OCHEM exam 2 Questions Answered Correctly Latest Update 2025-2026
ethers vs alcohols - Answers ethers do not contain an OH bond and have a R functional group
on both sides that is either on alkyl or an aryl. alchohols have an OH bond and can hydrogen
bond.
ethers have a much lower boiling point and are not reactive which means they can be a good
solvent. When ethers are reactive, they are cyclic due to ring strain such as epoxides
ethers from alcohols with alkenes (C=O double bond) 2Ps - Answers reagent: H+, alcohol
attacks the carbocation then deprotanates, forms a tertiary ether
ethers from primary alcohols 2Ps - Answers reagents: H2SO4, heat (acidic conditions, could be
H+), forms a symmetrical ether. Only works for symmetrical and only works with primary
alcohols becaus there would be mixtures
formation of unsymmetrical ethers from alkoxides (R-O negatively charged with a positive Na
metal on it or OH deprotantated from NA) and primary alkyl ahlides (williamson ether synthesis
mechanism - Answers reagent: alkyl halide and a strong base such as Na (starts as alcohol and
forms alkoxide)
first step: alkoxide forms from alcohol by being deprotanated by NaH, then H2 is realeased into
gas.
second step: The O then attacks the alkyl halide forcing the halide to leave and forming an ether,
then the Na and halide bond
formation of thiols from alkyl halides by SN2, then formation of thioethers fro alkyl thiolate and
primary/secondary alkyl halides mechanism - Answers SH- nucleophile breaks off a metallic and
attacks an alkyl halide to form a thiol (R-SH)
Then a strong base (NaOH) deprotanates the thiol to form a R-S (thiolate).
Thiolate attacks alkyl halide and breaks off the halide to form a thioether.
Since this is in two steps, it can be asymetrical since the thoiether can be formed from different
alkyl halides
Also, the alkyl halides can be primary or secondary since sulfur is a better nucleophile and a
worse base and therefore can undergo SN2 faster than E2.
formation of alkyl halides and alcohols (intermediate) from ethers and mineral acids (reverse
reaction from before) mechanism - Answers starting reactant: ether
reagents: excess mineral acids such as H-Br, H-I, H-Cl and heat
2 part mechanism
, first part: SN2, H-Br seperates, oxygen attacks Hydrogen forming an OH bond on the epoxide.
Then, bromine attacks the less stable (primary) R group and the R group leaves. In this reaction
a Br-R is formed and a more stable (tertiary) R group with an alcohol stays.
second part: SN1, oxygen attacks the hydrogen agian forming a H2O, then the water leaves
forming a carbocation and the Br- attacks forming a tertiary alkyl halide and water as the
byproduct.
This reaction does not stop in the middle at alcohol and will proceed to the end. An alcohol can
be formed in the reaction and stop as a final product if one of the R groups is sp^2 such as
benzene, then it stops because it can't undergo SN1 or SN2
epoxidation of alkenes with peracids, stereochemistry!! 2P - Answers reactant: alkene
reagent: mcPBA
when the alkene is cis, the epoxide is syn (both wedge or both dash)
when the alkene is trans the epxoide is also trans/anti (one wedge and one dash, there is a 1:1
ratio of enantiomers)
vicinal halohydrins stereochemistry - Answers if alkene and a halogen and hydrin reacts (Br2,
H2O) then the stereochemistry is anti, keep in mind however, what side it is on for cis and trans
because they are diastereomers of each other. both enantiomers are good 1:1
formation of epoxides from vicinal halohydrins - Answers reagent: strong base
strong base deprotanates the oxygen, negatively charged oxygen bond then attacks Br and
makes it leave. The Br and the O need to be anti, so watch for stereochemistry and where the
methyl groups are located. REMEMBER TO SWITCH TO ANTI if it looks like cis, one must be on
a wedge while the other is on a dash
strong nucleophiles - Answers organometallics (R-MgBr), CN, N3, OR, SR, LiAlH4
epoxide ring opening with strong nucleophile - Answers R-strong nucleophile such as MgBr
attacks the least hindered side of the ring and OH opens
second step: H3O workup step to donate a hydrogen to the negative oxygen atom the final
product is R-OH-R'. Remember that the functional group stays in the molecule after reaction.
Works for all epoxides . The stereochemistry for the OH group and the attacked group are
inverted to if it is attacked by a weak nucleophile
acid-catalyzed ring opening for tri substituted epoxide - Answers reagents: H2SO4 + R group.
OH bond is formed and dipoles are formed that break the ring. The side without the OH bond
after the ring opening is where the R group goes on to, remember that the stereochemistry is
anti for the R group and OH group are anti and the stereochemistry is inverted if it was for the
ethers vs alcohols - Answers ethers do not contain an OH bond and have a R functional group
on both sides that is either on alkyl or an aryl. alchohols have an OH bond and can hydrogen
bond.
ethers have a much lower boiling point and are not reactive which means they can be a good
solvent. When ethers are reactive, they are cyclic due to ring strain such as epoxides
ethers from alcohols with alkenes (C=O double bond) 2Ps - Answers reagent: H+, alcohol
attacks the carbocation then deprotanates, forms a tertiary ether
ethers from primary alcohols 2Ps - Answers reagents: H2SO4, heat (acidic conditions, could be
H+), forms a symmetrical ether. Only works for symmetrical and only works with primary
alcohols becaus there would be mixtures
formation of unsymmetrical ethers from alkoxides (R-O negatively charged with a positive Na
metal on it or OH deprotantated from NA) and primary alkyl ahlides (williamson ether synthesis
mechanism - Answers reagent: alkyl halide and a strong base such as Na (starts as alcohol and
forms alkoxide)
first step: alkoxide forms from alcohol by being deprotanated by NaH, then H2 is realeased into
gas.
second step: The O then attacks the alkyl halide forcing the halide to leave and forming an ether,
then the Na and halide bond
formation of thiols from alkyl halides by SN2, then formation of thioethers fro alkyl thiolate and
primary/secondary alkyl halides mechanism - Answers SH- nucleophile breaks off a metallic and
attacks an alkyl halide to form a thiol (R-SH)
Then a strong base (NaOH) deprotanates the thiol to form a R-S (thiolate).
Thiolate attacks alkyl halide and breaks off the halide to form a thioether.
Since this is in two steps, it can be asymetrical since the thoiether can be formed from different
alkyl halides
Also, the alkyl halides can be primary or secondary since sulfur is a better nucleophile and a
worse base and therefore can undergo SN2 faster than E2.
formation of alkyl halides and alcohols (intermediate) from ethers and mineral acids (reverse
reaction from before) mechanism - Answers starting reactant: ether
reagents: excess mineral acids such as H-Br, H-I, H-Cl and heat
2 part mechanism
, first part: SN2, H-Br seperates, oxygen attacks Hydrogen forming an OH bond on the epoxide.
Then, bromine attacks the less stable (primary) R group and the R group leaves. In this reaction
a Br-R is formed and a more stable (tertiary) R group with an alcohol stays.
second part: SN1, oxygen attacks the hydrogen agian forming a H2O, then the water leaves
forming a carbocation and the Br- attacks forming a tertiary alkyl halide and water as the
byproduct.
This reaction does not stop in the middle at alcohol and will proceed to the end. An alcohol can
be formed in the reaction and stop as a final product if one of the R groups is sp^2 such as
benzene, then it stops because it can't undergo SN1 or SN2
epoxidation of alkenes with peracids, stereochemistry!! 2P - Answers reactant: alkene
reagent: mcPBA
when the alkene is cis, the epoxide is syn (both wedge or both dash)
when the alkene is trans the epxoide is also trans/anti (one wedge and one dash, there is a 1:1
ratio of enantiomers)
vicinal halohydrins stereochemistry - Answers if alkene and a halogen and hydrin reacts (Br2,
H2O) then the stereochemistry is anti, keep in mind however, what side it is on for cis and trans
because they are diastereomers of each other. both enantiomers are good 1:1
formation of epoxides from vicinal halohydrins - Answers reagent: strong base
strong base deprotanates the oxygen, negatively charged oxygen bond then attacks Br and
makes it leave. The Br and the O need to be anti, so watch for stereochemistry and where the
methyl groups are located. REMEMBER TO SWITCH TO ANTI if it looks like cis, one must be on
a wedge while the other is on a dash
strong nucleophiles - Answers organometallics (R-MgBr), CN, N3, OR, SR, LiAlH4
epoxide ring opening with strong nucleophile - Answers R-strong nucleophile such as MgBr
attacks the least hindered side of the ring and OH opens
second step: H3O workup step to donate a hydrogen to the negative oxygen atom the final
product is R-OH-R'. Remember that the functional group stays in the molecule after reaction.
Works for all epoxides . The stereochemistry for the OH group and the attacked group are
inverted to if it is attacked by a weak nucleophile
acid-catalyzed ring opening for tri substituted epoxide - Answers reagents: H2SO4 + R group.
OH bond is formed and dipoles are formed that break the ring. The side without the OH bond
after the ring opening is where the R group goes on to, remember that the stereochemistry is
anti for the R group and OH group are anti and the stereochemistry is inverted if it was for the
