HS Chemistry - Carbonyl Functional Groups

Ketones

Overview of The Page

This page will cover:

• What are ketones?
• How are ketones named?
• What reactions do ketones participate in?
• How can the presence of ketones in a compound be tested for?

Carbonyl functional groups are functional groups that contain a C=O bond $i.e. there is at least one instance of a Carbon atom double-bonded to an Oxygen atom$. This does not mean, however, that all C=O bonds are carbonyl bonds. CO₂, for example, has the structural formula O=C=O, but the C=O bonds are not considered carbonyl because it is not an organic compound. Ketones are a type of carbonyl functional group, and this subpage will look at them.

Ketones are very similar to aldehydes. In aldehydes, the C=O is bonded to one Hydrogen atom and one other non-Hydrogen atom $i.e. it is at the end of the organic molecule's main chain$. In ketones, the C=O is bonded to two non-Hydrogen atoms $i.e. it is not at the end of the organic molecule's main chain$.

The formula for the aldehyde functional group is R₁COR₂, where R₁ represents the rest of the organic molecule before the ketone group, and R₂ represents the rest of the organic molecule after the ketone group. To form the name for an aldehyde, we use the same naming convention as we do for other organic molecules. The suffix ending for aldehydes is "-one", but the way it's added is different. Instead of removing the entire suffix, we take the suffix of the hydrocarbon and add "-\one" to the end. Thus, if we have the following molecule:

Where there is both an alkene functional group and a ketone group, we take the alkene suffix $"\-ene"$ and add "-one" to the end after removing the "e" at the end $giving us "\-enone"$. Thus, the molecule shown above is pent-3-en-2-one.

This raises an important question. Why did we start numbering from the right side of the molecule?

Why not the left side?

And why, when naming the molecule, did we note the position number of the ketone group?

Ketones are a type of carbonyl group, and carbonyl groups take precedence over alkene groups, so we start numbering from the side that gives the ketone group the smallest number. Additionally, since ketones are not present at the end of the main chain of an organic molecule, it is necessary to note their position in the organic molecule with a number.

Following that same logic, this:

Is pentan-2-one, or 2-pentanone. We can draw the Carbon atoms in the main chain in a single line if they are single-bonded to each other $and not double-bonded to each other like in alkenes$:

As you can see, the bond angles for the aldehyde functional group have not been retained - they are drawn at perpendicular angles to each other. It is important to note, however, that this does not change the bond angles between the atoms - it merely makes it easier to draw the molecule.

The skeletal formula of pentan-2-one would be:

So far, we've dealt with displayed formulae of molecules that show the ketone group as part of the main chain. What if we have a displayed formula of an organic molecule that shows the ketone functional group as a side chain?

In this organic molecule, not only is the ketone functional group shown as a side chain, but the longest chain of Carbon atoms is the chain that doesn't include the ketone group:

It may seem that we have to pick the longest Carbon chain shown above, which doesn't include the ketone group, as our main chain.

However, whenever there is a type of carbonyl group present in an organic compound, it takes precedence over the Carbon-Carbon double bonds and the Carbon-Carbon single bonds. Since ketones are a type of carbonyl functional group, they take precedence.

What does it mean for the ketone functional group to take precedence? It means:

1. The main chain of the organic compound is the longest Carbon chain that includes the ketone, unless there is another functional group present that takes precedence over the ketone. Carboxylic acids and aldehydes, for instance, will take precedence over ketones.

2. When we number the Carbon atoms in the main chain to determine positions, we start from the side that will give us the smallest number for the Carbon atom that is part of the ketone group.

3. Even if the organic compound includes an alcohol or alkene or some other functional group, the compound will be classified as a ketone unless there is some other functional group present which takes precedence over the ketone.

Thus, the main chain for the organic molecule shown above is:

In this case, since Carbon-Carbon single bonds are rotatable, the molecule can be re-drawn as:

Now its clear that the propyl group is the side chain, and we can number and name the molecule accordingly. The main chain is numbered in blue, and the side chain is numbered in red:

The organic molecule is 3-propylhexan-2-one.

Even if the side chain included an alkene functional group, like in the example below:

The main chain would still be the longest Carbon chain that includes the ketone, because the ketone functional group takes precedence over the alkene functional group.

However, if there are multiple possible main chains that include ketone functional groups, then there are a few criteria for determining which chain should be used as the main chain:

1. The chain with the most ketone groups will be used as the main chain.

2. If there are multiple such possible chains, then the chain with the most Carbon atoms will be used as the main chain.

When there are two ketone functional groups present in a molecule, the molecule is called a diketone. Diketone can be treated as aldehydes, except that when naming them, the suffix is "-dione" rather than "-one". This extends to triketones and further on, but those are not expected to be encountered at this level.

There are also organic molecules with both hydroxyl $OH$ groups and ketone groups. In these organic molecules, the ketone group takes precedence over the hydroxyl group, and the molecule is classified as an ketone. Thus, the suffix is "-one" $or "\-dione" if it is a diketone$. The hydroxyl group is added to the prefix of the organic molecule with the prefix "n-hydroxy-", where n is a number noting the hydroxyl group's position. Again, since ketones take precedence over hydroxyl groups, the numbering starts from the end that gives the ketone group the smallest number.

Thus, to name the following organic molecule:

We would number the main chain:

And the molecule would be named 6-hydroxyhexan-3-one.

Since the "hydroxy" part is a prefix, it follows the same rules as other prefixes - namely, that we have to indicate how many hydroxyl groups there are, and their positions, with the same rules that we have for other prefixes.

Thus, the following molecule:

Would be named 5,6-dihydroxyhexan-3-one. Organic molecules with both halogenoalkane and ketone groups are also treated the same way $and so are organic molecules with all three\!$.

Cyclic ketones are like cyclic aldehydes, but with the ketone group instead of the aldehyde group. When naming cyclic ketones, however, the suffix "-one" is used, with a prefix like "di" or "tri" appended to the beginning of the suffix to indicate the number of ketone groups present. The numbering starts from the ketone group $unless there is another functional group present that takes higher precedent$.

Thus, the following molecule:

Would be named 1,3-cyclobutanedione. Ketones attached to arenes are named the same way as cyclic ketones.

There are also organic molecules with both aldehyde and ketone functional groups. groups. In these organic molecules, the aldehyde group takes precedence over the ketone group, and the molecule is classified as an aldehyde. Thus, the suffix is "-al" $or "\-dial" if it is a dialdehyde$. The ketone group is noted in the prefix as "n-oxo", where n is a number noting the hydroxyl group's position. Again, since aldehydes take precedence over ketones, the numbering starts from the end with the aldehyde group.

There is also a "common naming system" for ketones, where the suffix "-ketone" is used instead. However, common names for ketones work differently than common names for aldehydes. Rather than use Greek letters for numbering the Carbon atoms, the common name for ketones is formed by taking the names of the two groups attached to the C=O bond in the ketone, ordering them alphabetically, and appending the word "ketone" at the end. For example, pentan-2-one:

Has a CH₃ group $a methyl group$ on one side of the C=O bond, and a CH₃CH₂CH₂ group $a propyl group$ on the other side of the C=O bond. Therefore, pentan-2-one's common name is methyl propyl ketone.

As another example, propan-2-one $also known as simply propanone or acetone$:

Has a CH₃ group $a methyl group$ on one side of the C=O bond, and a CH₃ group $a methyl group$ on the other side of the C=O bond. Therefore, propanone's common name is dimethyl ketone.

As a third example, hexan-3-one:

Has a CH₃CH₂ group $an ethyl group$ on one side of the C=O bond, and a CH₃CH₂CH₂ group $a propyl group$ on the other side of the C=O bond. Therefore, hexan-3-one's common name is ethyl propyl ketone.

A pattern emerges here. By using their common names, we can group ketones together. Methyl propyl ketone $pentan\-2\-one$ and dimethyl ketone $propanone$ are both types of methyl ketones. Methyl propyl ketone $pentan\-2\-one$ and ethyl propyl ketone $hexan\-3\-one$ are both types of propyl ketones.

Grouping ketones in this way using their common names can be very useful in identifying commonalities among different types of ketones.

Ketones can also be formed from, and participate in, certain reactions.

Ketone Reactions

• Forming Ketones Through Oxidation:

  • There are two ways in which aldehydes can be formed:

    1. Oxidizing an alkene using hot, concentrated acidified potassium permanganate $KMnO~4~$ solution. However, this requires that one of the Carbons in the double bond is bonded to two non-Hydrogen atoms. The products will include a ketone.

    2. Oxidizing a secondary alcohol using a hot, concentrated solution of acidified potassium permanganate $KMnO~4~$ or a hot, concentrated solution of acidified potassium dichromate $K~2~Cr~2~O~7~$. A secondary alcohol must be used, or else a ketone will not be formed. The products will be a ketone and H2O.

  • In both of these reactions, the alkene/secondary alcohol must be heated in a distillation apparatus with the acidified KMnO₄ or K₂Cr₂O₇. Unlike when producing an aldehyde by oxidizing a primary alcohol, the ketone that is being produced does not need to be distilled as it is formed, as the KMnO₄ or K₂Cr₂O₇ will be unable to further oxidize the ketone.

  • In both of these reactions, if KMnO₄ is used as the oxidizing agent, the KMnO₄ solution will lose its purple color and become colorless. If K₂Cr₂O₇ is used as the oxidizing agent, the K₂Cr₂O₇ solution will turn from an orange color to a green color.

  • When writing the oxidation reaction, we usually just show the Oxygen atoms from the oxidizing agent, rather than the whole oxidizing agent. We represent those Oxygen atoms as [O]. For example, the oxidation of 2-butanol, a primary alcohol, can be represented as:

    CH₃CH₂CHOHCH₃ + [O] → CH₃CH₂COCH₃ + H₂O

    Here, the secondary alcohol 2-butanol is oxidized to yield butan-2-one and water.

• Reducing Ketones:

  • Given that ketones can be formed through oxidation, they can also be reduced back when reacted with a reducing agent. Reducing a ketone will yield a secondary alcohol. The reaction method used depends on the reducing agent used:

    1. If sodium tetrahydridoborate $NaBH~4~, also known as sodium borohydride$ is the reducing agent, the ketone must be heated with a solution of NaBH₄ dissolved in water. The product will be a secondary alcohol.

    2. If lithium tetrahydridoaluminate $LiAlH~4~, also known as lithium aluminum hydride$ is the reducing agent, the LiAlH₄ must be dissolved in a dry ether at room temperature before being added to the ketone. The product will be a secondary alcohol.

  • When writing the reduction reaction, we usually just show the Hydrogen atoms from the reducing agent, rather than the whole reducing agent. We represent those Hydrogen atoms as [H]. For example, the reduction of butan-2-one can be represented as:

    CH₃CH₂COCH₃ + 2[H] → CH₃CH₂CHOHCH₃

    Here, the aldehyde butanal is reduced to the secondary alcohol 2-butanol.

• Addition Reaction With Cyanide Compounds:

  • Ketones can undergo addition reactions with hydrogen cyanide $molecular formula: HCN; structural formula: H−C≡N$. However, since HCN is a very toxic gas, it is not added directly. Instead, HCN is produced in the reaction apparatus by reacting another cyanide, like KCN, with an acid, like sulfuric acid $only a very dilute acid is needed to turn the KCN into HCN$. The newly-produced HCN will dissolve into the solution. Some of the HCN molecules will dissociate into cyanide $CN^\-^$ ions and H⁺ ions. The cyanide will then begin to react with the ketone in the reaction apparatus.

    • Note that cyanides like KCN are still toxic, but because they are a solid, they are safer than HCN, which is a gas at STP $Standard Temperature and Pressure$.

  • The aldehyde functional group contains a C=O bond. A CN^- ion, which is a nucleophile, can break this double bond. The Carbon atom in the C=O bond is slightly positively charged, as Oxygen is more electronegative than Carbon. Therefore, the CN^- ion is attracted to the Carbon atom and bonds with it. This causes the Carbon atom to break its double bond with the Oxygen atom so that it can form a bond with the cyanide ion. Since the Oxygen atom had a slightly negative charge before the bond was broken, it is left with an extra electron after the bond is broken $heterolytic fission$.

  • After the bond is broken, the Oxygen atom is left with a negative charge. It then attracts a H⁺ ion towards it and bonds with it, forming a hydroxyl group. The H⁺ ion could come from a HCN molecule $which would undergo heterolytic fission and supply the H^\+^ ion$, from the water, or from the acid.

  • The resulting product is a molecule that belongs to a class of molecules known as 2-hydroxynitrile molecules; after undergoing the reaction, the Carbon atom that was previously part of the carbonyl group now has a hydroxyl group and a nitrile group bonded to it. The nitrile group is the −C≡N group that was formerly part of the hydrogen cyanide molecule. To fully name the molecule, we can put the rest of the product organic molecule's name between the "2-hydroxy" and "nitrile" portions (e.g. 2-hydroxypropenenitrile, 2-hydroxy-4-chlorohexanenitrile, etc.). The reason for this is that "2-hydroxy" is the prefix, and "nitrile" is the suffix, that is common among all 2-hydroxynitrile molecules.

    • The addition reaction of pentan-2-one with HCN can be used to demonstrate this:

      CH₃CH₂CH₂COCH₃ + HCN → CH₃CH₂CH₂COHCH₃CN

      If we wanted to represent the molecules with their displayed formulae, we could show it as:

      

  • At first, this seems more complicated than the addition reaction of an aldehyde with HCN, because now we have a branched organic molecule as our product. However, fundamentally it's essentially the same.

    • Let's consider what went on in the addition reaction of an aldehyde with HCN. First, the CN^- ion formed a bond with the Carbon atom that is in the C=O bond, forcing it to break the double bond with the Oxygen atom. That left the Oxygen atom with an extra electron, as it is now only attached to the Carbon atom with a single bond, and it attracted a H⁺ ion to itself to form a hydroxyl group.

    • Now let's consider what's happening in the addition reaction of a ketone with HCN. First, the CN^- ion formed a bond with the Carbon atom that is in the C=O bond, forcing it to break the double bond with the Oxygen atom. That leaves the Oxygen atom with an extra electron, as it is now only attached to the Carbon atom with a single bond, and it attracts a H⁺ ion to itself to form a hydroxyl group.

  • The same thing happens in both scenarios. So why does the product look different?

    • In an aldehyde functional group, the Carbon atom in the C=O bond is bonded to one Hydrogen atom and one other non-Hydrogen atom. However, in a ketone functional group, that Carbon atom is bonded to two non-Hydrogen atoms.

    • When we added an aldehyde to HCN, the product molecule had a Carbon atom bonded to a nitrile group, a hydroxyl group, and a Hydrogen atom. This was the Carbon atom from the reactant's aldehyde group. However, since the Carbon atom in a ketone functional group isn't bonded to a Hydrogen atom, when we add a ketone to HCN, the product molecule will have a Carbon atom bonded to a nitrile group, a hydroxyl group, and one other group.

  • You might be wondering how we know what that other group is. It's one of the two groups bonded to the Carbon atom in the ketone group in the reactant molecule. But which one?

    • It'll always be the small one. The main chain of the product molecule will contain as many Carbon atoms as it can, all else being equal. That means that of the two groups originally bonded to the Carbon atom in the ketone group in the reactant molecule, the larger group $the one with a longer chain of Carbon atoms$ will be part of the product molecule's main chain, and the smaller group $the one with a shorter chain of Carbon atoms$ will become the product molecule's side chain.

  • We can now name our product molecule: 2-hydroxy-2-methylpentanenitrile.

    • Overall: Pentan-2-one + Hydrogen Cyanide → 2-hydroxy-2-methylpentanenitrile

    
    

  • An extra Carbon atom has been added to the main chain through this reaction. This makes this kind of reaction very useful in organic chemistry, as chemists often need to extend the main chain of an organic molecule, and repeating a reaction like this allows them to do so. The nitrile can then:

    1. Undergo hydrolysis with a dilute solution of HCl under reflux to become a carboxylic acid. The water molecule and dissociated Hydrogen ion react with the nitrile to form a carboxylic acid and an ammonium ion. At the end of the reaction, most of the organic molecule will left unchanged, and only the functional group at the end of the organic molecule will be different.

    2. Be reduced by sodium and ethanol to an amine. The [H] atoms react with the nitrile to form an amine. Unlike the hydrolysis reaction, there is only one product molecule here. Again, most of the organic molecule will be left unchanged, and only the functional group at the end of the organic molecule will be different.

  • When writing a reaction, to quickly determine the product's structural formula given the reactants, the Hydrogen atom will bond with the Oxygen atom in the ketone, and the CN will bond at the end of the main chain. The reaction between pentan-2-one and HCN is shown below, with the H and CN from the HCN colored in red to show where they bond:

    CH₃CH₂CH₂COCH₃ + HCN → CH₃CH₂CH₂COHCH₃CN

    This rule holds for aldehydes too.

• Haloform Reaction:

  • The common naming system for ketones named them by the groups attached to the Carbon atom in the C=O bond. Thus, it allowed us to group them together by the types of groups present on either side of the C=O bond. A ketone that had a methyl group on one side was a methyl ketone, regardless of what group was present on the other side. A group that had a propyl group on one side was a propyl ketone, regardless of what group was present on the other side. And so on. This allowed us to identify commonalities between different types of ketones.
  • Methyl ketones $ketones with a methyl group on one side of the C=O bond, which can be seen in the structural formula as CH~3~CO or COCH~3~$ can undergo a reaction known as the haloform reaction. The haloform reaction can transform a ketone into a carboxylic acid. The haloform reaction is carried out by placing and heating the organic compound in an alkaline solution where a hydroxide compound $e.g. KOH$ and a halogen molecule $e.g. Br~2~$ are present in excess.
  • The ketone is required to be a methyl ketone because there must be a methyl group $one Carbon and three Hydrogen atoms$ present on one side of the C=O bond for the reaction to occur.
  • Once the mixture is heated, the hydroxide ions will steal the Hydrogen atoms from the methyl group, leaving it with a negative charge. The halogen molecule will then split up and bond to the Carbon atom in the methyl group. At the end of this step, the Hydrogens in the methyl group have all been replaced by halogens. This is called halogenation.
  • After the halogenation process is complete, because the hydroxide compound is present in excess in the solution, a hydroxide ion will react with the methyl ketone after halogenation and effectively substitute out the halogenated methyl group $what actually happens is that the hydroxide ion attacks the C=O double bond and forms a single bond with the Carbon atom, and then the bond between the Carbon atom and the halogenated methyl group breaks as the Carbon atom reforms the double bond with the Oxygen atom$. This is why it is necessary to have the hydroxide compound present in excess in the solution.
  • After the halogenated methyl group breaks off, it has a negative charge. This allows it to "steal" the Hydrogen atom from the hydroxyl group now attached to the Carbon atom in the C=O bond. This completes the formation of the halogenoalkane. As the halogenoalkane produced in this reaction always has only one Carbon atom, with the Carbon atom being bonded to three halogen atoms, it is commonly referred to as a haloform.
  • The haloform reaction, for the most part, will only work with ketones, not aldehydes. However, the aldehyde ethanal, with the formula CH3CHO, will also undergo the haloform reaction, as it also contains a methyl group on one side of the C=O bond. The haloform test will also work with secondary alcohols where the Carbon bonded to the hydroxyl group is next to the Carbon atom at the end of the main chain $which can be seen in the structural formula as CH~3~CHOH, or as CHOHCH~3~$, as the secondary alcohol will be oxidized to form a methyl ketone. The reaction will then proceed to occur as expected.

Testing for Ketones

• 2,4-DNPH Test:

  • If we have an organic molecule, but we don't know whether it is an aldehyde or ketone or some other molecule, we can test it by mixing it with a solution of 2,4-dinitrophenylhydrazine $2,4\-DNPH, aka Brady's reagent$ in a test tube and then shaking it. 2,4-DNPH solution normally has a lime-green color, but if an aldehyde or ketone is present, an orange precipitate will be formed. If a precipitate doesn't form, then the organic molecule isn't an aldehyde or ketone.
  • A 2,4-DNPH molecule contains a NH₂ at its end. When an aldehyde or ketone reacts with it, the Oxygen atom in the carbonyl $C=O$ group reacts with the H`2~ to form H₂O. The aldehyde/ketone then bonds with the Nitrogen atom, and the products are H2O and a large molecule called a 2,4-dinitrophenylhyrdazone molecule. This molecule is seen as an orange precipitate. Note that the molecule's ending is "-one", not "-ine". To show the type of 2,4-dinitrophenylhydrazone molecule that is formed, we append the name of the original aldehyde/ketone to the beginning; for example, using hexanal would give us hexanal 2,4-dinitrophenylhydrazone. The overall reaction is a condensation reaction.
  • But what if we don't know the type of aldehyde/ketone that we originally had? By measuring the melting point of the precipitate and comparing it with known values, we can determine the type of aldehyde/ketone we originally had. In order to find the melting point, however, we first need to purify the precipitate crystals by washing them and recrystallizing them with a suitable solvent, like a mixture of ethanol and water for example. Once the crystals are purified, we can measure their melting point and compare it with known values of other 2,4-dinitrophenylhydrazones to find out which one we have, and by extension, what aldehyde/ketone we originally had.

• Fehling's Solution:

  • If we know that the organic molecule we have is either an aldehyde or a ketone $which can be proven if the molecule produces an orange precipitate when reacted with 2,4\-DNPH solution$, we can use Fehling's solution to see if it is an aldehyde. Fehling's solution contains Copper $Cu$ ions, which will form an oxide, seen as a red precipitate, if the organic molecule is an aldehyde.
  • However, Fehling's solution can't be stored directly, as the Copper compound in the solution will quickly break down. Therefore, it is formed when the test needs to be carried out by combining equal amounts of two stored solutions: Fehling's solution A, which has a blue color, and Fehling's solution B, which has no color. As would be expected, the product, Fehling's solution, has a light blue color.
  • To carry out the test, add the Fehling's solution to a test tube with the organic compound. If the organic molecule is an aldehyde, a red precipitate will form. If no precipitate forms, but the organic molecule reacted with 2,4-DNPH solution, it is a ketone. If it didn't react with 2,4-DNPH solution, it is neither an aldehyde nor a ketone.

• Tollens' Reagent:

  • If we know that the organic molecule we have is either an aldehyde or a ketone $which can be proven if the molecule produces an orange precipitate when reacted with 2,4\-DNPH solution$, we can use Tollens' reagent to see if it is an aldehyde. Tollens' reagent is a solution of silver nitrate $AgNO~3~$ in excess ammonia solution.
  • Excess ammonia solution is required to prevent the silver nitrate from reacting with the water, and to establish alkaline conditions $pH less than 7$ for the reaction to occur.
  • To carry out the test, add the Tollens' reagent to a test tube with the organic compound. If the organic molecule is an aldehyde, the silver $Ag^\+^$ ions will oxidize and form Ag₂O, seen as a mirror-like substance in the liquid. If no precipitate forms, but the organic molecule reacted with 2,4-DNPH solution, it is a ketone. If it didn't react with 2,4-DNPH solution, it is neither an aldehyde nor a ketone.

• Iodoform Test:

  • The iodoform test uses the haloform reaction to test for the presence of methyl ketones, CH₃CO or COCH₃ (although ethanal, with a formula of CH₃CHO, will also return a positive result). It can also test for the presence of a secondary alcohol where the Carbon bonded to the hydroxyl group is next to the Carbon atom at the end of the main chain $which can be seen in the structural formula as CH~3~CHOH, or as CHOHCH~3~$, as the alcohol will be oxidized to a methyl ketone in the reaction.
  • If a methyl ketone group or CH₃CHOH group exists, then tri-iodomethane $CHI~3~, aka iodoform$ will appear as a yellow precipitate.