Calculate percentage of SN1 product if (R)-2-chloro butane on reaction with NaOH/H2O and acetone gives 98 per cent inverted product.Correct answer is '4'. Can you explain this answer?

Question

Answer

Answer:

Given:
- Compound: (R)-2-chloro butane
- Reactants: NaOH/H2O and acetone
- Product: Inverted product
- Percentage of SN1 product: 98%

To find:
- Percentage of SN1 product

Solution:

Explanation:
- The given reaction is an SN1 reaction which involves the formation of a carbocation intermediate.
- In the reaction of (R)-2-chloro butane with NaOH/H2O and acetone, the chloride ion leaves to form a carbocation intermediate.
- The intermediate then reacts with the nucleophile (OH-) to give an inverted product.
- In an SN1 reaction, the stereochemistry of the starting material is lost during the formation of the carbocation intermediate, resulting in a racemic mixture of products.
- Therefore, if the product is 98% inverted product, then the percentage of SN1 product would be the remaining 2%.
- This means that only 2% of the product is formed via the SN1 mechanism, while the remaining 98% is formed via the SN2 mechanism.

Final Answer:
- Percentage of SN1 product = 2%

Answer

The SN1 mechanism is a two step reaction, in the first step LG leaves the site and creates a carbocation on which nucleophile attack from both (Front and Back) side, and the product formed is Racemic mixture i.e. 50% retention and 50% inversion.

if 98% that means 2% product are racemized will contain 2% retention and 2% inversion product. Hence the percentage of SN1 product will be 2+2 =4%.

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