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Dr. Rosenberg has been associated with the preparation of Schaum's Outline of College Chemistry since the third edition, published in LAWRENCE M. CHEMISTRY. Third Edition. David E. Goldberg, Ph.D. Professor of Chemistry. Brooklyn College. City University of New York. Schaum's Outline Series. This Schaum's Outline gives you: 1, fully solved problems; clear, concise explanations of all college chemistry concepts; and support for all the major.

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Schaum's Outline of Theory and Problems of Beginning Chemistry SCHAUM'S OUTLINE SERIES All trademarks are trademarks of their respective owners. Schaum's Outline of College Chemistry: 1, Solved Problems + 23 Videos ( 10th. Revised edition). Revised edition) PDF, remember to click the hyperlink listed. Organic Chemistry by Schaum'S OUTLINE - Ebook download as PDF File .pdf) or read book online. SCHAUM'S OUTLINE SERIES for CHEMISTRY is world-.

How many grams of each of the constituent elements are contained in one mole of a CH4, b Fe, c Ca3P2? How many atoms of each element are contained in the same amount of compound? Calculate the number of grams in a mole of each of the following common substances: a calcite, CaCO,; b quartz, Si02; c cane sugar, C12H; d gypsum, CaS What is the average mass in kilograms of a a helium atom, b a fluorine atom, c a neptunium atom?

How many moles of atoms are contained in a How many moles are represented by a A certain public water supply contained 0. How many molecules of CHC13 would be contained in a 0. Verify the law of multiple proportions for an element, X, which forms oxides having percentages of X equal to If the compound with These integers may be found by converting analytical weight composition data to the amount of each element, expressed in moles of atoms, contained in some fixed weight of the compound.

Consider a compound that analyzes Unlike static PDF Schaum's Easy Outline of Beginning Chemistry solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. You can check your reasoning as you tackle a problem using our interactive solutions viewer. Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available.

Our interactive player makes it easy to find solutions to Schaum's Easy Outline of Beginning Chemistry problems you're working on - just go to the chapter for your book.

Hit a particularly tricky question? The reaction proceeds through a bromonium ion [Problem 6. The addition, like that of Br2, is anti because H2O binds to the C from the side away from the side where the Br is positioned. The process whereby simple molecules, or monomers, are merged can continue, eventually giving high-molecular-weight molecules called polymers.

This reaction of alkenes is called chain-growth addition polymerization. The repeating unit in the polymer is called the mer. If a mixture of at least two different monomers polymerizes, a copolymer is obtained. See Steps 1 and 2 in Problem 6. This intermolecular H: Free-Radical Additions Problem 6. There are two degrees of unsaturation.

The second degree of unsaturation is a ring structure. The compound is a cycloalkene whose structure is found by writing the two terminal carbonyl groups facing each other: The 2 mol of H2 absorbed indicates two double bonds.

The third degree of unsaturation is a ring structure. Such delocalization stabilizes the allyl-type free radical. Dehydrohalogenation of RX 1. KOH a Hydrogenation 2. Dehydration of ROH Heterogeneous: There are five such constitutional isomers of C6H The double bond must be internal and is first placed between C2 and C3 to give three more sets of stereomers: The product is racemic: The double bond is reformed, and the two conformations produce a mixture of cis and trans isomers.

Addition is anti-Markovnikov. H as nucleophiles to give the two products. Inorganic reagents and solvents may also be used. Give Fischer projections and R, S designations for the products. Are the products optically active? There are two optically active diastereomers of 1,2,3-tribromobutane. H a hybride transfer from a tertiary position in an alkane. A positive simple chemical test is indicated by one or more detectable events, such as a change in color, for- mation of a precipitate, evolution of a gas, uptake of a gas, and evolution of heat.

Alkanes give none of these tests. One optically active stereoisomer is formed. Two optically active diasteriomers. One C picks off an H from the C of the other chain, to give an alkane at one chain end and an alkene group at the other chain end: These types of polymerizations also have stereochemical consequences. Problem 7.

Take each isomer of the parent hydrocarbon and replace one of each type of equivalent H by X. Classification is based on the structural features: These are geminal or gem-dichlorides. This effect is greatest for RF. Halogenation of alkanes with Cl2 or Br2 Section 4.

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Addition of HX to alkenes Section 6. HI does not undergo an anti-Markovnikov radical addition. The rate of the displacement reaction on the C of a given sub- strate depends on the nucleophilicity of the attacking base. Basicity and nucleophilicity differ as shown: The order in Problem 7.

Alternatively, the greater ease of distortion of the valence shell induced polarity makes easier the approach of the larger atom to the C atom. This property is called polarizability. The larger, more polarizable species e. The smaller, more weakly polarizable bases e. In nonpolar and weakly polar aprotic solvents, the salts of: Polar aprotic solvents solvate only the cations, leaving free, unencumbered anions.

The reactivities of all anions are enhanced, but the effect is more pronounced the smaller the anion. Hence, the order of Problem 7. Sulfonates are excellent leaving groups—much better than the halides. In order for the tail to become the head, the configuration must change; inversion occurs. Figure 7. In the SN1 TS, there is considerable positive charge on C; there is much weaker bonding between the attacking group and leaving groups with C.

Nu— or: In acid, ROH2 is first formed. The optical purities are as follows: The percentage of backside reaction is the sum of the inversion and one-half of the racemization; the percentage of frontside attack is the remaining half of the percentage of racemization. It is better able to react to give HS: SH see Prob- lem 7.

Show the transition state, indicating the partial charges. Solvolyses go by an SN1 mechanism. Relative rates of different reactants in SN1 reactions depend on the sta- bilities of intermediate carbocations.

See Problem 6.

The nitrite ion has two different nucleophilic sites: Role of the Solvent Polar solvents stabilize and lower the enthalpies of charged reactants and charged transition states. The more diffuse the charge on the species, the less effective the stabilization by the polar solvent. SH increases, for example: See Table 7. The phase-trans- fer catalyst, regenerated in the organic phase, returns to the aqueous phase, and the chain process is propagated.

Section 6.

Schaum's Outline of College Chemistry, Tenth Edition

Their features are compared and summarized in Table 7. A third mechanism, E1cb, is occasionally observed. Table 7. The observation of an isotope effect indicates that C —H bond-break- ing occurs in the rate-controlling step.

Use the wedge-sawhorse and Newman projections. TABLE 7. This gives the skeletal arrangement: The solvent is given above the arrow. Both carbocations react by two pathways: As you do this, keep asking what is needed to make what you want. Always try to use the fewest steps.

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The alco- hol is a poor choice, because it would have to be made from either of the alkenes and an extra step would be needed. Although free-radical chlorination of propane gives a mixture of isomeric propyl chlorides, the mixture can be dehydrohalogenated to the same alkene, making this partic- ular initial chlorination a useful reaction. Equilibrium is shifted to the right because NaI is soluble in acetone, while NaBr is not and precipitates.

NaI is soluble in acetone and NaCI is insoluble. This usually has little effect on SN2 reactions. Acetone has a low dielectric constant and is aprotic and favors SN2 reactions.

Its larger size also precludes SN2 reactions. The more substituted alkene is the major product because of its greater stability.

For either enantiomer, there are two conformers in which the H and Cl eliminated are anti to each other: Conformer I has a less crowded, lower-enthalpy transition state than conformer II.

If an electron-withdrawing group is adjacent to the positive C, it will tend to destabilize the carbocation. Arrows indicate withdrawn electron density. Each F has an unshared pair of electrons in a p orbital which can be shifted to —C — via p-p orbital overlap. Formation of 1-cyano—2-butene results from SN2 reaction at the terminal C. The rates are proportional to the products of the concentrations: Rate [0.

The remaining three compounds are differentiated by their reactivity with alcoholic AgNO3 solution. A less polar solvent than that in c favors E2. H2O is not basic enough to remove a proton to give elimination.

Use the following stereochemical results to suggest mechanisms for the two pathways: With no change in priority, inversion gives the S -RCl. Ether is too weakly basic to cause enough dissociation of HCl. This internal nucleophilic substi- tution SNi reaction proceeds through an ion pair and leads to retention of configuration. Problem 8. Only one of three p orbitals of C is hybridized.

The two unhybridized p orbitals pz and py are at right angles to each other and also to the axis of the sp hybrid orbitals. TABLE 8. The same situation prevails for the C—C bond in going from 4 to 6. Bonds to C therefore become shorter as the s character of the hybridized orbital used by C increases. Laboratory Methods of Preparation 1.

Dehydrohalogenation of vic-Dihalides or gem-Dihalides The vinyl alkenyl halide requires the stronger base sodamide NaNH2. In E2 eliminations, the more acidic H is removed preferably. To get this product, the less acidic H one of the CH3 group must be removed. The needed vic-dihalide is formed from propene, which is prepared from either of the alkyl halides.

Alkynes can add one or two moles of reagent but are less reactive except to H2 than alkenes.

Hydrogen 2. H2O Hydration to Carbonyl Compounds 5. Boron Hydride With dialkylacetylenes, the products of hydrolysis and oxidation are cis-alkenes and ketones, respectively. Dimerization 7. Nucleophiles Problem 8. The mechanism of electrophilic addition is similar for alkenes and alkynes. When HX adds to a triple bond, the intermediate is a carbocation having a positive charge on an sp-hybridized C atom: An addendum such as Br2 forms an intermediate bromonium-type ion: Such situations cause alkynes to be less reactive than alkenes toward Br2.

It is more stable and is formed more readily than the sp3-hybridized carbanion formed from a nucleophile and an alkene. How can this mixture be converted to pure a cishexene? Relatively pure alkene geometric isomers are prepared by stereoselective reduction of alkynes.

Acidity and Salts of 1-Alkynes [see Problem 8. Give any products and the reason for their formation. The products are the weaker acid propyne and the weaker base NH3. The precipitate shows an acetylene bond at the end of a chain with an acidic H. Conjugated diene, since it has alternating single and double bonds. Conjugated dienes are most stable, and cumulated dienes are least stable; under the proper conditions, allenes tend to rearrange to conjugated dienes.

This results in greater stability and decreased energy. Figure 8. Structure i has 11 bonds and makes a more significant contribution than the other two structures, which have only 10 bonds.

Since the contributing structures are not equivalent, the resonance energy is small. The molecular orbitals are considered to be stationary waves, and their relative ener- gies increase as the numbers of nodal points in the corresponding waves increase.

With an odd number of molecular orbitals, the middle- energy molecular orbital is nonbonding MOn. Molecular orbitals receive electrons in the order of their increas- ing energies, with no more than two of opposite spins in any given molecular orbital.

Note the simplification of showing only the signs of the upper lobes of the interacting p orbitals. The stationary waves, with any nodes, are shown superimposed on the energy levels. Indicate the relative energies of the molecular orbitals, and state if they are bonding, nonbonding, or antibonding. Note that the node of this MO n is at a C, indicated by a 0.

An MO n can be recognized if the number of bonding pairs equals the number of antibonding pairs or if there is no overlap. This cation adds the nucleophile at C2 to form the 1,2-addition product or at C4 to form the 1,4-addition product. The relative rates of formation of carbocations are: Reactions may shift from rate to thermodynamic control with increasing temperature, especially when the formation of the rate-controlled product is reversible.

Start from the allylic carbocation, the common intermediate. The reaction products are shown: The intermediate carbocation formed from the conjugated diene is allylic and is more stable than the iso- lated carbocation from the isolated diene.

It is noteworthy that although the conjugated diene is more stable, it nevertheless reacts faster. These are alternating sites. Industrial 1. Dehydration of Diols 1. Polar Additions 2. Dehydrohalogenation alc. KOH 3. Free-Radical Addition Dihalides: In Table 8. Relative reactivities are a 1, b 2, c 3, d 4, e 5. H-abstraction produces a resonance hybrid of two contributing structures having both 12C and 14C as equally reactive, free-radical sites that attack Br2. Therefore, acetylene is more easily hydrogenated and the process can be stopped at the ethylene stage.

In general, hydrogenation of alkynes can be stopped at the alkene stage. The addition of 2 mol of H2 excludes a cyclic compound. It may be either a diene or an alkyne, and the latter func- tional group is established by hydration to a carbonyl compound. The skeleton must be as established by the reduction product. Whenever the two substituents on C2 are different and the two substituents on C4 are different, the molecule lacks symmetry and is chiral.

Give the structures of compounds A through D. The precipitate with ammoniacal CuCl indicates that D is a 1-alkyne, which can only be 1-butyne. The reactions and compounds are: Although the ground-state enthalpy for the conjugated diene is lower than that of the isolated diene, the transition-state enthalpy for the conjugated system is lower by a greater amount see Fig. Ordinary ground-state O2 is a diradical, One bond could form, but the intermediate has two electrons with the same spin and a second bond cannot form.

When irradiated, O2 is excited to the singlet spin-paired state Singlet O2 reacts by a concerted mechanism to give the product. Cycloalkanes, having the general formula CnH2n, are isomeric with alkenes but, unlike alkenes, they are satu- rated compounds.

Two or more substituents are listed alphabetically and are assigned the lowest possible numbers. Polycyclic compounds have more than one ring; those with exactly two rings are bicyclic. The rings may merely be bonded to each other, as in Problem 9. Problem 9. The name is bicyclo[4.

The name is 1-cyclopropylmethylpentene. The name is 3-bromo-1,1-dimethylcyclohexane. Substituents on the largest bridge get the smallest numbers.

The name is 2,2,7,7-tetramethylbicyclo[2. Therefore, substituted cycloalkanes may be geometric isomers, as well as being enantiomers or meso compounds. Both geometric isomers form racemic mixtures. In 1-bromochlorocyclobutane, there are cis and trans isomers, but no enantiomers; C1 and C3 are not chiral, because a plane perpendicular to the ring bisects them and their four substituents.

The sequence of atoms is identical going around the ring clockwise or counterclockwise from C1 to C3. In these structural formulas, the other atoms on C1 and C3 are directly in back of those shown and are bisected by the indicated plane. They both have planes of symmetry. Thus, H unit de- creases with increasing size, which implies that H ring also decreases with increasing size. But a decreas- ing H ring means an increasing ring stability.

In short, stability increases with ring size. In terms of orbital overlap, the strongest chemical bonds are formed by the greatest overlap of atomic orbitals. For sigma bonding, maximum overlap is achieved when the orbitals meet head-to-head along the bond axis, as in Fig. Hence, the overlap must be off the bond axis to give a bent bond, as shown in Fig. Additional p character narrows the expected angle, while more s character expands the angle. Clearly, there are deviations from pure p, sp, sp3, and sp2 hybridizations.

Figure 9. Conformations of Cyclobutane and Cyclopentane Problem 9. The deviation from Actually, eclipsing strain is reduced because cyclobutane is not a rigid, flat molecule. Rather, there is an equi- librium mixture of two flexible puckered conformations that rapidly flip back and forth Fig. A trans-cyclohexene is too strained. Such a bridgehead C and the three atoms bonded to it cannot assume the flat, planar structure required of an sp2-hybridized C.

Compounds spiranes hav- ing a single C, which is a junction for two separate rings, are known for all size rings. However, the rings must be at right angles.

Three- and six-membered rings cannot be fused trans, since there is too much strain. Conformations of Cyclohexane Problem 9. Would one expect this if cyclohexane had a flat hexagonal ring? A flat hexagon would have considerable ring strain. Cyclohexane minimizes its ring strain by being puckered rather than flat. The two extreme conformations are the more stable chair and the less stable boat. In the chair form, Fig. Substituted Cyclohexanes 1.

Converting one chair conformer to the other also changes the axial bonds, shown as heavy lines in Fig. The equatorial bonds of Fig. Axial on the left; equatorial on the right. For methylcyclohexane, the conformer with the axial CH3 is less stable and has 7. This difference in energy can be analyzed in either of two ways: In Fig. The steric strain for each CH3—H 1,3-diaxial interaction is 3.

Gauche interaction. One gauche interaction is also 3. In general, a given substituent prefers the less crowded equatorial position to the more crowded axial position.

With chair conformers, the better way to determine whether substituents are cis or trans is to look at the axial rather than the equatorial groups. If one axial bond is up and the other is down, the isomer is trans; if both axial bonds are up or down , the geometric isomer is cis.

The equatorial bonds are also trans, although this is not obvious from the structure. In the cis-1,2-isomer, an H and CH3 are trans to each other Fig.

In the more stable conformer [Fig. In the trans isomer, one CH3 is axial and one equatorial Fig. Each enantiomer is ae and has only one conformer. Neither has configurational isomers. Each enantiomer has ee and aa conformational diastereomers. The cis isomer is more stable than the trans isomer by 7. Can you compare a cis- and transmethylbromocyclohexane? The cis substituents are ea.

Although CH3 is bulkier and has a greater e preference than has Br, the difference in preference is small and an appreciable number of molecules exist with the Br e and the CH3 a. At no time are there conformers with Br only in an a position. These iso- mers, therefore, cannot be used for this purpose. In practically all molecules of the cis isomer, Br is forced to be a. All molecules of the trans isomer have an e Br. When Br at C1 is cis, it has an e position; when it is trans, it has an a position.

These isomers can be used. Pertinent here is the intramolecular cyclization of polyenes an electrocyclic reaction. Intermolecular Cyclization In this method, two or occasionally more, open-chain compounds are merged into a ring. Examples are the common syntheses of cyclopropanes by the addition of carbene CH2 or substituted carbenes to alkenes Section 6. Write the equation for the reaction of: CCl2 and propene.

Ultraviolet-light-catalyzed dimerization of alkenes yields cyclobutanes in one step. A conjugated diene and an alkene form a cyclohexene. Exceptions are the cyclopropanes, whose strained rings open easily, and the cyclobutanes, whose rings open with difficulty. The larger rings are stable [see Problem 9. The two conformations of 1,3-butadiene are s-cis cisoid and s-trans transoid: Although s-trans is the more favorable conformer, reaction occurs with s-cis because this conformation has its double bonds on the same side of the single bond connecting them; hence, the stable form of cyclohexene with a cis double bond is formed.

Reaction of the s-trans conformer with ethene would give the impossibly strained trans-cyclohexene [Problem 9. As the s-cis conformer reacts, the equilibrium between the two conform- ers shifts toward the s-cis side, and in this way, all the unreactive s-trans reverts to the reactive s-cis conformer. Under these conditions, the strained three-membered ring opens.

Again, the three-membered ring opens. The strained four-membered ring opens, but a higher tem- perature is needed than in part c. The five-membered ring has no ring strain. Even the strained rings are stable toward oxidation. On the other hand, cyclopropanes are synthesized this way, yet they are the least stable cycloalkanes. Are these facts incompatible?

The relative ease of synthesis of cycloalkanes by intramolecular cyclization depends on both ring sta- bility and the probability of bringing the two ends of the chain together to form a C-to-C bond, thereby closing the ring.

This probability is greatest for the smallest rings and decreases with increasing ring size. The interplay of ring stability and this probability factor are as follows numbers represent ring sizes: Chains can also react intermolecularly to form longer chains.

This side reaction from collisions between different chains is minimized by carrying out the reac- tion in extremely dilute solutions. Intermolecular Reactions The rules state the following: Reaction occurs when the lowest unoccupied molecular orbital LUMO of one reactant overlaps with the highest occupied molecular orbital HOMO of the other reactant. Without ultraviolet light we have the situation indicated in Fig. The reaction is reversible. The Woodward-Hoffmann rule that permits the proper analysis of the stereochemistry is: The 0 is used whenever a node point is at an atom.

The allowed reaction occurs thermally as shown in Fig. The photoreaction is forbidden [Fig. In the structures that follow, dashed lines separate the isoprene units. The molecule is chiral; C1 and C6 are chiral centers. The molecule is chiral; C3 is a chiral center. The molecule is achiral. Indicate the stereo- chemistry where necessary, and account for the products. CH3Cl has only one Cl and is considerably less acidic. Trans-1,3- and cis-1,4- are ea.

An axial t-butyl group is very unstable, so that a twist-boat with a quasi- ee conformation Fig. C gives one, while D yields two, monocarboxylic acids.

B absorbs two moles of H2 and has two multiple bonds. A absorbs one mole of H2 and has a ring and a double bond; it is a cycloalkene. As a cycloalkene, A can form only a single product, a dicarbonyl compound, on ozonolysis. Since one molecule of B gives three carbonyl molecules, it must be a diene and not an alkyne. The cis isomer is D. In the mono- carboxylic acid G formed from C trans isomer , one Br is cis and the other trans with respect to COOH and there is only one isomer.

Draw their conformational structural formulas. For the trans isomer, only the rigid ee con- formation is possible structurally. As shown in Fig. Cis fusion is ea and the bonds can be twisted to reverse the a and e positions, yielding conformation enantiomers. Remember that vinyl chlorides are inert in SN2 and SN1 reactions. The uptake of H2, measured quantitatively, is 2 mol for 1 mol of the diene, but 1 mol for 1 mol of the cycloalkene.

However, it is like other cycloalkanes and differs from multi- ple-bonded compounds in not decolorizing aqueous KMnO4. On hydrogenation, A yields cyclopentane.

Organic Chemistry by Schaum'S OUTLINE

There are three disubstituted benzenes—the 1,2-, 1,3-, and 1,4-position isomers—designated as ortho, meta, and para, respectively. Problem All six C-to-C bonds have the identical length, 0. Is benzene the same as 1,3,5-cyclohexatriene? The bond lengths in 1,3,5-cyclohexatriene would alternate between 0.

The C-to-C bonds in benzene are intermediate between single and double bonds. Draw a conclusion about the relative reactivities of the two compounds. In computing the first column of Table Benzene is not 1,3,5-cyclohexatriene; in fact, the latter does not exist.

Benzene is less reactive than open-chain trienes toward all electrophilic addition reactions. Equations are written for the reactions so that their algebraic sum gives the desired reactant, products, and enthalpy.

The ease of addition of 1 mol of H2 is: When one molecule of benzene is converted to the diene, the diene is reduced all the way to cyclohexane by two more molecules of H2 before more molecules of benzene react.

If 1 mol each of benzene and H2 are reacted, the product is —13 mol of cyclohexane and —23 mol of unreacted benzene. Use these data to calculate the heat of combustion for C6H6 and the difference between this and the experimental value.

Compare the dif- ference with that from heats of hydrogenation. The contribution is calculated for each bond, and these are totaled for the molecule: Each C also has one electron in a p orbital at right angles to the plane of the ring.

They are therefore more delocalized, and this accounts for the great stability and large res- onance energy of aromatic rings. Figure These are indicated in Fig. Since benzene is cyclic, the stationary waves representing the electron clouds are cyclic and have nodal planes, shown as lines, instead of nodal points. See Problem 9. Thermal stability. In these reactions, the aromatic unsaturated ring is preserved. Resistance to oxidation by aq.

Unique nuclear magnetic resonance spectra.The MO levels are as follows: Designate products D,L and R,S. Acidity and Salts of 1-Alkynes [see Problem 8. A less polar solvent than that in c favors E2. The prefixes sec- and tert- before the name of the group indicate that the H was removed from a secondary or tertiary C, respectively. Are these facts incompatible? For sigma bonding, maximum overlap is achieved when the orbitals meet head-to-head along the bond axis, as in Fig.

If the compound with Problem 8. The molecule is achiral.

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