37
It is not the number of lone pairs that in any way explains basicity. Take a random sugar and it will have ten times the number of lone pairs (albeit on oxygen, not on nitrogen) without being significantly basic.
The problem is electronics. Nitrogen, being a rather electronegative atom, is able to draw the π electrons towards it well — to the extent ...
23
Both pyrrole and furan have a lone pair of electrons in a p-orbital, this lone pair is extensively delocalized into the conjugated pi framework to create an aromatic 6 pi electron system.
Where pyrrole and furan significantly differ is that,
in pyrrole there is an $\ce{N-H}$ bond lying in the plane of the ring and directed away from the ring
whereas in ...
20
Great question! It turns out that the rate of formation of the "expected" endo product is actually ~500 times faster than the rate of formation of the exo product. However, the Diels–Alder is a reversible reaction. In this case, the exo product is thermodynamically favored over the endo product by about $\pu{1.9 kcal/mol}$.
So, even though the predicted ...
18
The key point to understanding why fluorides are so reactive in the nucleophilic aromatic substitution (I will call it SNAr in the following) is knowing the rate determining step of the reaction mechanism. The mechanism is as shown in the following picture (Nu = Nucleophile, X = leaving group):
Now, the first step (= addition) is very slow as aromaticity is ...
16
Interesting question, more subtle than I realized!
Theoretical approach: I would turn to resonance and induction as the rationale for electronic properties. I think all of your resonance forms may contribute about equally to the resonance hybrid. This suggests that all four carbons would share roughly the same amount of electron density. However, we must ...
16
As you may know from diazotisation reactions, $\ce{HNO2}$ is a generator of the nitrosonium cation, $\ce{NO+}$. Electrophiles like these can react with the enol tautomer of hexanedione:
Following this, you tautomerise the nitroso compound to the oxime. The mechanism is exactly analogous to keto-enol tautomerism:
Note that I've elected to draw the oxime ...
14
The following figure shows the resonance structures we can draw to describe the intermediates produced by electrophilic at the 2- ($\alpha$) or 3- ($\beta$) positions in pyrrole.
image source
You can see that the intermediate produced by attack at the $\alpha$-position can be described by 3 resonance structures. Whereas attack at the $\beta$-position ...
14
It looks as if the NMR of morpholine is an AA′XX′ spectrum (the chemical shift difference is 0.80 ppm, or 320 Hz on your spectrometer, two orders of magnitude larger than the coupling constant).
Unlike linear AA′XX′ systems where the bonds can rotate, in the morpholine it very much has a fixed conformation.
In this fixed conformation, the protons are ...
14
Your Diels–Alder adduct is pretty much correct. Without any further evidence we cannot definitely say whether the exo or the endo adduct is formed (you have drawn the exo adduct, which generally arises via thermodynamic control). However, the stereochemistry of this adduct does not affect the final product, so it does not matter which one is formed.
By the ...
13
Orthocresol did a great job describing how to get to the isoxazole methyl ketone. However, nitrous acid can be an oxidant. The methyl ketone in the presence of mineral acid, which you need to get the nitrosyl cation, can tautomerize the methyl ketone to the enol which reacts with the cation to get an α-nitroso ketone. The nitroso group tautomerizes to ...
13
The @orthocresol link to the Zincke reaction is a good start to the mechanism. You are correct about the first step. Dimethylamine adds to the electrophilic pyridinium salt 1 as illustrated in 1 --> 5. The reaction is driven by loss of the poor nucleophile (decent leaving group), DNP-NH2. The (Z)-isomer 5 is in equilibrium with the (E)-isomer 6 but the ...
12
The correct answer to this question is more direct, and not listed in your items. The non-bonding electron pair in nitrogen 3 is in an orbital perpendicular to the π-bonding p orbitals of all other atoms in the imidazole ring. Thus, it does not have the appropriate geometry to overlap with other orbitals forming π-bonds, and does not participate in any ...
12
As correctly pointed out in the comments, the compound you present belongs to the class of sydnones. According to Hückel's rule, it should be aromatic, as the molecule is planar and contains $4n+2$ π-electrons.
However, calculations performed in 1998[1] that mainly looked at chemical shifts and inferred aromaticity from the magnitude of observed diamagnetic ...
12
That is really not a fair question and even with a computational approach the differences are more or less marginal. The approach by electron pusher is certainly the way to go, and here is the computations to back it up.
See if you can find the carbon with the highest electron density in the following plot showing the value of the electron density for the ...
12
TL;DR Most SNAr reactions proceed via concerted mechanisms without a discrete anionic intermediate. This is typically the case when there is a good leaving group (e.g. chloride or bromide) and if the ring is not extremely electron-poor.
Over the years, there has been increasing evidence of concerted nucleophilic substitutions at sp2-hybridised carbons. One ...
11
Both nitrogens in your drawing are roughly $\mathrm{sp^2}$ hybridized, therefore they each have a p-orbital and 3 $\mathrm{sp^2}$ hybdridized orbitals. All of the p-orbitals, including those on the 3 carbon atoms in the double bonds, are perpendicular to the plane of the screen (e.g. they project above and below the plane of the screen). So there are 5 p-...
11
From what I gathered in the comments you confused a few terms. Both compounds are monodentate, i.e. they can only coordinate one metal at a time. However, pyridine usually donates one electron pair, while pyrrole can donate three (to one metal) — I haven’t seen a structure where pyrrole does that, though.
The reason is pretty simple: Pyrrole is much like ...
11
The numbering of (R)-5-[(S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one has four parts:
1) Numbering of the furan ring
2) Numbering of the ethyl side chain
3) Stereodescriptors R and S
4) Italicized element symbols
The italic element symbol H denotes indicated or added hydrogen.
11
Only count the lone pairs/ pi-bonds/ groups which are participating in conjugation and ignore them in all other cases.
For example, in compound 2 (thiophene), there are two lone pairs on sulfur.
One lone pair (brown) is in a p-orbital, and hence participates in conjugation with the two π-bonds. The other lone pair (blue) is pointing outwards from the ring ...
10
The compound is anti-aromatic. While counting the number of π-electrons, you count the electrons which are delocalized over the ring. In this case the nitrogen lone pair is localised and does not participate in resonance. The nitrogen lone pair is in an $\mathrm{sp^2}$ orbital (red) which is orthogonal to the π system (blue):
So, the total number of π-...
10
Here is the proton nmr spectrum for pyridine.
Here is the proton nmr spectrum for furan
For comparison, benzene has a single proton nmr signal at 7.27 ppm. In both pyridine and furan there are different types of protons, so multiple signals are seen in both cases.
Here are the resonance structures for pyridine
and furan.
Pyridine removes electron ...
9
Resonance structures can be drawn for all aromatic compounds. Tropolone, furan and thiophene are considered to be aromatic compounds because
we can draw resonance structures
and they are cyclic compounds with 4n+2 (n=1) pi electrons
Here are the resonance structures we can draw for tropolone, the positive charge can be delocalized to every carbon in the ...
9
Unfortunately I cannot agree with the currently accepted answer. This 1,5-hydride shift that was invoked seems to have a very strained transition state (the cyclic, planar system means that the "1" and "5" atoms are quite far apart) and with all due respect, I am quite skeptical as to whether it would really work. Furthermore furan does not enjoy a ...
9
Consider 2,3-butanediol.
This compound has 2 stereocenters. The meso version of this compound is achiral. However, there are still two stereo centers. As drawn, the left one is (R) and the right one is (S). It is precisely because these two stereocenters are identical but opposite in configuration that leads the molecule to be meso.
A carbon center is ...
9
The traditional explanation for the reactivity of indole at C–3 is that attack at C–3 does not disrupt the aromaticity of the benzene ring in the cationic intermediate (see any organic chemistry book; for heterocycles a good book is Joule and Mills as has been suggested in the comments):
You have drawn many of the resonance structures for the intermediate, ...
8
A quick post so you can get started on thinking about this.
Remember, resonance is just a model. The electrons are delocalized all the time; the furan does not morph back and forth among the various structures.
You can do a "formal" evaluation of which resonance forms have the most contribution to the structure by looking at formal charges, and there are ...
8
Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$?
I doubt it and would rather expect a lot of tar to be formed here. Pyrrole tends to undergo acid-catalyzed polymerization!
For decades, acetyl nitrate, formed by addition of fuming nitric acid to acetic anhydride, is the reagent of choice.
Mononitration of pyrrole ...
answered Apr 22 '15 at 11:27
Klaus-Dieter Warzecha
41.2k7 gold badges84 silver badges147 bronze badges
8
The key to this question lies in realising that the Reimer–Tiemann reaction is not an electrophilic aromatic substitution but it is in fact
addition of a carbanion to a strongly electron-deficient carbene.
This was discussed by @KlausWarzecha and @ron in answers to a previous question.
Therefore the major product will be the one where the attack comes ...
8
Is this only because the overall structure will not be consistent, because sp-hybridization will make the COC line straight, and thus the OCC angles would have to be lesser than the carbons' sp2-hybridization would allow?
Yes, this is the correct answer. A very common misconception is that hybridisation can be used to predict the geometry, or that ...
8
No doubt it is aromatic.
The $4n+2$ rule is intended for a single cycle of π-electrons, which may possibly include multiple rings but can't have any interior atoms or pendant π- bonds (which would not fit in a cycle). Here, as Kartik indicates in a comment, there are evidently two such cycles and you have to consider them separately. One cycle has six π-...
Only top voted, non community-wiki answers of a minimum length are eligible
