Taken from the book of GOC by Dr. O.P Tandon, Himanshu Pandey, Dr. A.K. Virmani,
Can anyone elaborate on the mechanism?
$\begingroup$
$\endgroup$
10
-
1$\begingroup$ It seems like an interesting reaction, if it is correct; I've never seen alkanes reacting in such a manner. But surely acetone should undergo acid-catalysed aldol, it seems weird that the book has not mentioned that at all. $\endgroup$– Box Box Box BoxCommented Jul 28, 2021 at 13:32
-
2$\begingroup$ @TRC Under "saturated aliphatic hydrocarbons" $\endgroup$– NehaCommented Jul 28, 2021 at 13:40
-
2$\begingroup$ @AshishAhuja Oh yes, of course it can. I meant that the final product depicted here certainly isn't acetone's self-aldol product, rather it looks like a sort of condensation between the two compounds. $\endgroup$– TRCCommented Jul 28, 2021 at 13:40
-
4$\begingroup$ Due to lack of much complexity there is only really 1 plausible way to look at this, you have to generate a nucleophile to make any headway. This acidic media makes it impossible to do that. $\ce{H2SO4}$'s most extreme reaction is charring action (It never generates nucleophiles anyway). I don't see anyway to facilitate any further reaction. Self aldol looks way more plausible. $\endgroup$– napstablookCommented Jul 28, 2021 at 14:08
-
4$\begingroup$ This pubs.acs.org/doi/pdf/10.1021/ja01156a075 and this onlinelibrary.wiley.com/doi/10.1002/9780471678656.ch12 may help. $\endgroup$– RishiCommented Jul 28, 2021 at 18:24
|
Show 5 more comments
1 Answer
$\begingroup$
$\endgroup$
1
This was going to be a comment but it got too long.
TL;DR: - This is not an answer, rather a justification for why the question is (probably) wrong.
This paper (linked by @Rishi) gives us experimental evidence that under action of concentrated $\ce{H2SO4}$ hydrogens are exchanged from paraffins in the following fashion:
$$ \begin{align} \ce{(CH3)2CHCH3 + 2H2SO4 -> (CH3)C^{+}OSO3H^{-} + SO2 + 2H2O & \tag{R1}}\\ \ce{(CH3)3C+ + (CH3)3CH -> (CH3)3CH + (CH3)3C+ & \tag{R2}}\\ \end{align} $$
R1 is the slow step and chain initiating step. R2 is the chain propagating step (the hydride transfer here creates a chain of approximately 20 tert-butyl carbocation.
Conc. $\ce{H2SO4}$ is a well known dehydrating and oxidising agent and tert-butyl carbocation has considerable stability due to extensive hyperconjugation. So it is possible that the reaction mechanism is of the following manner:
From the paper by Otvos et.al.1:
Chains are terminated by some irreversible side reactions of carbonium ions leading to nonvolatile products. This picture would account for the lack of exchange between t-isobutane hydrogen atoms and sulfuric acid, since any molecule dissolving in acid, reacting and re-entering the vapor phase, would have received a new tertiary hydrogen from another isobutane molecule.
The chain termination step is where we run into a problem. Carbonyl group cannot act as a nucleophile at the carbon center. There is one way to do this but it is extremely unlikely
This is not a good mechanism due to the excess of $\ce{H^+}$ present. In fact, any mechanism for the given product wouldn't be satisfactory because any nucleophile produced at carbonyl carbon would be would be quickly protonated (tert-butyl carbocation can't compete with $\ce{H+}$ due to steric reasons). Also it is highly unlikely that $\ce{SO2}$ would participate in the reaction rather than escape to atmosphere (I suppose you can use sealed tubes).
Further problems are the much better suited side reactions such as acid-catalyzed aldol and attack of $\ce{C=O:}$ lone pair on the carbocation.
Citation
- The Behavior of Isobutane in Concentrated Sulfuric Acid1 J. W. Otvos, D. P. Stevenson, C. D. Wagner, and O. Beeck Journal of the American Chemical Society 1951 73 (12), 5741-5746 DOI: 10.1021/ja01156a075
-
$\begingroup$ I fixed some punctuation and citation. Feel free to roll back if you don't like the edit. Tip regarding formatting: Instead of spamming
~, use\begin{align}. It looks pleasant. $\endgroup$ Commented Aug 3, 2021 at 8:27