Following the approach of Weinberg's book to discuss the chiral symmetry, at a certain point he says
If the $\rm SU(2)\times SU(2)$ symmetry is exact and unbroken, then this would require any one-hadron state $|h\rangle$ to be degenerate with another state $\vec{X}|h\rangle$ of opposite parity and equal spin, baryon number and strangeness. No such parity doubling is seen in the hadron spectrum, so we are forced to conclude that if the chiral symmetry is a god approximation t all, then it must be spontaneously broken to its isotopic spin $\rm SU(2)$ subgroup.
where $\{\vec{X}\}_i$ is the charge associated to the conserved axial-current $\vec{A}^\mu = i\bar{q}\gamma^\mu\gamma_5\vec{t}q$.
Can anyone explain better why the symmetry is broken and why it is broken in the isotopic spin group and not in another group? Or maybe if anyone can suggest some notes.