CHEM 2211 1nd Edition Lecture 19Outline of Last Lecture I. midterm Outline of Current Lecture I. carbocation stability II. reaction of alkenes Current LectureCarbocation stability. Carbocations want to form on the most stable carbon they can find. With alkenes this is either carbon involved in the double bond.- Most stable o Tertiary carbocation (+CR3)o Secondary carbocation (+CHR2)o Primary carbocation (+CH2R)o Methyl carbocation (+CH3)- Least stable Reaction of alkenes Addition of hydrogen halides to alkenes.H3C CH3H3C CH3C=C + HBr CH-C BrH3C CH3H3C CH3The product would be the same no matter what C was chosen to put the H or the Br because the molecule is symmetrical and both are tertiary carbocations once the double bond is broken These 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.The process for adding a hydrogen halide will be the same no matter which halide it is HCl, HI, HBr or HF. You have to look at which C across the double bond will be a more stable carbocationand put the H on the opposite one and then the halide on that carbocation C. Unless the molecule is symmetrical across the double bond then there will be two different products you can form but one will be more in abundance than the other or one will not be ableto form at all, this is determined by the stability. Such as: H3C H3C H3CC=CH2+ HCl Cl C-CH3 (1) or CH - CH2 – Cl (2)H3C H3C H3CIn this case product 1 produced the more stable carbocation on the tertiary C. so it is the only product formed. It is produced faster than the other one because is it a more stable carbocation. The addition of water to an alkene Water cannot react with an alkene because it is to weakly acidic. To combat this problem H2SO4is added to the reaction over the arrow and acts as the electrophile needed. The product of this reaction is an alcohol. H2SO4CH3CH=CH2 + H2O CH3CHCH3
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