EK 1001 Orgo

[Synapsis]

Couple of questions.

First, here's the structure of aspartame.

A question asks what the hybridization on N8 (the amide nitrogen) is. One thing that my orgo professors hammered is that if an atom will donate (or receive) lone pairs through resonance, it must have an empty p orbital to do so with. Since nitrogen donates its lone pairs so well, and there will be resonance between the carbonyl carbon and the amide nitrogen, why is the nitrogen not sp2 hybridized? The answer says it's sp3.


Second, question 88 gives three molecules: (I) methane, (II) tetrachloromethane, and (III) ethanol. It asks which of the three molecules contains a polar bond.

A. II
B. III
C. I and III
D. I, II, and III

I would naturally pick II and III (not an option) because we normally think of hydrogen and carbon as having essentially the same electronegativity. In fact, most organic chemistry books and many sites say that the bond is nonpolar. How do I know when to account for the small electronegativity difference between carbon and hydrogen and when not to (other than using POE and knowing that since II and III surely have polar bonds, the answer must be D)?

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[Temperature101]

Couple of questions.

First, here's the structure of aspartame.

A question asks what the hybridization on N8 (the amide nitrogen) is. One thing that my orgo professors hammered is that if an atom will donate (or receive) lone pairs through resonance, it must have an empty p orbital to do so with. Since nitrogen donates its lone pairs so well, and there will be resonance between the carbonyl carbon and the amide nitrogen, why is the nitrogen not sp2 hybridized? The answer says it's sp3.


Second, question 88 gives three molecules: (I) methane, (II) tetrachloromethane, and (III) ethanol. It asks which of the three molecules contains a polar bond.

A. II
B. III
C. I and III
D. I, II, and III

I would naturally pick II and III (not an option) because we normally think of hydrogen and carbon as having essentially the same electronegativity. In fact, most organic chemistry books and many sites say that the bond is nonpolar. How do I know when to account for the small electronegativity difference between carbon and hydrogen and when not to (other than using POE and knowing that since II and III surely have polar bonds, the answer must be D)?

For question 88, I am not sure why EK includes methane....Every single orgo book considers the C---H bond as a non polar one....That must be a mistake on the part of EK...

For the other question, yes the nitrogen will donate the lone pair thru resonance, but I am sure why you think that matters. Resonance structure bond is somewhat an imaginary one, it does not exist in reality. May be someone else can give a clearer explanation...

 

[Synapsis]

For question 88, I am not sure why EK includes methane....Every single orgo book considers the C---H bond as a non polar one....That must be a mistake on the part of EK...

For the other question, yes the nitrogen will donate the lone pair thru resonance, but I am sure why you think that matters. Resonance structure bond is somewhat an imaginary one, it does not exist in reality. May be someone else can give a clearer explanation...

Right, I know that a resonance hybrid is a weighted average of the resonance forms, but that doesn't mean that the form drawn gives the accurate hybridization.

Even if we look at our nitrogenous bases, all of the nitrogens are sp2 hybridized - even the ones with lone pairs (the lone pairs reside in p orbitals, so they can participate in pi bonding and make the system aromatic). Since you can't make that pi bond (albeit in a resonance form) with a hybrid orbital, there must be an available p orbital. To double check this, I went to TPR orgo review. If you have it, check out page 80 question d and the answer.

In orgo 2, we also learn the amides are planar (just like in peptide bonds, which are planar). Well it can't be planar if the nitrogen is sp3. It must be sp2, right?

 

[Deadlifts]

Right, I know that a resonance hybrid is a weighted average of the resonance forms, but that doesn't mean that the form drawn gives the accurate hybridization.

Even if we look at our nitrogenous bases, all of the nitrogens are sp2 hybridized - even the ones with lone pairs (the lone pairs reside in p orbitals, so they can participate in pi bonding and make the system aromatic). Since you can't make that pi bond (albeit in a resonance form) with a hybrid orbital, there must be an available p orbital. To double check this, I went to TPR orgo review. If you have it, check out page 80 question d and the answer.

In orgo 2, we also learn the amides are planar (just like in peptide bonds, which are planar). Well it can't be planar if the nitrogen is sp3. It must be sp2, right?

VESPR model suggests the amide's N should be sp3 due to having 4 areas of electron density. However, that's not taking into account the partial pi bond across the peptide bond and resonance.

I'm pretty sure the general rule of thumb is that if a nitrogen is bonded to an sp2 carbon, it will be sp2 as well. However, in the drawing they didn't show resonance structure or even draw the lone pair so maybe that's why they claim sp3?

 

[DNA]

VESPR model suggests the amide's N should be sp3 due to having 4 areas of electron density. However, that's not taking into account the partial pi bond across the peptide bond and resonance.

I'm pretty sure the general rule of thumb is that if a nitrogen is bonded to an sp2 carbon, it will be sp2 as well. However, in the drawing they didn't show resonance structure or even draw the lone pair so maybe that's why they claim sp3?

No, what you said earlier was right. Don't second guess yourself . The actual structure for the molecule above takes into account all the major resonance forms. Technically, the structure depicted above doesn't even exist. What we see happening is a constant swaying electrons from the lone pair of electrons on the nitrogen which is then delocalized into the carbonyl, forming a partial double bond between nitrogen and the carbonyl carbon. In order for this delocalization to even occur, a P orbital is absolutely necessary. Therefore, nitrogen must be sp2 hybridized. At no point is nitrogen ever sp3 hybridized. Infact, being sp3 hybridized would hinder this delocalization from ever happening (due to lack of an available p orbital). This is clearly a mistake made by EK.

 

[DNA]

For all intents and purposes, we often treat the C-H bond as nearly nonpolar because the dipole moment is extremely small (~0.3 D compared to 1.56 for a C-Cl bond). Notice the word nearly. Anytime there is a difference in electronegativities, the bond is most certainly polar. Therefore, a C-H bond IS indeed polar (but weakly so). On the other hand, when you have two of the same atoms bonded to one another (ie. C-C or Cl-Cl), then the bond is strictly nonpolar (0 D) because they have the exact same electronegativites and are "pulling at each other with equal force".

I think the purpose of this question is for you to realize that C-H bonds ARE indeed polar, due to the slight differences in electronegativity. We often make that assumption that they're nonpolar because the difference is negligible. Will the polarity of a C-H bond have a significant effect on the physical properties of the molecule (ie. Boiling Point, Melting Point, etc.)? ...That's highly unlikely, especially when you might have extremely electronegative atoms like Oxygen and Chlorine. Hence, why we typically ignore the polarity of a C-H bond.

The MCAT will never explicitly ask you about the polarity of a C-H bond. You are more likely to see a question pertaining to the physical properties of the molecule (comparing various molecules and assessing which will have either the highest or lowest boiling point, for example).

 

[Synapsis]

No, what you said earlier was right. Don't second guess yourself . The actual structure for the molecule above takes into account all the major resonance forms. Technically, the structure depicted above doesn't even exist. What we see happening is a constant swaying electrons from the lone pair of electrons on the nitrogen which is then delocalized into the carbonyl, forming a partial double bond between nitrogen and the carbonyl carbon. In order for this delocalization to even occur, a P orbital is absolutely necessary. Therefore, nitrogen must be sp2 hybridized. At no point is nitrogen ever sp3 hybridized. Infact, being sp3 hybridized would hinder this delocalization from ever happening (due to lack of an available p orbital). This is clearly a mistake made by EK.

Thought so, that's exactly what I was saying. I hope no one goes with EK's answer. Always a scary thought when the ones teaching you make mistakes and you don't catch them.

Good, I'm glad they don't test the polarity of C-H bonds. The polarity's so small that it'd be really hard to test anyway. Thanks for the help!

 

[QrtrLifeCrisis]

From what I remember in Orgo 2, the amide bond has both sp2 and sp3 characteristic and depends on certain conditions, such as temperature. However, it spends most of its time in sp2. This would account for most of the planar secondary structures of amino acids. Conversely, esters also have both sp2 and sp3 characteristics, but spend most of their time in sp3. That EK question is very ambiguous, but like someone else said, just go with what is drawn. It is drawn as sp3.

The C-H bond is nonpolar. I would not consider it otherwise unless it was mentioned somewhere in a passage.

 

[Synapsis]

From what I remember in Orgo 2, the amide bond has both sp2 and sp3 characteristic and depends on certain conditions, such as temperature. However, it spends most of its time in sp2. This would account for most of the planar secondary structures of amino acids. Conversely, esters also have both sp2 and sp3 characteristics, but spend most of their time in sp3. That EK question is very ambiguous, but like someone else said, just go with what is drawn. It is drawn as sp3.

The C-H bond is nonpolar. I would not consider it otherwise unless it was mentioned somewhere in a passage.

But you can't "draw a bond as sp3". There's no such thing. One of the only times that you can look at something and immediately tell it's sp3 just from its drawing is if it's directly bonded to four things, like carbon in methane.

The nitrogens are sp2 in cytosine. If you look at the bottom nitrogen, by what you said, it's "drawn as sp3". But that's not real because even though it might look that way from how it's drawn, we know it's sp2 because that lone pair is in a p orbital, contributing to the aromaticity. And actually in this case, that nitrogen's lone pair would have essentially complete p orbital character due to the aromatic ring.

I always beware making that quick assumption because it's a really easy place to be tricked, especially on a test that's trying to trick you.

 

[QrtrLifeCrisis]

But you can't "draw a bond as sp3". There's no such thing. One of the only times that you can look at something and immediately tell it's sp3 just from its drawing is if it's directly bonded to four things, like carbon in methane.

The nitrogens are sp2 in cytosine. If you look at the bottom nitrogen, by what you said, it's "drawn as sp3". But that's not real because even though it might look that way from how it's drawn, we know it's sp2 because that lone pair is in a p orbital, contributing to the aromaticity. And actually in this case, that nitrogen's lone pair would have essentially complete p orbital character due to the aromatic ring.

I always beware making that quick assumption because it's a really easy place to be tricked, especially on a test that's trying to trick you.

You make a good point. However, aromatic compounds are different than aliphatic compounds. The lone pair electrons in cytosine are essentially locked into the aromatic ring because it provides greater stability to the molecule. Anyone who has an understanding of aromaticity knows that the bottom nitrogen is sp2 hybridized. Now, if you can find an example where an aliphatic nitrogen was sp2 hybridized, but was depicted as sp3, then I would concede the point.

Amides are a little more ambiguous. They can be depicted as either sp3 or sp2 and both would be correct. I was taught that the hybridization of nitrogen in an amide is somewhere in between. But the point has already been made. The EK question is flawed.

Now, if you really want to get deep...read this little blurb: http://chm233.asu.edu/notes/resonance/S2.html

 

[Synapsis]

You make a good point. However, aromatic compounds are different than aliphatic compounds. The lone pair electrons in cytosine are essentially locked into the aromatic ring because it provides greater stability to the molecule. Anyone who has an understanding of aromaticity knows that the bottom nitrogen is sp2 hybridized. Now, if you can find an example where an aliphatic nitrogen was sp2 hybridized, but was depicted as sp3, then I would concede the point.

Amides are a little more ambiguous. They can be depicted as either sp3 or sp2 and both would be correct. I was taught that the hybridization of nitrogen in an amide is somewhere in between. But the point has already been made. The EK question is flawed.

Now, if you really want to get deep...read this little blurb: http://chm233.asu.edu/notes/resonance/S2.html

I definitely agree with most of that, I just disagree that we can look at it and instantly call sp3 hybridization because of four centers of electron density. That's just not always true.

The same evidence could be used for the opposite point though: http://instruct.uwo.ca/chemistry/283g/Nifty Stuff/amides are flat B.htm

It just seems more correct to call it sp2 because it donates it's lone pairs so well and because without that p orbital, there's no delocalization at all. But the fact that a peptide bond (aliphatic) is planar screams that it must be sp2, otherwise that bond wouldn't be planar. That also tends to be the correct answer in most organic books and even a lot of MCAT books despite the fact that the actual answer is in between. The reason I posted the question is because it just seemed so odd and out of place

But whatever works best for ya.

 

[sazerac]

It is ironic that you guys keep comparing aspartame to a peptide bond, since aspartame IS a peptide bonding of aspartate and phenylalanine, with a methyl thrown in later for good measure.

It's a peptide bond. It's planar. The amide nitrogen is sp2 (the amine nitrogen is still sp3). EK is wrong. There's nothing ambiguous about this.

 

[loltopsy]

The hybridization of that amide nitrogen is sp2.5. It is in between sp3 and sp2. Pretty much any property of the amide nitrogen can be estimated by averaging the value for sp3 and sp2 together (ie bond angle). This is due to resonance.

Methane is non-polar because the C-H bond is not very polar, AND because methane is perfectly symmetrical. Tetrachloromethane is non-polar because TCM is perfectly symmetrical. The electron withdrawing effect of chlorine gets canceled out by another chlorine creating no net charge on the molecule meaning the bonds would be non-polar. TCM is a very non-polar solvent. Trichloromethane, however, breaks the symmetrical and becomes quite polar.

If the question is asking which molecule is polar, the answer would be 3 only. If the question was asking if individual bonds are polar, the answer would be 2 and 3. But since the C-H bond is SLIGHTLY polar, you can even argue that the answer 1 as well. The electronegativities of carbon and hydrogen are slightly different, so there is a slight polarity in the bond. Of course, this argument frees up any molecule to be called polar as long as the atoms are not binded to the same atoms.