Topics: Alcohol, Solubility, Chemical polarity Pages: 1 (323 words) Published: December 5, 2013
Benzophenone’s nonpolar nature makes it insoluble with water (which is polar), soluble with methanol (which is of intermediate polarity) and soluble with hexane (which is nonpolar). This is because generally, like dissolves like -- nonpolar will dissolve nonpolar but not polar, and vice versa. But note that this is an imperfect science -- you would expect hexane to dissolve faster and more completely than methanol (because of the differing polarities), but the opposite occurred. This may be because of benzophenone’s ketone group, which adds a slight polarity to the molecule, making it a little polar, although mostly nonpolar. Biphenyl, which is completely nonpolar, will have similar solubilities as benzophenone: insoluble with water (polar), partially soluble with methanol (intermediate polarity) and soluble with hexane (nonpolar). But with biphenyl (unlike with benzophenone) the solubilities are exactly as expected, because of biphenyl’s complete nonpolarity. But that explains any differences in solubilities between the two molecules. Polarity also played a role in the alcohol reactions: water, which is polar, will be insoluble with 1-octanol (which is nonpolar), insoluble with 1-butanol (also nonpolar), and soluble with methanol (polar). But it’s a little more complicated: the carbon chain of each of these organic compounds will always be nonpolar and therefore hydrophobic, but as these are alcohols, the -OH group, which is polar and therefore hydrophilic, comes into play as well. So it’s kind of a battle between the hydrophobic carbon chain and the hydrophilic alcohol group, and when the chain is long, as in octanol, it will overpower the alcohol. Conversely, in small alcohols like methanol (or ethanol etc.), where the carbon chain is much shorter, it will exert a smaller hydrophobic force and the overall molecule will therefore be hydrophilic. The hexane-alcohol reactions had the opposite solubilities, due to hexane’s nonpolar nature (and the rule of like...
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