Mechanisms and Intermediates - GRE Subject Test: Chemistry

This reaction is an electrophilic addition reaction with an alkene. This is one of many alkene addition reactions that can add an -OH group onto your starting material. The key aspect of an hydroboration-oxidation reaction is the anti-Markovinikov addition to the double bond. The -OH group should be on the least substituted of the two carbons that originate from the double bond. In light of this information, the answer is 2-cyclohexanol.

Electrophilic aromatic substitution occurs most rapidly when the aromatic compound has electron-donating groups attached. Due to their electron affinity, halogens are electron-withdrawing groups. Acetophenone and nitrobenzene both bear partial positive charges on the substituent directly attached to the benzene ring, which pulls electron density out of the ring as well, causing the reaction not to occur.

Anisole is the only compound with an electron-donating group, and is the correct answer. The lone pairs on the oxygen atom can be used to initiate new bonds.

The correct answer is sodium butoxide. Remember from your studies that for E2 reactions to be highly preferred they require a "big, bulky, base".

First eliminate ammonium for it is not a base and is the conjugate acid form of ammonia. Then we can eliminate methanol (it is not in its conjugate base form) because there are stronger bases among the answer choices.

Now we can look at the real bases. Sodium cyanide will yield the cyanide ion, which is an extremely strong nucleophile. Remember that the E2 competes with the SN2. This means that the cyanide ion will prefer performing the SN2 not the E2, so we can cross this answer choice off.

Finally we are down to 2 answer choices, potassium hydroxide and sodium butoxide. Though potassium hydroxide is a strong base, it does not come close to the bulkiness of sodium butoxide. Thus, our best choice is sodium butoxide.

Only nitrobenzene would be a meta-directing group for additional electrophilic aromatic substitution (EAS) reactions. Although the bromine of bromobenzene is an electron-withdrawing group, halogens are not meta-directors; therefore, additional EAS reactions with bromobenzene would result in ortho or para attached substituents.

Remember that ortho additions are adjacent to the first substituent, meta additions are two carbons displaced from the first substituent, and para additions are opposite the first substituent.

(Note that these reactions would take place much more slowly than if there was an electron-donating group attached).

When dealing with carbonyl compounds, remember that a carboxylic acid and all of its derivatives will undergo nucleophilic substitution. Aldehydes and ketones will undergo nucleophilic addition. Propanal is a three-carbon aldehyde, and will thus undergo nucleophilic addition.

Acetic acid is a carboxylic acid, methyl ethanoate is an ether, and ethanamide is an amide; each of these would undergo nucleophilic substitution.

The Wittig reaction involves a ketone or aldehyde reacting with a phosphorus ylide, a molecule with a negatively charged carbanion. The ketone will undergo nucleophilic addition and form a betaine. This intermediate will then form an alkene with a triphenylphosphine oxide being released. The Wittig reaction will form a mixture of both cis and trans isomers if the carbanion has two different substituents.

Wittig general reaction:

reactions, also known as unimolecular nucleophilic substitution reactions, occur in two steps. Here, we are concerned with the first and second (rate-determining) steps, in which the leaving group breaks off of the molecule to form a carbocation. Alkanes that form the most stable carbocations are most likely to undergo reactions. Tertiary carbocations are the most stable, followed by secondary. Primary and methyl carbocations are very unstable and unlikely to form at all. The tertiary alkane, , will form a very stable tertiary carbocation compared to the other answer choices.

The pKa value indicates how strong an acid is, and acid strength increases as pKa decreases. The side of a reaction with a lower pKa is going to dissociate more, pushing the equilibrium over to the other side. The equilibrium will thus lie on the side with the HIGHER pKa.

Since the pKa of acetic acid (4.76) is higher than the pKa of trifluoroacetic acid (0), the reaction will shift to the left to reach equilibrium.

A reducing agent is a substance that readily donates an electron to another substance. They consist primarily of elements with low electronegativities such as hydrogen. Therefore, the answer is (sodium borohydride) which consists primarily of hydrogen atoms.

The reason that the alkyl halide is preferred to be primary is because the mechanism for these reactions is SN2. SN2 indicates a substitution reaction that takes place in one step. A primary alcohol is preferred to prevent steric congestion caused by the simultaneous binding of the nucleophile and release of the leaving group. This reaction mechanism is faster because it omits the formation of a carbocation intermediate.

In contrast, SN1 reactions take place in two steps and involve the formation of a carbocation intermediate.

The more stable the carbocation, the lower the activation energy for reaching that intermediate will be. The more substituted a carbocation is, the more stable it is. The carbocation bonded to three alkanes (tertiary carbocation) is the most stable, and thus the correct answer.

Secondary carbocations will require more energy than tertiary, and primary carbocations will require the most energy.

The strong sulfuric acid protonates the hydroxyl group of compound B, resulting in the loss of water as a leaving group and the generation of a carbocation intermediate. Since this carbocation carbon is attached to three other carbons, this is a tertiary carbocation. It is bound to the phenyl substituent, a methyl group, and the branched carbon chain.

The correct answer is: carbon 1 because the intermediate will then be a secondary carbocation.

This is a case of addition across a double bond where stands for any halide. The first step in this reaction is the attack of by the double bond. This will create two intermediates, the first being the halide anion (so in our case ), the second being a carbocation on our compound at one of the two carbons that formerly shared the double bond.

If the hydrogen attached the carbon 2 then we would have a positive charge on carbon 1 and vice versa. A positive charge on carbon 1 is known as a primary carbocation (a carbon attached to 1 other carbon or function group), which is rarely if ever seen due to its overwhelming instability. A positive charge on carbon 2 is known as a secondary carbocation (a carbon attached to 2 other carbons or functional groups) and is much more stable than a primary carbocation.

We would want the more thermodynamically stable intermediate for our reaction to proceed, so we would want the positive charge on carbon 2 and the hydrogen attached to carbon 1.

The resonance form of this compound would put the positive charge on the carbon attached to the bromine. Unfortunately this carbon is already slightly positive due to the electron withdrawing effects of bromine due to its high electrophilicity. So this resonance form would be more unstable than a secondary carbocation due to the increased concentration of positive charge from bromine's electron withdrawal.

The three steps of a free radical chlorination reaction are, in order, initiation, propagation, and termination.

Free radicals are produced in the initiation and propagation steps. The termination steps combine any two free radicals formed in the reaction to produce a compound that has no unpaired electrons (free radicals).

Enolates are formed by an oxygen anion bound to an alkene carbon. Reactions II and III include an enolate intermediate, as shown in the mechanisms below, whereas reaction I is a simple SN2 reaction and does not include an enolate intermediate. Enolates are highlighted in red.

This reaction is an electrophilic addition reaction with an alkene. This is one of many alkene addition reactions that can add an -OH group onto your starting material. The key aspect of an hydroboration-oxidation reaction is the anti-Markovinikov addition to the double bond. The -OH group should be on the least substituted of the two carbons that originate from the double bond. In light of this information, the answer is 2-cyclohexanol.

Electrophilic aromatic substitution occurs most rapidly when the aromatic compound has electron-donating groups attached. Due to their electron affinity, halogens are electron-withdrawing groups. Acetophenone and nitrobenzene both bear partial positive charges on the substituent directly attached to the benzene ring, which pulls electron density out of the ring as well, causing the reaction not to occur.

Anisole is the only compound with an electron-donating group, and is the correct answer. The lone pairs on the oxygen atom can be used to initiate new bonds.

Only nitrobenzene would be a meta-directing group for additional electrophilic aromatic substitution (EAS) reactions. Although the bromine of bromobenzene is an electron-withdrawing group, halogens are not meta-directors; therefore, additional EAS reactions with bromobenzene would result in ortho or para attached substituents.

Remember that ortho additions are adjacent to the first substituent, meta additions are two carbons displaced from the first substituent, and para additions are opposite the first substituent.

(Note that these reactions would take place much more slowly than if there was an electron-donating group attached).

When dealing with carbonyl compounds, remember that a carboxylic acid and all of its derivatives will undergo nucleophilic substitution. Aldehydes and ketones will undergo nucleophilic addition. Propanal is a three-carbon aldehyde, and will thus undergo nucleophilic addition.

Acetic acid is a carboxylic acid, methyl ethanoate is an ether, and ethanamide is an amide; each of these would undergo nucleophilic substitution.