Question

4. Here is a counter-intuitive reaction. Use curved-arrows and add the structure of the anionic intermediate to the square brackets. (Hint: model your answer on the later steps of the mechanism). N-N HO HKOH ең олон нн NEN heat H20 + H2O Notes: first reaction occurs in one step, solvent in the dashed box may act as a proton donor in steps 2 and 4, and also allows the reaction to be heated to a high temperature 5. a) Based on either your knowledge of basicity or acidity, explain why the third step of the reaction in question 4 seems unlikely. Why? b) List three features of the step 3 that will promote the formation of products, particularly when the reaction is run in an open flask.

Answer 1

In the third step of the reaction there is a formation of secondary

1 tt NG N r N Step-1 KOH, A N -dats probm apalshachon Step 3 Bource (B) protum aq KOH is a sbompy abshact piotem (actdic) base e-y + NEN (D) -t Skp 4 OH HO

carbanion a(D)from our knowledge we know that two degree carbanion is quite unstable that is the carbanion is a strong base hence the conjugate acid (C) is a less acidic and therefore, the proton abstraction is harder or quite unfavorable....

Stability order of carbanion - methyl anion> primary> secondary> tertiary carbanion....

The features of the step 3 that will promote the formation of product-

1. The formation of the very strong N-N triple bonds that is nitrogen gas

2. (D) is a very strong base and proton source from the supplied diol to form a stable 5 membered product also promotes the reaction

3. Formation of water in that step also promote the reaction....