Official DAT Destroyer Q&A Thread

[densaugeo]

Hi guys, since I'm currently going through the DAT Destroyer and I'm sure many of you guys are as well I felt this thread would be helpful. I usually have various questions while going through Destroyer regarding why a certain answer choice is correct or other times I am in need of a more detailed explanation.

Instead of making multiple threads each time we have questions, I thought we could just post them here and anyone can chime in with an explanation or further clarification.

Please don't post entire questions as that would be a violation of copyright. You can ask specific questions regarding a particular problem in the Destroyers.

Hopefully others find this a useful thread. Thanks!

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[nextgendental]

@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.

The only way benzene would attack in this case is if it had FeCl3 in the mix.


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[nextgendental]

Or AlCl3


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[FutureDent20]

2016 Bio destroyer #292:
I am little confused on primary target of ADH in the kidney. I though ADH acts mainly on collecting duct whereas aldosterone acts on distal tubules?

 

[Ashish]

2016 Bio destroyer #292:
I am little confused on primary target of ADH in the kidney. I though ADH acts mainly on collecting duct whereas aldosterone acts on distal tubules?

I would say aldosterone targets ascending loop of henle more than distal convoluted tubules since this is where most salts are reabsorbed.

 

[rdub]

2016 Bio destroyer #292:
I am little confused on primary target of ADH in the kidney. I though ADH acts mainly on collecting duct whereas aldosterone acts on distal tubules?

Aldosterone does act on the distal tubules (as well as the collecting duct), but it functions to increase the absorption of sodium, not water (but water does passively follow along).

I know Cliffs APBio only mentions ADH acting on the collecting duct, but both wikipedia and DAT Destroyer say that it works on the distal tubules as well, so I'd go with that for sure.

 

[DAT Destroyer]

2016 Bio destroyer #292:
I am little confused on primary target of ADH in the kidney. I though ADH acts mainly on collecting duct whereas aldosterone acts on distal tubules?

Aldosterone acts primarily on the cells of the collecting duct. While ADH acts on the collecting duct and the Distal convoluted tubule.

Hope this helps!

 

[rdub]

Math Destroyer 2016 Test 3 #10

The solutions points out that arc AC is 180 degrees, which makes perfect sense because it is half of the circle. What I don't understand is how angle ABC is related to arc AC. How can we conclude that angle ABC is half of arc AC? Thanks

 

[akimaki]

I have a question for the Destroyer Orgo #120. Why is the proton labeled as E not so acidic?
I understand that D is vinylic, and not very acidic. But what does the solution mean that E is aldehylic and not so acidic? It's part of an aldehyde, which is quite acidic with a pka of around 18.
Thanks!

 

[Ashish]

I have a question for the Destroyer Orgo #120. Why is the proton labeled as E not so acidic?
I understand that D is vinylic, and not very acidic. But what does the solution mean that E is aldehylic and not so acidic? It's part of an aldehyde, which is quite acidic with a pka of around 18.
Thanks!

The fact that aldol condensation proceeds through alpha carbanion tells us that aldehydic Hydrogen is harder to deprotonate. Thats a really high pKa to be acidic. Low pKa means more acidic. I am sure pKa of alpha hydrogen is much less than 18. More so, the pKa of less sterically hindered alpha hydrogen the least of all.

 

[akimaki]

Math destroyer test 13 #19.

Arcsin(√2/2) should = π/4, since the angle it forms is 45 degrees.

Why does the answer show 3π/4? Is it because sin would be positive in the second quadrant also? But shouldn't the range for ArcSin be −π/2 ≤ θ ≤ π/2??
Any help to clear this up would be much appreciated!

 

[bigsmilenee]

Destroyer 2016 Orgo #52

I understand up until the last step where the NaOCH3 performs an SN2 on the secondary tosylate leaving group. I thought that basic oxides perform E2 on secondary leaving groups?
Thanks.

 

[Ashish]

Destroyer 2016 Orgo #52

I understand up until the last step where the NaOCH3 performs an SN2 on the secondary tosylate leaving group. I thought that basic oxides perform E2 on secondary leaving groups?
Thanks.

Since it is not a bulky base, it also does not go the Anti-zaitsev's route either since the more substituted carbon does not have a Hydrogen in this case.

 

[gangazi]

2016 dat destroyer orgo #123
which producces two diastreomers A and B?

I thought Br2 goes to anti addition, thus producing enantiomers...
so how do they become diastereomers? do you just flip the streochemistry on one of the streocenter after it produces enantiomers?
thank you!

 

[FutureDent20]

2016 dat destroyer orgo #123
which producces two diastreomers A and B?

I thought Br2 goes to anti addition, thus producing enantiomers...
so how do they become diastereomers? do you just flip the streochemistry on one of the streocenter after it produces enantiomers?
thank you!

So if you recall; multiple chiral centers with some that are inverted and other(s) that is/are not inverted refers to diasteriomers. If you notice in answer choice C stereochemistry around the ethyl group is not changing. if you assign R/S you get RSS and RRR (one stayed the same other two changed from SS to RR) hence diasteriomer. Hope it makes sense.

 

[DAT Destroyer]

2016 dat destroyer orgo #123
which producces two diastreomers A and B?

I thought Br2 goes to anti addition, thus producing enantiomers...
so how do they become diastereomers? do you just flip the streochemistry on one of the streocenter after it produces enantiomers?
thank you!

The minute you see 2 chiral carbons, the possibility of enantiomers are possible. Look closely at the solution where I have shown that once Br2 is added,across the double bond, they add anti...PLUS another carbon exits as shown..This carbon is also CHIRAL ...which makes possible products a diastereomeric pair relationship.

Hope this helps

Dr. Romano

 

[SmilezAllDay]

Just a quick question about lithium aluminum hydride. In question 171 orgo destroyer when we treat a nitrile with LiAlH4 it has to be in ether and then H2O because just LiAlH4 and then water would blow up all up in my face. I also noticed that when we reduce a carboxy acid into a primary alcohol we just use LiAlH4 and then water (as shown in the beginning reactions in the destroyer). In Question 259, last step of the synthesis, when we reduce the amide the reagents are just LiAlH4 and then water.

Maybe someone can clear this up for me? are there specific instances where we would be in ether, and other instances where we dont need ether???

 

[nextgendental]

Just a quick question about lithium aluminum hydride. In question 171 orgo destroyer when we treat a nitrile with LiAlH4 it has to be in ether and then H2O because just LiAlH4 and then water would blow up all up in my face. I also noticed that when we reduce a carboxy acid into a primary alcohol we just use LiAlH4 and then water (as shown in the beginning reactions in the destroyer). In Question 259, last step of the synthesis, when we reduce the amide the reagents are just LiAlH4 and then water.

Maybe someone can clear this up for me? are there specific instances where we would be in ether, and other instances where we dont need ether???

I'm pretty sure all you need to know is that water needs to be added as a second step AFTER you use LiAlH4. (*Always: 1) LiAlH4, 2) H20*) The fact that LiAlH4 is used with ether doesn't matter because it's just a solvent.


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[dan123Z4]

in the 2016 addition question #157 in the gen chem it says solubility is a physical change. can you please explain this? wouldn't it be a chemical change if it went from an NaCl to Na+?

 

[FutureDent20]

in the 2016 addition question #157 in the gen chem it says solubility is a physical change. can you please explain this? wouldn't it be a chemical change if it went from an NaCl to Na+?

If you're referring to dissolving table salt in water then yes it's a physical change since the identity of the atoms "composition" is unchanged. Here is another example of physical change, boiling water (liquid to Gas), it's still water but in its vapor form thus you didn't change the composition.
In contrast, when composition of the atoms are changed we refer to it as a chemical change. For example think about combustion (burning). When wood burns in presence of Oxygen it releases CO2(g) + H2O and tons of heat.

 

[DAT Destroyer]

in the 2016 addition question #157 in the gen chem it says solubility is a physical change. can you please explain this? wouldn't it be a chemical change if it went from an NaCl to Na+?

In a chemical change, we change the Identity of the substance. Solubility is a physical property because the chemical identity is not changed. Dissolve sugar into water......we can measure the solubility. It is still sugar, however !!!!

Hope this helps.

Dr. Romano

 

[FutureDent20]

2016 OC destroyer # 112 and 113:
What's the best way to know when to use phenyl and when to use benzyl when naming a compound?

 

[Neil Smoove]

My question is #283. It is stated above, but is answer A any different than E? I put A down and the only difference is that the bond from the CH2 to the benzene is "straight" 180 degrees instead of "angled". I am still confused on that.

 

[Neil Smoove]

My question is #283. It is stated above, but is answer A any different than E? I put A down and the only difference is that the bond from the CH2 to the benzene is "straight" 180 degrees instead of "angled". I am still confused on that.

I meant OChem section as well

 

[workofshart]

In Destroyer orgo #50 why wouldn't we count the other way so the flouro has a lower #? I never saw this in orgo so I don't understand what counting the "longest" way means.. help please!

 

[DAT Destroyer]

2016 OC destroyer # 112 and 113:
What's the best way to know when to use phenyl and when to use benzyl when naming a compound?

A phenyl substituent is simply when a benzene ring comes off a chain or exists as a discreet substituent. A benzyl group is a benzene ring with a CH2 group attached.

Dr. Romano

 

[FutureDent20]

A phenyl substituent is simply when a benzene ring comes off a chain or exists as a discreet substituent. A benzyl group is a benzene ring with a CH2 group attached.

Dr. Romano

Makes sense now. Thank you Dr. Romano!

 

[FutureDent20]

In Destroyer orgo #50 why wouldn't we count the other way so the flouro has a lower #? I never saw this in orgo so I don't understand what counting the "longest" way means.. help please!

I believe you are using the older Destroyer book. what year Destroyer are you using?

 

[workofshart]

I believe you are using the older Destroyer book. what year Destroyer are you using?

2015.. not that old lol

 

[FutureDent20]

2015.. not that old lol

Lol I didn't mean it that way. Can you explain the question? Because I have the 2016 destroyer and #50 is different for me

 

[T3DDIBEAR]

Can I get an explanation for orgo #80 (2016) please?

I am struggling trying to understand why it is C and not D.

 

[Dentoni]

Hi Dr. Romano and Nancy, I have a question: what is the difference between NaBH4, LiALH4, Na2Cr2O7 and PCC ? what are they and what do they specifically work on? Thank you for help.

 

[DAT Destroyer]

Can I get an explanation for orgo #80 (2016) please?

I am struggling trying to understand why it is C and not D.

It cannot...I repeat CANNOT be D......In choice D, the methyl group on the left would give us the most DOWNFIELD signal.....and it would be a SINGLET........we see a quartet, hence D is eliminated.

For a nice review on NMR, consult a text written by a PhD Organic chemist such as David Klein, Skip Wade, or John McMurray.

In choice C....the CH2 group is the most downfield, since it it most closely associated with the negative O,,,,,hence it would give a downfield quartet,,,,,and this exactly what is shown.......This was a slam dunk !!!!

Hope this helps.

Dr. Romano

 

[P0W3RL1FT3R]

pretty easy question but..
QR question 51:
why does 29.25 round to 30?
especially when one of the answer choices is 29

 

[T3DDIBEAR]

It cannot...I repeat CANNOT be D......In choice D, the methyl group on the left would give us the most DOWNFIELD signal.....and it would be a SINGLET........we see a quartet, hence D is eliminated.

For a nice review on NMR, consult a text written by a PhD Organic chemist such as David Klein, Skip Wade, or John McMurray.

In choice C....the CH2 group is the most downfield, since it it most closely associated with the negative O,,,,,hence it would give a downfield quartet,,,,,and this exactly what is shown.......This was a slam dunk !!!!

Hope this helps.

Dr. Romano

Thank you Dr. Romano

 

[rdub]

Question on 2016 Bio #504. A promoter site is where RNA polymerase binds to initiate transcription. Doesn't that mean it is also the start site for transcription?

 

[DAT Destroyer]

Question on 2016 Bio #504. A promoter site is where RNA polymerase binds to initiate transcription. Doesn't that mean it is also the start site for transcription?

Before RNA polymerase binds to the transcription start site, it must bind to the promoter. Promoter is located upstream of the transcription start site.
It sort of acts as a marker for the location of the transcription start site.
Hope this helps.

 

[droumayah]

For DAT destroyer version 2011, in the organic chemistry section (question 209) it asks which diene is most stable. I understand why it is narrowed down to C or D but it says C is more stable because it is more substituted when they are both equally substituted. Anyone know why? Thanks.

 

[Neil Smoove]

my question is Organic Chem #114. It claims A is aromatic, but isn't the Carbon on top sp3? We have 3 carbons attached as well as a lone pair which take up a p orbital wouldn't it? I need clarification please. Almost 3 weeks out and I must know this lol

 

[FutureDent20]

my question is Organic Chem #114. It claims A is aromatic, but isn't the Carbon on top sp3? We have 3 carbons attached as well as a lone pair which take up a p orbital wouldn't it? I need clarification please. Almost 3 weeks out and I must know this lol

IT is indeed aromatic because the lone pair on nitrogen is part of the ring (its in P orbital). For that reason when you're counting the domains you don't count the lone pair. You then end up with 6π electrons and a Sp2 hybridized carbon. Same thing with answer choice C, lone pair is part of the aromatic and we still have Sp2 hybridized carbon atom but based on hackles rule its anti-aromatic because it has 4π electrons. If this is still unclear i would highly suggest watching chads videos.

2,6,10, 14, 18, 22, 26 π electrons = aromatic
4, 8, 12, 16, 20, 24 π electrons = anti-aromatic

 

[DAT Destroyer]

my question is Organic Chem #114. It claims A is aromatic, but isn't the Carbon on top sp3? We have 3 carbons attached as well as a lone pair which take up a p orbital wouldn't it? I need clarification please. Almost 3 weeks out and I must know this lol

NO NO NO !!!! Focus on the ring........This is a pyrrole ring...every atom IN THE RING is sp2, the molecule is planar, cyclic, has 6 pi electrons, and is FULLY CONJUGATED !!!! Thus it is aromatic. The only sp3 carbon is the methyl that has nothing to do with the cyclic array of p orbitals. This is a HUGE question. Please consult an Organic text to reinforce your understanding of this concept. After you do, revisit this problem. The electrons in the ring delocalizes on each atom

Hope this help

Dr. Jim Romano

 

[linsanity2013]

.

 

[Neil Smoove]

NO NO NO !!!! Focus on the ring........This is a pyrrole ring...every atom IN THE RING is sp2, the molecule is planar, cyclic, has 6 pi electrons, and is FULLY CONJUGATED !!!! Thus it is aromatic. The only sp3 carbon is the methyl that has nothing to do with the cyclic array of p orbitals. This is a HUGE question. Please consult an Organic text to reinforce your understanding of this concept. After you do, revisit this problem. The electrons in the ring delocalizes on each atom

Hope this help

Dr. Jim Romano

Thank you for clarifying, Doc!

 

[emhl123]

DAT Destroyer 2016 Organic Chemistry #33

I am confused about this question. The most basic compound is the one that is the strongest base. The solution says that choices B,C,E all have electron withdrawing groups. However, in choices B,C, and D, can't the lone pairs on the nitrogen donate electron density to the aromatic ring, therefore increasing its basicity and making it a stronger base?
Thanks in advance.

 

[droumayah]

@linsanity2013 the answer says C I think. You are right that it will only go through a walden inversion not racemization.

 

[Fancy312]

Hi guys,

i was wondering if anyone can help me out with question on gen chem,# 64. molybdenum, Mo, is an element used in missile and aircraft parts. how many unpaired electrons are in Mo? i was just unsure why the 5s and 5d oribital were made as 5s1 and 4d5, instead of 5s2 4d4?

thanks again

 

[DAT Destroyer]

DAT Destroyer 2016 Organic Chemistry #33

I am confused about this question. The most basic compound is the one that is the strongest base. The solution says that choices B,C,E all have electron withdrawing groups. However, in choices B,C, and D, can't the lone pairs on the nitrogen donate electron density to the aromatic ring, therefore increasing its basicity and making it a stronger base?
Thanks in advance.

A base will use a lone pair of electrons and donate them to a proton. The minute you see an AROMATIC ring....realize that basicity will greatly be reduced because the benzene ring will take a part of the electron density. Thus we can eliminate b,c, d and e. You have missed a critical point....if an electron withdrawing group is present, and removes electron density from the Nitrogen atom...BASICITY WILL BE REDUCED...not increased. Choice a shows an amine with a single R group...devoid of any electron withdrawing group or aromatic ring, this would have the lowest pKb value, hence the strongest base !!! If you need further clarity on this, the Francis Carey and David Klein text do a wonderful job on further examples.

Hope this helps.

Dr. Romano

 

[ThePianist1]

Dr. Romano,

O-Chem Q#137: why is a benzene with a carbocation categorized as "horrible"? Isn't it still aromatic, which then stabilize the carbocation? Thanks in advance.

 

[DAT Destroyer]

Dr. Romano,

O-Chem Q#137: why is a benzene with a carbocation categorized as "horrible"? Isn't it still aromatic, which then stabilize the carbocation? Thanks in advance.

Horrible indeed !!!! This is a great question. It is unstable.....and rarely EVER seen. We come across this high energy specie in advanced organic chemistry when dealing with some diazonium ion mechanisms. The positive charge CANNOT be stabilized !!!! Recall all the orbitals on the benzene ring are in a parallel array, but when a carbocation forms...this new orbital will be in a DIFFERENT plane. !!!!!! This means there can be no resonance stabilization at all. Thanks for one hell of a GREAT QUESTION !!!!!! wow........

Hope this helps.

Dr. Romano

 

[linsanity2013]

.

 

[DentalNucleicAcid]

I found conflicting information regarding the normality of H2SO4. Chad says that if you have, say, 1 M of H2SO4 then you'll actually end up having a molarity (or normality) of slightly higher than 1 M because only the first H dissociates completely because H2SO4 is a strong acid. The second H will only partially dissociate because we are now starting with HSO4- which is no longer a strong acid, but rather a weak acid. This makes sense. But, I found a question in Destroyer that says that the normality of H2SO4 will actually be twice the molarity, not just slightly higher- implying that both of the H's dissociate completely, which doesn't make sense to me. I thought the only time when both of the H's/OH's dissociated completely was when we had a strong base with 2 OH's on it, like Ca(OH)2. Anyone?