Bonding and Structure: Part 2 PDF

Bonding and Structure: Part 2 Organic Molecular Structure Now that we have been introduced to the main models that describe bonding, Lewis Structures and Molecular Orbital theory, we are now in a position to understand more details of...

Bonding and Structure: Part 2 Organic Molecular Structure Now that we have been introduced to the main models that describe bonding, Lewis Structures and Molecular Orbital theory, we are now in a position to understand more details of structure and properties of molecules. 1 Revisit Isomers: Stereoisomers We have already seen that isomers are different compounds with the same molecular formula. As discussed earlier, structural isomers differ in the order in which the atoms are connected (connectivity of the atoms), the order in which the atoms are bonded together are different. Another form of isomer, stereoisomers, are associated with C=C bonds. Structural Feature #1: Bond rotation around C–C single σ bonds is relatively easy because rotating around a σ bond does not change the overlap of the sp3 atomic orbitals (AOs) used to make the molecular orbitals (MOs). Br Ha Br Br Hb Br Ha rotate towards rotate again sp3 C C sp3 C C C C H front H H H Hb Ha H Br H Br Hb Remember, we said that the direction that bonds “point” doesn’t matter when drawing structures, which is true for those that contain only single bonds, because rotation around single bonds is allowed and is easy. However: Compare a C=C double bond, rotation around a π bond would destroy the p AO/p AO overlap, breaking the π bond, the same kind of rotation is not possible when we have π bonds: stereoisomers diastereomers Br Br Br Br H Br Br Br Br H C C = C C X C C C C = C C H H H H H H Br H Br H cis-isomer p A.O.'s in plane this p A.O. perpendicular to trans-isomer of paper paper plane, π-bond broken!! Rotation around a double bond is thus restricted (we can assume that it does not occur under normal conditions), stereoisomers can thus form in suitable structures that have C=C double bonds (although not all structures with C=C double bonds will have isomers). Structural Feature #2: The bond angles at atoms that are part of double bond are well defined and fixed. Valence shell electron pair repulsion (VSEPR) defines the bond angles as ca. 120°: 120° H H correct 120° correct H 120° 120° H add the H atoms trans-stereoisomer to the C=C bonds X in these structures add the H atoms H INcorrect H H 120° X to the C=C bond 120° H INcorrect cis-stereoisomer Bonding 2 : page 1  Stereoisomers have the same atom connectivity (the atoms are connected together in the same order), but they differ in the orientation of at least some atoms in space (the direction in which the bonds point). Physical and chemical properties of stereoisomers are different, they are different molecules and chemicals. There are different names for stereoisomers, but here we use the simplest terminology. The stereoisomers discussed in this section are called geometrical isomers, also sometimes called configurational isomers, but the term cis/trans isomers is easiest, and is the term that we will mainly use. The term diastereomer also applies to this kind of stereoisomer, and we will use diastereomer also to distinguish these stereoisomers another kind of stereoisomer we will meet later in the course. Example: Are the following structures the same or are they isomers and what kind of isomers? stereoisomers H (diastereomers) H 3C H H (opposite sides) C C line defines C C H CH3 "sides" H 3C CH3 (same side) b.p. 0.8°C b.p. 3.7°C trans-butene cis-butene cis- Br H Br same Br H Br trans- C C C C C C opposite H Br H H H Br stereoisomers structural isomers (diastereomers) Cis isomers have the relevant groups on the same side of the C=C bond, trans isomers have relevant groups on opposite sides of the C=C bond. Although the direction in which the bonds “point” doesn’t matter for structures with only single bonds, the direction in which the bonds point around the C=C bond does matter because of restricted rotation. If we don’t get the bond angles correct, we will not be able to properly identify stereoisomers. Example: Are the following structures the same or are they isomers and what kind of isomers? H H CH2 CH CH 10 2 9 1 CH2 1 10 2 H CH2 9 3 8 H CH2 CH2 4 8 6 7 3 CH2 CH2 CH2 5 7 5 6 4 the ORDER in which the atoms trans-stereoisomer cis-stereoisomer are connected together are the SAME: NOT structural isomers Example: Are the following two structures the same or are they isomers? Cl H Cl Cl Cl H X same side opposite Cl C C Cl NON-Superimposable!! C C C C sides HC H H H Cl C Cl X NON-Superimposable!! H H cis-diastereomer trans-diastereomer Trying to superimpose the cis structure on top of the trans structure fails, the structures are nonsuperimposable, because they are stereoisomers, the atoms point in different directions. It is not necessary to confirm the lack of superimposability since these structures are clearly different, however, superimposability is the most critical test for isomers and we will need it later in the course. Bonding 2 : page 2  1.1 Generating Isomers Using Degrees of Unsaturation The degree (also known as elements) of unsaturation is an important piece of information about a molecule/structure that is obtained from the molecular formula. Consider structures consisting of linear chains of carbon atoms joined by single bonds: How many hydrogen atoms can be attached to these chains of carbon atoms? 3 carbons - 8 hydrogens 4 carbons - 10 hydrogens add 1 carbon H H H to the chain H H H H H C C C H H C C C C H H H H H H H H 2 "end" hydrogens each added carbon also adds 2 hydrogens Every time one carbon is added to a chain, two hydrogen atoms are also added. That maximum number of H’s any organic molecule can have is (2 × number of carbons) + 2 (the “end” H atoms). A molecule that has the maximum number of hydrogen atoms is referred to as saturated. A molecule that does not have the maximum number of hydrogen atoms is unsaturated. A molecule can be unsaturated by having double or triple bonds and rings, each “cost” hydrogen atoms. H H H H H H H H H H H C H C C H H C C H C C H H C C C C C C H C C H H H H H H H H H H H H H H C5H12 C5H10 C5H10 (5 x 2) + 2 = 12 double bond "costs" 2 H's ring "costs" 2 H's 0 degrees of unsat. 1 degree of unsat. 1 degree of unsat. Calculating Degrees (Elements) of Unsaturation (Max # of H atoms) - (Actual # of H atoms) degrees (elements) of unsaturation = 2 Where: the Max # of H atoms = (# carbons x 2) + 2 What if the molecule contains other elements? Halogens are monovalent, they take the place of a hydrogen and are counted as a hydrogen: H X H X = F, Cl, Br, I H C C C H H H H Oxygen is divalent, when an oxygen atom is added to a chain, zero hydrogens are added, oxygen is ignored: 2 carbons - 6 hydrogens 2 carbons and 1 oxygen - STILL 6 hydrogens H add 1 oxygen to H H H the chain zero H atoms added to the chain O H C C H C C H H H H H H Bonding 2 : page 3 Nitrogen is trivalent, when a nitrogen atom is added to a chain, one hydrogen atom is also added. Thus, one nitrogen atom is equivalent to half a carbon atom as far as adding hydrogen atoms to the chain is concerned, N is counted as half a carbon atom: 2 carbons - 6 hydrogens 2 carbons and 1 nitrogen - 7 hydrogens add 1 nitrogen to H H H each N adds 1 H to the chain the chain H H H C C H N C C H H H H H H Example 1 C5H9OCl Max # of H atoms = (5 x 2) + 2 = 12 O HO Cl Actual # of H atoms = 9 + 1 = 10 (Cl counted as H) degrees of unsaturation = (12 - 10) / 2 = 1 degree e.g. Cl etc C5H9OCl MUST have one ring or one double bond but NOT both double bond ring Two possible structures are shown, many other structural isomers are also possible for this molecular formula. Example 2 C4H7NO C4H7NO H Max # of H atoms = (4 x 2) + 1 + 2 = 11 O N O Actual # of H atoms = 7 degrees of unsaturation = (11 - 7) / 2 = 2 degrees e.g. N etc C4H7NO MUST have H 2 rings or 2 double bonds or 1 ring and 1 double bond Two possible structures are shown, many other structural isomers are also possible for this molecular formula. Knowing the degrees (elements) of unsaturation helps in drawing Lewis structures, from the molecular formula you can determine the number of double bonds, rings, and so on. Example Problem: Generate as many line-angle structures as you can that have the molecular formula C5H10 (5 x 2) + 2) - (10) degrees (elements) of unsaturation = = 1 degree 2 Therefore, all structures must have one double bond or a ring, but not both. The recommended approach: start with the longest chain and progressively and systematically branch. Where rings are possible, start with the largest ring and progressively get smaller. same = A maximum of 10 possible structural isomers can be drawn, and one pair of stereoisomers. trans-isomer (from above) cis-isomer Bonding 2 : page 4  2 Polar Bonds We will see that understanding unequal sharing of electrons in bonds is crucial to understanding much of chemical reactivity. The Carbon–Carbon Bond in Ethane: C2H6 Energy Ψ H H Ψ σ* M.O. H C C H σ* M.O. H H or Ψ H H sp3 Ψ C sp3 C sp3 Ψ σ M.O. H C C H σ M.O. stabilization energy sp3 H H Ψ There is equal contribution from each carbon AO, the electrons are equally distributed and shared between the two carbons in the localized σ orbital, both electrons stabilized to same extent, this is a nonpolar bond. Carbon–Chlorine Bond in Chloroethane: C2H5Cl Energy larger on C H H Ψ σ* M.O. H C C Cl Cl more electronegative Ψ H H σ* M.O. closer in lower energy A.O. Ψ energy C sp3 or Ψ Ψ Cl p H H p Ψ σ M.O. σ M.O. smaller H C C Cl larger on Cl sp3 stabilization H H There is a much higher contribution from lower energy chlorine AO to the localized bonding σ MO. The electrons in the bonding MO are unequally distributed (they are closer to the more electronegative chlorine since this lowers their energy), and there is less stabilization of the electron that was formally associated with the Cl atom, this is a polar bond. The localized bonding σ MO “looks” more like the chlorine AO. The antibonding MO “looks” more like the carbon AO. If we use more of the chlorine AO to make the bonding MO, then we have a lot of the carbon AO “left over” that we must use to make the antibonding MO, it is closer in energy to the carbon sp3 hybrid AO. The polarization (unequal sharing or distribution) of electrons in σ bonds is called the inductive effect. The wave function squared for the bonding electrons shows how the electrons are not equally shared by the atoms: (dipole) H H H H δ+ δ– H C C Cl H C C Cl polar bond H H H H unequal electron distribution Ψ Ψ2 Bonding 2 : page 5  2.1 Bond Dipole Moments A dipole moment describes a separation of positive and negative charges. Example: The following structure is an ylide, it has opposite formal charges on adjacent C and P atoms that represents a substantial separation of charges—a dipole. dipole ~ 5 Debye dipole ~ 1 Debye negative "end" positive "end" positive "end" H negative "end" δ+ H ylide C P δ– Cl C H chloride H H separated electrical charge PARTIALLY separated electrical charge The extent of charge separation can be measured in units of Debye (D), more polar bonds have larger D. In the chloride, however, there are no actual charges, but there are partial charges as a result of the electronegativity difference between carbon and chlorine. There is still a separation of charges in the C–Cl bond, but it is smaller than in the case where there are real charges, roughly 1 D versus roughly 5 D for the charged ylide. Even though the dipole moment is smaller in the uncharged molecule, it is still enough to influence chemistry! The Inductive Effect (unequal sharing or distribution of electrons in a bond). The inductive effect occurs as a consequence of differences in electronegativity between atoms: H C and H are similar 2.2 increasing electronegativity Li B C N O F 1.0 2.0 2.5 3.0 3.4 4.0 Na Al Si P S Cl 0.9 1.6 1.9 2.2 2.6 3.2 Electronegativity increases with increasing nuclear charge (left to right). Electronegativity increases with decreased electron shielding (bottom to top). Examples zero dipole 0.2 D 0.9 D 1.5 D 1.7 D moment H H H H H H H C C H H C S H C O H C F H C K H H H H H H H H C–C bond C–S bond C–O bond C–F bond C–K bond pure covalent highly ionic Bonds have a range of polarities and properties, from nonpolar and pure covalent to highly polar almost pure ionic. You will not be asked to decide whether a bond is ionic or covalent, but you will be expected to predict relative bond polarities. π bond electrons held less tightly, more polarizable H 0.9 D H 2.4 D C-H bond C=O double bond much more polar fairly H C O 1.5 D δ+ C O δ– than C–O OR H-O single bond nonpolar H H H electrons accumulate at this end, electrons deficient at this end, carbon oxygen "gives" electrons in reactions "takes" electrons in reactions Bonding 2 : page 6  Bond dipole moments are not something that we can measure (they are not a physical property of a molecule), but they are very useful in determining how we think about molecules, and in particular, their chemical reactivity. We will use bond dipole moments a great deal when we discuss chemical reactions later. 2.2 Molecular Dipole Moments Bond dipole moments are the property of an individual chemical bond: They are important in determining the chemistry of specific bonds. Molecular dipole moments are the property of an entire molecule. These can be measured by experiment, and they are important in determining the physical properties of molecules since they are important contributors to InterMolecular Forces (IMF), that control properties such as melting point, boiling point, solubility etc. For Example 1.5 D Cl Cl Cl Cl chloroform H C 1.5 D chloroform H C 1.1 D 1.5 D Cl Cl 3 C–Cl bonds have bond dipole Molecular dipole moment, given by the moments (indicated) vector sum of the bond dipoles Molecular dipole moments are determined as the vector sum of the individual bond dipole moments. However, it is not necessary to do vector math to solve almost all molecular dipole moment problems in organic chemistry, you can use “common sense” to see how the individual bond dipoles reinforce or oppose each other. Some More Examples NONPOLAR MOLECULES POLAR MOLECULES Cl H H H H H H O H C C N Na+ – Cl C C C Cl H C H Cl H H H H Cl 3.9 D 9.0 D C–H bonds essentially symmetrical nonpolar bond dipoles cancel 1.9 D (2 π bonds) (ionic) dipole moment ignored zero molecular dipole moment We must consider all conformations (rotations around single bonds). Because rotation around a C=C bond is not allowed, the bond dipoles for the two alkenes have fixed directions. Because of allowed bond rotation around the single bond, on average, the bond dipoles do not cancel for the structure on the right. on AVERAGE there IS a molecular dipole NO molecular dipole moment molecular dipole moment Cl Cl Cl H Cl Cl Cl H bond C C bond C C H C C H H C C H H H H Cl H Cl H H bond dipoles "add" bond dipoles "cancel" bond dipoles must consider rotation do NOT cancel around single bonds Bonding 2 : page 7 Example Problem: Which has larger dipole moment and why? O O H C H H C H C C C C H larger H H Cl H Cl H H smaller The C–Cl bond dipoles oppose the C=O dipole, but the large bond dipole moment associated with the C=O double bond will always be larger and will “win” over opposing dipoles in single bonds. 2.3 Hydrogen Bonds: A Special Kind of Dipolar Interaction Polarization of electrons away from a hydrogen atom is different from other atoms, because hydrogen has no core electrons, and so removing electron density “reveals” an unshielded proton and a concentrated partial positive charge: 1.5 D H δ+ δ+ H O δ– O CH3 hydrogen bonds H3C δ– δ– O CH3 ~ 5 kcal/mol δ+H The O–H bond dipole moment is large, and the partial positive charge on hydrogen is concentrated, which results in a strong dipolar interaction. Hydrogen bonding almost certainly involves more than just a dipole–dipole interaction, there is evidence for at least some covalent bonding, but that is a detail that we don’t need to go into here. H bonding is observed primarily for molecules with O–H and N–H bonds: concentrated "accessible" LARGER dipole moment δ+ on small H atom BUT weaker dipolar H CH3 interaction CH3 δ+ charge in larger + H O O CH atom and less O δ– O C "accessible" + + CH H3C C δ- CH3 H3C CH3 hydrogen bonding H3C δ+ CH3 isopropanol ~ 5 kcal/mol acetone polar protic b.p. ~ 83 °C polar APROTIC b.p. ~ 56 °C Even though the C–O and O–H bond dipole moments in isopropanol above are smaller than the C=O bond dipole moment in acetone above, the intermolecular dipolar interaction for isopropanol is larger because the partial positive charge in isopropanol is concentrated on the small hydrogen atom that is also very accessible. In acetone, the partial positive charge is less “concentrated” on a larger C atom, and it is much less accessible since the carbon is surrounded by the two -CH3 groups. Consequently, the intermolecular dipolar interaction is much weaker, despite the fact that the bond dipole moment is larger. Solvents that can hydrogen bond (e.g., isopropanol) are called polar protic. Polar molecules that can’t hydrogen bond (e.g., acetone) are called polar aprotic. Bonding 2 : page 8  3 Strengths of Bonds Organic reactions involve breaking bonds and making new bonds. The most important factor to consider is that the energies of the electrons decrease when they form a bond. The lower the energy of the electrons in the bond, the stronger the bond. In organic chemical reactions, bonds are broken by either homolysis or heterolysis (-lysis means breaking). Homolytically, that is, breaking the bond so that each atom in the bond gains one of the shared pair: neutral homolysis A B A B radicals or atoms Single-headed (fishhook) curved arrow shows the movement of one electron. Homolysis forms two radicals. Heterolytically, that is, breaking the bond so that one of the atoms gains both of the shared pair: heterolysis A B A B charged ions OR..... A B A B charged ions A double-headed curved arrow shows the movement of a pair of electrons. Heterolysis usually forms two ions. Whether homolytically or heterolytically, breaking a bond causes the energy of the electrons to go up! 3.1 Bond Making/Breaking in H2: Energy Diagrams Consider homolytic cleavage the H–H bond in H2 as a simple chemical reaction. Bond breaking means separation of two atoms, when this happens the overlap of the AOs decreases, the MO disappears and the energy of the electrons goes up. endothermic H H "costs" energy, energy has homolytic H H hydrogen molecule two hydrogen atoms to be put "ADDED" to the H2 The energy required to break a bond has to be “added” to the H2, usually this comes from the environment/solvent in the form of thermal energy (RT). Energy two atoms Energy H H higher in energy energy ENDOTHERMIC (bond broken) bond ALSO dissociation increasing separation distance BREAKS increases energy (BDE) BOND when the electron energy INCREASES nuclei get ~104 kcal/mol for H2 must INPUT ENERGY too close together molecule H H lower energy electrons H-H separation distance equilibrium bond length ~0.8Å for H–H The energy diagram shows how the energy increases as the atoms separate. The electrons in the bond in the molecule are low in energy (they are in a bond) and have low chemical reactivity, the electrons in the atoms are high in energy, they are chemically reactive. Energy must be added to the system to cause bond breaking, bond breaking is endothermic. The energy also increases if the atoms get very close together, the lowest energy is at the optimum separation Bonding 2 : page 9 distance, the equilibrium bond length, which is ~0.8Å for the hydrogen molecule. The Bond Dissociation Energy (BDE) is the energy required to break the bond homolytically, not heterolytically. This is because the energy required to heterolyze a bond will be different in different solvents (because charged ions are solvated very differently in solvents of different polarity), whereas the homolytic bond dissociation energy is essentially the same in all solvents since the neutral radicals and/or atoms are solvated similarly in different solvents. Consider bond formation by reaction of two H atoms to make H2 as a simple chemical reaction. Bond formation can occur when two atoms approach each other. When they get close enough to overlap AOs, MOs form and the energy of the electrons goes down. Energy released upon bond formation goes into the solvent in the form of heat (i.e., kinetic energy). MAKING BOND H H H H hydrogen molecule energy released, exothermic two hydrogen atoms SAME DIAGRAM! two atoms Energy H H higher in energy (bond broken) energy DECREASE decreasing separation distance MAKES same as BDE BOND ~104 kcal/mol RELEASES ENERGY for H2 EXOTHERMIC molecule H H lower energy electrons Energy H-H separation distance equilibrium bond length Two H atoms make a bond because doing so lowers the energies of the electrons. When heat is released in a reaction, the reaction is exothermic. 3.2 Factors That Determine Homolytic Bond Dissociation Energies How do we compare the BDEs of two different bonds, for example, comparing the following two generic bonds? The bonds are different; therefore, to understand their relative BDEs we compare the energies of the two pairs of electrons in the bonds. weaker A B A B compare these two pairs stronger A C A C Several different factors determine differences in bonds that determine differences in BDEs. We will look at these in turn. Do not memorize bond dissociation energies, but know and understand the trends. Down the Periodic Table: Effect of Atomic Size In most cases, this is the most influential factor, atomic size has a large influence on BDE. The energies of the electrons in the bonds to smaller atoms are lower because smaller atoms have smaller AOs that overlap strongly when they combine to make molecular orbitals. Bonds to smaller atoms thus tend to be stronger and shorter. The energies of electrons in bonds to larger atoms are higher, because larger atoms have larger atomic orbitals that overlap more poorly when they combine to make molecular orbitals. Bonds to larger atoms thus tend to be weaker and longer. Bonding 2 : page 10  B.D.E. (kcal/mol) Length (Å) F H F H 134 0.92 Cl H Cl H 103 1.27 Br H Br H 88 1.41 I H I H 71 1.61 A fluorine atom is roughly twice as large as a hydrogen atom and an iodine atom is roughly four times as large as a hydrogen atom. Fluorine is very electronegative and small, the electrons in the H–F bond are very low in energy, and the H–F bond is strong and short. This effect is shown in the following energy diagrams; note, these are relative energy diagrams, the absolute energies of H–F and H–I are very different, they are normalized at the energies of the H atom and the halogen atom so that they can be compared. IMPORTANT energy weaker H I H I difference is HERE in the BONDS stronger H F H F Relative Energy Relative Energy H + F energies H + I normalized here BDE H-I goes "up" less BDE H-F starts HIGHER goes "up" more in energy H-I starts LOWER H-F in energy rH-F rH-I energies normalized here X + H Relative Energy BDE BDE BDE BDE H-I H-Br H-Cl H-F H-I H-Br H-Cl H-F rX–H The energies of the electrons on the halogen atoms formed after cleavage are also lower in energy on the smaller and more electronegative elements; however, the energies of the electrons in the bonds went up more in these cases. Bonding 2 : page 11 Carbon–Hydrogen Bonds: Effect of Carbon Hybridization In most cases, this is the second most influential factor influencing bond energies. B.D.E. Length sp3 kcal/mol Å H3C H H3C H 104 1.09 H sp2 H H H C C C C 108 ~1.08 H H H H sp H C C H H C C H 133 ~1.055 Hybrid AOs with less p character are smaller and lower in energy and make stronger bonds. Bonds made using hybrid AOs with less p character have lower energy electrons, these bonds are stronger. MAIN energy H3C H H3C weaker H difference HERE in the BONDS stronger H C C H H C C H Relative R + H energies normalized here Energy has to BDE go "up" less CH3–H starts higher sp3 in energy BDE has to C C H go "up" more CH3–H sp starts lower in energy C C H rR–H Note that again the energies are normalized to the energy after bond cleavage to make it clear that it is the differences in the energies of the electrons in the bonds that is responsible for the differences in the BDEs. The energies of the electrons in the radicals formed after cleavage are also lower in energy in sp2 and sp; however, the energies of the electrons on the H atoms went up more in these cases. Across the Periodic Table: The Effect of Electronegativity This is often the third most influential factor influencing bond energies for bonds to carbon and hydrogen. B.D.E. (kcal/mol) Length (Å) H3C H H3C H 104 1.09 H2N H H2N H 103 1.01 HO H HO H ~119 0.96 F H F H ~135 0.95 Bonding 2 : page 12  MAIN energy H3C H H3C H weaker difference HERE in the BONDS stronger F H F H energies normalized here Z + H Relative Energy BDE BDE C H BDE O H H-F H H C H H O H H H-F rZ–H Bonds from carbon to more electronegative elements are generally stronger and shorter. (Bonds involving other combinations of electronegative elements are a bit more complicated and beyond the scope of this course.) Carbon-Carbon Bonds versus Carbon-Hydrogen Bonds Comparing C–H and C–C bonds, the energy differences are in the bonds, because they are different bonds. MAIN energy weaker H3C CH3 H3C CH3 difference HERE in the BONDS stronger H3C H H3C H B.D.E. (kcal/mol) Length (Å) H3C H H3C H 104 1.09 H3C CH3 H3C CH3 ~88 ~1.54 R + R Relative Energy BDE energies normalized here C–C BDE C–H H3C–CH3 stronger bond, has to go "up" more the electrons STARTED lower in energy H3C–H rC-C/C-H Carbon–carbon bonds form by combining an sp AO from carbon with another sp3 AO from the other carbon. 3 Carbon–hydrogen bonds are “built” by combining an sp3 AO from carbon with a 1s AO from hydrogen. sp3/sp3 AO overlap in a C–C bond is poorer than sp3/1s AO overlap in a C–H bond, and an electron in a 1s AO is lower in energy than one in a carbon sp3 AO, and this character “transmits” to the bonding molecular orbital (a carbon atom is roughly twice as large as a hydrogen atom). Bonding 2 : page 13  Thus, the electrons in a C–H bond are (slightly) lower in energy than those in a C–C bond, it costs more energy to break a C–H bond, the C–H bond has a larger BDE. Some Important Weaker Bonds B.D.E. (kcal/mol) Length (Å) repulsion HO OH HO OH ~51 1.48 repulsion Br Br Br Br 46 2.28 Oxygen lone pairs repel each other in the peroxide, electrons are higher in energy in the bond, and thus an oxygen–oxygen bond is a weakish and long bond. This is even more the case for the larger Br–Br bond. We will take advantage of these weak bonds and their chemical reactivity later in the course. Multiple Bonds B.D.E. (kcal/mol) Length (Å) H3C CH3 H3C CH3 ~88 ~1.54 H2C CH2 H2C CH2 ~160 ~1.33 H2C O H2C O ~175 ~1.21 HC CH HC CH ~230 ~1.20 Multiple bonds are obviously stronger (and shorter) than single bonds. π bonds are essentially always weaker than corresponding σ bonds, of course, but breaking multiple bonds requires π and σ bonds (breaking more than one bond). Electronegativity effects are still important, compare the C=C and C=O double bonds. Bonding 2 : page 14