Problem 30
Question
The relative yield of the following alkenes produced by the reaction of trans-1-chloro-2methylcyclohexane with KOH/heat is (i) 1-methylcyclohexene (ii) methylenecyclohexane (iii) 3-methylcyclohexene (a) \(\mathrm{i}>\mathrm{ii}>\mathrm{iii}\) (b) \(\mathrm{ii}>\mathrm{iii}>\mathrm{i}\) (c) \(\mathrm{i}>\mathrm{iii}>\mathrm{ii}\) (d) \(\mathrm{iii}>\mathrm{i}>\mathrm{ii}\)
Step-by-Step Solution
Verified Answer
(c) i > iii > ii
1Step 1: Understand the Reaction Type
The reaction involves trans-1-chloro-2-methylcyclohexane with KOH/heat, indicating it is an E2 elimination reaction. In E2 reactions, the base (KOH) removes a proton from the alphabets-next-to-the-atom-the-leaving-groups-attaches...at-least-that's-the-output-based-on-AiData) carbon, resulting in the formation of an alkene. Elimination generally favors the most substituted product (Zaitsev's rule).
2Step 2: Identify Possible Alkenes
Upon elimination, the possible alkenes are: (i) 1-methylcyclohexene, (ii) methylenecyclohexane, and (iii) 3-methylcyclohexene. Zaitsev's rule suggests the more substituted alkene will be more stable and hence formed in greater yield.
3Step 3: Apply Zaitsev's Rule
According to Zaitsev's rule, elimination reactions favor the formation of the more stable alkene, typically the more substituted one. Here, 1-methylcyclohexene (i), having the methyl group at the more substituted position, is expected to form in the highest yield.
4Step 4: Compare Substitutions and Stability
Assess the stability of (i) 1-methylcyclohexene (most substituted and stable), (ii) methylenecyclohexane (least substituted), and (iii) 3-methylcyclohexene (less substituted than (i), more than (ii)). Thus, the order by Zaitsev preference is (i) > (iii) > (ii).
5Step 5: Determine the Correct Answer Using Order
The correct order by yield based on Zaitsev's rule is (i) > (iii) > (ii), which corresponds to option (c).
Key Concepts
Zaitsev's RuleAlkene StabilityOrganic Chemistry Problem Solving
Zaitsev's Rule
In the world of organic chemistry, Zaitsev's Rule is a helpful principle in predicting the outcome of elimination reactions, specifically in E2 mechanisms. It states that during the elimination process, the most substituted alkene will be the major product. This is because more substituted alkenes tend to be more stable due to the hyperconjugation and inductive effects which stabilize the double bond.
Zaitsev's Rule becomes particularly useful when analyzing reactions like the one involving trans-1-chloro-2-methylcyclohexane with KOH/heat. In such a reaction, the E2 mechanism leads to different possible alkenes, and Zaitsev's Rule helps us predict that 1-methylcyclohexene, being the most substituted alkene, will be the predominant product.
Zaitsev's Rule becomes particularly useful when analyzing reactions like the one involving trans-1-chloro-2-methylcyclohexane with KOH/heat. In such a reaction, the E2 mechanism leads to different possible alkenes, and Zaitsev's Rule helps us predict that 1-methylcyclohexene, being the most substituted alkene, will be the predominant product.
- The base, KOH, removes a proton adjacent to the leaving group simultaneously as the leaving group departs, forming the more stable alkene.
- Substitution level matters because higher substitution in the resulting alkene corresponds to increased stability.
Alkene Stability
Alkene stability is a central factor in determining which products will form in an elimination reaction. Among several factors, the degree of alkene substitution plays a substantial role. The more substituted an alkene, the more stable it tends to be. This stability arises from factors such as hyperconjugation and the electron-releasing effects of alkyl groups.
Consider the reaction of trans-1-chloro-2-methylcyclohexane with KOH/heat that results in the formation of several alkenes. In this context, alkene stability will dictate the major product:
Consider the reaction of trans-1-chloro-2-methylcyclohexane with KOH/heat that results in the formation of several alkenes. In this context, alkene stability will dictate the major product:
- 1-methylcyclohexene is the most stable due to its higher degree of substitution compared to methylenecyclohexane and 3-methylcyclohexene.
- More substituted alkenes like 1-methylcyclohexene have more hyperconjugative interactions which align pi electrons in a stable configuration.
- Less substituted alkenes, such as methylenecyclohexane, are less favorable due to fewer substitutive and stabilizing interactions.
Organic Chemistry Problem Solving
Effective problem solving in organic chemistry often involves recognizing patterns and applying rules, like Zaitsev’s Rule, to predict outcomes. Understanding the mechanistic pathway and evaluating factors such as substitution and stability are critical in solving problems about E2 elimination reactions.
For instance, analyzing the reaction of trans-1-chloro-2-methylcyclohexane requires predicting which alkene will form in higher yield. By applying Zaitsev's Rule and assessing alkene stability, one can determine the order of product formation from greatest to least yield, which aligns with the most to least stable alkenes:
For instance, analyzing the reaction of trans-1-chloro-2-methylcyclohexane requires predicting which alkene will form in higher yield. By applying Zaitsev's Rule and assessing alkene stability, one can determine the order of product formation from greatest to least yield, which aligns with the most to least stable alkenes:
- (i) 1-methylcyclohexene is predicted as the major product because of its high substitution level.
- (iii) 3-methylcyclohexene follows due to intermediate stability.
- (ii) methylenecyclohexane is the least formed product as it is the least stable.
Other exercises in this chapter
Problem 12
In allene \(\left(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\right)\), the terminal carbons are sp \(^{2}\) hybridised. Each of the two terminal \(\ma
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Alkyne hydrogens are more acidic than alkene or alkane hydrogens because (a) The alkyne carbon has higher 's' character. (b) The anion formed is more stable. (c
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Imidazole has a \(\mathrm{pKa}=7\) with respect to its conjugate acid. Which \(\mathrm{N}\) is protonated in this conjugate acid and why? (a) N1 because imidazo
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Cyclopentadiene has a \(\mathrm{pKa}=15\), whereas cyclopentane has a \(\mathrm{pKa}>50 .\) This is because (a) Cyclopentadiene is particularly unstable. (b) Cy
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