[couldidothis]

Maybe I've been studying too long today, but when I try to make this into a terminal alkyne, I am itchy to use a "bulky" base to make it go Hoffman, forming the pi bonds toward the less substituted terminal end. The solutions manual's reagent (NaNH₃) seems like it cause the pi bonds to form toward the more substituted carbon, in Zaitsev, making 2-pentyne. Am I just having a mental block here, or is there a fact I'm missing, like when making an alkyne from a alkyl dihalide, it goes Hoffman-esque even without using a bulky base like t-buO^-?

Ohhh, is H₂N^- a bulky base?

[Extracted]

Maybe I've been studying too long today, but when I try to make this into a terminal alkyne, I am itchy to use a "bulky" base to make it go Hoffman, forming the pi bonds toward the less substituted terminal end. The solutions manual's reagent (NaNH₃) seems like it cause the pi bonds to form toward the more substituted carbon, in Zaitsev, making 2-pentyne. Am I just having a mental block here, or is there a fact I'm missing, like when making an alkyne from a alkyl dihalide, it goes Hoffman-esque even without using a bulky base like t-buO^-?

Ohhh, is H₂N^- a bulky base?

Are you sure it's not 1,2-dibromopentane?

[couldidothis]

Thank you for responding! Yes, I am sure...that would be better, wouldn't it? I do see I made an error in typing above-- the reagent per the sol. manual is NaNH2, of course, not 3.

[orgopete]

Just an opinion about Zaitsev vs Hoffman eliminations. I am not so convinced it is as straight forward as described. I find it somewhat difficult to believe the non-bonded electrons of KO-t-Bu cannot reach a secondary proton and therefore attack at the primary methyl groups. For example, 1-chlorododecane and t-butoxide gives 86% elimination to the alkene and 14% substitution. I find it preferable to argue from an acidity basis whether one should find the Zaitsev or Hoffman product.

If you have a bromide or iodide in an alcoholic solvent, the Zaitsev product usually predominates. If you replace the bromide with an chloride or fluoride, you can increase the amount of Hoffman product. If the leaving group is decidedly less basic, an ammonium salt, the product will be the Hoffman product. I interpret this that the C-X bond also has an effect on the product. I think if the electrons are pulled away from the carbon of the leaving group, then electrons of a neighboring carbon can approach that carbon. You may think of it like hyperconjugation. The carbons best able to donate electrons are the more electron rich or more substituted. This increases the acidity of a secondary or tertiary hydrogen. Elimination of it will give a Zaitsev product. Solvents that enable bond breakage, favor Zaitsev elimination reactions. These are the protic solvents like ethanol/ethoxide mixtures.

If bond breakage is reduced, that is if the bond to the leaving group is as not stretched, then deprotonation will occur from the most acidic hydrogens, in this case a methyl group. A base like NaNH2 is far more basic and acidity is likely a greater factor in the elimination than an early or partial bond breakage. If so, then it should attack the most acidic hydrogens, the CH3 group.

[couldidothis]

Wow, you are so great at explaining this! You should write a book...maybe some kind of Guide to Organic Chemistry Mechanisms?

Seriously, thanks so much, Pete. Wish you were my teacher.