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A big part of the confusion here is **how** we count “alkyl substituents” on the double bond.  At first glance, both 3‐methyl‐2‐hexene and 3‐methyl‐3‐hexene can _look_ like they each have three substituents, so why is one “more substituted” than the other?

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## 1.  Structures of the Two Alkenes

Let’s write them out more explicitly.  The difference is _where_ the double bond forms after dehydrating 3‐methyl‐3‐hexanol.

### **3‐Methyl‐2‐hexene**  
- The double bond is between C2 and C3.
- The skeleton is 
  \[
  \text{CH}_3 \;-\; \text{CH} = \text{C}(\text{CH}_3) \;-\; \text{CH}_2 \;-\; \text{CH}_2 \;-\; \text{CH}_3,
  \]
  where the italicized “(CH₃)” is the methyl substituent on C3.

### **3‐Methyl‐3‐hexene**  
- The double bond is between C3 and C4.
- The skeleton is 
  \[
  \text{CH}_3 \;-\; \text{CH}_2 \;-\; \text{C}(\text{CH}_3) = \text{CH} \;-\; \text{CH}_2 \;-\; \text{CH}_3,
  \]
  where again the italicized “(CH₃)” is the methyl substituent on C3.

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## 2.  Counting Substituents Properly

When chemists talk about an alkene being “mono‐, di‐, tri‐, or tetra‐substituted,” they mean:

- **Monosubstituted**:  One alkyl group attached to the pair of sp² carbons (total).  
- **Disubstituted**:  Two alkyl groups attached (total).  
- **Trisubstituted**:  Three alkyl groups attached (total).  
- **Tetrasubstituted**:  Four alkyl groups attached (total).

But the “main chain” piece bonded to the alkene doesn’t automatically count as a “substituent” if it’s _continuing_ the chain. We look carefully at **each sp² carbon** and see whether it’s attached to a hydrogen or to an alkyl group.

### **3‐Methyl‐2‐hexene** 
- **C2** is bonded to:  
  1. H  
  2. CH₃ (the “left” end of the chain)  
- **C3** is bonded to:  
  1. CH₃ (the 3‐methyl substituent)  
  2. –CH₂–CH₂–CH₃ (the “right” continuation of the chain)

Hence, in total, there are **3 alkyl substituents** on the double bond (CH₃ on C2, CH₃ on C3, and the propyl chain also on C3). This makes it a **trisubstituted** alkene.

### **3‐Methyl‐3‐hexene** 
- **C3** is bonded to:  
  1. –CH₂–CH₃ (the “left” part of the chain, which is effectively an ethyl)  
  2. CH₃ (the 3‐methyl substituent)  
- **C4** is bonded to:  
  1. –CH₂–CH₃ (the “right” part of the chain, effectively another ethyl)  
  2. H

So you might think, “That also looks like 3 total alkyl substituents.”  However, the usual classification lines up 3‐methyl‐3‐hexene as effectively “disubstituted” or a less stable tri‐substituted arrangement, because the substituents are **geminal** (both on the same sp² carbon, C3) and you still have an sp² carbon (C4) bearing a hydrogen.

In simpler terms, **3‐methyl‐2‐hexene** places alkyl groups on _both_ of the vinylic carbons in a way that allows more hyperconjugation and less steric strain.  **3‐methyl‐3‐hexene** localizes extra bulk on one carbon and leaves the other with a hydrogen.  Empirically (and via hyperconjugation arguments), this makes 3‐methyl‐2‐hexene the more stable (Zaitsev) product.

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## 3.  Zaitsev’s Rule and “Most Substituted” Usually Matches Highest Stability

Zaitsev’s rule says the alkene that ends up with more alkyl substitution on (or better _distribution_ across) the double‐bond carbons will be the major product.  Even though both final alkenes can appear “tri‐substituted” by a raw count, the one that puts those substituents in a more stable arrangement (spread across both sp² carbons) is **3‐methyl‐2‐hexene**.

That is why in practice, **3‐methyl‐2‐hexene** is the major (Zaitsev) product and **3‐methyl‐3‐hexene** is the minor product.

Sat, 11 Jan 2025 02:38 GMT