Given $J\_\alpha$ denotes the $1+\alpha$th term in the Jensen Hierarchy and $J=\bigcup\_{\alpha\in On}J\_\alpha$, are there any known large cardinals $\mathfrak{K}$ such that $\text{ZFC}+\mathfrak{K}$ is consistent and 
$\text{ZFC}+\mathfrak{K}\vdash J\neq L$? (assuming ZFC is consistent)



Large cardinals disproving J=L
Unfortunately, your question seems to be based on a misunderstanding of the $J$ hierarchy. There is no such large cardinal hypothesis.


The reason is that already ZFC and indeed much weaker theories such as KP prove $J=L$ outright. The point of the $J$ hierarchy was not to construct a different universe, but rather to reorganize the same constructible universe $L$ by means of a different stratification into levels, the $J\_\alpha$ hiearchy, so as to realize a better interaction with certain rudimentary set-theoretic constructions (see <http://en.wikipedia.org/wiki/Jensen_hierarchy#Rudimentary_functions>). This change in the hierarchy allowed certain observations to be made about the fine structure of the constructible universe, and for these reason, set theorists studying the fine structure of $L$ typically prefer to use the $J\_\alpha$ hierarchy.



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> > From Ronald Jensen, [Fine structure of the constructible hierarhcy](http://www.math.cmu.edu/~laiken/papers/FineStructure.pdf), ANNALS OF MATHEMATICAL LOGIC 4 (1972) pp. 229-308. 
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> > We have found it convenient to replace the usual $L\_\alpha$ hierarchy by a new hierarchy $J\_\alpha$. We define $J\_{\alpha+1}$ not as the collection of definable subsets of $J\_\alpha$, but as the closure of $J\_\alpha\cup\{J\_\alpha\}$ under a class of functions which we call "rudimentary". These are just the functions obtained by omitting the recursion schema from the usual list of schemata for primitive recursive set functions. In a sense they form the smallest class of functions $\mathcal{R}$ such that there is a smooth definability theory for transitive domains closed under $\mathcal{R}$. The main difference between the two hierarchies is that $J\_\alpha$ has rank $\omega\alpha$ rather than $\alpha$. However, the subsets of $J\_\alpha$ which are elements of $J\_{\alpha+1}$ are just the definable ones, $J\_{\alpha+1}$ is, so to speak, the result of "stretching" the collection of the definable subsets of $J\_\alpha$ upwards $\omega$ levels in rank without adding new ones. The exact correspondence between the two hierarchies is given by: $$J\_0 = L\_0 = \emptyset; \quad L\_{\omega+\alpha}=V\_{\omega+\alpha}\cap J\_{1+\alpha}.$$ 
> >  Thus $J\_\alpha=L\_\alpha$ whenever $\omega\alpha=\alpha$.
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So although $J\_\alpha$ is often not the same as $L\_\alpha$, nevertheless there is a class club of ordinals $\alpha$ where $J\_\alpha=L\_\alpha$, and so $J=L$ in the end.


