345 Created by Master Student Organic Chemistry Practice Test #3 1 / 30 Which diene would be more reactive in a Diels-Alder reaction and why? 2, because it can transform into the cis conformation 2, because it is in a trans conformation 1, because it is very electrophilic 1, because it can transform into a trans conformation 1, because it is in a cis conformation An important criterion for diene reactivity is for the diene to maintain a cis conformation. Cyclopentadiene (1) has the double bonds permanently cis oriented due to its cyclical nature, thus is the most reactive of the two. Re-drawing diene (2) into a cis conformation shows the steric hindrance experienced by the terminal methyl groups; diene 2 is unable to maintain the geometry necessary for a Diels-Alder reaction. 2 / 30 What is the product of the reaction between 1,3-dibutene and bromoethene? 4-bromocyclohexene 3-bromocyclopentene 3-bromocyclohexene No reaction occurs The electrons from one of the double bonds on the 1,3-dibutene create a new single bond. The other new single bond is created from the electrons in the double bond of the other reactant. These two new single bonds join the reactants to create a cyclic product. The electrons from the other double bond in the 1,3-dibutene move between carbon 2 and 3. Thus, the final product is a 6-carbon cycloalkene with a halogen substituent. 3 / 30 What reagent can be used to convert 2,2-dimethylbutan-1-ol into 2,2-dimethylbutanal? H2CrO4 PCC KMnO4 O3 LiAlH4 The conversion of 2,2-dimethylbutan-1-ol to 2,2-dimethylbutanal is an oxidation (alcohol -> aldehyde). Lithium aluminum hydride (E) is a strong reducing agent. Potassium permanganate (C) is used in the oxidation of alkenes to diols, and is generally not strong enough to further oxidize aliphatic alcohols. Ozone (D) is used in the oxidative cleavage of alkenes into the corresponding aldehydes or ketones. Chromic acid (A) is a harsh oxidant, oxidizing primary alcohols up to the corresponding carboxylic acid, and secondary alcohols up to the corresponding ketone. PCC (B, pyridinium chlorochromate) is a milder chromium-based oxidant used in the selective oxidation of primary alcohols to aldehydes. 4 / 30 Which of the following reaction conditions favours an SN2 mechanism? Protic solvent Strong base Weak base Strong nucleophile Tertiary alkyl halide substrate SN2 reactions undergo substitution via a concerted mechanism. Thus, no carbocation is formed, and an aprotic solvent is favored. Furthermore, tertiary substituted substrates have lowest reactivity for SN2 reaction mechanisms due to steric hindrance. 5 / 30 The absorption of what type of electromagnetic radiation results in electronic transitions? microwaves X-rays radio waves infrared light ultraviolet light Ultraviolet–visible spectroscopy (UV-Vis or UV/Vis) refers to absorption/reflectance spectroscopy UV-Vis spectral region. The absorption or reflectance in the visible range directly affects the perceived color of the chemicals involved. In this region of the electromagnetic spectrum, molecules undergo electronic transitions. Molecules containing π-electrons or non-bonding electrons (n-electrons) can absorb the energy in the form of ultraviolet or visible light to excite these electrons to higher anti-bonding molecular orbitals. The more easily excited the electrons (i.e. lower energy gap between HOMO and the LUMO), the longer the wavelength of light it can absorb. 6 / 30 The 1H NMR spectrum of an unknown compound shows the following signals: triplet, quintet and triplet (relative integrals 1:2:3, respectively). Which compound is consistent with this data? 1,1-dibromopropane 1,2-dibromopropane 1,3-dibromopropane 2-bromopropane 1,1-dibromoethane An approximated 1H NMR spectrum of 1,1-dibromopropane (CH3CH2CHBr2) is shown below: 7 / 30 All of the following are characteristics of a Wittig reaction except it results in the formation of a carbon-carbon double bond it produces a trialkylphosphine oxide or triarylphosphine oxide as a by-product it involves the reaction of a phosphonium ylide with a carbonyl it proceeds through a phosphaoxetane intermediate it results in the exclusive formation of trans double bonds The Wittig reaction involves the reaction of a phosphonium ylide (generated by treating a phosphonium salt with a strong base) with a ketone or aldehyde. PPh3R1+ketone−R2→HR1C=CR2H+PPh3O The reaction proceeds through a phosphaoxetane (4-membered ring containing both phosphorus and oxygen) intermediate to generate a new compound containing a carbon-carbon double bond, plus a phosphine oxide byproduct. It does not form trans double bonds exclusively; sometimes, a mixture of cis and trans isomers are obtained, and sometimes the cis isomer is the predominant product. 8 / 30 Which labeled proton in the following molecule is most acidic? A B C D E The stability of each resulting conjugate base should be considered; the more stable this is, the stronger the parent acid. For enolates, the greater the degree of charge delocalization (i.e. the more resonance structures that can be drawn) the stronger the acid. Proton Hb is most acidic due to the charge (and lone pair of electrons) on the conjugate base being delocalized over both carbonyl groups and the benzene ring. 9 / 30 The reaction of a Grignard reagent with ethylene oxide (oxirane) followed by work-up with dilute acid gives which of the following products? A tertiary alcohol A secondary alcohol A primary alcohol Ethanol No alcohol The reaction of a Grignard reagent with oxirane (a type of epoxide) in addition to the work-up with dilute acid will yield a primary alcohol solely because there is a work-up with dilute acid. This shows that there is an excess of hydrogen, which will yield to a primary alcohol versus a secondary or tertiary alcohol. 10 / 30 For carbonyls that are part of a conjugated π-network, the C=O stretch: has a higher frequency than the analogous C=O in a non-conjugated system is always located at 1715 cm–1 has a lower frequency than the analogous C=O in a non-conjugated system exhibits a much lower magnitude of peak intensity is identical to the analogous C=O stretch in a non-conjugated system All carbonyl compounds absorb in the IR region of 1760–1665 cm–1 owing to the stretching vibration of the C=O bond. This carbonyl band is distinct with a characteristic high intensity and therefore is quite useful for diagnostic purposes because few other functional groups absorb in this region. Conjugation with a double bond or benzene ring lowers the stretching frequency, usually by 30 to 40 cm–1. The stretching frequency of the conjugated double bond is also lowered and may be enhanced in intensity. An example of this phenomenon is provided 11 / 30 What is the product of the given reaction? I II III IV V Diels-Alder reactions create cyclohexene rings (eliminate III, IV, and V), and starting dienophile is trans (E conformation), so the product is E (Eliminate I). 12 / 30 Which position will be most favoured when adding chlorine to tert-butylbenzene? Meta Ortho and para Para Ortho Meta and para 13 / 30 Which of the following reaction conditions favours an SN1 mechanism? Strong nucleophile Weak base Strong base Weak electrophile Protic solvent SN1 reactions occur in two steps. First, the leaving group leaves, forming a carbocation. Next, the weak nucleophile attacks the carbocation (beware of rearrangements during this step). The protic solvent stabilizes the carbocation intermediate. 14 / 30 Which of the following statements best describes the structural relationship between cis-1,2-dibromocyclopropane and trans-1,2-dibromocyclopropane? The two compounds are mirror images of one another The two compounds can be separated by ordinary physical chemical separation methods The two compounds have the same melting points All of the above Cis and trans-1,2-dibromocyclopropane (pictured below) share a diastereomeric relationship; that is, the molecules are stereoisomers that are not enantiomers. Diastereomerism occurs when two or more stereoisomers (i.e. same connectivities) of a compound have different configurations at one or more (but not all) of the equivalent stereocenters and are not mirror images of each other. Accordingly, diastereomers are distinct molecules with different chemical properties and can be separated by ordinary physical means. Recall that enantiomers are mirror images of each other and are identical in all physical aspects except for their direction of rotation of plane-polarized light. 15 / 30 Which series of carbocations are arranged from most stable to least stable? Tertiary, secondary, primary, methyl Primary, secondary, tertiary, methyl Methyl, primary, secondary, tertiary Methyl, tertiary, secondary, primary The stability of a carbocation depends only on the solvent of the solution Tertiary carbocations are stabilized by the induction of nearby alkyl groups. Since the carbocation is electron deficient, it is stabilized by multiple alkyl groups (which are electron-donating). Less substituted carbocations lack stability. 16 / 30 Predict the major product of the given reaction. I II III IV Reaction of an alkene with osmium tetroxide (OsO4) results in syn addition of to hydroxyl groups across the double bond, resulting in a 1,2-diol. The correct answer is the option in which hydroxide groups are added to the previously double-bonded carbons on the same side of the ring. The correct answer is compound II. 17 / 30 Which of the following reagents can be used to create a E alkene from an alkyne? Na,H2O Na,NH3 H2,Lindlar′scatalyst H2,Pt None of these Metallic sodium in liquid ammonia creates solvated electrons which can convert an alkyne to an E alkene. The same will not happen when sodium is combined with water, where sodium reacts violently to create sodium hydroxide and hydrogen gas. Lindlar's catalyst is a poisoned catalyst used to form alkenes from alkynes, bud results in a Z conformation. Without the poisoned catalyst, an alkane will be formed. 18 / 30 Which of the following molecules would most readily undergo an SN2 mechanism? I II III IV V I− is a better leaving group than Cl− because it is a larger molecule and can distribute the negative charge over a larger area. SN2 works better with better leaving group and with less-substituted carbons (methyl > primary > secondary > tertiary) 19 / 30 What is the product of the reaction shown? I II III IV Step 1 converts the carboxylic acid into an ester. Step 2 adds two equivalents of Grignard reagent: one to turn the molecule into acetone, and a second one to turn it into tert-butoxide. There is no way to stop the reaction after the first addition of Grignard reagent. Step 3 will neutralize the base, leaving only t-butyl alcohol (I). 20 / 30 2-butone is reacted with NaBH4 to form a product. That product was then heated in acid (Heat/H3O+) to form a final product. What is the final product? 2-butene Butane 2-butanol None of these 2-butone is a carbonyl compound that can readily be reduced by NaBH4 into a secondary alcohol, 2-butanol. When 2-butanol is heated in acid, we get dehydration, which leads to 2-butene. 21 / 30 Methods of extraction are important for use in industry and research. One of the typical steps in an extraction is the use of either acid or base to treat an unknown mixture. This step is applied in order to: Change the water solubility of an unknown compound Change the boiling point of an unknown compound Change the molecular weight of an unknown compound Change the optical activity of an unknown compound Change the melting= point of an unknown compound Adding acid or base to a mixture of unknown compounds is designed to change the water solubility. The loss or gain of protons will make the compound become ionic in nature, and thus soluble in water. On the other hand, compounds that are already ionic may become neutrally charged, resulting in lower solubility in water. 22 / 30 Classify the type of reaction given above. Rearrangement Addition Elimination Substitution An addition reaction is a reaction in which the reactants react to combine and form one product. It is the opposite of an elimination reaction. In the reaction given, the reactants hydrobromic acid and propene combine to form the product 2-bromopropane. 23 / 30 Which of the following substituents is electron-withdrawing when added to a benzene ring? Phenyl NO2 NH2 OH None of these substituents are electron-withdrawing NO2 is the only electron-withdrawing substituent because it contains two electronegative oxygen atoms which pull electrons from the benzene ring towards itself. This effect is electron-withdrawing and makes the ring slightly positive in charge. All the other substituents are electron-donating groups, which activate the ring for electrophilic addition. 24 / 30 Which of the following is true for E2 reactions? Rate=k[Substrate][Base] The reaction is bimolecular The proton and the leaving group should be anti-periplanar A double bond is formed All of these All are true for E2 reactions. E2 reactions are bimolecular, with the rate dependent upon the substrate and base. Both E1 and E2 reactions generally follow Zaitsev's rule and form the substituted double bond. The stereochemistry for E2 should be antiperiplanar (this is not necessary for E1). 25 / 30 Which of the following characteristics does not reflect an SN2 reaction mechanism? Tertiary substrate Aprotic solvent Concerted mechanism Stereochemical inversion of the carbon attacked (backside attack) Strong nucleophile SN2 reaction mechanisms are favored by methyl/primary substrates because of reduced steric hindrance. No carbocation is formed via an SN2 mechanism since the mechanism is concerted; thus a strong nuclephile is used. 26 / 30 The addition of heat and acid to 2-oxocyclopentanecarboxylic acid produces: cyclopentene cyclopentanol cyclopentaldehyde cyclopentanecarboxylic acid cyclopentanone Beta-keto carboxylic acids undergo decarboxylation through ring formations. The carbonyl bond of the ketone will react with the alcohol group of the carboxylic acid to form a ring and release the carboxylic acid as CO2. The remaining enol will tautomerize into a carbonyl group again, leaving cyclopentanone. Notice a cyclopentaldehyde can’t even be a possible answer, aldehydes cannot exist in cyclo compounds. 27 / 30 Which of the following compounds is not a reducing agent? H2 NaBH4 LiAlH4 HCl PCC PCC is the only compound listed that is not a reducing agent. Pyridinium chlorochromate is a weak oxidizing agent and is often used to oxidize alcohols into carbony compounds. All of the other compounds are similar in that they function as reducing agents. 28 / 30 Quinine is well-known antimalarial drug from the bark of the cinchona tree. In principle, how many stereoisomers of quinine are possible? 2 4 8 16 32 Quinine has four centers of chirality (highlighted with asterisks). Thus, in principle, quinine has 24 (2 x 2 x 2 x 2 = 16) stereoisomers. 29 / 30 The above image undergoes an E1 elimination reaction in a lab. The researchers note that the major product formed was the "Zaitsev" product. Which of the following compounds did the observers see most abundantly when the reaction was complete? ABCD A B C D The Zaitsev product is the most stable alkene that can be formed. This is the major product formed in E1 elimination reactions because the carbocation can undergo hydride shifts to stabilize the positive charge. The most stable alkene is the most substituted alkene, and thus the correct answer. 30 / 30 What is the best explanation for why phenol (shown) directs substitution products ortho and para? Resonance structures exist that put slight negative charges on ortho and para positions. Resonance structures exist that put slight negative charges on meta positions. Resonance structures exist that put slight positive charges on meta positions. Resonance structures exist that put slight positive charges on ortho and para positions. None of the other answers Phenol contains the hydroxide group, which is an electron donor, puts electron density into the benzene ring. Resonance structures are drawn as follows Your score is The average score is 51% LinkedIn Facebook Twitter VKontakte 0% Restart quiz Previous Quiz Next Quiz
