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MCAT - Organic Chem

MCAT Organic Chemistry

QuestionAnswer
KMnO4 (cold, dilute, basic) -OH groups added to each side of double bond.
KMnO4 (hot, basic + acid wash) Non-terminal alkenes: Two carboxylic acid. Terminal alkenes: Carboxylic acid and carbon dioxide.
Ozonolysis (O3) in presence of double bond Monosubstituted alkene: two aldehydes Disubstituted alkene: ketone
Ozonolysis vs. KMnO4 Ozonolysis is more selective than KmnO4. While ozonolysis still cleaves the double bond in half, it only oxidizes the carbon to an aldehyde (or ketone if the starting molecule is disubstituted).
Ozonolysis + NaBH4 in presence of double bond Aldehyde or ketone products will be reduced with NaBH4 to alcohol. NaBH4 is a mild reducing agent.
What's special about terminal alkynes Terminal alkynes are fairly acidic. They can stabilize a negative charge fairly well.
NaNH2 and n-BuLi Strong Bases Can be used to add a terminal alkyne (which is acidic) to a new carbon skeleton.
Two methods to synthesize alkynes High temperature and strong base (not always practical). Add a terminal alkyne to an existing carbon skeleton in presence of a strong base.
Hydroboration B2H3 (often written as BH3) adds to a double bond to a double bond. It produces an alcohol. It follows anti-Markovnikov and syn orientation.
Two ways to reduce alkynes to alkenes Method 1: Lindar's catalyst (H2, Pd/BaSO4 with quinoline). Method 2: Sodium in liquid amonia (Na, NH3).
Lindar's Catalyst + Alkyne H2, Pd/BaSO4 with quinoline. Used to reduce alkynes. It stops at alkene stage. Cis-isomer.
Sodium in liquid amonia (Na, NH3) + Alkyne reduces the alkyne to alkene. Trans-isomer.
Two ways alkynes can be oxidized With Ozone and KMnO4
Difference between ozonolysis of alkenes and alkynes Alkenes: two aldehyde or ketones are produced Alkynes: two carboxylic acids are produced
Halogenation of aromatic rings Aromatic rings react with bromine or chlorine (not fluorine or iodine) in presence of a Lewis acid (FeCl3, FeBr3, AlCl3, AlFe3) to produce monosubstituted products. Fluorine produces multisubstituted products.
Benzene + Br2 vs. Alkene + Br2 Benzene = monosubstituted (Benzene-Br) Alkene = multisubstituted (Br-Alkene-Br)
Benzene + SO3/H2SO4 + Heat Benzene-SO3H
Benzene H2SO4 + HNO3 Benzene-N2O
TLC - the sample travels fartehr when _____. when it is similar to the solvent.
TLC - Generally, the more polar the compound, the (faster/slower?) it will travel. Slower. Only true if the solvent is not non-polar.
Gas chromatography separates compounds based on _____. based on their volatility, or boiling point. The higher the volatility (lower boiling point), the shorter the retention time. The lower the volatility (higher boiling point), the longer the retention time.
Ester + base (ex. RCOOR' + KOH) (Kaplan p. 183) Saponification: RCOO- +R'OH
Triaclyglycerol + NaOH (Kaplan p. 183) Saponification: Soap + Glycerol
2 Aldehydes + NaOH (Kaplan p. 144) Aldol (There is an alcohol group)
2 Aldehydes + NaOH + Heat (Kaplan p. 144) Aldol condensation reaction. The product is an aldehyde. The alcohol functional group is replaced with a double bond.
2 Esters + Base (Kaplan p. 184) Claisen condensation. This is the same as aldol condensation, but instead of two aldehyde we have two esters. The product attaches combines two esters (removes one middle oxygen).
Which one is nucleophile/electrophile/leaving group in Ethanoyl chloride + H2O (Kaplan p. 174) Nucleophilic acyl substitution: Water is nucleophile, the carbonyl carbon is electrophile, and Cl is the leaving group.
Organometallic Reagents (PR p. 181) Used to perform nucleophilic addition to a carbonyl carbon. An example would be grignard reagent: RMgBr. R has a partial negative charge (R- Mg+ Br-)
Organolithium: Ch3I + 2Li (PR p. 181) Ch3I + 2Li --> Ch3Li + LiI
Oxidizing Agent: Na2Cr2O7 Fully oxidizes primary and secondary alcohols. Primary alcohol becomes carboxylic acid and secondary alcohol becomes ketone.
Oxidizing Agent: PCC mild oxidizing agent.
Oxidizing Agent: CrO3 Strong oxidizing agent. Stronger than Na2Cr2O7.
Created by: salehjoon