OC DAT BOOSTER Missed Test Questions Flashcards
Why is CH3COCH2F more acidic than CH3COCH3?
electron withdrawing group
To determine which conjugate base is more stable, use the mnemonic CARDIO (Charge, Atom, Resonance Delocalization, Induction, Orbital). The acidic proton on both molecules is bound to a carbon atom located next to a carbonyl group. So, the charge, atom, and resonance delocalization will be the same for both molecules. The first point of difference in ‘CARDIO’ is inductive effects.
The conjugate base of CH3COCH2F is further stabilized by induction. The fluorine atom is an electron withdrawing group, so it will help disperse the negative charge of the conjugate base. Keep in mind fluorine is very electronegative, which makes it possible for it to pull electron density towards itself. CH3COCH3 does not have an electron withdrawing group, so it’s conjugate base will not be stabilized by inductive effects.
What signal will the protons of the methyl group of the following compound produce on a 1H NMR spectrum?
singlet
The number of 1H NMR signals is equal to the number of non-equivalent protons. Protons are equivalent when they exist in the same chemical environment. To determine the multiplicity of a given signal (splitting), we look for the number of non-equivalent neighboring protons plus 1. This is known as the n+1 rule.
The figure below shows the hydrogen atoms present on the given molecule.
The methyl group (blue) consists of three chemically equivalent hydrogen atoms. The neighboring carbon (red) is bound to two bromine atoms and another carbon atom. Note, the neighboring carbon (red) is NOT bound to any hydrogen atoms, so n=0 in the n+1 rule. As a result, the methyl group (blue) will produce a singlet because n+1 = 0+1 = 1.
Which of the following is true regarding the initiation step of free radical reactions?
In a free radical reaction, the formation of free radicals occurs in the initiation step. Since this is energetically unfavorable, the initiation step has the highest activation energy and is therefore the rate-determining step. The termination step is favorable (lower in energy) because the radicals combine to form a stable molecule.
Which compound will produce a large peak at approximately 1710 cm-1 on an IR spectrum?
ethanal
The question states that a large peak is observed at 1710 cm-1, which indicates that the compound contains a carbonyl group. Ethanal, being an aldehyde, is the only compound listed that contains a carbonyl group as shown below.
Which radical is most stable?
To determine which radical is the most stable, consider (1) substitution and (2) resonance delocalization. More substituted radicals are more stable because alkyl groups donate electron density to the carbon bearing the unpaired electron. Free radicals that are delocalized by resonance are even more stable due to the effects of resonance stabilization. Note: Vinyl radicals and phenyl radicals (radicals on alkene carbons) are extremely unstable.
Benzylic and allylic radicals are more stable than tertiary radicals due to the resonance stabilization. Tertiary radicals are more stable than secondary radicals, and secondary radicals are more stable than primary radicals (due to increasing substitution). Vinyl and phenyl radicals are less stable than primary radicals
Consider talose, an aldohexose sugar. What type of functional group will be formed when talose cyclizes?
hemiacetal
The cyclization of monosaccharides involves the primary alcohol at one end of the molecule adding to the aldehyde at the other end of the molecule. This will form a hemiacetal (shown below) because the product will have one -OH group and one -OR group. Note that the outcome of this reaction will be the same under acidic and basic conditions because the reaction involves an aldehyde and one alcohol.
Which of the following compounds is most acidic?
CHF2CH2SH
Acid strength is determined by conjugate base stability. To solve this problem, use the mnemonic CARDIO (charge, atom, resonance delocalization, induction, orbital). In all the molecules, the acidic proton is attached to oxygen or sulfur. A bigger atom will increase acidity because it can better accommodate the negative charge of the conjugate base. If we apply this concept to the question, we can narrow our options down to C and E because sulfur has a larger atomic radius than oxygen. Electron-withdrawing groups increase acidity because they stabilize the conjugate base via induction. More fluorine atoms means more induction and increased acidity. Option E has the most fluorine atoms (shown below), so it is the most acidic.
During an acid/base extraction, aqueous sodium hydroxide is added to a mixture of benzene and compound X in hexanes. Compound X is then isolated from the aqueous layer. Which of the following is most likely compound X?
In this scenario, both benzene and compound X are initially dissolved in the organic layer (hexanes). The addition of sodium hydroxide causes compound X to migrate to the aqueous layer.
Sodium hydroxide is a strong base, so it will deprotonate an acidic compound. To solve this problem, find the compound that can be deprotonated by NaOH. Option A, benzoic acid, is the only option that will be deprotonated by sodium hydroxide because it has an acidic hydrogen bonded to oxygen. The resulting negatively charged conjugate base (a benzoate ion) will move to the aqueous layer, while benzene (a nonpolar compound) will remain in the organic layer
therefore option a is the correct answer
Which of the following correctly describes the stereochemical configuration of the product from this reaction?
Which reagent will accomplish the following transformation?
Br2/FeBr3
In the structure below, which of the indicated groups is antiperiplanar to the OH group?
A
Which of the following molecules is most nucleophilic?
Nucleophilicity parallels basicity. Therefore, the most nucleophilic molecule will be the strongest or most reactive base. Negatively charged bases are more reactive than neutral ones, so we can eliminate Options B and D. If the negative charge of the base is stabilized by resonance, the base will be more stable and thus less reactive. Options C and E are capable of resonance, so they can also be eliminated. Option A is the strongest base and consequently the strongest nucleophile as shown below.
Which of the following functional groups is present in the product of the reaction?
amine
amide
What type of reaction is the following?
electrophilic aromatic substitution
The reaction shown is sulfonation of benzene. Recall that sulfonation is an electrophilic aromatic substitution (EAS) reaction. The general mechanism for an electrophilic aromatic substitution reaction is shown below. Remember that EAS reactions have a positively charged intermediate called a sigma complex.
What type of reaction is the following?
substitution
A substitution reaction is a reaction that replaces one atom/group with another atom/group. The reaction shown here is a radical substitution reaction. Free radical bromination replaces a hydrogen atom with a bromine atom via a radical intermediate.
Given the same molecular mass, which of the following has the highest boiling point?
Recall that hydrogen bonds are one of the strongest intermolecular forces. Molecules engage in hydrogen bonding when an H atom is directly bonded to an N, O, or F atom. Carboxylic acids have higher boiling points than alcohols because two carboxylic acid molecules can be involved in not one, but two hydrogen bonds (shown below). Aldehydes and ketones only contain a carbonyl group, so they engage in dipole-dipole interactions. Thus, aldehydes and ketones have weaker intermolecular forces than carboxylic acids. Benzene is nonpolar and thus only interacts via LDF and pi stacking.
What is the product of the reaction?
Which of the following compounds has the lowest boiling point?
A molecule’s boiling point is determined by the strength of its intermolecular forces. All the molecules are hydrocarbons, so they only engage in London dispersion forces (LDFs). Recall that as molecular weight decreases, a hydrocarbon will have weaker dispersion forces and consequently a lower boiling point. Looking at the option choices, we see that all the hydrocarbons contain six carbons except Option D, which contains five carbons. Because Option D has a lower molecular weight than the other molecules, it will have the lowest boiling point.
What is the relationship between the following pair of compounds?
Which of the following solvents would acetic acid be most soluble in?
Acetic acid is a carboxylic acid (CH3COOH). As a general rule, organic compounds dissolve best in a solvent with similar properties to the compound itself. We can remember this rule as “like dissolves like”. For example, nonpolar molecules such as n-hexane will dissolve best in nonpolar solvents such as benzene. On the other hand, polar molecules capable of hydrogen bonding, such as carboxylic acids, will dissolve best in polar solvents capable of hydrogen bonding.
The only solvent listed that is both polar and capable of hydrogen bonding is aniline
Which is the most stable conformation of cis-1-bromo-4-methylcyclohexane?
“Cis” implies that the two groups are on the same face of the ring, and 1,4 implies that the groups are on carbons 1 and 4. The two chair conformations for the given compound are shown in the figure below.
When substituents are in the axial position they experience diaxial interactions, a type of steric strain that reduces the molecule’s stability. The most stable chair conformation will have the least severe diaxial interactions. Ideally, both groups would be equatorial as this greatly reduces steric interactions.
However, based on the placement of the substituents, this is not possible. Of the two groups, the methyl group is the bulkier substituent because it has more atoms. As a result, the methyl group will experience more severe diaxial interactions in the axial position than the bromine atom. The most stable conformation will have the methyl group equatorial and the bromine atom axial.
Which reagent is needed for the following transformation?
When we compare the starting molecule to the product, we see that the ketone and the carboxylic acid have been reduced. LiAlH4 is a strong reducing agent, so it will reduce the carboxylic acid and the ketone to primary and secondary alcohols, respectively, as shown below.
Which of the following is the most appropriate solvent to use for a Grignard reaction?
Grignard reagents are highly basic compounds that can act as a nucleophile to attack carbonyl-containing compounds. An aprotic solvent must be used for reactions involving Grignard reagents because a protic solvent will ruin the Grignard reagent. This is because the Grignard reagent will act as a base to deprotonate the solvent rather than participate in the intended nucleophilic attack.
An example of a reaction involving a Grignard reagent is shown below for reference.
Aprotic solvents do not contain an acidic proton. To determine if a solvent is aprotic, first, draw its structure. If a solvent lacks hydrogen directly bonded to oxygen or nitrogen, it can be considered aprotic.
How many signals are present in the 13C NMR spectrum of cyclopentanone?
3
What type of reaction is shown below?
addition
Which compound has the highest boiling point?
The melting point and boiling point of a molecule are determined by the strength of its intermolecular forces. A molecule with strong intermolecular forces will have a high melting point and a high boiling point. Conversely, a molecule with weak intermolecular forces will have a low melting point and a low boiling point. There are four types of intermolecular forces that you must know for the
DAT: dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-ion interactions (ranked from weakest to strongest IMF). Option A has the highest boiling point because it engages in hydrogen bonding (shown below). Recall that hydrogen bonding occurs between molecules with an H atom bonded to an N, O, or F atom.
Which molecule can be used in the initiation step of a free radical reaction?
Br2 can be used in the initiation step of free radical reactions. When treated with heat or light, the bond between the two bromine atoms is broken, forming two bromine radicals as shown below. The bromine radicals go on to react with other molecules to propagate the formation of new radicals.
Which of these molecules contains the strongest electron-donating group?
Aldehydes, carboxylic acids and halogens are electron-withdrawing substituents. Aldehydes and carboxylic acids withdraw electron density from the ring via resonance. Halogens withdraw electron density from the ring due to their electronegativity.
Alkyl groups and amines are both electron-donating groups. Alkyl groups are mild electron-donating groups because they donate electron density via hyperconjugation. Amines, on the other hand, are strong electron-donating groups because they donate electron density through resonance (nitrogen lone pair delocalizes into ring).
Consider the following silica gel TLC plate of compounds A, B, C, and D. Which of the following statements is correct?
Thin layer chromatography (TLC) separates components based on differing polarities. The more polar the compound, the more it will interact with the silica on the TLC plate, and the less it will travel. The less polar the compound, the further it will travel on the plate.
The Rf value is equal to the distance the compound moved over the distance the solvent moved. Therefore, the compound that moves the furthest in TLC (the least polar compound) will have the largest retention factor, whereas the compound that moves the least in TLC (the most polar compound) will have the smallest retention factor.
Compound A traveled the furthest on the TLC plate. Because of this, compound A has the highest Rf and is the least polar compound.
Therefore, Option A is the correct answer.
Which would be the strongest nucleophile in a polar protic solvent?
Remember the following trends, as illustrated in the diagram below:
- Nucleophile strength increases from right to left in both polar protic solvents and polar aprotic solvents.
- Nucleophile strength increases down a group in a polar protic solvent but decreases down a group in a polar aprotic solvent.
Nucleophilicity increases from right to left due to decreasing electronegativity of the atom. More electronegative atoms hold electrons more tightly and are less prone to donating them in a nucleophilic attack.
Nucleophilicity increases down a group in a polar protic solvent due to decreased solvation. The larger the atomic radius, the weaker the solvation shell. Thus, the larger the atomic radius in a polar protic solvent, the more nucleophilic.
Additionally, negatively charged nucleophiles tend to be stronger than neutral nucleophiles. Of the nucleophiles listed, B, C and D are negatively charged. Because sulfur is lower on the periodic table than oxygen, it is more polarizable and thus more nucleophilic.
Which of the following is the least stable conformation of 2-methylbutane, looking down the C2-C3 bond?
Newman projections can exist in eclipsed or staggered conformations. The eclipsed conformation is less stable than the staggered conformation because overlapping atoms or groups of atoms leads to more strain in a molecule.
You can eliminate Options B and D because they depict butane, not 2-methylbutane. Options C and E depict eclipsed conformations, notice in C, that the two methyl groups are overlapping. This is not the case for Option E. Thus, Option C depicts a higher energy conformation of 2-methylbutane.
swern oxidation
swern oxidation
hydrogenation
hydrogenation
wolff-kishner
Dieckmann condensation
The reaction shown is ozonolysis with oxidative workup. The C-C double bond in the alkene will be split to form two C-O double bonds. The alkene carbon bonded to a carbon and a hydrogen will be converted to a carboxylic acid and the alkene carbon bonded to two carbons will be converted to a ketone. As shown below, if we number our carbons, we see that the carbonyls are separated by three carbons. Option B is the product of the reaction.
meta directors
ortho/para directors
How many possible stereoisomers exist for the following molecule?
2
A chiral center is an sp3 hybridized tetrahedral carbon attached to four different substituents. The compound shown has 1 chiral center. The maximum number of stereoisomers for any chiral compound is equal to 2n with n being the number of chiral centers. The compound can have the R or S configuration its one chiral center. As shown below, the total number of stereoisomers for the compound is 2.
Which of the following compounds CANNOT undergo a self-aldol condensation?
An aldol condensation is a reaction that combines two carbonyl compounds to form an α,ß-unsaturated carbonyl. In a self-aldol condensation, the two carbonyl compounds are identical. Therefore, the carbonyl compound that is deprotonated to form an enolate is identical to the compound that is attacked by an enolate. The question asks which of the compounds CANNOT undergo an aldol condensation — thus, we are looking for the compound that cannot be enolized. A compound with no protons bound to the alpha carbon cannot be enolized. Option C has no alpha protons, as shown below.
What is the product of the following reaction?
The NMR spectrum has five distinct chemical shifts, which indicates that the molecule has five chemically unique protons, as shown below. Additionally, the compound lacks a signal between 9 and 12 ppm. Thus, the compound is not an aldehyde or a carboxylic acid.
What is the first step in the following reaction?
Which of the following solvents would favor an SN2 reaction over an SN1 reaction?
The reaction shown is syn-dihydroxylation of an alkene because two OH groups are added to the same face of the alkene. The reagent for this reaction is cold, dilute KMnO4 in the presence of NaOH. OsO4 in the presence of NaHSO3/H2O could also be used, but it is not shown in the option choices.
Which pair of structures represents tautomers?
Keto-enol tautomerism arises from the movement of pi electrons and protons. It is defined as the equilibrium that occurs between the keto form (ketone or aldehyde) and the enol form (hydroxyl bonded to alkene carbon), as shown below. The pair of molecules in Option A are tautomers.
Therefore, Option A is the correct answer.
Which of the following is true regarding the initiation step of free radical reactions?
There are three steps in a free radical reaction: initiation, propagation, and termination. Initiation consists of one stable molecule splitting into two radicals (homolytic cleavage). Propagation consists of one radical reacting with a stable molecule to propagate the production of another radical. Termination consists of two radicals coming together to form a stable molecule.
Free radicals are unstable, which means their formation is energetically unfavorable. As a result, lots of energy is required to generate them. In a free radical reaction, the formation of free radicals (initiation step) has the highest activation energy and is the rate-determining step. On the other hand, in the termination step, radicals react to form stable molecules, making the reaction energetically favorable.
The pentavalent transition state is the highest energy configuration of atoms during an SN2 reaction. It consists of the electrophile, nucleophile (Nu), and leaving group (LG), as shown below:
The dashed lines represent partial bonds (bonds that are forming/breaking), and the delta +/- signs represent partial charges. There is a partial positive charge on the carbon atom because it is an electrophile, which, by definition, is electron-poor. On the other hand, the nucleophile has a partial negative charge because it is, by definition, electron-rich. The leaving group also has a partial negative charge because when it fully detaches from the molecule, it will be a negatively charged species.
Chlorine could be a leaving group or nucleophile, and the same can be said about bromine. However, regardless of their respective roles in the reaction, they both have partial negative charges
A separatory funnel contains water and dichloromethane. Dichloromethane is denser than water. If NH3 were added into this funnel, where would it be found?
In chemistry labs, the separatory funnel allows chemists to separate immiscible liquids. We are told that dichloromethane is denser than water, so the organic DCM layer will be found below the aqueous water layer. NH3 and water are both capable of donating and accepting hydrogen bonds, and thus ammonia will be located in
the aqueous (top) layer. On the other hand, dichloromethane is incapable of hydrogen bonding, so it will be confined to the organic layer. Be careful — usually, the organic layer is above the aqueous layer, however, DCM is denser than water (the density was stated in the question). Ammonia will be found at the top of the aqueous layer.
The product of the following reaction is a(n) _______.
Reduction increases the number of C-H bonds and decreases the number of C-O bonds. Oxidation decreases the number of C-H bonds and increases the number of C-O bonds. The reaction sequence consists of a reduction step, acidic workup, and oxidation step.
The ester has three C-O bonds. Following reduction by lithium aluminum hydride, two of those C-O bonds are transformed into C-H bonds. The aqueous acid is simply an acidic workup. Next, PCC oxidizes the alcohol to an aldehyde by converting one C-H bond to an additional C-O bond, as shown below.
Bromoethane treated with magnesium forms a Grignard reagent. Grignard reagents are strong nucleophiles that attack carbonyl carbons. Reaction of a ketone with a Grignard reagent generates a tertiary alcohol after acidic workup, as shown below.
Treatment of an alkene with ozone followed by (CH3)2S will accomplish ozonolysis with a reductive workup. This reaction will cleave the carbon-carbon double bond to form two carbonyl groups. In order to figure out how many carbons the ring consisted of, count the number of carbons from one carbonyl carbon to the other:
Which of the following molecules is capable of both donating and accepting hydrogen bonds?
Hydrogen bond donors have FH, OH, or NH bonds. Hydrogen bond acceptors have lone pairs on F, O, or N. The only molecule listed that is capable of both donating and accepting hydrogen bonds is ammonia, as shown below.
What is the name of the reaction that occurs?
The Dieckmann Condensation is an intramolecular Claisen condensation. Claisen condensation reactions occur between two esters. Because Dieckmann condensation is an intramolecular Claisen condensation, a cyclic product is formed, as shown below
In the first step, ethoxide deprotonates the alpha carbon of an ester. The resulting enolate attacks the carbonyl carbon of the other ester. This forms a cyclic, tetrahedral intermediate. The intermediate then collapses and a ketone is formed when the ethoxy group on the ester is knocked off as a leaving group, as shown below
Which of the following is best used to extract ethylamine from a solution of dichloromethane?
Ethylamine is capable of hydrogen bonding because it has a hydrogen atom directly bonded to nitrogen. Remember, “like dissolves like” so ethylamine will dissolve best in a polar solvent that can also form hydrogen bonds. Dichloromethane cannot form hydrogen bonds and is only slightly polar. Therefore, in a liquid-liquid extraction, ethylamine would be extracted from the organic layer (dichloromethane) into an aqueous layer made up of a solvent capable of hydrogen bonding.
CCl4, benzene, and hexanes are all nonpolar solvents, so they would not be capable of dissolving ethylamine. THF is a moderately polar solvent, but it is not capable of donating hydrogen bonds because it does not have a hydrogen atom directly bonded to O, N,, or F. In contrast, water is a polar solvent that is capable of hydrogen bonding due to the hydrogen atoms bonded directly to oxygen. Ethylamine will dissolve well in water, so water can function as the aqueous layer to extract ethylamine.
Which reagent would transform the following molecule to a molecule that does not rotate plane-polarized light?
Reaction with the strong oxidating agent KMnO4 will transform the hydroxyl group on the molecule into a carboxylic acid (COOH). This will make the molecule achiral because the chiral carbon will no longer be attached to four different groups. Achiral molecules do not rotate plane-polarized light, as shown below.
Which of the following dissolves better in water: CH3CH2OH or CH3(CH2)4OH?
Which of the following dissolves better in water: CH3CH2OH or CH3(CH2)4OH?
Which of the following structures is associated with the following 1H NMR spectrum?
The number of 1H NMR signals is equal to the number of non-equivalent protons. Protons are equivalent when they exist in the same chemical environment.
Chemical equivalence: To be considered equivalent, hydrogens must exist in the same chemical environment. In this problem, there are 8 hydrogen atoms. However, not all the hydrogens exist in different environments. The hydrogen atoms that are bound to the same carbon are chemically equivalent. Chemically equivalent hydrogen atoms are highlighted in the same color. In this example, even though there are 8 hydrogen atoms, they exist within 4 chemically different environments.
Splitting/multiplicity: To determine splitting, use the n+1 rule, where n is the number of hydrogen atoms attached to neighboring carbon atoms. For example, the group of hydrogen highlighted in pink are attached to 2 carbon atoms. The carbon atom to the right carries 2 hydrogens (blue) while the carbon on the left carries 3 hydrogens (pink). This would mean that the carbon atom in the middle has a splitting of 6, or a sextet.
Chemical shift: The chemical shift indicates how close a proton is to an electronegative atom. In the compound, the only electronegative atom is oxygen. So, the hydrogens that are closest to oxygen are shifted more to the left (downfield), while hydrogens that are farther away from the oxygen will be more upfield.
Molecule B produces four signals, and the multiplicity of each signal matches the given NMR spectrum, as shown below.
Therefore, Option B is the correct answer.
You may be inclined to think the molecules are stereoisomers, but they are actually identical. The molecules are achiral because they do not contain any stereocenters. Remember, a stereocenter (chiral center) is a tetrahedral carbon atom attached to four unique groups. None of the carbons are attached to four unique groups. Therefore, the molecules are identical as they are structurally equivalent, as shown below.
Therefore, Option A is the correct answer.
Although cyclohexane is often shown in a planar view (Haworth representation), that’s not how it exists in 3D space. Cyclohexane is lowest in energy when it adopts chair-like shape. Almost every group prefers being equatorial to being axial. If a group is axial, it experiences steric strain from bumping against the two other groups that are axial and pointing in the same direction. For this compound, the bulkier tert-butyl group must be in the equatorial position to minimize steric interactions. From there, we can conclude that the chloro group must be in the axial position due to the trans relationship between the two groups. The most stable conformation is Option B, as shown below.
This is a self-aldol condensation because the reaction occurs between two identical carbonyl compounds. In the first part of aldol condensation, an alpha proton on the aldehyde is removed to generate an enolate ion. The enolate then reacts with an aldehyde to form a β-hydroxy aldehyde or aldol. In the second part of aldol condensation, the aldol undergoes dehydration (elimination reaction) to form an α, β-unsaturated carbonyl or a conjugated enone, as shown below.
The maximum number of stereoisomers for any chiral compound is equal to 2n with n being the number of stereocenters (chiral centers). Therefore, 24 or 16 stereoisomers are possible for the compound.
There are three ways to hydrogenate an alkyne with three possible products: cis alkene, trans-alkene, or alkane. A cis alkene is formed with H2/Lindlar’s Catalyst. A trans-alkene (Option A) is formed with Na or Li in ammonia, NH3. An alkane is formed with two equivalents of H2 on metal catalyst such as Pd, Pt, and Ni. Trans hydrogenation of an alkyne (Option A) occurs via a free radical reaction, as shown below.
Therefore, Option A is the correct answer.
In an electrophilic aromatic substitution reaction (EAS), benzene acts as a nucleophile and attacks an electrophile. After this happens, the ring is in a cationic state (arenium ion). The reaction ends when the tetrahedral carbon intermediates are deprotonated to regenerate aromaticity. Option D shows the arenium ion intermediate of a generic electrophilic aromatic substitution reaction, as shown below.
Therefore, Option D is the correct answer.
Which functional group is present in CH3OCH3, CH3CH2OCH3 and CH3CH2CH2OCH3?
sp2
The longest carbon chain is seven carbons long, and the highest priority functional group is the ketone. Thus, the name of the parent chain is heptanone (“-one” suffix indicates that ketone is the highest priority group). We number the parent chain from right to left to give the ketone the lowest possible locant. The two substituents on the parent chain are the amine on carbon 5 and the hydroxyl group on carbon 6. By listing the substituents in alphabetical order, we get 5-amino-6-hydroxyheptan-2-one as the name of this compound, as shown below.
Therefore, Option B is the correct answer.
A strong base like ethoxide performs E2 on a secondary alkyl halide, as shown below. The rate of an E2 reaction is rate = k[substrate][base]. An increase in concentration from 2M to 6M means the concentration was tripled. If the concentration of the base is tripled, the rate will also triple.
Which of the following compounds would react with Br2/ CCl4 and make the brown color disappear?
To determine number of degrees of unsaturation in a molecule, count the number off rings and pi bonds present, where each ring or pi bond corresponds to one degree of unsaturation. In the given molecule, there are two rings and three pi bonds, adding up to a total of five degrees of unsaturation, as illustrated below.
Therefore, Option D is the correct answer.
Pyrrole is an aromatic compound. The nitrogen is sp2 hybridized because one of its lone pairs participates in aromaticity, as shown below. For the lone pair to participate in aromaticity, it must be in a p orbital. Therefore, nitrogen has 3 electron domains and is sp2 hybridized.
The conversion of a carboxylic acid to an ester in the presence of strong acid is known as Fischer esterification. In the reaction, the -OH in a carboxylic acid is replaced by the -OR group of an alcohol. An acid catalyst is needed for Fischer esterification, as shown below.
Therefore, Option B is the correct answer
If a Diels-Alder reaction occurs with a cyclic diene, the product will be a bicyclic compound. In the case of a bicyclic product, the major product of a Diels-Alder reaction is the endo product. Because the dienophile is cis, the two substituents of the dienophile are cis in the bicyclic product. The endo product has both substituents pointing in, towards the alkene, as shown below:
This is an SN2 reaction because an azide ion is a strong nucleophile, and the starting molecule (secondary alkyl halide) is not sterically hindered. The starting material has an (R) configuration to begin. SN2 reactions function via backside attack of the electrophile by the nucleophile. This results in inversion of stereochemistry. The final product of the reaction will be (S) due to inversion of stereochemistry, as shown below.
Therefore, Option B is the correct answer.
Phenol, benzaldehyde, and aniline (Options A, D, and E) are insoluble in water. They all have a phenyl group (benzene as a substituent group is called phenyl), which is completely nonpolar. Pentanoic acid (Option B) has a hydrocarbon chain longer than 4 carbons, so it is insoluble as well. Methyl ethanoate (Option C) is slightly soluble in water as esters engage in very little hydrogen bonding. However, aniline is different from the other options because the amine group can act as a base at low pH. When it becomes protonated in acid, aniline gains a net-positive charge that significantly enhances its solubility in water despite the nonpolar phenyl group.
Therefore, Option E is the correct answer.
Question #26
What is the C-N-C bond angle in triethylamine?
A. 90° [1%]
B. 107° [34%]
C. 109.5° [37%]
D. 120° [19%]
E. 180° [9%]
Incorrect
The correct answer is: B.
34%
Answered Correctly
0 min, 56 secs
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Aromatics & Bonding
Question Category
Concept
In a perfect tetrahedral structure consisting of a central atom bonded to four identical groups or atoms, the bond angle is 109.5°. A lone pair, however, occupies more space than bonds. As a result, the bond angles shrink with more lone pairs. A tetrahedral structure with one lone pair, as shown below, will have bond angles of 107°. Water, however, has two lone pairs on the oxygen, so the bond angles will be even smaller at 104.5°. The only value listed that is slightly smaller than 109.5° is 107°. The structure of triethylamine is shown below.
Column chromatography utilizes a stationary phase of silica gel (which contains hydroxyl groups) and a mobile phase which contains the compounds of interest dissolved in a nonpolar solvent or solvent mixture of intermediate polarity. The more polar groups a compound has, the more the compound will interact with the silica gel. Molecules that contain nitrogen, oxygen, or fluorine are able to act as hydrogen bond acceptors. Molecules that contain nitrogen, oxygen, or fluorine bound to hydrogen are able to act as both hydrogen bond donors and acceptors.
The more a compound interacts with the silica gel, the longer it takes for the compound to move through the column. Thus, column chromatography separates compounds based on differing polarities. The less a compound interacts with the silica gel (nonpolar compounds), the faster the compound will move through the column.
Therefore, Option C is the correct answer.
The reaction shown is hydroboration oxidation of an alkene. Hydroboration oxidation is an addition reaction that converts alkenes into alcohols in an anti-Markovnikov fashion. In the reaction, H and OH are added across the double bond such that OH is added to the less substituted carbon, and H is added to the more substituted carbon, as shown below.
fractional distillation
Distillation is the process of heating a liquid until it boils, then condensing and collecting the resultant hot vapors. The glassware pictured is a condenser. It cools and ultimately condenses vapors to liquid droplets that flow into the receiver flask, as shown below
vacuum filtration
vacuum filtration does not require a condenser. Vacuum filtration is a technique that separates particles of a mixture based on their physical size. Bigger particles make it through the filter and into the beaker, and larger particles remain stuck on the filter.
extraction
because extraction does not involve a condenser. Extraction is used to separate components of a mixture based on their relative solubilities in immiscible liquids.
crystallization
recrystallization does not require a condenser. This process purifies solids by dissolving an impure solid in a solvent at a high temperature, cooling the solution down, and separating pure solid crystals from the dissolved impurities through filtration. Recrystallization is based on the principle that solids are more soluble in hot liquids than they are in cool liquids.
rotary evaporation
This method is used to remove the solvent from a solution by rotating the flask rapidly to provide a greater surface area and for evaporation to occur. The rotary evaporator creates a low-pressure environment inside the flask to facilitate solvent evaporation. The vapors are then condensed by cooling coils and collected in a separate flask for recycling or proper disposal.
In electrophilic aromatic substitution reaction (EAS), the location of a new substituent on a benzene ring is determined by the existing substituent. Typically, electron-withdrawing groups act as meta directors, while electron-donating groups act as ortho/para directors.
In the given molecule, both substituents are activating groups. To determine the directing effect, we need to consider the strongest activating substituent. Hydroxyl groups are stronger activators than alkyl groups since they donate their lone pairs into the ring. Meanwhile, the methyl group only weakly activates the ring through hyperconjugation. Thus, the next substituent will be added to the ortho position relative to the hydroxyl substituent.
Refer to the table below for a list of common electron-withdrawing and electron-donating groups.
Ozonolysis cleaves carbon-carbon double or triple bonds down the middle. As a result, two carbonyl (C=O) containing compounds are formed. Ozonolysis of alkenes with a reductive workup produces aldehydes and ketones. Ozonolysis of alkenes with an oxidative workup produces carboxylic acids and ketones. Thus, ozonolysis does not produce an alkene.
ozonolysis with a reductive workup
ozonolysis with a oxidative workup
ozonolysis of a terminal alkyne
ozonolysis of an internal alkyne
Hydroxide acts as a base to deprotonate the hydroxyl group of the compound, causing an intramolecular SN2 reaction to occur. The result is a cyclic ether, as shown by the reaction mechanism below
For a compound to act as a base by the Bronsted definition, it must have a lone pair that can reach out and grab a proton. The only options listed that have a lone pair that can attack a proton are Options B, C and D.
Between Options B, C and D, trimethylamine (Option D) is the most basic because nitrogen is less electronegative than oxygen. Thus, the lone pair of nitrogen is most able to reach out and steal a proton from an acid.
because NH4+ cannot act as a base; it can only act as an acid. This is because unlike NH3, NH4+ does not have a lone pair on nitrogen to accept a proton.
amine
ketone
carboxylic acid
acetal
An electrophile is a reactant that is
electron-poor.
Electrophiles get attacked by nucleophiles, as shown below.
electron pair acceptor
An enolate can only be formed from one of the starting compounds because benzaldehyde has no alpha protons. After the ketone is deprotonated, the resulting enolate attacks the carbonyl carbon of benzaldehyde and forms an aldol. The aldol undergoes dehydration due to the presence of heat, giving the final thermodynamically stable α,β-unsaturated ketone (Option A), as shown below.
secondary aminie
When treated with KMnO4, primary alcohols are converted to carboxylic acids and secondary alcohols are converted to ketones. Carboxylic acids have higher boiling points than ketones because they are hydrogen bond donors and acceptors. The only primary alcohol listed is 1-propanol. Therefore, the product of the reaction between 1-propanol and KMnO4 will have the highest boiling point, as shown below.
A polar aprotic solvent is the best solvent for an SN2 reaction because it does not engage in hydrogen bonding. When a solvent forms hydrogen bonds with a nucleophile, solvent molecules surround the nucleophile, and this makes the nucleophile less nucleophilic. The polar protic solvent also stabilizes the carbocation intermediate that forms during an SN1 reaction. Remember: aprotic solvents favor SN2 reactions, whereas protic solvents favor SN1 reactions. The only polar aprotic solvent listed is acetone.
Butanamide (shown below) has the highest melting point as it forms more hydrogen bonds than the other compounds. Amides have high melting points because both the carbonyl group and N-H bonds engage in hydrogen bonding. Although butanoic acid, butylamine, and 1-butanol are also capable of forming hydrogen bonds, they do not do so to the same degree as butanamide.
The free radical addition of HBr to an alkene is shown in the question stem. Like regular HBr addition, this reaction adds an H and a Br, but it produces the anti-Markovnikov product. The reaction occurs via a free radical chain mechanism. In the initiation step, peroxides are used to form bromine radicals. In the propagation step, an alkyl bromide is produced when a bromine radical combines with an alkyl radical. As suggested by the name of the reaction, free radical addition is the most appropriate answer.
Option B is incorrect because free radical substitution occurs on alkanes. In free radical substitution, a hydrogen on an alkane is substituted with a bromine or chlorine. For reference, free radical bromination is shown below. In pic of this card.
Which of the following reagents is capable of transforming CH3CH2CH2CHO into CH3CH2CH2COOH?
A Diels-Alder reaction between ethyne (dienophile) and 1,3-butadiene (diene) forms 1,4-cyclohexadiene. The addition of mCPBA to 1,4-cyclohexadiene forms two epoxides because the molecule has two carbon-carbon pi bonds, as shown below.
To obtain the final product, the starting carboxylic acid must first be converted to an acid chloride. This is achieved using SOCl2, which replaces the hydroxyl group with a chlorine atom, forming an acid chloride. The resulting acid chloride is then reacted with a carboxylate ion in a nucleophilic acyl substitution reaction, which forms an acid anhydride as shown below. The use of a carboxylate ion in this step allows for the displacement of the chloride ion, which is a good leaving group.
In summary, the SOCl2 is used to facilitate the conversion of the carboxylic acid into a more reactive acid chloride, while the subsequent nucleophilic acyl substitution reaction with a carboxylate ion leads to the formation of the desired acid anhydride.
When a ketone is reacted with a primary amine in the presence of a catalytic acid (H2SO4), an imine is formed. In the first step of the mechanism, the carbonyl oxygen is protonated, which causes the carbonyl carbon to become more electrophilic. The amine nitrogen attacks the carbonyl carbon, and the pi electrons in the C-O double bond are pushed up onto the oxygen atom. To get rid of nitrogen’s positive charge, a proton is transferred from the amine nitrogen to the OH group. Next, a lone pair on the nitrogen moves down to form a C-N double bond, and a water molecule is kicked off. Finally, an imine is formed when the nitrogen is deprotonated, as shown below.
Like in 1H NMR spectroscopy, the number of 13C NMR signals is equal to the number of non-equivalent carbons. Carbons are equivalent when they exist in the same chemical environment. In total, we have 6 different signals (shown below) for this compound because there are carbon atoms existing in 6 different chemical environments.
What is the major product of the following reaction?
What is the major product of the following reaction?
Recall that neutral carbon has 4 bonds. In cyclohexane, each carbon is single bonded to 2 other carbons, so we can conclude that the remaining 2 bonds are to 2 hydrogens. Since all the carbons are single bonded to four distinct entities, every carbon has a tetrahedral geometry. Thus, the C-C-C bond angle is 109.5°, as shown below.
Phenol is benzene with a hydroxyl (—OH) substituent. Looking through all the options, we can see that all the compounds are disubstituted phenols. To determine the most acidic compound, we must shift out focus to the other substituents on each compound. Compound E contains a very strong electron-withdrawing group. The nitro (—NO2) group will withdraw electron density via resonance to stabilize the negative charge on the conjugate base,
We can assess the acidity of the compounds by looking at the stability of their conjugate bases. To compare the stability of conjugate bases, use the acronym CARDIO (charge, atom, resonance delocalization, induction, orbital hybridization).
Compound 2, Hydroiodic acid (HI) is the strongest acid. The large size of an iodine atom causes the negative charge of the conjugate base to be very stable. Additionally, the bond between hydrogen and iodine to be very weak.
Compounds 3 and 4 are both carboxylic acids. Their acidic proton is oxygen, and their base is resonance stabilized. However, compound 4 has a chlorine on the alpha carbon. Chlorine atoms are electronegative, and thus provide stabilization of the conjugate base via inductive effects. Therefore, compound 4 is more acidic than compound 3.
Finally, NaH is a salt of sodium and hydride. Hydride is a base, not an acid. Therefore, NaH is not acidic at all, but rather is incredibly basic.
The correct order of most acidic to least acidic: 2 > 4 > 3 > 1, as shown below.
Option A is incorrect because ClCH2COOH is more acidic than CH3COOH due to the inductive effects of the chlorine atom. In addition, HI is the more acidic compound due to the electronegative nature of iodine.
PCC is a weak oxidizing agent that converts a primary alcohol to an aldehyde and a secondary alcohol to a ketone. Therefore, the primary alcohol will be oxidized to form the aldehyde in Option B, as shown below. In order to oxidize the primary alcohol all the way to a carboxylic acid, a stronger oxidizing agent is needed.
Which of the following is an intermediate during the addition of ethyl magnesium bromide to 2-butanone?
The reaction being described is the reaction of a Grignard reagent with a ketone. Ethyl magnesium bromide adds to 2-butanone to form the alkoxide ion intermediate in Option C, as shown below. MgBr+ is a counterion.
Step 1 of the reaction sequence is a Diels-Alder reaction in which the conjugated diene reacts with a dienophile to yield a cyclohexene derivative. We note that since the dienophile has two methyl groups situated on the same carbon, this must be retained in the product. Thus, we can rule out molecules A and B.
mCPBA (meta-Chloroperoxybenzoic acid) is a peroxy acid, which is the reagent used for the epoxidation of alkenes. Epoxidation of the alkene formed from the Diels Alder reaction gives molecule C, as shown below.
Therefore, option C is the correct answer
Butan-1-ol is reacted with concentrated sulfuric acid in heat conditions. Which of the following is an intermediate of this dehydration?
An alcohol in the presence of concentrated sulfuric acid undergoes dehydration to form an alkene. In the first step of the reaction, the -OH group is protonated to create a good leaving group (water). Then, another alcohol removes a proton on an adjacent carbon, causing a new C-C pi bond to form and knocking off a water molecule as a leaving group, as shown below.
Which of the following molecules has the lowest pKa?
The stronger the acid, the lower its pKa. Thus, we are looking for the strongest acid. All the molecules listed are ketones. Thus, the acidic proton for each is the alpha proton of a ketone, which is resonance stabilized. To determine the most acidic compound, the different inductive effects of the various substituents must be analyzed, as shown below:
Option A is incorrect because a ketone contains a carbonyl group.
Option B is incorrect because an aldehyde contains a carbonyl group.
Option C is incorrect because an amide contains a carbonyl group.
Option D is incorrect because a carboxylic acid contains a carbonyl group.
As an aromatic compound, benzene is a Lewis base because it donates one of its electron pairs in an electrophilic aromatic substitution reaction
How many signals would you find on a 13C NMR spectrum of the following compound?
Like in 1H NMR spectroscopy, the number of 13C NMR signals is equal to the number of non-equivalent carbons. Carbons are equivalent when they exist in the same chemical environment. Due to the internal plane of symmetry of the molecule, it has four sets of chemically distinct carbon atoms, as shown below. Four signals will appear on a 13C NMR spectrum of this compound.
Azide ions are weakly basic but extremely nucleophilic. Therefore, the azide ion will perform SN2 on the substrate molecule by displacing bromo group with an inversion of stereochemistry, as shown below. Acetone is the solvent used in the reaction and does not actively participate in the reaction.
If the benzylic carbon of an alkylbenzene is bound to at least one hydrogen atom, treatment with KMnO4, OH– and H3O+ will result in oxidation to give a carboxylic acid. The rest of the substituent will be cleaved. In this case, the product has two carboxyl groups, so there must be two benzylic carbons on the substrate. The structure of Option E allows benzylic oxidation to occur twice because the two carbons on the benzene ring are directly linked to the cyclic alkyl group, as shown below.
Treatment of an alkyne with Na(s), NH3(l) produces a trans alkene. In the second step of the reaction, the alkene undergoes epoxidation in the presence of mCPBA (a peroxyacid). Epoxidation of a trans alkene results in an epoxide which has the alkyl groups trans to one another, as shown below. Note that the enantiomer is also observed, however, is not an option choice.
A carbon radical is sp2 hybridized with a single unpaired electron occupying the unhybridized p orbital. The formal charge of a carbon atom with three bonds and one unpaired electron is 0. Formal charge is calculated by the following formula:
Formal charge = (# of valence electrons of the atom) – (bonds + dots)
In this problem, 0 = 4 – (3 + 1).
Therefore, Option C is the correct answer.
