Exam (elaborations) TEST BANK FOR ORGANIC CHEMISTRY A Short Course 13t

Exam (elaborations) TEST BANK FOR ORGANIC CHEMISTRY A Short Course 13t This study guide and solutions book was written to help you learn organic chemistry. The principles and facts of this subject are not easily learned by simply reading them, even repeatedly. Formulas, equations, and molecular structures are best mastered by written practice. To help you become thoroughly familiar with the material, we have included many problems within and at the end of each chapter in the text. It is our experience that such questions are not put to their best use unless correct answers are also available. Indeed, answers alone are not enough. If you know how to work a problem and find that your answer agrees with the correct one, fine. But what if you work conscientiously, yet cannot solve the problem? You then give in to temptation, look up the answer, and encounter yet another dilemma–how in the world did the author get that answer? This solutions book has been written with this difficulty in mind. For many of the problems, all of the reasoning involved in getting the correct answer is spelled out in detail. Many of the answers also include cross-references to the text. If you cannot solve a particular problem, these references will guide you to parts of the text that you should review. Each chapter of the text is briefly summarized. Whenever pertinent, the chapter summary is followed by a list of all the new reactions and mechanisms encountered in that chapter. These lists should be especially helpful to you as you review for examinations. When you study a new subject, it is always useful to know what is expected. To help you, we have included in this study guide a list of learning objectives for each chapter—that is, a list of what you should be able to do after you have read and studied that chapter. Your instructor may want to delete items from these lists of objectives or add to them. However, we believe that if you have mastered these objectives—and the problems should help you to do this—you should have no difficulty with examinations. Furthermore, you should be very well prepared for further courses that require this course as a prerequisite. Near the end of this study guide you will find additional sections that may help you to study for the final examination in the course. The SUMMARY OF SYNTHETIC METHODS lists the important ways to synthesize each class of compounds discussed in the text. It is followed by the SUMMARY OF REACTION MECHANISMS. Both of these sections have references to appropriate portions of the text, in case you feel that further review is necessary. Finally, you will find two lists of sample test questions. The first deals with synthesis, and the second is a list of multiple-choice questions. Both of these sets should help you prepare for examinations. In addition, we offer you a brief word of advice about how to learn the many reactions you will study during this course. First, learn the nomenclature systems thoroughly for each new class of compounds that is introduced. Then, rather than memorizing the particular examples of reactions given in the text, study reactions as being typical of a class of compounds. For example, if you are asked how compound A will react with compound B, proceed in the following way. First ask yourself: to what class of compounds does A belong? How does this class of compounds react with B (or with compounds of the general class to which B belongs)? Then proceed from the general reaction to the specific case at hand. This approach will probably help you to eliminate some of the memory work often associated with organic chemistry courses. We urge you to study regularly, and hope that this study guide and solutions book will make it easier for you to do so. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. vi Introduction to the Student Great effort has been expended to ensure the accuracy of the answers in this book and we wish to acknowledge the helpful comments provided by David Ball (Cleveland State University) in this regard. It is easy for errors to creep in, however, and we will be particularly grateful to anyone who will call them to our attention. Suggestions for improving the book will also be welcome. Send them to: Christopher M. Hadad Department of Chemistry The Ohio State University Columbus, Ohio 43210 David J. Hart Department of Chemistry The Ohio State University Columbus, Ohio 43210 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1 1 Bonding and Isomerism Chapter Summary∗ An atom consists of a nucleus surrounded by electrons arranged in orbitals. The electrons in the outer shell, or the valence electrons, are involved in bonding. Ionic bonds are formed by electron transfer from an electropositive atom to an electronegative atom. Atoms with similar electronegativities form covalent bonds by sharing electrons. A single bond is the sharing of one electron pair between two atoms. A covalent bond has specific bond length and bond energy. Carbon, with four valence electrons, mainly forms covalent bonds. It usually forms four such bonds, and these may be with itself or with other atoms such as hydrogen, oxygen, nitrogen, chlorine, and sulfur. In pure covalent bonds, electrons are shared equally, but in polar covalent bonds, the electrons are displaced toward the more electronegative element. Multiple bonds consist of two or three electron pairs shared between atoms. Structural (or constitutional) isomers are compounds with the same molecular formulas but different structural formulas (that is, different arrangements of the atoms in the molecule). Isomerism is especially important in organic chemistry because of the capacity of carbon atoms to be arranged in so many different ways: continuous chains, branched chains, and rings. Structural formulas can be written so that every bond is shown, or in various abbreviated forms. For example, the formula for n-pentane (n stands for normal) can be written as: C H C H H H C H H C H H C H H H H or CH3CH2CH2CH2CH3 or Some atoms, even in covalent compounds, carry a formal charge, defined as the number of valence electrons in the neutral atom minus the sum of the number of unshared electrons and half the number of shared electrons. Resonance occurs when we can write two or more structures for a molecule or ion with the same arrangement of atoms but different arrangements of the electrons. The correct structure of the molecule or ion is a resonance hybrid of the contributing structures, which are drawn with a double-headed arrow (↔) between them. Organic chemists use a curved arrow ( ) to show the movement of an electron pair. A sigma (σ) bond is formed between atoms by the overlap of two atomic orbitals along the line that connects the atoms. Carbon uses sp3-hybridized orbitals to form four such bonds. These bonds are directed from the carbon nucleus toward the corners of a tetrahedron. In methane, for example, the carbon is at the center and the four hydrogens are at the corners of a regular tetrahedron with H–C–H bond angles of 109.5°. ∗ In the chapter summaries, terms whose meanings you should know appear in boldface type. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 Chapter 1 Carbon compounds can be classified according to their molecular framework as acyclic (not cyclic), carbocyclic (containing rings of carbon atoms), or heterocyclic (containing at least one ring atom that is not carbon). They may also be classified according to functional group (Table 1.6). Learning Objectives∗ 1. Know the meaning of: nucleus, electrons, protons, neutrons, atomic number, atomic weight, shells, orbitals, valence electrons, valence, kernel. 2. Know the meaning of: electropositive, electronegative, ionic and covalent bonds, radical, catenation, polar covalent bond, single and multiple bonds, nonbonding or unshared electron pair, bond length, bond energy. 3. Know the meaning of: molecular formula, structural formula, structural (or constitutional) isomers, continuous and branched chain, formal charge, resonance, contributing structures, sigma (σ) bond, sp3-hybrid orbitals, tetrahedral carbon. 4. Know the meaning of: acyclic, carbocyclic, heterocyclic, functional group. 5. Given a periodic table, determine the number of valence electrons of an element and write its electron-dot formula. 6. Know the meaning of the following symbols: δ+ δ– 7. Given two elements and a periodic table, tell which element is more electropositive or electronegative. 8. Given the formula of a compound and a periodic table, classify the compound as ionic or covalent. 9. Given an abbreviated structural formula of a compound, write its electron-dot formula. 10. Given a covalent bond, tell whether it is polar. If it is, predict the direction of bond polarity from the electronegativities of the atoms. 11. Given a molecular formula, draw the structural formulas for all possible structural isomers. 12. Given a structural formula abbreviated on one line of type, write the complete structure and clearly show the arrangement of atoms in the molecule. 13. Given a line formula, such as (pentane), write the complete structure and clearly show the arrangement of atoms in the molecule. Tell how many hydrogens are attached to each carbon, what the molecular formula is, and what the functional groups are. 14. Given a simple molecular formula, draw the electron-dot formula and determine whether each atom in the structure carries a formal charge. ∗ Although the objectives are often worded in the form of imperatives (i.e., determine …,write …, draw …), these verbs are all to be preceded by the phrase “be able to …”. This phrase has been omitted to avoid repetition. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Bonding and Isomerism 3 15. Draw the electron-dot formulas that show all important contributors to a resonance hybrid and show their electronic relationship using curved arrows. 16. Predict the geometry of bonds around an atom, knowing the electron distribution in the orbitals. 17. Draw in three dimensions, with solid, wedged, and dashed bonds, the tetrahedral bonding around sp3-hybridized carbon atoms. 18. Distinguish between acyclic, carbocyclic, and heterocyclic structures. 19. Given a series of structural formulas, recognize compounds that belong to the same class (same functional group). 20. Begin to recognize the important functional groups: alkene, alkyne, alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine, nitrile, amide, thiol, and thioether. ANSWERS TO PROBLEMS Problems Within the Chapter 1.1 The sodium atom donates its valence electron to the chlorine atom to form the ionic compound, sodium chloride. 1.2 Elements with fewer than four valence electrons tend to give them up and form positive ions: Al3+, Li+. Elements with more than four valence electrons tend to gain electrons to complete the valence shell, becoming negative ions: S2–, O2–. 1.3 Within any horizontal row in the periodic table, the most electropositive element appears farthest to the left. Na is more electropositive than Al, and C is more electropositive than N. In a given column in the periodic table, the lower the element, the more electropositive it is. Si is more electropositive than C. 1.4 In a given column of the periodic table, the higher the element, the more electronegative it is. F is more electronegative than Cl, and N is more electronegative than P. Within any horizontal row in the periodic table, the most electronegative element appears farthest to the right. F is more electronegative than O. 1.5 As will be explained in Sec. 1.3, carbon is in Group IV and has a half-filled (or halfempty) valence shell. It is neither strongly electropositive nor strongly electronegative. 1.6 The unpaired electrons in the fluorine atoms are shared in the fluorine molecule. +F F F F + heat fluorine atoms fluorine molecule 1.7 dichloromethane (methylene chloride) trichloromethane (chloroform) C Cl Cl H H or H C Cl Cl H ClC Cl H Cl or H C Cl Cl Cl Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4 Chapter 1 1.8 If the C–C bond length is 1.54 Å and the Cl–Cl bond length is 1.98 Å, we expect the C–Cl bond length to be about 1.76 Å: (1.54 + 1.98)/2. In fact, the C–Cl bond (1.75 Å) is longer than the C–C bond. 1.9 Propane H C H H C H H C H H H 1.10 Nδ+–Clδ– ; Sδ+–Oδ– The key to predicting bond polarities is to determine the relative electronegativities of the elements in the bond. In Table 1.4, Cl is more electronegative than N. The polarity of the S–O bond is easy to predict because both elements are in the same column of the periodic table, and the more electronegative atom appears nearer the top. 1.11 Both Cl and F are more electronegative than C. C F F Cl Cl δ+ δ– δ– δ– δ– 1.12 Both the C–O and H–O bonds are polar, and the oxygen is more electronegative than either carbon or hydrogen. C H H H O H 1.13 H C N H C N H C N 1.14 a. The carbon shown has 12 electrons around it, 4 more than are allowed. b. There are 20 valence electrons shown, whereas there should only be 16 (6 from each oxygen and 4 from the carbon). c. There is nothing wrong with this formula, but it does place a formal charge of –1 on the “left” oxygen and +1 on the “right” oxygen (see Sec. 1.11). This formula is one possible contributor to the resonance hybrid structure for carbon dioxide (see Sec. 1.12); it is less important than the structure with two carbon–oxygen double bonds, because it takes energy to separate the + and – charges. 1.15 Methanal (formaldehyde), H2CO. There are 12 valence electrons altogether (C = 4, H = 1, and O = 6). A double bond between C and O is necessary to put 8 electrons around each of these atoms. C O H H o O r C H H 1.16 There are 10 valence electrons, 4 from C and 6 from O. An arrangement that puts 8 electrons around each atom is shown below. This structure puts a formal charge of –1 on C and +1 on O (see Sec. 1.11). Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Bonding and Isomerism 5 C orO C O 1.17 If the carbon chain is linear, there are two possibilities: CC and H H H C H H C H H H HC C H C H H H H C H H But the carbon chain can be branched, giving a third possibility: CC H H C C H H HH HH 1.18 a. b. N H H C H H H Notice that the fluorine has three non-bonded electron pairs (part a) and nitrogen has one non-bonded electron pair (part b). 1.19 No, it does not. We cannot draw any structure for C2H5 that has four bonds to each carbon and one bond to each hydrogen. 1.20 First write the alcohols (compounds with an O–H group). H C O H H C H H H C H H and H C H H C O H C H H H H Then write the structures with a C–O–C bond (ethers). H C O H H C H H H C H H There are no other possibilities. For example, H C O H H C H H H C H H and C H H HC H H C O H H H are the same as C O H H H H C H H C H H Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 6 Chapter 1 They all have the same bond connectivities and represent a single structure. Similarly, H C O H H is the same as C H H C H H H HC H H H C H H O C H H 1.21 From left to right: n-pentane, isopentane, and isopentane. 1.22 a. C H H C H H C O H C H C H H C H H H H H H b. CC Cl Cl Cl Cl Notice that the non-bonded electron pairs on oxygen and chlorine are not shown. Non-bonded electron pairs are frequently omitted from organic structures, but it is important to know that they are there. 1.23 First draw the carbon skeleton showing all bonds between carbons. CC C C C C Then add hydrogens to satisfy the valency of four at each carbon. H H CH C H CH H C H H H CH C H H H H or CH H C CH 3 2 CH3 CH2 CH3 1.24 stands for the carbon skeleton Addition of the appropriate number of hydrogens on each carbon completes the valence of 4. 1.25 ammonia N H H H formal charge on nitrogen = 5 – (2 + 3) = 0 ammonium ion N HH H H + formal charge on nitrogen = 5 – (0 + 4) = +1 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Bonding and Isomerism 7 amide ion N H H – formal charge on nitrogen = 5 – (4 + 2) = –1 The formal charge on hydrogen in all three cases is zero [1 – (0 + 1) = 0]. 1.26 For the singly bonded oxygens, formal charge = 6 – (6 + 1) = –1. For the doubly bonded oxygen, formal charge = 6 – (4 + 2) = 0. For the carbon, formal charge = 4 – (0 + 4) = 0. C O OO 2– 1.27 There are 24 valence electrons to use in bonding (6 from each oxygen, 5 from the nitrogen, and one more because of the negative charge). To arrange the atoms with 8 valence electrons around each atom, we must have one nitrogen–oxygen double bond: O N OO – O N OO O N OO The formal charge on nitrogen is 5 – (0 + 4) = +1. The formal charge on singly bonded oxygen is 6 – (6 + 1) = –1. The formal charge on doubly bonded oxygen is 6 – (4 + 2) = 0. The net charge of the ion is –1 because each resonance structure has one positively charged nitrogen atom and two negatively charged oxygen atoms. In the resonance hybrid, the formal charge on the nitrogen is +1; on the oxygens, the charge is –2/3 at each oxygen, because each oxygen has a –1 charge in two of the three structures and a zero charge in the third structure. 1.28 There are 16 valence electrons (five from each N plus one for the negative charge). The formal charges on each nitrogen are shown below the structures. N NN NN –1 +1 –1 N 0 +1 –2 1.29 In tetrahedral methane, the H–C–H bond angle is 109.5°. In “planar” methane, this angle would be 90o and bonding electrons would be closer together. Thus, repulsion between electrons in different bonds would be greater in “planar” methane than in tetrahedral methane. Consequently, “planar” methane would be less stable than tetrahedral methane. 1.30 a. C=O, ketone; C=C, alkene; O–H, alcohol b. arene; C(=O)NH, amide; C–S–C, thioether, C(=O)O–H, carboxylic acid c. C=O, ketone d. C=C, alkene Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 8 Chapter 1 ADDITIONAL PROBLEMS 1.31 The number of valence electrons is the same as the number of the group to which the element belongs in the periodic table (Table 1.3). a. Cl b. c. S d. e. O f. K 1.32 a. covalent b. covalent c. covalent d. covalent e. ionic f. ionic g. ionic h. covalent 1.33 The bonds in sodium chloride are ionic; Cl is present as chloride ion (Cl–); Cl– reacts with Ag+ to give AgCl, a white precipitate. The C–Cl bonds in CCl4 are covalent, no Cl– is present to react with Ag+. 1.34 Valence Electrons Common Valence a. N 5 3 b. C 4 4 c. F 7 1 d. O 6 2 e. P 5 3 f. S 6 2 Note that the sum of the number of valence electrons and the common valence is 8 in each case. (The only exception is H, where it is 2, the number of electrons in the completed first shell.) 1.35 a. b. c. H C H H C H H C H H H d. e. H C H H C H H F f. 1.36 a. Fluorine is more electronegative than boron. b. H C F H H δ+ δ– Fluorine is more electronegative than carbon. c. O C O δ+ δ– δ– The C=O bond is polar, and the oxygen is more electronegative than carbon. d. Cl Cl Since the bond is between identical atoms, it is pure covalent (nonpolar). Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Bonding and Isomerism 9 e. F S F F F δ+ δ– F F δ– δ– δ– δ– δ– Fluorine is more electronegative than sulfur. Indeed, it is the most electronegative element. Note that the S has 12 electrons around it. Elements below the first full row sometimes have more than 8 valence electrons around them. f. H C H H H Carbon and hydrogen have nearly identical electronegativities, and the bonds are essentially nonpolar. g. O S O δ+ δ– δ– Oxygen is more electronegative than sulfur. h. H C O H H δ– δ+ H δ+ Oxygen is more electronegative than carbon, or hydrogen, so the C–O and O–H bonds are polar covalent. 1.37 The O–H bond is polar (there is a big difference in electronegativity between oxygen and hydrogen), with the hydrogen δ+. The C–H bonds in acetic acid are not polar (there is little electronegativity difference between carbon and hydrogen). The negatively charged oxygen of the carbonate deprotonates the acetic acid. 1.38 a. C3H8 The only possible structure is CH3CH2CH3. b. C3H7F The fluorine can replace a hydrogen on an end carbon or on the middle carbon in the answer to a: CH3CH2CH2F or CH3CH(F)CH3 c. C2H2Br2 The sum of the hydrogens and bromines is 4, not 6. Therefore there must be a carbon–carbon double bond: Br2C=CH2 or BrCH=CHBr. No carbocyclic structure is possible because there are only two carbon atoms. (In Chapter 3, Sec. 3.5, we will see that the relative orientations of the Br in the latter structure can lead to additional isomers.) d. C3H6 There must be a double bond or, with three carbons, a ring: CH CH3CH CH2 or 2 CH2 H2C e. C4H9Cl The carbon chain may be either linear or branched. In each case there are two possible positions for the chlorine. CH3CH2CH2CH2Cl CH3CH(Cl)CH2CH3 (CH3)2CHCH2Cl (CH3)3CCl f. C3H6Cl2 Be systematic. With one chlorine on an end carbon, there are three possibilities for the second chlorine. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 10 Chapter 1 C H Cl Cl H C H H C H H C H Cl H H C H Cl C H H C H Cl H H C H H C H Cl If one chlorine is on the middle atom, the only new structure arises with the second chlorine also on the middle carbon: C H H H H C Cl Cl C H H g. C3H8S With an S–H bond, there are two possibilities: CH3CH2CH2–SH (CH3)2CH–SH There is also one possibility in a C–S–C arrangement: CH3–S–CH2CH3 h. C2H4F2 The fluorines can either be attached to the same carbon or to different carbons: CH3CHF2 or FCH2CH2F 1.39 The problem can be approached systematically. Consider first a chain of six carbons, then a chain of five carbons with a one-carbon branch, and so on. CH3 CH2 CH2 CH2 CH2 CH3 CH3 CH2 CH2 CH CH3 CH3 CH3 CH2 CH CH2 CH3 CH3 CH3 CH CH CH3 CH3 CH3 CH3 CH2 C CH3 CH3 CH3 1.40 a. b. C H H C H H C C H C H C H H C H H H H H H H H c. H C H H H C H H N H C H H C H H d. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Bonding and Isomerism 11 e. H C Cl H C O H H H f. C O H H C H H C H H H C H H C H H H 1.41 a. CH3CH CH 2C(CH3) 2 b. C C C C C H H H H H H c. (CH3)3CCH(CH3)2 d. CH3CH2CH2CHCH2CHCH2CH2CH3 CH3 CH2CH3 e. CH3CH2OCH2CH3 f. (CH3)2CHCH2CH2CH3 g. C C C C C C H H H H H H H H H H h. H C 2C H2 CH2 O H2C 1.42 a. In line formulas, each line segment represents a C–C bond, and C–H bonds are not shown (see Sec. 1.10). There are a number of acceptable line structures for CH3(CH2)4CH3, three of which are shown here. The orientation of the line segments on the page is not important, but the number of line segments connected to each point is important. or or b. Bonds between carbon atoms and non-carbon atoms are indicated by line segments that terminate with the symbol for the non-carbon atom. In this case, the non-carbon atom is an oxygen. c. Bonds between hydrogens and non-carbon atoms are shown. or OH OH d. The same rules apply for cyclic structures. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 12 Chapter 1 or 1.43 a. 7 b. C7H14O c. H C H H H C O C H H C H H C H H C H H C H H 1.44 First count the carbons, then the hydrogens, and finally the remaining atoms. a. C10H14N2 b. C5H5N5 c. C10H16 d. C9H6O2 e. C6H6 1.45 a. CN– There are 10 valence electrons (C = 4, N = 5, plus 1 more because of the negative charge). C – N The carbon has a –1 formal charge [4 – (2 + 3) = –1]. Some might say that it would be better to write cyanide as NC–. b. HONO2 There are 24 valence electrons. The nitrogen has a +1 formal charge [5 – (0 + 4) = +1], and the singly bonded oxygen has a –1 formal charge [6 – (6 + 1)] = –1. The whole molecule is neutral. O H O N O (–1) (+1) c. H3COCH3 There are no formal charges. d. NH4 + The nitrogen has a +1 formal charge; see the answer to Problem 1.25. H (+1) N H HH e. HONO First determine the total number of valence electrons; H = 1, O = 2 x 6 = 12, N = 5, for a total of 18. These must be arranged in pairs so that the hydrogen has 2 and the other atoms have 8 electrons around them. H o O N O r O ONH Using the formula for formal ch