Give the S_N1 mechanism for the formation of 2-ethox...

Question

Give the S_N1 mechanism for the formation of 2-ethoxy-3-methylbutane, the unrearranged product in this reaction.

Accepted Answer

welcome back everyone draw an SN one mechanism for the self analysis of two I. Oto three methyl plantain in ethanol, resulting in an un rearranged product. So let's begin by drawing out our two I. Oto three methyl plantain. We have plantain, meaning we need a five carbon chain. So 12345. We have a iota group at carbon number two. So let's make carbon number two here and we have our iodine halogen and then we have a carbon number three, a methyl group. So we would just draw that there and then according to our prompt, we have a salve Alexis reaction with our solvent being ethanol, we have our ethanol and we want to write out an equilibrium era which I'll explain in a moment. So ethanol we can represent as E. T. O. H. Now the thing to note about ethanol is that ethanol is a weak base. So it's a weak nuclear file. Also because we see that it's actually a neutral molecule, it does not have a charge. And so its nuclear Phillip character is really only attributed due to the fact that it has these lone pairs on the oxygen atom and we can actually draw those in. And because again it's a weak nuclear file. We would therefore proceed with the SN one mechanism. We if we had a strong nuclear file on the other hand, we would say that this mechanism would be SN two because a strong nuclear file is going to directly attack our leaving group. Or sorry rather the the strong nuclear file will attack the carbon atom attached the leaving group in a backside attack which would result in a direct sN two substitution. But in this case, as we stated, we have a weak nuclear file being our ethanol. And so we would undergo an sn one mechanism as the prompt tells us so recognize that in our sn one we do have this equilibrium arrow because our next step is going to be recognizing that our iodine based on its high election negativity and large atomic size due to its position on the periodic table, is a good leaving group. So it's going to leave this bond on its own or leave with the electrons in the bond on its own. And we would have the formation of the following intermediate where we have a cargo caddy in here. And we should specifically recognize that this positively charged carbon is surrounded by two other carbon atoms. So we have the formation of a secondary carbo cat ion and then we also have our leaving group as just an ion being I minus. Now we want to recall that for the rate of an sN one mechanism, we have the fastest reaction rate associated with tertiary intermediates which have a faster rate than secondary intermediates. And then a fast rate than primary intermediates. And so because we have the formation of a secondary carb acadian intermediate here, we would see that this is not very favorable. And so that is why we have our equilibrium arrow because our reaction is going to want to proceed in both directions to remain stable since being a secondary intermediate is not as favored or preferred and is therefore less stable. So now we want to so far just label each part of our mechanism. Step one here we have the formation of our carbon copy on And then in step two we have our next step which I'll show as our nuclear filic attack. So we're going to take our molecule of ethanol over the equilibrium arrow again. And the nuclear Felix center which we stated is attributed from the lone pairs on oxygen are going to attack the electra Felix center on our carbon being this positively charged sorry on our structure being this positively charged carbon acadian here. So now with this attack of our electra Felix center we will show in our mechanism the next phase of our transition state where we have the following structure. We have our methyl groups still there but now we have in the place of our carbo Catalan a bond where we have our ethanol. So we have oxygen bonded to hydrogen and then bonded to each three c. and then H to see where we see that with this bond made, we now have three bonds on oxygen. And now oxygen is left with one lone pair, meaning it directly has attached 12345 valence electrons. Which is one less than it would prefer being in group six A. And so it has that plus one formal charge in the structure. And so now at this point we want to clean up the structure to get rid of that positive charge. So we'll have at this point our label of the deep rooted nation step of our mechanism where we'll take another molecule of ethanol and we'll still have our equilibrium arrow. So we have E. T. O. H. So our second molecule of ethanol is going to use its nuclear Felix center to attack that positive charge there to form a bond with the oxygen. Or sorry, not form a bond with the oxygen. It's going to attack the proton, the hydrogen rather since exactly as the steps states this is a deep throat nation. And so the bond with hydrogen is going to fall back onto the oxygen as a lone pair, leaving our final product in our next step where we have the following structure where we have O. C. H two ch three R. Methyl groups still there and then we have what's left over when the ethanol bonds with the proton as ch three ch 20 H two now and because oxygen now has and sorry not everything is visible here. So let's scoot this over. Well say plus plus this when we see that the oxygen atom is now bonded to a total of three atoms being two hydrogen and one carbon. This oxygen is now going to have a plus one charge here, which is not a part of our structure. So we're fine because we at least have our uncharged structure as our product. Now this is specifically going to be our product that is un rearranged and this is because not only the does the prompt ask us that, but we should recognize that in an sn one mechanism we have the capability of rearrangements and so recall that this carbon atom has a hydrogen bond that is implied here, so an implied hydrogen that is bonded to itself, which we can use to shift with the carbo Catalans position so that we would now have a carbo Catalan that is instead secondary rather than being or sorry, rather tertiary than being a secondary carbo Catalan as we have here and as we stated, a tertiary carbon Catalan is the most stable. And so if we were to show our rearranged product through that hydride shift, what we would have is the following product now in this tertiary location where we have our O ch two ch three bonded there. So this is our rearranged product which again was the result of a hydride shift. But for our final answer, we are confirming that we have our mechanism which showed the following steps the formation of the carbo Catalan, which we should specify as a slow step or the rate limiting step due to the fact that we saw we only were able to form a secondary carbo carry on through our leaving group leaving, which is not as favored. Our second step of our mechanism we will define as our nuclear filic attack. And then our third step is our deep throat nation, which is by our solvent ethanol. And then we are left with our un rearranged product as our final product of this mechanism. So what's highlighted in yellow are our key steps of this sN one mechanism to complete this example as our final answer with our un rearranged product. This is going to correspond to choice C in the multiple choice. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

Give the S_N1 mechanism for the formation of 2-ethoxy-3-methylbutane, the unrearranged product in this reaction.

Key Concepts

SN1 Mechanism

Carbocation Stability

Nucleophiles and Electrophiles

Watch next

Give the SN1 mechanism for the formation of 2-ethoxy-3-methylbutane, the unrearranged product in this reaction.

Key Concepts

SN1 Mechanism

Carbocation Stability

Nucleophiles and Electrophiles

Watch next

Give the S_N1 mechanism for the formation of 2-ethoxy-3-methylbutane, the unrearranged product in this reaction.