'A' Level Chemistry Problem Analysis: Similar Positive Iodoform Test Results For Alcohols And Carbonyl Compounds |
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Thought process: They give similar results because their mechanisms are similar, and in fact, identical beyond the initial 2 steps. In the mechanism for the alcohols with the CH3CH(OH)- group, the first step is the oxidation of the alcohol to an aldehyde or ketone (depending on the 'R' group). (Although one might think of potassium iodide as an electron donor (rather than a hydride donor like LiAlH4 in dry ether), and hence iodine merely accepts electrons in carrying out oxidation; but in the presence of hydroxide ions, the combination of hydroxide ions and iodine can oxidize by removing hydrogen atoms. Hydroxide ions remove protons, while the excess electrons are removed by the iodine.) First, an enol (ie. alkene alcohol) intermediate is formed when both the alpha and beta carbons (ie. the 1st and 2nd carbon atoms from the hydroxy functional group) are deprotonated, and the excess (pair of) electrons are removed by an iodine molecule (consequently reduced to a pair of iodide ions). A lone pair on the (hydroxy group) oxygen shifts down to form a pi bond with the alpha carbon, in turn creating the carbonyl group. To avoid violating the carbon's octet, the alkene pi bond nucleophilically attacks a (partially positively charged, due to an instantaneous dipole) iodine atom (of an iodine molecule). The formal charge on the (hydroxy group) oxygen is removed by the deprotonation of the hydroxy group. Alternatively, you can summarize the above mechanism by a redox equation that sees the alcohol (with the CH3CH(OH)- group) oxidized to a carbonyl compound (with the CH3CO- group), written as follows : From this point on (step 3), the mechanism is exactly identical to that of the mechanism for tri-iodomethane (iodoform) test for carbonyl compounds with the CH3CO- group. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The above content is contributed by Mr Heng, owner and 'A' Level Chemistry tutor at Bedok Funland JC. He also goes by the handle UltimaOnline on various online popular homework forums. |
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