PHA 208 Basic Organic Chemistry PDF

Bonding in Organic Compounds Chapter 1 1 Bonding in Organic Compounds CHAPTER SUMMARY Organic chemistry is the study of compounds of carbon....

Bonding in Organic Compounds Chapter 1 1 Bonding in Organic Compounds CHAPTER SUMMARY Organic chemistry is the study of compounds of carbon. This is a separate branch of chemistry because of the large numbers of organic compounds and their occurrences and applications. 1.1 Elements and Compounds – Atoms and Molecules Elements are the fundamental building units of substances. They are composed of tiny particles called atoms; atoms are the smallest particles of an element that retains the properties of that element. Atoms are composed of a positively charged nucleus that consists of protons (charge = +1, mass = 1) and neutrons (charge = 0, mass = 1). The nucleus is surrounded by negatively charged electrons that have negligible mass. Elements combine to form compounds. A molecule is the smallest particle of a compound that retains the properties of the compound; atoms bond to one another to form a molecule. 1 Chapter 1 Bonding in Organic Compounds 1.2 Electron Configuration A. Atomic Number and Atomic Mass The atomic number of an atom is the number of protons in the nucleus; this is equal to the number of electrons surrounding the nucleus in a neutral atom. The mass number is the number of protons plus neutrons in the nucleus. Isotopes are atoms with the same number of electrons and protons but different numbers of neutrons; they have the same atomic number but different mass numbers. The atomic mass of an element is the weighted average of the naturally occurring isotopes. B. Atomic Orbitals The space electrons occupy around an atomic nucleus is described by atomic orbitals. The most common orbitals in organic chemistry are s- orbitals, spherical orbitals with the atomic nucleus located in the center, and dumbbell shaped p-orbitals in which the nucleus is between the lobes. C. Filling Atomic Orbitals Orbitals exist in energy levels or shells (numbered 1-7). An atomic orbital can be occupied by 0, 1, or 2 electrons. Atomic orbitals are filled according to the Aufbau principle beginning with the lowest energy orbitals and proceeding to higher energy ones. The electron configuration of an atom is described by the orbitals occupied in each shell and the number of electrons in each orbital. D. Electron Configuration and the Periodic Table The periodic table of elements is organized according to electron configuration. Elements are placed in periods that are related to the outermost shell of electrons and in groups that are related to the number of electrons in the outer shell. All elements in a group have the same number of outer shell electrons (the same as the group number) and the same electron configuration except for the shell number (for example in Group IV, C is 2s22p2 and Si is 3s23p2; both outer shells have four electrons). E. Stable Octets The elements in Group VIII are especially stable; their outer shell configuration is known as a stable octet. 2 Bonding in Organic Compounds Chapter 1 1.3 Ionic Bonding and the Periodic Table A. Ionic Bonding, Electronegativity, Electron Configuration, and the Periodic Table Ionic bonding involves the complete transfer of electrons between two atoms of widely different electronegativities; charged ions are formed (one positive from the loss of electrons and one negative from the gain of electrons), both of which usually have a stable octet outer shell. The ionic bond results from the attraction between the positive cation and negative anion. Electronegativity is defined as the attraction of an atom for its outer shell electrons. Electronegative elements have a strong attraction for electrons and form anions in chemical reactions; electropositive elements have relatively weak attractions for electrons and form cations. B. Electron Dot Representation of Ions The electrons in the outer shell of an anion are represented by dots surrounding the element’s symbol. Anions have usually gained sufficient electrons to complete their outer shells. Cations have usually lost their outer shells, the next shell in becomes the new outer shell, a stable octet, and is not shown. 1.4 Covalent Bonding A. Covalent Bonding, Electron Configuration, and the Periodic Table Covalent bonds involve the sharing of electron pairs between atoms of similar electronegativites; in most cases one or both atoms obtain a stable octet outer shell of electrons. The most common valences in Groups I-IV of the periodic table result from the pairing of all outer shell electrons with outer shell electrons of other atoms; a stable octet results in Group IV, but Groups I-III have incomplete outer shells. The common valences of Groups V-VII result from the pairing of outer shell electrons with those of other atoms to form an octet. The predicted valences of Groups I-VII are 1,2,3,4,3,2,1 respectively. Electron dot formulas depict the outer shell of atoms in molecules showing bonding and non-bonding electron pairs. B. Covalent Bonding in Organic Compounds 3 Chapter 1 Bonding in Organic Compounds A single bond has one bonding pair of electrons; there are two bonding pairs (four electrons) in a double bond and three bonding pairs in a triple bond. The number of bonds formed by elements commonly found in organic compounds is: C - 4; N - 3; O, S - 2; H - 1; F, Cl, Br, I - 1. A carbon can have four single bonds, two double bonds, a double and two single bonds, or a triple and a single bond; all total four bonds. These bonds can be represented by electron dot or line bond formulas. C. Drawing Electron Dot Formulas In drawing electron dot formulas, one must use every atom in the molecular formula and satisfy the valence (the number of bonds formed) for each. A good procedure involves bonding together atoms with valences greater than one with single bonds, inserting double and triple bonds until all valences can be satisfied with the available monovalent atoms, and finally attaching the monovalent atoms. D. The Structural Nature of Compounds Ionic compounds are composed of positive and negative ions in a ratio that will provide an electrically neutral compound. The atoms of a covalent compound are attached to one another to form molecules. Dissolution of an ionic compound in water produces solvated ions whereas covalent compounds have solvated molecules. E. Polyatomic Ions and Formal Charge Polyatomic ions are charged species in which several atoms are connected by covalent bonds. The magnitude and location of the ion's charge is called the formal charge. The formal charge on an atom is equal to the group number of the atom on the periodic table minus the non-bonding electrons and half of the bonding electrons. F. Polar Covalent Bonds A polar covalent bond is composed of atoms with similar but not equal electronegativities. The more electronegative atom is partially negative and the other is partially positive. 1.5 An Orbital Approach to Covalent Bonding A. Sigma and Pi Covalent Bonds 4 Bonding in Organic Compounds Chapter 1 A covalent bond is formed by the overlap of two atomic orbitals each with one electron. There are two types: sigma and pi. A sigma bond involves the overlap of two atomic orbitals head-to-head in one position (such as two s-orbitals, an s and a p-orbital, or two p-orbitals). A pi-bond involves the overlap of parallel p-orbitals at both lobes. B. Electron Configuration of Carbon Bonding in carbon involves the promotion of a 2s electron to an empty 2p orbital thus creating four unpaired electrons, one in the 2s and one in each of the three 2p orbitals. This allows carbon to be tetravalent. C. Shapes of Organic Molecules The shapes of organic molecules are predicted using the following principle: atoms and non-bonding electron pairs attached to a common central atom are arranged as far apart in space as possible. If there are four surrounding groups, the shape is tetrahedral; with three, the groups protrude to the corners of a triangle (trigonal); and with two, the region is linear. D. Carbon with Four Bonded Atoms A carbon with four bonded atoms is sp 3-hybridized, tetrahedral, and has approximately 109o bond angles. The four atomic orbitals on carbon (an s and three p's) combine, through a process called hybridization, to form new orbitals with different geometric orientations. The four new sp3- orbitals are raindrop shaped and are oriented to the corners of a tetrahedron. All bonds to the carbon are sigma bonds. E. Carbon Bonded to Three Atoms A carbon with three bonded atoms is sp2-hybridized, trigonal, and has approximately 120o bond angles. There are three new sp2-hybrid orbitals directed to the corners of a triangle; these form sigma bonds with other atoms. The remaining p-orbital overlaps with a parallel p-orbital of an adjacent, sigma bonded atom to form a pi-bond and complete the double bond. F. Carbon Bonded to Two Atoms A carbon with two bonded atoms is sp-hybridized, linear, and has 180o bond angles. There are two new sp hybrid orbitals that are directed opposite to one another on a straight line; these form sigma bonds. The two remaining p-orbitals overlap with p-orbitals on a similarly hybridized atom to 5 Chapter 1 Bonding in Organic Compounds form two pi-bonds and complete the triple bond. Alternatively, the two p- orbitals can overlap with counterparts on two adjacently bonded sp2- hybridized atoms forming two double bonds. G. Bonding in Organic Compounds – A Summary A carbon with four bonded groups is tetrahedral, sp3-hybridized, and has 109.5O bond angles. A carbon with three is trigonal, sp2-hybridized, and has 120O bond angles. A carbon with two bonded groups is linear, sp-hybridized, and has 180O bond angles. A single bond is a sigma bond; a double bond is composed of one sigma bond and one pi-bond; a triple bond is one sigma and two pi-bonds. Triple bonds are stronger than double bonds and double bonds are stronger than single bonds. The opposite order describes relative bond lengths. 1.6 Bonding to Oxygen and Nitrogen An oxygen with two bonded atoms and two non-bonding electron pairs is sp3-hybridized and has two sigma bonds (single bonds). With only one bonded atom, the oxygen is sp2-hybridized and is involved in one sigma and one pi bond, a double bond. A nitrogen with three bonded atoms and one non-bonding electron pair is sp3-hybridized and has three sigma bonds (single bonds). With two bonded atoms the nitrogen is sp2-hybridized and involved in two single bonds (sigma) and a double bond (sigma and a pi). A nitrogen with only one bonded atom is sp-hybridized, has one single bond (sigma) and one triple bond (one sigma and two pi-bonds). CONNECTIONS 1.1: Diamond, Graphite, and Buckyballs 6 Bonding in Organic Compounds Chapter 1 SOLUTIONS TO PROBLEMS 1.1 Atomic and Mass Numbers Subtract the atomic number (the number of electrons; and protons) from the mass number (number of protons and neutrons) to get number of neutrons. (a-b) mass number atomic number electrons protons neutrons 12C 12 6 6 6 6 13C 13 6 6 6 7 35Cl 35 17 17 17 18 37Cl 37 17 17 17 20 (c) F 9, 19.0 S 16, 32.1 Al 13, 27.0 1.2 Electron Configuration Na 1s2 2s22p6 3s1 Mg 1s2 2s22p6 3s2 Al 1s2 2s22p6 3s23p1 Si 1s2 2s22p6 3s23p2 P 1s2 2s22p6 3s23p3 S 1s2 2s22p6 3s23p4 Cl 1s2 2s22p6 3s23p5 Ar 1s2 2s22p6 3s23p6 1.3 Electron Configuration and the Periodic Table Notice that the outer shells in all four periods are the same for each group except for the period number (1,2,3,4,5). K Ca Ga Ge As Se Br Kr 4s1 4s2 4s24p1 4s24p2 4s24p3 4s24p4 4s24p5 4s24p6 Rb Sr In Sn Sb Te I Xe 5s1 5s2 5s25p1 5s25p2 5s25p3 5s25p4 5s25p5 5s25p6 7 Chapter 1 Bonding in Organic Compounds 1.4 Electron Configuration and the Periodic Table The number of outer shell electrons is the same as the group number to which the element belongs. (a) H, 1 (b) Al, 3 (c) C, 4 (d) N, 5 (e) S, 6 (f) Br, 7 1.5 Electron Configuration and the Periodic Table He 1s2; Ne 2s22p6; Ar 3s23p6; Kr 4s24p6; Xe 5s25p6; Rn 6s26p6 1.6 Ionic Bonding : _ : + (a) Li. + :.F: Li + : : F (LiF) : 1s22s1 1s22s22p5 1s2 1s22s22p6 2+ 2- : : (b) Mg: +. O: Mg + : O: (MgO). : 1s22s22p63s2 1s22s22p4 1s22s22p6 1s22s22p6 1.7 Ionic Bonding (a) NaF (b) Mg(OH)2 (c) (NH4)2SO4 (d) Li2CO3 (e) CaO (f) CaCO3 (g) NaNO2 (h) KClO3 1.8 Covalent Bonding: Valences (a) 3 (b) 4 (c) 4 (d) 3 (e) 2 (f) 2 (g) 1 (h) 1 1.9 Electron Dot Formulas.... Cl........ H.... C..... H C. Cl............. O....C....O.. (d) H.. C..... (a).... (b). C.. O.. (c).Cl.... Cl.. H.. 1.10 Electron Dot Formulas of Polyatomic Ions A neutral atom will “own” the same number of electrons in its outer shell as its group number on the periodic table (that is half the bonding and all the non- bonding electrons). To determine the formal charge of an atom, subtract from the group number of that atom on the periodic table all the non-bonding electrons and half of the electrons in a bonding pair. 8 Bonding in Organic Compounds Chapter 1 C (4) - (0) - 1/2 (8) = 0 H.... _ H....H + H's (1) - (0) - 1/2 (2) = 0 H : C : O: H : C: N : H........ O (6) - (6) - 1/2 (2) = -1 H H H N (5) - (0) - 1/2 (8) = +1 1.11 Polar Bonds δ+ δ− δ+ δ− δ− δ+ δ+ δ− δ+ δ− δ+ δ− (a) C Br (b) C O (c) N H (d) C N (e) C O (f) C S 1.12 Bonding Picture of Propane H H Each carbon is tetrahedral, C H sp3-hybridized, H C C and has 109O bond angles H H H H 1.13 Bonding Picture of Propene H The carbons involved in the double bond are both trigonal, have 120O H C bond angles and are sp2-hybridized. H The other carbon is tetrahedral, has H 109O bond angles and is C C sp3-hybridized. H H 9 Chapter 1 Bonding in Organic Compounds 1.14 Bonding Pictures of Propyne and Propadiene (a) H The two carbons involved in the triple bond are both sp-hybridized, display C C C H linear geometry, and have bond angles H of 180O. The other carbon is sp 3- H hybridized, tetrahedral, and has 109O bond angles. (b) The middle carbon is involved in two double bonds, has two attached atoms H H isp-hybridization, a linear geometry, C C C and bond angles of 180O. The two end H H carbons are connected to three atoms and are sp2-hybridized, trigonal, and have 120O bond angles. 1.15 Orbital Picture of Bonding sp2 O H N C C C H O sp3 H sp sp sp3 sp2 1.16 Atomic and Mass Numbers: Section 1.2A See problem 1.1 for explanation. (a)127I 53 protons, 53 electrons, 74 neutrons; (b) 27Al 13 protons, 13 electrons, 14 neutrons; (c) 58Ni 28 protons, 28 electrons, 30 neutrons; (d) 208Pb 82 protons, 82 electrons, 126 neutrons. 1.17 Electron Configurations : Section 1.2B-D (a) Na, 3s1 ; (b) Mg, 3s2; (c) B, 2s22p1; (d) Ge, 4s24p2; (e) P, 3s23p3 (f) O, 2s22p4; (g) I, 5s25p5; (h) Kr, 4s24p6 10 Bonding in Organic Compounds Chapter 1 1.18 Electron Configurations: Section 1.2B-D (a) Fr (b) Sn (c) Cl (d) Mg (e) B (f) Se (g) He (h) Xe (i) As 1.19 Ionic Reactions: Section 1.3 (a) CaF2 Ca + F Ca2+ + F F F (b) Na2O Na Na+ 2- + O + O Na Na+ 1.20 Electron Dot Formulas: Section 1.4A-C F H H H (a) Cl C F (b) H C O H (c) H C N H (d) H S H Cl H H H H Cl Cl (e) H C C H O (f) C Cl (g) S C S (h) Cl C H H Cl C Cl (i) (j) H (k) (l) H H Cl O Cl H C S H O O H H C Cl H B H H H H (m) N N (n) H N (o) Cl Cl N H 11 Chapter 1 Bonding in Organic Compounds Cl (p) Cl Al Cl (q) H O N O 1.21 Electron Dot Formulas: Section 1.4A-C H a).H...H H.....H.... b)..H.H....H......H..... H.... C. H H. H..C.... H... H... C..C....C..C.. H.H... C...C... H....C..O.. H C..O....C. H H H H H H.... C.... H H H H H H H.....H....Cl... c)..H.H. H H H H d) H H. H..C.. C....C...Br......... H...C......Cl.....C...C......................C...C.. H.......Cl.. H...C..C....Cl. H H H H..Br.. H H H H H e) H H...........H...... H f) H H...... H... H.. C..H H...... H...C. C..C....N. H H..C....N..C.. H.. C.. C. H. H........H H H H H H H H..C. C.. H H 1.22 Electron Dot Formulas: Section 1.4A-C... a).H... H.. b)...H... H c) H.. O. H.H. d)... C...C H.............. C.....O.. H.. H. C....C.. C...... N......O. H. Cl. C.. C. H.. H...C... H H H...... e) H..O.......H f).H........... g).O......H.H.... H.. N.. C..... H. C. N.. C.. O... O..C...... C.. H.. N. H.. H. C... H H H... h).. H.H. H....H..... i) H.. H..... O.... j).H.... H H.... H...... C....O.. H. H..C... C...C. H. C..C....C....S... H.. C.. C...C...S. H.... H H H..O. H H H.... H 12 Bonding in Organic Compounds Chapter 1 1.23 Electron Dot Formulas: Section 1.4A-C H O O (a) Br C C O H (b) H S C N H H H 1.24 Electron Dot Formulas: Section 1.4A-C Place five carbons in a row and connect them with single bonds. It would take 12 hydrogens to satisfy the valences of all these carbons if all the carbon-carbon bonds are single. For each double bond you insert, you need two less hydrogens; three double bonds are needed. For every triple bond you insert, you need four fewer hydrogens; one triple and one double bond will work for this formula. H H H H H H H C C C C C H C C C C C H H C C C C C H H 1.25 Electron Dot Formulas and Formal Charge: Section 1.4E A neutral oxygen atom has six electrons. Ozone is neutral, O has three oxygen atoms, and thus a total of 18 electrons. The negative oxygen has three non-bonding and one bonding O O pairs; the oxygen "owns" seven and is thus negative. The positive oxygen has two non-bonding pairs and one bonding pair; it "owns" five electrons and is thus positive. 1.26 Formal Charge: Section 1.4E.H. + a) +CH3 b) CH3 c) -.. CH 3 d) CH3..O CH3 +.... - e) (CH 3)4N+ f) CH 3 N N.... g) CH 3O h) Br 13 Chapter 1 Bonding in Organic Compounds 1.27 Electron Dot Formulas and Formal Charge: Section 1.4E O O Carbonate Bicarbonate ion ion C C O O O O H 1.28 Polar Covalent Bonds: Section 1.4F Most bonds in organic compounds are considered polar except carbon- hydrogen and carbon-carbon bonds. ∂+ ∂- H H O ∂- ∂+ H S C C C ∂+ O H ∂+ H N H ∂- H ∂- ∂+ ∂+ 1.29-1.31 Bonding in Organic Compounds: Section 1.5 See section 1.5; there is a summary in 1.5G. Also see problems 1.12-1.15 and example 1.4. Problem 1.29: The carbons involved only in single bonds have four bonded groups and are sp3 hybridized, tetrahedral, and have 109o bond angles. The carbons that have one double bond have three bonded groups and are sp2 hybridized, trigonal, and have 120o bond angles. The carbons involved in triple bonds or two double bonds are sp hybridized, linear, and have 180o bond angles. Problem 1.30: All single bonds are sigma bonds. A double bond is a sigma and a pi bond. A triple bond is constructed of a sigma and two pi bonds. 14 Bonding in Organic Compounds Chapter 1 Problem 1.31: Bonding picture. pi bond sigma bond A A A A A A A A A A single bond double bond triple bond H H H (a) H H (b) C C H H C C H H C C H H H H H sp3, 109O, tetrahedral Two end carbons: sp3, 109O, tetrahedral 2 O Two middle carbons: sp , 120 , trigonal (c) The two end carbons have four attached H H groups, are sp3-hybridized, tetrahedral, and have 109O bond angles. The two middle H C C C C H carbons have two attached groups, are H H sp-hybridized, linear and have 180O bond angles. 15 Chapter 1 Bonding in Organic Compounds H (d) H H C C C C C H H H sp2 sp3 sp trigonal tetrahedral linear O 120 bond angles 109O bond angles O 180 bond angles 1.32-1.34 Bonding with Oxygen and Nitrogen: Section 1.6 Problem 1.32: Carbons, nitrogens, and oxygens involved only in single bonds have four groups that occupy space; four bonded groups with carbon, three bonded groups and a non-bonding electron pair with nitrogen, and two bonded groups and two non-bonding pairs with oxygen. All are sp3 hybridized, tetrahedral, and have 109o bond angles. Carbons, nitrogens, and oxygens with one double bond have three groups that occupy space; these are three bonded groups for carbon, two bonded groups and a non-bonding pair for nitrogen and one bonded group and two non-bonding pairs for oxygen. All are sp2 hybridized, trigonal, and have 120o bond angles. Carbons and nitrogens with a triple bond have only two groups that occupy space two bonded groups for carbon and one bonded group and a non-bonding electron pair for nitrogen. Both are: sp hybridized, linear, and have 180o bond angles. Problem 1.33: Single bonds are sigma bonds; double bonds are one sigma and one pi bond. Triple bonds are one sigma and two pi bonds. 16 Bonding in Organic Compounds Chapter 1 Problem 1.34: Bonding pictures. (a) H H H The carbons and nitrogen are sp3-hybridized, H C C N tetrahedral, and have bond angles tlhat are approximately 109O. H H H (b) H The carbon with three hydrogens is H sp3-hybridized, tetrahedral and has C 109O bond angles. The carbon and H nitrogen in the triple bond are both C N sp2-hybridized, trigonal and have bond angles of 120O. H H (c) H The carbon with three hydrogens is sp3 hybridized, tetrahedral, and has 109O H C C N bond angles. The carbon and nitrogen in the triple bond are both sp-hybridized H and linear; the carbon has 180O bond angles. (d) The carbons and oxygen each have four H H H space-occupying groups; four bonded atoms for each carbon and two bonded H C C O atoms and two non-bonding electron pairs for the oxygen. Tlhese atoms are all H H sp3-hybridized, tetrahedral, and have approximate bond angles of 109O. (e) The carbon with three hydrogens is H O sp3-hybridized, tetrahedral, and has 109O bond angles. The carbon and H C C oxygen in the double bond are both sp2-hybridized, trigonal, and have H approximate bond angles of 120O. H 17 Chapter 1 Bonding in Organic Compounds 1.35 Silicon: Section 1.4A-C Silicon was a logical choice of an element for the Star Trek episode about a very different life form. Silicon is just below carbon in Group IV of the periodic table, has the same number of outer shell electrons, and has some properties that are similar. It can bond to itself (though not as extensively as carbon) and, like carbon, it is tetravalent. 1.36 Molecular Shape: Section 1.5 In NH3, nitrogen has four groups that occupy space, three bonding pairs (hydrogens) and one non-bonding pair of electrons. As such the preferred geometry is tetrahedral and the nitrogen is sp3 hybridized. N s2 p1 p1 p1 hybridizes to (sp3)2 (sp3)1 (sp3)1 (sp3)1 in NH3 Surrounding boron are three space occupying groups, the three fluorines. Boron does not have an octet of electrons. Therefore it assumes a trigonal shape and is sp2 hybridized. B s1 p1 p1 p0 hybridizes to (sp2)1 (sp2)1 (sp2)1 p0 1.37 Bond Angles: Section 1.5 All four compounds have four pairs of electrons around the central atom. In CH4 they are all bonding pairs and relatively confined to the carbon-hydrogen bonds. Methane is a classic example of a tetrahedral molecule with 109o bond angles. In ammonia, NH3, there are three bonding pairs of electrons and one non-bonding pair. The non-bonding pair tends to occupy more space and repel the bonding pairs thus slightly compressing the bond angles; the result is 107o bond angles. In water there are two non-bonding electron pairs. These spacious pairs repel each other and the two bonding pairs thus further compressing the bond angles to 105o. 1.38 Reactivity: Section 1.4A-C The carbon in CH4 has a stable octet and all eight electrons are expressed as four bonding electron pairs. In NH3, the nitrogen has a stable octet but, since 18 Bonding in Organic Compounds Chapter 1 nitrogen is in Group V and has five outers shell electrons, there is a non-bonding electron pair remaining following formation of three bonds. This pair of electrons is available for sharing with electron-deficient species unlike the bonding pairs of CH4. Boron is in Group III of the periodic table. Since it has only three outer shell electrons it forms three bonds. However, it does not achieve a stable octet. Consequently, it is attracted to species that have electron pairs available for bonding (such as the N in NH3) because, in reacting, it can achieve a stable octet. H....H H.. H: N: H: H: B H :C......H H H Activities with Molecular Models 1. Make models of ethane (C2H6), ethene (C2H4), and ethyne (C2H2). These molecules illustrate sp3 (tetrahedral), sp2 (trigonal), and sp (linear) hybridizations respectively. Note the geometries and bond angles as you look at your models. (See textbook for models) 2. Make models of methane (CH4), formaldehyde (CH20), and hydrogen cyanide (HCN). Observe the geometries and bond angles at each carbon. Methane Formaldehyde Hydrogen Cyanide 19 Chapter 1 Bonding in Organic Compounds 3. Make models of methanol (CH4O) and formaldehyde (CH2O). Note the geometries and bond angles of both the carbons and oxygens in these molecules. (See textbook for models.) 4. Make models of CH5N, CH3N, and HCN. Note the geometries and bond angles at both the carbons and the nitrogens. 20 The Alkanes Chapter 2 2 THE ALKANES: STRUCTURE AND NOMENCLATURE OF SIMPLE HYDROCARBONS CHAPTER SUMMARY Organic compounds are classified according to common structural features that impart similar chemical and physical properties to the compounds within each group or family. 2.1 Hydrocarbons: An Introduction Hydrocarbons are composed only of carbon and hydrogen and fall into two major classes - saturated and unsaturated. Saturated hydrocarbons or the alkanes are entirely constructed of single bonds and have the general formula CnH2n+2. Unsaturated hydrocarbons include the alkenes (CnH2n) in which there is at least one carbon-carbon double bond; the alkynes (CnH2n-2) where there is at least one carbon-carbon triple bond; and aromatic hydrocarbons which appear to have double bonds but actually have a special structure that is discussed in Chapter 6. 2.2 Molecular and Structural Formulas - Isomerism Compounds are described by molecular formulas and structural formulas. Molecular formulas describe the kinds of atoms and numbers of 21 Chapter 2 The Alkanes each in a molecule. Structural formulas describe the bonding arrangements of the atoms, that is, what atoms are bonded to one another and by what kinds of bonds. Isomers are compounds with the same molecular formula but different structural formulas. Structural isomers (skeletal, positional, and functional) differ in the bonding arrangement of atoms; different atoms are attached to one another. In stereoisomerism the same atoms are bonded to one another but their orientation in space differs; conformational and geometric isomerism are forms of stereoisomerism presented in this chapter. 2.3 Skeletal Isomerism in Alkanes A. Isomers Isomers are different compounds with the same molecular formula but different structural formulas. Skeletal isomers differ in the arrangement of the carbons in a set of isomers; there are differences in the carbon skeletons. B. Drawing Structural Isomers The rules and procedures for drawing structural isomers are the same used for drawing electron dot formulas. Every atom in the molecular formula must be used and each atom must have its valence satisfied. To draw a structure, bond all atoms with a valence greater than one with single bonds. Attach monovalent atoms to the polyvalent ones until all valences have been satisfied. If there are insufficient monovalent atoms in the formula to accomplish this, insert double bonds, triple bonds or draw cyclic structures until it is possible to satisfy all valences. To draw isomers, vary the arrangements of atoms and bonds to form different molecules. C. Cycloalkanes The simplest cycloalkanes have the general formula CnH2n; they have two fewer hydrogens than the corresponding alkane and at least three of the carbons are arranged in a ring. 2.4 Representations of Structural Formulas In Chapter 1, electron dot formulas were used to describe covalent compounds. More condensed respresentations involve replacing the electron dots with lines (one for a single bond, two for a double bond, three for a triple bond), grouping hydrogens on a carbon, grouping identical carbons, and using stick diagrams in which each corner represents a carbon with sufficient hydrogens to satisfy the valence. 22 The Alkanes Chapter 2 2.5 Positional Isomerism Positional isomers differ in the position of a noncarbon group or of a double bond or triple bond. 2.6 IUPAC Nomenclature of Alkanes A. An Introduction to IUPAC Nomenclature Many organic compounds have informal common names but the accepted way of naming compounds is by the IUPAC system of nomenclature. B. Nomenclature of Continuous-Chain, Unbranched Alkanes: The Basis for Organic Nomenclature The base name of alkanes is derived from the Greek for the number of carbons in the longest continuous carbon chain (Table 2.1 of the text) followed by the suffix ane. The base name of cycloalkanes is based on the Greek for the number of carbons in the ring with the prefix cyclo and the suffix ane. C. Nomenclature of Branched-Chain Alkanes Branched-chain alkanes have a longer carbon chain, upon which the name is usually based, with attached shorter carbon chains. These shorter chains are called alkyl groups (Table 2.2 of the text) and are named by changing the ane (of the alkane name) to yl. The positions of alkyl groups are described with numbers; the longest carbon chain of an alkane is numbered starting at the end that gives the lowest number to the first substituent. Multiple numbers of identical alkyl groups are indicated with di, tri, tetra, etc. D. Nomenclature of Halogenated Hydrocarbons (Alkyl Halides) The prefixes fluoro, chloro, bromo, and iodo are used to indicate the presence of halogen in a molecule. 2.7 Conformational Isomerism Conformational isomers are isomers in which the spatial relationship of atoms differs because of rotation around a carbon-carbon double bond. Because the rotation is unrestricted in most cases, conformational isomers are constantly interconverting and are not isolatable. There are two extreme conformations. In the eclipsed conformation, atoms on adjacent carbons are lined up with one another and are as close together as possible; this is the least stable 23 Chapter 2 The Alkanes conformational arrangement. In the staggered conformation, atoms on adjacent atoms are staggered with one another and are as far apart as possible; this is the most stable conformational arrangement. Staggered and eclipsed forms are represented with sawhorse diagrams or Newman projections. Sawhorse diagrams are essentially stick structures highlighting the two carbons for which the conformations are being described. In a Newman projection the carbon- carbon bond is described by a circle with three bonds emanating from the center (the front carbon) and three from the perimeter (the back carbon). 2.8 Cycloalkanes - Conformational and Geometric Isomerism A. Structure and Stability Cycloalkanes are generally depicted with regular polygons though they actually exist in three-dimensional conformations. Cyclopropane and cyclobutane are less stable than other cycloalkanes since they are planar (cyclopropane) or nearly so (cyclobutane) and have internal bond angles significantly smaller (60o and 90o respectively) than the preferred tetrahedral angle (109o). The larger cycloalkanes are able to pucker out of planarity and assume tetrahedral angles. B. Conformational Isomerism in Cyclohexane Cyclohexane exists in two puckered conformations, the boat and chair forms, that have tetrahedral bond angles. The boat form is less stable and not preferred because of interactions between the two end or flagpole carbons and because the hydrogens on the other adjacent carbons are eclipsed. In the preferred chair form, atoms on adjacent carbons are staggered and there are no flagpole type interactions. There are two orientations of hydrogens in the chair conformation. Axial hydrogens are oriented directly above or below the "plane" of the ring in an alternating arrangement. Equatorial hydrogens protrude out along the perimeter of the ring. C. Drawing the Cyclohexane Chair In drawing the cyclohexane chair, keep in mind that there are four carbons in a plane. On one end there is a carbon oriented above the plane and on the other end there is a carbon oriented below the plane. Each carbon has an equatorial hydrogen oriented along the perimeter. There are three axial hydrogens on alternating carbons above the ring and three on the other carbons below the ring. 24 The Alkanes Chapter 2 D. Conformational Isomerism in Substituted Cyclohexanes In a monosubstituted cyclohexane, the substituent can be either in an equatorial or axial position. Equatorial positions are more spacious and in substituted cyclohexanes they are preferred. Cyclohexane rings flip between chair forms and establish an equilibrium. In the process of flipping, all equatorial positions become axial and all axial positions become equatorial. The equilibrium favors the chair in which substituents are equatorial. In monosubstituted cyclohexanes, the conformation in which the substituent is equatorial is favored. In disubstituted cyclohexanes where one group is axial and one equatorial, the equilibrium favors the chair form where the larger group occupies the more spacious equatorial position. E. Geometric Isomerism in Cyclic Compounds Disubstituted cycloalkanes exhibit geometric isomerism, a type of stereoisomerism. If both groups are on the same "side" of the ring (both up or both down) the isomer is termed cis. If they are on opposite "sides" (one up and one down) the isomer is trans. 2.9 Hydrocarbons: Relationship of Structure and Physical Properties The solid, liquid, and gas states of a compound differ in arrangements of molecules, not in molecular structure. In the solid state the molecules are orderly arranged and immobile with maximum intermolecular attractions. In the liquid state, molecules are mobile but still there are intermolecular attractions. In the vapor phase molecules are mobile and ideally there are no intermolecular attractions. For these reasons, solids have a constant shape and volume, liquids have a constant volume and variable shape, and gases assume the size and shape of the container. Physical properties of hydrocarbons are related to structure. A. Melting Point, Boiling Point, and Molecular Weight Melting points and boiling points generally increase with molecular weight within a homologous series (a series of compounds in which each succeeding member differs from the previous one by a CH2 group). B. Melting Point, Boiling Point, and Molecular Structure Branched chain hydrocarbons have less surface area and thus less opportunity for intermolecular attractions; as a result, their boiling points are lower than the straight chain isomers. However, their compact nature causes 25 Chapter 2 The Alkanes them to fit more easily in a crystal lattice and thus they generally have higher melting points. C. Solubility and Density Hydrocarbons are non-polar and insoluble in water and, because they are less dense, they float on the surface of water. CONNECTIONS 2.1: Petroleum SOLUTIONS TO PROBLEMS 2.1 Alkanes, Alkenes, and Alkynes H H H H H H H C C C H H C C C H H C C C H H H H H H H 2.2 Skeletal Isomers H H C H H H H H H H H C C C C H H C C C H H H H H H H H 2.3 Skeletal Isomers (a) All three of these structures are identical. In each, the longest continuous chain of carbon atoms is six with a CH3 group attached to the second carbon from the end. (b) The first and last structures are the same. In each, the longest continuous chain of carbons is six and there is a CH3 group bonded to the third carbon from the end. 26 The Alkanes Chapter 2 2.4 Skeletal Isomers Start with a chain of seven carbons. CH 3CH 2CH 2CH 2CH 2CH 2CH 3 Now draw isomers with six carbons in the longest chain and vary the position of the one-carbon chain. CH 3 CH 3 CH 3CHCH 2CH 2CH 2CH 3 CH3CH 2CHCH 2CH 2CH 3 Now draw five carbon chains and place one two-carbon chain or two one-carbon chains on it. If the two-carbon side chain is placed on either the first or second carbon, it merely extends the longest chain. However, placing it on the third carbon gives us another isomer. CH2CH 3 CH3CH 2 CHCH2CH3 Now attach two one-carbon chains to the carbon skeleton. CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CCH 2CH 2CH 3 CH 3CH 2 CCH 2CH 3 CH 3CHCHCH 2CH 3 CH 3CHCH 2CHCH 3 CH 3 CH 3 CH 3 Finally, draw a four carbon chain with three one carbon side chains. CH3 CH3 CH3C CHCH3 CH3 2.5 Skeletal Isomers of Cycloalkanes: Five cyclic compounds of C5H10. Start with a five-membered ring. Then use a four-membered ring with a one- carbon side chain. Finally, draw a three-membered ring with either one two- carbon side chain or two one-carbon side chains. CH3 CH2CH 3 CH3 CH3 CH3 CH3 27 Chapter 2 The Alkanes 2.6 Representations of Structural Formulas CH3 (CH3)2CH(CH2)3CHCH2C(CH3)3 2.7 Molecular Formulas from Structural Formulas (a) C13H8N2O2 (b) C8H12 (c) C15H29Br 2.8 Positional Isomers Two isomers of C2H4Br2 Br Br Br CH3CH CH2 CH2 Br Five isomers of C3H6BrCl Br CH 3CH 2CHBr CH3CH CH2 CH 2CH 2CH 2 CH3CH CH2 CH 3CCH 3 Cl Cl Br Cl Br Br Cl Cl 2.9 IUPAC Nomenclature (a) Names of compounds in Example 2.1 as they appear: hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane. (b) Names of nine isomers in Problem 2.4 as they appear: heptane; 2-methylhexane and 3-methylhexane; 3-ethylpentane; 2,2-dimethylpentane; 3,3-dimethylpentane; 2,3-dimethylpentane; and 2,4-dimethylpentane; 2,2,3-trimethylbutane c. Structures from names 28 The Alkanes Chapter 2 CH3 CH 2CHCH 3 CH3 CH2CH 3 CH3 CH 3CCH 2CHCHCHCH 2CHCH 3 CHCH 3 CH3 CH3 CH2CH 3 CH3 1-isobutyl-3-isopropylcyclopentane 5,6-diethyl-2,2,4,8-tetramethylnonane 2.10 Alkyl Halide Nomenclature Names of compounds in section 2.5 as they appear. 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, and 2-bromo-2-methylpropane. 2.11 Alkyl Halide Nomenclature (a) Names of compounds as they appear in problem 2.8a: 1,1-dibromoethane and 1,2-dibromoethane (b) Names of compounds as they appear in problem 2.8b.. 1-bromo-1-chloropropane, 1-bromo-2-chloropropane, 1-bromo-3-chloropropane, 2-bromo-1-chloropropane, and 2-bromo-2-chloropropane 2.12 Skeletal and Positional Isomerism CH 3CH 2CH 2CH 2CH 2Cl CH 3CH 2CH 2CHCH 3 CH 3CH 2CHCH 2CH 3 Cl Cl 1-chloropentane 2-chloropentane 3-chloropentane CH 3 CH 3 CH 2CHCH 2CH 3 CH 3CCH 2CH 3 Cl Cl 1-chloro-2-methylbutane 2-chloro-2-methylbutane 29 Chapter 2 The Alkanes CH 3 CH 3 CH 3CHCHCH 3 CH 3CHCH 2CH 2 Cl Cl 2-chloro-3-methylbutane 1-chloro-3-methylbutane CH 3 H3C C CH 2Cl 1-chloro-2,2-dimethylpropane CH 3 2.13 Newman Projections of Propane H CH 3 H3C H H Staggered Eclipsed H H H H H H H Most Stable Least Stable 2.14 Conformational Isomers of Butane CH3 H 3C H3C CH3 H CH3 H CH3 H H H H H H H H H H H 3C H H H H CH3 Least Stable Most Stable 2.15 Cyclohexane Chair: Axial and Equatorial Positions a) CH3 b) H c) Br H H CH3 H Br 30 The Alkanes Chapter 2 d) Br e) H H H Br Br Br H 2.16 Equilibrium between Chair Forms (a) The equatorial position is more spacious and the isomer more stable. ax Br more stable H eq eq Br H ax (b) The equatorial positions are more spacious and preferred in the equilibrium. ax CH(CH 3)2 ax more H eq H stable H eq CH 3 CH(CH 3)2 eq eq ax CH 3 H ax 31 Chapter 2 The Alkanes (c) The equilibrium favors the isomer in which the larger isopropyl group is in the more spacious equatorial position. ax CH(CH 3)2 more H3C ax stable H H eq eq eq eq CH 3 CH(CH 3)2 H ax H ax 2.17 Geometric Isomerism in Cyclic Compounds Br CH 3 Br H CH 3 H Br Br H CH 3 H H H CH 3 H H cis trans cis trans 1-bromo-2-methylcyclopentane 1-bromo-3-methylcyclopentane 2.18 Molecular Weights (a) 16 (b) 46 (c) 342 (d) 264 (11x12 + 17x1 + 2x14 + 2x16 + 32 +23) 2.19 Skeletal Isomerism: Sections 2.3 and 2.6 (a) C8H18: Start with an eight-carbon chain. Then systematically make the longest chain one carbon shorter. Arrange the remaining carbons in each case on the chain in as many different ways as possible without extending the length of the base chain. The following isomers are drawn in a logical, systematic order. CH3 CH 3CH 2CH 2CH 2CH 2CH 2CH 2CH 3 CH 3CHCH 2CH 2CH 2CH 2CH 3 octane 2-methylheptane CH 3 CH 3 CH 3CH 2CHCH 2CH 2CH 2CH 3 CH 3CH 2CH 2CHCH 2CH 2CH 3 3-methylheptane 4-methylheptane 32 The Alkanes Chapter 2 CH 2CH 3 CH 3 CH 3 CH 3CH 2CHCH 2CH 2CH 3 CH 3CCH 2CH 2CH 2CH 3 CH 3CH 2CCH 2CH 2CH 3 CH 3 CH 3 3-ethylhexane 2,2-dimethylhexane 3,3-dimethylhexane CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3CH CHCH 2CH 2CH 3 CH 3CHCH 2CHCH 2CH 3 CH 3CHCH 2CH 2CHCH 3 2,3-dimethylhexane 2,4-dimethylhexane 2,5-dimethylhexane CH 3 CH 3 CH 3 CH 3 CH 2 CH 3 CH 3CH 2CH CHCH 2CH 3 CH 3CH CHCH 2CH 3 CH 3CH 2 C CH 2CH 3 CH 2 CH 3 3,4-dimethylhexane 3-ethyl-2-methylpentane 3-ethyl-3-methylpentane CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3C CHCH 2CH 3 CH 3CCH 2CHCH 3 CH 3CH C CH 2CH 3 CH 3 CH 3 CH 3 2,2,3-trimethylpentane 2,2,4-trimethylpentane 2,3,3-trimethylpentane H3C CH 3 CH 3 CH 3CH 3 H3CCH CH CHCH 3 H3C C C CH 3 CH 3CH 3 2,3,4-trimethypentane 2,2,3,3-tetramethylbutane 33 Chapter 2 The Alkanes (b) three isomers of C9H20 with eight carbons in the longest chain: write the longest chain straight across. Don't put anything on the chain that would make it longer. CH 3 CH 3 CH 3CHCH 2CH 2CH 2CH 2CH 2CH 3 CH3CH 2CHCH 2CH 2CH 2CH 2CH 3 2-methyloctane 3-methyloctane CH 3 CH 3CH 2CH 2CHCH 2CH 2CH 2CH 3 4-methyloctane (c) 11 isomers of C9H20 with seven carbons in the longest chain: CH 2CH 3 CH 2CH 3 CH 3CH 2CHCH 2CH 2CH 2CH 3 CH3CH 2CH 2CHCH 2CH 2CH 3 3-ethylheptane 4-ethylheptane CH 3 CH 3 CH 3 CH3CCH 2CH2CH2CH2CH3 CH3CH2CCH 2CH 2CH 2CH 3 CH3CH2CH 2CCH 2CH2CH3 CH 3 CH 3 CH 3 2,2-dimethylheptane 3,3-dimethylheptane 4,4-dimethylheptane CH 3CH 3 CH 3 CH 3 CH 3 CH 3 CH 3CHCHCH 2CH 2CH 2CH 3 CH 3CHCH 2CHCH 2CH 2CH 3 CH 3CHCH 2CH 2CHCH 2CH 3 2,3-dimethylheptane 2,4-dimethylheptane 2,5-dimethylheptane CH 3 CH 3 CH 3CH 3 CH 3 CH 3 CH 3CHCH 2CH 2CH2CHCH 3 CH 3CH 2CHCHCH 2CH 2CH 3 CH 3CH 2CHCH 2CHCH 2CH 3 2,6-dimethylheptane 3,4-dimethylheptane 3,5-dimethylheptane (d) Eight isomers of C9H20 with five carbons in longest chain: Draw a five- carbon chain across the paper in a straight line. Arrange the remaining four 34 The Alkanes Chapter 2 carbons in as many ways as possible without making the chain longer. The ways to consider arranging the remaining four carbons are: a) one four-carbon chain; b) a three- and a one-carbon chain, c) 2 two-carbon chains; d) 1 two- and 2 one-carbon chains; and e) 4 one-carbon chains. Variations a) and b) are not usable as there is no way to place a four- or three-carbon unit on a five-carbon chain without extending the longest chain. CH3 CH3 CH3 CH2 CH2 CH3 CH2 CH3CH 2 C CH2CH 3 CH3CH 2CH CCH3 CH 3CHCHCHCH 3 CH2 CH3 CH3 CH3 CH3 3,3-diethylpentane 2,2-dimethyl-3-ethylpentane 2,4-dimethyl-3-ethylpentane CH3 CH2 CH3 CH3 CH3 CH3 CH3CH CCH2CH3 CH3C CCH2CH3 CH3C CH2 CCH3 CH3 CH3 CH3 CH3 CH3 CH3 2,3-dimethyl-3-ethylpentane 2,2,3,3-tetramethylpentane 2,2,4,4-tetramethylpentane CH3 CH3 CH3 CH3 CH3 CH3 CH3C CH CHCH3 CH3CH C CHCH 3 CH3 CH3 2,2,3,4-tetramethylpentane 2,3,3,4-tetramethylpentane (e) four isomers of C10H22 with nine carbons in the longest chain: CH3 CH3 CH 3CHCH 2CH 2CH 2CH 2CH 2CH 2CH 3 CH3CH 2CHCH 2CH 2CH 2CH 2CH 2CH 3 2-methylnonane 3-methylnonane 35 Chapter 2 The Alkanes CH3 CH3 CH 3CH 2CH 2CHCH 2CH 2CH 2CH 2CH 3 CH3CH 2CH 2CH 2CHCH 2CH 2CH 2CH 3 4-methylnonane 5-methylnonane (f) two isomers of C10H22 with only two alkyl groups on a six carbon chain: CH2CH 3 CH2CH 3 CH 3CH 2CCH 2CH 2CH 3 CH 3CH 2CHCHCH 2CH 3 CH2CH 3 CH2CH 3 3,3-diethylhexane 3,4-diethylhexane (g) six isomers of C10H22 with five carbons in the longest chain: CH3 CH2CH 3 CH3 CH2CH 3 CH3 CH2CH 3 CH 3CH CCH 2CH 3 CH 3C CCH 2CH 3 CH 3C CHCHCH 3 CH2CH 3 CH3 CH3 CH3 CH3 3,3-diethyl-2-methylpentane 3-ethyl-2,2,3-trimethylpentane 3-ethyl-2,2,4-trimethylpentane CH 2CH 3 CH 3 CH 3 CH 3 CH 3 CH 3CH 3 CH 3CH C CHCH 3 CH 3C C CHCH 3 CH 3C CH-CCH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 3-ethyl-2,3,4-trimethylpentane 2,2,3,4,4-pentamethylpentane 2,2,3,3,4-pentamethylpentane 36 The Alkanes Chapter 2 (h) the isomer of C13H28 with the shortest longest chain possible CH3 CH 3 CH3 CH 2 CH 3C C CCH 3 3,3-diethyl-2,2,4,4-tetramethylpentane CH CH3 CH 2 CH3 3 (i) five cyclic compounds of C5H10 CH3 CH2CH 3 CH3 CH3 CH3 CH3 Names in order: cyclopentane; methylcyclobutane; ethylcyclopropane 1,1-dimethylcyclopropane; 1,2-dimethylcyclopropane (j) twelve cyclic compounds of C6H12 CH3 CH2CH 3 CH3 CH3 CH3 CH3 CH3 CH3 Names in order: cyclohexane; methylcyclopentane; ethylcyclobutane; 1,1-dimethylcyclobutane; 1,2-dimethylcyclobutane; 1,3-dimethylcyclobutane. CH3 CH3 CH3 CH3 CH CH2CH 3 CH3 CH2CH 2CH3 CH3 CH2CH 3 CH3 CH3 CH3 CH3 Names in order: propylcyclopropane; isopropylcyclopropane; 1-ethyl-1-methylcyclopropane; 1-ethyl-2-methylcyclopropane; 1,1,2-trimethylcyclopropane; 1,2,3-trimethylcyclopropane. 37 Chapter 2 The Alkanes (k) five compounds of C8H16 that have a six-membered ring CH 2CH 3 H3C CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 Names in Order: ethylcyclohexane; 1,1-dimethylcyclohexane 1,2-dimethylcyclohexane; 1,3-dimethylcyclohexane; 1,4-dimethylcyclohexane (l) 12 isomers of C9H18 that have a six-membered ring CH 3 CH 2CH 2CH 3 CHCH 3 CH 3 H3C CH 2CH 3 CH 2CH 3 Names in Order: propylcyclohexane; isopropylcyclohexane; 1-ethyl-1-methylcyclohexane; 1-ethyl-2-methylcyclohexane CH 3 CH 3 H3C CH 3 H3C CH 3 CH 3 CH 2CH 3 CH 3 CH 2CH 3 Names in Order: 1-ethyl-3-methylcyclohexane; 1-ethyl-4-methylcyclohexane; 1,1,2-trimethylcyclohexane; 1,1,3-trimethylcyclohexane 38 The Alkanes Chapter 2 CH 3 CH 3 CH 3 H3C CH 3 CH 3 CH 3 CH 3 H3C CH 3 CH 3 CH 3 Names in Order: 1,1,4-trimethylcyclohexane; 1,2,3-trimethylcyclohexane; 1,2,4-trimethycyclohexane; 1,3,5-trimethylcyclohexane 2.20 Nomenclature of Alkanes: Section 2.6 Names are included with structures in Problem 2.l9. 2.21 Positional Isomers: Section 2.5 Each carbon that can have a hydrogen replaced with a chlorine is numbered. Identically numbered carbons produce the same isomer upon chlorination. 1 5 CH3 CH3 a) CH3CH 2CH 2CH 2CH 2CH 3 b) 1 CH 3CCH 2CH 2CHCH 3 1 2 3 3 2 1 2 3 4 5 CH3 3 isomers 1 5 isomers 1 5 1 1 Cl c) CH3 CH3 CH3 d) CH 3CHCH 2CHCH 2CHCH 3 1 1 1 2 3 4 3 2 1 1 isomer 5 isomers 2.22 Skeletal and Positional Isomerism: Sections 2.3 and 2.5 a) three isomers of C2H3Br2F H Br Br Br Br H H C C F H C C F H C C F H Br H H Br H 1,1-dibromo-1-fluoroethane 1,1-dibromo-2-fluoroethane 1,2-dibromo-1-fluoroethane 39 Chapter 2 The Alkanes (b) four isomers of C3H6Br2 Br CH 3CH 2CHBr CH 3CCH 3 CH 3CH CH 2 CH 2CH 2CH 2 Br Br Br Br Br Br Names in Order: 1,1-dibromopropane; 2,2-dibromopropane; 1,2-dibromopropane; 1,3-dibromopropane (c) twelve isomers of C4H8BrF Br Br CH 3CH 2CH 2CH CH 3CH 2CHCH 3 CH 3CH 2CHCH 2 CH 3CHCH 2CH 2 F F F Br F Br Names in order: 1-bromo-1-fluorobutane; 2-bromo-2-fluorobutane; 1-bromo-2-fluorobutane; 1-bromo-3-fluorobutane CH 2CH 2CH 2CH 2 CH3CH 2CH CH2 CH 3CHCH 2CH 2 CH3CH CHCH3 F Br Br F Br F Br F Names in order: 1-bromo-4-fluorobutane; 2-bromo-1-fluorobutane; 3-bromo-1-fluorobutane; 2-bromo-3-fluorobutane CH3 CH3 CH3 CH3 CH 3CHCHBr CH3 C CH2 CH3C CH2 CH2 C CH2 F F Br Br F F H Br Names in order: 1-bromo-1-fluoro-2-methylpropane; 1-bromo-2-fluoro- 2-methylpropane; 2-bromo-1-fluoro-2-methylpropane; 1-bromo-3-fluoro- 2-methylpropane (d) 6 isomers of C4H8Br2 with four carbons in the longest chain Br Br Br Br CH 3CH 2CH 2CH CH 3CH 2CCH 3 CH 3CH 2CHCH 2 Br Br 1,1-dibromobutane 2,2-dibromobutane 1,2-dibromobutane 40 The Alkanes Chapter 2 Br Br Br Br Br Br CH 3CHCH 2CH 2 CH 2CH 2CH 2CH 2 CH 3CHCHCH 3 1,3-dibromobutane 1,4-dibromobutane 2,3-dibromobutane (e) nine isomers of C5H10Br2 with four carbons in the longest chain For simplicity let us just show the required carbon skeleton and move the two bromines around systematically. First, place two bromines on the same carbon. C C Br C C C C C Br C C C C Br C C C C Br Br Br 1,1-dibromo-3-methylbutane 1,1-dibromo-2-methylbutane 2,2-dibromo-3-methylbutane Now put a bromine on carbon-1 and vary the position of the other bromine. C C C C C C C C C C C C C C C Br Br Br Br Br Br 1,2-dibromo-2-methylbutane 1,4-dibromo-2-methylbutane 1,3-dibromo-2-methylbutane Finally, draw isomers in which the bromines are on the middle two carbons and the two carbons on the other end. C Br C C C C C C C C C C C C C C Br Br Br Br Br 1,3-dibromo-2-ethylpropane 1,2-dibromo-3-methylbutane 2,3-dibromo-2-methylbutane 41 Chapter 2 The Alkanes (f) five isomers of C6H13Cl with four carbons in the longest chain. Again let us draw the carbon skeletons, there are two, and vary the position of the chlorine. C C C C C C C C C C C C C C C Cl C Cl Cl C C 1-chloro-2,2-dimethylbutane 1-chloro-3,3-dimethylbutane 3-chloro-2,2-dimethylbutane C C C C C C C C C C C C Cl Cl 1-chloro-2,3-dimethylbutane 2-chloro-2,3-dimethylbutane 2.23 Nomenclature of Halogenated Alkanes: Section 2.6D The names are with the structures in Problem 2.22. 2.24 Nomenclature of Alkanes: Section 2.6 (a) propane (b) decane (c) octane 2.25 Nomenclature of Alkanes: Section 2.6 (a) 4-methylnonane; (b) 5-propyldecane; (c) 2,5-dimethylhexane; (d) 4-ethyl-2-methylhexane; (e) 2,2,4,6-tetramethylheptane; (f) 6-propyl-2,4,4-trimethyldecane (longest chain is not written straight across the page); (g) 2-cyclobutylhexane; (h) 3,3,5,7-tetramethyldecane (note that the longest chain is not written straight across the page; the two fragments below are part of the longest chain.) (i) 2,2,3,3-tetramethylbutane; (j) ethylcyclopropane; (k) isopropylcyclopentane; (l) 1-butyl-4-t-butylcyclohexane; (m) 2,2-dimethylbutane; (n) 2,4-dimethylhexane; (o) 4-ethyl-2,2-dimethylhexane 42 The Alkanes Chapter 2 2.26 Nomenclature of Halogenated Alkanes: Section 2.6D (a) triiodomethane; (b) 2-bromo-3-methylbutane; (c) 1,3-dibromo-4-fluoro-2,4-dimethylpentane 2.27 IUPAC Nomenclature: Section 2.6 CH3 CH3 Cl Cl Cl (a) CCl 2F2 (b) CH 3CCH 2CHCH 3 (c) CH3 Cl CH3 CH2CH 3 CH Cl 3 Cl (d) CH 3CCH 2CHCH CHCH2CH3 CH3 CH2CH 2CH3 2.28 Conformational Isomers: Section 2.7 In each case, draw the compound, determine what three groups are on each of the carbons to be placed in the Newman projection, draw the Newman projection with the bonds for the front carbon emanating from the center of the circle and those of the back carbon coming from the perimeter, and put the three groups on each carbon. Rotate between staggered and eclipsed conformations to get the extreme forms. H H H H (a) OH H OH H C C OH H H H H H View H H H H front to back 43 Chapter 2 The Alkanes H H (b) HOCH 2 CH 2OH HO C C OH H H View front to back HO HO OH H OH OH H OH H H H H H H H H H H H HH H HO OH H CH3 H H (c) OH H OH H C C OH H H H H H View H front to back CH3H CH3 2.29 Conformational Isomerism in Substituted Cyclohexanes: Section 2.8 B-D In doing these problems, remember that equatorial positions are roomier than axial positions and substituents occupy equatorial positions preferentially when possible. If there are two groups on the cyclohexane chair, the conformation in which the larger group is equatorial, or, if possible, both groups are equatorial, is preferred. (a) H ax CH2CH 3 eq ethylcyclohexane 44 The Alkanes Chapter 2 (b) most stable chair forms of 1,2-; 1,3-; and 1,4-dimethylcyclohexane ax H H ax H ax CH3 CH3 CH3 eq eq H ax eq eq CH3 CH3 eq H ax CH3 eq H ax (c) least stable chair forms of compounds in part b CH3 ax CH3 ax CH3 ax H H H eq ax CH3 eq eq eq H H eq eq CH3 ax H ax CH3 (d) 1,2-dimethylcyclohexane with one group axial and one equatorial CH3 ax H eq CH3 eq H ax 45 Chapter 2 The Alkanes (e) most stable chair form of 1-butyl-3-methylcyclohexane with one group axial and one equatorial H ax ax CH2CH 2CH2CH3 eq the larger group CH3 is in the roomier equatorial eq H position 2.30 Conformational Isomerism in Substituted Cyclohexanes: Section 2.8 (a) bromocyclohexane Br H equatorial Br axial less stable more stable H (b) arrows show ring flipping Br Br H H H H Br Br Br Br H H one Br axial diaxial diequatorial less stable more stable one Br equatorial (c) 1-ethyl-3-methylcyclohexane CH3CH 2 CH3 H CH3 arrows show ring flip; axial groups become H equatorial and vice-versa CH2CH 3 H H more stable less stable both groups equatorial both groups axial 46 The Alkanes Chapter 2 ax CH2CH 3 ax eq H H eq H eq more stable - larger CH3 group in roomier equatorial position less stable- CH3 CH2CH 3 larger group eq ax H ax in crowded axial position (d) 1-ethyl-4-methylcyclohexane ring flip CH2CH 3 H H CH3 more stable - both groups equatorial H less stable- CH2CH 3 both groups axial CH3 H ring flip CH2CH 3 CH3 H more stable - larger H group in more spacious equatorial position CH3 less stable- CH2CH 3 larger group in more H crowded axial position H (e) 1,3,5-tribromocyclohexane Br more stable- H H H H ring less stable - two only one Br Br Br's in crowded in crowded H axial positions axial position; flip Br other two in Br Br Br spacious equatorial H 2.31 Conformational Isomerism in Cyclohexane: Section 2.8B The structures show a one-carbon bridge between the first and fourth carbons of the ring. In the boat form, the first and fourth carbons are directed toward one another and are easily tied together by the single bonds to the —CH2— bridge. However, in the chair form, these two carbons are so far removed from one 47 Chapter 2 The Alkanes another that they cannot be bridged by a single carbon. The two single bonds are not long enough nor can they be conveniently directed in the necessary geometry. CH3 CH 2 CH2 O CH3 H CH3 CH3 CH2 CH3 CH3 O CH3 O CH3 CH3 H Camphor Boat Form Chair Form 2.32 Conformational Isomerism CH 3 CH 3 In the first compound, H H both conformers have one methyl axial and one equatorial. They are the same. You CH 3 can bottle this isomer.It H3C identical structures H is the sole component H of the equilibrium. H CH 3 The diaxial conformer H3C H is in equilibrium with the diequatorial. The di- equatorial is virtually the exclusive component CH 3 of the equilibrium. The methyls diaxial methyls diequatorial H diaxial essentially H less stable more stable cannot exist. CH 3 2.33 Geometric Isomerism: Section 2.8E (a) 1,2-dimethylcyclopropane CH3 CH3 CH3 H cis trans H H H CH3 48 The Alkanes Chapter 2 (b) 1-bromo-3-chlorocyclobutane Br Cl Br H cis trans H H H Cl 2.34 Geometric Isomerism: Section 2.8E (a) 1,2-dimethylcyclohexane CH3 CH3 CH3 H trans cis H H H CH3 (b) 1,3-dimethylcyclohexane CH3 CH3 CH3 H cis trans H H H CH3 (c) 1,4-dimethylcyclohexane CH3 H CH3 CH3 trans cis H CH3 H H 2.35 Geometric Isomerism: Section 2.8E (a) 1,2-dibromo-3-chlorocyclopropane Br Cl Br H Br Cl Br Br H H H H Cl H H H H Br 49 Chapter 2 The Alkanes (b) 1,2,3-tribromocyclopropane Br Br Br Br Br H H H H H H Br (c) 1,2,4-tribromocyclopentane Br Br Br H Br Br Br Br H H H H Br H H H H Br (d) 2-chloro-1,4-dibromocyclopentane Br Br H Br Br H Br Br Cl Cl