Group Study Thread and Other Micellaneous Posts

[SDN]

This thread is for students to ask and answer each other's questions for the MCAT, DAT, OAT, and PCAT subjects. All posts from the other Q & A threads that are not either students' questions or moderators' replies will be moved to this thread. Students, please post your own answers to other students' questions here in this thread--or else in the main MCAT forum--rather than in the regular Q & A threads.

If you are a pre-health (MCAT, PCAT, DAT, OAT) test prep instructor, or you have a graduate-level background in one of the MCAT subjects, or you are a senior undergrad who scored well on the MCAT, and you would like to help answer questions for one of the regular Q & A threads, please PM Shrike or QofQuimica and let them know. Again, all undergrad students who are currently studying for any of these tests should NOT post answers to other students' questions in any of the regular Q & A threads. Thanks to everyone for your cooperation..

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[heyitscyndi]

what is the best way to remember how to determine if a reaction is SN1, SN2, E1, or E2? Here's what I have from class, but where do E1 rxns fit? How do you distinguish between SN1 and E1?


Alkyl halide 1 2 3
Nucleophile

Strong SN2 E2 (mostly) E2
SN2

Weak No rxn SN1 SN1


Thanx for the help!

 

[Nutmeg]

what is the best way to remember how to determine if a reaction is SN1, SN2, E1, or E2? Here's what I have from class, but where do E1 rxns fit? How do you distinguish between SN1 and E1?


Alkyl halide 1 2 3
Nucleophile

Strong SN2 E2 (mostly) E2
SN2

Weak No rxn SN1 SN1


Thanx for the help!

I had absolutely no idea what to make of your notes until I hit the quote button. Unfortunately, the SDN server will get rid of extra spaces, and the table formatting gets lost entirely. Anyone else trying to read this, hit the "quote" button and take a gander.

As for E1 versus Sn1: the Sn1 needs a nucleophile because you need to have an adding group that is more stable than the leaving group. By contrast, the E1 reaction requires a weak base because you need to deprotonate the carbocation to get the alkene (if the base is strong, you should expect E2, not E1).

I found this to be a pretty good chart for distinguishing.
http://ludwig.chem.selu.edu/sarah/CHEM266/SN1vsSN2vsE1vsE2.html

 

[eccles1214]

alright im getting overwhelmed with all these reactions some of which we didnt even talk about! Should i just memorize that this reagent does this and this does this? Like PCC ive never even heard of. Will the mcat tell you what the reagent does?

So what exactly is the secret -- if it is really a secret -- to doing well at OChem? I'm taking GenChem now, but as I walk the chemistry halls at my school, I can't help but notice the grade postings for OChem courses. A lot of Cs and Ds, a few Bs, and fewer As still.

Any advice on how to both understand OChem and go about studying the material?

 

[riceman04]

So what exactly is the secret -- if it is really a secret -- to doing well at OChem? I'm taking GenChem now, but as I walk the chemistry halls at my school, I can't help but notice the grade postings for OChem courses. A lot of Cs and Ds, a few Bs, and fewer As still.

Any advice on how to both understand OChem and go about studying the material?

Dont just memorize rxns. Understand why X and Y are occuring. Also, pay attention to general trends in rxs b/c they always appear.

And start studying for your first exam during your first week of class. What i mean by this is to review a little EVERY NIGHT and then when test time comes you will have reviewed everything at least once. Then you can just do lots and lots of practice problems.

hope this helps

 

[Caboose]

Hey daofu -
My forte is definitely not chem... I haven't actually found my forte... Anyway, I'm still on chapter one in the chemistry, but I pulled this off the EK site for you:

"By the same reasoning as for question 47, Kp for this reaction is the partial pressure of CO2 at equilibrium. Thus if the partial pressure of CO2 is greater than Kp the reaction will try to shift to the left...but there is no CaO in the container! So it can't shift left and equilibrium can't be reached. If the partial pressure of CO2 were less than Kp, the reaction would try to shift to the right--and succeed, because the beaker of calcium carbonate is still there. CaO would be formed in this reaction."


Oh, sorry, I'm awake! I'm awake! That chapter is next week.

Caboose.

 

[Caboose]

O.k., now my question:

Conjugated proteins = nonproteinaceous component havin' globs

Are cytochromes just examples of these class of proteins or are there other globs that require prosthetic shtuff too?

Just wondering.


Caboose

 

[vmp200]

Sorry didnt read your question before i replied.

 

[asmaa elaraby]

hi everybody, i'm new to the forums,i'm medical student,alex university,egypt and i guess i'll get lots of benifit from u guys

 

[gotgame83]

Maybe someone can help me clarify this. Regarding the force of an electric field and a magnetic field. I had this straight a few weeks ago but I seemed to forget it again.

When a negative charge is traveling perpendicular to an electric field it feels a force in the opposite direction then the electric field force vector is pointing. However If it is a postive charge then it feels a force in the same direction. Is that correct? How about magnetic field. If a postive charge goes through a magnetic field does it feel a force in the same directions as the magnetic field force vector is pointing, and would it be oppostive if it was a negative charge?

 

[QofQuimica]

hi everybody, i'm new to the forums,i'm medical student,alex university,egypt and i guess i'll get lots of benifit from u guys

Welcome to the forum, asmaa, and feel free to ask questions if you have any.

 

[Omar@JHU]

Hello, anybody out there who just took Kaplan 11R full length? Can anybody please convince me what the correct answer is to question 12 (in Physical Sciences Passage 2)? I think it's "A" and not "D". ("A" says 1.32 volts, which is what I got by adding the OXIDATION potential for copper, -.34V, with the REDUCTION potential for aluminum, 1.66V. On the other hand "D" is what we get when we add the REDUCTION pot for copper, .34V, with the REDUCTION pot for Aluminum, 1.66V). I think D is wrong and am confused at the discrepancy in Kaplan's answer key which says that "In this cell, Copper will be reduced and Aluminum will be oxidized, so reverse(??!!) the sign on
the Aluminum value from the table." In the table the only value given is the positive Al reduction value of 1.66V. I need help, anybody who has done this problem, can you help me? ThanKS!!!

 

[Omar@JHU]

Maybe someone can help me clarify this. Regarding the force of an electric field and a magnetic field. I had this straight a few weeks ago but I seemed to forget it again.

When a negative charge is traveling perpendicular to an electric field it feels a force in the opposite direction then the electric field force vector is pointing. However If it is a postive charge then it feels a force in the same direction. Is that correct? How about magnetic field. If a postive charge goes through a magnetic field does it feel a force in the same directions as the magnetic field force vector is pointing, and would it be oppostive if it was a negative charge?

Dear gotgame, Happily the equations are vector equations i.e. we can tell the direction of the force (+ =same direction; or - =opp. direction) by looking at the sign of the other side. Look at the equations, I've put vectors in bold.
First electric field. The equation is F = qE. You're correct, for pos. charges the force F is same direction as E and for neg. charges it is in the opp. or reverse direction, it points against the direction of the electric field.
Now for magnetic field. The equation here is F = qvBsinA. Here, F means the force felt by the charged particle, qv is the velocity vector of the particle, and B is the magnetic field vector of the magnetic field thru which the charged particle is traveling. The direction of the force is given by the Right Hand Rule, or RHR, which, if you have a right hand, is very easy. Extend your right hand forwards like you would in a handshake, but stiffen it so your fingers and thumb, joined, point straight ahead, and your palm faces to your left. Now raise up your thumb so it points vertically upwards. Now watch this. The RHR will give you the direction of the magnetic force felt by a charged particle in a magnetic field. Do it like this. Your fingers point in the direction of qv and your palm faces in the same direction as B, the magnetic field vector, and your thumb is showing you the direction of F, the force felt by the charged particle. Note that the three vectors are all perpendicular to each other. HOWEVER IF THE PARTICLE IS NEGATIVE the vector qv will point in the reverse direction. Make your hand flip towards yourself and make your palm face to the left again BECAUSE THE DIRECTION OF THE MAGNETIC FIELD HAS NOT CHANGED. Your thumb should now point DOWNWARDS and this means that the direction of the force on the negative particle is opposite to the direction of the force on the positive particle, if everything else stays the same. Hope this helps, Omar

 

[gotgame83]

Thanks a lot Omar! That really helps.

 

[GCT]

My compliments on the forum, very nice site. Does this forum have latex?

 

[DrChandy]

Hello, anybody out there who just took Kaplan 11R full length? Can anybody please convince me what the correct answer is to question 12 (in Physical Sciences Passage 2)? I think it's "A" and not "D". ("A" says 1.32 volts, which is what I got by adding the OXIDATION potential for copper, -.34V, with the REDUCTION potential for aluminum, 1.66V. On the other hand "D" is what we get when we add the REDUCTION pot for copper, .34V, with the REDUCTION pot for Aluminum, 1.66V). I think D is wrong and am confused at the discrepancy in Kaplan's answer key which says that "In this cell, Copper will be reduced and Aluminum will be oxidized, so reverse(??!!) the sign on
the Aluminum value from the table." In the table the only value given is the positive Al reduction value of 1.66V. I need help, anybody who has done this problem, can you help me? ThanKS!!!

http://forums.studentdoctor.net/showthread.php?p=2837931#post2837931

 

[dermchick]

A passage about an X-ray device states that electrons are accelerated from the cathode to the anode by an electric field. One question after the passage asks : What is the direction of the electric field that accelerates the electrons? The answer is FROM THE ANODE TO THE CATHODE. Can anyone explain? Is there an error?

 

[DrChandy]

A passage about an X-ray device states that electrons are accelerated from the cathode to the anode by an electric field. One question after the passage asks : What is the direction of the electric field that accelerates the electrons? The answer is FROM THE ANODE TO THE CATHODE. Can anyone explain? Is there an error?

For the MCAT, know that the direction of an electric field goes from positive to negative.

In the example above, electrons are accelerated from the cathode to the anode. Since electrons are negatively charged and are in this case accelerated towards the anode by an electric field, the anode should be positively charged.

Therefore, if the anode is positively charged, the electric field goes from anode to cathode.

 

[Sol Rosenberg]

to determine if a molecule is aromatic it has to satisfy Huckels Rule of 4n+1

How do you determine n?

n is any positive integer. Also, isn't it 4n+2?

So, benzene qualifies as an aromatic compound because it has 6 pi electrons. For n=1, 4n+2 = 6

 

[NontradICUdoc]

n is any positive integer. Also, isn't it 4n+2?

So, benzene qualifies as an aromatic compound because it has 6 pi electrons. For n=1, 4n+2 = 6

Sorry. Yes 4n+2. Up until now I have been counting the c=c and multiplying by 2 and then adding any extra electrons. I am not sure if I am doing this correctly. Please explain it to me.

 

[jp104]

Sorry. Yes 4n+2. Up until now I have been counting the c=c and multiplying by 2 and then adding any extra electrons. I am not sure if I am doing this correctly. Please explain it to me.

I believe n is the number of carbons with a pi orbital in a conjugated system. So if it looked like this: c-c=c-c=c-c that would be 6 carbons and 6 n.

Sorry if this is wrong info!

 

[QofQuimica]

I believe n is the number of carbons with a pi orbital in a conjugated system. So if it looked like this: c-c=c-c=c-c that would be 6 carbons and 6 n.

Sorry if this is wrong info!

No, this is not correct. Please see my post above. Also, I am going to move these posts to the student group study thread. If you all would like to continue studying together, you should do it there.

 

[jp104]

No, this is not correct. Please see my post above. Also, I am going to move these posts to the student group study thread. If you all would like to continue studying together, you should do it there.

I stand corrected Thanks QofQuimica!

 

[QofQuimica]

I stand corrected Thanks QofQuimica!

Your strategy will work in many cases, but you have to be careful because some rings have more or fewer atoms and are still aromatic. For example, furan, pyrrole, and cyclopentadiene anion are all n=1 aromatic compounds, but they only have five atoms in the ring. Likewise, cycloheptatriene cation is an n=1 aromatic with seven atoms in the ring. If you go by # of pi electrons, you won't go wrong.

P.S. Sorry but I can't draw the structures here, so if you don't know what these molecules I mentioned are, you'll have to look them up.

 

[Krazykritter]

is this the right thread? Ok I have a question about testesterone. Since it inhibits FSH secretion, wouldnt it inhibit spermatogenesis? Like those body builders who took too much testesterone. But why does the TPR book and another book say it also facilitates spermatogenesis. Is there something I'm missing here?

Also the same goes for estrogen right, that it inhibits FSH, which inhibits ovary. Is that what happens in a birth control pill?

Just a bit confused.

2) The birth control pill is generally giving progesterone so that the endometrial layers that the ovum would normally implant into is sloughed off. The woman still ovulates and the ova can be fertilized, but it inhibits implantation.

1) As for testosterone...it does facilitate spermatogenesis. FSH and LH are produced in the Ant. Pituitary. It is mainly LH that stimulates testosterone production. However, the inhibition that you are reading about is probably the negative feedback the testosterone has on FSH. Increasing Testosterone will inhibit FSH production so that you get less LH production and an eventual reduction in testosterone.

I could be wrong on #2, but it sort of makes sense if you look at the back ass-wards way the body regulates itself.

 

[Sol Rosenberg]

QofQuimica,


What is the splitting pattern for the second most downfield carbon 1-chloro-2-methyl propane?

I am getting confused as to whether the three hydrogens on the end carbon and the 2-methyl are equivalent.

My answer was 6 because they seem equivalent, where is my thinking wrong?


Thank you,


'Rambler

Those 6 H atoms are equivalent, but they are only split by the one hydrogen on the adjacent Carbon atom, so they will produce a (doublet, 6H) resonance

To answer your original question, it actually doesn't matter whether they are equivalent or not. The second most downfield Hydrogen is the Hydrogen attached to the #2 Carbon atom. It will get split by 8 hydrogens on adjacent carbons, so it will produce a (9-tet (I don't know the official term), 1H) resonance.

I think where you might've been getting confused is:

1. What makes the resonances more downfield (in this molecule, it is the proximity to an electron-withdrawing substituent -- the Cl atom). The two protons on the #1 carbon will resonate most downfield, followed by the proton on the #2 carbon, followed by the 6 methyl protons.

2. Splitting vs. Height (Integration.) The height of the peak is proportional to the number of equivalent protons, while the splitting (in simple n+1 splitting) is the number of protons on ADJACENT carbon atoms + 1

Hope that clears things up....

Jota

 

[Unknown3234]

Didn't see this thread at first but I have an easy question.

Which step (1 or 2) is the rate determining step? Thanks.

 

[Sol Rosenberg]

Didn't see this thread at first but I have an easy question.

Which step (1 or 2) is the rate determining step? Thanks.

2

 

[Unknown3234]

Even though the energy of activation is greater for the first step than it is for the second?

 

[Sol Rosenberg]

.

 

[Unknown3234]

No, of course you are right -- I'm not sure what I was thinking.

Trust, this why I asked I asked this question it was marked wrong on a test, I choose the first step.

this is response from the Professor:
Think about the intermediate, it could go to the left over a small hump or to the rate over large hump.

This is the Professor and it was only worth 2 points out of 110, so I just dropped it. But doesn't this response completely ignore the energy of activation of the first step?

 

[Sol Rosenberg]

Trust, this why I asked I asked this question it was marked wrong on a test, I choose the first step.

this is response from the Professor:
Think about the intermediate, it could go to the left over a small hump or to the rate over large hump.

This is the Professor and it was only worth 2 points out of 110, so I just dropped it. But doesn't this response completely ignore the energy of activation of the first step?

I knew, I was right the first time. The second part of the reaction has the highest energy transition state, therefore, for a given energy input, fewer molecules will be able to reach that transition state than for the first part of the reaction. That's why it is the rate-limiting step.

The activation energy for the second part of the reaction is the activation energy for the first + the activation energy for the second part, so the activation energy for the second part is not lower than that for the first part.

Make Sense?

Jota

 

[Caboose]

I got another one. Why are my diabetic friends overweight? I keep thinking of all the diabetics I know and they've all been a little overweight. They have decreased insulin activity, (through whatever mechanism), which means the cells aren't able to take in all that glucose and store it. They kind of look like they are storing it though. I feel so rude. Anyway, is this because of something else gone astray?

Caboose.

 

[Krazykritter]

I got another one. Why are my diabetic friends overweight? I keep thinking of all the diabetics I know and they've all been a little overweight. They have decreased insulin activity, (through whatever mechanism), which means the cells aren't able to take in all that glucose and store it. They kind of look like they are storing it though. I feel so rude. Anyway, is this because of something else gone astray?

Caboose.

Type II Diabetes is due to unresponsivity to insulin. That is why Type II diabetic will have higher blood sugars. They still make insulin, but there is some inability to make more when they need it so some do need daily insulin injections.

Now to the overweight question...Insulin is the major anabolic hormone of the body...meaning that it synthesizes substances instead of breaking them down (catabolism). Insulin increases the glycogenesis and increased deposition of adipose tissue.

So...insulin is ineffective at upregulating the glucose transporters that allow the glucose to be taken into the cells (resulting in high blood sugars). Increases anabolic activities that increase the deposit of fat through a different mechanism.

Type I, if you are interested, is inability to produce insulin due to pancreatic beta-cell activity.

--Hope I didn't miss anything, I am just writing from memory.

 

[Sol Rosenberg]

What is the source of the bouyant force? Is it electromagnetic in origin?

Also, does it do work when moving an object "up" just as gravity does in pulling it "down"?

Yes and Yes.

 

[christian15213]

k you're not understanding what chirality means? have you taken the class or have you read a book? chiral means (hand) mirror image... meaning if you looked in a mirror with your right hand and pressed it against the mirror it wouldn't be your left hand in the mirror it would be your mirror image of your right hand... this also means nonsuperimposible mirror image...

making sense yet? achiral means no mirror image i.e. meso or diasteromer... which is two chiral centers of different attachments on each chiral carbon...

 

[christian15213]

i have a last minute question about op/meta directors. i know most things are op, and i need to just remember which are meta (for some reason, this is the hardest thing EVER for me!)

is it safe to say carbonyl's are meta directors, or is there an exception to that? any help is appreciated.

sasha

ok, please go to your organic book... because this is something basic, but alas your book might not make it so basic... I will try to help...

ACTIVATING/DONATING are groups like NH2, OH,<•> CH3 WEAK o,p directors and halides are weak deacivators/ELECTRON WITHDRAWING... o,p directors...

now, your carbonyls cooh and NO2 CF3, etc are strong electron withdrawing M directors... get it... but, REMEMBER

the trick is this... whenever there is an OP DIRECTOR meaning ACTIVATING/DONATING GROUP including halides... they are the winners when there are two groups... Roles like Steric Hinderance will play a factor on which O OR P is directed too...

hope that helped...