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8a. (14 pts) Propose synthesis (~7 steps) for the formation of the product below. Show the product of each individual step for full credit:81. (2 pts) Provide an IU...

Question

8a. (14 pts) Propose synthesis (~7 steps) for the formation of the product below. Show the product of each individual step for full credit:81. (2 pts) Provide an IUPAC name for the product of problem 8a

8a. (14 pts) Propose synthesis (~7 steps) for the formation of the product below. Show the product of each individual step for full credit: 81. (2 pts) Provide an IUPAC name for the product of problem 8a



Answers

(a) Give the IUPAC name for $\textbf{A}$ and $\textbf{B}$. {b) Draw the product formed when $\textbf{A}$ or $\textbf{B}$ is treated with each reagent: [1] NaBH$_4$ , CH$_3$OH; [2] CH$_3$MgBr, then H$_2$O; [3] Ph$_3$P=CHOCH$_3$; [4] CH$_3$CH$_2$CH$_2$NH$_2$, mild acid; [5] HOCH$_2$CH$_2$CH$_2$OH, H$^+$.

So this problem is from organic chemistry and asked us to write the name of the product obtained when one Butin reacts with hydrogen and roomy. So we'll see the reaction and see the name of the product of paint. One. Between That is CH three CH two, C. H double bond CH two. When reacted with hydrogen correction takes place in the presence of catalyst where the hydrogen is added to double bond and produces the L cane ch three CH two. See it to see it three. The name of this product is building. The next one is reaction with broom in At the CH three Ch 2, C. H double bone CH two. When Romain is added to one beauty, the reaction occurs and the double born converts into single bond. Each of the bombing is added to the double bonded carbon atom. And the product formed have this structure CH three CH two ch we are. See it to be our. This product has the name one, two dia promo building. So these are the names of the products obtained when one beauty in reacts with hydrogen and broom in

Let's solve the problem from organic chemistry. This problem discusses the names of the organic compounds formed by the reaction of organic compounds. Here we are given a substance that is to Mikhail to Beauty, we have to write the name of the product when this compound reacts with chlorine as well as I. Odin. So we'll see the reaction for this. To Mitchell to Beauty can be represented as CH three. See having CH three double bond, Ch CH three. When chlorine is really added to the substance, the double bond is converted into a single bone and the double wanted cabinet. Um Each gets a chlorine atom. So the product of pain is this CH three C. Having CH three including then C. H. Including CH three. The name of this compound is too three. Dyke Larue two metal building than we have the reaction with iodine. We can write CH three. See having CH three double bond, CH CH three plus I. Odin. So the reaction occurs in the similar manner where the double bond is converted into single bond due to the addition of iodine attempt to each of the double wounded carbon atom. The product from this CH three see having metal group and I. Odin, ch having I. Odin, CH three. We can write the name of this compound as to three. They I do two metal butane. So these are the names of the compound obtained when two metal built to in reacts with chlorine and with I. O. D.

So we give the name of the product from hydrogenation for each of the following one Heck scene means it's hexane but with the Taliban. So if we get rid of the double bond is just X sane. Part B two method to a beauty in same case here. If you do, you still keep the methods, though, to fo and keep the method, and it just becomes butane instead of beauty. And because you get rid of that double bond, part C to canteen. Same thing here set of two instead of Pantene old competiting once we get rid of

This question is simply naming of organic compounds using the pack rules. I find it helpful to draw the entire structure rather than just using the shorthand technique. But you may prefer not to. This, then, is an Al cane. So we'll name it with an Al Cane with a substitution. Romain, the longest chain. It's going to have 12345 So this is a pen tain, and we have a metal group coming off of one of the carbons and a bro mean group coming off of another carbon. So we're going to use the we're gonna number it so that we get the smallest numbers. So we'll start right here where this is carbon 1234 as opposed to doing 12 and then five. We'll have a one and a four, so this will be one bro mo four metal plantain for the next one. We have a double bond and we have in a wage group, so we will name this as an alcohol and we have our double bond coming off of the first, the second carbon. So this will be a two protein, one all compound and for the last one. We have an Esther. We have a metal group here and then off of the last carbon on the three carbon chain. We have a mean group. So this is going to be meth. This is how we name Esther's. This group first and then the rest. So this is metal three amino. 123 We have the main group coming off. Three amino for Pana. Wait. Three metal three amino for pan away. If the amino group weren't there, this would just be metal. Propane await If this were a hydrogen.


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