Dr.Norris,
say you have a cyclopentane substituent on an alkene. It's just 2 carbons with a double bond & a cyclopentane attached to one. You react it with HBr and HOOH. The bromine radical bonds via the donation of an electron from the pi bond. Now, in the next step, is there a Bromine substituent, an extra electron, and a CH2 attached? or just a Bromine and an extra electron? & if the first, how does the hydrogen bond? because the carbon would already have 4 bonds...or can the radical only attach to the CH2 group, giving only one product? idk, sorry, I'm just really confused.

Dr. Norris, with regard to the E1, when you get to the carbocation( in the slide notes) you break off a C-H bond from one of the methyl groups to form the double bond and produce the product. Does it matter which bond you break? Also, the main reason why we use sulfuric acid and not HCL is to get the alkene? Also, is there only one product from that rxn. For instance, methylcyclohexanol has two products, is that because its a regioisomer?

nevermind Dr.Norris, I think I figured out the HBr/HOOH problem. I got a cyclopentane with a CH2-Br attached and a 1-Bromo-1-Methylcyclopentane.

Edited by author 10-22-2009 10:01 PM
Alan - In the second step of the E1 process (deprotonation), as long as the H is in the beta position then it can be removed to give an alkene. HCl (and HBr and HI) will give SN1 and not E1 since the conjugate bases (Cl, Br, I anions) are good nucleophiles; HSO4- is not and E1 occurs instead. Regioisomers are possible when you have more than one type of beta hydrogen.

So are all the reactions we've done exothermic as a whole?

Ouch - mostly, yes. The acid-catalyzed E1 is the main exception which is why we spent time discussing its relationship to the acid-catalyzed hydration of alkenes in Chapter 6.

Thanks...as far as drawing the last transition state of the E1, how do you show the forming of the double bond and the hydronium ion?

Alan - dashed line from H2O to the proton being removed, dashed line from that proton to the carbon it is breaking away from, then a dashed line to the adjacent carbon indicating the pi bond forming. Partial positive charges on the O of H2O and the carbon that was the carbocation.

I'm confused for number 10b on the 2006 exam why is the answer pentane instead of hexane

Amy - because the longest continuous carbon chain has 5 carbons in it and not 6 (?)

in the mechanism for the reaction where we had a cyclohexane with a D and a Br in anti formation that were already lined up to speed up the reaction, is this a concerted process? specifically E2?

Sara - indeed it is.

Dr. Norris,
For the reaction profile of an acid-catalyzed hydration of an alkene, at least on the flash card, why does it only have two transition states (assuming by there only being two hills) when it appears like there would be three looking at the mechanism?

Dr.Norris,
on the fall 07 exam, question 2, b, the answer key says electrophillic addition. wouldn't it be acid-catalyzed hydration?

Josh - that profile is a generic picture for an electrophilic addition (e.g. with HCl, HBr, H2O). There would be a third step in acid-catalyzed hydration which would be the deprotonation of the alkoxonium ion by solvent (water) to give the alcohol product.

Jennifer - acid-catalyzed hydration is an example of electrophilic addition.