What is the largest (most carbon atoms) alkane having heptane as its base name?
For example, 2,2,3,3-tetramethylbutane is the largest (most carbon atoms) alkane retaining butane as its base name.

  • 4
    $\begingroup$ You put as many side chains on it as possible without creating a new longest chain. Then you do the same with each of the side chains, and then you hope that the final molecule's center hasn't become so dense that it snaps in two. $\endgroup$ – Karl Jan 19 at 13:19
  • 1
    $\begingroup$ This is a subset of a very interesting theoretical question about how many isomers are possible for simple hydrocarbons discussed in this question: chemistry.stackexchange.com/q/16135/81 . Answers there also address questions about whether potential molecules are physically possible. $\endgroup$ – matt_black Jan 19 at 15:07
  • $\begingroup$ Thanks for the upvotes to my initial (stupid :-) ) comment: The surface of the molecule is densely packed, and the bonds in the center are stretched to the point where they rip. $\endgroup$ – Karl Jan 21 at 21:34

Any alkyl substituent of butane in position 2 or 3 cannot be longer than $\ce{CH3}$ since that would lead to a longer parent chain. And obviously, there cannot be any alkyl substituent at all in the first or the last position of the butane chain. Therefore, the largest structure based on a butane parent chain is 2,2,3,3-tetramethylbutane.

This principle can be expanded to a heptane parent chain. The maximum length for alkyl substituent chains are 0 for position 1 and 7, 1 for position 2 and 6, 2 for position 3 and 5, and 3 for position 4. Therefore, the largest theoretical structure based on a heptane parent chain is 3,3,5,5-tetra-tert-butyl-4,4-bis[3-(tert-butyl)-2,2,4,4-tetramethylpentan-3-yl]-2,2,6,6-tetramethylheptane ($\ce{C53H108}$).


  • 2
    $\begingroup$ I believe this thing is much too crowded to survive, but then again, that's not what the question was about. Other than that, you are right. $\endgroup$ – Ivan Neretin Jan 19 at 13:33
  • 5
    $\begingroup$ @IvanNeretin Agreed, the inner C–C bonds would probably be stretched to more than 2 Å. That’s why I wrote “largest theoretical structure”. $\endgroup$ – Loong Jan 19 at 14:08
  • 4
    $\begingroup$ Preferred IUPAC names is a good concept, but “tetrakis(tri-tert-butylmethyl)methane” is somewhat shorter and clearer ;) $\endgroup$ – mykhal Jan 19 at 17:56
  • 3
    $\begingroup$ It would seem I have found the structure for the maximal pentadecane in a completely different setting. I wonder if there's anything interesting that pops up from that unusual connection... Actually, it seems that if you take the infinite fractal and assign a single carbon atom to be a particular region size, then circle all regions of that size, you can recover the whole infinite family of maximal alkanes! $\endgroup$ – Nicolau Saker Neto Jan 19 at 23:54
  • 3
    $\begingroup$ @NicolauSakerNeto I also found big resemblance in the Cayley graph of the free group with 2 generators. As a non-chemist, but someone with interest in maths, I wonder whether there is any deeper connection between algebra and chemistry. If someone has any pointers for me, I am happy to take them. $\endgroup$ – ComFreek Jan 20 at 8:33

That's hard to tell, because already the next one after tetramethylbutane, tetra-tert-butylmethane (3,3-di-tert-butyl-2,2,4,4-tetramethylpentane) is so unstable it cannot exist. You want a parent chain that is even two atoms longer.

Actually it seems also removing one methyl group from $\ce{tBu_4}$methane doesn't make it sufficiently stable, so one has to be a bit more generous in weeding out branches for larger hyperbranched alkanes.

I believe that any permethylated (or per-n-alkylated) linear chain is principally possible, but adding more side-side chains to adjacent side chains will make the molecule snap in two.

So the biggest possible sum formula would be something based on

3,5-di-tert-butyl-2,2,3,4,4,5,6,6-octaamethylheptane = $\ce{C23H48}$

. You can surely make the inner methyl substituents longer,

3,5-di-tert-butyl-3,5-diethyl-4,4-di-n-propyl-2,2,6,6-tetraamethylheptane = $\ce{C29H60}$

, and then you can probably add 4 methyl to the nPr and Et to give iBu and i-Pr that's $\ce{C33H68}$. Any more and I'll bet it breaks, quite a way from the theoretical $\ce{C53H108}$.


Speaking of hypothetical structures, you can get, with little cheating, infinitely large “alkane” like poly(heptane-1,1-diyl)[1]

Fig.1 - structure formula      Fig.2 - structure formula 2

where $n=\infty$, i.e. with infinite degree of polymerization (or substituted branched ones like those in other, correct, answers, with even bigger infinity sizes).


  1. Alternative structure-based polymer name based on older organic chemistry nomenclature would be poly(1-heptylidene)
  • $\begingroup$ But in that case won't the parent be 1-chloro-4-(3-chloropropyl)infinitane? rather than dichlorosubstituted heptane? $\endgroup$ – Nandakumar U K Jan 20 at 0:26
  • 1
    $\begingroup$ @NandakumarUK I don't know if you are serious and where did you get chlorine. My name uses polymer nomenclature (see e.g. here), and it's a more systematic structure-based name, like poly(methylene) for more known “polyethylene” (source-based name). $\endgroup$ – mykhal Jan 20 at 6:52
  • $\begingroup$ @NandakumarUK .. but if you want to have fun non-polymer name, I'd name it rather like −∞,−∞+1,−∞+2,…,∞−2,∞−1,∞-infinitaheptylinfinitane. $\endgroup$ – mykhal Jan 20 at 7:17
  • $\begingroup$ Sir, i was referring to a choosable monomer before getting polymerized.; that is why i mentioned a chloro compound. If you consider the compound to be a polymer, there must be a monomer for sure; i wounder, which is that? $\endgroup$ – Nandakumar U K Jan 20 at 8:42
  • $\begingroup$ Again poly methylene can have odd or even number of carbon atoms in the chain, but poly ethylene can have only even number of carbon atoms in the chain; so could we consider them to be the same always? $\endgroup$ – Nandakumar U K Jan 20 at 8:44

The general answer is easy. First we compute the maximum number of carbon atoms $a_n$ in a side chain of length $n$ from bond to end. Its difference equation is $$a_{n+1}=3a_n+1$$ The $1$ is there to count the carbon that has the bond and the $3a_n$ counts the $3$ maximal side chains radiating from it. The homogeneous equation is $$a_{n+1,h}=3a_{n,h}$$ With general solution $$a_{n,h}=C\cdot3^n$$ And if we seek a particular solution of the form $$a_{n,p}=K=\text{constant}$$ We get the solution $$a_{n,p}=-\frac12$$ Since the above is a linear difference equation, the general solution is $$a_n=a_{n,p}+a_{n,h}=-\frac12+C\cdot3^n$$ We know that the solution for $n=1$ is a methyl group, so $$a_1=1=-\frac12+3C$$ So $C=1/2$ and $$a_n=\frac12\left(3^n-1\right)$$ Now consider $b_{2N+1}$, the maximum number of carbons in a molecule with maximum length $2N+1$ carbon chain. To make this we just substitute $4$ maximal side chains of length $N$ to a methane molecule to get $$b_{2N+1}=1+4a_N=2\cdot3^N-1$$ And to get $b_{2N}$ we substitute $6$ maximal sides chains of length $N-1$ to an ethane molecule to get $$b_{2N}=2+6a_{N-1}=3^N-1$$ Hmmm... I guess we could have arrived at the same result by just joining $2$ maximal side chains of length $N$ together :)

  • $\begingroup$ So for heptane as the longest chain N=3 (2N+1 carbon atoms), that gives 53, correct. Pentane 17, correct. Hexane 26, also right. Nice math! $\endgroup$ – Karl Jan 21 at 11:00
  • $\begingroup$ This equation was the one i always dreamt, attempted and miserably failed every time ;)...Kudos!!! $\endgroup$ – Nandakumar U K Jan 21 at 12:24

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.