# Iodine Mapping

Chemistry Level 3

$$\ce{Propyne}$$ is reacted in the following way:

• 1) $$\ce{NaNH_2}$$
• 2) $$\ce{BrCH_3}$$
• 3) 2 moles of $$\ce{HI}$$

The major product will be $$\ce{\underline{_},\underline{_}~-~diiodo~butane}$$. Fill in the blanked positions of the iodines.