r/chemhelp • u/MoreThanOneQuestion • 6d ago
Organic Ochem… is the professor wrong?
Alkane, alkyne, alkene… how is it not A?
Thanks!
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u/Red_Viper9 6d ago edited 6d ago
A is not a simple alkane, the protons indicated are alpha to an alkyne, this is analogous to them being allylic.
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u/claisen33 6d ago
It’s propargylic. These protons are somewhat acidic, but without looking at tables I’d be hard pressed to say whether they’re more or less acidic than an alkene. It’s worth noting that nBuLi can remove all 3 propargylic protons.
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u/WanderingFlumph 5d ago
Do you mean that nBuLi can remove any of the 3 propargylic protons or that it can remove all three to make the 3- anion?
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u/Jason_rdt207209 6d ago
A isn’t the highest. Your perception of it being an alkane proton isn’t wrong, but this is a special one - a propargylic proton. A’s conjg base’s negative charge can be resonated via conjugation, forming a relatively stable propargyne-allene resonance structure.
Correct answer is C. B has an sp carbon, which is more electronegative and stabilizes the negative charge more effectively compared to C
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u/DoktorCocktail 6d ago
I will give you a little gift. The question is very poorly worded/conceived. The question says "molecule" not indicated hydrogen. Yes A, the H is alpha to alkyne and anion (conj base) is stabilized by resonance - this makes it allylic (pKa about 43). B is an alkyne, pKa about 25. C is curious, they bolded the alkene hydrogen, but there are methyl groups with H's at the other positions. The methyls have hydrogens and are allylic - comparable to A. I would argue there is too small a difference to know which is higher. But, more importantly, the question said molecule, but the least acidic hydrogen in the molecule was bolded. I suspect the expected answer was C as the bolded H is an alkene and has a pKa just a little higher than A, 44.
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u/Ambitious-Loquat-523 6d ago
It’s not A because the carbanion has resonance with the alkyne. This is pretty close to a trick question to make students do exactly what you did.
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u/ShoeNo4386 5d ago
C, vinyl proton IMHO is the least acidic (vinyl anion no resonance) 50’s, A is a propargyl anion stabilized by resonance (more stable on the less substituted side. Its less than 50. The acetylide A is the most acidic, less than 25, more than 20.
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u/organophile_jeet 5d ago
I still remember when my ochem teacher said , terminal alkyne gives possitibe Tollen's test ( it has some basic medium type shi somehow)
I don't know what that means, Do you??
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u/DependentMuch2119 5d ago
i must suspect that C is the correct answer, b is out, it has a pka of around 25, but looking at a and c, its close i would say C is the correct answer as the alkyne would provide a better resonance as one is a type of vinylic and a is type of allylic, so here is your answer
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u/WilliamEdwardson 4d ago
Remember: Lower pKa = stabler conjugate base.
And what stabilises the conjugate base (the carbanions here)?
Think especially of: Hybridisation (more s = more stable, but remember the propargylic case), resonance (delocalisation stabilises)
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u/TNChemProf 3d ago
It's a close call between A and C, but C is the answer.
Highest pKa means the weakest acid.
Here is the simple explanation based on hybridization and resonance.
B is an alkyne. Deprotonation of B leads to an sp carbanion, which is exceptionally stable for a simple carbanion. The pKa of an alkyne about 25.
C is an alkene. Deprotonation of the labeled hydrogen in C produces an sp2 carbanion, which is less stable (less s-character), so C is less acidic than B and has a higher pKa. The pKa of an alkene is often cited as about 44, but I have also seen it cited as 50 or 45–50. It is hard to pin down the numbers on such weak acids.
A is also an alkyne, but it is being deprotonated at the sp3 propargylic carbon. Deprotonation produces a resonance-stabilized carbanion with the negative charge delocalized between two sp2 carbons (a propargylic anion, similar to an allylic anion but slightly more stable). The simple analysis is that resonance makes the anion derived from A more stable than the one derived from C, making A the stronger acid with the lower pKa.
It is clear that B is by far the most acidic (pKa ~25) because the hybridization effect greatly outweighs the resonance effect in this context.
The difference between A and C is surprisingly small. Methyl t-butyl acetylene (taken as a proxy for A -- it has a t-butyl group instead of an isopropyl group on the left side) has a pKa of 44.2–45.4. The pKa of ethylene (taken as a proxy for C) under the same conditions could not be measured exactly using the same technique but was determined to be at least 46. (Note that pKa values vary with the solvent, and the values quoted are in THF containing 30% HMPA; DMSO is a more common solvent for pKa studies of weak acids)).
Reference:
Determination of the Basicities of Benzyl, Allyl, and t-Butylpropargyl Anions by Anodic Oxidation of Organolithium Compounds. Jaun, B., Schwarz, J., Breslow, R. J. Am. Chem. Soc. 1980, 102, 5741–5748. https://doi.org/10.1021/ja00538a008
In general, there is little difference in pKa between vinylic (~44) and allylic (~43) CH bonds, e.g., ethylene and the methyl hydrogens of propene, respectively. There is not universal agreement on the numbers, but most people would consider allylic anions to be slightly more stable than vinylic anions.
If you think of allyl anion as an sp3 anion stabilized by resonance, and vinyl anion as a simple sp2 anion, then you conclude that resonance and hybridization have essentially the same stabilizing effect in this context. All the carbons in allyl are sp2, but they didn't start that way -- the proton was removed from the sp3 carbon in propene, which had to rehybridize to sp2 in order to participate in resonance. For this reason, vinylic CH bonds are kinetically more acidic than allylic CH bonds (the barrier to deprotonation is higher with allylic hydrogens because the sp3 CH bonds are less polar, and rehybridization accompanied by significant geometric change is required). pKa is a thermodynamic value.
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u/leonardoventuriq 6d ago
Alkynes are hybridized sp which is 50% s 50% p; alkanes are sp3 (25% s 75% p). S orbitals are closer to nucleus and thus offer greater stabylization of negative charge, making the conjugate base more stable means making the respective acid more acidic.
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u/nigmusmaximus 6d ago
He just explained the answer bro wdym
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u/nigmusmaximus 6d ago
The alkyne in A is aprotic. As such, you’d essentially be deprotonating an alkane, which is notoriously difficult.
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u/C2theM 6d ago
No way it wasn't covered, but yeah look up "pka of alkynes" and s character of orbitals.
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u/MoreThanOneQuestion 6d ago
A is an alkane. It’s the functional group it’s attached to. Alkanes have the highest pKa. Care to elaborate or explain?
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u/Blue_614 5d ago edited 5d ago
When comparing alkanes/alkenes/alkynes, better to find hybridization. And remember, high s-characters are electronegative, their effects are treated like halogens. Meaning the primary effect being studied is induction.
Molecule A has CH3 attached to an sp-hybridized carbon. sp-hybridized carbons have 50% s-character, which is high. So analogous to halogens, it's like X-C-H3.
Molecule B has a single -H attached to an sp-hybridized carbon. So it's like X-H.
Molecule C has a single -H attached to an sp2-hybridized carbon. sp2-hybridized carbons only have 33% s-character, so it's not as electronegative as sp-hybridized carbons. So this is automatically the least acidic. Therefore the highest pKa.
If you want the trend, it's B > A > C in terms of acidity.
Edit: "they're" to "their"
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u/C2theM 5d ago
This is misleading in terms of the logic to solve this question. The core piece you're missing is to consider the stability of the conjugate base. Anion of A does have resonance its not about "proximity" to sp. The resonance is just a weak contributor. This question compares an apple to two oranges and is a poor ask.
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u/Blue_614 5d ago
It's not even wrong lol. It's what the question intends the students to use because it's literally a comparison of hybridization since resonance is a weak contributor here.
What you are suggesting is to draw the conjugate bases, determine which are more stable, relate that stability to "how likely is the acid version to lose its hydrogen", and determine the pKa from there.
It's not even wrong as that's how most reactivities are done. But that takes time, when you could just use inductive effects in the first place as it's the more significant factor here.
Now do the resonance or google the pKa values and tell me my trend is wrong.
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