# Disiamylborane on alkyl halides

I am having problem predicting the last product in the following reaction:

$$\ce{CH#CH ->[Na/NH3 (l)][(Excess)] (P) ->[DCl] (Q) ->[(i)Sia2BH][(ii)CH3COOH] (R)}$$

I am getting an alkyl chloride as product $$\ce{Q}$$ and cannot understand how it will react with $$\ce{Sia2BH}$$. Or, I think I am doing some mistake in the formation of product $$\ce{P}$$. I do know that alkynes on reaction with $$\ce{Na/NH3}$$ shall give trans alkene (Birch reduction). Will the "excess" part affect the reaction differently?

\begin{align} \ce{CH#CH &->[Na/NH3 (l)][(Excess)] CH2=CH2} &(\ce{P}) \tag{1} \\ \ce{CH2=CH2 &->[DCl] CH2D-CH2Cl} &(\ce{Q}) \tag{2} \\ \ce{CH2D-CH2Cl &->[(i)Sia2BH][(ii)CH3COOH] ?} &(\ce{Q}) \tag{3} \end{align}

• I think the DCl step is the work up of the sodium/NH3 reduction step and will give the deuterated alkene – Waylander Jan 12 at 8:01
• That's quite an unclear question. It's more common for workup after every step to be assumed / implicit, and based on that your interpretation would be correct. But @Waylander's interpretation is the only one that makes step 3 sensible. – orthocresol Jan 13 at 3:21
• so the product P will be NaC≡CNa, and Q will be DC≡CD, and then Q as CHD=CHD? – Gautam Jan 13 at 4:02
• I did read today that Acetylene reacts with sodium in liq Ammonia or Sodamide to form sodium acetylides. I guess, NaC≡CNa shall be formed and the end product will be what @Waylander said. – Gautam Jan 13 at 4:05