Reagents & Products of Alkyne Reactions

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11 Terms

1
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Given: HCl, HBr, or I (1 eq)

  1. Keep same carbon chain or ring

  2. Change the triple bond to a double bond

  3. Attach Cl, Br, or I to the more substituted carbon

  4. Add H to the other carbon


2
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Given: HCl, HBr, or I (2 eq)

  1. Keep the same carbon chain or ring

  2. Change the triple bond to a single bond

  3. Draw two Cl, Br, or I atoms attached to the same carbon


3
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Given: Br2 or Cl2 (1 eq)

  1. Keep the same carbon chain

  2. Change the triple bond to a double bond

  3. Attach one Br or Cl to each carbon


4
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Given: Br2 or Cl2 (2 eq)

  1. Keep the same carbon chain or ring

  2. Change the triple bond to a single bond

  3. Draw two Br or Cl atoms on each of the original triple bond carbons


5
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Given: H2O, H+, Hg2+

  1. Keep the same carbon chain

  2. Replace the triple bond with a single bond

  3. Draw C=O on the more substituted carbon of the original triple bond


6
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Given: HgSO4, H2SO4, H2O

  1. Keep the same carbon chain

  2. Replace the triple bond with a single bond

  3. Draw C=O on the second carbon from the end


7
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Given: 1. BH3; 2. H2O2, OH–

  1. Keep the same carbon chain

  2. Change the triple bond to a single bond

  3. Draw C=O on the very last carbon of the original terminal alkyne

  4. The last carbon also has an H attached, making an aldehyde ( —CHO)


8
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Given: H2 with Pd, Pt, or Ni

  1. Keep the same carbon chain or ring

  2. Erase the two extra lines of the triple bond so only a single bond remains

  3. No other groups need to be drawn because the hydrogens are implied in the line structures


9
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Given: H2 with Lindlar Catalyst

  1. Keep the same carbon chain

  2. Change the triple bond to a double bond

  3. If the starting alkyne is an internal alkyne, draw the carbon groups attached to the double bond on the same side

Cis (Z) alkene

10
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Given: Na or Li with Liquid NH3

  1. Keep the same carbon chain

  2. Change the triple bond to a double bond

  3. For an internal acyclic alkyne, draw the carbon groups on opposite sides of the double bond

Trans (E) alkene

11
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Given: 1. NaNH2 2. R–X

  1. Keep the original carbon chain and triple bond

  2. Remove the H at the end of the triple bond

  3. Replace that H with the carbon group from R—X

  4. Do not include the halogen in the product

Result is a longer carbon chain