Given your data, an unpaired t test statistic (not assuming equal variances):
$$t = \frac{(\bar{x}_{\text{Vit C}} - \bar{x}_{\text{Sugar}})}{\sqrt{\left(\frac{s^{2}_{\text{Vit C}}}{n_{\text{Vit C}}}\right)+\left(\frac{s^{2}_{\text{Sugar}}}{n_{\text{Sugar}}}\right)}} = \frac{(2.775 - 11.55)}{\sqrt{\left(\frac{6.466829^{2}}{40}\right)+\left(\frac{5.44883^{2}}{40}\right)}} = -6.56,$$
with degrees of freedom (using Satherwaite's formula):
$$\nu = \frac{\left(s^{2}_{\overline{\text{Vit C}}} + s^{2}_{\overline{\text{Sugar}}}\right)^{2}}{\left(\frac{s^{4}_{\overline{\text{Vit C}}}}{n_{\text{Vit C}}-1} + \frac{s^{4}_{\overline{\text{Sugar}}}}{n_{\text{Sugar}}-1}\right)} = \frac{\left(1.0225^{2} + 0.8615^{2}\right)^{2}}{\left(\frac{1.0225^{4}}{40-1} + \frac{0.8615^{4}}{40-1}\right)} = 76$$
The $p$-value $= P(|T_{\nu=76}| \ge |-6.5|)=2.93\times10^{-9}$, so if
$H^{+}_{0}:\mu_{\text{Vit C}} - \mu_{\text{Sugar}}=0$; and $H^{+}_{\text{A}}:\mu_{\text{Vit C}} - \mu_{\text{Sugar}}\ne 0$ we would reject $H^{+}_{0}$ for even very tiny levels of $\alpha$.
Well, that's all well and good, but frequentist tests are biased to reject the null hypothesis, and we do not know if the rejection decision for $H^{+}_{0}$ was due to being overpowered for an effect size that we consider relevant. Fortunately, within a frequentist hypothesis testing framework, we can combine inference about equivalence using the two one-sided tests approach to draw conclusions that explicitly account for power and effect size.
Let's say that the minimum difference between $\bar{x}_{\text{Vit C}}$ and $\bar{x}_{\text{Sugar}}$ that we would accept as relevantly large is, say 2 (you might actually care about a different minimum relevant effect size), and let's call this number $\Delta$. We can then pose:
$H^{-}_{0}: |\mu_{\text{Vit C}} - \mu_{\text{Sugar}}|\ge \Delta$, which translates into two one-sided test null hypotheses:
$H^{-}_{01}: \mu_{\text{Vit C}} - \mu_{\text{Sugar}}\ge \Delta$, –or–
$H^{-}_{02}: \mu_{\text{Vit C}} - \mu_{\text{Sugar}}\le -\Delta$
We can reject $H^{-}_{0}$ by rejecting both the two one-sided t tests associated with $H^{-}_{01}$ and $H^{-}_{02}$:
$$t_{1} = \frac{\Delta - (\bar{x}_{\text{Vit C}} - \bar{x}_{\text{Sugar}})}{\sqrt{\left(\frac{s^{2}_{\text{Vit C}}}{n_{\text{Vit C}}}\right)+\left(\frac{s^{2}_{\text{Sugar}}}{n_{\text{Sugar}}}\right)}} = \frac{2 - (2.775 - 11.55)}{\sqrt{\left(\frac{6.466829^{2}}{40}\right)+\left(\frac{5.44883^{2}}{40}\right)}} = 8.05,$$
and
$$t_{2} = \frac{(\bar{x}_{\text{Vit C}} - \bar{x}_{\text{Sugar}}) + \Delta}{\sqrt{\left(\frac{s^{2}_{\text{Vit C}}}{n_{\text{Vit C}}}\right)+\left(\frac{s^{2}_{\text{Sugar}}}{n_{\text{Sugar}}}\right)}} = \frac{(2.775 - 11.55)+2}{\sqrt{\left(\frac{6.466829^{2}}{40}\right)+\left(\frac{5.44883^{2}}{40}\right)}} = -5.067072.$$
The $p$-values of both these test statistics are obtained in the right tail, so $p_{1} = P(T_{\nu=76} \ge 8.05) = 4.25\times 10 ^{-12}$ and $p_{2} = P(T_{\nu=76} \ge -5.067072) = 0.9999986$. Since we do not reject both $H^{-}_{01}$ and $H^{-}_{02}$, we do not reject $H^{-}_{0}$.
We can then draw conclusions based on both our test for difference (i.e. test of $H^{+}_{0}$) and our test for equivalence (i.e. test of $H^{-}_{0}$), and because we rejected the former but not the latter we conclude a relevant difference (given $\alpha$ and $\Delta$). Because the difference test statistic is negative, we can indeed conclude that the group treated with Vitamin C improves its performance on IQ tests significantly (and relevantly) worse than the group given Sugar.
Your comment "My stats teacher told me that sugar water is fine to use as a placebo with this virtual population, so let’s assume that it is not a sugar rush causing it." is telling: you have made a decision a priori to evaluating the evidence, so one wonders why you are bothering to attempt statistical inference, when you have already decided your answer?
