Question

Question #905a0

Answer 1

D) leaving group's dissociation rate from the molecule

Explanation:

I'll explain the possible choices and why they are important or not.

A) The nature of the leaving group is important. This is the substrate of the reaction, which is generally a haloalkane or an alkyl tosylate. Methyl and primary substrates will go through `S_"N"2` reactions. Tertiary substrates do not go through `S_"N"2` reactions.

B) The nature of the incoming group is important. This is the nucleophile of the reaction. Poor nucleophiles (e.g. water) cannot go through `S_"N"`2 reactions. Strong, unhindered bases (e.g. methoxy, `CH_"3"O^-`) go through `S_"N"2` or E2 reactions.

C) Size of groups other than leaving or and incoming groups is important. Generally speaking, if you have hindered molecules (mainly tertiary substrates and nucleophiles), you will generally see elimination reactions rather than substitution reactions. Anything highly branched may not go through substitution reactions. Remember, this is dependent on the substrate and/or nucleophile.

D) Leaving group's dissociation rate from the molecule is NOT important. The rate determining step in `S_"N"2` reactions depends on both the incoming and leaving groups. This will not change the nucleophilicity of the incoming or leaving groups, so the reaction cannot suddenly change from say `S_"N"`2 to E2.

E) The nature of the solvent is important. The solvent can react with the substrate (i.e. hydrolysis or solvolysis). Water, as stated in B, will not go through `S_"N"`2 reactions. Polar protic solvents tend to go through `S_"N"`1 or reactions, especially with good leaving groups. If fluoride (`F^-`) is the leaving group, it will not leave in polar protic solvents; however, it can leave in polar aprotic solvents.