Let $G$ be a semisimple group over a perfect field (at the moment I am interested in the case $k=\mathbb R$).
Then $G$ is an inner form of a quasi-split $k$-group $G_{\rm qs}$:
there exists a quasi-split form $G_{\rm qs}$ of $G$ and a 1-cocycle $c\in Z^1(k,\overline{G}_{\rm qs})$ such that
$G=\,_c G_{\rm qs}$, the inner twist of $G_{\rm qs}$ by the cocycle $c$, where $\overline{G}_{\rm qs}=G_{\rm qs}/Z_{\rm qs}$ and $Z_{\rm qs}=Z(G_{\rm qs})$.
Let $\xi=[c]\in H^1(k,\overline{G}_{\rm qs})$, the cohomology class of the cocycle $c$.
Let
$$\Delta\colon H^1(k,\overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$
denote the connecting map from the cohomology exact sequence
$$H^1(k,Z_{\rm qs})\to H^1(k,G_{\rm qs})\to H^1(k, \overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$
induced by the short exact sequence
$$ 1\to Z_{\rm qs}\to G_{\rm qs}\to \overline{G}_{\rm qs}\to 1.$$
By definition, the  *Tits class of $G$* is
$$t_G=\Delta(\xi)\in H^2(k,Z_{\rm qs}),$$
see The Book of Involutions, (31.7).
Note that $Z_{\rm qs}=Z:=Z(G)$.

> **Question 1.** Is there anywhere in the literature or in the Internet a table of $t_G$ for all simply connected absolutely simple ${\mathbb R}$-groups $G$?

For all connected Dynkin diagram except for ${\mathsf D}_{2m}$ (and also for ${\mathsf D}_{2m}$ when $G_{\rm qs}$ is not split)
we have either $H^2({\mathbb R},Z_{\rm qs})=0$ or $H^2({\mathbb R},Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}$.
From the cohomology exact sequence we see that $t_G=0$ if and only if $G$ is a *strong inner* form of $G_{\rm qs}$, that is, $\xi$ comes from $H^1({\mathbb R}, G)$.
I have a table of strong inner forms of $G_{\rm qs}$, and so I know when $t_G=0$; in all the other cases we have $t_G\neq 0$, hence $t_G=1+2{\mathbb Z}\in {\mathbb Z}/2{\mathbb Z}$.

However, for ${\mathsf D}_{2m}$, in the case when $G_{\rm qs}$ is split, namely $G_{\rm qs}\simeq {\bf Spin}(2m,2m)$, we have $H^2({\mathbb R}, Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}\times{\mathbb Z}/2{\mathbb Z}$.
I know that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, if $G\simeq{\bf Spin}(2m+4q, 2m-4q)$,
but I need an explicit formula for the cases  ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and the quaternionic form ${\bf Spin}^*(4m)$.
Thus Question 1 reduces to the following question:

> **Question 2.** What are the Tits classes for the simple ${\mathbb R}$-groups  ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and ${\bf Spin}^*(4m)$ ?

In order to formulate an answer to Question 2, one needs an explicit description of the center of the split ${\mathbb R}$-group ${\bf Spin}(2m,2m)$.
Such a description is given, for example, in [Brian Conrad's cheat sheet](http://www.math.stonybrook.edu/~jstarr/papers/Centers.pdf).