CHAPTER 3 Chemical Reactivity and Organic Reactions 33 > Figure 3.1 (d)A singlet carbene has three sp2HO's:two form two o bonds and the third holds the unshared pair of elec- (e)Atriple ized p AO's ha one Problem 3.3 Write formulas for the species resulting from the(a)homolytic cleavage and(b)heterolytic cleavage of the C- -C bond in ethane.CH and classify these species. aH,GCH→C..cH 3.3 Types of Organic Reactions 1.Displacement (substitution).An atom or group of atoms in a molecule or ion is replaced by another smald a single molecule.Addition frequently occurs at a double 3.Elimination.This is the a molecule.If th other than adjace produces a carbene. 4.Rearrangement,Bonds in the molecule are scrambled.converting it to its isomer 5.Oxidation -reduction (redox).These reactions involve transfer of electrons or change in oxidation num ber.A decrease in the number of H atoms bonded to Cand an increase in the number of bonds to other atoms such as C,O,N,Cl,Br,F,and S signals oxidation
(d) A singlet carbene has three sp2 HO’s; two form two σ bonds and the third holds the unshared pair of electrons. It also has an empty p AO perpendicular to the plane of the sp2 HO’s. See Fig. 3.1(d). (e) A triplet carbene uses two sp HO’s to form two linear σ bonds. Each of the two unhybridized p AO’s has one electron. See Fig. 3.1(e). Problem 3.3 Write formulas for the species resulting from the (a) homolytic cleavage and (b) heterolytic cleavage of the C⎯C bond in ethane, C2 H6 , and classify these species. CHAPTER 3 Chemical Reactivity and Organic Reactions 33 Figure 3.1 ( ) a H C : CH H C + CH ( 3 33 3 Ethane Methyl radicals → ⋅⋅ b) H C : CH H C 33 3 + :CH + – 3 Ethane Carbocation Car → banion 3.3 Types of Organic Reactions 1. Displacement (substitution). An atom or group of atoms in a molecule or ion is replaced by another atom or group. 2. Addition. Two molecules combine to yield a single molecule. Addition frequently occurs at a double or triple bond and sometimes at small-size rings. 3. Elimination. This is the reverse of addition. Two atoms or groups are removed from a molecule. If the atoms or groups are taken from adjacent atoms (β-elimination), a multiple bond is formed; if they are taken from other than adjacent atoms, a ring results. Removal of two atoms or groups from the same atom (α-elimination) produces a carbene. 4. Rearrangement. Bonds in the molecule are scrambled, converting it to its isomer. 5. Oxidation-reduction (redox). These reactions involve transfer of electrons or change in oxidation number. A decrease in the number of H atoms bonded to C and an increase in the number of bonds to other atoms such as C, O, N, Cl, Br, F, and S signals oxidation
34 CHAPTER 3 Chemical Reactivity and Organic Reactions Problem 3.4 The following represents the steps in the mechanism for chlorination of methane: Inititation Step ():::+energy→：d.+.a: Chlorine radicals H G)H :→H0+H Propagation Step Methyl radical (H H ·+:Gg→H:CG:+C H The propagation steps constitute the overall reaction.(a)Write the equation for the overall reaction.(b)What are 尚ne r ci p the inter ⑧usn3c8 e (c)Each step is homolytic.In Steps I and 3.Cl cleaves:in Step2,CHcleaves. (d)Step(3)involves the displacement of oneof Clbya-CH group.In Step:displaces aCH group from an H. (e)None. (f)H.C.+CH,-H.CCH,(ethane) Problem 3.5 Identify each of the following as(1)carbocations,(2)carbanions,(3)radicals,or(4)carbenes: (a)(CH,),C:(d)(CH)C: (8)C.H.CHCH, (b)(CH),C.(e)(CH)CH,CH,(h)CH,CH (e)(CH,),C*(f)CH,CH=CH ①)(c.fn.(2)(d).(3)b.(e.(g.(④(a.(h stitution.addition,elimination,rearrangement,or redox reactions (@)CH,=CH,+Br,-CH,BrCH,Br (b)C,H,OH HCI-C,H.CI H,O (e)CH,CHCICHCICH +Zn >CH,CH-CHCH,+ZnCl (d)NH(CNO H2NCNH2 (e)CH,CH CH,CH→(CH,CH H2 (H.+Bra BrCH:CH.CHaBr (g)3CH,CH0+2Mn0+OH△，3CH,C00+2Mm0,+2H,0 (△means heat.） /HCCI,,+OH→：CC,+H,0+CI CH2 (i)BrCHzCH2CH2Br+Zn -H2C-CH2+ZnBr2
Problem 3.4 The following represents the steps in the mechanism for chlorination of methane: 34 CHAPTER 3 Chemical Reactivity and Organic Reactions The propagation steps constitute the overall reaction. (a) Write the equation for the overall reaction. (b) What are the intermediates in the overall reaction? (c) Which reactions are homolytic? (d) Which is a displacement reaction? (e) In which reaction is addition taking place? (f) The collision of which species would lead to side products? (a) Add Steps 2 and 3: CH4 Cl2 CH3Cl HCl. (b) The intermediates formed and then consumed are the H3C· and radicals. (c) Each step is homolytic. In Steps 1 and 3, Cl2 cleaves; in Step 2, CH4 cleaves. (d) Step (3) involves the displacement of one of Cl2 by a ·CH3 group. In Step 2, displaces a ·CH3 group from an H. (e) None. (f) H3 C· ·CH3 H3 CCH3 (ethane) Problem 3.5 Identify each of the following as (1) carbocations, (2) carbanions, (3) radicals, or (4) carbenes: . . . . . Cl: . . . . . Cl: () ( ) : ( )( ) ( ) () ( ad g b CH C CH C: C H CHCH CH 3 2 3 3 6 5 3 3 − ) ()( ) ( ) () ( ) ( 33 3 3 3 C CH CH CH CH CH CH C 2 2 + ⋅ e h c f ) CH CH CH 3 (1) (c), (f). (2) (d). (3) (b), (e), (g). (4) (a), (h). Problem 3.6 Classify the following as substitution, addition, elimination, rearrangement, or redox reactions. (A reaction may have more than one designation.) (a) CH2 CH2 Br2 CH2BrCH2Br (b) C2 H5 OH HCl C2 H5 Cl H2 O (c) CH3CHClCHClCH3 Zn CH3 CHCHCH3 ZnCl2 (d) (e) CH3CH2 CH2CH3 (CH3)3CH (f) (i) ( ) CH CHO g 3 22 3 4 + 2MnO + OH 3CH COO 3 2 MnO H − − Δ − + + 2 2 2 O means heat.) ( ) HCCl + OH CCl H O Cl 3 ( : Δ h − − → ++
CHAPTER 3 Chemical Reactivity and Organic Reactions 35 oxidation number (ON)for C has change changed from 7-7 =0 to 7- 8=-1. (b)Substitution of a Cl for an OH. (c)Eli ton an (d)Rear ompound is re ent (iso (e)Rearrangement(isomerization). (f)Addition and redox.The Br's add to twoCatoms of the ring.These C's are oxidized,and the Br's are reduced. )Redox.CH,CHO s oxid山 ed and MnO is reduced. edox The tu from the n (i) s.giving a ring [see(c)l. 3.4 Electrophilic and Nucleophilic Reagents Reactions generally occur at the reactive sites of molecules and ions.These sites fall mainly into two categori e category h a high density b ise the (a)nas an un or (b)is th uch sites are called nucleophiles or electron-donor The nd cate ng more electronsor(b)is the+end of a polar bond.These electron-deficient sites are electrophilic,and the species possessing such sites are called electrophiles or electron-acceptors.Many reactions occur by coordi- nate covalent bond formation between a nucleophilic and an electrophilic site. Nu:+E-Nu:E Problem 3.7 Classify the following species as being (1)nucleophiles or (2)electrophiles.and give the rea on for you r cla sification:(a)HO ))())N()HC: (a carbanion).(i)SiF(Ag'(HC(a carbocation)(H.C:(a carbene).(m) (1)(a).(b).(e).(g).(h).and (m).They all have unshared pairs of electrons.All anions are potential nucle (2)(d)and (f are molecules whose central atoms (Band Al)have only six electrons rather than the more desirable octet;they are electron-deficient.(c).()and (k)have positive charges and therefore are elec tron-defici t.Mo t cations ar e potentia electrophiles.The Si in (i)can acquire more than eight elec trons by utilizng,for example SiF(an electrophile)+2:F: Although the Cin(has an unshared pair of electrons.()is electrophilic because the Chas only six electrons Problem 3.8 Why is the reaction CH,Br+OH- CH,OH Bra nucelophilic displacement? The:H-has unshared electrons and is a nucleophile.Because of the polar nature of the C-Br bond: r Cacts as an electrophilic site.The displacement of Brby OH-is initiated by the nucleophile HO Problem3.9 Indicate whether reactant(1)or(2)is the nucleophile orelectrophile in the following reactions (a)H,C=CH,(1)+Br,(2)-BrCH,-CH,Br
(a) Addition and redox. In this reaction, the two Br’s add to the two double-bonded C atoms (1,2-addition). The oxidation number (ON) for C has changed from 4 2(2) 2 −2 to 4 2(2) 1 −1; (ON) for Br has changed from 7 7 0 to 7 8 1. (b) Substitution of a Cl for an OH. (c) Elimination and redox. Zn removes two Cl atoms from adjoining C atoms to form a double bond and ZnCl2 (a β-elimination). The organic compound is reduced and Zn is oxidized. (d) Rearrangement (isomerization). (e) Rearrangement (isomerization). (f) Addition and redox. The Br’s add to two C atoms of the ring. These C’s are oxidized, and the Br’s are reduced. (g) Redox. CH3CHO is oxidized and MnO4 − is reduced. (h) Elimination. An H and Cl− are removed from the same carbon (α-elimination). (i) Elimination and redox. The two Br’s are removed from nonadjacent C’s, giving a ring [see (c)]. 3.4 Electrophilic and Nucleophilic Reagents Reactions generally occur at the reactive sites of molecules and ions. These sites fall mainly into two categories. One category has a high electron density because the site (a) has an unshared pair of electrons or (b) is the δ end of a polar bond or (c) has CC π electrons. Such electron-rich sites are nucleophilic and the species possessing such sites are called nucleophiles or electron-donors. The second category (a) is capable of acquiring more electrons or (b) is the δ end of a polar bond. These electron-deficient sites are electrophilic, and the species possessing such sites are called electrophiles or electron-acceptors. Many reactions occur by coordinate covalent bond formation between a nucleophilic and an electrophilic site. Nu: E Nu:E Problem 3.7 Classify the following species as being (1) nucleophiles or (2) electrophiles, and give the reason for your classification: (a carbanion), (i) SiF4, (j) Ag, (k) H3 C (a carbocation), (l) H2C: (a carbene), (m) . (1) (a), (b), (e), (g), (h), and (m). They all have unshared pairs of electrons. All anions are potential nucleophiles. (2) (d) and (f) are molecules whose central atoms (B and Al) have only six electrons rather than the more desirable octet; they are electron-deficient. (c), (j), and (k) have positive charges and therefore are electron-deficient. Most cations are potential electrophiles. The Si in (i) can acquire more than eight electrons by utilizing its d orbitals, for example: CHAPTER 3 Chemical Reactivity and Organic Reactions 35 Although the C in (l) has an unshared pair of electrons, (l) is electrophilic because the C has only six electrons. Problem 3.8 Why is the reaction CH3 Br OH− CH3OH Br− a nucelophilic displacement? The :O.. .. H has unshared electrons and is a nucleophile. Because of the polar nature of the C ⎯Br bond: C acts as an electrophilic site. The displacement of Br− by OH− is initiated by the nucleophile HO: .. . . . Problem 3.9 Indicate whether reactant (1) or (2) is the nucleophile or electrophile in the following reactions: (a) H2CCH2 (1) Br2 (2) BrCH2⎯CH2Br
36 CHAPTER3 Chemical Reactivity and Organic Reactions (b)CH.NH(1)+CHCOO-(2)-CH.NH.CHCOOH (e)CHC-9:()+AIC3(2) →CHC+AIC (dCH3CH=0(I)+:S0H(2)→CH3一CHSO3H (a)b)(c)(d) Nucleophile (1)(2)(1)(2) Electrophile (2)(1)(2)(1) 3.5 Thermodynamics The thermodynamics and therate whether the reaction proceeds.The thermodynamics of a system is described in terms of several imporant functions: (1)AE.the change in energy.equalsthe heat transferred to or from a system at constant volume AF= organic reactions are per than th medalamooTd nd ts,Hg △H=HP-HR ()③As is the c the greater the order,the smaller is S.For a reaction: of randomness.The more the randomness,the greater is (4)AG=G-G is the change in free energy.At constant temperature: △G-△H-T△S(T-Absolute temperature) For a reaction to be spontaneous,AG must be negative Problem 3.10 State whether the following reactions have a positive or negative AS,and give a reason for your choice. (@)H+H,C=CH2→H,CCH3 oH,CH△，H,C-aH=CH, (e)CH,COO-(aq)+Ho+(aq)-CH,COOH+Ho
(b) CH3 NH 3 (1) CH3 COO− (2) CH3 NH2 CH3 COOH 36 CHAPTER 3 Chemical Reactivity and Organic Reactions (a) (b) (c) (d) Nucleophile (1) (2) (1) (2) Electrophile (2) (1) (2) (1) 3.5 Thermodynamics The thermodynamics and the rate of a reaction determine whether the reaction proceeds. The thermodynamics of a system is described in terms of several important functions: (1) ΔE, the change in energy, equals qν, the heat transferred to or from a system at constant volume: ΔE qν . (2) ΔH, the change in enthalpy, equals qp , the heat transferred to or from a system at constant pressure: ΔH qp . Since most organic reactions are performed at atmospheric pressure in open vessels, ΔH is used more often than is ΔE. For reactions involving only liquids or solids, ΔE ΔH. ΔH of a chemical reaction is the difference in the enthalpies of the products, Hp , and the reactants, HR : ΔH HP HR If the bonds in the products are more stable than the bonds in the reactants, energy is released, and ΔH is negative. The reaction is exothermic. (3) ΔS is the change in entropy. Entropy is a measure of randomness. The more the randomness, the greater is S; the greater the order, the smaller is S. For a reaction: ΔS SP SR (4) ΔG GP GR is the change in free energy. At constant temperature: ΔG ΔH TΔS (T Absolute temperature) For a reaction to be spontaneous, ΔG must be negative. Problem 3.10 State whether the following reactions have a positive or negative ΔS, and give a reason for your choice. (a) H2 H2 CCH2 H3 CCH3 (c) CH3COO−(aq) H3O(aq) CH3COOH H2O (b) (d) (c)
CHAPTER 3 Chemical Reactivity and Organic Reactions 37 (a)Negative.Two molecules are changing into one molecule and there is more order(less randomness)in the (b)Positive.The rigid ring opens to give compounds having free rotation about the C-C single bond.There is now more randomness (S>S). (c)Positiv s form molecules. Problem 3.11 Predict the most stable state of H,O(steam,liquid,orice)in terms of(a)enthalpy,(b)entropy. and (c)free energy. (a)Gas→Liquid→Solid ud be exothermic processes and.therefore.ice has the least enthalpy.For this rea (b)SolidLiquid>Gas shows increasing randomness and therefore increasing entropy.For this reason, steam should be most stable. (c)Here the trends to lowest enthalpy and highest entropy are in opposition:neither can be used independently avored stale.Only G,v hich gives ance can be cannot be predicted until a calculation is made using the equation-TS. state,a fact whi 3.6 Bond-Dissociation Energies The bond-dissociation energy.AH.is the energy needed for the endothermic homolysis of a covalent bond A:BA.+B:AH is positive.Bond formation,the reverse of this reaction,is exothermic and the AH values more positive the Aff value,the stronger is the b ond.The e sum of all the (positive)values tor bond cleavages ps the sum of all the (negative)values for bd formations. Problem 3.12 Calculate AH/for the reaction CH+Cl CH,Cl HCI.The bond-dissociation energies in kJ/mol,are 427 for C-H.243 for Cl-Cl.339 for C-Cl.and 431 for H-CI. The values are shown under the bonds involved H,C-H+CI-CI-H,C-CI+H-CI 4+2想 339+-43=-100 formations (exothermic The reaction is exothermic.with AH=-100 kJ/mol. Problemu unshared electron pairs:(triple.double.and single bonds ond between at Compare the strengths of bonds b (a)Bonds are weaker between atoms with unshared electron pairs because of interelectron repulsion (b)Overlapping of porbitals strengthens bonds.and bond energies are greatest for triple and smallest for single bonds. 3.7 Chemical Equilibrium Every chemical reaction can proceed in either direction.dA+eB ton to a mic equ. eached nger cl ing place
(a) Negative. Two molecules are changing into one molecule and there is more order (less randomness) in the product (SP < SR). (b) Positive. The rigid ring opens to give compounds having free rotation about the C⎯C single bond. There is now more randomness (SP > SR). (c) Positive. The ions are solvated by more H2O molecules than is CH3COOH. When ions form molecules, many of these H2O molecules are set free and therefore have more randomness (SP > SR). Problem 3.11 Predict the most stable state of H2 O (steam, liquid, or ice) in terms of (a) enthalpy, (b) entropy, and (c) free energy. (a) Gas Liquid Solid are exothermic processes and, therefore, ice has the least enthalpy. For this reason, ice should be most stable. (b) Solid Liquid Gas shows increasing randomness and therefore increasing entropy. For this reason, steam should be most stable. (c) Here the trends to lowest enthalpy and highest entropy are in opposition; neither can be used independently to predict the favored state. Only G, which gives the balance between H and S, can be used. The state with lowest G or the reaction with the most negative ΔG is favored. For H2 O, this is the liquid state, a fact which cannot be predicted until a calculation is made using the equation G H TS. 3.6 Bond-Dissociation Energies The bond-dissociation energy, ΔH, is the energy needed for the endothermic homolysis of a covalent bond A:B A· ·B; ΔH is positive. Bond formation, the reverse of this reaction, is exothermic and the ΔH values are negative. The more positive the ΔH value, the stronger is the bond. The ΔH of reaction is the sum of all the (positive) ΔH values for bond cleavages plus the sum of all the (negative) ΔH values for bond formations. Problem 3.12 Calculate ΔH for the reaction CH4 Cl2 CH3 Cl HCl. The bond-dissociation energies, in kJ/mol, are 427 for C —H, 243 for Cl—Cl, 339 for C —Cl, and 431 for H—Cl. The values are shown under the bonds involved: CHAPTER 3 Chemical Reactivity and Organic Reactions 37 H C H Cl Cl H C Cl H Cl cleavage endo 3 3 427 243 —— —— ( + + + thermic) format + −+ ( )( ) 339 431 − ions exothermic ( ) = −100 The reaction is exothermic, with ΔH −100 kJ/mol. Problem 3.13 Compare the strengths of bonds between similar atoms having: (a) single bonds between atoms with and without unshared electron pairs; (b) triple, double, and single bonds. (a) Bonds are weaker between atoms with unshared electron pairs because of interelectron repulsion. (b) Overlapping of p orbitals strengthens bonds, and bond energies are greatest for triple and smallest for single bonds. 3.7 Chemical Equilibrium Every chemical reaction can proceed in either direction, dA eB f X gY, even if it goes in one direction to a microscopic extent. A state of equilibrium is reached when the concentrations of A, B, X, and Y no longer change even though the reverse and forward reactions are taking place.