Given a complex simple Lie algebra $\mathfrak{g}$ of rank $n\in\mathbb{N}$ with $n$ sufficiently large (say $n\ge10$), is there a way to determine whether $\mathfrak{g}$ contains a simple subalgebra of a prescribed type with rank "close" to $n$? For example, if $\mathfrak{g}$ is of type $B\_n$ with $n\ge10$, does $\mathfrak{g}$ contain a subalgebra of type $C\_{n-2}$?



Simple Subalgebras of Simple Lie Algebras
The answer to the question in your example is no in general: $B\_n$ does not contain $C\_{n-2}$ for large $n$. To see this, observe that $B\_n$ has an irreducible orthogonal representation $V$ of dimension $2n+1$. The Weyl dimension formula shows that for any simple Lie algebra, the dimension of an irreducible representation is the smallest for fundamental representations. 


[Edited] 


For large $n$, the smallest dimensional fundamental representation of $C\_{n-2}$ is the standard one, of dimension $2n-4$. 
Therefore the restriction of $V$ to $C\_{n-2}$ can only be some copies of the trivial representation and ONE copy of the standard representation $W$ of dimension $2n-4$. This means that the group of type $C\_{n-2}$ must preserve both a symplectic form and a quadratic form on $W$, which is impossible by Schur's lemma. Hence the restriction of $V$ to $C\_{n-2}$ can only be a sum of copies of the trivial representation which is 
impossible.


I had previously given a wrong reason (assuming that the rep of $B\_n$ had dimension $n$; thanks to @BS for pointing this out. 


