Chapter 1: Carbon Compounds and Chemical Bonds PDF

Chapter 1 Carbon Compounds and Chemical Bonds  Introduction  Organic Chemistry The chemistry of the compounds of carbon Taxol The human body is largely composed of organic compounds Organic chemistry...

Chapter 1 Carbon Compounds and Chemical Bonds  Introduction  Organic Chemistry The chemistry of the compounds of carbon Taxol The human body is largely composed of organic compounds Organic chemistry plays a central role in medicine, bioengineering, agriculture, and materials science etc. 2-Pentylpenicilin  Vitalism It was originally thought organic compounds could be made only by living things by intervention of a “vital force” Fredrich Wöhler disproved vitalism in 1828 by making the organic compound urea from the inorganic salt ammonium cyanate by evaporation: Chapter 1 2  Structural Theory of Organic Compounds  Valence Electrons  The electrons that surround the nucleus exist in shells of increasing energy and at increasing distances from the nucleus. The most important shell, called the valence shell, is the outermost shell because the electrons of this shell are the ones that an atom uses in making chemical bonds with other atoms to form compounds. How do we know how many electrons an atom has in its valence shell? We look at the periodic table. The number of electrons in the valence shell (called valence electrons) is equal to the group number of the atom. Carbon is in group 14- four valence electrons; Oxygen is in group 16- six valence electrons. The halogens of group 17- seven electrons. Carbon can form one or more bonds to other carbons Chapter 1 3  Structural Theory of Organic Compounds  Central Premises Between 1858 and 1861, August Kekulé, Archibald Scott Couper, and Alexander M. Butlerov, working independently, laid the basis for one of the most important theories in chemistry: the structural theory. Two central premises are fundamental:  1. The atoms in organic compounds can form a fixed number of bonds using their outermost shell (valence) electrons. Carbon is tetravalent; that is, carbon atoms have four valence electrons and can form four bonds.  Oxygen is divalent, and hydrogen and (usually) the halogens are monovalent: Chapter 1 4  Structural Theory of Organic Compounds  Central Premises  2. A carbon atom can use one or more of its valence electrons to form bonds to other carbon atoms: Chapter 1 5  Isomers The structural theory allowed early organic chemists to begin to solve a fundamental problem that plagued them: the problem of isomerism. Isomers are different molecules with the same molecular formula Many types of isomers exist Example  Consider two compounds with molecular formula C2H6O  These compounds cannot be distinguished based on molecular formula; however they have different structures  The two compounds differ in the connectivity of their atoms Number of Constitutional Isomers for Various Alkanes Molecular Formula Number of Constitutional Isomers C3H8 1 C4H10 2 C5H12 3 C6H14 5 C7H16 9 C8H18 18 C9H20 35 C10H22 75 C15H32 4,347 C20H42 366,319 C30H62 4,111,846,763 C40H82 62,481,801147,341 Chapter 1 6  Constitutional Isomers Constitutional isomers are one type of isomer They are different compounds that have the same molecular formula but different connectivity of atoms They often differ in physical properties (e.g. boiling point, melting point, density) and chemical properties Chapter 1 7  Three Dimensional Shape of Molecules Virtually all molecules possess a 3-dimensional shape which is often not accurately represented by drawings It was proposed in 1874 by van’t Hoff and le Bel that the four bonds around carbon where not all in a plane but rather in a tetrahedral arrangement i.e. the four C-H bonds point towards the corners of a regular tetrahedron Chapter 1 8  Chemical Bonds: The Octet Rule  Octet Rule Atoms form bonds to produce the electron configuration of a noble gas (because the electronic configuration of noble gases is particularly stable) For most atoms of interest this means achieving a valence shell configuration of 8 electrons corresponding to that of the nearest noble gas Atoms close to helium achieve a valence shell configuration of 2 electrons Atoms can form either ionic or covalent bonds to satisfy the octet rule Chapter 1 9  Electronegativity Electronegativity is the ability of an atom to attract electrons It increases from left to right and from bottom to top in the periodic table (noble gases excluded)  Fluorine is the most electronegative atom and can stabilize excess electron density the best Chapter 1 10  Dipole Moment Dipole moment is the amount of electrical charge × bond length and due to differences in electronegativity. Charge separation shown by Electrostatic Potential Map (EPM). Red indicates a partially negative region and blue indicates a partially positive region. Chapter 1 11  Molecular Dipole Moment The molecular dipole moment is the vector sum of the bond dipole moments. It depends on bond polarity and bond angles. Lone pairs or electrons contribute to the dipole moment. Chapter 1 12  Ionic Bonds When ionic bonds are formed atoms gain or lose electrons to achieve the electronic configuration of the nearest noble gas  In the process the atoms become ionic The resulting oppositely charged ions attract and form ionic bonds This generally happens between atoms of widely different electronegativities Example  Lithium loses an electron (to have the configuration of helium) and becomes positively charged  Fluoride gains an electron (to have the configuration of neon) and becomes negatively charged  The positively charged lithium and the negatively charged fluoride form a strong ionic bond (actually in a crystalline lattice) Chapter 1 13  Covalent Bonds Covalent bonds occur between atoms of similar electronegativity (close to each other in the periodic table) When the electrons are shared evenly the bond is said to be nonpolar. When the electrons are not shared evenly between the atoms, the resulting bond will be polar. Atoms achieve octets by sharing of valence electrons Molecules result from this covalent bonding Valence electrons can be indicated by dots (electron-dot formula or Lewis structures) but this is time-consuming The usual way to indicate the two electrons in a bond is to use a line (one line = two electrons) Chapter 1 14  Intramolecular Forces Strength of attractions between molecules influences the melting point (m.p.), boiling point (b.p.), and solubility of compounds. Classification depends on structure:  London dispersions  Dipole-dipole interactions  Hydrogen bonding Chapter 1 15  London Dispersions One of the Van der Waal forces. A temporary dipole moment in a molecule can induce a temporary dipole moment in a near by molecule. An attractive dipole-dipole interactive results for a fraction of a second. Main force in nonpolar molecules. Larger atoms are more polarizable. Branching lowers b.p. because of decreased surface contact between molecules. Chapter 1 16  Dipole-Dipole Interactions Dipole-dipole interactions result from the approach of two polar molecules. If their positive and negative ends approach, the interaction is an attractive one. If two negative ends or two positive ends approach, the interactions is repulsive. In a liquid or a solid, the molecules are mostly oriented with the positive and negative ends together, and the net force is attractive. Chapter 1 17  Hydrogen Bonding Strong dipole-dipole attraction. Organic molecules must have N-H or O-H to be able to hydrogen bond. The hydrogen from one molecule is strongly attracted to a lone pair of electrons on the oxygen of another molecule. O-H more polar than N-H, so alcohols have stronger hydrogen bonding. Chapter 1 18  Example Chapter 1 19  Solution Chapter 1 20  Writing Lewis Structures Atoms bond by using their valence electrons The number of valence electrons is equal to the group number of the atom  Carbon is in group 4A and has 4 valence electrons  Hydrogen is in group 1A and has 1 valence electron  Oxygen is in group 6A and has 6 valence electrons  Nitrogen is in group 5A and has 5 valence electrons To construct molecules the atoms are assembled with the correct number of valence electrons If the molecule is an ion, electrons are added or subtracted to give it the proper charge. If the structure is a positive ion (a cation), we subtract one electron for each positive charge. The structure is written to satisfy the octet rule for each atom and to give the correct charge If necessary, multiple bonds can be used to satisfy the octet rule for each atom Chapter 1 21  Example Write the Lewis structure for the chlorate ion (ClO3-)  The total number of valence electrons including the electron for the negative charge is calculated  Three pairs of electrons are used to bond the chlorine to the oxygens  The remaining 20 electrons are added to give each atom an octet Chapter 1 22  Example Write the Lewis structure for the chlorate ion (CH3F) Write the Lewis structure for the chlorate ion (C2H6) Chapter 1 23 The carbonate ion with 24 valence electrons and two negative charges must incorporate a double bond to satisfy the octet rule for every atom The organic molecules ethene (C2H4) and ethyne (C2H2) must also use multiple bonds to satisfy the octet rule for each atom Chapter 1 24  Exceptions to the Octet Rule The octet rule applies only to atoms in the second row of the periodic table (C, O, N, F) which are limited to valence electrons in the 2s and 2p orbitals In second row elements fewer electrons are possible Example: BF3 In higher rows other orbitals are accessible and more than 8 electrons around an atom are possible Example: PCl5 and SF6 Chapter 1 25  Formal charge Many Lewis structures are incomplete until we decide whether any of their atoms have a formal charge. First, we examine each atom and, using the periodic table, we determine how many valence electrons it would have if it were an atom not bonded to any other atoms. This is equal to the group number of the atom in the periodic table. For hydrogen this number equals 1, for carbon it equals 4, for nitrogen it equals 5, and for oxygen it equals 6. Next, we examine the atom in the Lewis structure, and we assign the valence electrons in the following way: We assign to each atom half of the electrons it is sharing with another atom and all of its unshared (lone) electron pairs. Chapter 1 26  Formal charge Then we do the following calculation for the atom: Formal charge = number of valence electrons -1/2 number of shared electrons – number of unshared electrons F = Z - (1/2)S -U where F is the formal charge, Z is the group number of the element, S equals the number of shared electrons, and U is the number of unshared electrons. It is important to note, too, that the arithmetic sum of all the formal charges in a molecule or ion will equal the overall charge on the molecule or ion. Chapter 1 27  Examples Ammonium ion (NH4)+  Nitrate ion (NO3)- Chapter 1 28 An atom will always have the same formal charge depending on how many bonds and lone pairs it has regardless of which particular molecule it is in For example, a singly bonded oxygen with 3 lone pairs will always have a negative charge and an oxygen with three bonds and one lone pair will always have a positive charge Knowing these forms of each atom is invaluable in drawing Lewis structures correctly and rapidly (See table next page) Chapter 1 29  Resonance Often a single Lewis structure does not accurately represent the true structure of a molecule The real carbonate ion is not represented by any of the structures 1,2 or 3 Experimentally carbonate is known not to have two carbon-oxygen single bonds and one double bond; all bonds are equal in length and the charge is spread equally over all three oxygens Chapter 1 30 The real carbonate ion can be represented by a drawing in which partial double bonds to the oxygens are shown and partial negative charge exists on each oxygen The real structure is a resonance hybrid or mixture of all three Lewis structures Double headed arrows are used to show that the three Lewis structures are resonance contributors to the true structure  The use of equilibrium arrows is incorrect since the three structures do not equilibrate; the true structure is a hybrid (average) of all three Lewis structures Chapter 1 31 One resonance contributor is converted to another by the use of curved arrows which show the movement of electrons  The use of these arrows serves as a bookkeeping device to assure all structures differ only in position of electrons A calculated electrostatic potential map of carbonate clearly shows the electron density is spread equally among the three oxygens  Areas which are red are more negatively charged; areas of blue have relatively less electron density Chapter 1 32 X-Ray studies have shown that carbon–oxygen double bonds are shorter than single bonds. The same kind of study of the carbonate ion shows, however, that all of its carbon– oxygen bonds are of equal length. One is not shorter than the others as would be expected from representations 1, 2, and 3. Clearly none of the three structures agrees with this evidence. In each structure, 1–3, one carbon–oxygen bond is a double bond and the other two are single bonds. None of the structures, therefore, is correct. How, then, should we represent the carbonate ion? One way is through a theory called resonance theory. This theory states that whenever a molecule or ion can be represented by two or more Lewis structures that differ only in the positions of the electrons, two things will be true: Chapter 1 33 This theory states that whenever a molecule or ion can be represented by two or more Lewis structures that differ only in the positions of the electrons, two things will be true: 1. None of these structures, which we call resonance structures or resonance contributors, will be a realistic representation for the molecule or ion. None will be in complete accord with the physical or chemical properties of the substance. 2. The actual molecule or ion will be better represented by a hybrid (average) of these structures. Chapter 1 34  Rules for Resonance:  Individual resonance structures exist only on paper  The real molecule is a hybrid (average) of all contributing forms  Resonance forms are indicated by the use of double-headed arrows  Only electrons are allowed to move between resonance structures  The position of nuclei must remain the same  Only electrons in multiple bonds and nonbonding electrons can be moved  Example: 3 is not a resonance form because an atom has moved  All structures must be proper Lewis structures Chapter 1 35 The energy of the actual molecule is lower than the energy of any single contributing form  The lowering of energy is called resonance stabilization Equivalent resonance forms make equal contributions to the structure of the real molecule  Structures with equivalent resonance forms tend to be greatly stabilized  Example: The two resonance forms of benzene contribute equally and greatly stabilize it Unequal resonance structures contribute based on their relative stabilities  More stable resonance forms contribute more to the structure of the real molecule Chapter 1 36  Rules to Assign Relative Importance of Resonance Forms A resonance form with more covalent bonds is more important than one with less  Example: 6 is more stable and more important because it has more total covalent bonds Resonance forms in which all atoms have a complete valence shell of electrons are more important  Example: 10 is more important because all atoms (except hydrogen) have complete octets Chapter 1 37 Resonance forms with separation of charge are less important  Separation of charge cost energy and results in a less stable resonance contributor  Example: 12 is less important because it has charge separation Forms with negative charge on highly electronegative atoms are more important  Those with positive charge on less electronegative atoms are also more important Chapter 1 38  Example The nitrate ion is known to have all three nitrogen-oxygen bond lengths the same and the negative charge spread over all three atoms equally Resonance theory can be used to produce three equivalent resonance forms  Curved arrows show the movement of electrons between forms  When these forms are hybridized (averaged) the true structure of the nitrate ion is obtained Chapter 1 39  Example Draw the important resonance forms for [CH3OCH2]+. Indicate which structure is major and minor contributor or whether they would have the same energy. Chapter 1 40  Example Draw the resonance structure the compound below. Indicate which structure is major and minor contributor or whether they would have the same energy. Chapter 1 41  Example Write the resonance structure that would result from moving the electrons as the curved arrows indicate. Be sure to include formal charges if needed. Chapter 1 42  The Structure of Methane and Ethane: sp3 Hybridization The structure of methane with its four identical tetrahedral bonds cannot be adequately explained using the electronic configuration of carbon Hybridization of the valence orbitals (2s and 2p) provides four new identical orbitals which can be used for the bonding in methane Orbital hybridization is a mathematical combination of the 2s and 2p wave functions to obtain wave functions for the new orbitals Chapter 1 43 When one 2s orbital and three 2p orbitals are hybridized four new and identical sp3 orbitals are obtained  When four orbitals are hybridized, four orbitals must result  Each new orbital has one part s character and 3 parts p character  The four identical orbitals are oriented in a tetrahedral arrangements  The antibonding orbitals are not derived in the following diagram The four sp3 orbitals are then combined with the 1s orbitals of four hydrogens to give the molecular orbitals of methane Each new molecular orbital can accommodate 2 electrons Chapter 1 44 Chapter 1 45 An sp3 orbital looks like a p orbital with one lobe greatly extended  Often the small lobe is not drawn The extended sp3 lobe can then overlap well with the hydrogen 1s to form a strong bond The bond formed is called a sigma (s) bond because it is circularly symmetrical in cross section when view along the bond axis Chapter 1 46 A variety of representations of methane show its tetrahedral nature and electron distribution  a. calculated electron density surface b. ball-and-stick model c. a typical 3-dimensional drawing Chapter 1 47  The Structure of Ethene (Ethylene) : sp2 Hybridization Ethene (C2H2) contains a carbon-carbon double bond and is in the class of organic compounds called alkenes  Another example of the alkenes is propene The geometry around each carbon is called trigonal planar  All atoms directly connected to each carbon are in a plane  The bonds point towards the corners of a regular triangle  The bond angle are approximately 120o Chapter 1 48 There are three s bonds around each carbon of ethene and these are formed by using sp2 hybridized orbitals The three sp2 hybridized orbitals come from mixing one s and two p orbitals  One p orbital is left unhybridized The sp2 orbitals are arranged in a trigonal planar arrangement  The p orbital is perpendicular (orthoganol) to the plane Chapter 1 49 Overlap of sp2 orbitals in ethylene results in formation of a s framework  One sp2 orbital on each carbon overlaps to form a carbon-carbon s bond; the remaining sp2 orbitals form bonds to hydrogen The leftover p orbitals on each carbon overlap to form a bonding p bond between the two carbons A p bond results from overlap of p orbitals above and below the plane of the s bond  It has a nodal plane passing through the two bonded nuclei and between the two lobes of the p molecular orbital Chapter 1 50 The bonding p orbital results from overlap of p orbital lobes of the same sign The antibonding p* orbital results from overlap of p orbital lobes of opposite sign  The antibonding orbital has one node connecting the two nuclei and another node between the two carbons The bonding p orbital is lower in energy than the antibonding orbital  In the ground state two spin paired electrons are in the bonding orbital  The antibonding p*orbital can be occupied if an electron becomes promoted from a lower level ( e.g. by absorption of light) Chapter 1 51 The s orbital is lower in energy than the p orbital  The ground state electronic configuration of ethene is shown Chapter 1 52  Restricted Rotation and the Double Bond There is a large energy barrier to rotation (about 264 kJ/mol) around the double bond  This corresponds to the strength of a p bond  The rotational barrier of a carbon-carbon single bond is 13-26 kJ/mol This rotational barrier results because the p orbitals must be well aligned for maximum overlap and formation of the p bond Rotation of the p orbitals 90o totally breaks the p bond Chapter 1 53  Cis-trans isomers Restricted rotation of groups joined by a double bond causes a new type of isomerism that we illustrate with the two dichloroethenes written as the following structures: These two compounds are isomers; they are different compounds that have the same molecular formula. These isomers have the same connectivity of atoms and differ only in the arrangement of atoms in space  This puts them in the broader class of stereoisomers We indicate that they are different isomers by attaching the prefix cis or trans to their names (cis, Latin: on this side; trans, Latin: across). Chapter 1 54  Cis-trans isomers The molecules below do not superpose on each other One molecule is designated cis (groups on same side) and the other is trans (groups on opposite side) Cis-trans isomerism is not possible if one carbon of the double bond has two identical groups Chapter 1 55  The Structure of Ethyne (Acetylene): sp Hybridization Ethyne (acetylene) is a member of a group of compounds called alkynes which all have carbon- carbon triple bonds  Propyne is another typical alkyne The arrangement of atoms around each carbon is linear with bond angles 180o Chapter 1 56 The carbon in ethyne is sp hybridized  One s and one p orbital are mixed to form two sp orbitals  Two p orbitals are left unhybridized The two sp orbitals are oriented 180o relative to each other around the carbon nucleus  The two p orbitals are perpendicular to the axis that passes through the center of the sp orbitals Chapter 1 57 In ethyne the sp orbitals on the two carbons overlap to form a s bond  The remaining sp orbitals overlap with hydrogen 1s orbitals The p orbitals on each carbon overlap to form two p bonds The triple bond consists of one s and two p bonds Chapter 1 58 Depictions of ethyne show that the electron density around the carbon-carbon bond has circular symmetry  Even if rotation around the carbon-carbon bond occurred, a different compound would not result Chapter 1 59  Bond Lengths of Ethyne, Ethene and Ethane The carbon-carbon bond length is shorter as more bonds hold the carbons together  With more electron density between the carbons, there is more “glue” to hold the nuclei of the carbons together The carbon-hydrogen bond lengths also get shorter with more s character of the bond  2s orbitals are held more closely to the nucleus than 2p orbitals  A hybridized orbital with more percent s character is held more closely to the nucleus than an orbital with less s character  The sp orbital of ethyne has 50% s character and its C-H bond is shorter  The sp3 orbital of ethane has only 25% s character and its C-H bond is longer Chapter 1 60 Chapter 1 61 Chapter 1 62 Chapter 1 63  The number of molecular orbitals formed equals the number of the atomic orbitals used  Hybridized orbitals are obtained by mixing the wave functions of different types of orbitals Four sp3 orbitals are obtained from mixing one s and three p orbitals  The geometry of the four orbitals is tetrahedral  This is the hybridization used in the carbon of methane Three sp2 orbitals are obtained from mixing one s and two p orbitals  The geometry of the three orbitals is trigonal planar  The left over p orbital is used to make a p bond  This is the hybridization used in the carbons of ethene Two sp orbitals are obtained from mixing one s and one p orbital  The geometry of the two orbitals is linear  The two leftover p orbitals are used to make two p bonds  This is the hybridization used in the carbons of ethyne Sigma (s) bonds have circular symmetry when viewed along the bond axis Pi (p) bonds result from sideways overlap of two p orbitals Chapter 1 64  Molecular Geometry: The Valence Shell Electron Pair Repulsion (VSEPR) Model This is a simple theory to predict the geometry of molecules All sets of valence electrons are considered including:  Bonding pairs involved in single or multiple bonds  Non-bonding pairs which are unshared Electron pairs repel each other and tend to be as far apart as possible from each other Non-bonding electron pairs tend to repel other electrons more than bonding pairs do (i.e. they are “larger”) The geometry of the molecule is determined by the number of sets of electrons by using geometrical principles Chapter 1 65  Methane The valence shell of methane contains four pairs or sets of electrons To be as far apart from each other as possible they adopt a tetrahedral arrangement (bond angle 109.5o) The molecule methane is therefore tetrahedral Chapter 1 66  Ammonia When the bonding and nonbonding electrons are considered there are 4 sets of electrons The molecule is essentially tetrahedral but the actual shape of the bonded atoms is considered to be trigonal planar The bond angles are about 107o and not 109.5o because the non-bonding electrons in effect are larger and compress the nitrogen-hydrogen bond Chapter 1 67  Water There are four sets of electrons including 2 bonding pairs and 2 non-bonding pairs Again the geometry is essentially tetrahedral but the actual shape of the atoms is considered to be an angular arrangement The bond angle is about 105o because the two “larger” nonbonding pairs compress the electrons in the oxygen-hydrogen bonds Chapter 1 68  Boron Trifluoride Three sets of bonding electrons are farthest apart in a trigonal planar arrangement (bond angle 120o)  The three fluorides lie at the corners of an equilateral triangle  Beryllium Hydride Two sets of bonding electrons are farthest apart in a linear arrangement (bond angles 180o) Chapter 1 69  Carbon Dioxide There are only two sets of electrons around the central carbon and so the molecule is linear (bond angle 180o)  Electrons in multiple bonds are considered as one set of electrons in total A summary of the results also includes the geometry of other simple molecules Chapter 1 70  Representations of Structural Formulas Dot formulas are more cumbersome to draw than dash formulas and condensed formulas Lone-pair electrons are often (but not always) drawn in, especially when they are crucial to the chemistry being discussed Chapter 1 71  Dash formulas Each dash represents a pair of electrons This type of representation is meant to emphasize connectivity and does not represent the 3- dimensional nature of the molecule  The dash formulas of propyl alcohol appear to have 90o angles for carbons which actually have tetrahedral bond angles (109.5o) There is relatively free rotation around single bonds so the dash structures below are all equivalent Constitutional isomers  Constitutional isomers have the same molecular formula but different connectivity  Propyl alcohol (above) is a constitutional isomer of isopropyl alcohol (below) Chapter 1 72  Condensed Structural Formulas In these representations, some or all of the dash lines are omitted In partially condensed structures all hydrogens attached to an atom are simply written after it but some or all of the other bonds are explicitly shown In fully condensed structure all bonds are omitted and atoms attached to carbon are written immediately after it For emphasis, branching groups are often written using vertical lines to connect them to the main chain Chapter 1 73  Representations of Structural Formulas Chapter 1 74  Bond-Line Formulas A further simplification of drawing organic molecules is to completely omit all carbons and hydrogens and only show heteroatoms (e.g. O, Cl, N) explicitly Each intersection or end of line in a zig-zag represents a carbon with the appropriate amount of hydrogens  Heteroatoms with attached hydrogens must be drawn in explicitly Chapter 1 75 Cyclic compounds are condensed using a drawing of the corresponding polygon Multiple bonds are indicated by using the appropriate number of lines connecting the atoms Chapter 1 76  Bond-Line Formulas or Line-Angle Formulas Chapter 1 77  Bond-Line Formulas Chapter 1 78  Bond-Line Formulas Chapter 1 79  Bond-Line Formulas Chapter 1 80  Three-Dimensional Formulas Since virtually all organic molecules have a 3-dimensional shape it is often important to be able to convey their shape The conventions for this are:  Bonds that lie in the plane of the paper are indicated by a simple line  Bonds that come forward out of the plane of the paper are indicated by a solid wedge  Bonds that go back out of the plane of the paper are indicated by a dashed wedge Generally to represent a tetrahedral atom:  Two of the bonds are drawn in the plane of the paper about 109o apart  The other two bonds are drawn in the opposite direction to the in- plane bonds but right next to each other Chapter 1 81 Trigonal planar arrangements of atoms can be drawn in 3-dimensions in the plane of the paper  Bond angles should be approximately 120o  These can also be drawn side-on with the central bond in the plane of the paper, one bond forward and one bond back Linear arrangements of atoms are always best drawn in the plane of the paper Chapter 1 82  Nucleophiles and Electrophiles Nucleophile: Donates electrons to a nucleus with an empty orbital. Electrophile: Accepts a pair of electrons. When forming a bond, the nucleophile attacks the electrophile, so the arrow goes from negative to positive. When breaking a bond, the more electronegative atom receives the electrons. Chapter 1 83