Galois cohomology class of a reductive group not coming from a torus Let $G$ be a (connected) reductive group over a perfect field $k$, and let $\xi\in H^1(k,G)$ be a cohomology class.
By a theorem of Steinberg (Serre, Galois cohomology, Appendix 1 to Chapter III, Theorem 11.1),
if $G$ is quasi-split, then there exists a $k$-torus $i\colon T\hookrightarrow G$ such that
$$\xi\in i_* \big(H^1(k,T)\big).$$

Question. What is an example of $k$, a non-quasi-split reductive $k$-group $G$, and  $\xi\in H^1(k,G)$, such that $\xi$ does not come from a $k$-torus?

Remark. Such $k$ cannot be $\Bbb R$, a $p$-adic field or a number field.
Indeed, let $G$ be a reductive $k$-group, not necessarily quasi-split.
If $k$ is $\Bbb R$ or a $p$-adic field, then there exists a $k$-torus  $i\colon T\hookrightarrow G$
such that the map
$$i_*\colon H^1(k,T)\to H^1(k,G) $$
is surjective.
If $k$ is a number field, then for any finite set $\Xi=\{\xi_1,\dots,\xi_n\}\subset H^1(k,G)$ there exists a $k$-torus
$i\colon T\hookrightarrow G$ such that $\Xi\subset i_*\big( H^1(k,T)\big)$.
For the last assertion, see M. Borovoi, Abelian Galois cohomology of reductive groups, Mem. Amer. Math. Soc. 132 (1998), no. 626,
Theorem 5.10.
 A: $\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$The  following example is due to Vladimir Chernousov (private communication).
Let $K={\Bbb Q}(x,y,x',y')$, where $x,y,x',y'$ are variables.
Consider the quadratic forms over $K$
$$ f= \la x,y,-xy\ra\qquad\text{and}\qquad f'=\la x',y',-x'y'\ra.$$
Here $ \la x,y,-xy\ra$ denotes the quadratic form $$f(t_1,t_2,t_3)=xt_1^2+yt_2^2-xyt_3^2.$$

Lemma 1. The quadratic form  over $K$
$$ f-f'=\la x,y,-xy, -x',-y', x'y'\ra$$
is anisotropic (does not represent 0 nontrivially).

Proof. The quadratic form $f-f'$ is monomial and multiplicity free
in the sense of Section 4 of Vladimir Chernousov and Jean-Pierre Serre,
Lower bounds for essential dimensions via orthogonal representations,
J. Algebra 305 (2006), no. 2, 1055–1070.
By their Proposition 5 on page 1061, the quadratic form $f-f'$ is anisotropic,
as required.
Consider the quaternion $K$-algebras
$$ D=(x,y)\qquad  \text{and} \qquad D'=(x',y').$$
Here $(x,y)$ denotes the 4-dimensional associative $K$-algebra with generators $i,j$ and relations
$$i^2=x,  \quad j^2=y,\quad    ij=-ji.$$
The reduced norm form for $D$ is isomorphic  to  $\ \la 1\ra-f=\la 1,-x,-y,xy\ra$.

Lemma 2. The algebraic $K$-groups $G={\rm PGL}(1,D)$ and $G'={\rm PGL}(1,D')$ have no isomorphic maximal tori.

Proof.
Assume for the sake of contradiction that $G$ and $G'$ have isomorphic maximal tori.
Then $D$ and $D'$ have isomorphic maximal subfields, say, $L\subset D$ and $L'\subset D'$.
Write $L=K(\sqrt{a})\subset D$ for $a\in K$.
This implies that there exists a pure quaternion $t_1 i+t_2 j+t_3ij$ such that
$$xt_1^2+yt_2^2-xyt
_3^2=a.$$
Thus the quadratic form $f$ represents $a$.
Similarly, from $L'=K(\sqrt{a})\subset D'$ (with the same $a$) we obtain that $f'$ represents $a$.
Therefore, the quadratic form $f-f'$ represents 0 nontrivially, which contradicts Lemma 1.

Theorem. For $G$ and $G'$ as above, let $c\in Z^1(K,G)$ is a 1-cocycle such that  $_c G\simeq G'$.
Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus of $G$.

Proof.
Assume for the sake of contradiction that there exists a maximal torus
$$ j\colon T\hookrightarrow G\quad\text{such that }\quad [c]\in j_* H^1(K,T).$$
Then there exists an embedding
$$j'\colon T\hookrightarrow {}_c G\simeq G'.$$
Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.
