Brtdku

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.

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 ($ce{C53H108}$).

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]

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).

Notes:

1. Alternative structure-based polymer name based on older organic chemistry nomenclature would be poly(1-heptylidene)

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
$$a_{n,h}=Ccdot3^n$$
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
$$a_n=a_{n,p}+a_{n,h}=-frac12+Ccdot3^n$$
We know that the solution for $n=1$ is a methyl group, so
$$a_1=1=-frac12+3C$$
So $C=1/2$ and
$$a_n=frac12left(3^n-1right)$$
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
$$b_{2N+1}=1+4a_N=2cdot3^N-1$$
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 :)