What would be the product when neopentyl chloride reacts with sodium ethoxide

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2-Methyl-2-butanol

Neo pentyl alcohol

Both 1 and 2

2-Methyl-2-butene

answer is D.

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Detailed Solution

CH3 CH3 − C CH3 C − CH2 − Cl

NaOH, H2O E1 mechanism
heat

⟶

CH3 CH3 − C CH3 C − CH2⊕

Carbocation rearrangement:

CH3 CH3 − C⊕ CH3 C = CH − CH3

Note: The tertiary carbocation rearranges to form a more stable structure through a 1,2-methyl shift.

Elimination:

CH3 − C = CH − CH3

Step 1: Ionization (Rate-determining step)

Step 2: Carbocation Rearrangement

The initially formed primary carbocation is unstable
A methyl group (CH3) shifts from the adjacent tertiary carbon to the carbocation center
This is a 1,2-hydride or alkyl shift
Results in a more stable tertiary carbocation with a double bond: (CH3)2C+-CH=CH3

Step 3: Elimination (Deprotonation)

Key Points:

E1 = Elimination Unimolecular (first-order kinetics)
Carbocation rearrangements are common in E1 mechanisms when a more stable carbocation can form
The reaction proceeds through a carbocation intermediate
Heat and base promote the elimination reaction
The mechanism shows formation of a conjugated system for added stability

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