Consider the reaction of 1-bromobutane with a large excess of ...

Question

Consider the reaction of 1-bromobutane with a large excess of ammonia (NH₃). Draw the reactants, the transition state, and the products. Note that the initial product is the salt of an amine (RNH₃⁺Br⁻), which is deprotonated by the excess ammonia to give the amine.

Accepted Answer

All right. Hello everyone. So this question is asking us to consider the S and two reaction of one chloro three methyl pentane with a large excess of ammonia. That's in age three, draw the reactants, the transition states and the products. So first, let's draw out the overall reaction. Our starting material is one chloro three methyl pentane. That's a five carbon parent chain with chlorine attached to carbon number one and a methyl substituent on carbon number three. So now chlorine is going to react with ammonia. That's an H three recall that the SN two reaction is a concerted backside attack. In this case, ammonia, specifically nitrogen is our nucleo and the lone pair of electrons on nitrogen is going to go ahead and attack the carbon atom that contains chlorine. This displaces chlorine at the same time or in the same step as a result, chlorine is replaced with nitrogen on carbon number one here. However, because nitrogen was neutral before this reaction in our intermediate nitrogen is going to have a positive charge. So an additional step is required to detonate the nitrogen that is attached to carbon at this point. So here another equivalent of ammonia can go ahead and detonate the nitrogen that is attached to carbon. So in this particular case, normally speaking, a neutron nucleo file is not going to go ahead and favor an S and two reaction. But ammonia is an exception. However, it is going to end up with a positive charge. So generally speaking, we would prefer to use the derotated neutral form as our final product. Hence why the second step was needed. Now, generally speaking, sn two reaction results in an inversion of stereo chemistry. However, chlorine was attached to a primary carbon that was not Cairo in the first place. So inversion does not change configuration at all. OK. So now now that we've drawn a scheme of all the steps in this reaction, let's go ahead and use this to illustrate the transition state in both cases because recall that the transition state is the in between, right, using partial bonds indicated with dashed lines, we're showcasing the bonds that are breaking and forming in that particular step. So in the case of the actual SN two step, a bond between carbon and nitrogen is forming and the bond between carbon and chlorine is breaking because of this, when drawing out my transition state, both of those aforementioned bonds should be represented as, as dashed lines. Notice how I'm making sure that nitrogen and chlorine are on opposite sides because of the SN two mechanism. So now recall that nitrogen ends up with a positive charge after the sn two step. So in my transition state, I would represent this with a partial positive symbol. And what I neglected to mention when talking about this reaction overall is that chloride is going to form as a by-product the chloride ion. So in my transition state, chlorine should have a partial negative charge. So this completes my transition state. So I'm going to enclose this in brackets and I'm going to have the double dagger symbol outside of the brackets. So now let's go ahead and repeat this step or rather the transition state process or our depot nation step. So at this point, chlorine is no longer attached to carbon. So I'm not going to represent it in this transition state. And nitrogen from the first equivalent of ammonia is still going to go ahead and stay intact instead. However, I would use a dash line to showcase the bond that's breaking between nitrogen and one of its protons. So that same hydrogen is establishing a new bond between the nitrogen from our additional equivalent of ammonia. So here the nitrogen attached to carbon before dep prote nation had a positive charge. So I would represent that with a partial positive charge. And the nitrogen from the molecule of ammonia that's going to deproteinate is going to have a positive charge after this Depot Nation step. So both nitrogens should have a partial positive charge and then I'm going to close this out with brackets and our double dagger symbol. So now lets go ahead and write this out in our final answer starting off with the reactants. So now in between the reactants and or intermediate, I'm going to go ahead and insert our first transition state. Then after this, I would go ahead and showcase the intermediate in which nitrogen has a positive charge as well as chloride as a by-product. This is where we introduce the dep protonation step. So now in between the intermediate and the final product, I'm going to go ahead and add our second transition state, make it smaller so it can fit as well. After the second transition state, I would go ahead and draw the structure of our final products as well as ammonium chloride, which is a by-product of the deportation process. So here let's go ahead and add some labels. First, we have the reactants followed by transition state one and then are intermediate, followed by transition state two and last, the are products and there you have it. So if you watch this video all the way through to the end, thank you so very much. And I hope you found this helpful.

Consider the reaction of 1-bromobutane with a large excess of ammonia (NH₃). Draw the reactants, the transition state, and the products. Note that the initial product is the salt of an amine (RNH₃⁺Br⁻), which is deprotonated by the excess ammonia to give the amine.

Key Concepts

Nucleophilic Substitution

Transition State

Deprotonation

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Consider the reaction of 1-bromobutane with a large excess of ammonia (NH3). Draw the reactants, the transition state, and the products. Note that the initial product is the salt of an amine (RNH3+Br−), which is deprotonated by the excess ammonia to give the amine.

Key Concepts

Nucleophilic Substitution

Transition State

Deprotonation

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Consider the reaction of 1-bromobutane with a large excess of ammonia (NH₃). Draw the reactants, the transition state, and the products. Note that the initial product is the salt of an amine (RNH₃⁺Br⁻), which is deprotonated by the excess ammonia to give the amine.