH ions? Thank you

 

[DAT Destroyer]

Gen Chem #238 2016 edition. Why does floride ion after titration reacts with water produce HF and OH ions? Thank you

Fluoride is a relatively strong base,,,,not very stabilized, thus will react in a hydrolysis reaction. RULE.....anions of strong acids such as I-....Br-.......Cl-....NO3-......ClO4-.....DO NOT REACT.....they are too weak.....thus do not react with water. Also....cations of strong bases such as Na+...K+....Li+...Rb +..... Cs+ also do NOT react......

Hope this helps.

Dr. Romano