Esters are more polar than ethers, but less so than alcohols. They participate in hydrogen bonds as hydrogen bond acceptors, but cannot act as hydrogen bond donors, unlike their parent alcohols and carboxylic acids.
This ability to participate in hydrogen bonding confers some water-solubility, depending on the length of the alkyl chains attached. Since they have no hydrogens bonded to oxygens, as alcohols and carboxylic acids do, esters do not self-associate.
Consequently, esters are more volatile than carboxylic acids of similar molecular weight. Esters are usually identified by gas chromatography, taking advantage of their volatility.
This peak changes depending on the functional groups attached to the carbonyl. Esters react with nucleophiles at the carbonyl carbon. The carbonyl is weakly electrophilic, but is attacked by strong nucleophiles such as amines, alkoxides, hydride sources, and organolithium compounds. The C-H bonds adjacent to the carbonyl are weakly acidic, but undergo deprotonation with strong bases. This process is the one that usually initiates condensation reactions.
The carbonyl oxygen is weakly basic less so than in amides , but can form adducts with Lewis acids. Boundless vets and curates high-quality, openly licensed content from around the Internet.
Because of their relationship with fats and oils, all of the acids above are sometimes described as fatty acids. Small esters have boiling points which are similar to those of aldehydes and ketones with the same number of carbon atoms.
Esters, like aldehydes and ketones, are polar molecules and so have dipole-dipole interactions as well as van der Waals dispersion forces. However, they do not form ester-ester hydrogen bonds, so their boiling points are significantly lower than those of an acid with the same number of carbon atoms.
Small esters are fairly soluble in water but solubility decreases with increasing chain length, as shown below:. The reason for this trend in solubility is that although esters cannot hydrogen bond with each other, they can hydrogen bond with water molecules. One of the partially-positive hydrogen atoms in a water molecule can be sufficiently attracted to one of the lone pairs on one of the oxygen atoms in an ester, forming a hydrogen bond.
Dispersion forces and dipole-dipole attractions are also present. Forming these intermolecular attractions releases some of the energy needed to solvate the ester. As chain length increases, the hydrocarbon portion forces itself between water molecules, breaking the relatively strong hydrogen bonds between water molecules without offering an energetic compensation; furthermore, the water molecules are forced into an ordered alignment along the chain, decreasing the entropy in the system.
This makes the process thermodynamically less favorable, and so solubility decreases. Fats and oils are not water soluble. The chain lengths are so great that far too many hydrogen bonds between water molecules must be broken; this is not an energetically profitable arrangement. Melting points determine whether the substance is a fat a solid at room temperature or an oil a liquid at room temperature. Fats normally contain saturated chains.
These allow more effective van der Waals dispersion forces between the molecules: more energy is required to separate the chains, increasing the melting point.
A greater number of double bonds, or degree of unsaturation , in the molecules results in a lower melting point, because the van der Waals forces are less effective. The efficacy of van derWaals forces depends on the ability of molecules to pack closely together. The presence of carbon-carbon double bonds in the chains disrupts otherwise tidy packing. Here is a simplified diagram of a saturated fat:. The hydrocarbon chains are, of course, in constant motion in the liquid, but it is possible for them to lie tidily when the substance solidifies.
If the chains in one molecule can lie tidily, that means that neighboring molecules can get close. That increases the attractions between one molecule and its neighbors and so increases the melting point. Unsaturated fats and oils have at least one carbon-carbon double bond in at least one chain.
Rotation about a carbon-carbon double bond is constrained, locking a permanent kink into the chain. This makes packing molecules close together more difficult. If the chains cannot pack well, the van der Waals forces will be less effective. Let's look at how to name this one over here on the right. Once again, we first start with our alkyl group coming off of this oxygen, and once again we have a methyl group.
Let's go ahead and write methyl here. When we're thinking about our carboxylic acid portion, this portion right here, I would draw the carboxylic acid over here on the left. If we remember how to name that, we have a cyclohexane ring. This part would be cyclohexane. Then we have a carboxylic acid coming off of that. Cyclohexanecarboxylic acid, kind of long here. Once again we're going to drop our ending. We're going to drop the -ic and the acid and add -ate. Cyclohexanecarboxylate would be the name.
Hopefully we'll have enough room over here. Methyl cyclohexanecarboxylate, let's see if I can squeeze this in.
Methyl cyclohexanecarboxylate, would be the name for our ester. Finally let's look at physical properties of esters. Here I have some different molecules. Let's look at our ester here. If we were to name this ester we would have an alkyl group, that's a methyl group, so we have methyl.
Then we think about the carboxylic acid portion. What kind of a carboxylic acid would this be? A two carbon carboxylic acid, that would be acetic acid.
We drop the ending and add -ate, that would be acetate. We could call this ester methyl acetate. Let's compare methyl acetate to some similar-sized molecules. Over here on the left we have 2-methylbutane.
If we just go ahead and number really fast, we see this is 2-methylbutane. Then over here on the right we have a four carbon alcohol. One, two, three, four. This would be 2-butanol. Let's compare methyl acetate to these other molecules in terms of boiling points.
Let's start over here with 2-methylbutane. When we think about boiling points we can think about our intermolecular forces. The only intermolecular forces present between these two non-polar [unintelligible] same molecule, 2-ethylbutane is a non-polar molecule, so the only forces present are London dispersion forces, which we know are the weakest.
It's the easiest to pull these two molecules apart. The boiling point of 2-methylbutane is approximately 28 degrees Celsius. We think about methyl acetate, we have some polarity here. This oxygen is partial negative, this carbon right here is a partial positive.
Something like methyl acetate, a small ester is moderately polar so we have a little bit of polarity here. Same thing for this molecule of methyl acetate. The attractive intermolecular force between these two molecules would be dipole dipole.
Dipole dipole we know is a stronger intermolecular force than London dispersion, so the boiling point of methyl acetate should be higher than the boiling point of 2-methylbutane. The boiling point turns out to be approximately 57 degrees Celsius.
It takes more energy, more heat to pull apart molecules of methyl acetate because dipole dipole is a stronger attractive force. Finally 2-butanol, the boiling point for 2-butanol is about 99 degrees Celsius because of the hydrogen bonding that's present. Hydrogen bonding right here, the strongest intermolecular force, so it takes increased energy to pull apart molecules of 2-butanol.
Methyl acetate in terms of boiling point is somewhat in between that for an alkane of a similar size and that of an alcohol of a similar size because of its polarity.
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