All Practice Problems

Problem # 711

Show two ways of preparing the alkene below via the Wittig reaction starting from triphenyl phosphine (PPh₃).

Is one route better than the other? Why?

Alkenes can be prepared from a combination of ylide and aldehyde or ketone. This is the Wittig reaction.

One carbon on the alkene comes from the carbon bonded to the PPh₃ on the ylide, and the other carbon comes from the carbonyl.

There are usually two different ways to make an alkene via the Wittig reaction. So is one way better than the other?

Yes. Ylides come from the S_N2 reaction of PPh₃ with an alkyl halide. Because it's an S_N2 reaction, you want to use the least substituted alkyl halide! (1º is better than 2º). So of the two reactions below that would result in the desired alkene, the top method is better because it involves the less bulky alkyl halide.

This principle also holds true for the Williamson ether synthesis (problem 703). Less substituted alkyl halides are better for S_N2 reactions.

Problem Details

MendelSet practice problem # 711 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): Wittig reaction

Problem # 712

Show what combination of amine and carbonyl would result in each imine or enamine.

Solution

For both imines and enamines, the sp² carbon comes from the carbonyl, and the nitrogen comes from the amine.

Problem Details

MendelSet practice problem # 712 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): imines, enamines

Problem # 713

When a carbonyl is treated with semicarbazide under acidic conditions an "imine" is produced called a semicarbazone.

Which of the two products below is the correct structure for a semicarbazone? Explain.

Solution

What this problem is really asking is "Which nitrogen on semicarbazide is more nucleophilic?"

The nitrogens bonded to the carbonyl on semicarbazide all have resonance forms, so their lone pairs aren't as available for nucleophilic attack.

The other nitrogen (on the left) is the most nucleophilic because it isn't involved in any resonance forms, so its lone pair isn't shared with another atoms.

Problem Details

MendelSet practice problem # 713 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

Question Type: Concept (explain why this is so..)

Keyword(s): nucleophilicity, imines

Problem # 714

Complete each synthesis below. All carbon sources must come from alkenes.

Each synthesis will involve protecting groups.

Solution

The protecting group most commonly used for aldehydes and ketones (in undergraduate orgo) is ethylene glycol.

It is put on using HOCH₂CH₂OH/H⁺ and removed with H₂O/H⁺.

For a), the reaction calls for the use of acetylide with an alkyl halide. But acetylides also react with carbonyls.

So before the alkyne is deprotonated using a strong base (such as NaNH₂), the carbonyl must be protected with ethylene glycol.

For b), the reaction calls for the addition of two equivalents of phenyl Grignard to the ester. The problem is that esters aren't as reactive as ketones (or aldehydes), so the Grignard would react with the ketone before it ever touched the ester! To prevent this, the ketone must be protected before Grignard is added.

Problem Details

MendelSet practice problem # 714 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): protecting groups, acetals and ketals

Problem # 715

α-D-glucose is shown below.

a) Is α-D-glucose an acetal, hemiacetal, ketal, or hemiketal?

b) Draw the carbonyl form of α-D-glucose.

Solution

a) You can spot the carbonyl carbon because it's the one with two oxygens bonded to it (oxidation state II).

The carbonyl carbon has an -OH (alcohol) bonded to it, and not an -OR (ether), so it's a hemiacetal or a kemiketal.

The carbonyl carbon also has a hydrogen bonded to it, so it must have come form an aldehyde (instead of ketone). So it's a hemiacetal. (remember: aldehyde -> acetal, ketone -> ketal).

b) The hemiacetal carbon's -OH (or -OR in the case of a ketal or acetal) is the oxygen that used to be the carbonyl C=O.

Problem Details

MendelSet practice problem # 715 submitted by Matt on July 23, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Carbohydrates

Question Type: Concept (explain why this is so..)

Keyword(s): acetals and ketals

Problem # 717

Rank each of the eight compounds A through H below in order of decreasing acidity (1 = most acidic).

Don't get intimidated! What are the differences between these compounds?

Consider electron withdrawing groups (EWG), resonance, hybridization, and the functional group of the acidic proton.

Solution

Carboxylic acids are much more acidic than alcohols (because of resonance in their conjugate bases), so A and D are the least acidic of the group. The conjugate base of D has resonance, so D is more acidic than A:

(everything) > D > A

Electronegative atoms increase acid strength (by stabilizing the conjugate base), and sp² carbons are more electronegative than sp³ carbons (due to higher s-character). So B (all sp³ carbons) will be the least acidic of the carboxylic acids, and H will be more acidic than B (two sp² carbons):

(everything) > H > B > D > A

Of the remaining 4 benzoic acids, C has the most EWG (two nitros) and so is the most acidic. F has no EWG and so is the least acidic. Both E and G have only one EWG and so will be similar in acidity, with E slightly more acidic than G because the EWG affect is strongest in the ortho position (closer to the action). So the overal order is:

(strongest acid) C > E > G > F > H > B > D > A (weakest acid)

Problem Details

MendelSet practice problem # 717 submitted by Matt on July 23, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength, carboxylic acids

Problem # 720

Base your answers to the three problems below on your knowledge of electron donating groups and electron withdrawing groups (EDG and EWG).

a) Based on the pK_a's listed below, Is formic acid more or less acidic than acetic acid? Propose an explanation why.

b) If acetic acid were added to a pH = 4.7 buffer solution, what percentage of it would be in its acetate (conjugate base) form?

c) Methoxy (-OCH₃) is usually considered an EDG. But based on the pK_a of methoxy acetic acid, do you think this is always the case? Explain.

Solution

a) The pK_a of formic acid is less than that of acetic acid, so formic acid is the stronger acid.

The pK_a difference is about ~1, and pK_a is logorithmic, so formic acid is ~10x as acidic as acetic acid.

Why? Because formate (formic acid's conjugate base) is more stable than acetate (acetic acid's conjugte base).

Why? Probably because of EDG. Alkyl groups such as methyl as weakly electron donating. Carboxylate is already negatively charged, and adding an EDG (the methyl group in acetate) makes it less stable, so acetate is a strong base, and acetic acid is the weaker acid. Formate has an hydrogen, which is neither EDG nor EWG, so doesn't have this problem.

b) The Henderson–Hasselbalch equal is pH = pK_a + log (A^-/HA). In this case, A^- is ^-OAc (acetate) and HA is HOAc (acetic acid). So plugging in the numbers:

pH = pK_a + log (^-OAc/HOAc)

4.7 = 4.7 + log (^-OAc/HOAc)

0 = log (^-OAc/HOAc),

so (^-OAc/HOAc) =1, and ^-OAc = HOAc.

So the ratio of acid to conjugate base (acetic acid to acetate) would be about 1:1.

The math above illustrates a rule you might have learned in general chemistry- when pH = pK_a, the concentration of acid equals the concentration of conjugate base.

c) The pKa of methoxy acetic acid (~3.6) is less than that of formic acid (~3.8), which means methoxy acetic acid is the strong acid, so methoxy acetate must be more stable than formate. This implies that methoxy must be somewhat electron withdrawing (an EWG).

This might be suprising because during the aromatic reactions chapter(s) (EAS) methoxy is considered an EDG. Oxygen is electron donating in terms of resonance. But remember that oxygen is also very electronegative. So in this case, it seems that the inductive EWG effect (electronegativity) outweighs the EDG effect.

Problem Details

MendelSet practice problem # 720 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength, carboxylic acids

Problem # 721

Rank each of the four compounds below in order of decreasing acidity (1 = most acidic).

Solution

The general rule is that the pK₁ or a dicarboxylic acid is lower (more acidic) than the pK_a of a regular carboxylic acid, and the pK₂ is a little higher (less acidic).

Why is this? Because carboxylic acids are electron withdrawing groups (EWG). They stabilize the conjugate base of an acid, making the acid more acidic. So Both D and A should be more acidic than C. Because A has its carboxylic acid groups closer together, the EWG effect is stronger, and it will be the most acidic.

B will be the least acidic of the group. Why? Because a carboxylate group has a negative charge, which is electron donating and destablizing to the conjugate base (is a molecule is already -1, it doesn't help to become -2.)

So the overall order is:

(strongest acid) A > D > C > B (weakest acid)

Problem Details

MendelSet practice problem # 721 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength, carboxylic acids

Problem # 722

Show how to prepare vinyl benzene from benzene.

Solution

This is a good synthesis to know, because vinyl benzene is a good starting point for many synthesis problems you will encounter down the road.

An ethyl group can be added to benzene via Friedel-Crafts alkylation.

To add that alkene, we you will have to do some sort of elimination reaction. The easiest way to do this is via bromination. Free radical bromination (with NBS or Br₂/hν) will add a bromine to the position of the most stable radical, which is the benzylic position (due to resonance).

Then adding a strong base like potassium tert-butoxide will do an E2 reaction to form the alkene.

Problem Details

MendelSet practice problem # 722 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): free radical halogenation, EAS mechanism, Friedel-Crafts, C-C bond formation

Problem # 723

Show how to prepare each compound from vinyl benzene.

How would you prepare the methyl ester of each compound?

Solution

Vinyl benzene is a common starting material and intermediate in second semester organic chemistry synthesis problems, so you should be familiar how to make vinyl benzene from benzene (see problem 722).

The trick to many synthesis problem is to realize how many carbons need to be added, as that will determine which route you use.

a) Adding zero carbons: alcohol then oxidation

All you need is an alcohol added to the last carbon, which you can oxidize to a carboxylic acid. So from the alkene do an anti-Markovnikov addition of H₂O via hydroboration, followed by oxidation using Jones reagent (aqueous chromium, Cr⁶⁺ in acid).

b) Adding one carbon: nitrile (cyanide, CN)

When you add one carbon that screams "cyanide!" The most common way to add CN is using NaCN with an alkyl halide in an S_N2 reaction.

Because you want to add the extra carbon (CN) to the end of the chain, you want to do an anti-Markovnikov addition using HBr and perioxides. Then add NaCN to form the nitrile.

To convert a nitrile to a carboxylic acid, hydrolyze under acidic conditions (add water with acid, H₃O⁺).

c) Adding two carbons: epoxide

When you need to add two carbons, ethylene oxide is the way to go. But ethylene oxide is an electrophile, so first you must transform vinyl benzene into a nucleophile, specifically, a Grignard reagent.

We want the Grignard part (MgBr) on the end, so do an anti-Markovnikov addition with HBr/peroxides followed by Mg⁰/ether. Then add the epoxide which adds two carbons with an alcohol on the end. Finally, oxidize using Jones reagent.