Some notation: $\varphi : A \longmapsto B$ is the attaching map of the $(n+1)$-cells. Here $A = \bigvee_j \mathbb{S}_j^n$ are the boundary of the $(n+1)$ disk(related) with relative $\varphi_j$ attaching map and $B = X^{n+1}$ (the $(n+1)-$th skeleton of some $CW$ called $X$, with $X^{n-1} = \left\lbrace * \right\rbrace$).
$X = X^{n+1}$ is $(n-1)$-connected by hypothesis. Assume you already know that we can consider $X = X^{n+1} = C_\varphi = B$(where $C_\varphi$ denotes the mapping cone)see here.
The question: is it true that $$\pi_n(M_\varphi,A) \simeq \pi_n(C_\varphi,*)?$$
I'd like to prove the latter using Blakers Massey in the following form
but taking $M_{\varphi} = C_\varphi \cup A$ I end up with an isomorphism between $\pi_i(M_\varphi,A) \simeq \pi_i(C_\varphi,*)$ just for $1 \leq i < n+n-2= 2(n-1)$
(Using the fact that ($\pi_i(A)$ is $(n-1)$-connected and $\pi_i(C_\varphi) = \pi_i(X)$ is $(n-1)$-connected by hypotesis).
It seems to me that the case $n=2$ is uncovered by this theorem.
The proof should also cover the case $n=2$ so can the Blakers Massey be applied properly in the case $n=2$ or an other argument should be used?
Any help, hint proof or reference would be appreciated.