Reagents Chapter 17 Flashcards
(47 cards)
KMnO4 + Aromatic Alkane
Carboxylic acid. Will replace the most acid H, or the group on the benzene ring.
KMnO4 + Aldehyde
Carboxylic acid anion. The KMnO4 will replace the H with O-. A later addition of an acid would protonate.
KMnO4 + Alcohol
Carboxylic acid. =O replaces the hydrogens of the Alcohol carbon.
Ag2O + Aldehyde
Carboxylic acid anion.
Ag(NH3)2+ + Aldehyde
Carboxylic acid anion.
CrO3 or H2CrO4 + Aldehyde
Carboxylic acid.
CrO3 or H2CrO4 + Alcohol
Carboxylic acid.
K2Cr2O7 + Alcohol
Carboxylic acid.
Alcohol + Carboxylic Acid
Ester. The H of the Alcohol combines with the OH of the Carboxylic acid and leaves as a H2O
Alcohol + Acid Chloride
Ester. The H of the Alcohol combines with the Cl of the Acid Chloride and leaves
Alcohol + Anhydride
Ester and/or Carboxylic Acid. Carboxylic on side that keeps its Oxygen, add OR to side that doesn’t.
Alcohol + Ester
Different Ester. Different Alcohol (?)
Alcohol + Amide
No reaction.
Amine (R2NH) + Carboxylic Acid
Amide.
Amine (R2NH) + Acid Chloride
Amide.
Amine (R2NH) + Anhydride
Amide + Carboxylic Anion. Splits at the Oxygen
Amine (R2NH) + Ester
Amide
RMgX or RLi + CO2
Carboxylic Acid
RMgX or RLi + Carboxylic Acid
Carboxylic Acid Anion and RH
NO REACTION
RMgX or RLi + Acid Chloride
3* Alcohol
RMgX or RLi + Anhydride
3* Alcohol
RMgX or RLi + Ester
3* Alcohol. The OR of the Ester leaves and could be protonated
RMgX or RLi + Amide
RH and Amide Anion. (R will steal the H on the nitrogen)
LiAlH4 + Carboxylic Acid
1* Alcohol
