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minutes of videosWrite structural formulas for all the constitutionally isomeric compounds having the given moleclular formula.

(c) C_{2}H_{4}Cl_{2}

So, now we've got C2H4Cl2 and guys we can just go ahead and start off with our IHD minus H, and here we need to worry about our halogen, right? So we know that hydrogens are equal to one but so are halides, okay? So we can say that halides are also equal to what? nitrogen, just going to throw this out, that is equal to negative one and oxygens are equal to zero, okay? So let's just go ahead and fill this in, so we've got two n, so that's 2 times 2 plus 2 minus, our H is four plus 2, right? since our halides count as one as well and this is all divided by 2 and I'm going to go ahead and shift this over a little bit, so I can have some room, technical difficulties, sorry about that, okay? And then we equal this, right? That's 2 times 2 plus 2, that's 6 minus 6, so that's going to be 0, so we've got no PI bonds and no rings. So, super easy, so now we just can go ahead and say, okay, we've got two carbons there and we can do 2 different arrangements with our two chlorines here and then we can also do one with a chlorine on one side and with on the other, okay? The rest, right? are hydrogens, so we know that we can say over here, on the Left CH3, CH, Cl2, right? in fact it looks pretty similar to this already and then this one over here we can say CH2, Cl and CH2 Cl, okay? so those are both perfectly valid. Alright guys, let me know if have any questions if not let's move on.

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