Let $L_n(\mathbb{Z})$ (resp. $L_n(\mathbb{Q})$) be the free Lie algebra over $\mathbb{Z}$ (resp. over $\mathbb{Q}$) with generating set $\{x_1,\dots,x_n\}$.

Let $\mathcal B$ be a $\mathbb{Q}$-basis of $L_n(\mathbb{Q})$ consisting of elements of the form $[x_{i_1},[x_{i_2},\dots,[x_{i_{r-1}},x_{i_r}]\dots]]$ for some $i_1,\dots,i_r\in\{1,\dots,n\}$. Is $\mathcal B$ automatically a $\mathbb{Z}$-basis of $L_n(\mathbb{Z})$?

Let $L(\mathbb{Z},n)$ (resp. $L(\mathbb{Q},n)$) be the free Lie algebra over $\mathbb{Z}$ (resp. over $\mathbb{Q}$) with generating set $\{x_1,\dots,x_n\}$.

Let $\mathcal B$ be a $\mathbb{Q}$-basis of $L(\mathbb{Q},n)$ consisting of elements of the form $[x_{i_1},[x_{i_2},\dots,[x_{i_{r-1}},x_{i_r}]\dots]]$ for some $i_1,\dots,i_r\in\{1,\dots,n\}$. Is $\mathcal B$ automatically a $\mathbb{Z}$-basis of $L(\mathbb{Z},n)$?