The linear free energy relationship is an empirical relationship in physical organic chemistry used to describe the variation of reaction rate constants or equilibrium constants with changes in substituent structure. The classical linear free energy relationship model is described by the Hammett equation: $\begin{array} { r } { \mathrm { l g } \left( k _ { \mathrm { R } } / k _ { \mathrm { H } } \right) = \rho \sigma } \end{array}$

Here, $k _ { \mathrm { R } } / k _ { \mathrm { H } }$ represents the ratio of the reaction rate constants for compounds with substituent R and without any substituent.  

Researchers established the $\sigma ^ { + }$ scale for different substituents $\mathrm { R }$ by studying the $\mathrm { S _ { N } } 1$ reactions of para-substituted benzyldimethylchloromethanes. They further investigated the mechanism of reaction 1 catalyzed by $\mathrm { R h } _ { 2 } ( \mathrm { I I } )$ complexes using the $\sigma ^ { + }$ scale. The $\sigma ^ { + }$ values for different substituents $\mathrm { R }$ and their relative reaction rates in a specific reaction are shown in the table below. Which of the following statements are correct?  

<MultiModal>![CCC1=CC(C)=CC(C2=CC=C([R])C=C2)=C1N loses a molecule of nitrogen gas under the catalysis of Rh catalyst to yield CCC1=CC(C)=CC2=C1NC3=CC([R])=CC=C23](/media/uploads/198c5724-6a6b-4d77-894c-0186b114cf3b.png)</MultiModal>  

| R     | OCH₃  | CH₃  | F     | H    | Cl    | CF₃  |
|-------|-------|------|-------|------|-------|------|
| σ⁺    | -0.78 | -0.31| -0.07 | 0    | 0.11  | 0.61 |
| \( k_R/k_H \) | 85/15 | 70/30 | 58/42 | 1    | 47/53 | 27/73 |
线性自由能关系是物理有机化学中用来描述反应速率常数或平衡常数随取代基结构变化的经验关系。经典的线性自由能关系模型由哈米特方程描述： $\begin{array} { r } { \mathrm { l g } \left( k _ { \mathrm { R } } / k _ { \mathrm { H } } \right) = \rho \sigma } \end{array}$

其中 $k _ { \mathrm { R } } / k _ { \mathrm { H } }$ 表示含取代基R和不含取代基的化合物反应速率常数之比。 

研究人员通过不同对位取代的苯基二甲基氯代甲烷的 $\mathrm { S _ { N } } 1$ 反应，建立了不同取代基 $\mathrm { R }$ 的 $\sigma ^ { + }$ 标度；并利用 $\sigma ^ { + }$ 标度对 $\mathrm { R h } _ { 2 } ( \mathrm { I I } )$ 配合物催化的反应1的机理进行研究，不同取代基 $\mathrm { R }$的 $\sigma ^ { + }$ 标度和在某反应中的相对反应速率如下表所示。下面的说法中哪些是正确的？

<MultiModal>![CCC1=CC(C)=CC(C2=CC=C([R])C=C2)=C1N在Rh催化剂的催化下失去一分子氮气，得到CCC1=CC(C)=CC2=C1NC3=CC([R])=CC=C23](/media/uploads/198c5724-6a6b-4d77-894c-0186b114cf3b.png)</MultiModal>

| R     | OCH₃  | CH₃  | F     | H    | Cl    | CF₃  |
|-------|-------|------|-------|------|-------|------|
| σ⁺    | -0.78 | -0.31| -0.07 | 0    | 0.11  | 0.61 |
| \( k_R/k_H \) | 85/15 | 70/30 | 58/42 | 1    | 47/53 | 27/73 |
$\rho=-0.87$, the reaction is an electrophilic aromatic reaction
$\rho=+0.87$, this reaction is an electrophilic aromatic substitution reaction.
$\rho=0.87$, this reaction is not an aromatic electrophilic reaction
$\rho=-2.0$, this reaction is an electrophilic aromatic reaction
$\rho=-2.0$, this reaction is not an aromatic electrophilic reaction
The determination of rate constant ratios for different substituents in this reaction can only be achieved by accurately measuring the rates of substrate consumption with different substituents and then taking the ratio.
None of the above statements are correct
At least two of the above statements are correct
Electron-donating substituents slow down the reaction.
$\rho=-0.87$，该反应是芳香亲电反应
$\rho=+0.87$，该反应是芳香亲电反应
$\rho=0.87$，该反应不是芳香亲电反应
$\rho=-2.0$，该反应是芳香亲电反应
$\rho=-2.0$，该反应不是芳香亲电反应
该反应的不同取代基的速率常数比值的测定只能通过准确测定不同取代基下底物消耗的速率做比来得到
以上说法均不正确
以上说法至少有两项是正确的
富电子的取代基会让反应变慢
Fitting $\begin{array} { r } { \mathrm { l g } \left( k _ { \mathrm { R } } / k _ { \mathrm { H } } \right)  } \end{array}$ against $\sigma ^ { + }$ yields $\rho=-0.87$

<Checkpoint>[$\rho=-0.87$](2)</Checkpoint>

Electron-donating substituents accelerate the reaction, but $\rho=-0.87$ is smaller than $-1$, thus this reaction cannot be an aromatic electrophilic substitution. If it were an aromatic electrophilic substitution, the rate enhancement by electron-donating groups would be greater than that in benzylic reactions.

<Checkpoint>[$\rho=-0.87$ is smaller than $-1$, thus this reaction cannot be an aromatic electrophilic substitution](1)</Checkpoint>

Note that the phenyl group in this reaction can be substituted on both sides. Therefore, connecting two different substituents to each side of the benzene ring conveniently provides the rate ratio between the two substituents, which is far more reliable than directly measuring substrate consumption rates. F is incorrect.

<Checkpoint>[The phenyl group can be substituted on both sides, thus connecting two different substituents to each side of the benzene ring conveniently provides the rate ratio between the two substituents](1)</Checkpoint>
用$\begin{array} { r } { \mathrm { l g } \left( k _ { \mathrm { R } } / k _ { \mathrm { H } } \right)  } \end{array}$对$\sigma ^ { + }$ 进行拟合，得到$\rho=-0.87$

<Checkpoint>[$\rho=-0.87$](2)</Checkpoint>

富电子的取代基会让反应变快，但是$\rho=-0.87$
比$-1$小，因此该反应不可能是芳香亲电取代反应。因为如果该反应是芳香亲电取代反应，给电子基团对反应速率的加快会比卞位反应更大。

<Checkpoint>[$\rho=-0.87$ 比$-1$小，因此该反应不可能是芳香亲电取代反应](1)</Checkpoint>

注意到，该反应的苯基两侧均可以连接取代基，因此在两侧连接两种取代基不同的苯环即可方便地得到两种取代基的速率之比，比直接测底物消耗速率要可靠很多。F错误。

<Checkpoint>[反应的苯基两侧均可以连接取代基，因此在两侧连接两种取代基不同的苯环即可方便地得到两种取代基的速率之比](1)</Checkpoint>