MCAT Organic Chemistry Quickies 1

No

sp2

When the free radical is inihibited by molecular oxygen. This forms a less reactive alkyl peroxy radical.

Both leading to a racemic mixture (if unsymmetrical). The halogen can attack from bottom or top of the sp2 hybdrized carbon.

methyl < 1 < 2 < 3

• Bromination - favors tertiary
• Chlorination - less selective because its greatly exothermic

• Negative charge on the least electronegative atom (C- > N- > O- > X-)
• Carbon is greatest nucleophile because it is not electronegative enough to hold the extra pair of electrons (-) as oxygen can so it is a highly reactive nucleophile

• Nucleophiles - have unshared pairs of electrons, partial negative, electron pair donors, lewis base/ligand
• N strength increases as negative charge increases
• N strength increases going down the period (F<Cl<Br<I)
• N strength increases going left in the period (OH < NH < C)
• basically, the less electronegative and more negative the charge, the stronger the nucleophile

Acids

• reactivity: CH3 > 1 > 2 >3
• stereochem: Inversion of stereochemistr due to backside attack
• kinetics: rate = k [nucleophile][electrophile] - second order
• solvent: polar aprotic
• rearrangement: no rearrangements b/c no carbocation formatoin
• favored conditions: strong, non-bulky nucleophile

• reactivity: 3 > 2 > 1 > methyhl
• sterochem: racemate
• kinetics: rate: k[electrophile] - carbocation formation
• solvent: polar protic - stabilize carbocation
• rerrangements: possible to make more stable carbocation
• favored conditions: non-basic, weak nucleophiles

No, becuase it does not have H attached to NOF atoms, but it can H bond with water and other atoms with H

sp2 because it can participate in resonance with the adjacent carbonyl group

• formation of carbocation b/c leaving group leaves first
• need beta hydrogens. if no adjacent hydrogens are available, eliminaiton cannot occur
• reactivity: 3 > 2 >>>> 1
• stereochemistry: most substituted favored. trans over cis
• kinetics: rate: k[haloalkane]
• solvent: protic
• rearrangemenst: possible!
• favoring conditions: weak base and high temp

• stereochem: anticoplanar or trans diaxial
• reactivity: 3 > 2 > 1
• reactoin rate: k[base][substrate]
• solvent: polar aprotic
• arrangements: nope
• favoring conditions: strong bulky bases, high temp

• The more stable the conjugate bas
• electron withdrawing groups and the ability to resonate stabilize the conjugate base
• electron donating groups would destabilize = stronger base, weak acid

• usually with pi bonds
• beaware of enantiomer formations and other stereochem such as hindrance
• markovnikov and anti

forms an alcohol

• hydroboration (1. BH3/B2H6, 2. H2O2)
• HBr addition with peroxides

Syn addition of diol

• 1) H2 with Ni, Pd, Pt = syn addition of H2
• 2) Alyne + lindlar catalyst with Pd = syn addition of H-H across pi bond resulting in cis - stops hydrogenation reduction of alkyne at alkene stage
• 3) Na, NH3 + alykne = trans alkene. does not reduce alkenes

Double bond to carbonyl gruop (ketone or aldehyde possible)

• 4n+2 = number of pi electrons
• n = whole number

• Ring activators
• Donate a pair of electrons
• R groups
• Halogens are O,P but deactivating

• Deactivators
• - NR3, N02, CN
• - things that pull electrons

keto

• need a hydrogen to remove. so if you have a tertiary alcohol, you cannot oxidize it
• some oxidizing agents include chromates, conc nitritic acid, oxygen

• requires alpha hydrogens
• results in alpha beta unsaturated ketone where the C=C bond is between the alpha and beta carbons
• the c=c is more electrophilic, so nucleophiles will attack c=c first then resonate to form enol which can tautoremize to a more stable keto form - this is called michael addition

carbonyl to alcohol even with good leaving group

• 
• remember, the oxygen and double bond can undergo keto-enol tauto (keto is favored)

thermodyanamic is most stable

Those that have carbonyl groups beta to the carboxylate

Ester

• Solubility
• ions tend to dissolve in the aq layer. As a result, you use acid bases
• NaOH would deprotenate a weak acid
• HCO3 would deprotenate a strong acid
• This allows one to separate a strong acid from strong base

Polar - allows nonpolar materials to travel up further while polar substancs would stick to the plate

Fractional allows for the separation of substances with similar boiling poitns

Conjugated systems

2 COOH and NH3 groups

The point where the amino acid acts as a zwitterion which has a net neutral charge. The pI is also the equivalence point.

• Detects carbonyl units of sugars
• Hemiacetals = positive test (plain glucose)
• Glycogen = weakly positive (only the end points are hemiacetals)
• Benedicts solution is an oxidized form of copper - this gets reduced by the sugars (Cu2+ to Cu+) - this oxidizes the sugar in return.
• The anomeric carbon in glucose gets oxidized.

• Alpha - OH pointing down
• Beta - OH pointing up
• Mutorotation = conversion b/w both (has to go through the chain formula)

phosphodiester

• gem - 2 X or 2 OH on same carbon
• vic - 2 X or 2 OH on adjacent carbons (nipple piercings