Exp 10. Nucleophilic Substitution Reactions PostLab Questions

1) You wish to substitute bromine in the following molecules with a nucleophile. Explain whether the given molecule would react by the S_N1 or S_N2 mechanism and explain why.

a) 1-bromo-methylcyclohexane

S_N1 - it is a tertiary alkyl halide, so it forms a stable carbocation

1) You wish to substitute bromine in the following molecules with a nucleophile. Explain whether the given molecule would react by the S_N1 or S_N2 mechanism and explain why.

b) 1-bromopropane

S_N2 - it is a primary alkyl halide, so there is little steric hindrance for backside attack

1) You wish to substitute bromine in the following molecules with a nucleophile. Explain whether the given molecule would react by the S_N1 or S_N2 mechanism and explain why.

c) 2-bromohexane

Either S_N1 or S_N2 - it is a secondary alkyl halide, so it can react by either mechanism. S_N1 forms a secondary carbocation, while S_N2 is possible because steric hindrance is moderate

2) Why does benzyl bromide react under both S_N1 and S_N2 conditions?

The benzyl carbocation is resonance stabilized, making S_N1 favorable. It also has low steric hindrance, so S_N2 can occur easily.

3) Why is bromobenzene unreactive under both S_N1 and S_N2 conditions?

Bromobenzene does not form a stable carbocation for S_N1, and the carbon bonded to bromide is sp² hybridized, preventing the backside attack needed for S_N2.

4) If bromocyclohexane reacts faster than cyclohexane in an S_N2 reaction, what could be the reason?

Bromine is a better leaving group than chlorine, so the C-Br bond breaks more easily, making the reaction faster.

5) Tertiary-butyl iodide reacts faster than t-butyl bromide via an S_N1 mechanism because iodide is a better leaving group than bromide. Why?

Iodide is larger and more stable after it leaves, so it forms more easily and speeds up the S_N1 reaction

6) To promote the S_N1 mechanism we used AgNO₃ in a polar, protic solvent. Why?

The polar, protic solvent stabilizes the carbocation, and Ag⁺ removes the halide by forming AgX, making carbocation formation easier.

7) The reaction of ClCH₂OCH₂CH₃ with ethanol (without catalyst) yields a di-ether in nearly quantitative yield. Provide the structure of the product. Mechanistically, provide two routes that account for the observed product.