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Q: polarity of benzaldehyde and benzoin ( Answered 5 out of 5 stars,   0 Comments )
Subject: polarity of benzaldehyde and benzoin
Category: Science > Chemistry
Asked by: shadowrider-ga
List Price: $3.00
Posted: 09 Apr 2004 22:24 PDT
Expires: 09 May 2004 22:24 PDT
Question ID: 327993
According to the Thin Layer Chromatography we did in class on silica
gel impregnated TLC plate, benzaldehyde is less polar than benzoin
(i.e., benzaldehyde traveled farther up the TLC plate and had a higher
Rf) but I thought the dual benzene rings on benzoin would make the
molecule less polar - can anyone explain????

Thank you very much!!!!!

Subject: Re: polarity of benzaldehyde and benzoin
Answered By: andrewxmp-ga on 10 Apr 2004 16:20 PDT
Rated:5 out of 5 stars
Hi Lee,

I think the problem here is arising from a confusion about what makes
one molecule more polar than another.  Although there is a second
benzene ring on benzoin, benzene rings themselves do not alter the
polarity of a molecule.  In fact, a benzene ring is completely
nonpolar, because it is symmetrical in both bond distance and
structure, as well as in it's electronic structure.  More detail about
why this is so can be found at:
[ ]

From the above page, the last paragraph pertains to benzene rings (as
they consist only of C and H atoms):
"When molecules begin to get very large it is necessary to examine the
polarities of either specific groups or combinations of groups. We can
make some simplifying assumptions, for example, molecules or portions
of molecules consisting of only C and H will be mostly nonpolar since
the bond polarities are quite small and the vectors generally cancel.?

So if the benzene rings do not influence overall molecular polarity,
then what does?  This is a property of the functional groups on the
molecule.  Different groups have larger dipole moment associated with
them, and thus contribute differently to the overall molecular
polarity.  A good chart summarizing the different functional groups
and their characteristic relative polarity can be found at the bottom
of the page at:
[ ]

In order of increasing polarity (which move slower on the TLC plate):
halogenated hydrocarbons
aromatic hydrocarbons
aldehydes and ketones
carboxylic acids

So with this in mind, let's look at the two molecules in question. 
Benzaldehyde is simply a benzene ring with an aldehyde group attached.
 In the aldehyde group, it is primarily the C=O double bond that has
the large dipole moment which contributes to the polarity of this
functional group.  The benzoin molecule has a C=O bond as well, but
bonded to two other carbons, thus it is a ketone.  So, roughly
speaking, both molecules up to this point have the same amount of
dipole-causing groups, and this is supported by the fact that on the
above chart ketones and aldehydes are considered about equal as far as
polarity goes.  The difference between benzaldehyde and benzoin,
however, is that benzoin has an additional alcohol (OH) group
attached.  This functional group contributes it's own dipole moment to
the molecule as a whole, thus making the molecule more polar than

I trust this has answered your question sufficiently.  If a
clarification is necessary, please request one before rating this
answer.  Thank you for coming to Google Answers!

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