What is the reason for the $ i \tau_2 $ - factor in the higgs coupling with up-type quarks? The quark mass term in the Standard Model Lagrangian looks like this:
$$ L = - \lambda_d \bar{Q}\phi d_R - \lambda_u \bar{Q} i \tau_2 \phi^* u_R $$
What is the reason for the  $ i \tau_2 $  - factor? What keeps me from writing undesirable things like  $$\lambda_u \bar{Q}  \phi u_R $$ ?
 A: The reason is that the $SU(2)$ invariant in $\mathbf{2}\otimes\mathbf{2}$ (or in their complex conjugate $\mathbf{2}^*\otimes \mathbf{2}^*$) is given by contracting the two $\mathbf{2}$ with the anti-symmetric $2\times 2$ matrix $\epsilon_{ab}$, as $i\tau_2$ is. In the case at hand the two $\mathbf{2}^*$ are $\bar{Q}$ and the $\Phi^*$. You could form another $SU(2)$ invariant by multiplying $\mathbf{2}^*\otimes \mathbf{2}$ (this time with the $2\times 2$ kronecker $\delta^a_{b}$) that is $\bar{Q}$ and $\Phi$. But you would not form a singlet under hypercharge by adding $u_R$ (which is $SU(2)$ singlet but hypercharged $2/3$). On the contrary, adding $d_R$ (which is $SU(2)$ singlet but hypercharge $-1/3$) you do get an invariant that is the down-type Yukawa interaction.
A: The Standard Model Yukawa interactions must be $SU(3)\times SU(2) \times U(1)_Y$ gauge invariant. The down-type Yukawa interaction is
$$
\mathcal{L} \supset -y_d \bar Q \phi d_R + \text{h.c.}.
$$
This is indeed gauge invariant. The $\bar Q d_R$ form a colour singlet ($3^* \times 3$), the $\bar Q \phi$ form an $SU(2)$ singlet ($2^*\times2)$, and the whole thing is neutral under $U(1)_Y$, because the quantum numbers are $-\frac13$, $1$ and $-\frac23$, for $\bar Q$, $\phi$ and $d_R$, which sum to zero.
Let's try writing a similar up-type Yukawa
$$
\mathcal{L} \supset^?_? -y_u \bar Q \phi u_R + \text{h.c.}
$$
Is it gauge invariant? No - it breaks $U(1)_Y$, because the quantum numbers are $-\frac13$, $1$ and $\frac43$, for $\bar Q$, $\phi$ and $d_R$, such that $Y=2$.
To fix this problem, we might try
$$
\mathcal{L} \supset^?_? -y_u \bar Q \phi^* u_R + \text{h.c.}
$$
This is $U(1)$ invariant because the hypercharge of $\phi^*$ is $-1$, so we have $-\frac13-1+\frac43=0$, but now it is no longer $SU(2)$ invariant ($2^*\times2^*$). Now we use the property that

$i\tau_2\phi^*$ trasforms in the same way under $SU(2)$ as $\phi$

Finally, we can write
$$
\mathcal{L} \supset -y_u \bar Q i\tau_2 \phi^* u_R + \text{h.c.}
$$
which is indeed fully gauge invariant.
