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Hello everyone.
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Today we're doing problem 6 .40 and this problem asks us to draw the transition state for each of the following reaction.
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So we see that we have cyclohexane attached to some sort of halide.
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This case is bromine and we're breaking it up to a carbocadion intermediate with a bromide anion.
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So remember the transition state is the point of a reaction where bonds are being made and broken.
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So your reactant bonds are breaking and your product bonds are forming.
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So in the transition state is the point of reaction where bonds are being made and broken.
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In this case, we know that the sigma bond between this carbon bromine must break.
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So i can draw the transition state and transition states are drawn in squared brackets.
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So our transition state is the reactant bonds breaking and the product bonds forming.
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Well, here there's no real products.
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We only have the reactant bonds breaking, which is the sigma bond, which is these dotted lines, and is attached to the bromine.
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And in this case, bromine is going to be getting an extra lone pair of electrons because of the heterolithic cleavage.
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So we know that there's going to be a delta minus sign here, while we're forming a carbocations with electron pore site at the carbons that's going to be delta positive here.
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So this is a transition state for this reaction.
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In this product, in this reaction scheme, we're given that we are going to use this chloride anion and nucleophilically attack this bromine center.
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And we're going to form our product here.
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So if i draw my transition state, remember, transition state is shown in squared brackets, and my starting material bonds are going to be breaking as my product bonds are forming.
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So there's nothing actually breaking here, but there's always going to be a formation of a new bond, which is going to be shown in dotted lines, attached to my chlorine nucleus, and you can actually put a double dagger here to indicate it's a transition state.
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So it's actually this double dagger, double positive sign actually means that it's a transition state.
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And if you recall, my bromine is going to get extra electrons because it's negatively charged.
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And my chlorine was also negative.
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So both of these are delta negative.
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If you go to the third example, we see that we're actually doing some sort of proton abstraction.
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So from neutral alcohol, we're going to negatively charged oxygen anion, and we're actually neutralizing this nitrogen product to make ammonia.
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So we are actually forming and breaking bonds here.
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So this is a good example.
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Remember your transition state on a squared brackets and you can draw your double -headed arrows to indicate the transition state...