Is there a $\mathbb{C}$ vector space structure over $\mathbb{R}$ with scalar multiplication as defined? To find is a $\mathbb{C}$ vector space structure over $\mathbb{R}$ with addition and scalar multiplication such that scalar multiplication has the following property:
$\cdot : \mathbb{C} \times \mathbb{R} \rightarrow \mathbb{R}$ s.t.
$\cdot : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is a usual multiplication
 in $\mathbb{R}$
 A: No. For any vector space $V$, any nonzero vector $v$, and any scalars $k$ and $l$, $kv=lv$ implies that $k=l$. Here, $kv$ already takes all possible values when $k$ ranges over $\Bbb R$, so there's nothing for (say) $iv$ to be.
A: Let $V$ be a real vector space. Using axioms of vector spaces you can prove that you only need to find out what $\imath \cdot v$ should be for $v\in V$. So let
$$
\phi(x) = \imath \cdot x
$$
Since $r_1 \cdot (r_2 \cdot v) = r_1r_2 \cdot v$ you get for $r_1 = r_2 = \imath$ the following  equality must hold
$$
\phi(\phi(v)) = -v
$$
for all $v \in V$. Moreover, again using vector space axioms you can prove that
$$
\phi(a\cdot x + b \cdot y) = a\phi(x) + b\phi(y)
$$
for all $a,b \in \mathbb{R}$ and all vectors $x,y \in V$. 
So existence of such a linear mapping $\phi: V \to V$ that squares to minus identity is equivalent to possibility of extending your real vector space structure to complex vector space structure. 
The rest is up to you. Hint: Suppose such a $\phi$ exists. What can you say about the dimension of $V$? 
A: This supposed vector space would be $1-$dimensional. Let $\{a\}$ be a basis. Then $i\cdot a = r =\frac{r}{a}\cdot a , \ \ r \in \mathbb{R}$ ↯.
