Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Rictiovarus on February 19, 2020, 02:48:35 AM
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Hello!
When some polymethylene dihalide reacts with diethyl malonate and a strong enough base, a single malonate will displace both halides on a single dihalide forming a ring. What are the main diving forces behind the dihalide reacting twice with the same molecule of malonate, and not say with two separate malonates forming a chain with a diester on each terminus? Does it boil down to the already close and locked in proximity of the malonate after the first displacement of a halide? Is there more to it? Thank you!
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This boils down to the rate of (i) the intramolecular (ring-forming) reaction being faster than that of (ii) the intermolecular (polymerisation) reaction.
The rate of (i) will be governed by the rate law d[ring]/dt = k[monoalylated intermediate], and is clearly first-order since the reaction is unimolecular.
The rate of (ii) will be d[polymer]/dt = k[monoalkylated intermediate][malonate] or k[monoalkylated intermediate][dihalide] - so this reaction is second order.
In the case of competing first order and second order reactions, one way to favour the first-order one (the cyclisation) is to run the reaction at high dilution, since the rate of the second order reaction will fall in proportion to the square of the concentration, while the first order reaction only scales linearly.
Another factor to consider is ring strain, since certain ring sizes (e.g. 8- and 9-membered) are unusually strained. In these cases you might observe more polymerisation and less cyclization than for, say, 1,5-dibromopentane, which would generate a 6-membered and unstrained cyclohexane product.
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This boils down to the rate of (i) the intramolecular (ring-forming) reaction being faster than that of (ii) the intermolecular (polymerisation) reaction.
The rate of (i) will be governed by the rate law d[ring]/dt = k[monoalylated intermediate], and is clearly first-order since the reaction is unimolecular.
The rate of (ii) will be d[polymer]/dt = k[monoalkylated intermediate][malonate] or k[monoalkylated intermediate][dihalide] - so this reaction is second order.
In the case of competing first order and second order reactions, one way to favour the first-order one (the cyclisation) is to run the reaction at high dilution, since the rate of the second order reaction will fall in proportion to the square of the concentration, while the first order reaction only scales linearly.
Another factor to consider is ring strain, since certain ring sizes (e.g. 8- and 9-membered) are unusually strained. In these cases you might observe more polymerisation and less cyclization than for, say, 1,5-dibromopentane, which would generate a 6-membered and unstrained cyclohexane product.
Thank you so much for the response!! I'm just a little confused on the high dilution principle—for in the synthesis, 1,5-dibromopentane was added drop-wise to a solution of diethyl malonate in ethanol (with sodium ethoxide). The only product that formed was the cyclic diester (diethyl cyclohexane-1,1-dicarboxylate) and no polymerization (or at least not a significant amount) occurred with two malonates and one dibromide. Does the addition of the dibromide drop-wise help to favour the first-order reaction? I'm rather confused because since the concentration of diethyl malonate is much higher than that of the dibromide (at least initially) at about 3.2 g in 20 ml EtOH, then there would be ample malonate to react. But perhaps I'm not quite understanding the unimolecular and bimolecular differences, it has been a few years since I've done kinetics.
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The ringclosure is very fast when five- or sixmembered ring is possible.