If $X$ is a topological space then the rational cohomology of $X$ carries a canonical $A_\infty$ structure (in fact $C_\infty$) with differential $m_1: H^\ast(X) \to H^{\ast+1}(X)$ vanishing and product $m_2: H^\ast(X)\otimes H^\ast(X) \to H^\ast(X)$ coinciding with the cup product.

If $X$ is a closed manifold then its cohomology algebra satisfies Poincare duality. This is a condition that refers only to the $m_2$ part of the $A_\infty$ structure. There are things more general than manifolds that satisfy rational Poincare duality, such as rational homology manifolds. In general, a space that satisfies rational Poincare duality is called \emph{rational Poincare duality space}.

Obviously, the statement that $X$ is a rational Poincare duality space places no additional restrictions on the $A_\infty$ structure beyond the condition on the product.

Question 1 Can one construct from the higher multiplications obstructions for a rational Poincare duality space to be rationally equivalent to a rational homology manifold, or a topological or smooth manifold?

Here is a second and somewhat related question. Poincare duality says that $H^\ast(X)$ is self-dual (with an appropriate degree shift). Thus the adjoints of the higher multiplication maps make the cohomology into an $A_\infty$ coalgebra (with appropriate adjustment of the grading).

Question 2 How does this $A_\infty$ coalgebra structure interact with the $A_\infty$ algebra structure on the cohomology?