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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[FeralisExtremum]

OC # 245 I was going through and why would a complex of 2,2- dimethylpentane have a higher melting point over stacking of the pentane. Also I was looking at my O-chem textbook and saw this

What you've shown in the textbook is for boiling point (not melting point), which the symmetrical branching = higher MP rule doesn't apply to. In the case of what you've shown, increased branching only diminishes the London dispersion forces, which lowers the boiling point.

 

[DAT Destroyer]

OC # 245 I was going through and why would a complex of 2,2- dimethylpentane have a higher melting point over stacking of the pentane. Also I was looking at my O-chem textbook and saw this View attachment 190637

As a general rule branching lowers the boiling point because there is less surface area to a branched molecule versus a straight chain. Melting points are not so straight forward. Branching that creates a high degree of symmetry will exhibit very high melting points. A good example will be 3,3-diethylpentane.

Hope this helps

Dr. Jim Romano

 

[Ame535]

I'm definitely going to be using this trick @orgoman22! Thank you.
I have another question (GC #114), would you please explain how Kw and temperature are related. I red the answer for this question, but I'm still confused.

 

[DAT Destroyer]

I'm definitely going to be using this trick @orgoman22! Thank you.
I have another question (GC #114), would you please explain how Kw and temperature are related. I red the answer for this question, but I'm still confused.

Kw is an equilibrium constant. We use this when dealing with water. As with any equilibrium constant in the temperature changes the value of this parameter.

Dr. Romano

 

[StudentDoc1234]

Just wanted to say thank you so much @orgoman22 for taking the time to answer these questions, it's been great help!

 

[DAT Destroyer]

Just wanted to say thank you so much @orgoman22 for taking the time to answer these questions, it's been great help![/QUOTE

It is my pleasure to give back to the SDN community. Glad I can help.

Dr. Romano

 

[Ame535]

For OChem 174 (Destroyer 2014),
I don't understand how there would be a 1,2-addition or 1,4-addition? Wouldn't the + charge form at c-2, or c-5?

Thanks

 

[DAT Destroyer]

For OChem 174 (Destroyer 2014),
I don't understand how there would be a 1,2-addition or 1,4-addition? Wouldn't the + charge form at c-2, or c-5?

Thanks

It the 2014 edition OC 174 had to do with sugars, perhaps you wrote the wrong question number. I will try to answer based on what you stated.

A 1, 2 and a 1,4 reaction compete depending on the temperature. The minute you see a conjugated system being added to by things like Br2. HBr, water, alcohols, etc....Think about these two reactions. Always try to form the most STABLE carbocation as the first move, and draw a resonance structure. The IUPAC numbers and the term 1,2 and 1,4 do not always MATCH !!!! Start calling number one the start of the double bond you first added to.

I hope this helps.... I will discuss this further if you like.

Dr. Romano

 

[Artdent97]

great thread! Thanks orgoman22 for answering questions, I will be posting some soon.

 

[Ame535]

It the 2014 edition OC 174 had to do with sugars, perhaps you wrote the wrong question number. I will try to answer based on what you stated.

A 1, 2 and a 1,4 reaction compete depending on the temperature. The minute you see a conjugated system being added to by things like Br2. HBr, water, alcohols, etc....Think about these two reactions. Always try to form the most STABLE carbocation as the first move, and draw a resonance structure. The IUPAC numbers and the term 1,2 and 1,4 do not always MATCH !!!! Start calling number one the start of the double bond you first added to.

I hope this helps.... I will discuss this further if you like.

Dr. Romano

Hi @orgoman22 ,
I am sorry I meant Q.147. Can you please elaborate on "The IUPAC numbers and the term 1,2 and 1,4 do not always MATCH"!
Also, is it because we should "Always try to form the most STABLE carbocation as the first move" we have the 1,2-addition at -40 degrees not 50?

I highly appreciate your help Dr.Romano.

 

[Anthony Bogdan]

In the Organic Chemistry Problems #44 (Which is incorrectly matched?). The solution says the answer is D (which I can see why) but I also think that B -the Aldol Condensation- is incorrect. I think the product of an aldol condensation would be an alpha beta unsaturated ketone, wouldn't it? It looks like the product shown was of the Aldol Addition, which is why I picked B, based on the technicality of the writing. Is that not correct also?

 

[DAT Destroyer]

The Aldol Product is fine. The product of an Aldol is a beta hydroxy aldehyde or ketone.... One doesn't have to heat it.........Heating would give the alpha-beta unsaturated compound......but the initial reaction was the ALDOL !!! Using acid for this reaction, usually will give the unsaturated compound. Consult any organic book, and look at the first few examples.....i.e Aldol prototypes to confirm this very important reaction !!!!

Hope this helps..

Dr. Romano

 

[DAT Destroyer]

Hi @orgoman22 ,
I am sorry I meant Q.147. Can you please elaborate on "The IUPAC numbers and the term 1,2 and 1,4 do not always MATCH"!
Also, is it because we should "Always try to form the most STABLE carbocation as the first move" we have the 1,2-addition at -40 degrees not 50?

I highly appreciate your help Dr.Romano.

Let us consider 2, 4-hexadiene.......we can do a 1,2 or a 1,4 addition.....correct ? However...... when you add say HBr.....we actually protonate C-2...and form a carbocation on C-3...then write a resonance form.....agree ? As you can see,,,,,1,2 and 1,4 addition does not have anything to do with the IUPAC numbers in this case,,,,,, !!! Focus more on understanding what you are doing.....Protonate first,,,,,to form the best carbocation.....then show a resonance form. The terms 1, 2 and 1, 4 are indeed misleading. If the H and Br are next to each other,,,,,this the 1,2 addition......the other being the1,4.

Hope this helps...

Dr. Romano

 

[Anthony Bogdan]

This isnt a destroyer question but I can't get it and the internet hasn't found me any solutions, if someone could help me with this QR question I'd really appreciate it ( I don't have the answer, I just have the answer choices)
If y = (x+2)/x, then which of the following is x?

A) 2/(y-1)
b) 2(y+1)
C) (2y+1)/y
D) 2y-1
E) y/(y+2)

 

[FeralisExtremum]

This isnt a destroyer question but I can't get it and the internet hasn't found me any solutions, if someone could help me with this QR question I'd really appreciate it ( I don't have the answer, I just have the answer choices)
If y = (x+2)/x, then which of the following is x?

A) 2(y-1)
b) 2(y+1)
C) (2y+1)/y
D) 2y-1
E) y/(y+2)

You want to approach this to isolate x, starting with:
y = (x+2)/x
Multiply both sides by x to get:
xy = x + 2
Move x to the other side of the equation:
xy - x = 2
Factor x out so it can be isolated:
x (y-1) = 2
Divide to isolate x:
x = 2/(y-1)

Did you happen to leave the / symbol out of option A), by any chance?

 

[Anthony Bogdan]

Yes I did actually, i just edited the post. Oh yeah duh! I did not even realize factoring x out haha, thank you

 

[wood11]

Hi,
Destroyer '12 version GC #128

I read the answer and I am still not understanding Normality vs. Molarity and when to use them. It has been a while since I took GC.

Thanks!

 

[Anthony Bogdan]

In Destroyer 2015 Organic Chemistry #77. I am curious why we do not divide the grams by the diatomic weight of Hydrogen and Oxygen?

 

[StudentDoc1234]

Hi,

For Destroyer 2015 Gchem #137, was choice C false because it stated that the 3rd IE of Mg was just slightly higher than its 2nd, as opposed to significantly higher?

 

[DAT Destroyer]

In Destroyer 2015 Organic Chemistry #77. I am curious why we do not divide the grams by the diatomic weight of Hydrogen and Oxygen?

Chemistry problem #77...... When doing Empirical formula calculations......we divide by the mass of the individual ATOM. In compounds such as alcohols, ethers, etc.......hydrogen and oxygen are not diatomic !!!!!! Hope this helps.

Dr. Romano

 

[DAT Destroyer]

Hi,

For Destroyer 2015 Gchem #137, was choice C false because it stated that the 3rd IE of Mg was just slightly higher than its 2nd, as opposed to significantly higher?

Yes.....the word slightly rendered this a False statement. Mg is in Group 2.....thus two electrons can be removed rather easily.....the THIRD would would be EXTREMELY High.....since a Noble Gas core would be breached !

Hope this helps.

Dr. Romano

 

[Ame535]

Hi,
For Destroyer'14 OChem Q.215, when do we use CH3OH and C2H5O-K+ ?

Thanks

 

[DAT Destroyer]

Hi,
For Destroyer'14 OChem Q.215, when do we use CH3OH and C2H5O-K+ ?

Thanks

In this problem......I needed to do an ELIMINATION reaction on a primary halide, thus I employed a large, sterically hindered strong base, t-butoxide. If C2H5O-K+,,,,potassium ethoxide reacted,,,,it would do SN2.........Water and alcohols such as CH30H would not react since the compound is a primary halide .

Keep up the good work

Dr. Jim Romano

 

[StudentDoc1234]

Hi,

For 2015 Ochem #62, can someone please explain why NH3 comes in with a positive charge, I was thinking because it has a lone pair that it's going to attack it with will make it a negative charge...

Thanks in advance!

 

[Ame535]

Hi,
Destroyer'14, I'm confused on when to use a pipet and when to use a buret.
In the Ochem section Q.201, to measure 15.6 ml of liquid the answer is to use a pipet. However, in Gchem section Q.206 to measure 12.3 ml of solution we use a buret?
What's the difference here?

Thanks

 

[Jackthesparrow]

Destroyer 2015 Ochem 245

would the answer be the same if the question was asking for the highest boiling point?

 

[DAT Destroyer]

Destroyer 2015 Ochem 245

would the answer be the same if the question was asking for the highest boiling point?

As a general rule branching will lower the boiling point because there is less surface area and therefore less intermolecular attractions. The highest boiling point would be less branched. Melting point is not so clear cut. If a molecule is highly symmetrical due to branching this will usually increase its melting point because symmetry allows it to fit into a crystal lattice more efficiently.

Hope this helps

Dr. Romano

 

[Jackthesparrow]

Destroyer 2015 Gchem Q. 196

Would you please explain, 'If acting as an acid this is the third dissociation step thus K3 = 1*10^-12'. I do not understand this because K3 = 1*10^-12 is for a base to me.
Also, how did you decide this solution acts as a base not an acid?

 

[DAT Destroyer]

Destroyer 2015 Gchem Q. 196

Would you please explain, 'If acting as an acid this is the third dissociation step thus K3 = 1*10^-12'. I do not understand this because K3 = 1*10^-12 is for a base to me.
Also, how did you decide this solution acts as a base not an acid?

This is a tricky problem.....but worth the effort to tackle this challenging question ! Since Ka values are given.....we must see whether this salt will be more favored to be acidic...or more favored to act as a base. We do this by removing the spectator and examine BOTH the acid and base reaction. As shown in the solution.......The acid reaction is the third ionization step......The base reaction......is the Kb of the second ionization step. Comparing the K values shows the Kb to be much larger than Ka. Thus,,,,,it prefers to act as a base !

Hope this helps

Dr. Romano

 

[Jackthesparrow]

This is a tricky problem.....but worth the effort to tackle this challenging question ! Since Ka values are given.....we must see whether this salt will be more favored to be acidic...or more favored to act as a base. We do this by removing the spectator and examine BOTH the acid and base reaction. As shown in the solution.......The acid reaction is the third ionization step......The base reaction......is the Kb of the second ionization step. Comparing the K values shows the Kb to be much larger than Ka. Thus,,,,,it prefers to act as a base !

Hope this helps

Dr. Romano

How do you know acid reaction is the third ionization step and base reaction is second ionization step by looking at the reaction?

 

[StudentDoc1234]

How do you know acid reaction is the third ionization step and base reaction is second ionization step by looking at the reaction?

For the acid reaction... you have to take HPO4-- and make it act like an acid by making it lose a proton. This is the same as the third ionization step as H3PO4 because on its third step its going to be losing its last proton, exactly like HPO4--.

For the base reaction... you have to take HPO4-- and make it act like a base by giving it a proton on the product side, which would be H2PO4-. Since this reaction, which is shown in the solutions page, is the reverse of the the ionization step of H2PO4 where it loses a proton, we would would just take that k and find its kb by dividing kw.

Not sure if my wording made any sense but hopefully it helped a bit.

 

[Lovelybond]

Dat destroyer 2015 General chemistry #114 how can you calculate to make a 1M solution only with the mass of CaCl2? without any given volume? Thank you so much!

 

[DAT Destroyer]

Dat destroyer 2015 General chemistry #114 how can you calculate to make a 1M solution only with the mass of CaCl2? without any given volume? Thank you so much!

Good question......1M...moles 1 mole in a liter of solution.........thus we have 1 liter !!!!

Hope this helps!

Dr. Jim Romano

 

[basserpa]

Destroyer '14 Bio #222
Which statement is false: spindle fibers made from microtubules and associated proteins/prophase = longest mitotic stage/genome = total DNA content/DNA is a highly conservative molecule/all are true

Solutions says "all are true, DNA is a highly conservative molecule that is passed down generation to generation" but if I were to see this on the real deal I would have picked "DNA is a highly conserved molecule" being the false entry. Every bio class I've taken has followed the convention of the "semi-conservative" model, as the "conservative" model was ruled out by the Meselson & Stahl experiment. Maybe this entry aimed to talk about the "high fidelity" of the DNA molecule?

 

[tooth_hurty]

Destroyer '15 Gen Chem #220

When using mcΔT = mcΔT to find specific heat of the metal, why do we use 100-40°C to find ΔT (final T-initial T) if 100°C was the initial temperature of the metal?

 

[basserpa]

Destroyer '15 Gen Chem #220

When using mcΔT = mcΔT to find specific heat of the metal, why do we use 100-40°C to find ΔT (final T-initial T) if 100°C was the initial temperature of the metal?

When finding ΔT, you are simply finding the magnitude of the change in temperature. In this case, 100° is 60° from 40°, just as 40° is 60° from 100°. Key word here is magnitude. If you change the units into Kelvin, this ΔT of 60° remains true.

I'm guessing that you probably found a negative value for specific heat. This simply wouldn't make sense, that would mean that your temperature would drop as you heat the metal. Try it yourself in the equation q=mcΔT.

 

[FeralisExtremum]

I would have picked "DNA is a highly conserved molecule" being the false entry.

There's an important difference between a molecule (in reference to its structure/sequence) being conserved - meaning that it doesn't get changed or altered much versus the word conservative to describe a method of replication. The words are similar but they have two different applications. Example: the amino acid sequence of cytochrome C is highly conserved over time. This doesn't have anything to do with replication - it is specific to the fact that the amino acid sequence has remained generally unchanged over time.

 

[super112]

DAT destroyer 2013 QR question 80.
The question reads angle of depression of boat out at sea is 6º but in the solutions it shows 6º to be in the bottom right corner of the triangle (hypotenuse is top left corner to bottom right). Is this an error?

 

[DAT Destroyer]

DAT destroyer 2013 QR question 80.
The question reads angle of depression of boat out at sea is 6º but in the solutions it shows 6º to be in the bottom right corner of the triangle (hypotenuse is top left corner to bottom right). Is this an error?

No it's correct. The other angle (84 degrees) is called the angle of elevation.

 

[super112]

No it's correct. The other angle (84 degrees) is called the angle of elevation.

Oh okay, I was looking at other pictures and just realized that I should have drawn a horizontal line to see it correctly, thanks!

 

[tbogdan89]

For 2015 Organic Chem Q 222: I understand why the T-ButylO- works best (because it's a sterically hindered bulky base) but I also noticed in the question that the only place to do the E2 reaction from is at the alpha carbon to the Br. Therefore wouldn't B,C, and E all produce the same product (least substituted alkene)

 

[DAT Destroyer]

For 2015 Organic Chem Q 222: I understand why the T-ButylO- works best (because it's a sterically hindered bulky base) but I also noticed in the question that the only place to do the E2 reaction from is at the alpha carbon to the Br. Therefore wouldn't B,C, and E all produce the same product (least substituted alkene)

The only way to do an elimination reaction on a primary halide in a significant yield would to be use the sterically hindered base t-butoxide. The other choices b, d, and e, would have given the SN2 product not the E2 product. Choice a would not react since alcohols are too weak as nucleophiles on primary halides.

Hope this helps

Dr. Romano

 

[StudentDoc1234]

For 2015 Gchem #262: I'm having trouble trying to figure out why it would be okay to assume the solvent to be 1 mole.

 

[DAT Destroyer]

For 2015 Gchem #262: I'm having trouble trying to figure out why it would be okay to assume the solvent to be 1 mole.

Molality is moles of solute per kg of solvent........since no exact amount of solvent always assume 1 liter.

Hope this helps

Dr. Romano

 

[DrNoCavity]

Ochem Odyssey Pg. 106: What is the Kinetic Isotope Effect with regards to E2 mechanism?

 

[DAT Destroyer]

Ochem Odyssey Pg. 106: What is the Kinetic Isotope Effect with regards to E2 mechanism?

The Kinetic Isotope Effect is considered to be one of the most essential and sensitive tools for the study of reaction mechanisms. A C-D bond is more difficult to break than a C-H bond.....as such.....this can tell us what mechanism is operating. If a reaction is E2, for example.....we have a single step.....thus the breaking of a C-H and a C-D bond would be quite different in the transition state. A kinetic isotope effect would provide direct evidence for the E2 mechanism. The E1 reaction would not show this,,,since a carbocation is formed,,,,,and represents the slow step. Thus....no Kinetic Isotope Effect.

Hope this helps

Dr. Romano

 

[StudentDoc1234]

2015 Bio #475: Isn't the answer supposed to be E? The solution states that the promoter site determines where transcription begins... so isn't that the same thing as "a start site for DNA transcription"?

Edit: Maybe it's supposed to say RNA transcription instead, I just never thought there was a difference in saying DNA or RNA transcription.

 

[DAT Destroyer]

2015 Bio #475: Isn't the answer supposed to be E? The solution states that the promoter site determines where transcription begins... so isn't that the same thing as "a start site for DNA transcription"?

Edit: Maybe it's supposed to say RNA transcription instead, I just never thought there was a difference in saying DNA or RNA transcription.

The problem is correct. In genetics, a promoter is a region of DNA that initiates transcription of a particular gene. Promoters are located near the transcription start sites of genes, on the same strand and upstream on the DNA (towards the 5' region of the sense strand). Promoters can be about 100–1000 base pairs long.

Hope this help

 

[StudentDoc1234]

The problem is correct. In genetics, a promoter is a region of DNA that initiates transcription of a particular gene. Promoters are located near the transcription start sites of genes, on the same strand and upstream on the DNA (towards the 5' region of the sense strand). Promoters can be about 100–1000 base pairs long.

Hope this help

Got it thank you!

 

[StudentDoc1234]

Molality is moles of solute per kg of solvent........since no exact amount of solvent always assume 1 liter.

Hope this helps

Dr. Romano

Sorry but I was trying to redo the problem and got confused... if the mole fraction of the gas is 2x10^-18 wouldn't that mean it's equal to (moles of gas/moles of H2O + moles of gas)? And if that is correct, and we assume water to be 1 mole, shouldn't we factor that 1 mole of water into the mole fraction equation to get moles of gas to be different than 2x10^-18?