Reaction of alkane with alkene in acidic medium

[![question figure](https://i.stack.imgur.com/8BUdu.jpg)](https://i.stack.imgur.com/8BUdu.jpg)


This question was asked in jee mains 2022 june 25 shift 1. I have never seen such a kind of reaction before and also could not find it anywhere on internet. But I guessed that the proton will add to primary carbon of the double bond and form tert-butyl carbocation while isobutane will release proton to form isobutyl carbanion. They will then add to each other to form the product D. This is why I marked it.


However the released answer key shows that option B is correct. One possible thought that came in my mind was that bulky bases like tertbutoxide ion make Hoffman product major because they can't attack middle carbon due to steric hindrance. Same case happen here also with nucleophile. So, instead primary carbocation would form and get attacked to form that product in option B.


However I am not sure of myself. My query is not regarding this specific question but asking in general wether such kind of reactions with alkane and alkene actually happen? What is their mechanism? And most importantly how to decide wether a given base or nucleophile is bulky or not?




This reaction is actually called as
"alkylation of alkenes". If I had known *this* term, I could have directly searched it on internet but won't have asked it on SE ;)


<https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.10%3A_Alkylation_of_Alkenes>


This website shows that according to the scientist P.D. Bartlett, a carbocation can react with a hydrocarbon with tertiary carbon by transferring hydride (H-). This is reminiscent of the familiar cannizaro reaction where also this hydride transfer takes place. A chain reaction occurs leading to formation of tertbutyl cation adding with isobutene to form a carbocation which then extracts hydride from isobutane, and then the cycle continues.


I have found similar explanation by coaching channels on YouTube, like this one.(watch from 17:33)
<https://youtu.be/T6TUROPO4eM>
Though it is in hindi,but you can still see the structures


The catalyst required could be any acid like sulphuric or hydrofluoric acid etc but this website says that this particular reaction can also occur without catalyst at high temperatures (and is just lesser feasible economically). [https://www.sciencedirect.com/topics/physics-and-astronomy/alkylation#:~:text=Conventional%20paraffin%20(alkane)%E2%80%93olefin,yielding%202%2C3%2Ddimethylbutane](https://www.sciencedirect.com/topics/physics-and-astronomy/alkylation#:%7E:text=Conventional%20paraffin%20(alkane)%E2%80%93olefin,yielding%202%2C3%2Ddimethylbutane).


As expected there is no carbanion formation. The reaction mechanism is therefore quite similar to free radical halogenation in the sense that there is an initiation step and then propagation step although there is no radical involved.


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