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Introduction to Organic Chem ch. 4 #31

Posted: Sat Mar 04, 2017 9:51 pm
by Michael Lonsway 3O
For the electrophilic addition reaction of hydrogen bromide, HBr, to propene, CH3CHCH2, producing 2-bromopropane, CH3CHBrCH3, write the rate laws for each elementary step and the overall rate law. Explain your answer for the overall rate law.

How can we determine which step is the slow step to create the rate law? I'm confused since we weren't given what the hw reaction looks like.

Re: Introduction to Organic Chem ch. 4 #31

Posted: Sat Mar 04, 2017 10:18 pm
by Christopher Reed 1H

In step one of the reaction two bonds are broken: pi bond in c=c double bond in propENE and H-Br sigma bond.

In step two of the reaction not bonds are broken, rather one is formed.

Since more bonds were broken in step one than step two, the activation energy of step 1 will be greater than step 2: Ea1 > Ea2

Since Ea1 > Ea2, the rate constant (think miles per hour) for step 1 will be smaller than that of step 2: k1 < k2.

How did I know it was two steps? I don't have a definitive answer, but I was just basing it off examples I've seen and previous problems. I figured that in the first step the double bond would attack the partially charged hydrogen and the electron lost in HBr bond being broken would go to Br and make it Br-. This leaves you with a positively charged carbon and a bromide. In previous examples, the joining of these two happened in a separate step. So it was mainly intuition and patterns that got me there, not any hard rules.

I'm sorry I can't give you a firm answer on that part of your question, but perhaps someone else can?

Re: Introduction to Organic Chem ch. 4 #31

Posted: Fri Mar 10, 2017 12:28 am
by edward_qiao_3I
The reaction is two steps because it creates intermediates. The slower step is determined by which one takes more energy, which in this case means breaking more bonds. Step one breaks one pi and one sigma bond, while step two only forms bonds, so step one takes more energy