Draw the bond-line structure for:
2-ethoxypropane
Draw the general structure of an ether functional group.
R–O–R
What is the name of a three-membered cyclic ether?
Epoxide
Oxirane is also acceptable.
What does mCPBA do to an alkene?
It converts an alkene into an epoxide.
C=C + mCPBA → epoxide
Draw the major organic product:
CH₃CHO + NaBH₄ → ?
CH₃CH₂OH — ethanol
CH₃—CH₂—OH
Name the following molecule using IUPAC nomenclature:
CH₃–CH₂–CH₂–CHO
Butanal
Give the IUPAC name of this ether:
CH₃CH₂–O–CH₂CH₂CH₃
1-ethoxypropane
What is the major product of this reaction?
Cyclohexene + mCPBA → ?
1,2-Epoxycyclohexane (cyclohexene oxide)
What reagents are needed to convert an alkyl bromide into a Grignard reagent?
R–Br → R–MgBr
Mg⁰ in ether
Draw the major organic product:
2-propanol + CrO₃ → ?
Acetone (propanone)
CH₃–CH(OH)–CH₃ → CH₃–C(=O)–CH₃
This tests the alcohol oxidation material from your notes, which distinguishes oxidation behavior based on the type of alcohol.
O
║
CH₃ — C — CH₃
A molecule contains both an alcohol and a ketone.
Which functional group gets the outer suffix, and what happens to the other functional group in the IUPAC name?
Predict the major organic product:
CH₃CH₂O⁻ Na⁺ + CH₃CH₂CH₂Br → ?
Also name the type of reaction used to make the ether.
CH₃CH₂–O–CH₂CH₂CH₃
Product: 1-ethoxypropane
🧪 Game 1 — Epoxides — $300
What the players see
Predict the major product and show where the nucleophile attacks:
An unsymmetrical epoxide is treated with:
1. NaCN
2. H₂O
Which epoxide carbon does CN⁻ attack?
CN⁻ attacks the LESS substituted carbon.
The final product is a β-hydroxynitrile: the CN group ends up on the less substituted carbon, while the epoxide oxygen becomes OH after protonation.
This follows the strong-nucleophile epoxide-opening rule in your notes: attack occurs at the less substituted carbon through an SN2-type backside attack, followed by protonation.
Which reagent from your notes would oxidize a primary alcohol to an aldehyde without continuing to a carboxylic acid?
R–CH₂OH → R–CHO
Swern oxidation
Your nucleophilic-addition notes specifically cover Swern oxidation alongside alcohol oxidation and strong vs. weak oxidation conditions.
Draw the final organic product:
CH₃CHO
1. CH₃MgBr
2. H₃O⁺
2-propanol
CH₃–CH(OH)–CH₃
OH
|
CH₃ — CH — CH₃
Draw the bond-line structure for:
(E)-3-methylpent-2-enoic acid
with the higher-priority groups across the C2=C3 double bond on opposite sides (E).
Predict the product and show the stereochemistry:
A chiral secondary alkyl bromide reacts with NaOCH₃.
What type of reaction occurs, and what happens to the configuration at the carbon that originally held Br?
SN2 → ether + inversion of configuration
The OCH₃⁻ replaces Br, and the stereocenter undergoes Walden inversion.
An unsymmetrical epoxide is treated with:
1. H⁺
2. CH₃OH
Which carbon does CH₃OH attack, and what two functional groups are present in the final product?
CH₃OH attacks the MORE substituted carbon.
Final product contains:
–OCH₃ on the more substituted carbon
–OH on the other carbon
Which reagent would reduce an aldehyde or ketone into an alcohol?
NaBH₄, mCPBA, or CrO₃?
Also state whether the reaction is oxidation or reduction.
NaBH₄ — Reduction
Draw the major organic product:
CH₃–C(=O)–CH₃
1. CN⁻
2. H₃O⁺
OH
|
CH₃ — C — CH₃
|
CN
Write the complete IUPAC name. Include stereochemistry.
NH₂
|
HOOC—CH=CH—CH—CH₃
(E)-4-aminopent-2-enoic acid
Predict the final organic products:
CH₃CH₂–O–CH₂CH₂CH₃
1. Excess HI
2. Heat
Draw the organic products.
CH₃CH₂–I + CH₃CH₂CH₂–I
An unsymmetrical epoxide is treated with:
1. CH₃MgBr
2. H₂O
Draw the major product and indicate which epoxide carbon CH₃⁻ attacks.
he carbon nucleophile attacks the LESS substituted carbon.
The final product has:
Choose the reagents needed for this transformation:
Alkene → Epoxide → Alcohol with a NEW carbon–carbon bond
Fill in the blanks:
Alkene → [ _____ ] → Epoxide → [ _____ ] → Alcohol with a new C–C bond
1. mCPBA
2. Grignard reagent (RMgBr), then H₂O/H⁺ workup
So:
Alkene → mCPBA → Epoxide → 1. RMgBr 2. H₂O/H⁺ → substituted alcohol
Draw the final major organic product:
CH₃CHO + excess CH₃OH
H⁺
CH₃–CH(OCH₃)₂ + H₂O
The aldehyde is converted into an acetal.
OCH₃
|
CH₃ — CH — OCH₃