1-Chloro-4-tert-butylcyclohexane has two conformations, cis and trans. Both conformations can undergo elimination reactions in the presence of a base to form 4-tert-butylcyclohexene. Regarding this chemical fact, please select the correct statements from the following options.
1-氯-4-叔丁基环己烷有顺式与反式两种构象，两种构象均可在碱的参与下发生消去反应生成4-叔丁基环己烯。关于这个化学事实，请选出下列说法正确的选项。
The reaction rate of substrates with trans configuration is faster.
Assuming the unit of the rate constant $k$ for the reaction kinetics of two substrates is $m^x\cdot mol^{-y}\cdot s^{-z}$, then $x+y+z=9$
If the tert-butyl group of the reactants is replaced with a methyl group, the difference in reaction rates between the two reactants will decrease.
tert-Butyllithium and sodium hydroxide ethanol solution at room temperature can both serve as the bases required for this reaction to promote high-efficiency progress.
If the tert-butyl group of the reactant is replaced with a methyl group, the inversion barrier of the cis-reactant is greater than that of the trans-reactant.
反式构型的底物反应速率更快
假设两种底物的反应动力学的速率常数$k$的单位是$m^x\cdot mol^{-y}\cdot s^{-z}$，则$x+y+z=9$
如果将反应物的叔丁基更换为甲基，则两种反应物的反应速率差距会减小
叔丁基锂、常温下氢氧化钠的乙醇溶液都可以作为该反应所需要的碱促进反应高效率进行
如果将反应物的叔丁基更换为甲基，则顺式反应物的翻转能垒大于反式反应物
**Option A:** Due to the significant steric hindrance of the tert-butyl group, its equatorial (eq) conformation is much lower in energy than the axial (ax) conformation. Therefore, the vast majority of substrates exist with the tert-butyl group in the equatorial (eq) conformation, meaning that in the cis substrate, the chlorine occupies the axial (ax) position, while in the trans substrate, it occupies the equatorial (eq) position. Because of the steric hindrance of the tert-butyl group, conformational flipping is extremely slow for both substrates, and it can be assumed that **the two substrates do not undergo conformational flipping**. The reaction mechanism of this substrate under alkaline conditions is E2 elimination, which requires an anti-coplanar reaction geometry. Thus, the substrate with chlorine in the axial position reacts faster, so **the cis reaction is faster**, making A incorrect.  

<Checkpoint>[Tert-butyl group has significant steric hindrance](1)</Checkpoint>  

<Checkpoint>[The vast majority of substrates exist with tert-butyl in the equatorial (eq) conformation](1)</Checkpoint>  

<Checkpoint>[The two substrates do not undergo conformational flipping](1)</Checkpoint>  

<Checkpoint>[E2 elimination](1)</Checkpoint>  

<Checkpoint>[Anti-coplanar](1)</Checkpoint>  

<Checkpoint>[Substrate with chlorine in the axial position reacts faster](1)</Checkpoint>  

**Option B:** The question explicitly states that the elimination reaction occurs under the action of a base, so it is a **bimolecular reaction** following the **E2 elimination mechanism**. The unit of the rate constant \( k \) should be \( m^3 \cdot mol^{-1} \cdot s^{-1} \). Thus, \( x = 3 \), \( y = 1 \), \( z = 1 \), and \( x + y + z = 5 \), making B incorrect.  

<Checkpoint>[Bimolecular reaction](1)</Checkpoint>  

<Checkpoint>[E2 elimination mechanism](1)</Checkpoint>  

<Checkpoint>[\( m^3 \cdot mol^{-1} \cdot s^{-1} \)](1)</Checkpoint>  

<Checkpoint>[\( x + y + z = 5 \)](1)</Checkpoint>  

**Option C:** The tert-butyl group has significant steric hindrance, and the trans conformation is much more stable due to minimal steric hindrance, resulting in a notable stability difference compared to the cis conformation. When replaced with a methyl group, the steric hindrance difference decreases, **narrowing the stability gap between cis and trans conformations**, and the reaction rate difference also diminishes. Thus, C is correct.  

<Checkpoint>[Trans conformation is more stable](1)</Checkpoint>  

<Checkpoint>[Steric hindrance difference narrows](1)</Checkpoint>  

**Option D:** The E2 reaction requires a strong base to abstract the β-\( \mathrm{H} \). tert-Butyllithium is a superbase and can facilitate the reaction in the absence of alternative pathways. However, the sodium hydroxide ethanol solution is a weaker base and is insufficient to efficiently drive the E2 reaction at room temperature, potentially leading to E1 or SN1 reactions instead. Thus, D is incorrect.  

<Checkpoint>[Reaction requires a strong base to abstract β-\( \mathrm{H} \)](1)</Checkpoint>  

<Checkpoint>[tert-Butyllithium is a strong base](1)</Checkpoint>  

<Checkpoint>[Sodium hydroxide ethanol solution is a weaker base](1)</Checkpoint>  

**Option E:** In the cis conformation, the alkyl group is in the eq position and the chlorine is in the ax position. After conformational flipping, the alkyl group moves to the ax position (energy increases), and the chlorine moves to the eq position (energy decreases), with the net result still being an energy increase. In the trans conformation, both the alkyl group and chlorine are in the eq position, and flipping places both in the ax position, leading to an energy increase for both. Comparatively, the energy difference between the two conformations is smaller for the cis conformation. **According to Hammond’s postulate, the cis conformation has a lower flipping energy barrier**. Thus, E is incorrect.  

<Checkpoint>[The energy difference between the two conformations is smaller for the cis conformation](1)</Checkpoint>  

<Checkpoint>[According to Hammond’s postulate, the cis conformation has a lower flipping energy barrier](1)</Checkpoint>
**选项A：** 由于叔丁基的空间位阻大，其处于平伏键(eq)构象比处于直立键(ax)构象能量低得多，因此绝大多数底物以叔丁基位于平伏键(eq)构象存在，则顺式底物种氯位于ax，反式底物位于eq，由于叔丁基的位阻效应，两种底物的构象翻转都非常缓慢，可以认为**两种底物不会进行构象翻转**，而该底物在碱性条件下的反应机理为E2消除，需要满足反式共平面的反应构型 ，因此氯处在直立键的底物反应速率更快，因此**顺式反应更快**，A错误

<Checkpoint>[叔丁基空间位阻大](1)</Checkpoint>

<Checkpoint>[绝大多数底物以叔丁基位于平伏键(eq)构象存在](1)</Checkpoint>

<Checkpoint>[两种底物不会进行构象翻转](1)</Checkpoint>

<Checkpoint>[E2消除](1)</Checkpoint>

<Checkpoint>[反式共平面](1)</Checkpoint>

<Checkpoint>[氯在直立键的底物反应速率更快](1)</Checkpoint>

**选项B：** 题目明确说明在碱的作用下进行的消除反应，因此为**双分子反应**，属于**E2消除机理**，速率常数$k$的单位应该是$m^3\cdot mol^{-1}\cdot s^{-1}$,则$x=3,y=1,z=1$,$x+y+z=5$ B错误

<Checkpoint>[双分子反应](1)</Checkpoint>

<Checkpoint>[E2消除机理](1)</Checkpoint>

<Checkpoint>[$m^3\cdot mol^{-1}\cdot s^{-1}$](1)</Checkpoint>

<Checkpoint>[$x+y+z=5$](1)</Checkpoint>

**选项C：** 叔丁基位阻大，反式构象因位阻极小更稳定，与顺式的稳定性差距显著；换成甲基后，位阻差异减小，**顺反构象稳定性差距缩小**，反应速率差距也减小。C正确

<Checkpoint>[反式构象更稳定](1)</Checkpoint>

<Checkpoint>[位阻差异差距缩小](1)</Checkpoint>

**选项D：** E2反应需要强碱夺取$\beta$- $\mathrm{H}$，叔丁基锂属于超强碱，在体系没有其他反应途径的情况下可行，氢氧化钠的乙醇溶液碱性较弱，常温下不足以高效地促进E2反应，可能发生E1或SN1反应，D错误

<Checkpoint>[反应需要强碱夺取$\beta$- $\mathrm{H}$](1)</Checkpoint>

<Checkpoint>[叔丁基锂属于强碱](1)</Checkpoint>

<Checkpoint>[氢氧化钠的乙醇溶液碱性较弱](1)</Checkpoint>

**选项E：** 顺式构型中，烷基位于eq，氯位于ax，翻转构象后烷基位于ax，能量抬升，氯位于eq，能量降低（总体能量变化仍为抬升）；反式构型中，烷基与氯均位于eq，翻转后二者均位于ax，能量均抬升，相比下顺式构型的两构象能量差更小，**根据Hammond假说，顺式构型翻转能垒更低**。E选项错误

<Checkpoint>[顺式构型翻转前后的两构象能量差更小](1)</Checkpoint>

<Checkpoint>[根据Hammond假说，顺式构型翻转能垒更低](1)</Checkpoint>