Chapter 18 Source Answers and Examples

Step 1: Identify the base chain.

The base chain is the longest chain. For this molecule, the entire molecule comprises the base chain.

Step 2: Determine the chain prefix.

Our molecule has 8 carbon atoms, therefore we have a molecule that begins with $oct-$.

Step 3: Determine the chain suffix.

This molecule has three double bonds, so our suffix is $-triene$.

Step 4: Number the carbon atoms.

Numbering from the right to the left gives us our earliest point for a double bond. Therefore, our numbered molecule is:

Step 5: Number the double bonds (if any).

According to our numbering scheme, our double bonds are on carbon numbers 1, 4, and 6.

Step 6: Compile name.

With a prefix $oct-$ and a suffix $-triene$, our full molecule name is $1,4,6\text{-}octatriene$. The a was added simply to provide a link between the two consonants in the name.

$1,4,6\text{-}octatriene$

Step 1: Draw the base chain.

In this name, the base chain is at the end, the $1,3\text{-}cyclopentadiene$. This means we have a cyclohexane of 5 carbons, with two double bonds:

Step 2: Add the side-chains.

Our first side chain is an $ethyl$ group on the first carbon:

Next, we have two $methyl$ groups on the fifth carbon. We add both to the same carbon, and wind up with our final molecular structure:

Step 1: Identify the base chain.

We can see that the longest chain is the cyclical part of the molecule. Therefore, our base chain is $cyclopentane$.

Step 2: Identify the highest priority functional group.

The functional groups on our base chain are two ethers, a halogen, an alcohol, and a ketone. Of this, the ketone is the highest priority.

Step 3: Number the carbons in the base chain.

We have a cyclical base chain, which means we can start from any carbon. Therefore, we set our highest priority functional group (the ketone) to be on carbon #1. From there, we can go clockwise or counter-clockwise.

To decide, we look for the next highest priority functional group.

The next highest priority is the alcohol (treat ethers similar to side-chains). This means that we need to minimize the number of the carbon the alcohol group is on. Going clockwise, that number is 3. Counterclockwise gives us 4. Therefore, we number the carbon atoms clockwise:

Step 4: Determine base chain name.

We know that our base chain is $cyclopentane$. We need to change that suffix, though. Our highest priority functional group was a ketone. The ketone is on carbon number 1. Therefore, our base chain’s full name is:

$1\text{-}cyclopentanone$

Step 5: Catalogue the other functional groups.

We have 4 other functional groups, two ethers, a chlorine, and an alcohol group. Assign their carbon numbers:

$\begin{matrix} Ether\ A: 2&Alcohol:3\\ Chlorine: 4&Ether\ B:5 \end{matrix}$

Step 6: Determine prefixes from functional groups.

Our first ether is itself an ether. Therefore, we start from the outside and replace each $-ane$ with $-oxy$. On the outside we have a 4 carbon chain: $butoxy$. Following that, we have a 2 carbon chain: $ethoxy$ , which is ethered with our main chain. This makes this functional group have a prefix: $butoxtethoxy$.

Going around, next we have the alcohol. The prefix for this is $hydroxy-$.

Chlorine’s prefix is $chloro-$

The last ether is a simple ether with methane. This makes this prefix $methoxy-$

Step 7: Order the prefixes alphabetically.

The prefixes are:

$butoxyethoxy, hydroxy, chloro, methoxy$

Ordering those, we get:

$butoxyethoxy, chloro, hydroxy, methoxy$

Step 8: Assemble the name.

By coupling the carbon numbers with each prefix, as we have decided previously.

This gives is our molecules name:

$\begin{matrix}2\text{-}butoxyethoxy\text{-}4\text{-}chloro\text{-}3\text{-}hydroxy\text{-}\\5\text{-}methoxy\text{-}1\text{-}cyclopentanone \end{matrix}$

Step 1: Draw each molecule.

We can start with the easiest one, $hexane$. This is simply a saturated 6 carbon chain:

$Propanoic\ Acid$ is a carboxylic acid with a base of $propane$:

$Methyl\ proponoate$ is an ester with $methane$ and $propane$:

Finally, $2,3\text{-}dimethylbutane$ is a $butane$ chain with two $methyl$ groups attached to it:

Step 2: Determine dominant IMF.

$Methyl\ propanoate$ has some oxygen, but none are attached to any hydrogen, so we don’t have the full suite needed for H-bonding. The oxygen do induce a dipole on the molecule, though, allowing the development of D-D bonding.

$Propanoic\ acid$ has a couple of oxygen atoms, and one is even bonded to a hydrogen! This means we have both parts necessary for H-bonding to occur, making H-Bonds the dominant bonding for this molecule.

$2,3\text{-}dimethylbutane$ is a branched aliphatic, with no functional groups aside from the side-chains. Therefore, we don’t have anything except for ID-ID bonding.

In $hexane$, we don’t have any atoms that can be used in H-bonding, and there aren’t any dipoles created across the molecule. This means that ID-ID bonding is the dominant force.

Step 3: Rank the molecules

We can start by ranking the molecules in order of dominant bonding strength:

$Propanoic\ acid>Methyl\ propanoate>hexane$

$hexane = 2,3\text{-}dimethylbutane$

Now we just need to pick between $hexane$ and $2,3\text{-}dimethylbutane$. Both have very similar molecular weights, so we have to look at the structure. As a branched aliphatic, $2,3\text{-}dimethylbutane$ has little opportunity to mingle with its peers, unlike $hexane$. This means that the tight-knit $hexane$ molecules have a higher IMF than $2,3\text{-}dimethylbutane$. The final rankings are:

$Propanoic\ acid >Methyl\ propanoate>hexane>2,3\text{-}dimetyhlbutane$

Step 1: Draw each molecule.

We can start with the easier ones, $octane$ and $hexane$. These two are simple saturated hydrocarbons:

Next we can draw up the halogenated $2\text{-}chloropentane$, which is a $pentane$ molecule with a chloride group on the second carbon:

The branched aliphatic, $3\text{-}methylpentane$ is easy to draw:

The last two are each a form of $pentane$ with a single functional group.

$2\text{-}pentanol$ has an alcohol group:

Step 2: Group the atoms into ‘likely soluble’ or ‘likely insoluble’

We can group the pure hydrocarbons easily into the ‘likely insoluble’ group. Next, we see that $2\text{-}pentanone$ and $2\text{-}pentanol$ can participate in H-bonding, making them ‘likely soluble’. As for the $2\text{-}chloropentane$, the chlorine gives the molecule a bit of a dipole, but with a 5-carbon chain to polarize, it remains ‘likely insoluble’.

Step 3: Rank the molecules in each group.

Starting with the ‘likely soluble’

We have $2\text{-}pentanone$ and $2\text{-}pentanol$. Both have compounds that permit H-bonding, but $2\text{-}pentanol$ is the only one that can act as both an H-Donor and an H-acceptor. As their structures are nigh-identical otherwise, we can say that $2\text{-}pentanone$ will be less soluble.

$2\text{-}pentanol > 2\text{-}pentanone$

Move on to the ‘likely insoluble’

Of our four molecules in this group, only one has any sort of dipole. It is also tied for the shortest, making $2\text{-}chloropentane$ the choice for most soluble. The remaining are all pure hydrocarbons. We see that $octane$ is unbranched and the longest, making it the least soluble of the group. Remaining is $3\text{-}methylpentane$ and $hexane$. As we know branching tends to increase the solubility, and as that is the only difference in our last two molecules, we can say that $hexane$ is the lesser soluble. This makes our overall ranking:

$\begin{matrix} 2\text{-}pentanol>2\text{-}pentanone>2\text{-}chloropentane>\\ 3\text{-}methylpentane>hexane>octane \end{matrix}$