The longest carbon chain is  carbons long (thus ""), and the double bond makes it an alkene (thus ""). The highest priority functional group is the double bond, and the carbon chain must be numbered such that this functional group is given the lowest number carbon. Counting from left to right puts the double bond between carbons  and , which is lower than counting from right to left , therefore the numbering goes from left to right, and the molecule is thus a "." The substituents are a bromine on carbon   and a chlorine on carbon  . IUPAC rules state that the substituents are listed in alphabetical order when naming (thus "").

Finally, stereochemistry must be taken into account. On carbon , the bromine is of higher priority than the methyl group (when taking atomic number into account, per the Cahn-Ingold-Prelog rules), and is located "above" the double bond. On carbon , the chlorine group is higher priority, and is located "below" the double bond. Because the higher priority groups are across the double bond from each other, the molecule is given the "E" designation.

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

In this particular case we have:

An Ester in the middle as shown here:

a Methyl group shown here: 

and a Butyl group attached to the ester:

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 Acid derivatives including Esters have the highest priority then carbonyls then alcohols, amines, alkenes, alkynes, and alkanes, so in this case the Ester 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 carbonyl of the ester to be carbon number #1, so let's start there and number the carbon chain.

You should get something like this:

Notice there are two sixes. The reason why is because there are two possible pathways for the carbon numbering to continue, but both are equivalent meaning no matter what we do there is a 5-methyl group and the carbon chain is 6 carbons long.

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 an ester compound is -oate, so in this case we have a hexanoate, which can also be written as hexan-1-oate, but this isn't needed as it as the ester is at carbon 1.

We also have a methyl group at the 5-carbon so in this gives us:

5-methylhexanoate.

However, we aren't done as we haven't named the substituent on the other side of the ester. Let's first count the number of carbons it has. Since this chain has 4 carbons it is a butyl group, as according to IUPAC the chain on the side farthest from carbonyl carbon of the ester is named as a substituent and placed in front of the name of the compound.

This makes our final answer Butyl 5-methylhexanoate.

Now let's go over the other answer choices and why they are wrong:

1) Butyl 5-methylhexenoate is almost correct except for the fact it says Butyl 5-methylhexenoate. The "en" indicates there is an alkene (double bond) in the compound, and since there isn't this can't be the right answer.

2) Butyl 2-methylhex-6-anoate is wrong because the ester group isn't assigned the highest priority. In IUPAC nomenclature you want to assign the highest priority functional group the lowest number possible in the carbon chain.

3) Butyl 2-methylhex-6-enoate is wrong for a mix of the reasons in the previous 2 answers. It says Butyl 2-methylhex-6-enoate in it, and the compound doesn't have an alkene. It also makes the mistake of not making the ester group (the highest priority functional group) have the lowest number possible in the carbon chain, so this can't be right either.

4) 1-butoxy-5-methylhexanone is wrong because it interprets the ester as being a ketone and an ether group instead of an ester.

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 carboxylic acid shown here:

An alkene in the middle of the carbon chain:

A methyl group:

and a ketone towards the end:

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 Acid derivatives including Esters have the highest priority then carbonyls (in this case the ketone) then alcohols, amines, alkenes, alkynes, and alkanes, so in this case the Ester 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 carbonyl of the carboxylic acid to be carbon number #1, 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 an carboxylic acid compound is -oic acid, so in this case we have a decanoic acid, which can also be written as decan-1-oic acid, but this isn't needed as it as the carboxylic acid is at carbon 1.

Next notice that we have an alkene that's part of the main chain, since it is part of the main chain we include it in the base name, so we must change our name from decanoic acid to dec-4-enoic acid because the lowest it can be numbered is #4, however since the highest priority groups on the alkene are facing opposite to each other it is an E (Entgegen) alkene, so we can name it (4E)-decenoic acid

We also have a methyl group at carbon 5. This gives us:

(4E)-5-methyldecenoic acid.

Finally we have a ketone as a substituent, and a ketone as a substituent is called an oxo, so it becomes 9-oxo.

Now we must order our substituents alphabetically. Thus it becomes

(4E)-5-methyl-9-oxodecenoic acid which is our final answer.

Now let's go over the wrong answers:

1) (4E)-9-oxo-5-methyldecenoic acid is wrong because the substituents aren't ordered alphabetically.

2) 5-methyl-9-oxodecanoic acid is wrong because it says decanoic acid when there is an alkene present.

3) 9-oxo-5-methyldecanoic acid is wrong because it says decanoic acid when there is an alkene present, and because the substituents aren't ordered alphabetically.

4) 10-carboxy-5-methyldecan-2-one is wrong because the carboxylic acid group isn't highest priority and it omits the alkene in this compound.

The carbon chain is  carbons long (thus ""), and the presence of a double and triple bond make it necessary to put both the "" suffix (to designate the double bond) and the "" suffix (to designate the triple bond) in the name of the molecule. The carbon chain is numbered from left to right, putting the alkyne and alkene functionalities between carbons  and , respectively. To reduce ambiguity when there are multiple multiple bonds, it's advisable to put the locand of the multiple bond directly before the suffixes, rather than before the parent chain. When ordering the multiple bonds in the name of the molecule, the alkene suffix comes before the alkyne suffix, and the final "" in the suffix "" is dropped. Thus, the molecule is "."

Finally, stereochemistry must be taken into account. On carbon , the carbon chain is of higher priority than the implied hydrogen (when taking atomic number into account, per the Cahn-Ingold-Prelog rules), and is located "above" the double bond. On carbon , the alkyl group is higher priority, and is located "below" the double bond. Because the higher priority groups are across the double bond from each other, the molecule is given the "E" designation.

The longest carbon chain is  carbons long (thus ""), and the double bond makes it an alkene (thus ""). The highest priority functional group is the double bond, and the carbon chain must be numbered such that this functional group is given the lowest number carbon. Counting from either direction puts the double bond between carbons  and the molecule is thus a "." The substituents are a bromine on carbon  when counting from right to left, and  from left to right. The carbon chain is numbered so substituents are given the lowest numbers possible. Thus, the carbon chain is numbered from right to left, and the molecule is .

Finally, stereochemistry must be taken into account. On carbon , the carbon chain is of higher priority than the implied hydrogen (when taking atomic number into account, per the Cahn-Ingold-Prelog rules), and is located "below" the double bond. On carbon , the alkyl group is higher priority, and is located "above" the double bond. Because the higher priority groups are across the double bond from each other, the molecule is given the "E" designation.

The systematic IUPAC nomenclature of ethers (molecules of the structure ) is such that the smaller of the two alkyl groups is labeled as an "alkoxy" substituent of the larger chain. In the molecule shown, the longer chain has  carbons and does not contain any multiple bonds or functional groups. Therefore the parent chain is "." The ether functionality is thus called a "" group (a carbon chain one carbon long, "" attached to an oxygen "" which is attached to the parent chain by said oxygen), and is a substituent of the parent chain. Since there are multiple places the  group can attache to the parent chain, a locand must be specified in this case (thus "."

The carbon chain is four carbons long (thus ""), and the presence of a double bond makes it an alkene (thus ""). Numbering the carbon chain from left to right puts the double bond between carbons  and , while numbering from right to left puts the double bond between carbons  and . Because the alkene is the most important functionality in this molecule, the carbon chain is numbered such that the double bond gets the lower numbers (thus, the carbon chain is numbered from right to left). Because there are multiple possible positions of the double bond it is necessary to put a locand on the lower numbered carbon that is involved in the double bond. In this case, it would be called "."

The longest carbon chain is  carbons long (thus ""), and the lack of a double bond makes it an alkane (thus ""). The carbon chain must be numbered giving the substituents the lower numbers. Counting from right to left puts the bromine and methyl groups on carbons  and  respectively, which is lower than the designations counting from left to right ; therefore, the numbering goes from right to left. IUPAC rules state that the substituents are listed in alphabetical order when naming (thus ")."

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

We have an amide shown here:

We have an ethyl group shown here:

and Finally we have an alcohol (hydroxy) group 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 Acid derivatives including Amides have the highest priority then carbonyls then alcohols, amines, alkenes, alkynes, and alkanes, so in this case the Amide 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 carbonyl of the amide to be carbon number #1, 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 an amide compound is -amide, so in this case we have a hexanamide, which can also be written as hexan-1-amide, but this isn't needed as it as the amide is at carbon 1.

Next we have an ethyl group at carbon 4, which gives us:

4-ethylhexanamide.

Finally we have an alcohol group at carbon 5, which as a substituent is referred to as an hydroxy.

so we have a 5-hydroxy group.

Now we must order the substituents in alphabetical order to follow IUPAC rules, so our final answer is 4-ethyl-5-hydroxyhexanamide.

Now let's go over the wrong answer choices:

1) 6-carboxamide-3-ethylhexan-2-ol is wrong because it interprets the amide as being lower priority than the alcohol.

2) 4-ethyl-5-hydroxypentanamide is incorrect because the longest carbon chain is six carbons long, not 5.

3) 3-ethyl-2-hydroxyhexan-6-amide is wrong because the amide group isn't the lowest number it can possibly be, as according to IUPAC rules the highest priority functional group should have the lowest possible numbering.

4) 3-ethyl-2-hydroxypentan-5-amide is wrong because the amide isn't the lowest number it can be and the longest carbon chain is 6 carbons, not 5.

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:

An amide shown here:

A Bromine (bromo) group shown here:

A Fluorine (fluoro) group shown here:

and A chlorine (chloro) group 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 Acid derivatives including amides have the highest priority then carbonyls then alcohols, amines, alkenes, alkynes, and alkanes, so in this case the amide 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 that the amide is attached to to be #1, so let's start there and number the carbon chain, and then we go in alphabetical order for the substituents, so we make bromo a lower number than chloro.

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:

Since this is a cyclic compound we name the compound cyclohexanecarboxamide, as in IUPAC nomenclature you name the ring normally then put carboxamide at the end to show it has an amide attached to it.

Next we number the substituent alkyl halides so that they are in alphabetically order and the one that is nearest to the beginning of the alphabet has the lowest priority, so we have a 3-bromo, 4-fluoro, and 5-chloro.

Putting this together and ordering the substituents named in alphabetical order, we get 3-bromo-5-chloro-4-fluorocyclohexanecarboxamide as our final answer.

Now let's go over the wrong answers:

1) 3-bromo-5-chloro-4-fluorocyclohexanamide is wrong because in IUPAC nomenclature we must name the ring normally and then put carboxamide after it.

2) 3-bromo-5-chloro-4-fluorohexanamide is wrong because it ignores the fact that this compound is cyclic (a ring), and would be the correct name if this were a straight chain compound.

3) 5-bromo-3-chloro-4-fluorocyclohexanecarboxamide is wrong because the substituents aren't numbered in alphabetical order so that the lowest number alkyl halide is closest to the beginning of the alphabet.

4) 5-bromo-3-chloro-4-fluorocyclohexanamide is wrong because the substituents aren't numbered in alphabetical order so that the lowest number alkyl halide is closest to the beginning of the alphabet. It is also wrong because in IUPAC nomenclature we must name the ring normally and then put carboxamide after it.