Taken from the book of GOC by Dr. O.P Tandon, Himanshu Pandey, Dr. A.K. Virmani,
Can anyone elaborate on the mechanism?
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this communityTaken from the book of GOC by Dr. O.P Tandon, Himanshu Pandey, Dr. A.K. Virmani,
Can anyone elaborate on the mechanism?
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
~
, use \begin{align}
. It looks pleasant.
$\endgroup$
Aug 3, 2021 at 8:27