Transistors replacement
I wanted to add a few thoughts about your schematic, itself, as well. There are a few questions and problems that I see (as a hobbyist.)


You don't specify what your input signal looks like. Without that information it is difficult to say much at all about the circuit. The gain shown is about \$100\times\$, which may be too high, too low, or just right. No way to tell. But for now, let's assume it is just right and move on.
Assuming you really do need that gain and that the BJTs are driving close to the rails (let's say no closer than \$1V\$, though, so that they stay out of saturation), they each will be delivering about \$P_{load}=\frac{V_{peak}^2=5V^2}{2\cdot 8\Omega}\approx 1.5 W\$ into the load speaker. Since optimal delivery of power into a load requires a source impedance that is about the same, an efficient linear amplifier design can at best hope to deliver \$1.5 W\$ by itself dissipating the same amount. So each of your output BJTs will be picking up at least \$P_{bjt}=\frac{1}{2}\cdot 1.5W = 750mW\$. This will require packaging that is NOT a TO-92, but instead a TO-220 style.
Each BJT might be expected to supply \$I_{C_{peak}}=\frac{V_{peak}=5V}{8\Omega}=625mA\$. You also need to make sure that the selected BJTs are specified to supply \$I_C\ge 1A\$. \$I_C=2A\$ would be even better to see on the datasheet.
The base current for each BJT will be, because they are staying out of saturation (per my earlier estimation above), on the order of a factor of \$\beta \ge 50\$ lower. So you must expect that the opamp itself can supply at least \$20mA\$ at its output, just to be safe. The LM741 has a short-circuit output of about \$25mA\$. So you are near the margins here. But I think it will be okay because although I estimated \$\beta \approx 50\$ here, you may get as much as \$3-5\times\$ better than that depending on the output BJT and its operating point. So let's assume it is okay for now.
The LM741 also has a problem reaching its rails. Let's say your LM741 is required to supply \$I_{peak}=10mA\$ to your BJTs. Assuming it can reach to the rails (it cannot even come close to that), this implies an effective load of about \$\frac{6V}{10mA}=600\Omega\$. However, the datasheet says that even with a lighter load of only \$2k\Omega\$ the best you can be absolutely sure of is that it can come within about \$5V\$ of a \$15V\$ rail, or about \$10V\$. They do say that it typically can get closer -- perhaps within \$2V\$ of the rails. But no matter how you look at it, this opamp isn't going to come close to your rails. So this means you cannot expect to see even \$V_{peak}=5V\$ on your speakers. If we go with a drop of \$3V\$ at the opamp output, plus another diode drop for the BJT emitter, you are looking at a maximum \$V_{peak}=2V\$ or so. This suggests only \$\frac{1}{4}W\$ of output. And an even smaller input signal acceptance, given the gain programmed into your circuit. On the good side, this means that your BJTs could just be TO-92 packaged parts and you can use small signal devices there (assuming they can still provide \$\frac{2V}{8\Omega} = 250mA\$ of collector current.) But only if you changed your voltage rail to the BJTs. If you can't (and I have no reason to believe you can), then those BJTs will still dissipate the same as before. So forget the small signal BJT entirely without changing your whole design and goals. This means you might want to look for a different, rail to rail output type, opamp.
Now that we've decided that your output is sorely limited to a much smaller output power, your circuit's very significant cross-over distortion will now be terrible and, I'm pretty sure, quite noticeable. You really need a different arrangement to drive those two BJTs, stuck with that LM741. Here I'm thinking of a pair of diodes (optionally tied thermally to your BJTs) and using the opamp to sink a current mirror's current through them.
You may want to include a small capacitor to help stabilize your voltage divider's mid-point value. Given the substantial input currents used by the LM741, your resistor divider impedance will probably have an impact on the output, as well. I may consider putting a cap there.


I can provide a more specific arrangement for you, based on your existing idea but where these details are accounted. But given that I have no idea what you are using to feed the amplifier as a signal, no idea what kind of 12V power source you are using, and no idea what kind of output power you want, it's really hard to do much more than the above commentary.

ADDED:

Just by way of reference, here's a first improvement step. It's not complete. But it's better. I've left off some of the specific quantities or device numbers, as that's not my point. But I've used specific values to get across some ideas:







simulate this circuit &ndash; Schematic created using CircuitLab

The current mirror uses a pair of BJTs that are thermally linked (in the same package) as a BCV62. With the value of \$R_{set}=3.3k\Omega\$ (probably the largest value you should try, as it produces the lowest viable set current you should use), the current available for the output BJT bases will be about \$3.2mA\$ or so. This is enough to keep the diodes with active drops, as well. (Those diodes should probably be thermally coupled to the two output BJTs, as well.) The opamp should be one capable of getting down to within \$\frac{1}{2}V\$ of its (-) rail, or better. (The upper rail has a lesser requirement due to the two diodes.)

This isn't a design. It's just a direction to head to help increase your output power to the load and to reduce (or nearly eliminate, if the BJTs can both be kept ON throughout it) cross-over distortion. It's gain here is set to \$\frac{R_{fb}}{R_{in}}=10\$. This is not what your original circuit shows. It's just a value I chose, for now.