Sodium acetylide, from deprotonation

Now let s draw the forward scheme. The 3° alcohol is converted to 2-methylpropene using strong acid. Anti-Markovnikov addition of HBr (with peroxides) produces l-bromo-2-methylpropane. Subsequent reaction with sodium acetylide (produced from the 1° alcohol by dehydration, bromination and double elimation/deprotonation as shown) produces 4-methyl-1-pentyne. Deprotonation with sodium amide followed by reaction with 1-bromopentane (made from the 2° alcohol by tosylation, elimination and anfi -Markovnikov addition) yields 2-methyl-4-decyne. Reduction using sodium in liquid ammonia produces the E alkene. Ozonolysis followed by treatment with dimethylsulfide produces an equimolar ratio of the two products, 3-methylbutanal and hexanal. 

Isomerization also results when sodium amide is used as the base in the double dehydrohalogenation. All possible triple-bond isomers are formed, but sodium amide is such a strong base that it deprotonates the terminal acetylene, removing it from the equilibrium. The acetylide ion becomes the favored product. When water is added to quench the reaction, the acetylide ion is protonated to give the terminal alkyne. 

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