Question

3. (Solubility and equilibrium) Explain the following changes in equilibrium observed in experiments The explanation must consider all compounds and all of major factors. (5 pts CH CI + CHONa CHOCHE in water nearly 100% yield in hexane nearly 0 yield 4. Basicity and equilibrium) Carralete the following reaction you must show more

Answer 1

Protic solvents contain hydrogen bonds, thus O-H or N-H bonds. Examples are water, ethanol (alcohols in general), acids and primary or secondary amines (tertiary amine contains no N-H bond).

Aprotic solvents contain no hydrogen bonds, but it still needs to be polar for substitution or elimination reactions. Because the transition state is polar at least and possibly also the starting substance and/ or product, a nonpolar solvent like hexane won’t produce any reaction like this as the transition state isn’t stabilized. Examples of polar aprotic solvents are dimethylformimide (DMF), dimethylsulfuroxide (DMSO), ethylacetate and dichloromethane. (These molecules contain halides or oxygen to make it polar in general but no O-H or N-H bonds).

In Sn2 reactions (or E2), the nucleophile needs to react with the starting substance and this is the rate determining step. If the solvent is protic then it will stabilize the nucleophile so much that it basically won’t attack. Thus an aprotic solvent is needed, but remember that it still needs to be polar to stabilize the stransition state.

In Sn1 reactions (or E1), the rate determining step is forming the carbocation. It doesn’t matter if the nucleophile is stabilized by a protic solvent as the carbocation attracts the nucleophile enough to still make it attack (the carbocation has a formal charge while the starting substance is a bit polar at most, in general). Thus, it doesn’t matter if the solvent is protic or aprotic (but always polar).