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Q6.14(vi):

Write the structure of the major organic product in each of the following reactions: 

Solution :

Initial Reaction Parameters

The given reaction involves 2-bromobutane ($CH_3CH_2CH(Br)CH_3$) reacting with sodium ethoxide ($NaOCH_2CH_3$) in ethanol ($CH_3CH_2OH$). This is a classic dehydrohalogenation reaction occurring via an $E2$ mechanism [Per the Mechanism of Elimination Reactions].

Step 1: Identification of Reactants and Mechanism

  • Substrate: 2-bromobutane is a secondary alkyl halide.
  • Reagent: Sodium ethoxide ($NaOCH_2CH_3$) is a strong base and a strong nucleophile.
  • Conditions: In the presence of a strong base and heat, secondary alkyl halides undergo $E2$ (Bimolecular Elimination) reaction [Reference: Zaitsev's Rule].

Step 2: Applying Zaitsev's Rule

Zaitsev's Rule states that in dehydrohalogenation reactions, the alkene formed in the greatest amount is the one that corresponds to the removal of the hydrogen from the $\beta$-carbon having the fewest number of hydrogen atoms [Theoretical Basis: Thermodynamic stability of substituted alkenes].

Pathway Beta Carbon Product Classification
Path A (Removal of H from $C_1$) $CH_3$ group But-1-ene Monosubstituted (Minor)
Path B (Removal of H from $C_3$) $CH_2$ group But-2-ene Disubstituted (Major)

Step 3: Mechanism Visualization

CH3 CH2 CH(Br) CH3 Major Product: CH3-CH=CH-CH3

Step 4: Final Reaction Equation

The elimination of $HBr$ from the $C_2$ and $C_3$ positions leads to the formation of the more substituted internal alkene, but-2-ene, which exists as a mixture of cis and trans isomers [Per Stereochemical considerations of E2 mechanism].

$CH_3CH_2CH(Br)CH_3 + C_2H_5ONa \xrightarrow{C_2H_5OH, \Delta} CH_3CH=CHCH_3 + C_2H_5OH + NaBr$

Final Solution: The major organic product is but-2-ene ($CH_3CH=CHCH_3$).


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