1 / 27

Bond Making and Bond Breaking

Addition and elimination reactions are exactly opposite. A  bond is formed in elimination reactions, whereas a  bond is broken in addition reactions. Bond Making and Bond Breaking.

naiara
Download Presentation

Bond Making and Bond Breaking

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Addition and elimination reactions are exactly opposite. A  bond is formed in elimination reactions, whereas a  bond is broken in addition reactions.

  2. Bond Making and Bond Breaking • A reaction mechanism is a detailed description of how bonds are broken and formed as starting material is converted into product. • A reaction can occur either in one step or a series of steps.

  3. Regardless of how many steps there are in a reaction, there are only two ways to break (cleave) a bond: the electrons in the bond can be divided equally or unequally between the two atoms of the bond.

  4. Homolysis and heterolysis require energy. • Homolysis generates uncharged reactive intermediates with unpaired electrons. • Heterolysis generates charged intermediates.

  5. To illustrate the movement of a single electron, use a half-headed curved arrow, sometimes called a fishhook. • A full headed curved arrow shows the movement of an electron pair.

  6. Homolysis generates two uncharged species with unpaired electrons. • A reactive intermediate with a single unpaired electron is called a radical. • Radicals are highly unstable because they contain an atom that does not have an octet of electrons. • Heterolysis generates a carbocation or a carbanion. • Both carbocations and carbanions are unstable intermediates. A carbocation contains a carbon surrounded by only six electrons, and a carbanion has a negative charge on carbon, which is not a very electronegative atom.

  7. Three reactive intermediatesresulting from homolysisandheterolysis of a C – Z bond

  8. Radicals and carbocations are electrophiles because they contain an electron deficient carbon. • Carbanions are nucleophiles because they contain a carbon with a lone pair.

  9. Heterolytically cleave each of the carbon-hetratom bonds and label the organic intermediate as a carbocation or carbanion a) carbocation b) carbanion

  10. Bond formation occurs in two different ways. • Two radicals can each donate one electron to form a two-electron bond. • Alternatively, two ions with unlike charges can come together, with the negatively charged ion donating both electrons to form the resulting two-electron bond. • Bond formation always releases energy.

  11. Relative stabilities of carbocations

  12. Relative stability of radicals

  13. Bond Dissociation Energy • The energy absorbed or released in any reaction, symbolized by H0, is called the enthalpy change or heat of reaction. • Bond dissociation energy is the H0 for a specific kind of reaction—the homolysis of a covalent bond to form two radicals.

  14. Because bond breaking requires energy, bond dissociation energies are always positive numbers, and homolysis is always endothermic. • Conversely, bond formation always releases energy, and thus is always exothermic. For example, the H—H bond requires +104 kcal/mol to cleave and releases –104 kcal/mol when formed.

  15. Comparing bond dissociation energies is equivalent to comparing bond strength. • The stronger the bond, the higher its bond dissociation energy. • Bond dissociation energies decrease down a column of the periodic table. • Generally, shorter bonds are stronger bonds.

  16. Which has the higher bond dissociation energy? • H-Cl or H-Br b) c)

  17. Bond dissociation energies are used to calculate the enthalpy change (H0) in a reaction in which several bonds are broken and formed.

  18. Bond dissociation energies have some important limitations. • Bond dissociation energies present overall energy changes only. They reveal nothing about the reaction mechanism or how fast a reaction proceeds. • Bond dissociation energies are determined for reactions in the gas phase, whereas most organic reactions occur in a liquid solvent where solvation energy contributes to the overall enthalpy of a reaction. • Bond dissociation energies are imperfect indicators of energy changes in a reaction. However, using bond dissociation energies to calculate H° gives a useful approximation of the energy changes that occur when bonds are broken and formed in a reaction.

  19. Calculate H for each of the following reactions, knowing H of O2 and O-H = 119 kcal/mol, H of C-H = 104 kcal/ml and H of one C=O = 128 kcal/mol. a) Bonds Formed Bonds Broken C-H = 4 x 104 kcal/mol = 416 kcal/mol C-O = 2 x -128 kcal/mol = -256 kcal/mol O-O = 2 x 119 kcal/mol = 238 kcal/mol O-H = 4 x -119 kcal/mol = -476 kcal/mol H = 416 + 238 = +654 kcal/mol H = -256 + -476 = -732 kcal/mol H = 654 + -732 kcal/mol = -78 kcal/mol

More Related