What is the product of the reaction of an ester with excess acetylide ion followed by the addition of pyridinium chloride?

Hey everyone. And welcome back to another video. Today, we are going to solve another problem when the following ester is treated with excess Aite Ron and then Perini chloride. What type of product is expected? We're given the structure of the which is methyl ethanolate. And we're given for answer choices. A a secondary alcohol B A tertiary alcohol C A ketone and D A carboxylic acid. Now we are going to redraw the starting material and we are going to think about that reaction. So we have our right. So we are going to draw an a satellite ion in a general case that will be an group R bonded to our triple bonded carbon. And then we would have a negative charge at the terminal S P carbon. So now the very first step is that because we have an electrolytic position within the carbon group, we will observe a nucleic attack followed by the cleavage of the pi bond. Then the negative charge on oxygen can actually re reform the pi bond, right? The lone pair can be brought back into a pi bond because we have a good leaving group which is metox. Also recall that a satellite is a good nu file and it is a better nu file than metox side because the negative charge is on a less electron negative carbon as opposed to more electron negative oxygen. So carbon more readily donates electrons into the bond formation. That's how we form our first intermediate. So we are going to get our carbon group. And now, instead of metox side, we actually have our triple bond and then we have our, our group because the problem did not specify what sort of a satellite we're using. Exactly. Now, the next step because we use an excess of a satellite will be another nucleic a tag because once again, we have our carbonyl group, we attack the carbonyl group, we're breaking the pie bond and now there is no good leaving group nearby. So we are going to form an oxide anion and we can simply draw our tetrahedral intermediate. Now, so that would be oxide, we're forming a negative charge on oxygen. Then we have our metal group and then we have our triple bonded carbons. So let's draw the two alk groups. OK. And now essentially the very last step as the problem suggests, we are going to use perineum chloride. Let's remember that podium chloride essentially corresponds to a six member Drang where we have nitrogen, right? And essentially in nitrogen, we will have a proton, we have a positive charge on nitrogen because it forms a total of four bonds, chloride is not bonded directly, right. We are just have, we just have a salt compound. And the most important part is actually nitrogen, which has a proton that it can donate. We're using podium chloride for the proton transfer to al oxide to form an alcohol instead of any strong acid because we essentially have triple bonded carbon atoms. And we do not want to interfere or perform any reaction with our Alkins. So we want to make sure that we're using a weak acid to proton al cox side. And as a result, we can actually draw our final product. Now, so oxide becomes an alcohol after the proton transfer. And we may notice that we get an alcohol with three alco substitutions which is a tertiary alcohol. And therefore our correct answer is choice B a tertiary alcohol is formed in the given reaction. Thank you for watching.