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# 化学代写|有机化学代写organic chemistry代考|E2 Eliminations of H/Het and the E2/SN2 Competition

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## 化学代写|有机化学代写organic chemistry代考|Substrate Effects on the E2/SN2 Competition

Tables 4.1 and 4.2 summarize the typical substrate effects on the chemoselectivity of E2
vs SN2 reactions. These substrate effects are so pronounced because NaOEt was used as
base. As a reasonably strong base and a quite good nucleophile, NaOEt is able to convert a fair portion of many elimination substrates into SN2 products (see Section 4.4.2).

Table 4.1 gives the chemoselectivity of E2 eliminations from representative bromides of the type Rprim¬Br, Rsec¬Br, and Rtert¬B r. The fraction of E2 product increases in this sequence from 1 to 79 and to 100% and allows for the following generalization:
E2 eliminations with sterically unhindered bases can be carried out chemoselectively (i.e., without a competing SN2 reaction) only starting from tertiary alkyl halides and sulfonates. To obtain an E2 product from primary alkyl halides and sulfonates at all or
to obtain an E2 product from secondary alkyl halides and sulfonates exclusively, one must change the base (see Section 4.4.2).

Table 4.1 also allows one to identify reasons for the chemoselectivities listed therein.
According to Equation 4.3, they equal the ratio kE2: of the two competing reactions. According to Table 4.1, this ratio increases rapidly in the sequence Rprim¬Br,Rsec¬Br, and Rtert¬Br because the values become smaller and smaller in this order (for the reason, see Section 2.4.4) while the kE2 values become larger and larger.
There are two reasons for this increase in the kE2 values. The first reason is a statistical factor: the number of H atoms located in b-position(s) relative to the leaving group and that can be eliminated together with it is 3, 6, and 9, respectively, in the three bromides discussed. One can adjust the gross kE2 values of Table 4.1 for this factor by converting them to kE2-values per single H atom in a b-position. However, these numbers—“kE2 (per b H atom)” in Table 4.1¬still increase in the series Rprim¬Br →Rsec¬Br → Rtert¬Br. This second effect is due to the fact that the E2 eliminations considered lead to C“C double bonds with an increasing number of alkyl substituents: EtBr results in an unsubstituted, iPrBr in a monosubstituted, and tert-BuBr in a disubstituted alkene. The stability of olefins, as is well known, increases with the degree of alkylation. A certain fraction of this stability increase becomes noticeable in the transition state of E2 eliminations in the form of product-development control: The rate constant kE2 (per b H atom) is therefore smallest for the most unstable E2 product (ethene) and largest for the most stable E2 product (isobutene).

## 化学代写|有机化学代写organic chemistry代考|Base Effects on the E2/SN2 Competition

Chemoselective E2 eliminations can be carried out with sterically hindered, sufficiently strong bases. Their bulkiness causes them to attack at an $\mathrm{H}$ atom at the periphery of the molecule rather than at a $\mathrm{C}$ atom deep within the molecule. These bases are therefore called nonnucleophilic bases. The weaker nonnucleophilic bases include the bicyclic amidines DBN (diazabicyclononene) and DBU (diazabicycloundecene). These can be used to carry out chemoselective E2 eliminations even starting from primary and secondary alkyl halides and sulfonates (Figure 4.16).
$\beta$-Eliminations of epoxides lead to allyl alcohols after release of a high-energy (cf. Section 2.3) alkoxide ion as a leaving group. For this reaction to take place, the strongly basic bulky lithium dialkyl amides LDA (lithium diisopropyl amide), LTMP (lithium tetramethyl piperidide) or LiHMDS (lithium hexamethyl disilazide) shown in Figure $4.17$ are used. As for the amidine bases shown in Figure 4.16, the bulkiness of these amides guarantees that they are nonnucleophilic. They react, for example, with epoxides in chemoselective E2 reactions even when the epoxide contains a primary $\mathrm{C}$ atom that is easily attacked by nucleophiles (see, e.g., Figure 4.17).

## 化学代写|有机化学代写ORGANIC CHEMISTRY代考|A Stereoelectronic Effect on the E2/SN2 Competition

(4-tert-Butylcyclohexyl)trimethylammonium iodide and potassium-tert-butoxide ( $\mathrm{KO} t \mathrm{Bu}$ ) can undergo both an E2 and an $\mathrm{S}{\mathrm{N}}$ 2 reaction (Figure 4.18). However, the reactivity and chemoselectivity differ drastically depending on whether the cis- or the trans-configured cyclohexane derivative is the substrate. The cis isomer gives a $90: 10$ mixture of $\mathrm{E} 2$ and $\mathrm{S}{\mathrm{N}} 2$ products in a fast reaction. The trans isomer reacts much more slowly and gives only the substitution product. According to Equation 4.3, these findings mean that $k_{\mathrm{E} 2, \text { cis }}: k_{\mathrm{S}{\mathrm{N}} 2 \text {, cis }}=90: 10$ and $k{\mathrm{E} 2, \text { trans }}: k_{\mathrm{S}{\mathrm{N}}^{2} \text {,trans }}<1: 99$. A plausible assumption is that for the respective substitution reactions $k{\mathrm{S}{\mathrm{N}} 2, \text { cis }} \approx k{\mathrm{S}{\mathrm{N}} 2 \text {, trans. }}$. This, in turn, means that the opposite chemoselectivities of the two reactions in Figure $4.18$ arise almost exclusively from the fact that $k{\mathrm{E} 2, \text { cis }} \gg k_{\mathrm{E} 2, \text { trans }}$.

How this gradation of the $k_{\mathrm{E} 2}$ values comes about is easily understood by considering Figure $4.19$ and realizing the following: in the most stable transition state of an E2 elimination the half-broken ${ }^{\alpha} \mathrm{C} \cdots$ Het bond and the likewise half-broken ${ }^{\beta} \mathrm{C} \cdots$ $\mathrm{H}$ bond are oriented parallel. This is because the hybridization change at ${ }^{\alpha} \mathrm{C}$ and ${ }^{\beta} \mathrm{C}$ from $s p^{3}$ to $s p^{2}$ hybridization has started in the transition state. As a result, $2 \mathrm{p}_{\mathrm{z}}$-like AOs are being formed as ${ }^{\alpha} \mathrm{C}$ and ${ }^{\beta} \mathrm{C}$ and their increasingly effective overlap establishes the $\pi$ bond. In other words, coplanar ${ }^{\alpha} \mathrm{C} \cdots$ Het and ${ }^{\beta} \mathrm{C} \cdots \mathrm{H}$ bonds stabilize the transition state of an E2 elimination through a $\pi$-like interaction. Interactions of this type are possible in the most stable transition state of syn-eliminations as well as in the most stable transition state of anti-eliminations. When a substrate can assume both of these transition state geometries, the anti-elimination is always preferred. This is a consequence of the lower steric hindrance of the anti-transition state, in which the substituents at ${ }^{\alpha} \mathrm{C}$ vs ${ }^{\beta} \mathrm{C}$ are nearly staggered. The sterically more hindered syn transition state has a nearly eclipsed structure. In addition, there are reasons to believe that orbital overlap is better in the transition state of an anti-elimination.

## 化学代写|有机化学代写ORGANIC CHEMISTRY代考|SUBSTRATE EFFECTS ON THE E2/SN2 COMPETITION

kE2 值增加的原因有两个。第一个原因是一个统计因素：位于 b 位的 H 原子的数量s在所讨论的三种溴化物中，相对于离去基团，可以与它一起消除的分别是 3、6 和 9。可以通过将表 4.1 中的总 kE2 值转换为 b 位上每个 H 原子的 kE2 值来调整表 4.1 中的总 kE2 值。然而，这些数字——“kE2p和rbH一种吨这米” 在表 4.1 中，Rprim-Br →Rsec-Br → Rtert-Br 的系列仍然增加。第二个效果是由于考虑到的 E2 消除导致 C”C 双键具有越来越多的烷基取代基：EtBr 导致未取代，iPrBr 导致单取代，叔丁基溴导致二取代烯烃。众所周知，烯烃的稳定性随着烷基化程度的增加而增加。这种稳定性增加的一部分在产品开发控制形式的 E2 消除的过渡状态中变得明显： 速率常数 kE2p和rbH一种吨这米因此对于最不稳定的 E2 产品是最小的和吨H和n和和最大的最稳定的 E2 产品一世s这b在吨和n和.

## 化学代写|有机化学代写ORGANIC CHEMISTRY代考|BASE EFFECTS ON THE E2/SN2 COMPETITION

b-释放高能后，环氧化物的消除导致烯丙醇CF.小号和C吨一世这n2.3醇盐离子作为离去基团。为了使该反应发生，强碱性大体积二烷基酰胺锂 LDAl一世吨H一世在米d一世一世s这pr这p是l一种米一世d和, LTMPl一世吨H一世在米吨和吨r一种米和吨H是lp一世p和r一世d一世d和或 LiHMDSl一世吨H一世在米H和X一种米和吨H是ld一世s一世l一种和一世d和如图4.17被使用。至于图 4.16 中所示的脒碱基，这些酰胺的庞大性保证了它们是非亲核的。例如，它们在化学选择性 E2 反应中与环氧化物发生反应，即使环氧化物含有初级C容易被亲核试剂攻击的原子s和和,和.G.,F一世G在r和4.17.

## 化学代写|有机化学代写ORGANIC CHEMISTRY代考|A STEREOELECTRONIC EFFECT ON THE E2/SN2 COMPETITION

4−吨和r吨−乙在吨是lC是Cl这H和X是l三甲基碘化铵和叔丁醇钾$ķ这吨乙在$可以同时经历 ( $\mathrm{KO} t \mathrm{Bu}$ ) can undergo both an E2 and an $\mathrm{S}{\mathrm{N}}$ 2 reaction (Figure 4.18). However, the reactivity and chemoselectivity differ drastically depending on whether the cis- or the trans-configured cyclohexane derivative is the substrate. The cis isomer gives a $90: 10$ mixture of $\mathrm{E} 2$ and $\mathrm{S}{\mathrm{N}} 2$ products in a fast reaction. The trans isomer reacts much more slowly and gives only the substitution product. According to Equation 4.3, these findings mean that $k_{\mathrm{E} 2, \text { cis }}: k_{\mathrm{S}{\mathrm{N}} 2 \text {, cis }}=90: 10$ and $k{\mathrm{E} 2, \text { trans }}: k_{\mathrm{S}{\mathrm{N}}^{2} \text {,trans }}<1: 99$. A plausible assumption is that for the respective substitution reactions $k{\mathrm{S}{\mathrm{N}} 2, \text { cis }} \approx k{\mathrm{S}{\mathrm{N}} 2 \text {, trans. }}$. This, in turn, means that the opposite chemoselectivities of the two reactions in Figure $4.18$ arise almost exclusively from the fact that $k{\mathrm{E} 2, \text { cis }} \gg k_{\mathrm{E} 2, \text { trans }}$.

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