My Problems

Submitted by Matt

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List of user's published problems.

Problem # 518

The alcohol below is protonated and contains an oxygen with a positive charge. Using curved arrows, show the two "legal moves" that result in a neutral oxygen.

Solution

Protonation is a common first step in mechanisms. When you have a protonated alcohol, it's likely that either:

The alcohol will deprotonate to form a neutral alcohol (right), or
The alcohol will leave as a neutral molecule to form a carbocation.

Either one is possible. If you're not sure which one happens, remember that when writing out mechanisms, never go backwards- don't draw structures you have drawn before.

Problem Details

MendelSet practice problem # 518 submitted by Matt on June 30, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): dehydration, alcohols

Problem # 669

In your own words, what is the major difference in the addition of a Grignard reagent to an oxidation state III carbonyl (ester/acid chloride) versus an oxidation state II carbonyl? (aldehyde/ketone)

Solution

Oxidation state III carbonyls (esters, acid chlorides) contain a built-in leaving group (such as ^-OR or ^-Cl) and so undergo nucleophilic acyl substitution reactions. The product is another carbonyl.

Oxidation state II carbonyls (aldehydes and ketones) do not contain a built-in leaving group and so undergo nucleophilc acyl addition reactions (instead of substitution). The product is an alcohol.

Because Grignards react with all carbonyls- esters and aldehydes/ketones- esters and acid chlorides will react twice with Grignards: once in a Nuc Acyl Sub mechanism to form a ketone, which will then react with another equivalent of Grignard in a Nuc Acyl Add mechanism to form an alcohol.

Problem Details

MendelSet practice problem # 669 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition

Problem # 1286

For the molecule shown below, indicate the hybridization (sp³, sp², sp, etc.) of atoms A through H, and the bond angles of X, Y, and Z.

Solution

The trick to this problem is to look for implicit hydrogens- hydrogens (protons) that are not drawn in, but assumed to be there because there are no charges.

Also remember these rules:

If an atom has no multiple bonds, the total number of "things" (bonds and lone pairs) surrounding the atom is probably four, so it's sp³ hybridized (s¹ + p³ = 4).
If an atom has one double bond, the total number of "things"surrounding the atom is probably three, so it's sp² hybridized (s¹ + p² = 3).
If an atom has one triple bond or two double bonds, the total number of "things" surrounding the atom is probably two, so it's sp hybridized (s¹ + p¹= 2).

A sp² This oxygen has one double bond and two lone pairs. Three things total = sp².

B sp² This carbon has one double bond and two single bonds, to a carbon and an oxygen (so two bonds). Three things total = sp².

C sp³ This nitrogen has one long pair, two implicit hydrogens (two bonds), and one single bond to a carbo (another bond). Four things total = sp³.

D sp³ This carbon has one implicit hydrogen and three single bonds. Four things total = sp³.

E sp³ This oxygen has two lone pairs and two bonding pairs. Four things total = sp³.

F sp² This carbon has one double bond (one bond), one implicit hydrogen (another bond), and is bonded to one other carbon (a third bond). So three bonds in total. Three things total = sp².

G sp This carbon has a triple bond (one bond) and a single bond (another bond). So two things in total = sp.

H sp This nitrogen has a lone pair and a triple bond. Two things total = sp.

For angles, remember that electrons repel each other, and so will try to be as far apart from each other as possible:

Best way to separate two things is a straight line (180°).
Best way to separate three things is a "trifecta" (120°).
Best way to separate four things in three dimensional space is a tetrahedron (109.5°).

X 120° The carbon between the two oxygens is sp²hybridized, so the shape is trigonal planar and the bond angle is 120°.

Y 180° The carbon bonded to the nitrogen is sp hybridized, so the shape is linear and the bond angle is 180°.

Z 109.5° The carbon is sp³ hybridized (the most common type of carbon). So the shape is tetrahedral and the bond angle is 109.5°.

Problem Details

MendelSet practice problem # 1286 submitted by Matt on October 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Bonding, Geometries, and Hybridization

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

Keyword(s): hybridization, bond angles

Problem # 1281

Allylic and benzylic halides tend to undergo both S_N1 and S_N2 substitution reactions at a faster rate than their alkyl counterparts.

For example, both allyl chloride and benzyl chloride undergo S_N2 reaction at a faster rate than propyl chloride.

The same holds true for SN1 reactions: a 2° allyl or benzyl halide undergoes S_N1 reaction faster than a 2° alkyl halide. Explain.

Solution

There are separate explanations for SN1 and SN2.

S_N1: The S_N1 mechanism involves a carbocation intermediate, and both allylic and benzylic carbocations have resonance, which increases the stability of their carbocations, and speeds up the rate of S_N1 reaction.

S_N2: This explanation is less obvious, and is probably only mentioned in passing in your orgo textbook, if at all.

The pi systems present in allylic and benzylic halides are able to overlap with the pi orbitals of the nucleophile and the leaving group in the 5-coordinate transition state of an S_N2 reaction. This orbital overlap lowers the energy of the transition state (which stabilizes it), and so increases the rate of S_N2 reaction.

There are other possible explanations as well (Which also vary by textbook). For one, an allylic halide is less sterically hindered than an alkyl halide (less hydrogens sticking out in 3D).

Also, the carbons on the double bond of allyl and benzyl compounds are sp² hybridized, and so are more electronegative than sp³ carbons. So the sp²carbons pull alway some electron density from the alpha carbon (the carbon attached to the leaving group), making it more electrophilic, and thus increasing S_N2 reactivity.

Problem Details

MendelSet practice problem # 1281 submitted by Matt on October 1, 2011.

Subject: Organic Chemistry

Subject Topic(s): Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control), Benzene Side Chain Reactions and the Benzylic Position

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

Keyword(s): E1 mechanism, SN2 mechanism, allylic position, benzylic position

Problem # 1280

The structure of 2-methylbutane is shown below.

a) Draw the structures of all possible monochloro products resulting from the free-radical chlorination of 2-methylbutane.

b) Based on statistics alone, what do you expect the major product to be? Is this the same structure as the expected major product? Explain.

c) How would the relative yield of the products differ if bromine was used instead of chlorine?

Solution

a) There are five different carbons in 2-methylbutane, so hypothetically, you should get five different products, 20% A, 20% B, 20% C, 20% D, and 20% E.

But notice that A and B are the same product! So only four products are possible.

b) Based on statistics alone, we would expect product A (or B) to be the major product, encompassing 40% of all the products (20% A + 20% B).

But this is not what happens in real life because A is primary, and for free-radical halogenation reactions, more substituted products are favored. So the major product will be C, which is tertiary.

This happens because the intermediate is a radical, and radical stability goes 3° > 2° > 1°.

c) Chlorine reacts faster than bromine, and so is less selective than bromine.

For example, we know that the 3° halide will be the major product. But with chlorine, the proportion might turn out to be ~70% 3° , 20% 2°, and 10% 1°.

But since bromine reacts more slowly, you will get better selectivity. The proportion of products will might be something like ~95% 3° , 4% 2°, and 1% 1°.

While on this topic of reactivity, it's interesting to note that iodine reacts too slowly to be useful, and fluorine reacts so violently it's dangerous!

Problem Details

MendelSet practice problem # 1280 submitted by Matt on October 1, 2011.

Subject: Organic Chemistry

Subject Topic(s): Radicals (Stability, Halogenation)

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

Keyword(s): free radical halogenation

Problem # 1287

Rank each set of compounds in order of decreasing boiling point (1 = highest boiling point):

a) ethane, n-octane, n-pentane

b) n-butane, 1-butanol, 1-chlorobutane.

c) n-octane, 2-methylheptane, 2,5-dimethylhexane

(Note that the n- prefix before an alkane just means that it's one chain, without any branching.)

Solution

Remember the rules of what gives compounds a higher boiling point (and melting point):

Van der Waals Forces

More electrons = more Van der Waals interactions = higher boiling point

So in general, the more "stuff" in a molecule (the higher the molecular weight), the higher its b.p. Example: C₁₀H₂₂ will have a higher b.p. than C₅H₁₂.

How to remember this trend? What do you grill with? propane gas. (C₃H₈)

What's in a lighter? butane liquid (C₄H₁₀)

Hydrogen Bonding

Compounds that can hydrogen bond have higher b.p./m.p/ than those that don't.

Compounds can hydrogen bond is they have a N, O, or F bonded directly to an H.

In organic chemistry, the best examples are alcohols (ROH) and amines (RNH₂).

Branching

Branching makes it harder for molecules to pack together, which makes it harder to form Van der Waals interactions, and so tends to lower b.p./m.p.

a) (highest b.p.) n-octane > n-pentane > ethane (lowest b.p.)

Reason: octane has the most "stuff" (higher molecular weight).

b) 1-butanol > 1-chloroethane > n-butane

Reason: 1-butanol can hydrogen bond. 1-chloroethane has a higher molecular weight than n-butane.

c) n-octane > 2-methylheptane > 2,5-dimethylhexane

Reason: Branching. n-octane has no branching. 2,5-dimethylhexane has the most branching. Notice that each compound has the same molecular formua- C₈H₁₈. (and therefore the same weight).

Problem Details

MendelSet practice problem # 1287 submitted by Matt on October 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Intermolecular Forces and Functional Groups

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

Keyword(s): boiling point, intermolecular forces

Problem # 757

Propose a synthesis to accomplish each transformation. The only carbon sources allowed are alkenes and NaCN.

Solution

a) The starting alkene has 4 carbons, and the product has 5 carbons and a nitrogen. So you have to add 1C and 1N. This screams "add cyanide!"

Based on the location of the alcohol on the product, it looks like the way to go is to add NaCN to an epoxide. This puts the -CN and the -OH in the right positions.

b) The best way to prepare 2º and 3º amines is via reductive amination. Working backwards, the product can be prepared from a 4-carbon aldehyde (butyraldehyde) and a 3-carbon amine (propyl amine).

Performing hydroboration on the butene results in the anti-Markovnikov alcohol, which can be oxidized to the aldehyde using PCC.

To prepare propyl amine from an alkene, add HBr/peroxides to make the bromide (anti-Markovnikov), and then add NaN₃ follow by H₂/Pd.

Problem Details

MendelSet practice problem # 757 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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

Keyword(s): nitriles, reductive amination

Problem # 756

For each amine below, show all Hofmann elmination products.

If more than one product is formed, predict which one will be the major product.

Solution

The Hoffman elimination first turns an amine into a better leaving group by methylating it with excess methly iodide, and then eliminates the methylated amine using base (such as ^-OH fomed from Ag₂O in H₂O). The products are alkenes.

This reaction is similar to an E2 elimination in that a beta proton is ripped off, and the alkene forms between alpha and beta. One notable difference from the E2 is that the less substituted product is favored, so Hoffman products are usually anti-Zaitsev.

a) There are three different types of beta protons, so three possible products can form. Product A is the least substituted alkene, so it's the major product.

b) There is only one possible product. The Hoffman elimination can't be done on aryl hydrogens (benzene rings).

Problem Details

MendelSet practice problem # 756 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): Zaitsev's rule, Hoffman elimination

Problem # 755

The nitrosyl cation is shown below. Also shown are several proposed resonance arrows, only one of which is correct.

Draw the resonance forms that would follow from each set of arrows, and include formal charges. Which one is the correct resonance form? Explain your reasoning.

Solution

When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).

Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.

Problem Details

MendelSet practice problem # 755 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Diazonium Chemistry (NaNO2, Sandmeyer, Nitrosyl)

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

Keyword(s): resonance, Lewis structures

Problem # 754

Show how each amine can be prepared from a carbonyl and an amine via reductive amination.

Solution

Reductive amination is a two step process. First, an imine or enamine is formed from a carbonyl and an imine.

Then, a the imine (or enamine) is reduced to an amine (using reducing agents such as H₂, NaBH₄, NaBH₃(CN), NaBH(OAc)₃, etc.). For practice on forming imine and enamines, see problem 712).

Problem Details

MendelSet practice problem # 754 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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

Keyword(s): imines, enamines, reductive amination

Problem # 753

Rank the amines W through Z below in order of decreasing basicity (1 = most basic). Explain your reasoning.

Solution

Electron withdrawing groups (EWG) reduce electron density and reduce basicity. Resonance delocalizes electron density and so also decreases basicity (the electrons are less available to pick up a proton).

Electron donating groups (EDG) add electron density and increase basicity.

Alkyl groups are weakly EDG (induction), and hydrogen is neither EDG nor EWG. So cyclohexylamine (compound X) is a stronger base than ammonia (compound W).

Aniline (compound Y) has resonance, so it is less basic than compound X, which does not have resonance.

N-acyl aniline (compound Z) has more resonance forms than aniline, so it will be the least basic.

So the overall over is:

(most basic) X > W > Y > Z (least basic)

Problem Details

MendelSet practice problem # 753 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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

Keyword(s): base strength, amine bases, EDG and EWG

Problem # 752

Rank the amines A through D below in order of decreasing basicity (1 = most basic). Explain your reasoning.

Solution

sp² carbons are more electronegative (electron withdrawing) than sp³ carbons due to increased s-character. Electron withdrawing groups (EWG) make acids more acidic and bases less basic.

For this reason, piperidine (compound A) is the more basic than pyridine (compound B):

(more basic) A > B (less basic)

Nitro (-NO₂) is an EWG, so 4-nitropyridine (compound C) will be less basic than compound A.

Dimethylamine is an electron donating group (EDG), so it adds electron density to pyridine and increases basicity. So dimethylamino pyridine (DMAP, compound D) will be more basic than compound A:

(more basic) D > B > C (less basic)

So how does compound A compare with compound D? There is no way of predicting this based on EWG/EDG rules alone. After all, what should be more basic: piperidine, which has sp³ carbons (increases base strength), or DMAP, which has sp² carbons (decreases base strength) but also an EDG (increases base strength) ?

There's no way to tell. But it turns out that Piperidine is a little more basic than DMAP. So the overal order is:

(more basic) A > D > B > C (least basic)

Problem Details

MendelSet practice problem # 752 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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

Keyword(s): base strength, amine bases, EDG and EWG

Problem # 750

Let's work through a Robinson annulation.

Work backwords to determine the starting materials needed to produce each intermediate below, then show a mechanism for the overall reaction.

Solution

A Robinson annulation is just several intramolecular reactions in a row:

1. Michael addition of an aldehyde and methyl vinyl ketone (butenone).

2. Intramolecular aldol condensation to form the β-hydroxyl carbonyl (3-hydroxy cyclohexanone).

3. Elimination to form the α,β-unsaturated carbonyl (cyclohexenone).

Problem Details

MendelSet practice problem # 750 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Condensation Reactions (Aldol, Claisen, Dieckmann, Robinson)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): carbonyl alpha substitution, Robinson annulation

Problem # 749

Show how to prepare each compound below from propanal. I've marked the "cuts" for you.

Solution

a) Adding to alpha position, so this must have come from the combination of an enolate and an electrophile.

b) Adding to the beta position, so this must have come from a Michael addition (also called a conjugate addition).

To carry out a Michael addition, the first step is to convert propanal to an alpha, beta unsaturated carbonyl.

To do this, perform an alpha bromination with (Br₂/H₃O⁺) and then do an E2 elimination (tBuOK/heat).

Then add a Gilman reagent to the alpha,beta unsaturated aldehyde to add a carbon-carbon bond at the beta position.

Problem Details

MendelSet practice problem # 749 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork), Carbonyl Conjugate Additions (Michael Addition, Soft Nucleophiles)

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

Keyword(s): carbonyl alpha substitution, Michael addition

Problem # 748

Alpha bromination is usually carried out under acidic conditions via the enol intermediate.

Alpha bromination is uncontrollable under basic conditions, which goes through the enolate intermediate. Let's explore why.

a) Rank each carbonyl below in order of decreasing alpha-proton acidity (1= most acidic). Explain.

b) based on a), why does the reaction below lead to polyhalogenation?

Solution

a) Bromine is an electron withdrawing group (EWG), which makes nearby protons more acidic. So the carbonyl with two bromines is the most acidic.

b) To perform alpha bromination, the enolate (or enol) must first be formed, which then attacks Br₂.

The problem with doing this under basic conditions is that each successive bromination leads to a carbonyl that is more acidic, and so forms an enolate even more easily. So once this reaction starts, it's difficult to control.

Enols are less nucleophilic than enolates, so this react can be controlled under acidic conditions. (Br₂/H3O⁺ instead of Br₂/NaOH).