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Deterine the reactant that required t0 accomplish the following transformation: Enter Ihe SMILES string lor the compound_HCI H-Oheat...

Question

Deterine the reactant that required t0 accomplish the following transformation: Enter Ihe SMILES string lor the compound_HCI H-Oheat

Deterine the reactant that required t0 accomplish the following transformation: Enter Ihe SMILES string lor the compound_ HCI H-O heat



Answers

For each pair of reactants given, write the net ionic equation for the molecule-formation reaction that will occur. $$\mathrm{KNO}_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq})$$

So we have one mole of iron metal F. E. And that's a solid, is reacting with Aquarius hydrochloric acid to form iron to chloride and hydrogen gas. That's a bs f e c L two and then gashes hydrogen. We put a two hour in front of hcl, then everything is balanced. Just make the ceo look more like A. C. L. There we go. And um You evolved 89.1 killer jewels of heat. So that means the entropy change associated with this reaction is going to be negative, 89.1 killer jewels since it says evolved, that means eggs a thermic and so he is negative.

So in this example, all we need to do is cleaves the molecule at the end of the CC bond that is bound to the SB two carbons and then simply add a halogen to my fast example here. We have just cleaved here at this position, and then we've simply added or halogen as we said, which was all bro mean, we've got one more example which is label be for us on. Where we have cleaved is simply hair. We've added idea, which is also a halogen, and then our other component has maintained the double bond that is present.

In hack reactions The all group off the Vienna Click Hey, Light or the air I'll Hey, light will replace hydrogen that we're bonded to an SP two carbon on an AL Keen. So as a result, our product hair has formed a new CC bond which is joined by two SP two carbons. So in my fast example, this is our new bond and then we would break it down into its respective components. We've simply added a halogen to one off our component, which is our vinyl. Hey, light here and then we just have a proton on the other. In our second example, we've also identified where that new bond has been formed. Where again we haven't a royal hey light now, But all we need to do is add an halogen. They're considered very good leaving groups. So in reactivity, it is very easy to for have a coupling reaction while you're needing to expel SAT and groups and use a halogen, for example because they are very willing to leave and you don't need much forcing conditions with that. And then our other component just has that proton attached

In this problem, the the action will happen. Something like this. Just look at it carefully. I'm writing the reaction here, but this place one more This plus one mall, actual CH two, CH two oh, add in pageants. So SCL gas will give the product ID this compound, which I am writing here. Just look at it carefully. We'll give the product ID this component here. It is Who or and this. So according to the option here, according to the option. In this problem option we eat, option B. Eat That extends it.


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