Show how you would synthesize the following alcohol by reducing a...

Question

Show how you would synthesize the following alcohol by reducing appropriate carbonyl compound.

(c) 2-methylhexan-3-ol

Accepted Answer

Hey everyone. How can three metal pen tend to all be synthesized by reducing a carbon compound. Let's think about this alcohol. According to the problem. We have three metal pens hands to you. Oh, so we can just draw that alcohol now pence. Inal right, that's what it says. We have a parent with five carbon atoms. So let's draw it at five carbon atoms. The position of the alcohol group is at carbon number two and position number three contains a metal group. So this is how our alcohol looks like. We want to synthesize it out of a carbonell compound by reducing it. If we want to use a reduction reaction, we may recall that there are two reducing agents to reduce carbon compounds. So we are generally familiar with any of them. Right. In this course we are interested in number one sodium world High Tribe which is a mild reducing agent. Right? And be lithium aluminum high Tribe, which is a strong one. If you notice we are actually forming a secondary alcohol and you may recall that we may reduce key tones. So we need to have a key tone using either a lithium aluminum hydride or sodium laurel. High Tribe. And reducing key zones where these regions would lead us to the formation of an alcohol. So we would say we would get our secondary alcohol, that's how it's done. So if we recall this basic approach, right, we understand that reduction of ketones produces secondary alcohols. We understand that. Okay, we just want to form a key tone out of it. So our starting material would be key tone. We can just draw our five member chain and turn this alcohol and say keystone. So that be our carbon group, there will be an adjacent metal group. And we are done. We have our starting material. The reduction of this ketone will lead to the formation of a secondary alcohol at the same position. Now we already know what reducing agents we can use and we can use any of them lithium aluminum hydride, which is a strong one, or sodium boron hydride, which is a mild one. Now if you ask me, I would just go with sodium boron hydride. There is no difference which one of them we use but there's no necessity to use a strong one. When a mild one can see the same job, right? So if you think about this reaction, we are going from right to left and our starting material is the following key zone, which is then reduced with either a lithium aluminum hydrate or sodium or hydrate. To give us this alcohol, let's label this is our reaction scheme. Keep in mind that program from right to left because we started by Drawing our final product 1st. So we kind of reverse it right. If you want to look at the reaction, you can just redraw it with our key tone being the starting material, treating it with any of these two regions and then getting our secondary alcohol. That said the correct answer choice to the small poachers question is B, however, if you have any questions, don't hesitate to leave your comment down below and thank you for watching.

Show how you would synthesize the following alcohol by reducing appropriate carbonyl compound. (c) 2-methylhexan-3-ol

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