S N 1 and S N 2 Mechanisms

CH 2 R—CHX—CH

1. S N 1 and S N 2 Mechanisms

The two major mechanisms of nucleophilic displacement are outlined in Table 7.1.

Problem 7.10 Give the three steps for the mechanism of the S_N1 hydrolysis of 3° Rx, Me₃CBr.

Problem 7.11 Give examples of the four charge-types of S_N2 reactions, as shown in the first line of Table 7–1.

Problem 7.12 (a) Give an orbital representation for an S_N2 reaction with (S)- RCHDX and :Nu^, if in the tran-sition state, the C on which displacement occurs uses sp²hybrid orbitals. (b) How does this representation ex-plain (i) inversion and (ii) the order of reactivity 3° 2°  1°?

(a) See Fig. 7.1.

(b) (i) The reaction is initiated by the nucleophile beginning to overlap with the tail of the sp³hybrid orbital hold-ing X. In order for the tail to become the head, the configuration must change; inversion occurs. (ii) As H’s on the attacked C are replaced by R’s, the TS becomes more crowded and has a higher enthalpy. With a 3°

RX, there is a higher ΔH^‡and a lower rate.

Problem 7.13 (a) Give a representation of an S_N1 TS which assigns a role to the nucleophilic protic solvent mol-ecules (HS:) needed to solvate the ion. (b) In view of this representation, explain why (i) the reaction is first-order;

TABLE7.1

(ii) R^reacts with solvent rather than with stronger nucleophiles that may be present; (iii) catalysis by Ag^takes place; (iv) the more stable the R^, the less inversion and the more racemization occurs.

(a)

(b) (i) Although the solvent HS: appears in the TS, solvents do not appear in the rate expression. (ii) HS: is already partially bonded, via solvation, with the incipient R^. (iii) Ag^has a stronger affinity for X^than has a solvent molecule; the dissociation of X^is accelerated. (iv) The HS: molecule solvating an unstable R^is more apt to form a bond, causing inversion. When R^is stable, the TS gives an intermediate that reacts with another HS: molecule to give a symmetrically solvated cation:

which collapses to a racemic product:

The more stable is R^, the more selective it is and the more it can react with the nucleophilic anion Nu^.

Figure 7.1

Problem 7.14 Give differences between S_N1 and S_N2 transition states.

1. In the S_N1 TS, there is considerable positive charge on C; there is much weaker bonding between the attacking group and leaving groups with C. There is little or no charge on C in the S_N2 TS.

2. The S_N1 TS is approached by separation of the leaving group; the S_N2 TS by attack of :Nu^–or :Nu.

3. The ΔH^‡of the S_N1 TS (and the rate of the reaction) depends on the stability of the incipient R^. When R^is more stable, ΔH^‡is lower and the rate is greater. The ΔH^‡of the S_N2 TS depends on the steric effects. When there are more R’s on the attacked C or when the attacking :Nu^–is bulkier, ΔH^‡is greater and the rate is less.

Problem 7.15 How can the stability of an intermediate R^in an S_N1 reaction be assessed from its enthalpy-reaction diagram?

The intermediate R^is a trough between two transition-state peaks. More stable R^’s have deeper troughs and differ less in energy from the reactants and products.

Problem 7.16 (a) Formulate (CH₃)₃COH  HCl  (CH₃)₃CCl  H₂O as an S_N1 reaction. (b) Formulate the reaction CH₃OH  HI  CH₃I  H₂O as an S_N2 reaction.

Problem 7.17 ROH does not react with NaBr, but adding H₂SO₄forms RBr. Explain.

Br^, an extremely weak Brönsted base, cannot displace the strong base OH^. In acid, RO^H₂is first formed.

Now, Br^displaces H₂O, which is a very weak base and a good leaving group.

Problem 7.18 Optically pure (S)-()-CH₃CHBr-n-C₆H₁₃has [α]²⁵_D 36.0°. A partially racemized sample having a specific rotation of 30° is reacted with dilute NaOH to form (R)-()-CH₃CH(OH)-n-C₆H₁₃ ([α]²⁵_D 5.97°), whose specific rotation is 10.3° when optically pure. (a) Write an equation for the reaction using projection formulas. (b) Calculate the percent optical purity of reactant and product. (c) Calculate percent-ages of racemization and inversion. (d ) Calculate percentpercent-ages of front-side and back-side attack. (e) Draw a con-clusion concerning the reactions of 2° alkyl halides. (f ) What change in conditions would increase inversion?

Bromide = + (100%) = 83% Alcohol = +

30 36

 5 9



– . 77

10 3 100 58



– . ( %)= %

(b) The percentage of optically active enantiomer (optical purity) is calculated by dividing the observed specific rotation by that of pure enantiomer and multiplying the quotient by 100%. The optical purities are as follows:

(c) The percentage of inversion is calculated by dividing the percentage of optically active alcohol of opposite configuration by that of reacting bromide. The percentage of racemization is the difference between this per-centage and 100%.

Percentage inversion =58%

83%

Perce

(100%)=70% n

ntage racemization = 100% – 70% = 30%

Percentage backside reaction = 70% + ¹₂(30 ))% % (

= 85 30

Percentage frontside reaction = ¹₂ %%)=15%

CH₃

H C Br C₆H₅

CH₃

H

C C₆H₅ + HBr

HO 2% inversion

98% racemization CH₃

H

C C₆H₅ + HBr

CH₃O 27% inversion

73% racemization

CH

3OH

H2O

(d) Inversion involves only backside attack, while racemization results from equal backside and frontside at-tack. 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.

(e) The large percentage of inversion indicates chiefly S_N2 reaction, while the smaller percentage of racemiza-tion indicates some S_N1 pathway. This duality of reaction mechanism is typical of 2° alkyl halides.

(f ) The S_N2 rate is increased by raising the concentration of the nucleophile—in this case, OH^. Problem 7.19 Account for the following stereochemical results:

H₂O is more nucleophilic and polar than CH₃OH. It is better able to react to give HS:---R^δ---:SH (see Prob-lem 7.13), leading to racemization.

Problem 7.20 NH₃reacts with RCH₂X to form an ammonium salt, RCH₂NH^₃X^. Show the transition state, indicating the partial charges.

N gains δ as it begins to form a bond.

Problem 7.21 H₂CCHCH2Cl is solvolyzed faster than (CH₃)₂CHCl. Explain.

Solvolyses go by an S_N1 mechanism. Relative rates of different reactants in S_N1 reactions depend on the sta-bilities of intermediate carbocations. H₂CCHCH2Cl is more reactive because

is more stable than (CH₃)₂CH^. See Problem 6.35 for a corresponding explanation of stability of an allyl radical.

Problem 7.22 In terms of (a) the inductive effect and (b) steric factors, account for the decreasing stability of R^:

(a) Compared to H, R has an electron-releasing inductive effect. Replacing H’s on the positive C by CH₃’s dis-perses the positive charge and thereby stabilizes R^.

(b) Steric acceleration also contributes to this order of R^stability. Some steric strain of the three Me’s in Me₃C — Br separated by a 109° angle (sp³) is relieved upon going to a 120° separation in R^with a C using sp²hybrid orbitals.

Carbocations have been prepared as long-lived species by the reaction RF  SbF₅ R^ SbF^6. SbF₅, a covalent liquid, is called a superacid because it is a stronger Lewis acid than R^.

Problem 7.23 How does the S_N1 mechanism for the hydrolysis of 2-bromo-3-methylbutane, a 2° alkyl bro-mide, to give exclusively the 3° alcohol 2-methyl-2-butanol, establish a carbocation intermediate?

The initial slow step is dissociation to the 2° 1,2-dimethylpropyl cation. A hydride shift yields the more stable 3° carbocation, which reacts with H₂O to form the 3° alcohol.

Problem 7.24 RBr reacts with AgNO₂to give RNO₂and RONO. Explain.

The nitrite ion

has two different nucleophilic sites: the N and either O. Reaction with the unshared pair on N gives RNO₂, while RONO is formed by reaction at O. [Anions with two nucleophilic sites are called ambident anions.]