Ch 9 Alkynes

“-yne”

suffix for 1 triple bond

“-adiyne”

suffix for two triple bonds

“-atriyne”

suffix for three triple bonds

priority

name ending (functional group) depends on

carbonyl> -OH (-ol)> -NH2 > double bond (-en) = triple bond unless at same locant

highest to lowest priority functional groups

Naming rules

keep numbers low. Alkenes and alkynes have same priority, unless they have same locant, then double bond is higher priority than triple

Forward (towards products)

higher pKa in the products then the reaction moves

Backwards (towards reactants)

higher pKa in the reactants then the reaction moves

poison (Pd w/ CaCO₃ + PbO₂) can take an alkyne to an alkene. Hydrogenation. H₂/Lindlar

Lindlar’s catalyst is

strong base. Good at removing a proton from a terminal alkyne, also good at executing elimination reactions. Makes an alkyne with 2 eq. (done three times) another 1 eq puts the Na ion on the terminal end of the alkyne

NaNH₂ and NaH

Hydrogenation. Hydrogen added to triple bonds to yield a trans alkene (Na/NH₃), cis alkene (H₂/Lindlar), or an alkane/no double or triple bond (H₂/Pd, Pt, or Ni)

Na/NH₃, H₂/Lindlar catalyst, H₂/Pd, Pt, or Ni

hydrohalogenation. Mark addition of H and Br in the anti-addition (opposite sides of the double bond). Trans alkene. Pi complex

HBr/(1 eq)

hydrohalogenation. Mark addition of H and Br twice.

HBr/(2 eq)

hydrohalogenation. Anti-Mark addition of H and Br (H goes on least substituted, Br on most substituted). (E) and (Z) result. 2 Products (looks like cis and trans but is E and Z). Free radical Mechanism. 

HBr/ROOR

Hydration. Acid catalyzed. Mark addition of OH and H. H₂O adds, then the enol undergoes tautomerization to the carbonyl.

H₂SO₄/H₂O, HgSO₄

Hydration. Hydroboration/oxidation. Anti- Mark addition of OH and H. H₂O adds syn, then the enol undergoes tautomerization to the carbonyl.

1.R₂BH (or 9-BBN) / 2. H₂O₂, OH^-

Halogenation. Anti addition. Makes a trans alkene. Pi complex, first addition.

Br₂/(1 eq)

Addition of Br₂, twice. Final product yields compound with 4 Brs attached. pi complex, first add; bridged intermediate, bromonium ion, 2^nd addition

Br₂/(2 eq.)

O3 followed with reduction. keep C-H bonds. O3 cleaves the double bond. Count carbons carefully. Yields carbonyl and a 2H-C=O

1.O₃ / 2. DMS or Zn/H₂O