Question

Problem set for Competitive Nucleophiles

1) Why does water always enter the bottom of the condenser and exit from the top?

2) Why are boiling stones added to the solvent undergoing reflux?

3) Why are any large clumps of ammonium salts broken down into a powder?

4) Why is it necessary to perform the competing nucleophiles reactions under acidic conditions?

5) Which is the better nucleophile, fluoride or iodide ion? Explain this in the terms of the nature of the fluoride ion and iodide ion.

6) Explain why the product is washed with sodium bicarbonate solution (Procedure A). Support your answer with a chemical equation.

7) What is removed when the organic layer sits over anhydrous sodium sulfate (Procedure A)?

8) What is the purpose of reflux for the competitive nucleophiles experiment? Be explicit.

9) When the following optically active alcohol is treated with HBr, a racemic mixture of alkyl bromides is obtained. Draw a mechanism of this process, and explain the stereochemical outcome

Answer 1

1. As water will enter from the top it will reach the hottest vapours fast. Moreover this will allow the liquified sample to be collected from below. The mechanism acting here is the counter current mechanism and this will avoid all the thermal stresses.

2. Boiling stones are small insoluble stones with a porous structure. These pores will entrap air and will make theformation ofsolvent bubbles possible. This will ensure even boiling. In addition these will helpinavoiding bumping of the solution and unnecessary solvent losses.

3. If large lumps are present, the surface area for interaction of reactants will be low. Hence inorder to increase the surface area, the lumps are broken into powder.

4. Since thenucleophiles are competing in nature, they will have poor leaving group property. Inorder to make the leaving group a good one, acidic conditions can be used and this will lead to easier reaction.

5. In polar protic solvents, I^- is a better nucleophile than F^-. This is because F^- has small size and high charge density. As a esult it will be easily solvated. In case of I^-, it has large size and will be less solvated. Hence inorder to attack as a nucleophile I^- will have less solvent cages around it than F^-. Hence it will be the better nucleophile.

6.