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Q6.10(iii):
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (iii) 2,2,3-Trimethyl-3-bromopentane.

Solution :

Step 1: Structural Representation of 2,2,3-Trimethyl-3-bromopentane

The chemical structure of 2,2,3-trimethyl-3-bromopentane is represented by a central pentane chain with a bromine atom on $C_3$, a methyl group on $C_3$, and two methyl groups on $C_2$. The condensed structural formula is:

$CH_3-C(CH_3)_2-C(Br)(CH_3)-CH_2-CH_3$

C1 C2 C3 C4 C5 Br

Step 2: Analysis of $\beta$-Hydrogen Atoms

Dehydrohalogenation via the $E_2$ mechanism requires the presence of $\beta$-hydrogen atoms relative to the carbon bearing the halogen ($C_3$). We must identify all available $\beta$-carbons:

  • $\beta_1$ Carbon ($C_2$): This carbon has 0 hydrogen atoms attached (it is quaternary: bonded to $C_1$, two methyls, and $C_3$). Thus, no alkene can form between $C_2$ and $C_3$.
  • $\beta_2$ Carbon ($C_4$): This carbon ($CH_2$) has 2 hydrogen atoms. Elimination here leads to the formation of a double bond between $C_3$ and $C_4$.
  • $\beta_3$ Carbon (Methyl on $C_3$): The methyl group attached to $C_3$ has 3 hydrogen atoms. Elimination here leads to the formation of a double bond between $C_3$ and the methyl substituent.

Step 3: Prediction of Alkenes

Based on the available $\beta$-hydrogens, two products are possible:

  1. Product A: 2,2,3-Trimethylpent-3-ene (via elimination of $H$ from $C_4$).
  2. Product B: 2,2,3-Trimethyl-3-methylenepentane (or 2,2-dimethyl-3-ethylbut-1-ene structure logic; specifically 2,2,3-trimethylpent-2-ene isomer).

Considering the structure $CH_3-C(CH_3)_2-C(CH_3)=CH-CH_3$ and $CH_3-C(CH_3)_2-C(=CH_2)-CH_2-CH_3$.

Step 4: Identification of the Major Product

[Per Saytzeff’s Rule], the major product in dehydrohalogenation is the more highly substituted alkene (the one with the most alkyl groups attached to the double-bonded carbons) because it is thermodynamically more stable.

Alkene Degree of Substitution
2,2,3-trimethylpent-3-ene Tri-substituted
2,2,3-trimethyl-3-methylenepentane Di-substituted

The tri-substituted alkene (2,2,3-trimethylpent-3-ene) possesses greater hyperconjugative stabilization compared to the di-substituted terminal alkene.

Final Solution: The alkenes formed are 2,2,3-trimethylpent-3-ene and 2,2,3-trimethyl-3-methylenepentane. The major product is 2,2,3-trimethylpent-3-ene.


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