CHEM 3331 1nd Edition Lecture 17Outline of Last Lecture I. AlkenesII. Elements of unsaturationIII. NomenclatureIV. UsesV. StabilityVI. Physical propertiesOutline of Current Lecture I. Formation of alkenesa) E2b) Vicinal dibromidec) E1d) Dehydration of alcoholsII. Industrial methods of formationCurrent LectureI. Formation of alkenesa) E2 reactionA strong base attaches to a periphery hydrogen causing a double bond to form and pushing our leaving group off of the molecule. To avoid competition with Sn2 we want to use a strong bulky base and if possible have a tertiary substrate.With a strong small base we get the product we want but we also form some Sn2 products. The size of the base can also determine the major product produced. Potassium t-butoxide major product is a double bond on the first carbon. Potassium ethoxide will produce a major product with a double bond on the second carbon. For E2 reaction we want the antiperiplanar conformation between our leaving group and a hydrogen atomThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.b) Vicinal dibromideThis reaction is driven by the formation of I-Br. I- attaches to a bromine creating a positive charge and a need for a double bond and pushes the remaining bromine off to form a double bond between the two carbons where the bromines were detached. This is also done with zinc which can pick up both bromines. c) E1As we know the E1 reaction creates a carbocation by the removal of our leaving group. It hten forms a double bond. d) Dehydration of AlcoholsWe introduce an acid, such as H2SO4. This acid donates a hydrogen to the OH group. Resulting in O(+)H2 and HSO4-. The water is removed leving a positive charge on the carbon the OH groupwas attached to. This H20 then bonds to an adjacent hydrogen atom creating a double bond between carbons and H3O+. II. Industrial methods of formationWe have already talked about fracturing. A single long chain of hydrocarbons is broken into two smaller chains of hydrocarbons.Another method is dehydrogenation. An alkane is exposed to heat and a catalyst to form H2 andan alkene. We break two sigma bonds for a sigma bond in H-H and a pi bond in C=C. this is energetically unfavorable. However, we can increase the temperature to achieve a negative ΔG, to achieve spontaneity. Our ΔS is greater than zero because we go from one molecule to two molecules in the
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