ORGANIC CHEMISTRY EXAM 2 TEXTBOOK SUMMARY NOTES
QUESTIONS AND ANSWERS
williamson ether synthesis - CORRECT ANSWER✅✅basic conditions, can make symmetric or
unsymmetrical ethers by an SN2 reaction between an alkoxide anion (R'O-) and an alkyl halide (Rx)
Williamson ether mechanism - CORRECT ANSWER✅✅Alkyl halide is attacked by alkoxide anion; halide
leaves and O- attaches to carbon where halide left.
condensation reaction to make ethers - CORRECT ANSWER✅✅under acidic conditions, symmetric
ethers can be formed from an alcohol
condensation mechanism to form an ether - CORRECT ANSWER✅✅Alcohol is protonated by H2SO4, in
a proton transfer step; then in an sn2 step, the OH2+ group leaves while another alcohol attacks the
carbon where the OH2+ leaves. The last step is a deprotonation step of where the alcohol attacked to
form the ether
PBr3 and PCl are used for - CORRECT ANSWER✅✅converting primary and secondary alcohols into alkyl
halides stereospecifically by back to back SN2 reactions. The configuration of the alkyl halide is opposite
that of the initial alcohol
alpha halogenation - CORRECT ANSWER✅✅can occur at an alpha carbon of aldehydes and ketones
under either acidic or basic conditions. happens when it is treated with a molecular halogen
alpha halogenation under acidic conditions - CORRECT ANSWER✅✅stops after a single halogenation
because it takes a lot of energy to do another and would form. a base (electron withdrawing effects
from the halide destabilize the positive charge so that's why it only happens once)
alpha halogenation under basic conditions - CORRECT ANSWER✅✅multiple halogenation occur
mechanism for alpha halogenation under basic conditions - CORRECT ANSWER✅✅step 1: the base
deprotonates the alpha carbon, producing a nucleophilic enolate anion
step 2: the enolate anion attacks the molecular halogen in an SN2 reaction (can happen twice for
polyhalogenation)
, mechanism for the halogenation of a ketone under acidic conditions - CORRECT ANSWER✅✅step 1: O
is protonated
step 2: alpha hydrogen is removed by -OAc
step 3: Cl2 breaks as one gets attacked by nucleophilic carbon where the H is removed and the other
halide is the leaving group
step 3L proton transfer to remove OH proton to make a ketone
epoxide opening under neutral/basic conditions - CORRECT ANSWER✅✅nucleophilic attacks the less
highly alkyl substituted carbon of the epoxide ring
epoxide opening under acidic conditions - CORRECT ANSWER✅✅protonation occurs first; then sn2 step
where the nucleophile attacks the more highly alkyl-substituted carbon of the ring because the
additional alkyl group is electron donating and helps stabilize the developing positive charge
epoxide production from a halohydrin (X-C-C-OH) - CORRECT ANSWER✅✅under basic conditions, the
hydroxyl group is deprotonated and the strongly nucleophile O- then displaces the halide leaving group
in an sn2 step
diazomethane - CORRECT ANSWER✅✅CH2N2
diazomethane uses - CORRECT ANSWER✅✅converts a carboxylic acid (RCO2H) into a methyl ester
(RCO2CH3)
diazomethane conversion of carboxylic acid to methyl ester mechanism - CORRECT ANSWER✅✅step 1;
the C atom of diazomethane is protonated by the acidic proton of carboxylic acid.
step 2: the carboxylate anion RCO2- is e- rich and acts as the nucleophile in the subsequent SN2
reaction. The protonated form of diazomethane (CH3) is electron poor and acts as the substrate. The LG
is N2
Hofmann product - CORRECT ANSWER✅✅the least highly substituted alkene product
QUESTIONS AND ANSWERS
williamson ether synthesis - CORRECT ANSWER✅✅basic conditions, can make symmetric or
unsymmetrical ethers by an SN2 reaction between an alkoxide anion (R'O-) and an alkyl halide (Rx)
Williamson ether mechanism - CORRECT ANSWER✅✅Alkyl halide is attacked by alkoxide anion; halide
leaves and O- attaches to carbon where halide left.
condensation reaction to make ethers - CORRECT ANSWER✅✅under acidic conditions, symmetric
ethers can be formed from an alcohol
condensation mechanism to form an ether - CORRECT ANSWER✅✅Alcohol is protonated by H2SO4, in
a proton transfer step; then in an sn2 step, the OH2+ group leaves while another alcohol attacks the
carbon where the OH2+ leaves. The last step is a deprotonation step of where the alcohol attacked to
form the ether
PBr3 and PCl are used for - CORRECT ANSWER✅✅converting primary and secondary alcohols into alkyl
halides stereospecifically by back to back SN2 reactions. The configuration of the alkyl halide is opposite
that of the initial alcohol
alpha halogenation - CORRECT ANSWER✅✅can occur at an alpha carbon of aldehydes and ketones
under either acidic or basic conditions. happens when it is treated with a molecular halogen
alpha halogenation under acidic conditions - CORRECT ANSWER✅✅stops after a single halogenation
because it takes a lot of energy to do another and would form. a base (electron withdrawing effects
from the halide destabilize the positive charge so that's why it only happens once)
alpha halogenation under basic conditions - CORRECT ANSWER✅✅multiple halogenation occur
mechanism for alpha halogenation under basic conditions - CORRECT ANSWER✅✅step 1: the base
deprotonates the alpha carbon, producing a nucleophilic enolate anion
step 2: the enolate anion attacks the molecular halogen in an SN2 reaction (can happen twice for
polyhalogenation)
, mechanism for the halogenation of a ketone under acidic conditions - CORRECT ANSWER✅✅step 1: O
is protonated
step 2: alpha hydrogen is removed by -OAc
step 3: Cl2 breaks as one gets attacked by nucleophilic carbon where the H is removed and the other
halide is the leaving group
step 3L proton transfer to remove OH proton to make a ketone
epoxide opening under neutral/basic conditions - CORRECT ANSWER✅✅nucleophilic attacks the less
highly alkyl substituted carbon of the epoxide ring
epoxide opening under acidic conditions - CORRECT ANSWER✅✅protonation occurs first; then sn2 step
where the nucleophile attacks the more highly alkyl-substituted carbon of the ring because the
additional alkyl group is electron donating and helps stabilize the developing positive charge
epoxide production from a halohydrin (X-C-C-OH) - CORRECT ANSWER✅✅under basic conditions, the
hydroxyl group is deprotonated and the strongly nucleophile O- then displaces the halide leaving group
in an sn2 step
diazomethane - CORRECT ANSWER✅✅CH2N2
diazomethane uses - CORRECT ANSWER✅✅converts a carboxylic acid (RCO2H) into a methyl ester
(RCO2CH3)
diazomethane conversion of carboxylic acid to methyl ester mechanism - CORRECT ANSWER✅✅step 1;
the C atom of diazomethane is protonated by the acidic proton of carboxylic acid.
step 2: the carboxylate anion RCO2- is e- rich and acts as the nucleophile in the subsequent SN2
reaction. The protonated form of diazomethane (CH3) is electron poor and acts as the substrate. The LG
is N2
Hofmann product - CORRECT ANSWER✅✅the least highly substituted alkene product
