What would be the product(s) when 2-iodobutane is heated with tertiary butoxide?

The following question from *Black Book Organic Chemistry IIT JEE Advanced Level Papers* [1]:



> 
> Choose the correct option(s) among the following about **[P]**:
> 
> 
> [![2-iodobutane + 2‐methylpropan‐2‐olate](https://i.stack.imgur.com/pmXSu.png)](https://i.stack.imgur.com/pmXSu.png)
> 
> 
> A) Two C−H bonds in **[P]** are involved in hyperconjugation.  
> 
> B) **[P]** cannot show stereoisomerism.  
> 
> C) Hydrogenation of **[P]** gives mainly *n*-butane.  
> 
> D) Monochlorination of **[P]** gives 3
> 
> 
> 


What is the product when 2-iodobutane is heated with tertiary butoxide? Is it 1-butene, 2-butene, or both?


Bulky bases generally favor the Hoffmann product, but my teacher told that in this specific case having much less hindrance Zaitsev product dominates. I am not able to completely digest it how this happened and is this correct or not.


Answer key: C, D.


### Reference


1. Ashish Mishra; Dr Ramu Petakamsetty. *Black Book Organic Chemistry IIT JEE Advanced Level Papers*; Blue Rose Publishers, **2022**. ISBN 978-93-5611-183-7.




As described in most introductory organic chemistry textbooks (my usual reference is Loudon), E2 elimination to form an alkene typically produces the more substituted alkene as the major product. This outcome is the Saytzeff/Zaitsev product (roman spelling varies depending on source).


The primary exception is if a sterically hindered ("bulky") base (the standard example is t-butoxide) is used, in which case the most accessible protons are removed, leading to a terminal alkene. Similarly, if the leaving group is quite large (trialkyl amines are typical examples), the terminal alkene is favored even for small bases. This outcome is the Hofmann product.


An exception to the exception is the case of large halide leaving groups. It has been observed that the percentage of 2-alkene (ie more substituted) product in eliminations using t-butoxide as base increase when one changes the leaving group from chloride to bromide and from bromide to iodide. For the 2-halobutanes specifically, the ratio of 1-butene to 2-butene (*cis* or *trans*) products is approximately 2:1 for 2-chlorobutane, 1:1 for 2-bromobutane and 1:2 for 2-iodobutane[1]. Competing explanations were put forth for these observations (sometimes referred to as the "Ingold-Brown controversy"), and I'm not aware of a consensus even today.


[1] Brown, HC and Klimisch, RL (1966) *J Amer Chem Soc* **88:** 1425


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