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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 the Substrate

The reactant is 2,2,3-trimethyl-3-bromopentane. The molecular structure is a pentane chain with a bromine atom at $C_3$, two methyl groups at $C_2$, and one methyl group at $C_3$.

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

C1 C2 C3 C4 C5 CH3 CH3 Br CH3

Step 2: Identification of $\beta$-Hydrogens

Dehydrohalogenation via an $E2$ mechanism requires the abstraction of a $\beta$-hydrogen by the ethoxide ion ($C_2H_5O^-$). We analyze the adjacent carbons to the $C_3$ (bearing the bromine):

  • $\beta_1$ (at $C_2$): $C_2$ has no hydrogen atoms because it is bonded to four carbons (quaternary carbon). Therefore, no elimination can occur toward $C_2$.
  • $\beta_2$ (at $C_4$): $C_4$ is a $-CH_2-$ group. It contains two hydrogen atoms. Removal of one $H$ results in a double bond between $C_3$ and $C_4$.
  • $\beta_3$ (at the $C_3$ methyl group): The methyl group attached to $C_3$ contains three hydrogen atoms. Removal of one $H$ results in a double bond between $C_3$ and the methyl carbon.

Step 3: Prediction of Alkenes

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

  1. Product A: 2,2,3-trimethylpent-3-ene (Double bond between $C_3$ and $C_4$).
  2. Product B: 2,2,3-trimethyl-3-methylenepentane or 3,3,4,4-tetramethylpent-1-ene analogue (Correction: 2,3,3-trimethylpent-1-ene by abstraction from the $C_3$ methyl group).

Step 4: Determining the Major Product via Saytzeff’s Rule

According to [Saytzeff's Rule], the major product in an elimination reaction is the more substituted alkene (the one with the most alkyl groups attached to the double-bonded carbons).

Alkene Substituents on $C=C$ Stability
2,2,3-trimethylpent-3-ene Tri-substituted Higher
2,3,3-trimethylpent-1-ene Mono-substituted Lower

The tri-substituted alkene is more stable due to the inductive effect and hyperconjugation of the attached alkyl groups [Per the Stability of Alkenes series].

Final Solution: The alkenes formed are 2,2,3-trimethylpent-3-ene and 2,3,3-trimethylpent-1-ene. The major product is 2,2,3-trimethylpent-3-ene.


More Questions from Class 12 Chemistry Haloalkanes and Haloarenes Exercises


CBSE Solutions for Class 12 Chemistry Haloalkanes and Haloarenes


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