Predict the major products of the following reactions.
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Question

Predict the major products of the following reactions.

(a) 2,4-dinitrochlorobenzene + NaOCH₃

Accepted Answer

Hey everyone and welcome back to another video. In today's lesson, we are going to identify the correct major product formed when for nitro Bmo benzene reacts with sodium dioxide. First of all, if you drove for nitro bromo benzene, we understand that this would be a benzene ring with two substitutions. As it implies, there is a nitro group and there's also a bromo supposition. So that's where nitro that's position number four, meaning the bromo supposition would be at position number one. Okay, so we got our structure and we understand that the reaction takes place in the presence of sodium with oxide. It's just sufficient to draw teeth oxide itself because sodium is an ion spectator, as you recall from general chemistry. So that's pretty much everything that we need to know. Now the first thing we keep in mind is that nitro group over here is an electron withdrawing group. Which means that this carbon blooming bond becomes weaker than usually because this bond becomes weaker, it's just easier to get rid of blooming out of our structure so we can perform in nuclear attack at this position. Let's simply take a look at the mechanism of this reaction to correctly predict the product. So the negative charge on the stock side would attack this electrolytic position. And then due to the presence of the adjacent double bond, we would be able to transfer these pi electrons in the form of a lone pair on the adjacent carbon atom. So we made to our first intermediate that we got our first intermediate would be that bromo substitute. Now we have bonded s oxide. So you can say that's our O. E. T. Thinking about the positions of our double bonds there too that we had initially. There's no more double bond over here. Instead we are getting our negative charge because there's a lone pair on that carbon. Now, of course we have resonance stabilization because their pi electrons within the ring and the nitrile group itself. So let's see which residents structures we get. First of all, we may notice that we may use these electrons and form a double bond a disposition and transfer these pi electrons onto the adjacent carbon. So our next resident structure would be this one we made through our ring. Then we have two double bonds of these positions bromo group is oxy. Now, for the purposes of this problem, we may already expand the nitrile group bonds and we can say that this is an 001, oxygen is negatively charged. Nygren has a formal positive charge. Right? And then we have a negative charge at this position over here. So let's indicate that and we continue our residents. The next resident structure would essentially imply that these are the presence of that nitro group. We can place these electrons in the form of a newly formed pi bon and break the adjacent pipe on over here, placing the pi electrons onto oxygen. Right? So this will be our next residence form. Let's draw it out. Starting with the ring itself, there are still two double bonds at the same positions, bromo. It's oxy. Now here we are forming a double bond to that nitrogen and there are two oxygen atoms with negative charges. Okay, so now what's happening next? We can essentially take the lone pair of box. General restore the double bond and then we can shift these pi electrons in the into the formation of that carbon carbon double bond. Right? But that would require us to break this adjacent pi bond and place the result on electrons in the form of a lone pair on that carbon. So that's how we get our next residence form. And we can show it as follows. So if we draw our structure, we will notice that there will be a negative charge over here. Now, wonderful bond over here here, that original bromo group and keith oxy. And then we have our nitro group. We don't need to expand it anymore because we're not interested in the bonds. The bonding is back to the original bonding now within the nitro group. Now, if you think about it bromo group is a good leaving group and this is where it takes place. We can essentially use the slum pair to restore our benzene ring, followed by the loss of the leaving group bromo group is a good leaving group and that would give us our final product. We can just say that based on that structure. Our final product would look as follows. This is our benzene ring. There's no more bromo substitute anymore. All that is left is the ith oxy group itself. And then at position number four we have our nitro group. That said this would be our major product. If if we look at the answer choices we want side and find that product within our answers and we want to find the one that has Nitro group and anyth oxy group, positions one and four. Not this one not this one that has promo This one is the one that we are interested in. Right? So it has anyth oxy group. We just used an abbreviation that's oh et because there's oxygen bonded to an ethnic group. So that's Toxie. And with respect to each oxy group. At position number four, we have our nitro group. That's that we may conclude that the correct answer choice to the small cultural question is C. However, if you have any questions, you're welcome to leave your comment down below in the comment section and thank you for watching

Predict the major products of the following reactions.
(a) 2,4-dinitrochlorobenzene + NaOCH₃

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Activating and Deactivating Groups

Regioselectivity in EAS Reactions

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