Suggest carbonyl compounds and reducing agents that might be used...

Question

Suggest carbonyl compounds and reducing agents that might be used to form the following alcohols (a) octan-1-ol

Accepted Answer

Hey everyone and welcome back to another video. In today's lesson, we are going to identify which carbon compound and the reducing agent can be used to synthesize benzel alcohol. Now, as the problem suggests, we are forming mental alcohol. And the most important thing is to understand that this is a primary alcohol and the reason why it's primaries because the alpha carbon, that is a record bond to the alcohol group has one AL kill neighbor. So this is a primary alcohol and we want to use a reducing agent. So we want to perform a reduction reaction. Reduction reaction essentially goes from left to right, just as the arrow suggests. Right. But what if you think about your starting material, you have to suggest what kind of starting material you can use in this reaction. So if you go in reverse, you would use oxidation, meaning you essentially want to oxidize your primary alcohol into your starting material to get that starting material that can participate in reduction. Now, let's recall that oxidation of primary alcohols. That's what we have. Producers alga hides. So this piece of information is really vital for us. If we understand that oxidation of a primary alcohol gives us an AL to hide, we are good. So how do we turn an AL to hide out of it? We are going to remove this hydrogen and that hydrogen on that adjacent carbon atom to get our double bond. So that said, we're just creating that carbon auction double bond, getting rid of this hygiene and that hygiene while one of the implicit hye jin's stays right, there's still one more left. There are initially two of them and we are going to remove this one and that one form our carbon steel and that's it. So let's draw the resultant at the hide. Now we are going to draw our benzene ring, There is one carbon atom that carbon atom now turns into a carbon steel group, C O double bond. And as we said before, we are removing this hygiene as well. So there is one hygiene left and this is indeed our Ben's Aldo hide. So if we use this out of the hide, we can easily turn it into an alcohol using a reduction reaction. This is our primary alcohol as we previously discussed, oxidation of primary alcohols produces Aldo heights. And this is what we are seeing here, right? Meaning the reduction of Aldo heights produces primary alcohol. So you can think in reverse. So we can also add this as our general statement. That may be helpful for you reduction of Aldo heights. We can say our ch O produces primary alcohols are C O H. See that would be how many hydrogen atoms. So that would be a judge too. Okay, so this will be our general form of the alcohol forms. And we see that this scheme indeed works for us. We notice that we are starting with an AL to hide. We're reducing it into a primary alcohol. We essentially just want to think about what set of regions we can use here. So, if we're performing a reduction reaction, we may recall that their t major reducing agents. One of them is lithium aluminum hydride, which is strong and it reduces the majority of carbon compounds. and number two is sodium borrow hydride, which is mild and it only reduces Aldo hides and keystones. Because we are starting with an elder Hyde, it's it's really sufficient to use sodium carbohydrate. But we can also use a strong oxidizing agent because it reduces everything. So we can say lithium aluminum hydride might might be our Regent or sodium Borough high Tribe. You can choose any of the two. That's perfectly fine. However, in my opinion, if it's just sufficient to use a mild reducing agent, I would just go with a mild reducing agent. It's definitely not worth using a strong reducing agent when you can use a mild one and it's already sufficient to perform that reaction well. So it's perfectly fine. Once again, to use any of the two, but I would go with sodium carbohydrate. Any of the two options are correct and we may label these our final answers Ben's Aldo hide is our starting material and the reaction conditions are lithium aluminum hydrate or sodium world hydrate to perform that reduction of an alga hide into a primary alcohol, meaning the correct answer choice to the small pool choice question is C however, if you have any questions, don't have states, leave your comment down below in the comment section and thank you for watching.

Suggest carbonyl compounds and reducing agents that might be used to form the following alcohols (a) octan-1-ol

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