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CHM2210 Test 3 Dr Hilinski Study Guide Potential Energy Diagrams The point of maximum potential energy encountered by the reactants as the proceed to products is called the transition step The difference between the reactants and the transition state is known as the activation energy A plot of potential energy vs reaction coordinate Draw the transition state for each of the reactions in the mechanism Use dashed lines to show bonds that are breaking and forming in the reaction Include partial and partial where they belong When carbocations are formed they have a high energy level so place them at the top of the graph If the reaction in the mechanism is endo thermic it will be product like and placed high on the graph If the reaction in the mechanism is exo thermic it will be reactant like Alkyl groups attached to the carbocation make it more stable The electro atom pulls electrons from the Carbon so if you have to choose the most stable compounds choose the one with less electronegativity The most stable Tertiary carbocation Secondary Primary methyl Only electrons that are Beta to the positively charged Carbon can stabilize it by hyper conjugation Beta ones are the ones that surround the positive carbon directly Rate Laws The rate of reaction of alcohols is Tertiary carbocation Secondary Primary methyl Arrhenius equation for activation energy determines how fast the reaction will happen k Ae E RT a SN1 E1 rate law for secondary and tertiary Rate k RO H2 SN2 E2 rate law for secondary and tertiary Rate k RO H2 x OH groups are bad leaving groups H2O groups are good leaving groups Naming Alkenes Highest Priority OH C C R X Count the longest Carbon chain through the double bond In a ring you start at the double bond and count giving the substituents the lowest number put the ene in the name of the ring Free Radical Substitution Reaction Don t use Fluorine or Iodine because they are too unreactive Bromine always needs heat and light Chlorine can use either heat or light Mechanism Initiation the Cl2 or Br2 goes through bond hemolysis and splits in half Propagation a The Cl Br takes a H from the most likely spot tertiary secondary primary and forms acid and a radical a molecule with 1 free electron b The radical and the remaining Cl Br combine to form the target molecule Termination propagation a Cl2 or Br2 is reformed b The 2 radicals join together c The radical and the Cl Br bond together most of this was formed in Phenols are often free radical inhibitors tert butyl cation has an Sp2 Carbon and the anion has an Sp3 Carbon because the anion has an electron pair forming another sigma orbital Chlorination and Bromination When giving all possible mono chlorination bromination products be careful to not give a repeat To calculate the of each product formation you divide the number of possible hydrogen s it can replace by the total number of spots the Cl or Br can bind to Chapter 5 E and Z Labeling E trans Z cis When adding E or Z to an alkene you have to determine the highest priority and lowest priority group on each side of the double bond To determine the priority you label what the first carbon is attached to and if it is tied you progress outward until the first point of difference Rules 1 Higher atomic number takes precedence over lower 2 When two atoms attached to the same carbon of the double bond are identical compare the atoms attached to these two on the basis of their atomic numbers 3 Work outward from the point of attachment 4 If an atom has a double bond you count what ever it is bonded to twice Cis and Trans is not possible when a carbon in the double bond of the alkene has two identical substituents NEVER use E or Z to describe substituents on a cyclic compound Physical Properties A polar molecule with a dipole moment has a higher boiling point then a non polar The more substituents on the double bond the more stable it is giving it a lower bp and molecule with no dipole moment heat of combustion Alkenes 8 carbons are more stable in the cis form Alkenes between 8 and 11 Carbons are more stable in the trans form Alkenes 11 carbons are equally stable in each form In cyclohexene there is a half chair confirmation and the substituents go from equatorial and axial to pseudoaxial and pseudoequatorial Preparation of Alkenes Elimination Reactions Zaitsev s rule elimination reactions the major reaction product is the alkene with the more highly substituted more stable double bond This most substituted alkene is also the most stable Regioselective when a reaction can proceed in more than one direction but one of the directions is preferred Stereoselective a reaction with a single starting material that yields 2 or more stereoisomeric products but one of them is produced in greater amounts Dehydration of Alcohols H2SO4 is the common acid used Tertiary alcohols dehydrate at the lowest temperatures Mechanism for E1 1 The OH on the alcohol grabs an H from H3O to form an oxonium ion and water 2 Water breaks off from the oxonium ion to form a carbocation 3 A hydrogen from the carbon without the positive charge is taken by water and its electrons go to forming the double bond to make an alkene and H3O Mechanism for E2 1 The OH on the alcohol grabs an H from H3O to form an oxonium ion and water 2 Water steals a proton from the alcohol allowing its electrons to form a double bond and water also breaks off of the alcohol making an alkene H3O and water If the carbon skeleton of the product looks different then the reactant then a rearrangement could have occurred The carbocation in a rearrangement can either rearrange or give an alkene Rearrangements occur to create a more stable carbocation secondary is usually made into tertiary Dehydrohalogenation The loss of a hydrogen and a halogen from an alkane to form an alkene The preparation happens with a strong base Beta elimination predominates in the direction that leads to the more highly substituted alkene It is regioselective and sterioselective forming the more stable stereoisomer cis or trans When a ring is less than 10 carbons the reaction produces cis In an E2 mechanism the main things taking place are 1 Base Hydrogen bond forming 2 C H bond breaking 3 C C formation from the H s electrons it leaves behind 4 C X bond breaking Stereoelectronic effect when one special arrangement of electrons is more stable than another There is stereoelectronic preference for the anti coplanar arrangement of proton and leaving group in E2 reactions A C D

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FSU CHM 2210 - Study Guide

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