The reaction of an amine with an alkyl halide gives an ammonium s...

Question

The reaction of an amine with an alkyl halide gives an ammonium salt.R3N amine  + R'—X alkyl halide —> R3(N+)—R' X- ammonium salt The rate of this SN2 reaction is sensitive to the polarity of the solvent. 1. Draw an energy diagram for this reaction in a nonpolar solvent and another in a polar solvent. 2. Consider the nature of the transition state, and explain why this reaction should be sensitive to the polarity of the solvent. 3. Predict whether it will be faster or slower in a more polar solvent.

Accepted Answer

welcome back everyone. The rate of the sN two reaction given below changes with polarity of the solvent. Explain this sensitivity towards polarity of the solvent and draw an energy diagram for this reaction to show whether it will be faster in a more polar or less polar solvent. So we need to draw out the transition state in order to determine how the solvent would impact this reaction. So we're going to begin by drawing out our reactant. We have nitrogen bonded to two methyl groups, so that's H three C. And then here we would have stereo chemistry where we have a dash bond to a methyl group and then a wedge bond to hydrogen where the nitrogen has its three bonds and lone pair on itself. So this nitrogen we recognize is a neutral reactant and it's reacting with methyl bromide. So we have And we'll draw H3 C bonded to roman. So we want to recognize that this second reactant is a alcohol lead substrate. We recognize browning is in our halogen group in group seven a on the periodic table. So that's why we have an alcohol. Hey lead and for our products, what is going to occur is we recall that in an SN two reaction. This occurs in one step where our nuclear filic attack, which is going to occur, where we have more negative character from our loan electrons here are going to attack the carbon atom to form a bond with carbon and so carbon will have one too many bonds, meaning it's going to have to collapse its bond with bro mean, as a lone pair onto bro mean, And so as a result, we have the following transition state where we have the bond with nitrogen and carbon being formed Again, carbon is bonded to three hydrogen atoms. Nitrogen still has its bonds to the two metals that is bond forming. And then we have a bond breaking between carbon and our bro mean hey lied. So this dashed bond is our bond sorry, bond broken and the first dash bond is our bond formed. So for our official product, what we're going to have is and this is our transition state. What we're going to have is now nitrogen bonded to the third methyl group here where it's still bonded to its original first two methyl groups and still bonded to that hydrogen atom on a wedge. And our second product is going to be our hey lied now as a bromine atom with four sets of lone pairs on itself. And we recall that since it's in the Halogen Group, seven a on the periodic table, it should prefer to have seven valence electrons. But now it has a total of eight. So it has a formal charge of -1 since it has one too many electrons on itself. Now, what we should recognize is that in our products, nitrogen no longer is bonded according to its bonding preference where it originally was happy with its three bonds and its single lone pair to be neutral and stable, it now has a total of four bonds. And so because this nitrogen has four bonds on itself where it would prefer to have directly attached five valence electrons to itself, it now has directly attached only 1234 valence electrons. So that leaves one less than it would prefer. So would have a formal charge of plus one. Where are booming as we wrote out has a formal charge of -1. Now, what we can recall is that in our transition state we have these partial charges characterizing the nitrogen and bromine atom. So the nitrogen will have a partial positive charge because as a product it has a fully positive formal charge on itself and the bromine atom is going to have a partially negative formal charge. Or sorry, a partially negative dipole charge here. Since as a product it has a formal charge of minus one. Now, as we stated, because we have these partial charges. This is going to be a polar transition state where we begun with just a neutral nuclear file reacting here. Now we have a charged on this nitrogen that was originally neutral. So we have a more charged polar transition state in comparison to our nuclear filic reactant. And we want to recall that polar substrates are going to be stabilized through dipole dipole interactions. So stabilized by dipole dipole interactions, meaning we would need a polar solvent. So polar solvent needed to accomplish this dipole dipole stabilization of our transition state. And if we use a polar solvent to stabilize our transition state so that it becomes neutral. Again, this is going to be favored by the reaction since a stabilized transition state will result in lower energy. And so that means that the therefore the rate of the reaction is enhanced in a polar solvent. And so we just need to draw an energy diagram to prove our point here. So we're going to recall that in an energy diagram on our Y access. We have the depiction of our energy for the reaction and it's going to be increasing as we go further up and then we have our reaction coordinates. So how the reaction proceeds, which increases as we go towards the right. What we're going to begin with is describing our reaction coordinate in a non polar solvent. So as we stated, if we do not have a non polar solvent or if we use a non polar solvent to stabilize our transition state, we actually won't be able to stabilize our transition state. Since we saw that our transition state is polar and charged. And so we would actually have a reaction coordinate where our transition state, which occurs at the hump of our graph here is going to be high in energy. So this is our reaction coordinate if we have a non polar solvent used to dissolve our reactant and now showing whether we use a polar solvent to dissolve our reactant. our reaction coordinate is going to be stabilized in the transition state, since we would stabilize the charge state, the charged polar transition state through a polar solvent. That would cause the reaction rate to be enhanced, meaning the reaction is going to occur faster, so less energy is needed and therefore we would have lower energy and we have the hump here of our transition state at lower energy. So this is in reference to a polar solvent characterized by lower energy needed to start it and corresponding to a faster reaction. So this entire energy diagram is going to complete this example as our final answer for how the reaction would proceed if we used a polar solvent versus a non polar solvent in our S N. Two reaction. So this is going to correspond choice. Be in the multiple choice as our final answer. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

Key Concepts

SN2 Reaction Mechanism

Transition State Theory

Solvent Polarity and Reaction Rate

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