Organic Chemistry Question Thread

[QofQuimica]

All users may post questions about MCAT, DAT, OAT, or PCAT organic chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what organic topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
-general, MCAT-level organic
-particular MCAT-level organic problems, whether your own or from study material
-what you need to know about organic for the MCAT
-how best to approach to MCAT organic passages
-how best to study MCAT organic
-how best to tackle the MCAT biological sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

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If you really know your organic, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the Organic Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university organic chemistry TA teaching experience. In addition, I teach organic chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 15 on BS, 43 overall.

P.S. If you shorten "organic chemistry" to "orgo," not only will I not answer your questions, but during the BS section, your test form will backside attack you with a zillion strong nucleophiles (via the SN2 mechanism, of course).

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the BS section of the MCAT, and 36 overall.

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[QofQuimica]

Which is a stronger acid? Sodium Carbonate Na2CO3 or Sodium Acetate CH3-C-O-Na (there's also a double-bond O on the central C)

In a problem like the above, it's not a simple case of looking at the periodic table to determine which has greater acidity (or is it?) I believe acidity is partially based on how 'available' electrons are for removal from an atom, but how exactly does one determine that availability?

Is this a trick question? Because sodium carbonate and sodium acetate are both bases, not acids. Since you are comparing two oxygen-containing bases, you would judge their relative strengths on pKa values. Higher pKa values correspond to stronger bases and weaker acids. Acetic acid has a pKa of about 5, and bicarb (dissociating to carbonate) has a pKa of about 10. Thus, acetic acid is a stronger acid than bicarb is, and carbonate is a stronger base than acetate is.

 

[pandora123]

Hi

When quarternary ammonium salts undergo Hofmann elimination to form alkenes, why are hofmann products the major product and Saytzeff products the minor product? Does the reaction involve formation of carbocations? I always tend to think that the most stable product should form.

Thanks,

 

[QofQuimica]

Hi

When quarternary ammonium salts undergo Hofmann elimination to form alkenes, why are hofmann products the major product and Saytzeff products the minor product? Does the reaction involve formation of carbocations? I always tend to think that the most stable product should form.

Thanks,

It's basically a steric effect. The LG for a Hofmann elimination is a quaternary ammonium salt, which is a large, bulky group. Since the mechanism for Hoffman elimination is an E2 and not an E1, there is no carbocation intermediate (which means no rearrangements to the more substituted Zaitzev products). Remember that for an E2 reaction, you must have the LG and the proton arranged in an antiperiplanar conformation. Because the ammonium LG is so bulky, it interferes with the more substituted proton being able to achieve this position, which in turn prevents the Zaitzev product from forming easily. However, one of the less substituted protons will usually be able to achieve an anti relationship with the ammonium LG, allowing the Hofmann product to form.

 

[naiominonna]

Sorry, if this is a repeat question that you answered before but I am confused about Distillation

it says: that distillation requires boiling contaminated water and collecting steam generated in a seperate vesssel where it condenses as purified water.

CONTAMINANTS THAT DO NOT EVAPORATE ARE LEFT BEHIND AND CAN BE DISPOSED OF.

so distillation would be most effective for removing which of the following from water?

AsH3

AsF3

ASCl3

AS Br3=correct answer why

Thanks

 

[Lests55]

Sorry, if this is a repeat question that you answered before but I am confused about Distillation

it says: that distillation requires boiling contaminated water and collecting steam generated in a seperate vesssel where it condenses as purified water.

CONTAMINANTS THAT DO NOT EVAPORATE ARE LEFT BEHIND AND CAN BE DISPOSED OF.

so distillation would be most effective for removing which of the following from water?

AsH3

AsF3

ASCl3

AS Br3=correct answer why

Thanks

Hi there. A higher molecular wt. means a higher melting/boiling point (all things being equal). The intermolecular attractions are stronger because of the increased polarizability. So the heaviest compound will be the least likely to boil, and thus be most likely to be left behind.

Hey Q, thanks for all the help when I was studying! I'll be happy to help out on here this summer if needed.

 

[QofQuimica]

Hi there. A higher molecular wt. means a higher melting/boiling point (all things being equal). The intermolecular attractions are stronger because of the increased polarizability. So the heaviest compound will be the least likely to boil, and thus be most likely to be left behind.

Hey Q, thanks for all the help when I was studying! I'll be happy to help out on here this summer if needed.

Great, please do. Any fan of Jean-Luc is a friend of mine.

 

[2DEG]

Great, please do. Any fan of Jean-Luc is a friend of mine.

After all these years and a few gray hair, I still love Star Trek, Next gen like no other program on TV.

I was wondering if I could get some help on Organic Chem though. I took Organic I online and I definitely feel the lack of understanding on many basic things.

If these questions were answered already, please let me know..Thanks

1) In determining enantiomer, diastereoisomer, sometimes you have to rotate a sigma bond. And many times, what seemed like two different stereoisomer, after you rotate it, they are actually the same molecule. My problem is when I am reading a book and they are talking about this rotation and then they show four elements in different planes. I just cannot grasp it. I wish it was a true 3D view. But I cannot mentally rotate a bond and assign the planes to each of the four things attached to the chiral C.

I am not sure if I asked the question right..but I hope someone will understand my question.


2) What is an unhybridized p orbital? And more importantly, how do you know if there is an unhybridized p orbital by looking at a lewis structure (molecular schematic)?

Organic is my favorite subject for sure, but I wish I had a bit more solid understanding of the basics.

Thanks,
S.

 

[QofQuimica]

After all these years and a few gray hair, I still love Star Trek, Next gen like no other program on TV.

I was wondering if I could get some help on Organic Chem though. I took Organic I online and I definitely feel the lack of understanding on many basic things.

If these questions were answered already, please let me know..Thanks

1) In determining enantiomer, diastereoisomer, sometimes you have to rotate a sigma bond. And many times, what seemed like two different stereoisomer, after you rotate it, they are actually the same molecule. My problem is when I am reading a book and they are talking about this rotation and then they show four elements in different planes. I just cannot grasp it. I wish it was a true 3D view. But I cannot mentally rotate a bond and assign the planes to each of the four things attached to the chiral C.

I am not sure if I asked the question right..but I hope someone will understand my question.


2) What is an unhybridized p orbital? And more importantly, how do you know if there is an unhybridized p orbital by looking at a lewis structure (molecular schematic)?

Organic is my favorite subject for sure, but I wish I had a bit more solid understanding of the basics.

Thanks,
S.

1) Don't feel bad; stereochemistry is one of the toughest concepts for many students to grasp. Do you have a model kit? If not, get a hold of one, and practice doing your stereochemistry exercises using the model kit until you are able to do them in your head without the model kit. Some people have a better intuitive sense of spatial orientation than others do, but anyone can learn to see spatial orientation in their heads with practice.

2) When you hybridize the orbitals of carbon atoms, you can choose to use one, two, or all three p-orbitals. This gives you sp, sp2, or sp3 hybrid orbitals, respectively. If you form sp or sp2 hybrids, then the remaining p-orbitals that are "left over" are the unhybridized p-orbitals. In contrast, if you form sp3 hybrids, then there are no p-orbitals remaining; they're all hybridized. Any time you have multiple bonds, you will have at least one unhybridized p-orbital on each atom of the double bond (two of them for triple bonds or cumulenes). Some other species, such as carbocations and carbanions, are also sp2 hybridized, and therefore they have an unhybridized p-orbital as well.

 

[2DEG]

Thanks a lot. Someone gave me a really nice set of model kit. It was collecting dust in my office. I am gonna finally use it. I think this is exactly what I needed.

And also, thanks for explaining unhybridized p orbital thing. Wonderful explaination.

Have a pleasant day.

1) Don't feel bad; stereochemistry is one of the toughest concepts for many students to grasp. Do you have a model kit? If not, get a hold of one, and practice doing your stereochemistry exercises using the model kit until you are able to do them in your head without the model kit. Some people have a better intuitive sense of spatial orientation than others do, but anyone can learn to see spatial orientation in their heads with practice.

2) When you hybridize the orbitals of carbon atoms, you can choose to use one, two, or all three p-orbitals. This gives you sp, sp2, or sp3 hybrid orbitals, respectively. If you form sp or sp2 hybrids, then the remaining p-orbitals that are "left over" are the unhybridized p-orbitals. In contrast, if you form sp3 hybrids, then there are no p-orbitals remaining; they're all hybridized. Any time you have multiple bonds, you will have at least one unhybridized p-orbital on each atom of the double bond (two of them for triple bonds or cumulenes). Some other species, such as carbocations and carbanions, are also sp2 hybridized, and therefore they have an unhybridized p-orbital as well.

 

[Creightonite]

reviewing my organic at this point.

Some questions?

1) What is the difference between melting and bioling point? In my head they are very similar

2) Why does alkane branching increase the melting point but descreases the boiling point? Something about stacking for thje melting point? Why does it decrease the bioling point? Thank you.

 

[QofQuimica]

reviewing my organic at this point.

Some questions?

1) What is the difference between melting and bioling point? In my head they are very similar

2) Why does alkane branching increase the melting point but descreases the boiling point? Something about stacking for thje melting point? Why does it decrease the bioling point? Thank you.

1) MP is for phase changes between solids and liquids; BP is for phase changes between liquids and gases. So they are similar in that both describe the temperature at which a phase change occurs for a given pressure, but they describe different phase conversions.

2) It depends on how symmetrical the alkane is with regard to MP, because this affects how it stacks. For BP, it depends on dispersion forces. I wrote a post explaining about this in detail in the organic chemistry explanations thread. If you haven't read it yet, you might want to check it out.

Edit: Here is the link: http://forums.studentdoctor.net/showpost.php?p=2877597&postcount=3

 

[Creightonite]

still struggling with Sn1, sn2, E1, E2... IS there an easy and good way to determine these things?

 

[QofQuimica]

still struggling with Sn1, sn2, E1, E2... IS there an easy and good way to determine these things?

Heh, unfortunately the answer to your question is, "it depends." I am going to go ahead and write an explanations post for this, because you aren't the first person who wanted some guidelines. I'll post the link in here after I'm done or you can just check in the organic explanations thread later.

 

[QofQuimica]

Heh, unfortunately the answer to your question is, "it depends." I am going to go ahead and write an explanations post for this, because you aren't the first person who wanted some guidelines. I'll post the link in here after I'm done or you can just check in the organic explanations thread later.

Ok, here is the link to the organic mechanisms decision post. Hope this helps.

 

[Creightonite]

one more question. Kaplan book says that to determine the "R" or "S" of the chiral carbon you have to consider the atom directly attached to the chiral carbon, not the whole group. If it is true then methyl group would be greater by mass then the CH2 of the ethyl group. But they give an ethyl group a higher priority over methyl. Anyhow, may question so priority will go like this: Propyl>ethyl> methyl, right

 

[Lests55]

one more question. Kaplan book says that to determine the "R" or "S" of the chiral carbon you have to consider the atom directly attached to the chiral carbon, not the whole group. If it is true then methyl group would be greater by mass then the CH2 of the ethyl group. But they give an ethyl group a higher priority over methyl. Anyhow, may question so priority will go like this: Propyl>ethyl> methyl, right

Right on, brother. You are looking for the highest wt. ONE atom that is connected. So far your example of methyl vs. ethyl. Sine they both have the same atom (the first carbon), you see what the heaviest next atom connected is. For the methyl group it is hydrogen, for the ethyl group, it is carbon. You keep on going for the propyl group too.

Another tidbit, if you have a double bond it counts as "two" carbon bonds and would have higher priority than a methyl group (I think this is ethylene, maybe it is acetylene). So -C=CH2 > -C-CH3

 

[QofQuimica]

one more question. Kaplan book says that to determine the "R" or "S" of the chiral carbon you have to consider the atom directly attached to the chiral carbon, not the whole group. If it is true then methyl group would be greater by mass then the CH2 of the ethyl group. But they give an ethyl group a higher priority over methyl. Anyhow, may question so priority will go like this: Propyl>ethyl> methyl, right

You don't assign priority by mass; you do it by ATOMIC NUMBER. A single fluorine atom will have priority over a million-atom long carbon chain, even though the C chain has a much greater mass. If the first atoms out of two substituents are identical (as they are for methyl and ethyl), then you must go out to the second atoms attached to those and compare them. You keep going out until you reach a point where there is a difference, and then you give priority to the one with the higher atomic number. So yes, ethyl would be higher priority than methyl, because the methyl C is connected to 3 Hs, while the ethyl C1 is connected to two Hs and one C. Likewise, if you're comparing propyl and ethyl, you'd get out to C2 and the same thing would happen: the ethyl C2 is connected to 3 Hs, and the propyl C2 is connected to 2 Hs and one C.

 

[legobikes]

Haha that diene avatar rules.

In that nucleophilicity/basicity post you say:

One way to gauge basicity is to use the periodic table. In the periodic table, basicity decreases from left to right (i.e., CH3- > NH2- > OH- > F-), and it also decreases going down a group (i.e., F- > Cl- > Br- > I-). The left to right trend occurs because the less electronegative elements on the left side of the table are less capable of stabilizing the negative charge present on the base.

I don't understand this. Shouldn't basicity increase going from left to right, if the less electronegative elements are less capable for stabilizing the negative charge on a base?

 

[laneyj34]

Haha that diene avatar rules.

In that nucleophilicity/basicity post you say:


I don't understand this. Shouldn't basicity increase going from left to right, if the less electronegative elements are less capable for stabilizing the negative charge on a base?

No. Think about electronegativity as how well an atom can handle a negative charge by itself, or as how much an atom likes to have a negative charge on it. So, if an atom is very electronegative, it is much better equipped to handle a bare negative charge, or to handle a majority of the negative charge as in a polar molecule, than an atom that is not very electronegative. So, the very electronegative atoms, like oxygen, or flourine, can have a bare negative charge on them, and be relatively stable (i.e. not need a proton to stablize the negative charge), as compared to an atom that is not as electronegative (which has a strong desire to get rid of it's negative charge by bonding to a proton, for example). To sum up, basically the more electronegative an atom is, the happier it will be with a full negative charge on it (this does not mean that the negative charge will stay on it and not react, it just means that it can handle having a full negative charge better than a less electronegative atom can).

Make any sense? Hope it helped.

 

[legobikes]

Make any sense? Hope it helped.

Got it, thanks. The more electronegative elements are better at handling the negative charge and so not as reactive when acting as bases, they can fare ok without accepting protons/donating electrons.

 

[legobikes]

More questions...


Regarding the actual difference between the four mechanisms:

Sn1 gets you R-Nu, racemic mixture if the substrate was chiral to begin with
Sn2 gets you R-Nu, flipped chirality and optical activity if the substrate was chiral

E1 gets you an alkene (how does this compare to to E2?)
E2 gets you an alkene ("halide ion _anti_ to removed proton leaves"? and the more substituted double bond is preferentially formed)

Basically, what are any stereochemistry issues in the elimination reactions?


Also, I know this has been done to death on here but what alkene/alkyne reactions, besides E1/2, do we need to know?

 

[QofQuimica]

Haha that diene avatar rules.

In that nucleophilicity/basicity post you say:


I don't understand this. Shouldn't basicity increase going from left to right, if the less electronegative elements are less capable for stabilizing the negative charge on a base?

No. Greater basicity means that the species is LESS capable of stabilizing that negative charge. The less electronegative elements are more basic because they are more "desperate" to gain that proton back and become neutral. They would rather be in the form of their conjugate acids. If you look at the pKas of the conjugate acids of the four compounds in that series I gave as an example, HF has a pKa of about 5 or so, H2O is about 16, HN3 is about 35, and CH4 is approximately 50. (Remember that the greater the pKa, the stronger the base and the weaker the conjugate acid.)

If you remember this general rule, it will serve you well in organic chemistry: less stability = greater reactivity (and vice versa). The least stable species is going to be the most reactive, while the most stable one will be the least reactive.

P.S. Thanks for appreciating my avatar. I drew it myself.

 

[QofQuimica]

More questions...


Regarding the actual difference between the four mechanisms:

Sn1 gets you R-Nu, racemic mixture if the substrate was chiral to begin with
Sn2 gets you R-Nu, flipped chirality and optical activity if the substrate was chiral

E1 gets you an alkene (how does this compare to to E2?)
E2 gets you an alkene ("halide ion _anti_ to removed proton leaves"? and the more substituted double bond is preferentially formed)

Basically, what are any stereochemistry issues in the elimination reactions?


Also, I know this has been done to death on here but what alkene/alkyne reactions, besides E1/2, do we need to know?

Everything you've said looks good. The main stereochemistry issues in elimination reactions are the following:

1) For E1, you must be able to form a carbocation. (If you have a primary alkyl halide, it will not undergo E1.) So your E1 product will lose its stereochemistry if the carbon with the leaving group on it was chiral. Most of the rules that apply for Sn1 apply for E1 also; they have the same rate-determining step. E1 reactions tend to form the more substituted (Zaitzev) product over the less substituted (Hoffman) product.

2) For E2, there are very few steric constraints, except that your leaving group and proton must be able to orient themselves in an antiperiplanar fashion as you mentioned. There usually is not a problem with the molecule being able to orient itself; see my explanations post on the mechanisms in the Organic Chemistry Explanations Thread for the most important exception to this. Again, the Zaitzev product tends to be favored unless its formation is prevented by steric hindrance.

I would also be familiar with addition reactions for the MCAT. You are not going to be tested heavily on alkenes, but they do have a way of showing up indirectly whether you want them to or not; organic chemistry and biochemistry are just full of them.

 

[Cloudcube]

Hi Everyone! This really is an awesome thread and a great help.

I was wondering if anyone would be able to help me with aromaticity? Could someone please take the time to explain, in detail, how aromaticity works. I feel like I'm missing a lot of stuff. For something that has groups attached to it (like N or O) I find it much more difficult to tell whether it's aromatic. I realize this is probably a simple question, but I'd really appreciate it if someone took the time to answer.

Thanks!

 

[QofQuimica]

Hi Everyone! This really is an awesome thread and a great help.

I was wondering if anyone would be able to help me with aromaticity? Could someone please take the time to explain, in detail, how aromaticity works. I feel like I'm missing a lot of stuff. For something that has groups attached to it (like N or O) I find it much more difficult to tell whether it's aromatic. I realize this is probably a simple question, but I'd really appreciate it if someone took the time to answer.

Thanks!

There are four considerations that you need to think about, regardless of whether heteroatoms are present. Use this list like a flowchart:

1) Is the compound cyclic? If not, it's definitely not aromatic. If it is cyclic:

2) Does every atom in the ring have a p-orbital on it (i.e., is every atom in the ring sp2- or sp-hybridized?) If not, it's definitely not aromatic. If every ring atom does have a p-orbital on it:

3) Is the compound planar (flat)? If not, it's definitely not aromatic. If it is planar (or if you're not sure):

4) Does the compound obey Huckel's Rule? Huckel's Rule states that every aromatic compound should have 4n + 2 pi electrons, where n is any integer equal to zero or greater. Most aromatic compounds that you are familiar with have an n of 1, which means they have six pi electrons. Benzene is an example of this. If the compound has 4n electrons, then it is not aromatic; in fact, it is an especially UNstable compound called an antiaromatic compound. Cyclobutadiene is an example of this where n = 1.

Heteroaromatic compounds (rings that contain atoms besides carbon) may seem tricky, but you should treat them just like you would a ring of all carbons. The heteroatom lone pairs may or may not be part of the ring. You can tell whether they are based on whether those electrons are necessary in order for the compound to obey Huckel's Rule. For example, pyridine (six-membered aromatic ring containing nitrogen) has a nitrogen whose lone pair is NOT part of the aromatic system. This is because pyridine already has three double bonds, providing it with six pi electrons. If the nitrogen lone pair were also included, that would make eight pi electrons, which would make the compound antiaromatic. In contrast, pyrrolidine (five-membered aromatic ring containing nitrogen) has a nitrogen that IS part of the aromatic system, because there are only two double bonds in the molecule. If the nitrogen lone pair were not included, then pyrrolidine would have four pi electrons, making it antiaromatic.

This analysis explains why pyridine is an excellent base, but pyrrolidine isn't. The lone pair on the pyridine nitrogen is not part of the aromatic system, so it is free to accept a proton without disrupting the molecule's aromaticity. In contrast, pyrrolidine cannot accept a proton without disrupting its aromaticity. Thus, pyrrolidine is not a very good base.

Note that aromatic compounds can also be cationic or anionic, as long as they obey the four rules I've listed above. Two common examples you may run into are cycloheptatriene cation (seven-membered ring with three double bonds and one carbocation) and cyclopentadiene anion (five-membered ring with two double bonds and one carbanion).

 

[Cloudcube]

This helps so much! I feel like I actually really understand it! A lot of your posts make me actually think about Chemistry, unlike a textbook, which is exactly what I need.

Thanks for taking the time to help me out!

 

[RAD11]

This is an equation from an Organic Chemistry passage (Kaplan FL #6). The question asks: "The boron atom in the alkylborane shown in Equation 2 has which of the characteristics? I. It has 6 valence electrons; II. It is sp2 hybridized; III. It has a formal charge of +1." Their answer is I and II.

I know that alkylborane is the the one labeled n-octylBpin. My question is doesn't boron only have 3 valence electrons since it is in Group III?

 

[RAD11]

This is from an Organic Chemistry passage (Kaplan FL #6). The question asks how many nitrogen atoms are sp2 hybridized? Their answer is 3. I don't understand how the nitrogen on the right hand side and the one on the botton are sp2 hybridized as well? I thought they're sp3 since they're bonded to 3 other atoms plus both have a lone pair of e-, making both tetrahedral; thus, sp3 hybridized.

 

[QofQuimica]

This is an equation from an Organic Chemistry passage (Kaplan FL #6). The question asks: "The boron atom in the alkylborane shown in Equation 2 has which of the characteristics? I. It has 6 valence electrons; II. It is sp2 hybridized; III. It has a formal charge of +1." Their answer is I and II.

I know that alkylborane is the the one labeled n-octylBpin. My question is doesn't boron only have 3 valence electrons since it is in Group III?

It does as a free atom, but if it's an alkylborane, it isn't a free atom anymore. Boron is one of the many elements that regularly disobey the octet rule. When a boron atom has three substituents bound to it, there are six valence electrons: three from the boron, and one each from every substituent.

 

[QofQuimica]

This is from an Organic Chemistry passage (Kaplan FL #6). The question asks how many nitrogen atoms are sp2 hybridized? Their answer is 3. I don't understand how the nitrogen on the right hand side and the one on the botton are sp2 hybridized as well? I thought they're sp3 since they're bonded to 3 other atoms plus both have a lone pair of e-, making both tetrahedral; thus, sp3 hybridized.

Here they are testing whether you understand that a nitrogen atom that is able to delocalize into a pi system with its lone pair will sp2-hybridize in order to do so. You should have seen this phenomenon when you learned about peptide chemistry. If you make a peptide (amide) bond between two amino acids, the C-N bond between the nitrogen and the carbonyl carbon has some double bond character, which happens because that nitrogen atom is sp2-hybridized so that its lone pair can delocalize into the carbonyl. This is an analogous situation. (If you don't remember this, look up the peptide bond section of your organic book and review it.)

 

[deleted106503]

It does as a free atom, but if it's an alkylborane, it isn't a free atom anymore. Boron is one of the many elements that regularly disobey the octet rule. When a boron atom has three substituents bound to it, there are six valence electrons: three from the boron, and one each from every substituent.

So to summarize to see if I got this right: # of valence electrons of an atom not in a free atom form = number of bonds x 2e- per bond + # of free valence e-? I.e. The oxygen in water has 2 bonds, so 2 pairs of shared valence electrons, plus 2 pairs of non-shared valence electrons, summing to 8 valence electrons. This is the right logic, correct?

It's always the basics that mess me up the most (shakeshead).

 

[QofQuimica]

So to summarize to see if I got this right: # of valence electrons of an atom not in a free atom form = number of bonds x 2e- per bond + # of free valence e-? I.e. The oxygen in water has 2 bonds, so 2 pairs of shared valence electrons, plus 2 pairs of non-shared valence electrons, summing to 8 valence electrons. This is the right logic, correct?

It's always the basics that mess me up the most (shakeshead).

You got it. Oxygen almost always obeys the octet rule; the only example I can think of offhand where it doesn't sometimes is for a molecule like NO. That molecule has 11 valence electrons and so either N or O must have an incomplete octet.

 

[RAD11]

Hey Q,

Thanks a lot for your help.

 

[Orthodoc40]

Just remembered a question that came up while taking April's test.
Something I'd never seen before in a drawing is a squiggly line - like, instead of a solid wedge indicating coming out of the page, or a dashed wedge indicating going into the page, there was this squiggle, and then - I don't know what the compound was I just remember wondering... What the heck does a squiggly line mean??

 

[deleted106503]

Just remembered a question that came up while taking April's test.
Something I'd never seen before in a drawing is a squiggly line - like, instead of a solid wedge indicating coming out of the page, or a dashed wedge indicating going into the page, there was this squiggle, and then - I don't know what the compound was I just remember wondering... What the heck does a squiggly line mean??

Maybe the emission of a particle??? Was it a fission thing?

 

[RAD11]

Was it an ozonolysis reaction? Squiggly line for breaking the double bond? That's how my professor always used to write it on the board.

C=C + 1) O3 then 2) Zn2, H2O ----> C=O + O=C

 

[Orthodoc40]

No the squiggly line was indicating some atom was there. Except instead of using like, the solid black wedge and the dashed line wedge that you see indicating coming out of the page or going into the page, it was this squiggly line. I saw it the other day in an organic passage in the PR science workbook, too - which made me remember it came up on the test in April...?!

 

[QofQuimica]

No the squiggly line was indicating some molecule was there. Except instead of using like, the solid black wedge and the dashed line wedge that you see indicating coming out of the page or going into the page, it was this squiggly line. I saw it the other day in an organic passage in the PR science workbook, too - which made me remember it came up on the test in April...?!

It means that the stereochemistry isn't being shown. So a wedge means the bond is coming out at you, a dash means it's going into the page, and a squiggle means that it could go either way.

 

[Orthodoc40]

It means that the stereochemistry isn't being shown. So a wedge means the bond is coming out at you, a dash means it's going into the page, and a squiggle means that it could go either way.

Oh wow - thanks! So, it could go either way, as in "ha ha, we're not going to tell you!" or as in, "it isn't going to matter in order to answer the questions" or as in, "it will behave either way, depending on something you either know or don't know..."?

Q - wow, you are the organic QUEEN (and my hero!)!! Thanks!

 

[abcxyz0123]

I was just wondering, does the phrase "like dissolves like" apply when it comes to acidity and basicity, and not just polarity? For example, will a compound with an acidic functional group always be soluble in a weakly acidic solution? And will a compound with a basic group always be soluble in a weakly basic solution? Or is it the other way around?

Thank you for the help!

 

[QofQuimica]

Oh wow - thanks! So, it could go either way, as in "ha ha, we're not going to tell you!" or as in, "it isn't going to matter in order to answer the questions" or as in, "it will behave either way, depending on something you either know or don't know..."?

Usually it means that you can have either configuration. It doesn't mean that it will behave either way; remember that you can't convert between R and S without breaking bonds.

 

[QofQuimica]

I was just wondering, does the phrase "like dissolves like" apply when it comes to acidity and basicity, and not just polarity? For example, will a compound with an acidic functional group always be soluble in a weakly acidic solution? And will a compound with a basic group always be soluble in a weakly basic solution? Or is it the other way around?

Thank you for the help!

No, it doesn't. Acids don't dissolve in acidic solutions; they dissolve in basic solutions. Likewise, bases dissolve in acidic solutions.

 

[Cloudcube]

I've been looking through my Kaplan book and 13C NMR doesn't much sense. I've been told that 13C NMR is pretty much the same as H-NMR only 13C-NMR doesn't have the coupling/splitting. In the book I'm studying from, there's a picture of a propane that shows a 13C NMR Spectrum. They show 1,1,2-trichloropropane and have four peaks (in the spin-decoupled spectrum) and then have a picture of the spin-coupled spectrum with 2 peaks at the first (going from left to right), 2 peaks at the second, 4 peaks and the thrid, and four peaks and the fourth. I'm wondering how they get that. I thought I understood it until I saw that picture and realized it wasn't all that clear. The first peak isn't right at zero so I'm thinking it's not the TMS peak.

Thanks in advance for all your help!

 

[QofQuimica]

I've been looking through my Kaplan book and 13C NMR doesn't much sense. I've been told that 13C NMR is pretty much the same as H-NMR only 13C-NMR doesn't have the coupling/splitting. In the book I'm studying from, there's a picture of a propane that shows a 13C NMR Spectrum. They show 1,1,2-trichloropropane and have four peaks (in the spin-decoupled spectrum) and then have a picture of the spin-coupled spectrum with 2 peaks at the first (going from left to right), 2 peaks at the second, 4 peaks and the thrid, and four peaks and the fourth. I'm wondering how they get that. I thought I understood it until I saw that picture and realized it wasn't all that clear. The first peak isn't right at zero so I'm thinking it's not the TMS peak.

Thanks in advance for all your help!

Ok, for the decoupled one, you are seeing one peak for each unique carbon. You have three unique carbons in that molecule, so there should be three peaks. The fourth one is probablyTMS if it's at 0 ppm.

For the coupled one, what you are seeing is the effect of coupling a carbon with the protons attached to it, and you follow the n + 1 rule just like you do for proton NMR. So the first C has one proton on it, and it shows up as a doublet. The middle C has one proton as well, and it shows up as a second doublet, further upfield (toward 0) because it only has one Cl on it. The third C has three protons on it, and it shows up as a quartet the furthest upfield because it has no Cl on it. Again, the fourth one is probably TMS. TMS is tetramethylsilane, so you'd expect those methyls to be quartets that show up around 0.

 

[deleted106503]

^Thanks! I didn't realize that the n+1 rule also applied to the doublet peaks. I only really used it when I had the 6 low peaks in a row.

 

[mcat_study]

hello
what do we need to know about C13 NMR

what ppms are we supposed to know?

please help

 

[QofQuimica]

hello
what do we need to know about C13 NMR

what ppms are we supposed to know?

please help

What's with all the CMR questions all of a sudden??? All right, this will be the next organic explanations post I write, but I can't do it right now. You don't need to memorize the ppms for CMR for the MCAT. But if you want to know a couple, the most helpful ones are probably carbonyls, which come around 175ish, and aromatics, which come just upfield of that in the 120-160 range.

Here's a pic of a C-13 of methyl salicylate, which is a compound that is aromatic and also contains a carbonyl:

The one on the very far left is the carbonyl, and the next six are from the aromatic ring. The one on the far right is the methyl group.

 

[mcat_study]

What's with all the CMR questions all of a sudden??? All right, this will be the next organic explanations post I write, but I can't do it right now. You don't need to memorize the ppms for CMR for the MCAT. But if you want to know a couple, the most helpful ones are probably carbonyls, which come around 175ish, and aromatics, which come just upfield of that in the 120-160 range.

Here's a pic of a C-13 of methyl salicylate, which is a compound that is aromatic and also contains a carbonyl:

The one on the very far left is the carbonyl, and the next six are from the aromatic ring. The one on the far right is the methyl group.

thank you

 

[Cloudcube]

Ok, for the decoupled one, you are seeing one peak for each unique carbon. You have three unique carbons in that molecule, so there should be three peaks. The fourth one is probablyTMS if it's at 0 ppm.

For the coupled one, what you are seeing is the effect of coupling a carbon with the protons attached to it, and you follow the n + 1 rule just like you do for proton NMR. So the first C has one proton on it, and it shows up as a doublet. The middle C has one proton as well, and it shows up as a second doublet, further upfield (toward 0) because it only has one Cl on it. The third C has three protons on it, and it shows up as a quartet the furthest upfield because it has no Cl on it. Again, the fourth one is probably TMS. TMS is tetramethylsilane, so you'd expect those methyls to be quartets that show up around 0.

I just want to make sure I have this straight. For C-NMR we just see how many hydrogens are attached to that particular Carbon and add 1. For H-NMR we would look at the Carbon and see how many neighbours that carbon has and add one. Yes? So, for example, 1,1,2-tricholoropropane (first carbon has two Cl's, one H; second carbon has 1 Cl and 1 H; last Carbon has 3 H's) we would say that the first Carbon has 2 peaks because it's neighbour has 1 H add one, second has the 1 H from the first Carbon plus the 3 H's from last Carbon and add 1 for a total of 5 peaks and the third one has 2 peaks as well from the H's of the middle Carbon. Yes?

Thanks!

 

[deleted106503]

Wait, I thought for C-NMR they show you the carbons (odd # isotope, C 13 for example). Because in that graph, the benzene has 6 peaks, and benzene has 6 carbons. Carbonyl has 1 carbon. I hope I'm looking at this right

Also, how can we tell the number of carbons in a molecule if say every carbon has another carbon just like it (same chemical shift). So the graph would only show each TYPE of carbon, not how many of each type if a few are the same type (substituents and orientation)? Will there be more info given?