If 3-chloro 3 – methyl hexane is treated with $C_2{H}_{5}ONa/C_2{H}_{5}OH, E_2$ Elimination reaction takes place predominantly. How many isomeric alkenes would be formed?

Step 1. Substrate
2-chloro-3-methylpentane is treated with sodium ethoxide (C₂H₅ONa) in ethanol.

Step 2. Mechanism
E2 elimination requires:

• Strong base (EtO⁻).

• Anti-periplanar β-hydrogen relative to the leaving group (Cl).
  Hydrogen from a β-carbon is abstracted, Cl leaves, and a double bond forms.

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Step 3. Products
Three alkenes are formed:

1. 2-methylpent-2-ene (major product, more substituted alkene → Zaitsev product).

2. 3-methylpent-2-ene (possible due to β-H elimination from the other β-carbon).

3. 3-methylpent-1-ene (minor, less substituted → Hofmann product).

4.  

Step 4. Selectivity

• Sodium ethoxide favors Zaitsev’s rule.

• Major product = 2-methylpent-2-ene (most stable, internal alkene).

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Final Answer:
The reaction is an E2 elimination of 2-chloro-3-methylpentane with EtONa/EtOH, yielding mainly 2-methylpent-2-ene, with minor 3-methylpent-2-ene and 3-methylpent-1-ene.