Organic chemistry section test (3)

Extraction with a strongly acidic aqueous solution, followed by extraction with a weakly acidic aqueous solution, followed by distillationB. Extraction with a strongly acidic aqueous sol

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Organic Chemistry Subject Test 3

1 Which of the following alkyl halides would be the least

reactive substrate in a Williamson ether synthesis?

5 Starting with a molecule of benzyl alcohol ΦCH2OH,

we wish to form ΦNH2 (aniline) Which of the

following syntheses should be used? [Φ = phenyl

7 What is the major product of the reaction below?

? heat

H

H 3 C

CH3

E No reaction will occur

8 Predict the product of the following reaction:

(CH 3 ) 3 C

C Br

E Either A or B, depending on reaction conditions

9

H +

H 2 O HCN

OH

H

D

O OCH 3

CH2 OHOH

E

O OH

CH2 OHOH

11 If the compound pictured below undergoes E2

elimination, what is the most likely product?

B Extraction with a strongly acidic aqueous solution, followed by extraction with a weakly acidic aqueous solution, followed by distillation

C Extraction with a strongly acidic aqueous solution, followed by extraction with an even stronger acidic aqueous solution, followed by distillation

D Extraction with a strongly basic aqueous solution, followed by extraction with a weakly acidic aqueous solution, followed by distillation

E Extraction with a weakly basic aqueous solution, followed by extraction with a strongly basic aqueous solution, followed by distillation

15 Which one of the following compounds can be most

easily converted to a Grignard reagent?

17 A general formula for para-substituted benzoic acids is

shown below The Ka of a particular para-substituted

benzoic acid will be smallest when X is which of the following?

The products of this reaction are

A ortho to the hydroxyl group

B para to the hydroxyl group

C meta to the hydroxyl group

D A and B

E A, B, and C

20 Which of the following compounds will give a single

E None of the above

21 To prepare a primary alcohol with a Grignard reagent,

the Grignard reagent must react only with

22 Which of the following compounds would be expected

to be the most basic?

23 Which of the following compounds would not give a

positive result under the iodoform test?

24 A compound produces an infrared spectrum with a sharp

peak at approximately 2950 and 1700 cm–1, as well as a

cm–1 The substance yields a negative result under

Tollens’ test This substance is most likely a(n)

25 Which of the numbered hydrogens in the molecule

pictured below is most acidic?

B Br is more electronegative than F

D FCH2COO- is a stronger base than BrCH2CH2COO-

E F- is a better leaving group than Br-

28 The organic acid pictured below will be most acidic when X is which of the following?

X COOH

29 What is the major product of the reaction below?

30 Which of the following aromatic compounds will

undergo electrophilic substitution primarily in the ortho and para positions?

THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE

ORGANIC CHEMISTRY SUBJECT TEST 3

secondary alkyl halides; these compounds all undergo SN2 reactions more readily than does the tertiary alkyl halide of choice C

In order for a reaction to produce a racemic mixture, the product formed must possess at least one chiral center and its enantiomers must form in a one-to-one ratio The product of the addition reaction in choice C is 1,2-dibromobutane This compound has one chiral center, at carbon number 2, and its R and S enantiomers should be expected to form in equal amounts in the absence of any special effort to produce either in enantiomeric excess

as the permanganate oxidatively cleaves the alkene double bond, while the reaction in choice E will yield the Markovnikov addition product 2-chloropropane which is likewise achiral

The synthesis of cyclopropane via addition to the double bond in an alkene is accomplished by reaction with

a neutral molecule (Notice the lone pair of electrons on the carbon atom.) This reactive intermediate is most commonly formed by the heat- or light-induced decomposition of diazomethane, as shown in choice E The lone pair of electrons on the carbene intermediate are used together with the _ electrons of the alkene to form the two new single bonds in the cyclopropane product (An alternative route to cyclopropane synthesis involves reacting diiodomethane and a zinc-copper couple to generate a carbene-like species called a carbenoid which will add a methylene group to the double bond.)

The compound in choice D is pyrimidine itself, a heterocyclic analog of benzene containing two nitrogen atoms, positioned as shown Pyrimidine and its derivatives are essential components of nucleic acids such as DNA and RNA

Choice A is almost, but not quite, the general structural formula for a steroid; it differs from the standard steroidal ring structure in that the ring on the far right has six carbons instead of five Choices B and E, known as quinoline and pyridine respectively, each have only one nitrogen atom and thus fail to meet the definition of a pyrimidine Choice C is purine, another essential component of nucleic acids

Since many synthetic pathways might be available to effect any given transformation, it makes most sense to check the choices offered Choice A will yield the desired product: permanganate oxidizes primary or benzylic alcohols to the corresponding carboxylic acids in the first step The benzoic acid formed by this oxidation will, in the presence of aqueous ammonia, form an ammonium carboxylate salt which, upon heating, will produce an amide in the second step Finally, treatment of the benzamide formed in the second step with hypobromite will produce an amine with one less carbon This reaction is known as a Hofmann rearrangement, and it will convert benzamide into aniline, the desired product, in the third step

Choice B is incorrect Phosphorus tribromide, step one, will convert an alcohol into an alkyl bromide, in this case benzyl bromide Treatment of this benzyl halide with sodium cyanide, step two, will produce a nitrile via an SN2 pathway Finally, hydrogen over a nickel catalyst will reduce the nitrile to a primary amine; the primary amine formed by this series

of steps will have two carbons between the phenyl ring and the amino group, and thus is not aniline Incorrect choice C, like B, produces benzyl bromide after the first step Benzyl bromide should be inert to ammonium and acid in the absence

of any nucleophilic species, thus no reaction should be expected in the second step Choices D and E also do not lead to the formation of aniline

The electrophilic addition of bromine to an alkene proceeds through a cyclic bromonium ion intermediate

Configuration-wise, the addition is anti Regiospecificity also dictates that the geminal dihalide shown in choice B cannot

possibly be a product of this reaction Complete scratchwork would show that choices A and D should form with equal probability, and therefore that choice E is correct:

H Br

generation of one lone achiral meso compound An enantiomeric pair must therefore be the product of the reaction By this

faster method of reasoning it is not necessary to figure out which product or products will form, but only that more than one will

This reaction is an example of benzylic oxidation Substituted benzene derivatives are easily oxidized by

permanganate or dichromate to benzoic acid derivatives if there is at least one hydrogen attached to a benzylic carbon (The first step in the reaction is the abstraction of a benzylic hydrogen to give a benzylic radical, thus making the hydrogen atom necessary.) The reactant shown has three benzylic carbons, one of which, the carboxylic acid carbon at the top of the structure, is already completely oxidized The other two, the benzylic carbon in the alkyl group on the right and that in the alkenyl group on the left, can both be oxidized to carboxylic acid groups, yielding the tricarboxylic acid compound shown

in choice C Choices A and D can be dismissed for several reasons, among which is that they suggest reduction of the COOH group in the reactant to a CH3 group in the choices; permanganate is far and away not a reducing agent Choices A and D also show the conversion of the alkene side chain to a diol; this mild oxidation would occur only if the permanganate used were cold, dilute, and dissolved in basic solution One might more quickly eliminate choices A and D by noticing that they differ only by rotation around a carbon-carbon single bond; they are thus equivalent choices Choice B is wrong because it shows the selective oxidation of the alkyl side-chain only, when in fact both side-chains will be oxidized under the conditions indicated in the reaction shown

This question is testing our understanding of fundamental organic reaction mechanisms The reactant shown is a secondary alkyl halide, thus it is susceptible to nucleophilic attack via either an SN1 or SN2 pathway It also has a proton α

to the bromine atom (on the methyl group), so it is capable of undergoing elimination to form an alkene Sodium hydroxide

is a strong base which can facilitate E2; the hydroxide ion can also behave as a nucleophile and lead to an SN2 reaction under the proper conditions The preference for either E2 or SN2 can be enhanced by the appropriate choice of reaction conditions, justifying choice E as the credited choice

This question is testing our knowledge of organic synthesis, and requires that we work through the steps shown to predict the final product (Partial knowledge of the reagents shown would, however, allow for the elimination of some of the answer choices For instance, noticing that the final step in the synthesis includes acidification allows us to eliminate choice D since this product would be protonated in acidic solution.) Let’s go through the steps:

KMnO4 is a strong oxidizing agent which will oxidize most alkylbenzenes, including the ethylbenzene starting material here, to benzoic acid After the first step we thus have benzoic acid, which can now react with thionyl chloride, SOCl2, in the second step Thionyl chloride is most commonly used to convert a carboxylic acid to the corresponding acyl chloride; the product of the second step is thus benzoyl chloride Acyl halides are especially reactive toward nucleophilic acyl substitution, the reaction by which the halide is replaced by any of a large number of suitable nucleophiles while the carbonyl C=O bond remains intact The third step of the synthesis in this question should therefore result in the substitution

of the chloride by cyanide from the HCN and the product of this step should be the α-ketonitrile derivative of benzoic acid The cyano group of nitriles is easily hydrolyzed in aqueous acid; the final step of the synthesis should thus be expected to convert the α-ketonitrile to the α-ketoacid shown in choice A The sequence of reactions is as follows:

C 2 H 5 KMnO4

C

O OHSOCl2

Cl

O

C HCN

10 D

By definition, an acetal must contain a carbon atom bonded to two -OR groups and a hydrogen atom Acetals might thus be thought of as geminal diethers The carbon atom on the right side of the ring in choice D is bonded to the OCH3group on the far right and to the O atom in the ring which is, in turn, bonded to the rest of the ring; a third substituent would

be a hydrogen atom which is not shown by convention Choice D does therefore meet the criteria for classification as an acetal Choice A is a carboxylic acid, while choice B is a b-diketone Choices C and E can be classified as hemiacetals, but not as acetals Hemiacetals, by definition, contain a carbon atom which is bonded to one -OR group, one -OH group, and a hydrogen atom

11 C

E2 elimination occurs in the anti configuration, requiring that the proton abstracted be in a position trans to the

leaving group As drawn, only the proton labeled as Hd is in the proper position for anti elimination, i.e., trans to the

chlorine atom in the reactant, and thus Hd must be absent from the final product This qualifies choice C as the only

possible choice

12 A

This question is testing our understanding of organic separation techniques, specifically extraction, as well as our knowledge of the relative acidity of the three compounds shown The first compound, aniline, is the most basic of the three;

it has an amino group in which the nitrogen has a lone pair of electrons available for donation to a suitable Lewis acid The

amino group nitrogen atom in the second compound, para-nitroaniline, is rendered less basic due to the

electron-withdrawing nature of the nitro group; with decreased electron density, it is less eager to donate electrons, i.e., to act as a Lewis base Lastly, the third compound, nitromethane, is not basic at all Applying this relative order of basicity to the extraction, now, requires the understanding that acidic extraction will make organic bases soluble in the aqueous phase via protonation of the base to form a soluble ionic salt Therefore choice A is the best choice Extraction with a sufficiently weak aqueous acid will protonate only the most basic compound, aniline, and allow it to transfer, in protonated form, to the aqueous phase A second extraction with a stronger acid will then protonate the more basic of the two compounds

remaining in the chloroform layer, para-nitroaniline, and allow it to enter the aqueous phase The third compound would

then remain alone in the organic layer, and the separation could be completed by distillation

If a strong enough acid were used first, both aniline and nitroaniline would be protonated and transferred to the aqueous layer, and the separation would not be effective Extraction with a base would not effect a separation at all, since the first two compounds, as bases themselves, are inert to base

13 C

The reaction shown is an example of a Grignard reaction One should predict that the Grignard reagent used, benzyl magnesium bromide, will act as a nucleophile and attack the carbonyl carbon in the substrate, phenyl benzoate, forming the tetrahedral intermediate shown below

C O CH2

O -

After formation of the tetrahedral intermediate, a pair of electrons on the negatively charged oxygen can move down

to the carbon atom of the original carbonyl group, restoring the carbon-oxygen double bond and displacing the phenoxide leaving group Once the carbonyl group has been restored, it is susceptible to further nucleophilic attack by a second