Chemistry - Largest alkane having a given alkane as its base name

Solution 1:

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 ($\ce{C53H108}$).

Solution 2:

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 (search for Hellmann, Beckhaus, Rüchardt "Thermolabile Hydrocarbons", e.g. hexamethylpentane is stable at RT), 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 $\ce{C53H108}$.

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Solution 3:

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

    

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).

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Notes:

1. Alternative structure-based polymer name based on older organic chemistry nomenclature would be poly(1-heptylidene). However the “longest chain” rule seems to applied differently in polymer nomenclature, so somewhat more “preferred name” could be poly(hexylmethylene).

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Solution 4:

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 :)