 Continuing your journey of comparing the rates of various substrates towards an SN2 reaction, let's just get a bit more interesting and compare the stability of the attacking nucleophile and the leaving room. Let's do it. Which of the following would undergo substitution via SN2 mechanism when treated with Cl-ion? Okay, there is SN2 mechanism happening. The incoming nucleophile is chloride ion for each of them. Try doing this problem yourself and then we'll do it together. Okay, the leaving group in the first case is OH- while in the second it's I- and the third it's NH2- Wait, I know, I know, I- has the largest size. So the larger the size of the ion, the more spread out the charges, the lesser is the charge per unit area and the more stable the ion is. So could the answer be option B? Wait, wait, wait. Here we are not supposed to compare the leaving group amongst themselves. Look at the question carefully. They ask us that which of the following would actually undergo substitution when treated with Cl-ion? So what we really need to do here is compare the stabilities of the attacking nucleophile and the leaving group. If the leaving group is more stable than the attacking nucleophile, why would it leave? Wouldn't it just attack back instead and just make the attacking nucleophile leave? So what we really need to compare is OH- and Cl-, I- and Cl- and NH2- with Cl-. In the first case chloride ion has a larger size and therefore the larger the size of the ion the more spread out the charges, the more stable is the anion. Oh wait, the attacking nucleophile is more stable alone as compared to the leaving group. OH- wouldn't want to leave? Nah. Similarly, in the third case, nitrogen is smaller than chlorine, right? So the negative charge would be more stable on the chlorine atom. NH2 would also not want to leave as NH2-ion. What are we left with? Let's try and compare I- and Cl-. Iodide ion has a larger size than the chloride ion. Okay, the negative would be more stable on the iodine atom. And therefore the less stable chloride ion can attack and expel or replace the more stable iodide ion. And therefore the one that would actually react with chloride ion would be the second one. Let's go to the last question for the day. Here, which amongst the following will not react via SN2 mechanism when treated with OH-? Focus on not reacting. Okay, here we don't care about comparing these leaving groups with each other. But instead we care about comparing them individually with the attacking nucleophile. If the leaving group is less stable individually as compared to the attacking nucleophile, it wouldn't want to leave. So try doing it yourself first before we attempt it together. Okay, so we need to compare OH- and SH-, OH- and NH-CH3- where the negative charges on the nitrogen atom and OH- and Br-. In the first case, I see how the negative charge on the sulphur would be more spread out as compared to the negative charge on the oxygen atom. Because sulphur has a larger size, so SH- would be more stable than OH-. The leaving group being more stable than the attacking nucleophile. The leaving group would leave. Similarly in the third case, the bromine atom has a larger size than the oxygen atom. The negative charge would be more stable on it. So the bromide ion being more stable than the attacking nucleophile OH- ion would happily leave when the substrate is attacked by OH- ion. They both belong to the same period. Hey, electronegativity. Oxygen is more electronegative than nitrogen. That means it can keep the electron density towards itself in a much better way as compared to the nitrogen atom. So the attacking nucleophile is able to keep the electron density in a more stable way as compared to the leaving group. The leaving group wouldn't want to leave and therefore the one substrate that does not undergo SN2 reaction when treated with OH- would be the second one.