Which reactant in each of the following pairs undergoes an eli...

Question

Which reactant in each of the following pairs undergoes an elimination reaction more rapidly? Explain your choice.

a. Comparison of two reactants, (CH3)3CCl and (CH3)3CBr, for elimination reaction rates with CH3O− as the base.

b. Two molecular structures showing elimination reactions with CH3O- as a reactant, comparing two pairs of reactants.

Accepted Answer

Hey, everyone. And welcome back, which alky Hali from each of the given pairs reacts more rapidly in an elimination reaction. Provide a short explanation of your answer. Part A, we are comparing one Chloro propane and one bromo propane and part B. We are comparing chair confirmations of two chloro 13 dimethyl cycle hexane. Let's start with part A and we noticed that the main difference is that if we think about an elimination reaction where we are removing our beta hydrogen followed by the loss of the living group, let's consider elimination E two, right? Because we have primary substrates. So the base comes in captures the proton, then we are forming our Pyon followed by the loss of the leaving group. And that's exactly what we observed for the second structure. So we have a very similar mechanism for the formation of our alkin. The main difference is that we have our chloral group in the first structure and our bromo group in the second one. Let's remember that with an increase in atomic radius, the leaving group becomes better when we consider halogens, right. So if we specifically consider chlorine, bromine and iodine, according to the periodic table, if we go down the group, the atomic radius R increases. And that means this strength of the living group increases as well, right. So we can say that the strength of the leaving group also increases. So we can draw another error representing a correlation between an increase in radius and an increase in strength of the living group. So that means bromine because it has a greater radius than chlorine is a better living group. So, Bromine is a better living group than chlorine. And because the leaving group affects the rate of an elimination reaction, the better the leaving group, the faster the reaction rate, meaning structure two would correspond to a more rapid reaction, right? Basically, because it has a better living group, chlorine has a smaller radius. It's a worse living group than Bromine. And therefore, Bromine being a better living group with produce an alky halite that undergoes a faster elimination reaction. So we're going to choose structure two or basically one bromo propane as our answer. The reason why it undergoes a faster elimination reaction is because it has a better living group. Now, for b we have a slightly different scenario because if we notice our alpha position, which has our leaving group, that's chlorine, right? The structural arrangement is different in terms of spatial arrangement. But thinking about the linkage of our metal groups, they are bonded to the same carbon atoms. So if we enumerate them 123123, both structures correspond to 13 dimethyl structures, right? And at position number two, we have our Chloro substituent. So the linkage is exactly same, the leaving group is exactly the same meaning for this one. Because we're considering our spatial arrangement, then we have to recall our N C co plan or requirement for an elimination reaction E two. And the reason why we're considering an E two reaction is because both substrates are secondary substrates. So we don't need to worry about an E one reaction and E one reaction is preferred with tertiary substrates. So an anti coplanar requirement which is a characteristic of an E two reaction is important. So if we consider the two structures, we noticed that the first structure has two beta hydrogens that are c to chlorine, right. Hydrogen at position one is pointing up. Chlorine at position two is pointing down and hydrogen at position three is pointing up. So both hydrogens are anti to our leaving group. Now, for the second structure hydrogen at position number one is equatorial, right? It's not axial anymore and chlorine is axial is winding down. So we have sin arrangement right. Let's remember that if one axial group is pointing up, the other one is pointing down that then that is our an arrangement. But if we have an axial group pointing down and hyen being in an equatorial position nearby, that means we have sin arrangement. And similarly hyen at position three is also to chlorine, right? Because it's actually the methyl group that is anti to chlorine. So that means the first structure can, can undergo and E two reaction, the second one cannot, right. And the reason why it cannot undergo an E two reaction is because we don't have any anti hydrogens to our leaving group, which is chlorine at the beta carbons. And that's it, we can state that for B the first structure will undergo a more rapid reaction. And basically, the reason is because it has beta hydrogens in and see co planner arrangement. OK. While the second one does not have those ac hydrogens to our chloral group. And that means the second compound doesn't even undergo an elimination reaction. When we consider E E two, that would be it. We are ready to label the first structure for part B as our final answer. Thank you for watching. And I hope to see you in the next video.

Which reactant in each of the following pairs undergoes an elimination reaction more rapidly? Explain your choice.
a.
b.

Key Concepts

Elimination Reactions

Leaving Group Ability

Steric Hindrance

Watch next