Let's start with some simple calculations based only upon a few starting points:
- \$R_{_\text{C}}=6\:\text{k}\Omega\$
- \$A_v=100\$
- \$V_{_\text{CC}}=12\:\text{V}\$
- \$v_{_\text{PP(IN)}}=20\:\text{mV}\$
All of the above are taken from your schematic.
It's easy to compute \$v_{_\text{PP(OUT)}}=20\:\text{mV}\cdot A_v=2\:\text{V}\$. In addition, \$V_{\text{CE}_{_\text{MIN}}}=1\:\text{V}\$ in order to avoid saturation. So at best you only have about \$9\:\text{V}\$ left over. Let's assign most of that to \$R_{_\text{C}}\$ such that \$R_{_\text{C}}\cdot I_{_{\text{C}_Q}} = 8\:\text{V}\$ or that \$I_{_{\text{C}_Q}}=1.\overline{3}\:\text{mA}\$. But ignoring \$r_e^{\:'}\approx 20\:\Omega\$ for now, this means that with \$R_{_\text{E}}=60\:\Omega\$ we find \$V_{_{\text{E}_Q}}=80\:\text{mV}\$!
There is no possible way to design for that, assuming real parts that you can grab out of a bag. Even a \$50\:\text{mV}\$ difference between one BJT and another would mean more than a 50% shift in the quiescent operating point! And a 50% shift in the wrong direction means that \$R_{_\text{C}}\cdot I_{_{\text{C}_Q}} = 12\:\text{V}\$ instead and there's no room left over for anything else. The entire power supply is gobbled up. And things are likely to be much worse.
And I haven't even gotten to \$r_e^{\:'}\$'s impact on the voltage gain -- which is substantial -- or the fact that ambient temperatures from one place to another in the world can vary widely enough to completely ruin the design, even assuming every single BJT from a bag were identical (which they are not.)
For any chance at all in just getting parts out of a bag and having a circuit operate reasonably similarly over both variations in ambient temperature and also variations in BJT parts, you will need \$V_{_{\text{E}_Q}}\ge 500\:\text{mV}\$ (and likely still more than that to be safer.) But this means that \$R_{_\text{C}}\cdot I_{_{\text{C}_Q}}\ge 50\:\text{V}\$!!! So we are looking at a very large power supply rail and a BJT that can stand off that voltage, too! (2N5551?) We are already now starting to think in terms of \$I_{_{\text{C}_Q}} \approx 10\:\text{mA}\$, too! The only nice thing about all this is that \$r_e^{\:'}\approx 2.5\:\Omega\$ now and that means that \$R_{_\text{E}}\approx 56\:\Omega\$ could work here.
Let's do that design and see where it goes. New specs are:
- \$A_v=100\$
- \$R_{_\text{C}}=6\:\text{k}\Omega\$
- \$R_{_\text{E}}=56\:\Omega\$
- \$I_{_{\text{C}_Q}} = 10\:\text{mA}\$
- \$V_{\text{CE}_{_\text{MIN}}}\ge 4\:\text{V}\$
- \$v_{_\text{PP(IN)}}=20\:\text{mV}\$
- \$v_{_\text{PP(OUT)}}=2\:\text{V}\$
- \$T_{_\text{MAX}}=55^\circ\text{C}\$ and \$T_{_\text{MIN}}=-20^\circ\text{C}\$,
From this, I find \$V_{_\text{CC}}\ge 65\:\text{V}\$ to be good enough. I'll use a 2N5550 BJT for this, as it has sufficient \$V_{_\text{CEO}}\$.
I don't know the \$\beta\$ of any given part, but the above datasheet says that when \$I_{_{\text{C}_Q}} = 10\:\text{mA}\$ then \$\beta_{_\text{MIN}}=60\$. So that means we design for that base current. Also, I think that we should use, based upon the charts in that datasheet and considering the temperature range at hand, that \$V_{_{\text{BE}_Q}}\approx 750\:\text{mV}\$.
We can find that \$V_{_{\text{E}_Q}}=56\:\Omega\cdot 10\:\text{mA}\cdot\frac{1+\beta=60}{\beta=60}\approx 570\:\text{mV}\$ and that \$V_{_{\text{C}_Q}}=65\:\text{V}-6\:\text{k}\Omega\cdot 10\:\text{mA}= 5\:\text{V}\$.
\$R_1\$ and \$R_2\$ should be rated to pass \$10\times\$ the base current for a stiff biasing pair. So \$R_2=\frac{570\:\text{mV}+750\:\text{mV}}{10\cdot\frac{10\:\text{mA}}{\beta=60}}= 864\:\Omega\$ and \$R_1=\frac{65\:\text{V}-\left(570\:\text{mV}+750\:\text{mV}\right)}{\left(10+1\right)\cdot\frac{10\:\text{mA}}{\beta=60}}\approx 34.73\:\text{k}\Omega\$.
Dropping down to a standard value I'd get \$R_2=820\:\Omega\$ and given that I expect \$\beta\$ to be a bit better than the absolute minimum I'd choose to round upwards so that \$R_1=39\:\text{k}\Omega\$.
We are still cutting things thin because \$V_{_{\text{E}_Q}}\approx 570\:\text{mV}\$ and this still isn't really a good enough margin (I'd like at least twice this much and actually more like four or five times more.) But at least we've a chance in the design.
Let's see what LTspice says:

Okay. I consider that a good result. I still feel this is a bit dicey. And I'd still expect some trouble if I were to put this onto a protoboard. (I've not accounted for bulk impedances for the 2N5550 at the emitter, for example, and I don't have the voltage margins I'd like at the quiescent emitter. But at least I feel I'd have a shot at it.)
As you can see, the voltage gain is very close to what's desired, too!
But I think you can now see that trying to get \$A_v=100\$ in a single BJT stage is 'difficult,' at best.
I haven't yet talked about THD and distortion or the use of global NFB. These are whole other topics. But I think the point remains. With only \$12\:\text{V}\$ to work with and only a single BJT stage to work with, high voltage gain isn't in the cards.