Concept: Concept: General Mechanism7m
In this page we're going to discuss another product that forms when carbonyls react with alcohols and that's called acetals. So guys, in general I'm just going to say a few facts and then we're going to move straight into the mechanism. The first fact I want you to know is that acetals once formed are actually stable in base so if you want to keep it an acetal for a long time, keep it in a neutral to basic solution however they're easily hydrolysed back to carbonyls using acid and that makes sense guys because remember that this is a reversible reaction, it's acid-catalysed, so it makes sense that if you use acid you're going to go back to the carbonyl and it's going to be in equilibrium. Now if you want to specifically make an acetal that not only has R R O R O R but it's actually cyclic meaning it forms a ring, then you're going to have to use a diol because the diol is going to have carbons in the middle that are going to link together and so for example the diol that I would need here would be a 1, 2 ethendiol because as you can see I've got my O, 1, 2, 3, 4, I've got my O, 2 carbons and an O and hat's what I have here. 1, 2, 3, 4. So I would need a diol to make my cyclic acetal. Now let's just go straight into the mechanism for acetals and what we'll find is that it actually the only way to get to an acetal is to use the acid-catalysed mechanism. The base-catalysed mechanism is fine if you want to get to a hemiacetal but it's not going to take it all way to a full acetal. To get to a full acetal, you're going to need to use the acid-catalysed mechanism.
So that being said let's actually start back from the beginning and just make it hemiacetal and then I'm going to show you guys how a hemiacetal can be made into an acetal. So at this point guys you guys should be pros. We know that the first step is going to be protonation, just you know I'm just reiterating here this is the same exact mechanism that we did for hemiacetals. So now I've got my resonant structure, I know that this positive on the O can resonate to the bottom so I could have a positive at the bottom as well. That shows how my carbon has been made hyperelectrophilic and the alcohol can do a nucleophilic addition. So I do my nucleophilic addition, I'm just going to put here a nucleophilic addition, and then, I'm sorry and then this was protonation, and then my last step was to do a deprotonation with my conjugate base. Now in this case my conjugate base was alcohol but whatever your conjugate base is just depends on which acid you use. Not a big deal. So then I would do this, this is my deprotonation step and we know that what we would get if drawn exactly the way we had it before was O H O R H H or how I like to draw it O H O R H H plus my catalytic acid. So we're done with the hemiacetal part, how do we turn this into an acetal? Guys, it's almost the same exact mechanism it's just going to happen again. So now that we have our hemiacetal, we have to figure out, we know that we're going to protonate again so we're basically going to go through a protonation, deprotonation step again but what do we protonate? Do we protonate the O H group or the O R group. How do I know which one? Well to be honest it could be either one. The O H could protonate or the O R could protonate but the direction of your reaction depends on which one you protonate. If you want to go towards the acetal, then that means you're trying to get rid of water. So then you should protonate the O H. If we're trying to go back to the original carbonyl, then you're trying to get rid of the O R so you can reform the carbonyl. So in this case I'm trying to go forward so I'm going to react with my O H. Does that make sense guys? Literally I'm choosing to use the O H because I'm trying to draw the forward reaction but if you were drawing the backwards reaction here then we would use the O R. This is going to give me a compound that looks like this so I'm going to have O H 2 positive, H, H, O R.
Now before in my other structure for hemiacetal I had a resonant structure. In this case I don't need one because the water is a great leaving group, it's just going to take off by itself making a full carbocation. So we could again say that this was protonation I'm sorry yeah protonation and then this would be you know that forming formation of leaving group and then I would get, I'm trying to use red here sorry, and then I would get nucleophilic attack, nucleophilic addition. So then I would get my nucleophilic attack of the O to the C and that's going to give me my tetrahedral carbon in the middle. At which point, what's going to be the last step? Deprotonation guys, you've got this. So then I deprotonate and I get my final compound that looks like O R O R H H or the way that I like to draw it, O R O R H H and you get your catalytic acid. Okay guys? So really the only, it's a lot of steps but it makes it easier the fact that you're doing the same thing twice. Also the only really tricky part is that middle step with the hemiacetal, figuring out which one to protonate. Once you figure out which one to protonate, moving forward isn't hard at all. So guys, I hope that made sense now you guys know how to do a full acetal mechanism just so you know this is one of the most highly tested mechanisms in all of organic chemistry so it's definitely one you want to commit to memory, it's definitely one that you could be asked to draw to some capacity or know about to some capacity on your exam. Alright, so that's it for this topic. Let's move on to the next one.
Concept: Practice 1: Acetal Formation2m
Hey guys, let's take a look at the following question. So here it says provide the chemical steps necessary for the following synthesis. So what do we have here? We're starting out with a ketone and to this ketone, what am I changing it into? Well we should notice here is that that carbonyl carbon is no longer a carbonyl carbon and instead what does it have? It has two O R groups. So we say to ourselves what do I call a carbon that has two O R groups? We call that an acetal group or in this case because it came from a ketone we call it a ketal group.
So then we say to ourselves, how do I change a ketone into a ketal group? And remember going back to the earlier videos all we have to say is to change this ketone into that group there we'd have to use two moles of that alcohol form over your acid-catalyst. So you could use H plus or H 3 O plus, they mean the same thing and then what would happen here? We get our two O R groups getting added. So we go from a ketone to ketal group.
Again the key to synthesis is figure out what's your functional group at the end. Usually there's two at most three different ways to get there. We figured out that it was a ketal and we know the only way to get to a ketal is to use two moles of an alcohol with an acid-catalyst on the ketone and then that way we got our final answer.
Concept: Practice 2: Acetal Cap Formation2m
Hey guys, let's take a look at the following question. So here it says provide the chemical steps necessary for the following synthesis. So what we're beginning with here is a ketone and what we're ending with is a carbon with this cap that we have on it.
Now remember, if you're observant we say that this is called a ketal cap. So how do we make these caps? Remember in order to make a ketal cap I have to use a diol. What the heck does this diol look like? Well the diol had this portion here connected to it so just make that the diol and don't forget your acid-catalyst on the bottom because this requires an acid-catalyst in order to form the cap. So what we get here now is we get that as our final answer.
So that is the key to this question, knowing what our functional group is at the end is key to knowing what reagents we need to get there. Again there's only a handful of ways to get to certain types of functional groups and in this case to get a ketal cap, we have to use a diol with an acid-catalyst.
Concept: Practice 3: Acetal Retrosynthesis3m
Hey guys, let's take a look at the following question. So here it says determine the starting materials based on the acetal group present. Now technically this is a ketal group but acetal group, ketal group, very much similar. Now to answer a question like this this is a retro synthesis question we're going backwards. We have the product, we have to determine the starting material. To answer a question like this you look for the carbon that is connected to both O R groups. That carbon there is the only one attached to two O R groups. Here is an oxygen with a carbon so that's an O R group and it's connected to this O R group.
Now here all we've got to do is first locate carbon connected to two O R groups. Two, what you're going to do here is you're going to cut the carbon oxygen bonds. So we're going to cut here and cut here. Three, you're going to add an H to each oxygen and then finally you're going to add a double bond O to the carbon and if you notice, we're adding hydrogens to each oxygen so that's two hydrogens here and we're adding, that's two hydrogens and we're adding one oxygen to the carbon so what are we really adding? We're adding water because remember to take to, get rid of the acetal group or the ketal group you just use acidified water. So we get at the end as our final answer would be this. So right there, that would be our answer at the end. Knowing these steps the key to being able to go retroactively backwards to find our starting material.
List the following six mechanistic terms (nucleophilic addition, deprotonation, protonation, proton transfer, dehydration, nucleophilic addition) in order of the correct mechanistic sequence in the six step conversion of cyclopentanone to 1,4-dioxaspiro[4.4]nonane as shown below.
Provide the missing product. Show only one most preferred product. Consider only monosubstitution for EAS where appropriate.
For the reaction below, draw the structure of the appropriate compound in the box provided. Indicate stereochemistry where it is pertinent.
Propose a detailed mechanism for the following transformations. Show all possible resonance forms that contribute to this reaction pathway.
Which of the following compounds is an acetal?
E) None of these
What is the product of the reaction shown below (Hint: hydrate formation under acidic conditions)?
Propose an electron push mechanism for the following transformation. Be sure to use the correct arrow formalism and correct formal charges
Draw the product of the reaction of 3-heptanone and two equivalents of 2-propanol.
Which of the following is an acetal?
Draw the mechanism for the following reaction. Draw all the arrows to indicate movement of the all electrons, write all lone pairs, all formal charges, and all products for each step.
Complete the mechanism for the following acetal formation reaction. Be sure to show arrows to indicate movement of all electrons, write all lone pairs, all formal charges, and all the products for each step. Remember, I said all the products for each step. IF A NEW CHIRAL CENTER IS CREATED MARK IT WITH AN ASTERISK AND WRITE RACEMIC IF APPROPRIATE . Do not draw arrows to indicate how one contributing structure relates to the other.
Which is the main product that can be isolated from the reaction shown?
Predict the major product for the following reaction paying attention to the regio- and stereochemistry.
Provide a mechanism for the following transformation. Show all important flows of electrons, charges and intermediates.
The product from the reaction of 4-methyl-2-pentanone and excess methanol with H2SO4 as a catalyst would be
Predict the major product for the following reaction, paying attention to the regio- and stereochemistry where appropriate.
Predict the major product(s) in the following reaction.
Which acetal is derived from a ketone?
Draw the structural formula of the major organic product(s) in the boxes provided for the following reaction.
The compound shown below is a wasp pheromone. Draw the major product formed when this compound is hydrolyzed in aqueous acid.
Consider the three constitutional isomers of dioxane (C 4H8O2), shown below.
One of these constitutional isomers is stable under basic conditions, as well as mildly acidic conditions, and is therefore used as a common solvent. Another isomer is stable under basic conditions, but undergoes hydrolysis under mildly acidic conditions. The remaining isomer is extremely unstable and potentially explosive. Identify each isomer and carefully explain the properties of each compound.
Provide a detailed reaction mechanism for the following transformation.
Suggest a reasonable mechanism for the following spirocyclic acetal formation. Use curved arrows to show electron flow.
What products result from treating the compound shown with aqueous acid?
(a) Acetone and 1,3-propanediol
(b) Propanedial and 2-propanol
(c) Propanedial and 2,2-propanediol
(e) None of the above