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 (b) 1-cyclohexylpropan-1-ol

Accepted Answer

Everyone and welcome back to another video. In today's lesson, we are going to identify which carbonell compound and the reducing agent can be used to synthesize one, fennel Ethan one. Oh now if you look at the structure form, this is a secondary alcohol, right? Because the alpha position that is directly bonded to the alcohol group has to alcohol substations, one is a muscle group and another one is a final substitution. Now, what we're going to do because as the problem suggests we're going to use a reduction reaction. We're going to reduce our starting material, which is the unknown to a secondary alcohol. We want to make sure that we're thinking in reverse. What if we oxidize it, recall that oxidizing secondary alcohols produces key songs. The general reaction scheme can look as follows, suppose that you have a secondary alcohol with two alcohol suspicions R one and R two. Suppose you're oxidizing it using a strong oxidizing agent and that's perfectly fine. You can use a mild one as well. But the most important part is that if you oxidize it, you get a kid zone. So you're essentially removing this hygiene and you're forming a double bond with oxygen and these the original al kill groups. So you're removing a total of two hydrogen atoms, one from here and another from here and then forming that double bond. So why don't we do that? We can essentially remove that hygiene form a double bond at this position and remove that hydrogen, add the position in red, which is an implicit hydrogen to form our kids own. Well done. So if we understand that we can simply draw our starting material as benzene ring Bonded to carbon # one, which is now a key tone. Right? So we are going to add one metal group as well. And then this is our car panel group that we have formed during our oxidation. So this makes sense. Right. We have turned our secondary alcohol into its own. Everything that we want to do now is just to form our alcohol and we want to suggest an appropriate reducing agent for that transformation. So if we want to form our alcohol which is our secondary alcohol, we want to think about what, what what kind of a reducing agent we can use. And there are two reducing agents that we are familiar with. One of them is lithium aluminum hydride which is strong. Right? And another one which is mild that would be or sodium world hydrate which is mild. Now there is no difference which one you use actually because key zones can be reduced with both of them. sodium World hydride definitely reduces Allah heights and key tones to alcohol's listening. Um, aluminum hydride reduces nearly all of the carbon compounds. Right? So it's a much stronger reducing agent while sodium hydride is more selective towards carbon compounds that it can reduce. So it doesn't really matter which one you use. I would personally choose a milder one because there's no necessity to use a strong one when a milder one does the job well. So we can simply say that there are two possibilities for this reaction to take place. We are starting with a key tone on the left hand side, and we are using either lithium aluminum hydrate or sodium boron hydride to perform that reduction reaction from a key tone to the corresponding secondary alcohol, meaning the correct answer choice to this multiple choice question is D 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 (b) 1-cyclohexylpropan-1-ol

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