Organic compounds and solubility (1 Viewer)

erucibon

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Explain why these organic compounds have very different solubilities in water.
C4H10 - 0.061 g/L
C4H9OH - 73.0 g/L
C4H8O - 27.5 g/L

How would I answer this?
Thanks
 

jazz519

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You need to understand what solubility is related to

Solubility means that when we combine two solutions they form a homogenous solution

This occurs when ''like dissolves like''

I.e. non-polar dissolves non-polar
and polar dissolves polar

So then you also need to be able to determine what makes something non-polar and what makes something polar:
Non-polar compounds are ones that are symmetrical in the horizontal and vertical axes
Polar compounds are ones that are asymmetrical in EITHER axes (i.e. it can be symmetrical in horizontal and then assymetrical in the vertical which would make it polar)
Screen Shot 2019-07-30 at 6.36.04 pm.png
So if we look at our structures, the C4H10 is symmetrical in both axes therefore is non-polar
The C4H9OH is symmetrical in horizontal axes, BUT is not symmetrical in vertical axes therefore is polar
The C4H8O is symmetrical in horizontal axes, BUT is not symmetrical in vertical axes therefore is polar

Now you have to look at the thing you are trying to dissolve it in. In this case we have water, which is polar because it is asymmetrical

Therefore, because like dissolves like dissolves, this means that C4H9OH and C4H8O will have a greater solubility in water compared to the non-polar C4H10

This however, won't explain why the C4H9OH is more soluble in water. Recall that compounds that have OH, NH or HF groups can also form hydrogen bonds

Because water has an OH and the C4H9OH has an OH they can form hydrogen bonding as well to increase the dissolving ability, which is the reason why it is more soluble than the C4H8O
 

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