0 of 27 completed
Select at least one problem.
No problems selected.
| Topic | Quick rule |
|---|---|
| SN2 | Strong nucleophile, low steric hindrance, inversion |
| SN1 | Carbocation, rearrangements possible |
| E2 | Strong base, anti-periplanar, heat helps |
| E1 | Carbocation elimination, often competes with SN1 |
| Carbonyl addition | Aldehydes and ketones usually add nucleophiles |
| Acyl substitution | Carboxylic acid derivatives exchange leaving groups |
| Aromaticity | Cyclic, planar, fully conjugated, $4n+2$ pi electrons |
Difficulty
What is the hybridization and approximate geometry of the carbonyl carbon in acetone, $(CH_3)_2CO$?
Solution
The carbonyl carbon has three electron groups, so it is $sp^2$ hybridized.
Its geometry is trigonal planar, with bond angles of about 120 degrees.
How many sigma bonds and pi bonds are in $CH_3-CH=CH-CH_3$?
Solution
Count the bonds carefully:
$8$ C-H sigma bonds
$3$ C-C sigma bonds
$1$ pi bond from the double bond
So the molecule has $11$ sigma bonds and $1$ pi bond.
In $HOCH_2CH_2CHO$, which functional group determines the suffix, and how should the other group be named?
Solution
The aldehyde is the highest-priority functional group, so it determines the suffix.
The suffix is $-al$, and the hydroxyl group is named as a hydroxy substituent.
1-propanol and 2-propanol have the same molecular formula, $C_3H_8O$.
What type of isomers are they?
Solution
They are constitutional isomers because the atoms are connected differently.
The molecular formula is the same, but the connectivity is not.
A stereocenter has priorities $1 = OH$, $2 = CH_2CH_3$, $3 = CH_3$, and $4 = H$.
If the $H$ group points away from you and the order $1 \to 2 \to 3$ is clockwise, what is the configuration?
Solution
With the lowest-priority group pointing away, a clockwise $1 \to 2 \to 3$ path is $R$.
So the configuration is $R$.
At an alkene, the highest-priority groups on each double-bond carbon are on opposite sides.
Is the alkene $E$ or $Z$?
Solution
If the highest-priority groups are on opposite sides, the alkene is $E$.
If they were on the same side, it would be $Z$.
Between acetic acid and ethanol, which is the stronger acid? Briefly explain why.
Solution
Acetic acid is the stronger acid.
Its conjugate base, the carboxylate ion, is resonance-stabilized, so it is much more stable than the ethoxide ion.
In the reaction of hydroxide ion with methyl bromide, identify the nucleophile and the electrophile.
Solution
The nucleophile is hydroxide, $OH^-$, because it donates an electron pair.
The electrophile is the methyl carbon attached to bromine, because that carbon is attacked as bromide leaves.
A tertiary alkyl bromide is dissolved in methanol with no strong base.
Which substitution mechanism is most likely?
Solution
SN1 is most likely.
The tertiary substrate can form a relatively stable carbocation, and methanol is a polar protic solvent that supports ionization.
Compute the degree of unsaturation for $C_6H_{10}O$.
Solution
Use the DBE formula:
For $C_6H_{10}O$:
So the degree of unsaturation is $2$.
Difficulty
A molecule has both a ketone and an alcohol, such as $CH_3COCH_2CH_2OH$.
Which group controls the suffix, and from which end should numbering start?
Solution
The ketone has higher naming priority than the alcohol, so it controls the suffix.
The suffix is $-one$, and numbering should start from the end that gives the carbonyl carbon the lowest possible locant.
The alcohol is then named as a hydroxy substituent.
Why is an amide much less basic than an amine?
Solution
The nitrogen lone pair in an amide is delocalized into the carbonyl by resonance.
Because that lone pair is less available to accept a proton, amides are much less basic than amines.
Rank these from most acidic to least acidic:
terminal alkyne, alkene, alkane
Solution
The order is:
An sp carbon has more s-character, so its conjugate base is more stable.
Propene undergoes hydroboration-oxidation.
What alcohol is formed?
Solution
Hydroboration-oxidation gives anti-Markovnikov addition.
So the OH group ends up on the less substituted carbon, and the product is 1-propanol.
A secondary alkyl bromide is treated with a strong base and heat.
What mechanism is favored, and what geometric requirement must be met?
Solution
E2 is favored because the substrate is secondary, the base is strong, and heat promotes elimination.
For E2, the leaving group and the beta hydrogen must be anti-periplanar.
In methylcyclohexane, which chair conformation is more stable, with the methyl group axial or equatorial?
Solution
The equatorial methyl conformation is more stable.
An axial methyl group experiences 1,3-diaxial strain, so it is less favorable.
An IR spectrum shows a broad O-H stretch and a strong C=O band.
What class of compound is suggested?
Solution
This pattern suggests a carboxylic acid.
Carboxylic acids show both a carbonyl stretch and a very broad O-H stretch.
A mass spectrum has M and M+2 peaks in about a 3:1 ratio.
Which halogen is likely present?
Solution
Chlorine is likely present.
A 3:1 M to M+2 pattern is characteristic of chlorine, while a roughly 1:1 pattern suggests bromine.
Difficulty
Acetone reacts with $CH_3MgBr$, then water.
What product forms?
Solution
The methyl group adds to the carbonyl carbon, and workup protonates the alkoxide.
The product is tert-butanol, also called 2-methyl-2-propanol.
What ester forms when acetyl chloride reacts with methanol?
Solution
Acyl chlorides undergo nucleophilic acyl substitution, and methanol replaces chloride.
The product is methyl acetate.
Ethanal undergoes base-catalyzed self-aldol addition before dehydration.
What product is formed at the addition stage?
Solution
An enolate from one ethanal molecule adds to another ethanal molecule.
The addition product is 3-hydroxybutanal, a beta-hydroxy aldehyde.
When a glucose-like aldehyde cyclizes by reaction with an internal OH group, what functional group is formed, and what is the new carbon called?
Solution
A cyclic hemiacetal is formed.
The former carbonyl carbon becomes the anomeric carbon.
An unknown has formula $C_4H_8O$, DBE $1$, a strong IR band near $1720\ \text{cm}^{-1}$, and a 1H NMR signal near 9-10 ppm.
What functional group is most likely?
Solution
The formula gives one degree of unsaturation, and the IR band shows a carbonyl.
The 9-10 ppm proton is characteristic of an aldehyde hydrogen, so the functional group is an aldehyde.
Difficulty
A secondary carbocation forms next to a tertiary carbon in a polar protic solvent.
What rearrangement can occur before nucleophilic attack, and why?
Solution
A hydride shift can occur to move the positive charge to the more stable tertiary carbon.
Carbocations rearrange when a shift gives a more stable intermediate before the nucleophile attacks.
Two elimination products are possible. One forms faster, but the other is more stable and dominates when the reaction is reversible and heated.
Which control is operating, and which product wins?
Solution
This is thermodynamic control.
When the reaction can equilibrate, the more stable product dominates, so the more substituted alkene is favored.
Classify each ring:
A planar, cyclic, fully conjugated ring with 6 pi electrons.
A cyclic, conjugated ring with 4 pi electrons.
Solution
The first ring is aromatic because it is cyclic, planar, fully conjugated, and has $4n+2$ pi electrons.
The second ring is antiaromatic if it is planar, because it has $4n$ pi electrons.
Why are peptide bonds relatively stable and weakly basic?
Solution
A peptide bond is an amide linkage between amino acids.
Its nitrogen lone pair is delocalized into the carbonyl by resonance, which gives the C-N bond partial double-bond character and makes the carbonyl less reactive.
That same resonance makes the nitrogen much less basic.