Solutions manual to accompany organic chemistry

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The solutions manual to accompany Organic Chemistry provides fully-explained solutions to problems that accompany each chapter of the second edition of the book. The solutions manual to accompany Organic Chemistry provides fully explained solutions to all the problems that are featured in the second edition of Organic Chemistry.

FEATURES - Detailed worked solutions to all of the problems in the text - Brief explanations describing the purpose of each problem and the rationale behind the solutions - Helpful notes in the margin highlighting important principles and directing students to further information in the chemical literature Intended for students and instructors alike, the manual provides helpful comments and friendly advice to aid understanding, and is an invaluable resource wherever Organic Chemistry is used for teaching and learning.

Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide. All the 2p orbitals 2px, 2py, 2pz are of equal energy, and each is singly occupied before any is doubly occupied. The same holds for the 3p orbitals. The additional electron enters the 2pz orbital. The covalent bond in hydrogen fluoride arises by sharing the single electron of hydrogen with the unpaired electron of fluorine.

Combine H 1. Each carbon is surrounded by eight electrons. Thus, C2F4 has 36 valence electrons. The octet rule is satisfied for carbon only if the two carbons are attached by a double bond and there are two fluorines on each carbon.

The pattern of connections shown below left accounts for 12 electrons. The remaining 24 electrons are divided equally six each among the four fluorines. The complete Lewis structure is shown at right below. Three carbons contribute 12 valence electrons, three hydrogens contribute 3, and nitrogen contributes 5, for a total of 20 valence electrons.

Since the octet rule is satisfied for carbon, add the remaining two electrons as an unshared pair on nitrogen below right. From Table 1. When bonded to carbon, hydrogen and lithium bear a partial positive charge, and carbon bears a partial negative charge. Conversely, when chlorine is bonded to carbon, it bears a partial negative charge, and carbon becomes partially positive. In this group of compounds, lithium is the least electronegative element, chlorine the most electronegative.

The structure shown has 13 covalent bonds, and so all the valence electrons are accounted for. The molecule has no unshared electron pairs. The constitution at the left below shows seven covalent bonds accounting for 14 electrons. The remaining 12 electrons are divided equally between the two chlorines as unshared electron pairs.

The octet rule is satisfied for both carbon and chlorine in the structure at the right below. It, too, has 26 valence electrons, and again only chlorine has unshared pairs. There are 26 valence electrons, and 24 of them are accounted for by the covalent bonds in the structural formula. The remaining two electrons complete the octet of nitrogen as an unshared pair below right.

Either negatively charged oxygen can serve as the donor atom. The four electron pairs surround boron in a tetrahedral orientation. The [email protected] H angles are These four bonds are directed toward the corners of a tetrahedron. Thus azide ion is linear. Since the double bond in carbonate ion is treated as if it were a single unit, the three sets of electrons are arranged in a trigonal planar arrangement around carbon.

Water has a dipole moment. Methane has no dipole moment. Methyl chloride has a dipole moment. Formaldehyde has a dipole moment. Hydrogen cyanide has a dipole moment. Each [email protected] bond corresponds to overlap of a half-filled sp3 hybrid orbital of nitrogen and a 1s orbital of hydrogen. O is directly bonded to three other atoms two hydrogens and one oxygen.

It is sp2-hybridized. Ketene has two carbons in different hybridization states. One is sp2-hybridized; the other is sp-hybridized.

The carbons of the double bond are sp2-hybridized. O] are sp3-hybridized. The C? O carbon is sp2-hybridized. The carbons in acrylonitrile are hybridized as shown: sp2 H2C 1.

X Y Unshared electron pair contributes 2 electrons to electron count of X. Unshared electron pair contributes 2 electrons to electron count of Y. Triple bond contributes half of its 6 electrons, or 3 electrons each, to separate electron counts of X and Y. Nitrogen, as before, is electrically neutral. A neutral carbon has 4 valence electrons, and so carbon in this species, with an electron count of 5, has a unit negative charge.

There are two negatively charged carbon atoms in this species. Here again is a species with a neutral nitrogen atom. Oxygen, with an electron count of 5, has 1 less electron in its valence shell than a neutral oxygen atom. Carbon monoxide is a neutral molecule. Atom X has an electron count of 4, corresponding to half of the 8 shared electrons in its four covalent bonds. Each atom Y has an electron count of 6; 4 unshared electrons plus half of the 4 electrons in the double bond of each Y to X.

As in part a , each oxygen is electrically neutral. Sulfur contributes 6 valence electrons and phosphorus 5 to the anions. Each oxygen contributes 6 electrons. The double negative charge in sulfate contributes 2 more, and the triple negative charge in phosphate contributes 3 more.

Oxygen has an electron count of 5. The species as a whole has a unit positive charge. An electron count of 5 is one more than that for a neutral carbon atom. This species has 1 less electron than that of part b. None of the atoms bears a formal charge. The species is neutral. Its only electrons are those in its three covalent bonds to hydrogen, and so its electron count is 3. This corresponds to 1 less electron than in a neutral carbon atom, giving it a unit positive charge.

Its formal charge is 0, and the species is neutral. Two unshared electrons contribute 2 to the electron count of carbon. H C H Half of the 4 electrons in the two covalent bonds contribute 2 to the electron count of carbon.

Hydrogen and chlorine each can form only one bond, and so the only stable structure must have a carbon—carbon bond. Of the 20 valence electrons, 14 are present in the seven covalent bonds and 6 reside in the three unshared electron pairs of chlorine.

There are 18 valence electrons in C2H3Cl, and the framework of five single bonds accounts for only 10 electrons. Six of the remaining 8 are used to complete the octet of chlorine as three unshared pairs, and the last 2 are used to form a carbon—carbon double bond. The problem states that all three fluorines are bonded to the same carbon, and so one of the carbons is present as a CF3 group. The other carbon must be present as a CHBrCl group. Connect these groups together to give the structure of halothane.

Since each carbon bears one chlorine, two ClCF2 groups must be bonded together. Place unshared electron pairs on nitrogen and oxygen so that nitrogen has an electron count of 5 and oxygen has an electron count of 6. These electron counts satisfy the octet rule when nitrogen has three bonds and oxygen two. They differ only in the arrangement of their electrons. They are therefore resonance forms of a single compound. All resonance forms of a particular species must have the same net charge.

In this case, the net charge on A, B, and C is 0. Both A and B have the same number of covalent bonds, but the negative charge is on a more electronegative atom in B nitrogen than it is in A carbon. Structure B is more stable. Structure B is more stable than structure C. Structure B has one more covalent bond, all of its atoms have octets of electrons, and it has a lesser degree of charge separation than C. The carbon in structure C does not have an octet of electrons.

Structures A and B contain a positively charged nitrogen. None of the structures contain a positively charged oxygen. Structure A contains a negatively charged carbon.

None of the structures contain a negatively charged nitrogen. Structures B and D contain a negatively charged oxygen. All the structures are electrically neutral. Structure B is the most stable. All the atoms except hydrogen have octets of electrons, and the negative charge resides on the most electronegative element oxygen.

Structure C is the least stable. Nitrogen has five bonds 10 electrons , which violates the octet rule. Structure C is incorrect. CH3 1.

Notice that the location of formal charge has changed, but the net charge on the species remains the same. Sharing the lone pair gives an additional covalent bond and avoids the separation of opposite charges. Sulfur dioxide, therefore, has 18 valence electrons. The resulting Lewis structure has 10 valence electrons around sulfur. It is a valid Lewis structure because sulfur can expand its valence shell beyond 8 electrons by using its 3d orbitals.

In each of the parts to this problem, the more electronegative atom is partially negative and the less electronegative atom is partially positive. Electronegativities of the elements are given in Table 1. H b d O Cl H Chlorine is more electronegative than iodine. I Oxygen is more electronegative than hydrogen. I The direction of a bond dipole is governed by the electronegativity of the atoms involved. Fluorine therefore attracts electrons away from chlorine in FCl, and chlorine attracts electrons away from iodine in ICl.

Hydrogen chloride has a polarized bond but is a covalent compound. Sodium chloride has a larger dipole moment. The measured values are as shown. Its individual [email protected] bond dipoles cancel. It has no dipole moment. The carbon—hydrogen bond in CHCl3 reinforces it. CHCl3 therefore has a larger dipole moment.

The atoms of this species are in a trigonal pyramidal arrangement. Remember: A neutral carbon has four bonds, a neutral nitrogen has three bonds plus one unshared electron pair, and a neutral oxygen has two bonds plus two unshared electron pairs. Halogen substituents have one bond and three unshared electron pairs. Two of these compounds, b and c , have the same molecular formula and are isomers of each other.

Compounds a and d in Problem 1. OH a b d Carbon is sp2-hybridized when it is directly bonded to three other atoms. Compounds f , g , and j in Problem 1. If the half-filled 2px, 2py, and 2pz orbitals are involved in bonding to H, then the unshared pair would correspond to the two electrons in the 2s orbital. The following orbital is a bonding orbital.

It involves overlap of an s orbital with the lobe of a p orbital of the same sign. Both the positive lobe and the negative lobe of the p orbital overlap with the spherically symmetrical s orbital. The bonding overlap between the s orbital and one lobe of the p orbital is exactly canceled by an antibonding interaction between the s orbital and the lobe of opposite sign.

You should use Learning By Modeling for these exercises. Write the electronic configuration for each of the following: a Phosphorus b Sulfide ion in Na 2S A NH2 Write a second Lewis structure that satisfies the octet rule for each of the species in Problem A-2, and determine the formal charge of each atom. Which of the Lewis structures for each species in this and Problem A-2 is more stable?

Write a correct Lewis structure for each of the following. Be sure to show explicitly any unshared pairs of electrons. What is the molecular formula of each of the structures shown? Clearly draw any unshared electron pairs that are present. Which compound in Problem A-5 has a Only sp3-hybridized carbons b Only sp2-hybridized carbons c A single sp2-hybridized carbon atom A Account for the fact that all three sulfur—oxygen bonds in SO3 are the same by drawing the appropriate Lewis structure s.

The cyanate ion contains 16 valence electrons, and its three atoms are arranged in the order OCN. Write the most stable Lewis structure for this species, and assign a formal charge to each atom. What is the net charge of the ion? Free Global Shipping. Weininger and Frank R. This book provides the complete answers to all the problems in the textbook and also contains several study features to help broaden and strengthen the knowledge of the material presented in each chapter.

These features are applied in the organization of the manual, including Study Hints, New Mechanisms, Reactions, and Answers to Problems. Give the common name and the systematic name for each. Classify each of the following reactions according to whether the organic substrate is oxidized, reduced, or neither. How many methyl groups are there in this compound?

How many isopropyl groups? Give the IUPAC name for each of the following alkyl groups, and classify each one as primary, secondary, or tertiary. Write a balanced chemical equation for the complete combustion of 2,3-dimethylpentane.

Write structural formulas, and give the names of all the constitutional isomers of C5H10 that contain a ring. Each of the following names is incorrect. Give the correct name for each compound.

Which C8H18 isomer a Has the highest boiling point? The compound shown is an example of the broad class of organic compounds known as steroids. What functional groups does the molecule contain? Given the following heats of combustion in kilojoules per mole for the homologous series of unbranched alkanes: hexane , heptane , octane , nonane , estimate the heat of combustion in kilojoules per mole for pentadecane. Choose the response that best describes the following compounds: 1 a b c d 2 3 4 1, 3, and 4 represent the same compound.

All the structures represent the same compound. Following are the structures of four isomers of hexane. Which of the names given correctly identifies a fifth isomer? Which of the following structures is a 3-methylbutyl group? Which of the following substances is not an isomer of 3-ethylmethylpentane?

Which of the following compounds is not a constitutional isomer of the others? Which of the following reactions requires an oxidizing agent? The energy diagram resembles that of ethane in that it is a symmetrical one. This strain is, however, less than the van der Waals strain between the methyl groups of butane, which makes the activation energy for bond rotation less for propane than for butane.

If A is axial as specified in the problem, X must therefore be equatorial. X A X and A are gauche. The most stable conformation of 1-tert-butylmethylcyclohexane has an axial methyl group and an equatorial tert-butyl group. The less stable isomer has the higher heat of combustion. Ethylcyclopropane has more angle strain and is less stable has higher potential energy than methylcyclobutane.

Rewrite the structures as chair conformations to see which substituents are axial and which are equatorial. It is more stable than trans1,3,5-trimethylcyclohexane shown in the following , which has one axial methyl group in its most stable conformation. Draw a chair conformation of cyclohexane, add an equatorial tert-butyl group, and then add the remaining substituent so as to give the required cis or trans relationship to the tert-butyl group.

In cistert-butyl3-methylcyclohexane the C-3 methyl group is equatorial. Compare the molecular formulas of the compounds given to the molecular formula of spiropentane. Other pairs of bond cleavages are possible and lead to the same conclusion. The name tells us that there is a fivecarbon bridge and a two-carbon bridge. The 0 in the name bicyclo[5. The structure can be written in a form that shows the actual shape of the molecule or one that simply emphasizes its constitution.

One-carbon bridge Three-carbon bridge One-carbon bridge d Bicyclo[3. Three-carbon bridge Three-carbon bridge 3. The three bonds are arranged in a trigonal pyramidal manner around each nitrogen in hydrazine H2NNH2. Conformation c is the least stable; all its bonds are eclipsed. As shown at left, C-2 bears three methyl groups, and C-3 bears two hydrogens and a methyl group. The most stable conformation is the staggered one shown at right. All other staggered conformations are equivalent to this one.

Sight along this bond. In one left , the methyl group at C-3 is gauche to both of the C-2 methyls. In the other right , the methyl group at C-3 is gauche to one of the C-2 methyls and anti to the other.

The hydrogens at C-2 and C-3 may be gauche to one another left , or they may be anti right. CH3 and one anti CH3. CH3 relationship is more stable than the one with two gauche CH3. CH3 relationships. The more stable conformation has less van der Waals strain. All the eclipsed conformations are equivalent and represent potential energy maxima. The potential energy diagram of 2-methylbutane more closely resembles that of butane than that of propane in that the three staggered forms are not all of the same energy.

Similarly, not all of the eclipsed forms are of equal energy. The most stable conformation has the bonds of the methyl group and its attached carbon in a staggered relationship. The methyl group in structure B has its bonds and those of its attached ring carbon in a staggered relationship. The methyl groups are rather close together in A, resulting in van der Waals strain between them.

In B, the methyl groups are farther apart. Van der Waals strain between cis methyl groups. CH3 Methyl groups remain cis, but are far apart. In each one the carbon chain is unbranched. The Newman projections represent different staggered conformations of the same molecule: in one the hydrogens are anti to each other, whereas in the other they are gauche.

The compounds differ in the order in which the atoms are connected. They are constitutional isomers. Although the compounds have different stereochemistry one is cis, the other trans , they are not stereoisomers. Stereoisomers must have the same constitution. In the structure on the left, the methyl is axial and the ethyl equatorial. The orientations are opposite to these in the structure on the right.

The two structures are ring-flipped forms of each other—different conformations of the same compound. The methyl and the ethyl groups are cis in the first structure but trans in the second.

The two compounds are stereoisomers; they have the same constitution but differ in the arrangement of their atoms in space. Remember, chair—chair interconversion converts all the equatorial bonds to axial and vice versa. Here the ethyl group is equatorial in both structures. The two structures have the same constitution but differ in the arrangement of their atoms in space; they are stereoisomers. They are not different conformations of the same compound, because they are not related by rotation about CGC bonds.

In the first structure as shown here the methyl group is trans to the darkened bonds, whereas in the second it is cis to these bonds. CH3 H Methyl is trans to these bonds. H CH3 Methyl is cis to these bonds. Thus, cyclopropane, even though it is more strained than cyclobutane or cyclopentane, has the lowest heat of combustion.

All these compounds have the molecular formula C7H They are isomers, and so the one with the most strain will have the highest heat of combustion. Their heats of combustion decrease with decreasing number of carbons, and comparisons of relative stability cannot be made. The three-membered ring in bicyclo[5. The third hydrocarbon, bicyclo[4. H Bicyclo[4. The substituents at C-3 are two hydrogens and a tert-butyl group.

The substituents at C-4 are the same as those at C The most stable conformation has the large tert-butyl groups anti to each other. H H C CH3 3 H H C CH3 3 Anti conformation of 2,2,5,5-tetramethylhexane b The zigzag conformation of 2,2,5,5-tetramethylhexane is an alternative way of expressing the same conformation implied in the Newman projection of part a.

Draw a chair conformation of cyclohexane, and place an isopropyl group in an equatorial position. H 1 CH CH3 2 3 Notice that the equatorial isopropyl group is down on the carbon atom to which it is attached. Add a methyl group to C-3 so that it is also down. One substituent is up and the other is down in the most stable conformation of trans-1isopropylmethylcyclohexane.

Begin as in part c by placing an isopropyl group in an equatorial orientation on a chair conformation of cyclohexane. To be cis to each other, one substituent must be axial and the other equatorial when they are located at positions 1 and 4 on a cyclohexane ring.

Next, add methyl groups to C-3 and C-4 so that they are cis to each other. There are two different ways that this can be accomplished: either the C-3 and C-4 methyl groups are both up or they are both down.

Compare them by examining their conformations for sources of strain, particularly van der Waals strain arising from groups located too close together in space. Both groups are equatorial in the cis stereoisomer of 1-isopropylmethylcyclohexane; cis is more stable than trans in 1,3-disubstituted cyclohexanes.

H 5 CH CH3 2 H CH3 CH CH3 2 1 3 CH3 cisIsopropylmethylcyclohexane more stable stereoisomer; both groups are equatorial c The more stable stereoisomer of 1,4-disubstituted cyclohexanes is the trans; both alkyl groups are equatorial in transisopropylmethylcyclohexane. All its methyl groups are equatorial in its most stable conformation. The most stable conformation of the second stereoisomer has one axial and two equatorial methyl groups.

All its methyl groups are equatorial, but one of the methyl groups is axial in the most stable conformation of the second stereoisomer. The first one, however, has a close contact between its axial methyl group and both methylene groups of the two-carbon bridge.

The second stereoisomer has repulsions with only one axial methylene group; it is more stable. It exists in a conformation with only one axial methyl group, while the trans stereoisomer has two axial methyl groups in close contact with each other. The trans stereoisomer is destabilized by van der Waals strain. Structure B has more van der Waals strain than A because two pairs of hydrogens shown here approach each other at distances that are rather close.

An oxygen atom is present in the six-membered ring, and we are told in the problem that the ring exists in a chair conformation. Remember, ring flipping transforms all axial substituents to equatorial ones and vice versa.

The two structures differ with respect to only one substituent; they are stereoisomers of each other. Parts a through e are concerned with positions 1, 4, 7, 11, and 12 in that order. The following diagram shows the orientation of axial and equatorial bonds at each of those positions. At C-4 the bond that is cis to the methyl groups is axial up. At C the bond that is trans to the methyl groups is equatorial down.

At C the bond that is cis to the methyl groups is equatorial up. Analyze this problem in exactly the same way as the preceding one by locating the axial and equatorial bonds at each position.

It will be seen that the only differences are those at C-1 and C CH3 a e 11 Both methyl groups are up. At C-4 the bond that is cis to the methyl groups is equatorial up. At C-7 the bond that is trans to the methyl groups is axial down. Statement 1 is false. Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Manual.

Sight along the C-1, C-2 bond the chlorine is attached to C Write the structure of the most stable conformation of the less stable stereoisomer of 1-tertbutylmethylcyclohexane. Draw the most stable chair conformation of this substance. Consider compounds A, B, C, and D. Which two are stereoisomers of one another? Which one has the highest heat of combustion? Which one has the stereochemical descriptor trans in its name? Draw clear depictions of two nonequivalent chair conformations of cisisopropyl4-methylcyclohexane, and indicate which is more stable.

Which has the lower heat of combustion, cisethylmethylcyclohexane or cisethyl4-methylcyclohexane? The hydrocarbon shown is called twistane. Classify twistane as monocyclic, bicyclic, etc. What is the molecular formula of twistane? Sketch an approximate potential energy diagram similar to those shown in the text Figures 3.

Does the form of the potential energy curve more closely resemble that of ethane or that of butane? Draw the structure of the sulfur-containing heterocyclic compound that has a structure analogous to that of tetrahydrofuran. Which of the listed terms best describes the relationship between the methyl groups in the chair conformation of the substance shown?

CH3 CH3 a b B Eclipsed Trans c d Anti Gauche Rank the following substances in order of decreasing heat of combustion largest B smallest. Which of the following statements best describes the most stable conformation of trans1, 3-dimethylcyclohexane? Compare the stability of the following two compounds: A: cisEthylmethylcyclohexane B: transEthylmethylcyclohexane a A is more stable. What, if anything, can be said about the magnitude of the equilibrium constant K for the following process?

What is the relationship between the two structures shown?