Organic Chemistry Practice Problems and Problem Sets
All Practice Problems
Write out a mechanism for the reaction below using curved arrows. Be sure to include formal charges.
Ethers react with 2 equivalents of H-X to form water and two equivalents of alkyl halide.
In this case, the ether was cyclic, so the ring had to open up.
The reaction can go through either an SN1 or SN2 mechanism. Since this was a primary ether, it will go through an SN2 mechanism (the carbocation is too unstable for the reaction to go SN1).
MendelSet practice problem # 700 submitted by Matt on July 21, 2011.
The acid-catalyzed condensation of alcohols to form ethers is reversable; ethers can be hydrolyzed back to alcohols. How can the direction of this equilibrium be controlled to preferentially form ethers?
To push an equilibrium to one side, add starting material and remove product. This is Le Chatelier's principle from general chemistry.
So to push this reaction to the right and form ether, add alcohol and remove ether and water as they form.
To push this reaction to the left and form alcohol, add water to ether and remove alcohol as it forms.
How do you "remove something as it forms?" Alcohols and ethers have (relatively) low boiling points, and can be removed by hooking up a vacuum line and condenser to your reaction. The ether (or alcohol) boils off under the reduced pressure, and then recondenses in a separate piece of glassware. (Sort of like in distillation.)
Water has a relatively high boiling point and so is difficult to remove under reduced pressure. To remove water, molecular sieves are used. They're like tiny sponges that only absorb water (and not other solvents), removing it from the reaction.
MendelSet practice problem # 701 submitted by Matt on July 21, 2011.
Show how to prepare each compound starting from propylene oxide.
(Propylene oxide image below courtesy of Wikipedia.)
Epoxides can open up in two different ways.
To add a nucleophile to the less substituted side of an epoxide, use basic conditions. This is done in #2 below.
To add a nucleophile to the more substituted side of an epoxide, use acidic conditions. This is done is #1 below.
Why do the conditions matter? Epoxides have two electrophilic carbons. Normally nucleophiles will preferentially attack the less substituted carbon, as they do in SN2 reactions. Recall that SN2 reactions usually happen with strong nucleophiles- that is, negative charges (basic conditions).
When an epoxide reacts under acidic conditions, the transition state has carbocation character, and so it's sort of like an SN1 reaction. That is, instead of less substituted carbons being favored due to less steric bulk, more substituted carbons are favored do to a more stable carbocation. So acidic conditions cause an epoxide to open up on the more substituted side.
MendelSet practice problem # 702 submitted by Matt on July 21, 2011.
Show two ways to prepare the ether below from a combination of an alcohol and an alkyl halide via the Williamson ether synthesis.
Is one way better than the other? Why?
The Williamson ether synthesis takes place in two steps. First an alcohol is deprotonated to form a strong nucleophile (RO-, this step isn't shown in the image below). Then the alkoxide (negative alcohol) attacks an alkyl halide in an SN2 reaction.
So this problem is really asking, which step of conditions is most favorable for an SN2 reaction?
Recall that SN2 reactions compete with E2 reactions. If the nucleophile is too basic, or if there is too much bulk, it will go E2 instead of SN2. (See problem 560 for a full explanation of these competition reactions)
Below, the top combination uses the less substituted (1º) alkyl halide, and so is the best for an SN2 reaction.
The bottom reaction uses a bulkier (2º) alkyl halide, and will probably give a higher percentage of E2 side reaction.
MendelSet practice problem # 703 submitted by Matt on July 21, 2011.
Carbonyls are in equilibrium with their hydrate forms. This equilibrium happens in both acid and base.
Let's go through this equilibrium under basic conditions. Draw a mechanism using curved arrows for each reaction below.
Remember that under basic conditions, most species are either neutral or negatively charged, and rarely positively charged. So your structures will contain either ROH or RO-, but not ROH2+.
a) Carbonyl to Hydrate
Notice that no oxygen is ever positive during these basic mechanisms (always negative or neutral).
b) Hydrate to Carbonyl
The mechanisms in this problem and problem 706 are the most common mechanisms you will draw during second semester organic chemistry, and so it's a good idea to draw them out a few times.
In a), the nucleophile attacks the carbonyl carbon, and the double bond goes "up" to form a tetrahedral (sp3) carbon.
In b), one of the oxygen atoms acts as leaving group, and a lone pair on the other oxygen comes "down" to reform the double bond (and the sp2 carbon).
You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!
a) Carbonyl to Hydrate (basic conditions) "UP"
b) Hydrate to Carbonyl (basic conditions) "DOWN"
MendelSet practice problem # 705 submitted by Matt on July 21, 2011.
Carbonyls are in equilibrium with their hydrate forms. This equilibrium happens in both acid and base.
Let's go through this equilibrium under acidic conditions. Draw a mechanism using curved arrows for each reaction below.
Remember that under acidic conditions, most species are either neutral or positively charged, and rarely negatively charged. So your structures will contain either ROH or ROH2+, but not RO-.
a) Carbonyl to Hydrate (acidic)
b) Hydrate to Carbonyl (acidic)
The interconversion between a carbonyl (sp2 carbon) and a tetrahedral intermediate (sp3 carbon) is the most common mechanism you will encounter in second semester organic chemistry.
You should be familiar drawing it under both acidic (this problem) and basic (problem 705) conditions.
In a), the carbonyl "goes up" to form a tetrahedral intermediate.
In b), an oxygen "comes back down" to reform the carbonyl and kick off a leaving group.
You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!
a) Carbonyl to Hydrate (acidic) "UP"
Because this reaction takes place under acidic conditions, the carbonyl must protonate before the nucleophile attacks, to prevent oxygen from ever being negative (ROH instead of RO-).
b) Hydrate to Carbonyl (acidic) "DOWN"
The leaving group must protonate before it leaves, so it doesn't leave as a negative molecule (H2O instead of HO-).
MendelSet practice problem # 706 submitted by Matt on July 22, 2011.
The overall mechanism for imine formation is shown below. (This isn't a real mechanism, just an outline)
Use curved arrows to draw the full mechanism for imine formation under acidic conditions. (I've added outlines of the intermediate structures for you to use as a guide). This mechanism is similar to that in problem 706 (carbonyl hydrate equilibria).
This reaction takes place under acidic conditions, so oxygen should never be negative (only neutral or positively charged).
Acidic mechanisms tend to be a bit of pain, because you have to include many proton transfer steps (protonation and deprotonation).
MendelSet practice problem # 707 submitted by Matt on July 22, 2011.
The overall mechanism for Fischer esterification is shown below. This isn't a real mechanism, just an outline.
Methanol (the nucleophile) attacks the carbonyl carbon, forming a tetrahedral intermediate, which then loses a water to reform the carbonyl. This mechanism is called nucleophilic acyl substitution.
Use curved arrows to draw a full mechanism for this reaction. I've included structures for you to use as a guide.
This reaction takes place under acidic conditions, so the mechanism you draw will be similar to those in problem 706.
Acidic mechanisms only appear complicated because they contain several proton transfer steps.
Nucleophilic acyl substitution mechanisms have only three real steps- the "up, down, and kick."
First, the nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate (the "up").
Then the carbonyl reforms (the "down") and a leaving group leaves (the "kick").
MendelSet practice problem # 708 submitted by Matt on July 22, 2011.
Rank the carbonyls A-D below in order of decreasing electrophilicity (reactivity with nucleophiles).
(1 = Most reactive). Explain your reasoning.
The carbonyl carbon is electrophilic because it has a partial positive charge.
Are there any groups that make a carbon more positive? Yes, electron withdrawing groups (EWG). They pull away electron density, which increases electrophility. So C is the most reactive.
Conversely, electron donating groups (EDG) add electron density, and so make the carbonyl carbon less positive, and less electrophilic. Alkyl groups (carbon chains) are mildly EDG so ketone B will be less reactive than C.
Hydrogen is neither a EDG or EWG, so the aldehyde A will be in between the B and C. Also, the hydrogen is very small, so the carbonyl carbon is easily to get to (less steric bulk to block an attack).
Ester D is the least reactive because it has a resonance (the lone pair on the oxygen gets involved).
So overall, the order form most reactive to least reactive is C > A > B > D.
MendelSet practice problem # 710 submitted by Matt on July 22, 2011.