I am looking for an extension $F/\mathbb{Q}$ with the following properties:

- $F/\mathbb{Q}$ is Galois with $\mathrm{Gal}(F/\mathbb{Q}) \simeq A_4$.
- $F$ is totally real.
- The prime $2$ has full decomposition group.
- For every odd prime $p$ that ramifies in $F/\mathbb{Q}$, the primes above $p$ all have decomposition group of odd order. (In other words, every such prime is totally split in the biquadratic sub-extension cut out by the subgroup $V_4$ of $A_4$).

Using John Jones' number fields database http://hobbes.la.asu.edu/NFDB/ I found 7670 totally real quartic extensions of $\mathbb{Q}$, each of which with normal closure a totally real $A_4$-extension of $\mathbb{Q}$. Using Magma, I found that 792 of these also have full decomposition group at 2. But then none of these also satisfy the last condition.

Is there some reason why no such extension exists, or is it just the case that I haven't looked hard enough?

Note that if no such extensions exist, the reason cannot be purely local in nature. For example, let $F$ be the Galois closure of $x^4 - 108x^2 - 136x - 8$. Then $F/\mathbb{Q}$ is a totally real $A_4$ extension with full decomposition group at $2$. The other primes dividing the discriminant are 17 and 241. The primes above 241 behave in the way that I want, but those above 17 do not. It appears that something similar always happens, i.e., there is always at least one odd prime dividing the discriminant that doesn't behave in the way that I want.

Also note that is there is only one $A_4$-extension of a $p$-adic field (for any $p$), and this is an $A_4$-extension of $\mathbb{Q}_{2}$. That means that no odd prime can have full decomposition group and that there is no local obstacle to the condition on $2$. Moreover, the unique $A_4$ extension of $\mathbb{Q}_2$ has a cubic unramified subextension, so any $A_4$ extension of $\mathbb{Q}$ with full decomposition group at 2 must also be ramified at (at least) one odd prime.