Identifying Carbonyl Compounds - Organic Chemistry

This question tests ones knowledge of carboxylic acid derivatives and how different groups effect reactivity. The order is 3, 1, 4, 2 and this is because of the electronic effect. Acid chloride (3) is first because of its electron withdrawing group making it most reactive. (1) is a weakly donating group and (4) is a normal strength donating group and (2) is a good electron donating group.

Ketones are carbonyls with two R-groups attached the the carbonyl carbon.

This is an acid halide, were X is any halogen (group VII). They are the most reactive of all the carboxylic acid derivatives.

Carboxylic acids are composed of one R goup and a hydroxy group bound to the carbonyl carbon. The hydroxyl group gives the molecule acidic properties.

This is a hemiketal. Hemiketals are formed from the reaction of a ketone with an alcohol. The alcohol attacks the carbonyl carbon, reducing the double bond and adding a hydroxyl group.

Ketals have two -OR groups bound to a central carbon. The R groups attached to the oxygens are not necessarily the same. Ketals result from the reaction of a ketone with two equivalents of alcohol.

Acid anhydrides are relatively reactive, since they have a fairly good leaving group (the conjugate base of a carboxylic acid) \[COO-\]. They are formed by reacting two carboxylic acids via dehydration synthesis.

This is an amide. It is very unreactive since its leaving group is the conjugate base of an amine \[HNR-\].

Esters are relatively unreactive since the leaving group is an alkoxide ion (OR-), which is the conjugate base of an alcohol (a relatively weak acid).

The correct answer is D.

The molecule pictured above is an aldehyde because it contains an oxygen double bonded to the last carbon on the carbon chain. A contains a amide group, B contains a ketone, and C contains a carboxylic acid group.

To answer this question, it's important to recall what sigma bonds and pi bonds are. Sigma bonds are the result of overlapping s orbitals, and are contained in all bonds. Pi bonds, on the other hand, are the result of p orbitals overlapping.

Double bonds between atoms consist of one sigma bond (from overlapping s orbitals) and one pi bond (from overlapping p orbitals). Triple bonds, though not shown in this molecule, consist of one sigma bond and two pi bonds.

In the molecule shown in this question, we can recognize the following bonds:

1 carbon-oxygen double bond, which is equal to one sigma bond and one pi bond

2 carbon-carbon single bonds, which equates to two sigma bonds

6 carbon-hydrogen single bonds, which equals six sigma bonds

Adding these up, we obtain a total of nine sigma bonds and one pi bond.

Aldehydes are carbonyls with one R-group and one hydrogen attached to the carbonyl carbon.

A carbonyl functional group takes the form of . This leaves the carbonyl carbon to form two remaining bonds with various substituents. If one of these substituents is a hydrogen, the molecule is an aldehyde (the carbonyl group is at the end of the carbon chain). If neither substituent is a hydrogen, the molecule is a ketone (the carbonyl group is not at the end of the carbon chain).

An acid contains the carboxylic acid functional group, which takes the form of . Carbonyl groups do not contain the .

Amino acids contain the carboxylic acid functional group () and an amine functional group (). The description in the question does not fit this molecule.

The molecule's longest carbon chain has 5 carbons (thus, "pent-"), and the lack of carbon-carbon double bonds makes it an alkANE (thus "pentan-"). The presence of a carbonyl () group that is not at the end of the chain means that the molecule is a ketone (thus "pentanone"). Numbering the carbon chain so that the carbonyl is on the lowest possible carbon will put the carbonyl on carbon 2 (Thus "2-pentanone").

The molecule's longest carbon chain has 4 carbons (thus, "but-"), and the lack of carbon-carbon double bonds makes it an alkANE (thus "butan-"). The presence of a carbonyl () group that is not at the end of the chain means that the molecule is a ketone (thus "butanone"). There is no other way to make a ketone on a carbon chain consisting of 4 carbons, and therefore a locant is not needed when naming this compound.

The prefix "acet-" or "aceto-" indicates a group. This group by itself is known as the "acetate" anion, and its protonated acidic form is known as "acetic acid." This is a very common organic acid and it is important to know the name of this acid.

The prefix "form-" indicates a group. This group by itself is known as the "formate" anion, and its protonated acidic form is known as "formic acid." This is a very common organic acid and it is important to know the name of this acid.

When naming an organic compound by the IUPAC, it's best to first start by identifying the functional groups present.

In this particular case we have:

A ketone here:

and an ethyl group in the middle:

Next, we should identify what functional group has the highest priority, as that will form the base name of the compound:

According to IUPAC convention, Carboxylic Acids and their derivatives have the highest priority then carbonyls (ketones in this case) then alcohols, amines, alkenes, alkynes, and alkanes, so in this case the ketone group has the highest priority and therefore makes up the name of the base compound.

Next, we want to number the longest carbon chain with the highest priority functional group with the lowest number. In this case this means we want the ketone to be carbon number #2, which is the lowest number it can possibly be so let's start there and number the carbon chain.

You should get something like this:

Now that we have numbered the carbon chain we can begin our naming.

Let's start with the base name:

According to IUPAC convention the base name for a ketone compound is -one, so in this case we have a octan-2-one.

We also have an ethyl group at carbon five. This gives us 5-ethyloctan-2-one, which is our final answer.

Now let's look at the wrong answer choices:

1. 5-propylheptan-2-one is wrong because the longest carbon chain has 8 carbons and following the path of the ethyl group as part of the main chain prevents naming being done for the longest possible carbon chain.

2. 5-ethyloctanal is incorrect because it mistakes the ketone for an aldehyde as the suffix -al refers to aldehydes which are RCHOs that can only be carbon 1.

3. 5-propylheptanal is wrong because the longest carbon chain is 8 carbons long, not 7, and this compound isn't an aldehyde.

4. 4-ethyl-octan-2-one is wrong because the ethyl is at carbon 5 and not carbon 4.

When naming an organic compound by the IUPAC, it's best to first start by identifying the functional groups present.

An aldehyde as shown here:

And two ethyl substituents as shown here:

Next, we should identify what functional group has the highest priority, as that will form the base name of the compound:

According to IUPAC convention, Carboxylic Acids and their derivatives have the highest priority then carbonyls (in this case the aldehyde) then alcohols, amines, alkenes, alkynes, and alkanes, so in this case the aldehyde group has the highest priority and therefore makes up the name of the base compound.

Next, we want to number the longest carbon chain with the highest priority functional group with the lowest number. In this case this means we want the carbon on the ring attached to the aldehyde to be carbon number #1, so let's start there and number the carbon chain around the ring.

You should get something like this:

Now that we have numbered the carbon chain we can begin our naming.

Let's start with the base name:

According to IUPAC convention when we have an aldehyde attached to the ring normally (in this case cyclopentane) and put carbaldehyde after it, so in this case the base name is cyclopentanecarbaldehyde.

Next we wish to number ring such that the ethyl substituents have the lowest numbers possible, this is done by making ethyl group next to carbon #1, the second carbon (2-ethyl), then the second ethyl group is 4-ethyl.

Since there are two ethyls the substituents are 2,4-diethyl, so the final name of the compound is 2,4-diethylcyclopentanecarbaldehyde.

Now let's go over the wrong answers:

1. 2,4-diethylcyclopentanal is incorrect because according to IUPAC naming conventions we name the ring normally (cyclopentane) and then put carbaldehyde after its normal name.

2. 2,4-diethylpentanal is incorrect because this compound is a ring and not a straight chain.

3. 3,6-diethylcyclopentanecarbaldehyde is incorrect because the substituents can have lower numbers by making the ethyl group next to the first carbon, 2-ethyl.

4. 3,6-diethylcyclopentanal is wrong because the substituents can have lower numbers by making the ethyl group next to the first carbon, 2-ethyl. Also according to IUPAC naming conventions we name the ring normally (cyclopentane) and then put carbaldehyde after its normal name.

The longest carbon chain is carbons long (thus ""), and the lack of double bonds makes it an alkane (thus ""). The highest priority functional group is the carboxylic acid group, and the suffix of the IUPAC name should reflect the highest priority functional group (thus ""). Because the carboxylic acid group is assumed to lie on carbon number there is no need to designate a locand for this functional group. The other substituents of this molecule are the methyl groups on carbons and when numbering the chain from left to right. Thus, the name is .