An unknown alcohol with a molecular formula of C₇H

Question

An unknown alcohol with a molecular formula of C₇H₁₄O was oxidized to an aldehyde with HOCl. When an acidic solution of the alcohol was distilled, two alkenes were obtained. The alkene formed in greater yield was determined to be 1-methylcyclohexene. The other alkene formed the original un-known alcohol when treated with BH₃/THF followed by H₂O₂, HO^-, and H₂O. Identify the unknown alcohol.

Accepted Answer

Hello, everyone. And welcome back. Our next problem since alcohol X with the molecular formula of C six H 12 0 was oxidized to an aldehyde using NAOCL and CH three coo H at zero °C when this alcohol was distilled in H two H two. So 42 alkenes were formed. The major alkane product is one methyl cyclops, one, the minor alkene product formed the starting alcohol when reacted with number one bh three slash thfn number 2 H2O 2 ho minus and H2O. What is the identity of alcohol? X? Well, this one is definitely a scary one to look at, but if we just work our way through step by step, we will get to it. So I've underlined each of the sort of major pieces of information and then we'll highlight in blue as we deal with each thing that will help us keep track as we go through. So the first and easiest step to start with, we have a molecular formula, so we can calculate our degree of saturation or index of hydrogen deficiency that will tell us does our structure have a ring? Does it have double bonds? So that formula for index of hydrogen deficiency is open parentheses, two C plus two plus N minus H minus X low parentheses. And then all divided by two and their C represents a number of carbons. N number of nitrogens H hydrogens X halogens. So in this case, that equals put brackets here. So we have two multiplied by the number of carbons which is six plus two, no nitrogen. So no nitrogen factor there minus the number of hydrogens, which is 12 minus the number of halogens which is zero. So no, nothing there close brackets and then divided by two G equals two, multiplied by 612 plus two is 14 minus 12 is two. So we get two divided by two equals one. So we know we have either one ring or double bond. However, we note that in our whole long description here, we talk about um warming alkenes and it doesn't mention that we have an alkene in our starting product. So it doesn't mean that for sure we don't, but probably means we can start with the assumption that there's a ring. If that doesn't work out, we can go back and try an alkene. But given that our products are described as alkenes, it seems like they would mention that if we started with one. So let's start by assuming we have a ring, highlight, ring, one ring and now we've dealt with the molecular formula. So I'm going to highlight that in blue, we're done with that piece of information. So what would give us one ring and six carbons and then an alcohol? Well, before we draw that, I want to add in one more piece of information, I know that my alcohol was oxidized to an aldehyde. Well, an alcohol oxidized to an aldehyde must be a primary alcohol because secondary alcohols get oxidized to ketones. Tertiary alcohols can't get oxidized at all. And indeed, when we look at the reaction, it says it's oxidized to an aldehyde using sodium hypochlorite and acetic acid at zero degrees, which is a set of reagents that generates hypo chloric acid, which will partially oxidize our alcohol. Recall that if you completely oxidize an a primary alcohol, you'll end up with um a carboxylic acid. So this confirms this. So I'm going to highlight in blue oxidized to an aldehyde because we've used that piece of information to tell us it's a primary alcohol. So let's go back to our potential structures, one ring, six carbons, primary alcohol. So we know that this means the alcohol group cannot be a substituent on the ring or it would be a secondary alcohol. So it must be on a branch. So that means it can't be a six carbon ring. The largest ring, it can be is a five carbon ring. So let's draw that potential structure. So five carbon ring with a branch, including one carbon and then roh on the end being a primary alcohol or we could have a four carbon ring with a branch containing two carbons and then an alcohol or even a three carbon ring with a branch containing three carbons, 123 and a primary alcohol. So those are potential structures that fit the information we're given, I'm going to put little ors in between each of these. So I remember that. So now let's move on to our next piece of information when this primary alcohol was distilled in hydro sulphuric acid, two alkanes were formed and the major alkane product is one methyl cyclops one E. So let's draw that. We're calling that distillation in hydro sulphuric acid is a dehydration reaction with an acid catalyst. So let's draw our major product here. So we have cyclops one in. So we have R five carbon ring, we have a double bond within the ring from carbons 1 to 2. And then we also have also have a methyl group on carbon one. Well, this is very interesting because it doesn't look like our primary alcohol. However, there must be, so there must be some sort of rearrangement going on, but it does tell us what our original structure is because we have this five carbon ring structure. So that tells us that our first structure must be the correct alcohol. So I'm just going to put a little blue circle around this again, since we have a five carbon ring in the alkene we generate. So now I highlight in blue, the major alkene product and its structure. And the last piece we need to deal with is it says the Miner Alkin product formed the starting alcohol when reacted with and then the reagents were given with this bh three thf and then hydrogen peroxide hydroxide and water that is a hydro bore oxidation reaction. So let's think about what's going on here. So we have our primary alcohol which we now know is that cyclops ring, oops uh ring is a little bit deformed, looking redraw that with that alcohol group. And then we treat it with hydro sulphuric acid, it's distilled. So we know there's going to be heat involved. Well, this is going to be an E two reaction since we have a primary alcohol. So the primary alcohol E two reaction. So that means there can't be any Carbocaine rearrangements initially um since it's an E two, so we'll get rid of this beta hydrogen that's on the ring. And that will generate cycling ring with that methyl group having the double bond, not within the ring, but we have this double bond to the carbon. That was um the, the substituent. So that's our first initial product, but we know it's not the major alkane. So what is happens here? Well, we still have acid in this mixture on hydros acid and heat. So in that situation, we can actually generate a carbo cion and that will form the most stable position. So that will give us the positive charge on the carbon one in the ring and then break the double bonds. We have just a methyl group there. And then again, since we have water present due to our acid, the water can deproteinate that beta carbon within the ring and end up generating the major alkene product. We're told about where there's now a double bond within the ring from that carbon one to carbon two position with the methyl group on carbon one. So that's how we get generate this product. This is our major alkene product because it's more stable, the double bond is more substituted. I'm just going to scroll up at this point to get a little more space here. So we'll say major alkane more substituted and I will highlight that in blue. And then our initial product will be our minor alkene product again because that's a less substituted alkene. So let's go back to our initial conditions and we've dealt with the fact that, well, we haven't dealt with the fact yet. So now what does it say about the minor alkane, the minor alkane product formed the starting alcohol when it participates in this hydro bore oxidation. So let's draw our minor alkene go up a little bit more. We'll just leave that line there, which was that cyclen ring with the double bond on the outside there. And what do we get with hydro beration oxidation? So the straw the regents there. Well, we get the thin addition of hydrogen and hydroxide across our double bond with the oh group on the less substituted carbon. So that will give us cyclops, ring, double blond is gone and the oh group is on the less substituted carbon out the one out on the substituent branch. So what was the goal of that reaction? When we go back to our problem? It said that when we do this hydro beration oxidation, we formed the starting alcohol is this are positive, starting alcohol. Yes, it is. So let's go ahead and highlight this in yellow as this is our final answer. The alcohol is the most important functional group here. So this will actually be cyclo pencil methanol. And as you can see, that's what we posited back up when we drew our potential structures of our primary alcohols on a molecule with six carbons. So this fits all the conditions we're given. We used our index of hydrogen deficiency to tell us we had a ring, we knew that we had a primary alcohol with those six carbons which gave us a five carbon ring with alcohol. On that substituent branch, we knew that it produced two alkanes with this um hydrogens hydrosol acid, excuse me. And heat, which makes sense an E two reaction giving the minor alkane and then the presence of heat and acid allowing this formation of carbo Canion and the formation of a more stable major alkane. And then we knew that our minor alkene product could go through Hyde operation oxidation to give back the original alcohol. So again, that gives us cyclopentolate of alcohol. X. See you in the next video.

Key Concepts

Oxidation of Alcohols

Alkene Formation and Distillation

Hydroboration-Oxidation

Watch next