For a nice topological space $X$, let $X_\mathbb{Q}$ denote the rationalization of $X$. I am reading along in Hatcher's notes on spectral sequences and I have a question that is presumably very simple, but I am a total noob with these things. Using the Serre spectral sequence for cohomology, Hatcher proves that we have a fibration $K(\mathbb{Q}, 4k-1) \to S_\mathbb{Q}^{2k} \to K(\mathbb{Q}, 2k)$.
Hatcher then claims that we can see from the existence of this fibration what the Postnikov tower for $S_\mathbb{Q}^{2k}$ must look like. I can see that the first $2k-1$ terms of the Postnikov tower are points, the $2k$ term is $K(\mathbb{Q},2k)$ with the above fibration as the map for $S^{2k}_{\mathbb{Q}}$, and all the terms from the $2k$ term to the $4k-2$ term are just $K(\mathbb{Q}, 2k)$, but what is the $4k-1$ term of the Postnikov tower? Is it just $K(\mathbb{Q}, 2k) \times K(\mathbb{Q}, 4k-1)$? This has all of the correct homotopy groups at least.
Recall that the Postnikov tower $(P_n)$ of $X$ must have, in stage $4k-1$, a fibration $K(\pi_{4k-1}(X),4k-1) \to P_{4k-1}\to P_{4k-2}$.
With $X= S^{2k}_\mathbb Q$, $P_{4k-2} = K(\mathbb Q,2k)$ as you pointed out, and so we must have a fibration $$K(\mathbb Q,4k-1) \to P_{4k-1}\to K(\mathbb Q,2k)$$.
Does that $P_{4k-1}$ now remind you of someone else, who could rightfully live in the Postnikov tower ?