Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

When we look at the given reaction, we are presented with a tertiary alkyl halide, specifically ethyl 2-chloro-2-methylbutanoate. This substrate is reacting with sodium ethoxide ($\text{NaOEt}$) in the presence of heat ($\Delta$).

The first critical decision we must make is determining the reaction pathway. Sodium ethoxide is a strong base and a good nucleophile. However, the substrate is a sterically hindered tertiary halide, which makes the S${}_{\text{N}}$2 substitution pathway virtually impossible. Furthermore, the presence of heat is a classic indicator that an elimination reaction will dominate. Therefore, we are dealing with an E2 elimination mechanism.

In an E2 elimination, the strong base must abstract a proton from a $\beta$-carbon (a carbon adjacent to the $\alpha$-carbon bearing the leaving group). Let's map out our options by locating all the $\beta$-carbons in our substrate.

The central $\alpha$-carbon is bonded to three other carbon groups: 1. A methyl group ($-{\text{CH}}_3$) 2. An ethyl group ($-{\text{CH}}_2{\text{CH}}_3$) 3. An ester group ($-{\text{COOCH}}_2{\text{CH}}_3$)

The ester group's adjacent carbon is a carbonyl carbon, which possesses no $\beta$-hydrogens. This leaves us with two viable pathways for elimination:

Pathway 1: The ethoxide ion abstracts a proton from the methyl group. The electrons collapse to form a double bond, and the chloride ion leaves. This results in a double bond at the terminal position, yielding the less substituted alkene: ${\text{CH}}_2=\text{C}({\text{CH}}_2{\text{CH}}_3){\text{COOCH}}_2{\text{CH}}_3$. This is known as the Hofmann product.

Pathway 2: The ethoxide ion abstracts a proton from the $-{\text{CH}}_2-$ of the ethyl group. This leads to a double bond forming inside the carbon chain, yielding a more substituted alkene: ${\text{CH}}_3\text{CH}=\text{C}({\text{CH}}_3){\text{COOCH}}_2{\text{CH}}_3$. This is known as the Saytzeff product.

Now we must decide which of these two competing products will be the major one. This is where Saytzeff's Rule comes into play.

Saytzeff's rule states that in an elimination reaction, the most highly substituted alkene will be the major product because it is thermodynamically more stable. The stability arises from hyperconjugation; more alkyl groups attached to the double bond mean more $\sigma$-bonds can overlap with the empty ${\pi }^{\ast }$-orbital of the alkene, lowering the overall energy of the molecule.

Comparing our two products, the internal alkene (Pathway 2) is trisubstituted, while the terminal alkene (Pathway 1) is only disubstituted. Therefore, the internal alkene is significantly more stable and will be formed as the major product.

Final Answer: The major product is ${\text{CH}}_3\text{CH}=\text{C}({\text{CH}}_3){\text{COOCH}}_2{\text{CH}}_3$, which corresponds to option (b).