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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?

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    $\begingroup$ Firstly, isobutane will never just form an anion by itself. Never. Loss of the proton must be a concerted reaction with the nucleophilic attack. Then, the bulky tertiary carbocation cannot attack the central tertiary carbon atom of the isobutane. Sterically bad on both sides. So IF a reaction takes place, B must be the product, primary cation attacks tertiary C. $\endgroup$
    – Karl
    Commented Jul 8, 2022 at 8:47
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    $\begingroup$ I doubt any reaction takes place. Is there any literature precedent? $\endgroup$
    – Waylander
    Commented Jul 8, 2022 at 9:33
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    $\begingroup$ Correction: What I described would be an electrophilic attack of course. Probably just as unlikely as the spontaneous deprotonation, but at least the acid would make sense. A carbanion CANNOT exist in the presence of acidic protons long enough for ANY reaction other than just grabbing one of those acidic protons. $\endgroup$
    – Karl
    Commented Jul 8, 2022 at 12:54
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    $\begingroup$ I do not recognize this set of reaction conditions to be any typical in an undergraduate chemistry course (I TA-ed for an honors two semester sequence in college so I am quite familiar still). I would sooner believe there was a typo in this question than believe that any of these given molecules can result from mixing these reagents together. $\endgroup$
    – Jeff Gustafson
    Commented Jul 8, 2022 at 14:37
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    $\begingroup$ @JeffGustafson Well, this is definitely not something many academics care about, a reaction between two gases. I actually had a prof who gave a two-semester industrial chemistry lecture, but I think that's pretty rare. There was no exam or anything, not part of the curriculum. $\endgroup$
    – Karl
    Commented Jul 9, 2022 at 14:01

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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.

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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  • $\begingroup$ Very analogous: the cationic polymerisation of propene. thieme-connect.de/products/ebooks/lookinside/10.1055/…. With methylpropene, the post-addition tertiary cation is so sterically hindered that it can do nothing but accept a hydrogen. $\endgroup$
    – Karl
    Commented Jul 9, 2022 at 13:46
  • $\begingroup$ @Karl I am unable to acess this $\endgroup$
    – Ritil
    Commented Jul 10, 2022 at 4:26
  • $\begingroup$ I wonder if there is also a chain termination step? $\endgroup$
    – Ritil
    Commented Jul 10, 2022 at 4:31
  • $\begingroup$ One immediate thought came into mind that when the octyl carbocation was formed, is not it possible that it is added to one more molecule of isobutene electrophilically to give a kind of dodecyl cation which after abstracting hydride forms dodecane isomer? And why can't this process continue even longer, just like polymerization? Can there be some some reason (for example steric hindrance) to show that such a reaction will be either kinetically or thermodynamically unfavorable? $\endgroup$
    – Ritil
    Commented Jul 10, 2022 at 4:36
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    $\begingroup$ Look here: pubchem.ncbi.nlm.nih.gov/compound/… That's a terribly distorted molecule. Accepting a hydride is much better entropically. But yea, the polymerisation works at low temperatures. Oligomers at 0°C, high Mw's only at -100. de.wikipedia.org/wiki/Polyisobutylen $\endgroup$
    – Karl
    Commented Jul 10, 2022 at 10:54

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