Chemical Properties of Organics

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:

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