quiz 1a answers - university of...

1A ANSWER KEY

2.222 In-class Quiz #1 Friday January 18, 2002. CIRCLE the correct response for each of the following questions.

1. Which of the following alkenes would react most rapidly with HBr in CH2Cl2?

Since protonation of the alkene is rate limiting, the alkene that will react fastest will be the one that forms the most stable intermediate cation. As the textbook points out, the activation energy for this step will correlate with cation stability. Note that delocalization of the positive charge onto the aromatic ring is a highly stabilizing factor.

2. Which one of compounds A-E below would be the major product of the following reaction?

This reaction involves a 1,2-migration of a methyl group to convert the initially formed secondary cation into the more stable tertiary cation. Note that structure A also arises from a 1,2-shift, but that it is disfavoured by the ring strain in the cyclobutane ring. Structures D and E are less favourable since they arise from secondary cations.

BrO

A B C D E

H2SO4, H2O ?

A B C D E

OH

OH

OH

OH

OH

1A ANSWER KEY

3. Which of the following is NOT a feasible reaction?

As we discussed in class, the preparation of a geminal dibromide from an alkyne is not really possible in water because the dibromide will react further to form a ketone under those conditions. Your textbook doesn’t actually say that the reaction can be done in water, but in Figure 10.8 it shows the participation of H3O+, which implies the presence of water. I disagree with this. All the other reactions are ones we have discussed – notice that D is an addition of singlet dichlorocarbene, generated in situ by base treatment of chloroform.

Br2, CH2Cl2

Br

H

H

Br

A

B

C

D

E

HCl, H2O

OH

CH3C C CH3 HBr, H2OCH3C CH2 CH3

Br Br

NaOH (aq.), Bu4N OHCHCl3

Cl

ClH

1) Hg(OAc)2, H2O2) NaBH4

OH

1B ANSWERS


1. Which of the following alkenes would react SLOWEST with HBr in CH2Cl2?

The initial protonation step of the reaction is rate limiting, and its activation barrier is correlated with the stability of the resulting cation. The leaststable cation will be the one substituted with an electron-withdrawing group, and a ketone is a very good EWG. All the other alkenes are substituted with H and/or donor groups.


This question is basically the same as Exercise 10.7a in the textbook. Protonation of the alkene forms a secondary cation, which rearranges to a more-stable tertiary cation by a 1,2-shift of a methyl group. Note that structures D and E are impossible to obtain under these conditions, while structure B is extremely unlikely since the conjugate base of sulfuric acid is a poor nucleophile.

O

A B C D E

H2SO4, H2O ?OH OSO3H

HO

OH

OH

A B C D E

1B ANSWERS

3. Only ONE of the following IS a feasible reaction. Which one is it?

Acid-catalyzed hydrolysis of a terminal alkyne to a ketone proceeds with heating as shown here, or at room temperature in the presence of mercuric acetate. Notice that reaction B looks a bit like examples we have seen, but that there is no driving force for a 1,2-shift when the ring is NOT aromatic! Reaction D would require a two-step mechanism consistent with a triplet carbene, but these conditions form singlet dichlorocarbene.

I2, CH2Cl2

I

H

H

I

A

B

C

D

E

HCl, H2O

OH

CH3CH2C CHH2SO4, H2O,

heatCH3CH2C CH3

O

NaOH (aq.), Bu4N OHCHCl3

1) Hg(OAc)2, H2O2) NaBH4 HO

Cl Cl

HH

1C ANSWERS


1. Which of the following alkenes would react with HBr in CH2Cl2 to give a CHIRAL product?

In all cases (except A) the bromide ion will attack the more-substituted position and the proton will end up on the other carbon of the alkene. The resulting bromides from A, B, D and E all have a plane of symmetry passing through the Br-substituted position, but in the case of C the molecule is not symmetrical. Of course, the chiral bromides obtained from C are racemic.


Reaction will occur fastest at the most-substituted alkene group, and of course will give the Markovnikov product. Note that if the reaction were left long enough, both alkenes would react to give a diol, but I didn’t include that option in this question.

Br

A B C D E

H2SO4, H2O ?

A B C D E

OH

HO

HOOH

HO

HBr

Br

H

Br

H

if addition is syn

if addition is anti

1C ANSWERS

3. Which ONE of the following DOES NOT involve electrophilic addition to an alkene?

The reaction in D proceeds by protonation of oxygen, loss of water, and isomerization of the double bond to the internal position. Although it is disfavoured by formation of a primary cation, the product is more stable than the reactant because the alkene group is more-substituted.

ICl, CH2Cl2A

B

C

D

E

HCl, H2O

CH3CH2C CHH2SO4, H2O,

heatCH3CH2C CH3

O

1) Hg(OAc)2, H2O2) NaBH4

Cl

I

OH

OH

OH H2SO4, H2O, heat

OH

2A ANSWER KEY

2.222 In-class Quiz #2 Friday February 1, 2002. CIRCLE the correct response for each of the following questions.

1. What is the Major Product of the following reaction?

This Freidel-Crafts Alkylation proceeds via the more-stable secondary cation formed from the alkene on treatment with the Lewis Acid AlCl3. Notice that this is the same Markovnikov selectivity that you would expect in an acid-catalyzed addition to the alkene – in fact this reaction actually IS an acid-catalyzed addition from the alkene’s point of view!

2. Which set of reagents will perform the following transformation?

A. Zn(Hg)/HCl (aq.)

B. Na (s), NH3 (l), EtOH C. H2 (g), Pd (cat.), AcOH

D. 1) Hg(OAc)2, H2O; 2) NaBH4

E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)

Selective reduction of the alkyne by dissolving metal always gives the trans alkene product. Option A is a Clemmensen Reduction, which would remove the carbonyl, while option C would hydrogenate both the carbonyl and the alkyne. The alkyne would be reduced all the way to an alkane. Options D and E would result in the formation of ketone products via intermediate enols.

+ AlCl3 ?

A B C D E

ClOH

O

?O

H

H

2A ANSWER KEY

3. Which set of compounds could be used to prepare the following molecule using the conditions specified?

This is a Diels-Alder cycloaddition. Recall that this type of reaction always occurs between a diene and a dienophile (alkene), and that the dienophile should be substituted with an electron-withdrawing group. Also, recall that the product will be a cyclohexene ring, in which the double bond is located between what were the C2 and C3 positions of the diene.

Option A shows conditions for a Simmons-Smith cyclopropanation, but the product is NOT a cyclopropane – note that the bridging CH2 group is attached across the ring, not to two adjacent carbons from one of the alkenes. Option E describes an oxidative cleavage of the alkenes, which would form carboxylic acids, but it would cleave BOTH alkene groups, not just the exocyclic one.

heat?CO2CH3

A B

C D

E

CO2CH3

+ Et2Zn CH2I2+

+CO2CH3

CO2CH3

+ CH2H2C

H3CO2C+

+ KMnO4

CO2CH3

+CO2CH3

H3CO2C

H3CO2C

HO2C

HO2C

CO2H

CO2HHO2C

HO2C

2A ANSWER KEY

4. What product is obtained from the following 2-step sequence of reactions?

The N,N-dimethylamino group is a strongly activating ortho/para director. Thus, Freidel-Crafts acylation will occur primarily para. The second reaction will be influenced by both the dimethylamino and the acyl groups, but both these groups favor nitration at the same position – ortho to the amine and meta to the ketone.

5. Which set of reaction conditions would perform the following transformation?

A. 1) Hg(OAc)2, H2O; 2) NaBH4

B. H2SO4, H2O

C. OsO4

D. 1) O3, CH2Cl2; 2) Zn, AcOH

E. 1) B2H6, THF; 2) H2O2, NaOH (aq.) Only Hydroboration/Oxidation can add a hydroxyl group to an alkene in an anti-Markovnikov fashion. Option A and Option B would both form the tertiary alcohol, while Option C would form a DIOL. Option D is an oxidative cleavage, which would chop off the terminal CH2 group and form a ketone in its place.

N

AlCl3

O

Cl HNO3, H2SO4 ?

A B C D E

N

NO2

O

N

NO2

O

N

NO2

O

N

NO2O

N

O

NO2

?OH

2B ANSWER KEY



Remember that you CANNOT carry out a Freidel-Crafts Alkylation at a primary position because the primary cation is too unstable. It rearranges via a 1,2-shift mechanism to a more stable secondary or tertiary ion, depending on the exact structure you are working with. In this case, the secondary ion is formed, and reacts to produce isopropylbenzene.


A. Zn(Hg)/HCl (aq.)

B. Na (s), NH3 (l), EtOH

C. H2 (g), Pd (cat.), Quinoline D. 1) Hg(OAc)2, H2O; 2) NaBH4

E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)

Quinoline is a “catalyst poison” that decreases the ability of the Pd to reduce alkenes. Thus the alkyne can be hydrogenated to a cis alkene – recall that all hydrogenations give SYN addition of H2 across the double bond being reduced.

+ FeBr3 ?

A B C D E

OH

Br

Br

O

?O

H

H

2B ANSWER KEY


This is a Diels-Alder cycloaddition. Recall that this type of reaction always occurs between a diene and a dienophile (alkene), and that the dienophile should be substituted with an electron-withdrawing group. Also, recall that the product will be a cyclohexene ring, in which the double bond is located between what were the C2 and C3 positions of the diene. Options A and B both will produce mixtures of several Diels-Alder adducts, but not the desired structure. Option C gives only one adduct, but not the one we want. Option E leads to no reaction, because there is no diene!

heat?

A B

C D

E

CN

CN

+

CN

+

CN

+

CN

+

CN

+CH2

CH2

CN

CN

NC

etc.

NC

etc.

CN

NO REACTION

2B ANSWER KEY


The N,N-dimethylamino group is a strongly activating ortho/para director. Thus, nitration occurs at the para position. The second reaction is controlled by both the nitro and amino groups, and both favor reaction at the same site, ortho to the amine and meta to the nitro group.



B. H2SO4, H2O

C. OsO4

D. 1) O3, CH2Cl2; 2) Zn, AcOH E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)

This reaction requires the cleavage of the alkene in order to form the ketone product. The only set of reagents from the list that can do that is Option D, ozonolysis followed by reductive workup. The other reagent sets all lead to various addition reactions forming alcohol products.

N

AlCl3

O

ClHNO3, H2SO4 ?

A B C D E

N ON

NO2

O

N

NO2

O

N

NO2O

N

O

NO2 NO2

?O

2C ANSWER KEY



This is a straightforward Freidel-Crafts Alkylation. Note that the electrophile that is formed when 2-bromobutane is treated with FeBr3 is a stable secondary cation, and there is no rearrangement pathway to improve upon it.


A. Zn(Hg)/HCl (aq.) B. Na (s), NH3 (l), EtOH

C. H2 (g), Pd (cat.), Quinoline

D. 1) Hg(OAc)2, H2O; 2) NaBH4

E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)

This is a Clemmensen Reduction of the aryl ketone. Option B describes a dissolving metal reduction, which in this case would actually reduce the benzene ring, although we did not discuss this aspect of this reaction – as far as you have been told it would not reduce the alkene or the ketone. Option C is a hydrogenation using a poisoned catalyst – which reduces alkynes to cis alkenes, but won’t touch ketones. Options D and E are addition reactions leading to alcohols.

+ FeBr3 ?

A B C D E

OH

Br

Br

O

?

2C ANSWER KEY


This is a Diels-Alder cycloaddition. Recall that this type of reaction always occurs between a diene and a dienophile (alkene), and that the dienophile should be substituted with an electron-withdrawing group. Also, recall that the product will be a cyclohexene ring, in which the double bond is located between what were the C2 and C3 positions of the diene. This example comes right out of lecture and the textbook. Notice that Option E describes conditions appropriate for a Simmons-Smith cyclopropanation, but that the organic molecule is actually an AROMATIC compound – it would not react!

heat?

A B

C D

E

O

O

O

HH

+O

OO

CO2HHO2C +

+O

OO

+O

O

O

Et2Zn + CH2I2

+O

OO

O

O

O

HO2C

CO2H

O

O

O

NO REACTION

2C ANSWER KEY


Bromine is a weakly deactivating ortho/para director. Thus, Freidel-Crafts Acylation will occur at the para position. The second reaction might actually be a bit difficult with two deactivators on the ring, but it would occur at the position favored by both of them – ortho to the bromine and meta to the ketone.



B. H2SO4, H2O

C. OsO4

D. 1) O3, CH2Cl2; 2) Zn, AcOH

E. 1) B2H6, THF; 2) H2O2, NaOH (aq.) The product clearly arises from the SYN addition of H2O across the alkene, in an anti-Markovnikov sense. This result could only come from a hydroboration/oxidation sequence. Option A and B would both give Markovnikov addition to form the tertiary alcohol. Option C forms a vicinal diol, while Option D leads to oxidative cleavage of the alkene to form a ketone and an aldehyde group!

Br

AlCl3

O

Cl HNO3, H2SO4 ?

A B C D E

Br

O

Br

NO2

O

Br

NO2

O

Br

NO2O

Br

O

NO2

NO2

? OH

H

3A ANSWER KEY

2.222 In-class Quiz #3 Monday March 4, 2002. CIRCLE the correct response for each of the following questions.

1. Which of the following pairs of reagents could react to form an ENAMINE product?

O

O

+

HNA

NH

O

+OB

O

+N

CH3

C

O

+

HND

O

O

+

HN

E

O

Enamines are formed when secondary amines react with enolizable ketones or aldehydes. Note that when primary or secondary amines react with esters or other carboxylic acid derivatives, the product is an amide. This arises from substitution of the leaving group, not of the carbonyl oxygen. Tertiary amines cannot form adducts with carbonyl compounds.

2. Which of the following carbonyl compounds is most acidic at its C-α position?

O

H

O

H

O

H3CO

O

H

O

A B C D E

As discussed in class and in the textbook, aldehydes are more acidic at the C-α position than either ketones or esters. Also as discussed in class, we find that acidity parallels the stability of the conjugate base, and therefore any structural feature that stabilizes the anion by delocalization will increase the acidity of the carbonyl compound.

3A ANSWER KEY

3. Which of the following benzene derivatives would be nitrated most rapidly by HNO3/H2SO4?

CH3 Br OH CN OAc

A B C D E

Hydroxyl is a strong π-donor and is thus a powerful activating group for aromatic substitution. The alkyl groups are weak activators, as are ester groups when linked via their oxygen. Halogens are weak deactivators, and cyano is a moderately strong deactivator.

4. Which of the following reactions would NOT proceed as written? A mild acidic workup can be assumed to follow each proposed reaction.

H3C OCH3

O2 CH3CH2MgBr

ether, 25 oC H3C CH2CH3

HO CH2CH3

A

H3C OH

O 2 CH3CH2MgBr

ether, 25 oCH3C CH2CH3

HO CH2CH3

B

H3C CH3

O CH3CH2MgBr


HO CH3

C

H3C H

O CH3CH2MgBr


HO H

D

H3C SCH3

O 2 CH3CH2MgBr


HO CH2CH3

E

You cannot use Grignard reagents in the presence of acidic functional groups!

3A ANSWER KEY

5. Which of the compounds listed below would be a major product of the following two-step reaction sequence?

OCH3

1) CH3COCl,AlCl3, CH2Cl2

2) NaOH (aq.), 3 I2

OCH3

CH2COOH

OCH3

CH2COOH

OCH3

COOH

OCH3

COOH

OCH3

COCH3A B C D E

Friedel-Crafts acylation will occur either ortho or para to the methoxy group, forming a methyl ketone. Methyl ketones are converted to carboxylic acids by the iodoform reaction.

3B ANSWER KEY


1. Which of the following pairs of reagents could react to form an ENAMINE product?

OHH3C

O

+CH3

HN

H3CA

NH2H3C

O

+ CH3

O

H3CE

CH3H3C

O

+

CH3

N

H3C

CH3

D

CH3H3C

O

+CH3

HN

H3CB

+CH3

HN

H3CC

H3C O CH3

O O

Enamines are formed when secondary amines react with enolizable aldehydes or ketones. Carboxylic acids simply protonate amines and are not nucleophilically attacked by them. Secondary amines will react with anhydrides to form amides, while the final two options both lead to no reaction.

2. Which of the following carbonyl compounds is most acidic at its C-α position?

O

H

O

H2N

O

H3CO

O

Br H

O

A B C D E

Aldehydes typically have pKas for removal of the C-α hydrogens on the order of 16-17 as we mentioned in class. Ketones are less acidic, and carboxylic acid derivatives even less. Amide and ester pKa values at C-α are about 25. The addition of an electron withdrawing group such as a halogen will certainly increase the acidity of an ester but not by 8 orders of magnitude.

3B ANSWER KEY

3. Which of the following benzene derivatives would be nitrated most rapidly by HNO3/H2SO4?

CH3 Br OCH3CNOAc

A B CDE

H3C H3C H3C H3CH3C

All these rings have a mildly activating methyl group, plus one other group which is either activating or deactivating. The methoxy group is the strongest activator, and thus this compound will be the one nitrated fastest. The sequence of letters attached to the answers was incorrect as you can see – don’t worry about that.

4. Which of the following reactions would NOT proceed as written? A mild acidic workup can be assumed to follow each proposed reaction.

H3C OCH3

O2 CH3MgBr

ether, 25 oC H3C CH3

HO CH3

A

H3CH2C OCH3

O 2 CH3MgBr

ether, 25 oCH3C CH3

HO CH2CH3

B

HO CH2CH3

O 2 CH3MgBr


HO CH3

C

H3C H

O 2 CH3MgBr

ether, 25 oCH3C CH3

HO H

D

H3C SCH3

O 2 CH3MgBr

ether, 25 oCH3C CH3

HO CH3

E

You cannot use Grignard reagents in the presence of acidic functional groups!

3B ANSWER KEY



2) NaOH (aq.), 3 I2

N

N

CH2COOH

N

HOOCH2C

N

COOH

N

COOH

N

COCH3

A B C D E

Friedel-Crafts acylation will occur ortho or para to the strongly activating dimethylamino group. The resulting methyl ketone is cleaved to the carboxylic acid by the iodoform reaction.

3C ANSWER KEY


1. Which of the following pairs of reagents could react to form an IMINE product?

CH3H3C

O

+ CH3

HN

H3C

A

NH2H3C

O

+ CH3

O

H3CD

CH3H3C

O

+

CH3

N

H3C

CH3

C

CH3H3C

O

+ CH3H2NE

+ CH3H2NB

H3C Cl

O

Imines are formed when primary amines react with ketones or aldehydes. Secondary amines will form enamines if the carbonyl compound is enolizable. Tertiary amines do not form adducts with carbonyl compounds, while the reaction of a primary or secondary amine with an acid chloride will lead to an amide.

2. Which of the following carbonyl compounds is least acidic at its C-α position?

O

H

O

H2N

O

H3CO

O

H

O

Br

A B C D E

The acidity at C-α in carbonyl compounds was discussed in class and also in the textbook. We noted that carboxylic acid derivatives were all less acidic than ketones or aldehydes, and that this paralleled their decreased positive charge character at the carbonyl carbon. Amides are much less positively charged than are esters because the nitrogen is a better π donor than is oxygen.

3C ANSWER KEY

3. Which of the following benzene derivatives would undergo nitration by HNO3/H2SO4 the slowest?

CH3 Br OCH3CNOAc

A B CDE

H3C H3C H3C H3CH3C

All these compounds have at least one weakly activating methyl group, so their relative reactivities will be controlled by whether the other group is an activator or a deactivator. The cyano group is the strongest deactivator in this list of compounds. NB: the letters attached to each compound are out of sequence – don’t worry about that.

4. Which of the following reactions would proceed as written? A mild acidic workup can be assumed to follow each proposed reaction.

H3C OH

O2 CH3MgBr

ether, 25 oC H3C CH3

HO CH3

A

H3CH2C OCH3

O 2 CH3MgBr

ether, 25 oCH3C CH3

HO CH2CH3

B

H3C

O CH3MgBr

ether, 25 oCH3C

HO CH3

C

H3C OCH3

O CH3MgBr

ether, 25 oCH3C CH3

O

D

H3C SCH3

O CH3MgBr

ether, 25 oCE

OH

CH3

OH

CH3

H3C CH3

O

In A and C, the carbonyl compound contains an acidic functional group. Grignard reagents are incompatible with acidic groups.In D and E, the problem is that you cannot get a single addition to an ester or a thiolester – the ketone that is formed is more reactive and will add a second Grignard.

3C ANSWER KEY



2) NaOH (aq.), excess I2

O

OCH3

A B C D E

OCH3

HO2C

OCH3

CO2H

O OHO2C OCH3

HO2C

HO2C OCH3

O

OCH3

O

O

Friedel-Crafts acylation will occur either ortho or para to the methoxy group, and meta to the ketone group. Any of these structures could be derived from the initial acylated product. However, the second step will convert both methyl ketones to carboxylic acids, and so only response C is correct.

4A NAME: STUDENT NO:

2.222 In-class Quiz #4 Friday March 15, 2002. CIRCLE the correct response for each of the following questions.

1. Which of the following proposed reactions is CORRECT?

O

H3C(CH3)2CHMgBr

CuIEther

(workup with NH4Cl aq.)

O

H3C

CH3

CH3

O

H3CLDA

THF, -78 oCthen CH3CH2I

O

H3C

H3CH2C

H3CO

O

OCH3

O

CH3NaOCH3CH3OH

CH3

OCH3

OO

O

H

O

+

NaOHH2O, heat

O

Br

O

OCH3

O

Mg, ether


O

OCH3

OH

A

B

C

D

E

The first reaction is correct – conjugate addition of a Grignard in the presence of copper(I) iodide. The second reaction should generate the kinetic enolate, but the product shown would arise from alkylation of the thermodynamic enolate. In C, the product shown lacks the acidic hydrogen between the two carbonyls – recall that enolization is what provides the driving force for this reaction to go to completion. Reaction D is a crossed aldol, but you would not get enolization of the ketone in the presence of the aldehyde. In reaction E, the conditions would suggest formation of a Grignard reagent, but you can’t form a Grignard from a molecule that contains a C=O group.

4A

2. Which of the following anions is the strongest base?

CH3O (CH3)2N O

O

HO

O

A B C D E

The dialkyl amide ion shown in B is the most basic. Recall that we use the diisopropylamide ion in LDA to quantitatively deprotonate carbonyl compounds – this indicates that such ions are significantly more basic than enolates (option E). The carboxylic acid in C is the conjugate base of a moderately strong acid, and is thus a weak base. Hydroxide (D) and methoxide (A) have similar basicities, and are the conjugate bases of water and methanol respectively. Alcohols are quite a bit more acidic than amines, which we can understand in terms of oxygen’s greater electronegativity over nitrogen.

In terms of pKa, the following table may be helpful. Remember that the strongest acid corresponds to the weakest conjugate base and vice versa:

Compound Type pKa Carboxylic Acids 4 to 5 Alcohols 17 to 20Water 14 Carbonyl α-positions 17 to 25Amines ca. 42

3. What set of reagents would be required to complete the following transformation?

HO HOH

H3CO OCH3

H

O

?

A. LiAlH4, THF then H2O.

B. KMnO4, H2O/NaOH.

C. SOCl2, heat.

D. H2SO4, H2O.

E. NaOCH3, CH3OH.

This reaction is taken from the textbook, see page 613. Acetals are hydrolysed to aldehydes by aqueous acid.

4A

4. Which of the following reaction processes involves cationic intermediates?

A. Epoxidation of alkenes by Peroxycarboxylic acids.

B. Formation of cyclopropyl rings using metal carbenoids.

C. Hydrogenation of alkenes in the presence of Pt catalysts.

D. Nitration of benzene with nitric acid and sulfuric acid.

E. Reduction of ketones to secondary alcohols with NaBH4.

Cationic intermediates are usually associated with multistep reactions occurring under acidic conditions. Epoxidation by peroxyacids is a concerted reaction which has no intermediates. Cyclopropanation by carbenoids (Simmons-Smith reaction) is also concerted. Hydrogenation on a catalyst is basically a type of free radical reaction, while reduction by borohydride involves basic, anionic intermediates. Electrophilic aromatic substitution (D) proceeds via the NO2

+ ion and the arenium ion intermediates.

5. Which of the following organic compounds could be used in making a Wittig Reagent (a phosphonium ylide)?

BrP

O

P

OA B C D E

Wittig reagents are prepared from an alkyl triphenylphosphonium ion, which is made by SN2 displacement of a primary or secondary alkyl halide by triphenylphosphine (B). The tertiary halide (A) cannot undergo this reaction. Acetone (C) may be attacked by a Wittig reagent, but it is not involved in making the actual reagent. Triphenylphosphine oxide (D) is a product from Wittig reagents, as is the alkene (E).

4B NAME: STUDENT NO:


1. Which of the following proposed reactions is INCORRECT?

O

H3C

Ether(workup with NH4Cl aq.)

O

H3C

CH3

O

H3CLDA

THF, -78 oC

then CH3CH2I

O

H3C

H3CO

O

OCH3

O

NaOCH3CH3OH

OCH3

OOH

O

H

O

+

NaOHH2O, heat

O

Br

O

OCH3

O

Zn, ether


O

OCH3

OH

A

B

C

D

E

(CH3CH2)2CuLi

CH2CH3

Option C is incorrect – the product would not arise from Dieckmann cyclization of a diester but instead would have to come from a reaction involving an aldehyde as electrophile. All of the other reactions are straightforward examples of processes covered in class and/or in the text.

4B

2. Which of the following compounds is the weakest Bronsted acid?

CH3OH (CH3)2NHOH

O

H2O

O

A B C D E

The weakest acid corresponds to the strongest conjugate base. See the answer to question 2 on Quiz 4A for details.


H3COH3CO H

H3CO OCH3

H

O

?O O

A. NaBH4, H2O.

B. CH3OH, H2SO4 (cat.).

C. NaOCH3, CH3OH.

D. H2SO4, H2O.

E. SOCl2, heat.

The product is an acetal, the result of nucleophilic attack on an aldehyde by two molecules of alcohol. This reaction is only favourable under acidic conditions. See the textbook pages 609-613.

4. Which of the following reaction processes does not involve cationic intermediates?

A. Addition of Br2 to alkenes in CH2Cl2 solvent.

B. Nitration of benzene with nitric acid and sulfuric acid.

C. Conversion of terminal alkynes to methyl ketones using Hg(OAc)2/H2SO4.

D. Epoxidation of alkenes by Peroxycarboxylic acids.

E. Diazotization of arylamines with nitrous acid.

Epoxidation by peracids is a concerted cycloaddition process that has no intermediates at all. All other reactions are multi-step processes. Addition of Br2 involves the cyclic bromonium ion, a 3-centre, 2-electron cation. Nitration of benzene is an electrophilic aromatic substitution that proceeds via an arenium ion. Hydrolysis of alkynes to ketones in the presence of Hg(II) involves several cationic species, including a cyclic mercurinium ion. Diazotization with nitrous acid involves several cationic species as well.

4B

5. Which of the following organic compounds could be the product of a Wittig Reaction?

Br

P

O

A B C D E

Br

Wittig reactions form alkenes by the reaction of a phosphonium ylide with a carbonyl.

4C NAME: STUDENT NO:


1. Which of the following proposed reactions is INCORRECT?

O

H3C

Ether(workup with NH4Cl aq.)

HO

H3C

O

H3C

O

H3C

O

O

NaOCH3CH3OH

O

O

H

O

+

NaOHH2O, heat

O

O

OH

A

B

C

D

E

(CH3CH2)2CuLi

Br2, PBr3 Br

O

OH

H3CH2C

NaOH/H2O

CH3CH2I

(workup with H2O)

Reaction B is incorrect – lithium dialkyl cuprates are soft nucleophiles that add 1,4 to α,β-unsaturated ketones.

2. Which of the following compounds is the weakest base?

CH3O (CH3)2NHO

O

Cl

O

A B C D E

Chloride ion is the conjugate base of hydrochloric acid, an extremely strong acid. Thus, chloride is the weakest base of this set. See the answer to question 2 on Quiz 4A for details.

4C


H3COH3CO OHH

O

?O O

A. NaBH4, H2O.

B. CH3OH, H2SO4 (cat.).

C. H2NNH2, KOH, heat.

D. Zn, HCl.

E. H2 (g), Pt.

The reaction shown is a reduction, formally involving the addition of a hydride nucleophile to the aldehyde. Sodium borohydride will reduce aldehydes or ketones, but will have difficulty reducing esters. Option B is not a reduction process. Options C and D are specific reactions for the complete removal of a C=O group, leading to a CH2 group. Option E will not reduce saturated aliphatic aldehydes, although it could reduce a benzylic carbonyl group. 4. Which of the following reaction processes involves bridged cationic intermediates?

A. Addition of Br2 to alkenes in CH2Cl2 solvent.

B. Friedel-Crafts acylation of aromatic rings.

C. Conversion of terminal alkynes to geminal dibromides with HBr.

D. Conversion of alkenes to vicinal cis diols using OsO4.

E. Diazotization of arylamines with nitrous acid.

Addition of Br2 to alkenes proceeds through a cyclic bromonium ion, in which a bromine atom bridges the two alkene carbons in an electron-deficient 3-centre, 2-electron structure. The other reactions do not involve bridged structures – in fact reaction D does not involve cationic intermediates at all.

5. Which of the following compounds could react with a Wittig Reagent?

NH2

O

P

O

A B C D E

O

OCH3

Wittig reagents react with aldehydes or ketones to form alkenes.

5A ANSWER KEY

2.222 In-class Quiz #5 Monday April 1, 2002. CIRCLE the correct response for each of the following questions.

1. What would be the expected product of the following reaction?

O O

KOH (aq.)

(acid workup) ?

CHOCOOH

COOH COOHO

O

A B C D E

OHOH

This is a Benzilic acid rearrangement, in which a 1,2-diketone is converted to an α-hydroxyacid on treatment with hydroxide ion. In fact, this example is identical to the one presented in the textbook and in class, except that the two phenyl rings are linked by an additional bond.

2. Which of the following compounds could NOT result from a C–C bond forming reaction of an electrophile with a malonate ester enolate?

H3CO OCH3

O O

H3C CH3

H3CO OCH3

O O

H

H3CO OCH3

O O

H3CO OCH3

O O

H3CO OCH3

O O

H

A B C D E

Options A-C are simply alkylation products – resulting from SN2 attack of an enolate on appropriate primary or secondary alkyl halides. Option E would arise from the Knoevenagel condensation, which is just a special case of the Aldol condensation. Option D could not be formed by enolate chemistry because you CANNOT displace a leaving group from a phenyl ring in an SN2 process or any other nucleophilic attack involving an enolate.

5A ANSWER KEY

3. Which of the following compounds could arise from alkylation of the KINETIC enolate of an unsymmetrical ketone?

O

H3C CH3

O

CH3

O

H3CCH3 O OH

A B C D E

The kinetic enolate is formed by irreversible deprotonation of a ketone at low temperature by a very strong base. It will be formed on the less-substituted side of the ketone. Option B could not arise from enolate chemistry of the type we have been discussing. Options C and D are the result of thermodynamic enolates, while Option E is simply irrelevant. It would arise from a Claisen rearrangement and not from a ketone enolate at all.

4. What would be the product of the following rearrangement reaction, an example of the Hofmann Rearrangement? (HINT: Look closely at the reaction conditions.)

O

NH2

Br2NaOCH3/CH3OH ?

O

NH2

Br

NH2HN

O

OCH3NC

OBr

A B C D E

In the Hofmann Rearrangement, an amide is transiently converted to an N-bromoamide. This rearranges to an isocyanate (Option A). If the reaction is conducted in aqueous NaOH, the isocyanate is attacked by OH- to form an unstable carbamic acid, which loses CO2 to form an amine (Option C). However, in this example the solvent and base are methanol and sodium methoxide – there is no water present. Thus, isocyanate A will be attacked by methoxide to form the carbamate D, which cannot decarboxylate. The hint was directing you to think about the mechanism and the role of the base in this reaction.

5A ANSWER KEY

5. Which set of reagents would be the best choice to complete the following reaction?

O

H

OH?

a. NaOH/H2O (acid workup)

b. (CH2=CH)2CuLi, Ether (acid workup)

c. [(C6H5)3P+–CH3]Br–/n-BuLi/THF

d. CH2=CHMgBr, Ether (acid workup)

e. i) SOCl2; ii) CH2=CH2; iii) LiAlH4/THF (acid workup)

The reaction shown is nucleophilic addition to the carbonyl group, by a 1,2-addition pathway. This requires a HARD nucleophile such as a Grignard reagent.

5B ANSWER KEY



O

?

COOH

COOH

O

O

A B C D E

CH3CO2H

H2O2/H2SO4

OH COOH O

O

This is a Baeyer-Villiger rearrangement, which converts a ketone into an ester. Option A is a reduction product, whereas these reaction conditions are clearly oxidizing. Option B and C both have more carbons than the starting material and are thus impossible under these conditions. Note that Option D would require epoxidation of an aryl ring, with destruction of the aromatic system, which will NOT happen with peroxycarboxylic acid reagents.

2. Which of the following compounds could result from an intramolecular condensation reaction of a diester, under basic conditions?

A B C D E

O

OO

OOCH3

OOCH3

OOCH3

OHO

O

OCH3

H3C

When a diester undergoes an intramolecular condensation reaction (a Dieckmann condensation), the product is a β-ketoester. The enolate formed from one ester group attacks the carbonyl group of the other, displacing an alcohol and forming a C-C bond. Moreover, the process is driven by the highly exothermic deprotonation of the ketoester product and thus products such as Option A are NOT observed. Options C and E would both have to arise from a crossed Aldol between an ester and an aldehyde. Option B is not a condensation product at all.

5B ANSWER KEY

3. Which of the following compounds could arise from alkylation of the THERMODYNAMIC enolate of an unsymmetrical ketone?

O

H3C CH3

O

CH3

O

H3CO OH

A B C D E The thermodynamic enolate is formed by equilibrium (reversible) deprotonation of a ketone by a relatively weak base. It will be formed on the more-substituted side of the ketone, so that the enolate C=C bond will be more highly substituted. Note that Option A would arise from the kinetic enolate (see Quiz 5A), while Option C would not come from an unsymmetrical ketone.

4. What would be the product of the following rearrangement reaction, an example of the Beckmann Rearrangement?

N

?

NC

O

A B C D E

OH

H2SO4, Heat

NH

O

NH2

O

NH

O

In the Beckmann rearrangement, the group anti to the OH in the starting oxime will be the one that migrates, to form an amide product.


O

CH3

O?CH3






The product arises from 1,4-addition of a nucleophile to the starting enone. This requires a SOFT nucleophile such as a lithium dialkyl cuprate.

5C ANSWER KEY



?

A B C D E

H2O/H2SO4

HOOH

CHOO

OOHCH2OH CH2

This is an example of a Pinacol Rearrangement. Note that the reaction leads to contraction of the relatively strained cycloheptyl ring into the less-strained cyclohexyl system.

2. Which of the following compounds could result from an intramolecular condensation reaction of a diketone, under basic conditions?

A B C D E

O

OO

OO O

OH

O

Options A and C would arise from reactions of a ketone enolate with an aldehyde, while option D is the result of attack of the ketone enolate on an ester. Option B is not the product of a condensation of a diketone, although it could arise from attack of a ketone enolate on an ester.

5C ANSWER KEY

3. Which of the following compounds COULD NOT arise from alkylation of a beta-ketoester, followed by decarboxylation?

O

H3C CH3

O

CH3

O

H3C

O

A B C D E

H3C CH3O

You cannot obtain tetrasubstituted carbon centres from the acetoacetic ester synthesis. When decarboxylation occurs, the C-C bond to the carboxyl group is replaced by a C-H bond.

4. What would be the product of the following rearrangement reaction, an example of the Claisen Rearrangement? (HINT: Look at each possibility carefully!).

A B C D E

O

heat ?OO O

O

OH

Options A and E are identical to the starting compound, although drawn in different orientations. In the Claisen rearrangement, an allyl vinyl ether is converted to an unsaturated carbonyl compound – the driving force is the conversion of a pair of C-O single bonds to the C=O double bond. Only Option B has the correct functional group pattern.


O

CH3

CH2?CH3






This is a Wittig reaction – conversion of a ketone into an alkene.

quiz 1a answers - university of...

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