CHEM 110 Lecture Notes - Lecture 7: Reaction Coordinate, Activation Energy, Exothermic Process

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15 Jun 2018

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CHEM 110: P7 CHAP 21: ORGANIC REACTIONS

Mechanisms and Energy

Concerted Reaction: one-step mechanism, reactants are converted directly to

products (potential energy vs. reaction coordinate)

z activation energy

activation energy

Stepwise:

Intermediate

Practice Using Curved Arrows:

Exothermic (substitution rxn)

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CHEM 110 Full Course Notes

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Concerted reaction: one-step mechanism, reactants are converted directly to products (potential energy vs. reaction coordinate) z. Hydrogenation of oils: oils are unsaturated fats (they have double bonds) they contain alkenes, kink in the chain and can"t pack tightly, lower intermolecular forces, lower bp/mp makes liquid! Fatty acids: intermolecular forces are stronger, higher melting/boiling points. Hydrogenation the addition of hydrogen to an unsaturated compound. *slide* net result= alkene turns to an alkane. Hydrogen sticks to metal catalyst, forms a bond, becomes weak enough for so that the alkene can take the hydrogens from the metal. Take the hydrogen"s away, turns into a trans double bond. Margarine is not used by hydrogenation anymore due to production of trans fats; which are extremely indigestible (partial hydrogenation trans fats) *slide* made from individual repeat units of the same molecule. Ex: polyethylene made from individual chain reactions of ethylene. Addition polymerization: chain-growth polymerization making polymers (alkene on alkene on alkene)

Related Questions

1. provide a mechanim that shows how the two products are formed. use curved arrows to indicate the movement of electrons.

2. Rank the following free radicals from least stable 4 to most stable 1.

61"(aq) + BrO3(aq) + 6 H+(aq) â 3 12(aq) + Br(aq) + 3 H2O lodide ions (I), react with bromate ions (BrO) with acid (H') to produce iodine (12), bromide (Br) and water. Based on the reaction, 13 particles (six I's, one BrO 's and six Hs) all must somehow collide, hold on to each other, and react. The probability is highly unlikely, a more realistic scenario is for the reaction to take place in a series of collisions, each one involving two or three particles, each one resulting in part of the reaction taking place. This is one of the purposes behind the study of rate laws establishing the sequence of events occurring during the transformation of reactants into products. This sequence of events is called a reaction mechanism. The order of the reaction of I, BrO and H will be calculated in this experiment. The results from this experiment will be used to calculate the energy of activation in the next experiment. The reaction has the rate law: The initial rate method will be used to determine the order of each reactant (the values of x, y, and z). In this method, the reaction is clocked as it proceeds for only a small portion of the total time it would take to reach completion. The time the reaction is measured during the first 5% or so of the reactants being consumed. Each experimental trial will change the concentrations of one of the other concentrations remain unchanged. compounds while the Rxn l : 6 I(aq) + BrO3(aq) + 6 H+(aq) â 3 12(aq) + Br(aq) + 3 H2O The timing method you will use is based on the properties of one of the products of the reaction, the I2 molecule. Two reactions that the 2 molecule can undergo in this experiment are: (a fast reaction) Rxn 2: 3 12 (aq) + 3 starch â deep blue complex Rxn 3: 3 12(aq) + 6 S,032-(aq) â 6 I(aq) + 3 S,062-(aq) (a very-fast reaction) (The stoichometry of Rxn's 2 and 3 have been adjusted to match the 3 moles of h produced in Rxn 1.) The reaction to be studied is rxn. As the reaction proceeds it creates iodine (2). The iodine created will quickly react with S2Os (thiosulfate ion) via rxn 3. Rxn 1 wil continue to produce iodine and the iodine will quickly react with thiosulfate al of the thiosulfate has reacted. Once all of the thiosulfate has reacted, the iodine produced will now react with starch via rxn 2. The reaction with starch produces a blue color. The rate of the reaction can now be timed and triggered by the appearance of a blue color and can be directly calculated as Î[S20321 The rate of Rxn 1 can now be correlated to the rate of Rxn3

Nucleophilic Substitution â Synthesis of an Alkyl Halide

Introduction

Alkyl halides can be prepared from alcohols by reactions with hydrogen halides (HCl, HBr, or HI) via nucleophilic substitution. In this type of reaction, the nucleophile displaces a leaving group from a carbon atom of an organic substrate (here the alcohol once protonated). Both electrons of the new bond to the carbon are provided by the nucleophile while the leaving group departs with both electrons of its bond to the substrate. There are two limiting mechanisms that are most common in describing nucleophilic substitution reactions at sp3 carbon centres : (1) a direct displacement (SN2) process and (2) a process that generates an intermediate carbocation (SN1). Which mechanism is a better description depends on the structure of the molecule containing the leaving group, the nature of the leaving group, the size of the nucleophile, and the polarity of the solvent. This experiment is a near-perfect example of an SN1 substitution. H3C OH CH3 H3C + HCl H3C Cl CH3 H3C + H2O (Equation 1) In this reaction, you will prepare 2-chloro-2-methylpropane (t-butyl chloride) from 2- methyl-2-propanol (t-butyl alcohol) using HCl as the strong acid and chloride source. Note that HCl first activates the alcohol in an acid/base reaction before the SN1 reaction proceeds. The final product is characterized by its yield, boiling point, and density. The presence of a halide atom is confirmed with a silver nitrate test.

1) State which of the electrophiles given below will react preferentially by i) SN1, ii) by SN2, or iii) capable of reacting by either of the two mechanisms depending on the given conditions. How can you affect those conditions to favour SN1 or SN2? Reason your predictions based on the structures of the compounds. Br-CH3, Br-CH2CH3, Br-CH(CH3)2, Br-C(CH3)3, Br-CH2-C5H6; C5H6 = phenyl

2) Rank the nucleophiles given below based on high, mediocre, and low nucleophilicity and reason why based on their structures. HO-CH3, H2N-CH3, KO-CH3, KO-C(CH3)3

3) Name two solvents that are commonly used for SN2 reactions. because they considerably slow down SN1 reactions.

4) Can you ever have only SN2 or only SN1?

5) Iodine is a better leaving group than bromine. But iodine is a better nucleophile than bromine. Why is that?

6) Using orbitals, draw the transition state of the SN2 reaction between sodium cyanide and bromomethane. Which orbitals are interacting with which? Why is the bond breaking? Why is the orientation important? Is there another angle besides backside attack that could theoretically work based on orbital geometry? Why doesnât this happen?

7) Name four reagents you can use to dry solvent. Why do we use magnesium sulfate in this lab instead of them?

8) Draw a mechanism with all the proper arrows for the reaction from this experiment.

9) What are we removing with the water wash? Why donât we just add the sodium bicarbonate directly to the reaction mixture instead of doing a water wash?

CHEM 110 Lecture Notes - Lecture 7: Reaction Coordinate, Activation Energy, Exothermic Process

CHEM 110 Full Course Notes

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