When H and OH add to the alkyne, an enol is formed, which rearranges to form a carbonyl (C=O) group:

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1 Next Up: Addition of, : The next two reactions are the Markovnikov and non-markovnikov additions of and to an alkyne But you will not see alcohols form in this reaction! When and add to the alkyne, an enol is formed, which rearranges to form a carbonyl (=) group: enol This particular type of rearrangement is called a tautomerization and thus forms aldehydes or ketones in these reactions: enol Tautomerization always favors the carbonyl form (look at the equilibrium arrows above) presumably because the electrons in the pi system are more stable closer to the electronegative oxygen atom. 6. Mercury-catalyzed hydration (1X Addition of, ) -Markovnikov 2 S 4, 2 gs 4 Addition of, in a Markovnikov fashion always produces an enol, which then undergoes tautomerization to form a ketone. These products always seem harder to draw. It may be easier for you if you draw the intermediate enol (physically draw or mentally note!), placing the on the more substituted end. When you tautomerize (to draw the final product), the carbonyl (=) always forms where the was attached. 1

2 2 S 4, 2 gs 4 2 S 4, 2 gs 4 Terminal alkynes always produce methyl ketones (why are they called that?) and symmetrical alkynes will always produce the same ketones. egiochemistry doesn t matter because symmetrical internal alkynes are equally substituted with the same group. Markovnikov doesn t matter 2 S 4, 2 gs 4 As with the addition of X, asymmetrical alkynes are a poor choice for starting material. Because the two ends are equally substituted with one alkyl group, they are equally good at stabilizing the carbocation intermediate and Markovnikov cannot be easily discerned the reaction produces two different ketone products. Bad eaction! gs 4, 2 2 S 4 More Examples: 2 S 4, 2 gs 4 2 S 4, 2 gs 4 2, 2 S 4 2

3 1. g(ac) 2, 2 2. NaB 4 7. ydroboration of Alkynes (1X Addn of, ) Non-Markovnikov This reaction commonly uses disiamylborane instead of B 3 to prevent multiple additions to the alkyne pi system (more sterics slow down addition process to the TW pi bonds of alkyne). B Addition of disiamylborane to a terminal alkyne in a Non-Markovnikov fashion always produces an enol that tautomerizes to form an aldehyde. 1. disiamylborane 2. Na, 2 2 Tautomerization of an enol: 1. disiamylborane 2. Na, 2 2 For the addition to internal alkynes, we can use B 3 since there are already sterics in the molecule surrounding both ends of the alkyne. Symmetrical Internal Alkyne will always produce a single ketone product due to symmetry. egiochemistry doesn t matter since its symmetrical and addition in either manner forms the same four-membered cyclic transition state. 3

4 1. B 3 2. Na, 2 2 As with the mercury-catalyzed hydration, hydroboration of an asymmetrical internal alkyne is problematic. Because both ends are equally substituted but not symmetrical, the reaction cannot distinguish the Non-Markovnikov transition state, resulting in two different ketone products. Bad eaction! Try these: 1. B 3 1. B 3 2. Na, Na, Disiamylborane 2. Na, Disiamylborane 2. Na, B 3 2. Na, xidative leavage of Alkynes -Severs both pi bonds as well as the sigma bond -Always form arboxylic Acids (or 2 ) -Use either reagent and get SAME results: KMn 4, 2 or Zn, 3 + For Internal Symmetrical Alkynes both ends form same carboxylic acid: Zn, 3 + 4

5 For Internal Asymmetrical Alkynes forms two different carboxylic acids KMn 4, 3 + For Terminal Alkynes one end has only a hydrogen atom attached. In the oxidative cleavage process, that carbon will be oxidized to its full extent forming 2 (or carbonic acid, if you like) Zn, 3 + KMn 4, Zn, Zn, 3 + KMn 4, 3 + And Finally: 9. Acetylide Anion Formation and eaction Powerful - bond forming reaction increases length of carbon chain - uses Acetylene (- -) or any terminal alkyne Watch this general reaction: In hapter 9: Acetylide Anion eaction In hapter 18: Williamson Ether Synthesis 5

6 ' ' In hapter 21 (EM 332): ' ' In hapter 22 (EM 332): Bronsted-Lowry Acid-Base Theory: Acid + Base onj Base + onj Acid pka = 25 + NaN 2 Na + N 3 pka = 35 Alkene protons (vinylic protons) have a pka of about 45 and alkane protons have a pka of 60! Why are terminal alkynes acidic? (Why is the conjugate base stable?) B: + B The anion that forms is contained in an sp hybridized orbital, the shortest, roundest hybrid orbital of all the hybrids, and orbital that will hold the electron-pair closest to the positive carbon nucleus, stabilizing the anion. Acetylide anions are nucleophiles and they react with electrophiles. The electrophiles of choice for this section are alkyl halides. Acetylide anions can be alkylated (addition of alkyl groups, either a methyl group, - 3, or any primary alkyl group, - 2.) 6

7 NaN 2 Na + N 3 E + E Alkyl halides are electrophiles, due to the polarity of the -X bond: δ + δ - l eaction with alkyl halide will result in addition of carbon chain to alkyne, where the ydrogen atom once was: Na Br NaBr 1. NaN 2, N Br 1. NaN 2, N 3 2. l 1. NaN 2, N 3 2. I 7