Organic Chemistry Question Thread

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QofQuimica

QofQuimica

Seriously, dude, I think you're overreacting....
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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

********

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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rogerwilco said:
I used the same thought process on that very same question the first time around. Then it suddenly dawned on me that carbons in an aromatic ring can't be chiral.
Click to expand...
Right. Only the red marked atom is chiral. 🙂

Get off SDN and go get a good night's sleep. Good luck tomorrow, everyone. 🙂
 
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I'd just like to say thanks to QofQuimica and everyone else that helped for answering all my questions! It really helped out a lot today!
 
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Are alcohols too volatile a solvent to use instead of ketones in reactions involving an allylic molecule and a phenyllithium?
 
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tracy34 said:
Are alcohols too volatile a solvent to use instead of ketones in reactions involving an allylic molecule and a phenyllithium?
Click to expand...
No. But they are too acidic. 😉
 
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does anyone know what the MCAT Student Manual wants from us here?:

Kinetic onctrol versus thermodynamic control of a reaction
(this on the 3rd pg under kinetics line)

does this mean that kinetics deals with how an rxn got to equil. and that thermodynamics deals with equilibrium?

I really dont know, plz help!
 
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comoseshama said:
does anyone know what the MCAT Student Manual wants from us here?:

Kinetic onctrol versus thermodynamic control of a reaction
(this on the 3rd pg under kinetics line)

does this mean that kinetics deals with how an rxn got to equil. and that thermodynamics deals with equilibrium?

I really dont know, plz help!
Click to expand...
Yes, in a nutshell. A reaction under kinetic control is one that forms the fastest product, even if that product isn't the most stable. A reaction under thermodynamic control forms the most stable product, even if it takes longer. One example you are probably familiar with is double bond formation after elimination reactions. Remember how you can have the Zaitzev product versus the Hoffman product? The Zaitzev product is the thermodynamic product: it is more substituted and more stable. The Hoffman product is the kinetic product: it forms faster, but it is less stable than the Zaitzev product.
 
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I just started organic this semester, so this is a really simple question.
In C6H10, are all 6 carbons in the same plane? When I used a model set to build the moleclue, it appears like one is out of the plane of the other 5? Is this true, or are they assumed to be all in the same plane?
 
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t2oo5 said:
I just started organic this semester, so this is a really simple question.
In C6H10, are all 6 carbons in the same plane? When I used a model set to build the moleclue, it appears like one is out of the plane of the other 5? Is this true, or are they assumed to be all in the same plane?
Click to expand...

That depends on what isomer you're talking about.
 
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t2oo5 said:
I just started organic this semester, so this is a really simple question.
In C6H10, are all 6 carbons in the same plane? When I used a model set to build the moleclue, it appears like one is out of the plane of the other 5? Is this true, or are they assumed to be all in the same plane?
Click to expand...
It depends. You can have several molecules all having that same molecular formula. These are called isomers. Some of the isomers are coplanar, and others aren't. You probably haven't covered isomers yet, but after you do, this concept will make more sense to you.
 
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QofQuimica said:
It depends. You can have several molecules all having that same molecular formula. These are called isomers. Some of the isomers are coplanar, and others aren't. You probably haven't covered isomers yet, but after you do, this concept will make more sense to you.
Click to expand...

Q, you think it's too early to tell him about conformational isomers? 😀
 
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Dave_D said:
Q, you think it's too early to tell him about conformational isomers? 😀
Click to expand...
And people think it's EASY to teach organic chemistry. 🙄 Luckily, it won't be a problem for a fully unsaturated molecule. 🙂
 
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QofQuimica said:
It depends. You can have several molecules all having that same molecular formula. These are called isomers. Some of the isomers are coplanar, and others aren't. You probably haven't covered isomers yet, but after you do, this concept will make more sense to you.
Click to expand...

Well.. How about if I try to describe it? I know what an isomer is, but we havent covered it in class..sorry for being a little ambiguous about the structure. Well, C6H10, cyclic molecule. One C=C doublebond, the rest of the carbons attached by a single bond. So basically benzene, but instead of three C=C double bonds and resonance, only one C=C double bond. Do you know what I mean now?
Heres my problem. When I build the molecule with my nice molecular model set, it looks as if the one carbon is not in the same plane as the other 5 carbons. Its hard to explain. If anyone knows what I mean, or can affirm whether or not all the C's are planar, the help is appreciated! Thanks!
 
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t2oo5 said:
Well.. How about if I try to describe it? I know what an isomer is, but we havent covered it in class..sorry for being a little ambiguous about the structure. Well, C6H10, cyclic molecule. One C=C doublebond, the rest of the carbons attached by a single bond. So basically benzene, but instead of three C=C double bonds and resonance, only one C=C double bond. Do you know what I mean now?
Heres my problem. When I build the molecule with my nice molecular model set, it looks as if the one carbon is not in the same plane as the other 5 carbons. Its hard to explain. If anyone knows what I mean, or can affirm whether or not all the C's are planar, the help is appreciated! Thanks!
Click to expand...

Ok, so you know the specific molecule which turns out to be cyclohexene. I'm just pointing that out since a molecule with a formula of C6H10 is one of the following

Has a triple bond
Has one double bond and one ring
Has 2 rings(I'm thinking dicyclopropane but I'm not sure if that's real)
Has 2 double bonds.

Anyway back to cyclohexene. I found it here
some 3d molecules

Looking at that I've got to go with no, it doesn't have 5 carbons in one plane. Q, is this site accurate in this case?
 
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Dave_D said:
Ok, so you know the specific molecule which turns out to be cyclohexene. I'm just pointing that out since a molecule with a formula of C6H10 is one of the following

Has a triple bond
Has one double bond and one ring
Has 2 rings(I'm thinking dicyclopropane but I'm not sure if that's real)
Has 2 double bonds.

Anyway back to cyclohexene. I found it here
some 3d molecules

Looking at that I've got to go with no, it doesn't have 5 carbons in one plane. Q, is this site accurate in this case?
Click to expand...
Right-o. Cyclohexene will not be planar, t2005. In fact, cyclohexene has very different physical and chemical properties versus benzene.
 
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1. How many isomers are possible for isomeric pentanols (C5H12OH)?

2. Are there any known relationship between the number of possible isomers of a certain compound and the number of C or O or any element for that matter?

Danke! Thanks!🙂
 
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pezzang said:
1. How many isomers are possible for isomeric pentanols (C5H12OH)?

2. Are there any known relationship between the number of possible isomers of a certain compound and the number of C or O or any element for that matter?

Danke! Thanks!🙂
Click to expand...
1). Assuming they must be straight chain alcohols, you have three pentanol isomers possible: 1-pentanol, 2-pentanol, and 3-pentanol. If you're allowed to have some branched ones, you can also have some methyl-substituted butanols (2-methyl-2-butanol, 2-methyl-1-butanol, 3-methyl-1-butanol, and 3-methyl-2-butanol).

2) I don't know. Probably. But my mind doesn't work that way; I just sit and draw them all out. 🙂
 
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QofQuimica said:
1). Assuming they must be straight chain alcohols, you have three pentanol isomers possible: 1-pentanol, 2-pentanol, and 3-pentanol. If you're allowed to have some branched ones, you can also have some methyl-substituted butanols (2-methyl-2-butanol, 2-methyl-1-butanol, 3-methyl-1-butanol, and 3-methyl-2-butanol).

2) I don't know. Probably. But my mind doesn't work that way; I just sit and draw them all out. 🙂
Click to expand...

Q, isn't 1 a trick question? I mean pentane has 12 hydrogens and the formula he has is C5H12OH. Isn't the answer Zero?
 
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What would determine the various organic compounds' solubility in toluene and methanol? Toluene is nonpolar so any nonpolar organic compounds such as two cyclohexanes connected by a single bond or any symmetric org compounds are soluble?
How about in methanol? Will the alcohol group in methanol attract organic compounds with polarity or with other OH's? In which solution will Succinic acid be soluble?

Thanks for your help!
 
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Dave_D said:
Q, isn't 1 a trick question? I mean pentane has 12 hydrogens and the formula he has is C5H12OH. Isn't the answer Zero?
Click to expand...
No. Oxygens don't affect the unsaturation count like nitrogens do. Pentanol isomers also have twelve H's. 🙂
 
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pezzang said:
What would determine the various organic compounds' solubility in toluene and methanol? Toluene is nonpolar so any nonpolar organic compounds such as two cyclohexanes connected by a single bond or any symmetric org compounds are soluble?
How about in methanol? Will the alcohol group in methanol attract organic compounds with polarity or with other OH's? In which solution will Succinic acid be soluble?

Thanks for your help!
Click to expand...
You've got the general idea. Like dissolves like, so polar molecules will dissolve in a polar solvent, and nonpolar ones will dissolve in a nonpolar solvent. I wrote a post about polar versus nonpolar in the organic explanations thread.

In answer to your second question, no, Ph means a phenyl group attached to something as a substituent. That is different than phenol, which is a benzene having an OH on it.

I thought you took the Aug. MCAT! Are you re-studying even before scores come out?
 
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QofQuimica said:
You've got the general idea. Like dissolves like, so polar molecules will dissolve in a polar solvent, and nonpolar ones will dissolve in a nonpolar solvent. I wrote a post about polar versus nonpolar in the organic explanations thread.

In answer to your second question, no, Ph means a phenyl group attached to something as a substituent. That is different than phenol, which is a benzene having an OH on it.

I thought you took the Aug. MCAT! Are you re-studying even before scores come out?
Click to expand...

Thanks QofQuimica. And yes I took the MCAT but I cancelled it after the test because I wasn't entirely feeling well. I am taking again on April. Meanwhile another question....

In the liquid-liquid extraction experiment, you separate two different solutes based on their solubility in two different solvents. There are extraction of an organic mixture with HCl, NaHCO3 and NAOH. I know that during the extraction, there will be a buildup of carbon dioxide because of the mixing sodium bicarbonate and strong acid. What other compounds/gas would build so that we will need to vent the solution while mixing? How will any heat be produced? I am not sure how to predict release of heat without any free-energy number or enthalpy/entropy... Any help would be appreciated!
Thanks again!!!
 
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pezzang said:
Thanks QofQuimica. And yes I took the MCAT but I cancelled it after the test because I wasn't entirely feeling well. I am taking again on April. Meanwhile another question....

In the liquid-liquid extraction experiment, you separate two different solutes based on their solubility in two different solvents. There are extraction of an organic mixture with HCl, NaHCO3 and NAOH. I know that during the extraction, there will be a buildup of carbon dioxide because of the mixing sodium bicarbonate and strong acid. What other compounds/gas would build so that we will need to vent the solution while mixing? How will any heat be produced? I am not sure how to predict release of heat without any free-energy number or enthalpy/entropy... Any help would be appreciated!
Thanks again!!!
Click to expand...
It's always a good practice to vent frequently while performing an extraction. You should assume that any extraction will need to be vented. Even if you're just washing your organic layer with water or brine, your organic solvent is probably volatile, and like you said, heat can be an issue. (You will commonly get heat produced when you mix an acid with a base.) So for example, if you're using diethyl ether as your organic layer, you really need to vent, and often, due to the volatility of the ether. The heat from your hands is enough to cause it to vaporize.

I always would start an extraction by capping the sep funnel, turning it upside down, and venting it first before any shaking to "test" the pressure. Make sure to point it at an angle toward the back of the hood and away from you or other people. If it vents audibly, give it one or two gentle shakes, then vent again. Repeat as many times as necessary until you no longer hear the hiss of escaping gas. As the pressure starts decreasing, you can shake harder and vent less often between shakes.

Carbon dioxide is probably the most common gas that you'd produce while extracting with bicarb. But you could conceivably produce hydrogen gas too, if you had certain metals in acid, for example. Again, if you follow my technique above, you won't have to worry about a sep funnel popping on you because you didn't consider some source of heat or gas production.

Hope this helps, and good luck in April. 🙂
 
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<<<<<<<<--------------------------------------

Please refer to my avitar picture on the left for the skeletal structure I am talking about. I think it's a tricky question. The question asks which bond that connects C and O is shorter. I think there can be no resonance because you can't just move electrons due to octet. The carbon in the hexane (the right middle one) won't be able to have octet. Do you agree with me that there can be no resonance and hence the double bond to oxygen will be shorter than the single bond to oxygen? I am really anxious about this so please let me know!!! Thanks!
 
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QofQuimica said:
No. Oxygens don't affect the unsaturation count like nitrogens do. Pentanol isomers also have twelve H's. 🙂
Click to expand...

Wow, I'm slow at getting back to this. Anyway I still think it's a typo. The formula is C5 H12 OH so this compound has 13 hydrogens when it should have 12. Wouldn't C5 H11 OH actually be the correct formula?(Unless we're talking about a penta-coordinate carbon in Sn2 reaction.)

Damn I'm not even currently in college and I still think about this stuff😀
 
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pezzang said:
<<<<<<<<--------------------------------------

Please refer to my avitar picture on the left for the skeletal structure I am talking about. I think it's a tricky question. The question asks which bond that connects C and O is shorter. I think there can be no resonance because you can't just move electrons due to octet. The carbon in the hexane (the right middle one) won't be able to have octet. Do you agree with me that there can be no resonance and hence the double bond to oxygen will be shorter than the single bond to oxygen? I am really anxious about this so please let me know!!! Thanks!
Click to expand...
Sorry, but that structure DOES have a resonance form. It's basically an enolate ion. You can make a double bond and put a lone pair on the far C of the double bond. That being said, I agree with you that the permanent C=O should be shorter, b/c its bond order is higher (i.e., a full 2 instead of one point something).
 
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Dave_D said:
Wow, I'm slow at getting back to this. Anyway I still think it's a typo. The formula is C5 H12 OH so this compound has 13 hydrogens when it should have 12. Wouldn't C5 H11 OH actually be the correct formula?(Unless we're talking about a penta-coordinate carbon in Sn2 reaction.)

Damn I'm not even currently in college and I still think about this stuff😀
Click to expand...
Heh, yeah, you're right. I was thinking H12 total; didn't notice that it was H12OH. 🙂
 
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QofQuimica, thanks for your help. Do you know any website where they have a good summary and list of reactions and synthesis? (Like reactions/equations showing how nerol would turn into alpha-terpineol when H+ was put in) Thanks!
 
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pezzang said:
QofQuimica, thanks for your help. Do you know any website where they have a good summary and list of reactions and synthesis? (Like reactions/equations showing how nerol would turn into alpha-terpineol when H+ was put in) Thanks!
Click to expand...
I assume you're asking about this for class, because synthesis is basically not covered on the MCAT at all. What I recommend you do if you're really interested in learning synthesis is to get a hold of a book and workbook called "Organic synthesis: the disconnection approach" by Stuart Warren. If you work through the exercises in the book and workbook, it will teach you how to do retrosynthesis and come up with synthetic plans in a logical way. The first chapters cover easy molecules, and it gets harder and harder until you're doing some pretty complex molecules. You'd definitely ace all of your synthesis tests if you work through these books. Even doing the first few chapters would help you. See if your school library has them and check them out.

If you're actually asking about learning to write mechanisms (and your question makes it sound like maybe you are), you should get yourself a mechanism book and do the same thing. I used "Electron Flow in Organic Chemistry" by Paul Scudder, but there are some others too. I've seen several people on SDN say they liked "Pushing electrons" by Daniel Weeks, but for the record I've never read that book, so I can't really recommend it myself.
 
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hi, QofQuimica. i have question about the Br2 addtion to alkene
i know that the first answer is right, but not sure the rest..pls help
 

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Henn said:
hi, QofQuimica. i have question about the Br2 addtion to alkene
i know that the first answer is right, but not sure the rest..pls help
Click to expand...
Your answer looks good to me. Whenever you do an electrophilic addition to a double bond like this, you always should consider the possibility of rearrangements to more stable carbocations occuring after the electrophile adds. However, since Br2 preferentially forms a cyclic bromonium ion once the electrophilic Br adds, we wouldn't predict that any rearrangements would occur in this case. None of the other double bonds is in the correct location to give you that bromination pattern except for the first one.
 
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I have a lab this week and have to understand the procedure and exp.("dehydration of a tertiary alcohol") before teh actual experiment. we will be using IR spectroscopy. The procedure asks to think about the quesitons below but I can't understand them.. my lab book only has a brief summary of oh group having a broad band above 1500 cm^-1 but nothing else. any help would be appreciated, QofQuimica! Thanks!!

(a) The O-H portion of the spectrum changes with concentration. As the
solution becomes more dilute, the broad band centered at roughly 3300
cm-1 sharpens. Explain.
(b) The O-H stretching band for tert-butyl alcohol is much sharper than that for
methyl alcohol. Explain.
 
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pezzang said:
I have a lab this week and have to understand the procedure and exp.("dehydration of a tertiary alcohol") before teh actual experiment. we will be using IR spectroscopy. The procedure asks to think about the quesitons below but I can't understand them.. my lab book only has a brief summary of oh group having a broad band above 1500 cm^-1 but nothing else. any help would be appreciated, QofQuimica! Thanks!!

(a) The O-H portion of the spectrum changes with concentration. As the
solution becomes more dilute, the broad band centered at roughly 3300
cm-1 sharpens. Explain.
(b) The O-H stretching band for tert-butyl alcohol is much sharper than that for
methyl alcohol. Explain.
Click to expand...

Intermolecular forces impact stretching frequencies. Alcohol molecules tend to hydrogen bond with each other, which impacts the O-H stretch. Hydrogen bonding is kind of like adding an extra "mass" onto the hydroxy group "spring", so a hydrogen-bonded O-H stretches at a lower frequency than a non-hydrogen bonded (free) molecule. The difference is significant. The attraction of the neighboring oxygen "pulls" on the hydrogen of the adjacent hydroxy group electrostatically, which lengthens the O-H bond and lowers the frequency of vibration. Capeesh?

Here's the kicker: not all molecules are hydrogen bonded, and/or hydrogen-bonded at the same distance and orientation (the strength of hydrogen-bonds are sensitive to these factors), at any given point in time, so the O-H stretch absorption band for the entire solution broadens as a kind of "blurred" average of all the different O-H stretches of the alcohol molecules in their various states.

For alcohols, the strength of the hydrogen bond is roughly about 5 kcal/mol; therefore, you would probably get fairly rapid association, dissociation and re-orientation of the hydrogen bond at room temperature in a particular bonding network. The IR machine usually isn't able to resolve all these different temporal vibrational states (the changes in hydrogen bonding between molecules occur in picoseconds), so what you get when you do IR on this hydrogen-bonding phenomenon is something akin to exaggerated, overlapped photographic blurring at low shutter speed. While this occurs to some degree normally, hydrogen bonding adds a larger range of vibrational states and frequencies, resulting in significant broadening of the IR stretching band. Cool?

What things might interfere with hydrogen bonding, you ask? Well, how about steric factors? If an alcohol molecule, or group, is sterically hindered, it wouldn't be able to hydrogen bond as effectively or as favorably. So what would that do to the O-H stretch absorption band for the solution? It would tend to narrow it, right (i.e., make it sharper)?

What other factors could influence hydrogen bonding? How about concentration? If you dilute an alcohol solution with a solvent (non-polar), you would increase the mean distance between the alcohol molecules, consequently decreasing the prevalence of hydrogen bonding in the solution at any given point in time, I would think. Alcohol molecules need to be in relatively close proximity to each other in order to hydrogen bond, no? With increasing dilution, the solution becomes less and less dense with alcohol. Thus, the absorption band of the O-H stretch would tend to narrow as the solution becomes more dilute.

I'm no expert, but after I read your questions, I started to think about the answers because I was curious about them myself. Take them with a grain of salt... But I think this should help get you started on answering the questions above.

Q, did I miss anything?
 
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spicedmanna said:
Q, did I miss anything?
Click to expand...
I think I'm going to retire and turn this thread over to you. 👍 Would you mind if I edited your post a bit for uniformity and added it to the organic explanations thread? Also, would you be interested in writing some more posts like it?
 
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QofQuimica said:
Would you mind if I edited your post a bit for uniformity and added it to the organic explanations thread?
Click to expand...

I don't mind at all. Feel free to edit it and add it to the explanations thread.

QofQuimica said:
Also, would you be interested in writing some more posts like it?
Click to expand...

Yes, I'd be glad to write more posts like it. I enjoy helping out when I can, and Organic chemistry was among my most favorite premedical subjects. 🙂
 
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Wow, thanks I really appreciate it. It was so insightful... and thanks to QofQuimica as well.

spicedmanna said:
Intermolecular forces impact stretching frequencies. Alcohol molecules tend to hydrogen bond with each other, which impacts the O-H stretch. Hydrogen bonding is kind of like adding an extra "mass" onto the hydroxy group "spring", so a hydrogen-bonded O-H stretches at a lower frequency than a non-hydrogen bonded (free) molecule. The difference is significant. The attraction of the neighboring oxygen "pulls" on the hydrogen of the adjacent hydroxy group electrostatically, which lengthens the O-H bond and lowers the frequency of vibration. Capeesh?

Here's the kicker: not all molecules are hydrogen bonded, and/or hydrogen-bonded at the same distance and orientation (the strength of hydrogen-bonds are sensitive to these factors), at any given point in time, so the O-H stretch absorption band for the entire solution broadens as a kind of "blurred" average of all the different O-H stretches of the alcohol molecules in their various states.

For alcohols, the strength of the hydrogen bond is roughly about 5 kcal/mol; therefore, you would probably get fairly rapid association, dissociation and re-orientation of the hydrogen bond at room temperature in a particular bonding network. The IR machine usually isn't able to resolve all these different temporal vibrational states (the changes in hydrogen bonding between molecules occur in picoseconds), so what you get when you do IR on this hydrogen-bonding phenomenon is something akin to exaggerated, overlapped photographic blurring at low shutter speed. While this occurs to some degree normally, hydrogen bonding adds a larger range of vibrational states and frequencies, resulting in significant broadening of the IR stretching band. Cool?

What things might interfere with hydrogen bonding, you ask? Well, how about steric factors? If an alcohol molecule, or group, is sterically hindered, it wouldn't be able to hydrogen bond as effectively or as favorably. So what would that do to the O-H stretch absorption band for the solution? It would tend to narrow it, right (i.e., make it sharper)?

What other factors could influence hydrogen bonding? How about concentration? If you dilute an alcohol solution with a solvent (non-polar), you would increase the mean distance between the alcohol molecules, consequently decreasing the prevalence of hydrogen bonding in the solution at any given point in time, I would think. Alcohol molecules need to be in relatively close proximity to each other in order to hydrogen bond, no? With increasing dilution, the solution becomes less and less dense with alcohol. Thus, the absorption band of the O-H stretch would tend to narrow as the solution becomes more dilute.

I'm no expert, but after I read your questions, I started to think about the answers because I was curious about them myself. Take them with a grain of salt... But I think this should help get you started on answering the questions above.

Q, did I miss anything?
Click to expand...
 
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So I am curious if esters always have to have a form "O=C-O-C"

For example,
O
||
a) CH3CH2 C CH2CH3

O
|
b) CH3CH2O S OCH2CH3
|
O

c)
O
|
CH3CH2 S CH2CH3
|
O


d)
OCH2CH3
|
CH3CH2O B OCH2CH3

e)
CH2CH3
|
CH3CH2 P CH2CH3
|
O
Will any of the choices listed above esters despite not having typical "O=C-O-C"?

So will O=C-C or any forms other than O=C-O-C be used for esters?
 
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pezzang said:
So I am curious if esters always have to have a form "O=C-O-C" ?
Click to expand...

Technically, no. The form you list above is a carboxylate ester, the form you see most often in organic chemistry. In general, in introductory organic chemistry, esters, as a class, are oxy-acids where the proton of the -OH group of the oxy-acid is replaced by an alkyl (or organic group), usually via reaction of the oxy-acid with an alcohol. The oxy-acid in the ester doesn't have to be organic compound (for example, phosphoric or sulfuric acids).

There are also thio- (sulfur) analogs... Sulfur is similar to oxygen, so you can easily imagine a thio-acid like a thiocarboxylic acid (O=C-SH) forming a thioester (O=C-S-R).
 
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i learned in class last week that primary OH from R-OH cannot be released in a reaction with H+ (strong acid) to have R- because primary carbocation is unlikely (unstable). So my questions are:
1. Is it correct?
2. How about R-Br in a reaction with H+? Will Br be released? If not, will using Ag2+ or K+ make a difference?

Thank you everyone! Have a nice day! 🙂
 
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So are you saying that all the choices I listed above (a - e) are classified as esters?

spicedmanna said:
Technically, no. The form you list above is a carboxylate ester, the form you see most often in organic chemistry. In general, in introductory organic chemistry, esters, as a class, are oxy-acids where the proton of the -OH group of the oxy-acid is replaced by an alkyl (or organic group), usually via reaction of the oxy-acid with an alcohol. The oxy-acid in the ester doesn't have to be organic compound (for example, phosphoric or sulfuric acids).

There are also thio- (sulfur) analogs... Sulfur is similar to oxygen, so you can easily imagine a thio-acid like a thiocarboxylic acid (O=C-SH) forming a thioester (O=C-S-R).
Click to expand...
 
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pezzang said:
So are you saying that all the choices I listed above (a - e) are classified as esters?
Click to expand...

No. If I am interpreting your drawings correctly, I don't think choices (a), (b), (c), or (e) are esters as they appear to be drawn in your post. Choice (a) looks like a ketone, choices (b) and (c) don't look like valid structures to me, and choice (e) isn't an oxy-acid derivative. Choice (d) looks like it could be a tri-ester of boric acid, B(OEt)3, triethyl borate; therefore I'd say choice (d) could be classified as an ester.

I'm thinking that your drawings got butchered by the justification of the text editor and also have errors. Perhaps you drew structure (b) incorrectly, and it was actually supposed to be (O=)2S(OEt)2, in which case it would be an diester of sulfuric acid, diethyl sulfate; if that is what you meant, then choice (b) could be classified as an ester, too. I think perhaps you meant to draw structure (c) as (O=)2S(Et)2, in which case it would be called 1,1'-sulfonyldiethane, and it is not an ester. Lastly, perhaps you meant to draw structure (e) as O=P(Et)3, in which case it would be called triethylphosphine oxide, and it is not an ester.

In general, I think you misunderstand what an ester is. As I mentioned before, it is a derivative of an oxy-acid, an acid that contains (-OH). When the oxy-acid is esterified, the proton on the (-OH) group is replaced by an alkyl group. Let's stick with simpler examples, shall we?

An example of an oxy-acid would be sulfuric acid (H2SO4). An example of an ester of sulfuric acid would be:

(EtO)SO3H, ethyl sulfate

Another example of an oxy-acid would be phosphoric acid (H3PO4). An example of an ester of phosphoric acid would be:

(EtO)PO3H2, ethyl phosphate

Yet another example would be boric acid (B(OH)3). An example of an ester of boric acid would be:

(EtO)B(OH)2, ethyl borate

A very common example of an oxy-acid would be a carboxylic acid (RC(=O)OH). An example of an ester of carboxylic acid would be:

(EtO)C(=O)CH3, ethyl acetate

A thioester would look like this:

(EtS)C(=O)CH3, s-ethyl ethanethioate
 
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chpzz said:
i learned in class last week that primary OH from R-OH cannot be released in a reaction with H+ (strong acid) to have R- because primary carbocation is unlikely (unstable). So my questions are:
1. Is it correct?
2. How about R-Br in a reaction with H+? Will Br be released? If not, will using Ag2+ or K+ make a difference?

Thank you everyone! Have a nice day! 🙂
Click to expand...

You are on the right track. Primary, and even secondary, alcohols generally do not form carbocation intermediates in reaction with acid, primarily due to thermodynamic considerations (as you stated, a primary carbocation is considerably unstable). They can undergo slow nucleophilic substitution with a mineral acid, however. Remember that -OH is a poor leaving group, but becomes a much better one when it is protonated by an acid, creating an oxonium ion intermediate (R-OH2+). Halide ion (X-), the conjugate base of a mineral acid, is a very decent nucleophile and can then displace water via an SN2 mechanism, resulting in an alkyl halide (R-X) substitution product and water. This is not the best way to synthesize an alkyl halide from a primary alcohol, however. It would be better, and more common, to react a primary (or secondary) alcohol with phosphorus tribromide (PBr3) or thionyl chloride (SOCl2).

Tertiary alcohols do react readily with mineral acids and dilute aqueous acid solutions. Tertiary carbocations are the intermediates in both cases. For the reaction of tertiary alcohols with mineral acids, the mechanism can proceed through an SN1 pathway. Like before, the hydroxyl group is a poor leaving group, but after protonation, it becomes a much better one. Following the SN1 pathway, water leaves the oxonium ion (loss of water), resulting in a relatively stable tertiary carbocation. The halide nucleophile then attacks the carbocation, resulting in a tertiary alkyl halide. In dilute acid solution like dilute aqueous sulfuric acid, however, acid-catalyzed elimination is the favored pathway, rather than substitution because the conjugate base of sulfuric acid is a pretty weak nucleophile. The elimination reaction occurs through the E1 mechanism.

In response to your second question, primary alkyl halides (R-X) are generally unreactive to acids (as you have written it above). The type of reactions that alkyl halides, as a class, tend to undergo, as studied in general organic chemistry, are nucleophilic substitution and elimination reactions with an appropriate nucleophile; exactly which type of reaction, substitution or elimination, occurs, and by what kind of mechanism (SN1, SN2, E1, or E2) it undergoes, depends on steric factors, reaction conditions, leaving group, and nucleophile present. Primary alkyl halides do not generally form carbocations by the same reasoning that primary alcohols do not. They can react with strong nucleophiles like cyanide CN-, alkoxide (RO-), or hydroxide ions, etc., through an SN2 mechanism to form the resultant substitution product; this type of reaction is sensitive to steric factors (for example, a beta-substituted primary alkyl halides are sterically hindered from back-side attack).

I do not think adding silver ion or potassium ion will make a significant impact on the reactivity of primary alkyl halides with acid. Maybe you are thinking of some other kind of reaction? Adding a small amount of silver ion to a tertiary alkyl halide, for example, undergoing an SN1 reaction, or solvolysis-type reaction, with an appropriate nucleophile, can catalyze the reaction. Perhaps this is what you are thinking about? The silver (or mercury) ion can help facilitate the dissociation of the halide leaving group in an SN1-type reaction. Obviously, when, say, silver nitrate is used to catalyze the reaction, the SN1 product is favored; it seems like primary or secondary alkyl halide may undergo solvolysis in the presence of silver nitrate given the potential for rearrangement to form a more stable carbocation intermediate. I couldn't tell you if it is appreciable or practical, however. I don't know much about this at all. Q, any comments about this? In general, primary alkyl halides do not typically undergo SN1 reactions because primary carbocations are particularly unstable intermediates, as discussed above.

These are sweeping generalization, however, just to give you an idea of what is going on here. You will need to study the exact conditions, mechanism, and reactions for more specific and accurate results.
 
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Ok, I am confused with this really simple concept. In lab, we are going to be asked to make an equimolar solution of two reagents. How do I figure out how much of each to use. I get the whole idea of coming up with a molar ratio, however, once I get that molar ration, what exactly does it tell me? I need to make a 50mL solution, and my molar ratio is close to .95mol x/1mol y. Thanks!
 
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t2oo5 said:
Ok, I am confused with this really simple concept. In lab, we are going to be asked to make an equimolar solution of two reagents. How do I figure out how much of each to use. I get the whole idea of coming up with a molar ratio, however, once I get that molar ration, what exactly does it tell me? I need to make a 50mL solution, and my molar ratio is close to .95mol x/1mol y. Thanks!
Click to expand...

I'm not sure I understand your question, but I'll give it a shot; it's been awhile since I completed any labs. I forgot a lot about laboratory technique. If I don't get it right, perhaps someone will step in with the correct answer. I'm confused by whether you want two separate solutions that are equimolar in concentration to use for a reaction sequence (or titration), or one solution with equimolar amounts of A and B. Maybe it won't matter for your lab.

In essence, when you are saying one solution is equimolar to another in concentration, you are saying that they have the same molarity (number of moles of given solute per total liters of solution). Does that make sense? But the question is, how are these two reagents related in your reaction scheme, or sequence? More information is required to answer your question accurately, procedure-wise. However, I'll approach it generally.

The first step in your process should be calculating the respective molecular weights of your two reagents, A and B (let's call them mA and mB), you can do that by using a periodic table and the molecular fomulae of the two compounds. You will use the molecular weights and the number of moles (nA and nB) required to find out how many grams (gA and gB) of each you will need to mix into solution. Then, you will look in your lab manual for the number of moles of the limiting reagent required to generate the amount of desired product, or the molarity that is indicated (they should give you one of the two, or some way to determine them). You need to have some reference point for the number of moles required, however it is that it is indicated.

Let's assume that the manual indicates you need x moles of A; the mass of A required to make the solution is then determined by mass of A = (mA * nA) = X(mA). So by definition of equimolar (the ratio of nA to nB = 1), you will need also need x moles of B; the mass of B required to make the solution is then determined by mass of B = (mB * nB) = X(mB). Now you have the mass of each, A and B, required to make two separate solutions of equimolar concentration, or make one solution with equimolar amounts of each.

To make two separate solutions of equimolar concentration, just measure the masses indicated above and place in two separate volumetric flasks (or other appropriate container). Fill both with dH2O (or whatever solvent is indicated) to the same final volume. Now you have two solutions that are equimolar in concentration. Depending on your reaction, you will use/mix together whatever volume of each you require into the reaction mixture to generate the amount of product or intermediate you want.

If your lab experiment simply wants you to use equimolar amounts of A and B for the reaction sequence, just measure in the masses indicated in the fourth paragraph into a single flask and mix with the desired, or appropriate volume of solvent.

Hope this helps.
 
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Wow, SM you are really smart...
anyways I'm learning about Grignard reagent and am aware that it reacts with all protons more acidic than alkenes or alkanes and any carbonyl becomes alcohol. But I am not sure what would happen to the following problems:

1)
R-Mg-X -----------------------------------> ???????
````````(i)``O
````````````||
````````H3C-C-C-OEt
(ii) H+

My guess is that it will be something like this:

````OH
````|
H3C-C-OCH2CH4 + OH-MgX
````|
````R


2) R-Mg-X ----------------------------------> ???
i) C2H2O (cycloethane with an oxygen in the ring and one hydrogen coming out from each carbon)
ii) H+

My guess:
Something will change with the O in the cyclomethane due to its two pairs of lone bonds but not sure....

3) R-Mg-X ------------------> ??
i) CO2
ii) H+

My guess:
``````R
``````|
```OH-C-OH
``````|
``````R
Is it reasonable to think that there will be two same reactions for this case? (That is, =O(carbonyl group) changing to -OH )
Please let me know if each one is right.
As always, thanks SM!🙂
 
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chpzz said:
Wow, SM you are really smart...
anyways I'm learning about Grignard reagent and am aware that it reacts with all protons more acidic than alkenes or alkanes and any carbonyl becomes alcohol. But I am not sure what would happen to the following problems:

1)
R-Mg-X -----------------------------------> ???????
````````(i)``O
````````````||
````````H3C-C-C-OEt
(ii) H+

My guess is that it will be something like this:

````OH
````|
H3C-C-OCH2CH4 + OH-MgX
````|
````R


2) R-Mg-X ----------------------------------> ???
i) C2H2O (cycloethane with an oxygen in the ring and one hydrogen coming out from each carbon)
ii) H+

My guess:
Something will change with the O in the cyclomethane due to its two pairs of lone bonds but not sure....

3) R-Mg-X ------------------> ??
i) CO2
ii) H+

My guess:
``````R
``````|
```OH-C-OH
``````|
``````R
Is it reasonable to think that there will be two same reactions for this case? (That is, =O(carbonyl group) changing to -OH )
Please let me know if each one is right.
As always, thanks SM!🙂
Click to expand...
1) You're adding a Grignard to an ester, so you would get a double addition there. What you've shown is the first step. In other words, once the first R group has added, the O- formed from the carbonyl will kick the ethoxy group out, reforming the carbonyl in the process. That carbonyl can then be attacked by a second R- group.

2) That's not a cycloethane; that's an epoxide. It's a type of cyclic ether, and because of its small size, its bonds are constrained, meaning it is very reactive to attacks by nucleophiles like Grignards. When you react it with a Grignard, it will open up the ring. But it doesn't attack on oxygen; remember, since the Nu has its own lone pair, it doesn't want another lone pair from the oxygen. It wants to attack somewhere that's electron deficient. That means it will attack one of the carbons.

3) Nope, not quite. Once the first R group adds to CO2, you've got a deprotonated carboxylic acid on your hands. You won't be able to add another equivalent of R- to it because the carboxylic acid that you've formed is already carrying a negative charge.

It looks to me like you need to work on your understanding of nucleophiles and electrophiles. Keep in mind whenever you do a reaction that you always want to have a nucleophile (something that is electron rich) react with an electrophile (something that is electron poor). Hope this helps, and best of luck to you. 🙂
 
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I have some questions about synthesis.

1) 1-butyne -> (+/-) 3,4-dipentanol

2) Trans-3-hexene -> cis-3-hexene

3) CH3CH2OH -> 2-butanol

For each synthesis, what should be the reagents (multiple steps probably)? Any help would be greatly appreciated!
 
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pezzang said:
I have some questions about synthesis.

1) 1-butyne -> (+/-) 3,4-dipentanol

2) Trans-3-hexene -> cis-3-hexene

3) CH3CH2OH -> 2-butanol

For each synthesis, what should be the reagents (multiple steps probably)? Any help would be greatly appreciated!
Click to expand...
Come on, pezzang, you've been a member long enough to know that we aren't going to do your HW for you. You need to at least make an honest effort. Whenever you're doing synthesis, start with the product, and try to work your way backward to the starting material. (This is called retrosynthesis.) So for example, you have 3,4-dipentanol. What can you make a diol from? Think of a few ideas, and keep going until you get back to your butyne. You're almost certainly not going to have to do more than a few steps at a sophomore organic level.
 
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