Quimica Organica Spellbook

Question 1

Mass spectrometry involves the production and detection of:

A) positively charged species

B) negatively charged species

C) neutral species

D) positively and negatively charged species

Answer 1

(A). Mass spectrometry uses high energy to break the covalent bonds in molecular compounds and the instrument only detects the positively charged species resulting in this fragmentation process.

Question 2

Which of these statements best explains why an aromatic alkyl halide is most reactive?

A) the ether product formed from it is the most thermodynamically stable

B) the activation energy needed to form a carbocation from it is the lowest

C) it is the least sterically hindered to nucleophilic attack by ethanol

D) the ether product formed from it is more soluble in ethanol

Answer 2

B. The lower the reaction’s activation energy is, the faster its rate will be.

Question 3

What are the conditions necessary for a molecule to be antiaromatic?

Answer 3

• Must be cyclic
• Must be planar
• Must have a completely conjugated system of p-orbitals in the ring of the molecule
• Must have 4n pi electrons (4, 8, 12, etc…)*

* The number of double bonds multiplied by 2.

Question 4

What is the reaction kinetics associated with S_N1 and S_N2 reactions?

Answer 4

S_N1: First order kinetics k[Substrate]

S_N2: Second order kineitcs k[Nu][Substrate]

Question 5

How many isoprene units make up the structure of the terpene compound caryophyllene?

Answer 5

3 terpene units. All terpene compounds are composed of different numbers of isoprene molecules. Each isoprene molecule contains 5 carbons. The compound shown is composed of 3 isoprene units because it contains 15 carbons.

Question 6

Which of the following is not true about the reaction solvent used (THF) to prepare the Grignard reagent?

A) it conains an ether functional group

B) it is capable of hydrogen bonding

C) it helps solvate the grignard reagent as it forms

D) it is a cyclic compound

Answer 6

(B) THF is a cyclic saturated ether. Ethers cannot hydrogen bond.

Question 7

What would be the strongest expected intermolecular interaction between sodium acetate (the solute) and water (the solvent) in a solution?

A) hydrogen bonding

B) dipole-dipole interaction

C) ion-dipole interaction

D) dispersion forces

Answer 7

(C) The solute is the salt of acetic acid and being ionic would completely ionize when dissolved in water–the solvent. The sodium and acetate ions would be surrounded and seperated by polarized water; thus the interactions between the solute and solvent species are ion-dipole.

Question 8

The λ_max UV/VIS for trans-β-carotene (a yellow compound with elevent conjugated bonds) would be closest to:

A) 300 nm

B) 350 nm

C) 400 nm

D) 500 nm

Answer 8

(C) Because this is a yellow compound, it absorb radiation in the violet (~400 nm) part of the visible region of the electromagnetic radiation.

Question 9

Which is more stable: cis-1,3-dibromocyclohexane or trans-1,3-dibromocyclohexane?

Answer 9

cis-1-3-dibromocyclohexane because the bromines can both be equatorial but in trans, they cannot.

Question 10

A compound was produced as a cis/trans isomer mixture. What would probably be the best laboratory method for separation this mixture?

a) recrystallization
b) extraction
c) chromatography
d) distillation

Answer 10

(C) their boiling points are likely to be similar, so separation by chromatography techique would exploit the small differences in the isomers affinity for a carefully selected stationary phase..

Question 11

The Lucas test distinguishes between the presence of primary, secondary, and tertiary alcohols based on reactivity with a hydrogen halide. The corresponding alkyl chlorides are insoluble in Lucas reagent and turn the solution cloudy at the same rate that they react with the reagent. The alcohols A, B, and C are solvated separately in Lucas reagent of hydrochloric acid and zinc chlroide. If alcohols are primary, secondary, and tertiary, respectively, what is the order of their rates of reaction from fastest to slowest?

Answer 11

C, B, A. S_N1 reactions are much faster than S_N2 reactions, which would only react in the primary alcohol. So because this is S_N1, it will go from most stable (C) to least stable (A).

Question 12

What are the reaction kinetics for S_N1 and S_N2?

Answer 12

S_N1: rate = k[S]

S_N2: rate = k[S][Nucleophile]

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Question 55

Which of the following alkanes will have the lowest heat of combustion per CH₂ group?

A) cyclopropane

B) cyclobutane

C) cyclohexane

D) cycloheptane

Answer 55

(C). The potential energy of a cycloalkane is a function of its torsional, steric, and ring strain. When it undergoes combustion, cyclohexane yields less than heat per methylene because it is initially at a lower potential energy per methylene.

Question 56

Cholesteriol is a precursor for which of the following hormones:

A) Insulin

B) Gastrin

C) Thyroxine

D) Estrogen

Answer 56

(D) Cholestrol is a structural component of plasma membranes and a precursor of steroid hormones.

Question 57

The most likely function of an ebulliator is to:

a) keep the condensed vapors cool in the recieving flask
b) promote the establishment of a high vacuum in the system
c) prevent superheating of the liquid to be distilled
d) provide an outlet when the pressure inside the system becomes too high.

Answer 57

(C). Ebulliators introduce small air bubbles into the system which break surface tension of the liquid being heated to prevent superheating and bumping.

Question 58

Which of the following experimental modifications will most likely improve the degree of separation between limonene and (+)-carvone.

A) Heating the distillation flask at a slower rate

B) Using a vacuum source that can achieve a lower pressure inside the distillation apparatus

C) Cooling the condenser with ice water

D) Using a shorter fractionating column

Answer 58

(A). The separation of two liquids takes place in the fractionating column as two liquids vaporize and condense, with the lower boiling liquid distilling first. If the fractionating column is shortened, the liquid will vaporize and condense fewer times and degree of separation will worsen. Cooling the condenser will have no effect on separation. Creating a lower pressure inside the distilling apparatus will lower the boiling points of both liquids and narrow their difference in boiling point, making it necessary to lengthen the column. Heating at a slower rate will allow both liquids more time in the fractionating column to increase the frequency of vaporization.

Question 59

If a leak develops in the vacuum distillation apparatus, the boiling points of the two components of caraway seed oil will:

a) both increase
b) both decrease
c) both remain the same
d) become more similar

Answer 59

(A) The boiling point of a liquid is the temperature at which vapor pressure of the liquid equals the surface pressure. The normal boiling point is measured at 1 atm pressure. The vapor pressure of a liquid increases with increasing temperature. Hence, the boiling point of a liquid decreases as the pressure on the surface of the liquid is decreased. If a leak develops, the surface pressure will increase as will the boiling points.

Question 60

The aldol self-condensation of acetone is an equilibrium that favors acetone over its condensation product. Which of the following experimental modifications is most likely to shift the position of equilibrium toward Product A?

A) Using only a catalytic amount of NaOH

B) Using only a catalytic amount of acetone

C) Removing Product A as it is formed

D) Increasing the reaction temperature to the boiling point of acetone

Answer 60

(C) Le Chatelier’s Principle. Heating acetone to a boiling point would remove the reactant and shift equilibrium to the left.

Question 61

The product of an aldol self-condensation of acetone was treated with hot sulfuric acid to dehydrate it into the new product. When a drop of Br₂ in CCl₄ is added to the new product, the resulting solution will be:

A) colorless because the new product has no carbon-carbon double bonds

B) colorless because the new product has carbon-carbon double bonds

C) red because the new product has no carbon-carbon double bonds

D) red because the new product has carbon-carbon double bonds

Answer 61

(B) The red bromine will disappear as bromine adds to the double bond.

Question 62

Which of the following compounds will give a positive iodoform test?

I. Acetone

II. The product of self-aldol condensation of acetone

III. The product of self-aldol condensation of acetone treated with hot sulfuric acid to result in the corresponding dehydration product

Answer 62

I, II and III. All three compounds are methyl ketones.

Question 63

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Question 64

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Question 65

Benzene, phenol, acetoacetic acid, and ethanamine are all mixed together in a separatory funnel. If hydrochloric acid is added, which of these chemicals would be protonated?

Answer 65

Ethanamine is the only basic compound out of these four, so it would be protonated by hydrochloric acid.

Question 66

Which of the following compounds is expected to have the lowest vapor pressure?

A

CH₃OCH₃

B

CH₃CH₂OH

C

CH₃CH₂CH₃

D

CH₃Cl

Answer 66

(B). The -OH hydroxyl groups will form hydrogen bonds with one another, which are the strongest of the intermolecular forces. This hydrogen bonding provides a strong attractive force for the molecules, increasing the amount of energy required to evaporate the substance, thus decreasing the vapor pressure.

Question 67

Toluene is treated with excess concentrated nitric acid, HNO₃. Given that CH₃ is an ortho/para director, what is the product of the reaction?

Answer 67

The first addition of NO² will occur directly adjacent, or directly opposite to the carbon that is bonded to the CH₃. From the carbon bonded to the NO₂, every other carbon is at the meta position, as well as being either ortho- or para- to the carbon bonded to the CH₃. This means that an NO₂ will be added to every other carbon on the benzene ring, producing trinitrotoluene.

Question 68

Oxidative stress damages cell membranes, proteins, and other cellular structures. Antioxidant defenses such as ascorbate play a key role in reducing the effects of oxidative stress. An example of antioxidant defenses is the ascorbate-glutathione cycle which helps to process H₂O₂ to H₂O in the cell, according to the following simplified scheme: The red couplings indicate enzymes which catalyze the oxidation of one molecule and result in the subsequent reduction of another molecule. Glutathione is a tripeptide with a cysteine in the middle. The cysteine contains the sulfhydryl group -SH which is converted to a disulfide bridge, forming GSSG, a molecule consisting of two covalently bonded glutathione tripeptides. Which of the following best describes the reaction taking place from two GSH molecules to GSSG?

A. oxidation

B. reduction

C. esterification

D. transpeptidation

Answer 68

(A). First, you can work from left to right in the diagram presented to you. Hydrogen peroxide is an oxidizing agent, thus it is reduced in its reaction with ascorbate. The result is dehydroascorbate, the oxidized version of ascorbate. In order to replenish ascorbate in the cell, glutathione must reduce ascorbate, i.e. be oxidized to GSSG.

Question 69

Oxidative stress damages cell membranes, proteins, and other cellular structures. Antioxidant defenses such as ascorbate play a key role in reducing the effects of oxidative stress. An example of antioxidant defenses is the ascorbate-glutathione cycle which helps to process H₂O₂ to H₂O in the cell, according to the following simplified scheme: The red couplings indicate enzymes which catalyze the oxidation of one molecule and result in the subsequent reduction of another molecule. This cycle plays a key role in the degradation of the superoxide radical, one of many types of free radicals that play an important role in biological systems. The superoxide radical is converted to hydrogen peroxide via the enzyme superoxide dismutase. Which of the following best describes the classification of superoxide dismutase?

A. hydrolase

B. synthase

C. oxidoreductase

D. isomerase

Answer 69

(C). Oxidoreductases, as implied by the name, catalyze reactions involving the transfer of electrons. A free radical is an excellent target for an oxidoreductase to produce a less reactive, and thus less damaging molecule. Hydrolases catalyze reactions by the addition of water, synthases carry out synthesis reactions, for example, ATP synthase. Isomerases produce isomers of compounds; an isomerase could thus have no effect on the superoxide radical since it has only one isomer.

Question 70

Oxidative stress damages cell membranes, proteins, and other cellular structures. Antioxidant defenses such as ascorbate play a key role in reducing the effects of oxidative stress. An example of antioxidant defenses is the ascorbate-glutathione cycle which helps to process H₂O₂ to H₂O in the cell, according to the following simplified scheme: The red couplings indicate enzymes which catalyze the oxidation of one molecule and result in the subsequent reduction of another molecule. A biologist designs an experiment in which GSSG, dehydroascorbate, all necessary enzymes, and excess NADPH and H₂O₂ are present. She also adds pyridine, a compound which binds and inactivates glutathione. After a long time, which of the following statements describes the contents of the experiment? Assume that H₂O₂ is stable in solution unless otherwise transformed to H₂O via the ascorbate-gluathione cycle, and that pyridine inactivates all glutathione immediately upon formation.

A. there is no H2O2 left, no dehydroascorbate left, and no NADPH left

B. there is no H2O2 left, no ascorbate left, and no NADPH left

C. all the H2O2 is left, no GSSG is left, and all the NADPH is left

D. all the H2O2 is left, no GSSG is left, and no NADPH is left

Answer 70

(D). The rightmost reaction will proceed to completion, leaving behind no GSSG or NADPH. Because pyridine has inactivated GSH, the dehydroascorbate cannot be converted to ascorbate, and thus there is no ascorbate to drive the conversion of H2O2 to H2O. Thus, all of the H2O2 will be left, and no GSSG or NADPH will be left.

Question 71

Oxidative stress damages cell membranes, proteins, and other cellular structures. Antioxidant defenses such as ascorbate play a key role in reducing the effects of oxidative stress. An example of antioxidant defenses is the ascorbate-glutathione cycle which helps to process H₂O₂ to H₂O in the cell, according to the following simplified scheme: The red couplings indicate enzymes which catalyze the oxidation of one molecule and result in the subsequent reduction of another molecule. Catalase is another important enzyme which degrades H₂O₂ into less harmful byproducts. The catalase test is used to help differentiate bacterial species. In the catalase test, a drop of H₂O₂ is placed on a colony that is being tested. Which of the following would identify a catalase-positive colony

A. blackening of the colony

B. formation of bubbles

C. absorption of H₂O₂ to form a large mass

D. H₂O₂ will have no observable effects on a catalase-positive colony

Answer 71

(B). The degradation of H₂O₂ does not result solely in water. O₂ is a byproduct of this reaction (as you can determine by checking the stoichiometry of the reaction H₂O₂ –> H₂O). A bacterial colony which is catalase-positive will quickly degrade H₂O₂, resulting in the formation of O₂ bubbles. These bubbles are readily visible to the naked eye when H₂O₂ is placed on a catalase-positive colony.

Question 72

Rank the following compounds in order of the most acidic to the least acidic.

I. p-chlorobenzoic acid

II. benzoic acid

III. p-methylbenzoic acid

IV. p-nitrobenzoic acid

Answer 72

IV > I > II > III

The acidity of each of these compounds can be determined by the strength of the electron withdrawing properties of each functional group. The more electron-withdrawing a functional group, the more acidic the compound will become (since electron density will be drawn further from the hydrogen, making it more easily donated, hence more acidic). The nitro (NO2) group is the most electron-withdrawing, therefore ‘IV’ is the most acidic. The methyl (CH3) group is the only one listed that is electron-donating, hence III is the least acidic. Chlorine is electron withdrawing, more so than hydrogen, hence ‘I’ is more acidic than ‘II’

Question 73

How can 3-ethyl-2-pentanone be synthesized starting with acetoacetic ester?

A) By an alkylation of the enolate with isobutyl iodide

B) By successive alkylations with methyl iodide

C) By successive alkylations with ethyl iodide

D) By alkylation of the dianion with isopropyl iodide

Answer 73

(C) All other choices produce compounds different from 3-ethyl-2-pentanone

Question 74

The saponification product is acidified to convert:

A) an ester into an acid

B) an acid into a salt

C) a salt into an acid

D) an ester into a salt

Answer 74

(C) Saponification is the hydrolysis of an ester using aqueous hydroxide. The saponification product is a carboxylate salt which is acidified to the corresponding carboxylic acid.

Question 75

At room temperature, Compound A was insoluble in water and soluble in dilute HCl. It was also observed that Compound A slowly dissolved in refluxing NaOH to result in the formation of two new compounds, one of which was a compound with an amine functional group and a carboxylic acid functional group. Compound A most likely contains which of the following functional groups?

A) Ether

B) Ester

C) Ketone

D) Aldehyde

Answer 75

(B) The treatment of Compound A with hot NaOH produces two new compounds, ethanol and the amine/carboxylic acid product. Since the basic hydrolysis of Compound A produces an alcohol and a carboxylic acid.

Question 76

A sample of 4-aminobenzoic acid (m.p 188-189) is contaminated with a small percentage of camphoric acid. What can be determined about the melting point of this mixture?

A) The mixture will have a broad melting rage above 189 degrees Celsius

B) The mixture will have a broad melting range below 188 degrees Celsius

C) The mixture will have a sharp melting point of 186 degrees Celsius

D) The mixture will have a sharp melting point of 189 degrees Celsius

Answer 76

(B) An impurity in an organic solid lowers the melting point of the solid and broadens the temperature range of melting.

Question 77

Compound C has a melting point of 187-190 degrees Celsius and was formed when the remaining NaOH mixture was carefully acidified. Compound C was optically inactive and soluble in both dilute acid and base. Compound C most likely contains which of the following functional groups:

A) Carboxylic acid and amide

B) Amine and amide

C) Amide and nitro

D) Amine and carboxylic acid

Answer 77

(D) Carboxylic acids are soluble in dilute base and amines are soluble in dilute acid. Amides and nitro compounds are not soluble in dilute acid or base.

Question 78

The compound attached is the β-derivative of the D-glucuronic acid. The α-D-glucuronide differs in configuration from the β-derivative at:

A) C-1

B) C-2

C) C-3

D) C-4

Answer 78

(A) When cyclic hemiacetals and acetals are formed, the ring carbon atom derived from the carbonyl group in the open-chain form of a monosaccharide is called the anomeric carbon. The alpha and beta prefixes are employed to define the stereochemistry of the C-OR bond at the anormeric carbon atom.

Question 79

The compound formed by replacing the oxygen atom between the two carbonyl groups in acetic anhydride with an –NH- group is classified as an:

a) amide
b) imide
c) imine
d) enamine

Answer 79

(B) An amide is a N attached to a carbonyl. An imide is a N attrached to two carbonyls. An imine is an enol with a carbon double bonded to a N. An enamine is hella complicated.

Question 80

Equation 3 can be reversed by:

A) heating only

B) acidification only

C) heating followed by acidification

D) acidification followed by heating

Answer 80

(D) The correct reaction sequence to be used to reverse Equation 3 is shown below:

Question 81

The first step in the mechanism of the reaction shown in Equation 1 is:

A) protonation of a carbonyl oxygen atom by a hydrogen atom from the amino group

B) protonation of a carbonyl oxygen atom by a hydrogen atom from the water

C) attack at a carbonyl carbon atom by the lone pair of electrons on then nitrogen atom of the amino group

D) attack at a carbonyl oxygen atom by the lone pair of electrons on the nitrogen atom of the amino group

Answer 81

(C) The initial step in the mechanism involves nucleophilic attack by the lone electron pair on the nitrogen atom in RNH₂ on a carbonyl carbon atom in the substrate as shown below:

Question 82

According to the Cahn-Ingold-Prelog priority rules, which group bonded to the chiral carbon atom (x) in Figure 1 has the highest priority?

A) The H atom

B) The ring system

C) The NHCHO group

D) The isopropyl group

Answer 82

[C] The atoms directly bonded to the chiral carbon atom (x) are H, C, C, and N. Nitrogen has the highest atomic number and therefore the NHCHO group has the highest priority.

Question 83

What is the maximum possible number of stereoisomers of Compound 1?

A) 4

B) 8

C) 16

D) 32

Answer 83

[C] The number of stereoisomers can be calculated using the formula 2^N where n is the number of stereocenters in the molecule. Compound 1 has 4 stereocenters.

Question 84

Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?

A) isopropyl chloride

B) methyl chloride

C) sec-butyl chloride

D) tert-butyl chloride

Answer 84

[D] Tertiary alcohols react much more quickly with HCl than do other types of alcohols.

Question 85

If the ester shown below were hydrolyzed in acidic H₂¹⁸O, which product would be expected to contain ¹⁸O?

Answer 85

[A] Protonation of carbonyl oxygen occurs first to make acetic acid.

Question 86

Which of the following statements describes the solubility properties of fatty acid salts?

A) They are soluble in polar media only

B) They are soluble in nonpolar media only

C) They can partially dissolve in both polar and nonpolar media

D) They are completely insoluble in both polar and nonpolar media

Answer 86

[C] A fatty acid salt contains a long hydrocarbon chain which is insoluble in nonpolar solvents. The salt also contains the charged group which is soluble in polar solvents.

Question 87

How much sodium hydroxide is needed to completely saponify a triacylglycerol?

A) a catalytic amount because OH- is continuously being regenerated during saponification

B) 1/3 of an equivalent because each OH- ion reacts to form three fatty acid salts

C) One equivalent because each OH- reacts to produce one molecule of glycerol

D) Three equivalents because one OH- ion is required to saponify each of the three fatty acid groups

Answer 87

[D] One hydroxide ion is required to hydrolyze one ester linkage of a tricylglycerol molecule. With three ester linkages, we need three equivalents of sodium hydroxide.

Question 88

When glycerol reacts with three different fatty acids, how many stereogenic centers does the product triacylglycerol contain?

A) 0

B) 1

C) 2

D) 3

Answer 88

[B] Only carbon 2 in the resulting triacylglycerol is attached to four different groups

Question 89

The most effective way to remove triethylamine during the workup of an organic reaction would be to extract the reaction mixture with aqueous:

A) sodium bicarbonate

B) sodium bisulfite

C) sodium sulfate

D) hydrochloric acid

Answer 89

[D] Triethylamine would most likely be soluble in an organic solvent. Washing with aqueous hydrochloric acid will result in the formation of triethylamine hydrochloride which is water soluble and removed with water wash.

Question 90

Which of the following compounds could contribute to the nitrogen required in the growth medium?

A) Glucose

B) Glycerol

C) Glycine

D) n-Hexanoic acid

Answer 90

[C] Only glycine contains nitrogen.

Question 91

The four five-carbon carbohydrates shown below illustrate principles of carbohydrate nomenclature. Another five-carbon carbohydrate is xylulose. Which of the following statements applies to xylulose?

I. It is an isomer of deoxyribose

II. It is an isomer of ribose

III. It is an isomer of xylose

A) I only

B) II only

C) I and III only

D) II and III only

Answer 91

[D] Ribose, ribulose, and xylose are all isomers with the same molecular formula. Deoxyribose has a different molecular formula.

Question 92

Which of the following structures might represent a compound that might replace CH₃COCl in Step 3?

Answer 92

[D] Acetic anhydride can be used as an acetylating agent and could be substituted for acetyl chloride.

Question 93

What is the configuration of the stereogenic center in Compound 1 and Compound 2, respectively?

A) R and R

B) R and S

C) S and R

D) S and S

Answer 93

[A] The stereogenic center in Compound 1 has groups in descending order of priority: OH, CO₂CH₃, Ph orientated in a counterclockwise direction. The stereogenic center in compound 2 has groups in descending order of priority: OH, C(OH)Ph₂, and Ph) oriented in a counterclockwise direction with a group of Hydrogen pointed out making it R.

Question 94

Compound 1 is a stronger acid than Compound 2 because the anion of Compound 1 is better stabilized by:

A) resonance effect

B) dehydration

C) an inductive effect

D) hydrogen bonding between OH and CO₂^-

Answer 94

[D] The carboxylate ion formed from Compound 1 can be stabilized by hydrogen bonding with the hydroxyl group. The carboxylate ion formed from compound 2 is not in a suitable orientation to hydrogen bond with the hydroxyl group.

Question 95

The physical properties of 1,3-propanediol should be most similar to those of:

A) 1-propanol

B) 1,2-ethanediol

C) 1-propanethiol

D) 1,2-ethanedithiol

Answer 95

[B] 1,3-Propanediol is a 3 carbon diol and its physical properties should be most like those of other diols. 1,2-Ethanediol is a 2-carbon diol.

Question 96

If Figure 4 is repeated with HCl and the compound shown below, which of the following compounds is not a direct product (without rearrangement)

Answer 96

[A] The outcome of the second experiment involves the acid-promoted S_N1 substitution of an –OH group by halide ions. Substitution occurs through a carbocation intermediate and as a consequence both stereochemical outcomes for –OH substitution by Cl- will result in options C and D. Furthermore, elimination is possible by a loss of proton by a neighboring carbon atom. This reaction sequence required to produce the structure in A requires formation of an intermediate carbocation by a hydride shift to produce the carbocation.

Question 97

If 2-pentanol replaces 1-petanol in the reaction in Figure 3, the rate of substitution is less because:

A) The C-O bond in 2-pentanol is stronger than the C-O bond in 1-pentanol

B) There is a competing elimination reaction that slows the rate of substitution

C) There is more steric hindrance at the oxygen atom in 2-pentanol than in the 1-pentanol, making protonation less likely

D) There is more steric hindrance at the 2-position of 2-pentanol than at the 1-position of 1-pentanol.

Answer 97

[D] Figure 3 depicts the acid promoted S_N2 reaction of halide ions with 1 pentanol. The reaction rate depends on the energy difference between the starting material and the transition state for the reaction which includes the energy of halide to carbon bond formation

Question 98

In thin-layer chromatography using silica gel plates, the analytes can interact with the silica gel through hydrogen bonding. Which of the following classes of compounds would most likely have the strongest interaction with silica gel?

A) Alcohols

B) Carboxylic Acids

C) Esters

D) Ketones

Answer 98

[B] Hydrogen bonding results form dipole-dipole interaction between a relatively acidic, covalently-bound hydrogen atom and an electronegative element. Whereas silica gel consists of a siloxane matrix (O-Si-O bond) the surface of silica gel contains residual silic acid hydroxyl groups that can enter hydrogen bonding.

Alcohols and carboxylic acids contain both an electronegative atom and a relatively acidic O-H hydrogen. Carboxylic acids are more acidic than alcohols, hence, the O-H protons in the former are capable of entering into more effective hydrogen bonding.

Question 99

The disaccharides (+)-maltose and (+)-cellbiose are composed of two D-glucose subunits. The structural differences are a result of:

A) α- versus β- glucoside linkages

B) Chair versus boat conformations in the glucose subunits

C) Constitutional isomerism

D) 5 versus 6 membered rings in the glucose subunits

Answer 99

[A] The structures of (+)-maltose and (+)-cellobiose differ by virtue of the nature of their respective glucoside (full acetal) linkages