CHEM 303

Organic Chemistry II

Problem Set V

Chapter 16

Answers

1) A and B are isomeric dicarbonyl compounds of the molecular formula C5H8O2.

The 1H-NMR spectrum of A contains a singlet at δδδδ2.05 ppm, and another singlet at

δδδδ5.40 ppm. The 1H-NMR spectrum of B contains three signals: a singlet at δδδδ2.30

ppm, a triplet at δδδδ1.10 ppm, and a quartet at δδδδ2.70 ppm. Suggest structures for A

and B.

From the molecular formula, we find that both A and B have 2 degrees of

unsaturation, which is taken up with the two carbonyls (given information).

Compound A therefore, must be fairly symmetric. The singlet at δ2.05 is pretty

indicative of a CH3 next to a carbonyl, and the singlet at δ5.40 looks like it could be next

to two electron-withdrawing groups. All this makes me think we have 2,4-pentanedione,

CH3C(O)CH2C(O)CH3 here.

Compound B having a triplet at δ1.10 tells me that we have a methyl group next

to a methylene, which itself seems to be near an electron-withdrawing group. The quartet

verifies this assumption. The singlet at δ2.30 is probably a methyl next to an electron-

withdrawing group. All told, I think we have 2,3-pentanedione, CH3C(O)C(O)CH2CH3

2) Linamarin is a potentially toxic substance that releases hydrogen cyanide in

aqueous acid. Suggest a mechanism for this reaction.

O

HO

HO

H

OH

O

OH

NC

Well, here’s the mechanism.

O

HO

HO

H

OH

O

OH

NC

H+O

HO

HO

H

OH

O

OH

NC

H

O

HO

HO

H

OH

HO

OH

NC

+H2O

H OH2

HCN +

O

H2O

O

HO

HO

H

OH

OH

O

H

H

H2O

O

HO

HO

H

OH

OH

OH

Essentially, it is just the hydrolysis of an acetal, with the added benefit of one of

the resultant alcohols being a cyanohydrin, which itself can hydrolyze in acid.

3) The preparation of each of the following compounds requires protection of a

carbonyl functionality at some stage of the synthesis. Provide feasible sequences to

construct these molecules using the indicated starting materials and any other

necessary reagents.

a)

OH

H

O

OCH3

H

O

O

from

OH

H

O

OCH3

H

O

O

HOCH2CH2OH

H+

OCH3

H

O

OO

1) LiAlH4

2) H3O+

b)

C C CH2

H2C

O

H

from phenylacetyleneand 3-bromopropanal

This one is pretty straight-forward:

C C CH2

H2C

O

H

Br H

O

Br H

OO

HOCH2CH2OH

H+

C CH C CNaH

+

1) hexane

2) H3O+

c)

H

O

H

O

Cl

and benzyl chloride

from

Again, fairly straight-forward:

H

O

H

O

Cl

HOCH2CH2OH

H+

H

Cl

OO

1) t-BuLi

2) CuBr

3) PhCH2Cl

4) H3O+

The first two steps make the Gilman reagent (lithium dialkylcuprate), which

allows for alkyl halide coupling, and then we hydrolyze off the protecting group.

4) When the epoxide shown below was treated with CH3O- followed by an acidic

work-up, a compound of the molecular formula C6H12O3 was formed as the major

product. Its 1H-NMR spectrum exhibited signals at δδδδ1.07 (t, 3H), 2.60 (q, 2H), 3.41

(s, 6h), 4.50 (S, 1H) ppm, and a strong absorption was seen at 1730 cm-1 in its IR

spectrum. Propose a structure for this molecule, and explain how it could have been

formed under these reaction conditions i.e. propose a mechanism

O

H3CO

1) CH3O-

2) H3O+

C6H12O3

The IR peak at 1730 cm-1 tells us right away that we have a carbonyl group;

looking at the starting material, I’m going to go out on a limb (not really) and say we

have a ketone. Now to work on structure, and then the mechanism.

The NMR data suggests a fairly normal methyl next to a methylene (δ1.07, t, 3H).

The methylene is next to the methyl (no kidding!) and an electron-withdrawing group,

which is no doubt the carbonyl. Thus far we have: CH3CH2C(O). The 6H singlet at 3.41

is two methyl groups on an electronegative atom, so this signal is the methoxyls. Finally

a down-field 1H singlet is a methine near two electronegative atoms, or O2CH. Putting it

all together, we get:

H3CO

H3COO

Now to the mechanism:

O

H3CO

H3CO

H3COO

-OCH3

H3CO

H3COO

H OH2

H3CO

H3COOH

5) An oxidation reaction was performed on 3-phenyl-1,2-propanediol, with the

intention of producing the corresponding dicarbonyl compound. Upon examination

of the reaction product by 13C-NMR, it quickly became apparent that only one

carbonyl carbon was present. Using the 13C and

1H NMR data provided, suggest a

reasonable structure for the observed product.

OH

OH

oxidation

1H-NMR: δ9.22δ9.22δ9.22δ9.22 (s, 1H), 7.83 (d, 2H), 7.3-7.4 (m, 3H), 6.4 (broad s, 1H), 6.2 (s, 1H)

ppm 13C-NMR: δδδδ 122.8, 128.7, 129.3, 130.4, 133.5, 148.7, 188.3 ppm

Well, the 1H-NMR clearly shows an aldehyde hydrogen (δ9.22, s, 1H) plus the usual

array of aromatic protons. The 13C-NMR tells us we have eight different carbon atoms, so

there must be a bit of symmetry operating. The shift at δ188.3 is confirming of the

presence of a carbonyl. What is revealing is that there are no chemical shifts indicative of

any alkyl positions! All the 1H shifts are those of vinyl and/or aryl protons, and the same

can be said for the 13C shifts. The total number of hydrogens accounted for from the

1H-

NMR is 8; there were 12 in the original compound, and two would be expected to have

been lost per oxidation. Therefore, it looks as if both oxidations worked! Now the

question becomes: where is the second carbonyl? If we look at the structure we expected

O

O

H

we see an α-keto-aldehyde. This may not tell us much, but gives us a hint in explaining

two things from the 1H-NMR spectrum: the broad singlet at δ6.4, and the singlet at δ6.2.

The broad singlet seems like an –OH group, and the (sharper) singlet upfield of the broad

one is right in the vinyl range. Putting these together into a sensible structure, we obtain:

O

O

H

H

which, as shown, can engage in intramolecular hydrogen bonding to stabilize the enol.

The extension of conjugation doesn’t hurt, either.

6) The SN2 reaction of (dibromomethyl)benzene, C6H5CHBr2, with NaOH yields

benzaldehyde rather than (dihydroxymethyl)benzene, C6H5CH(OH)2. Explain with

a good mechanism.

Let’s take the process stepwise, and perhaps we can see what is happening. The

first substitution (probably) occurs as expected:

Br Br

-OH

Br OH

Now if we look at what we have, it is an alcohol (which contains an acidic

hydrogen by definition) on a carbon with a good leaving group attached. The next

reaction is not a substitution, but an acid-base reaction leading to an elimination reaction.

Here’s the mechanism:

Br O

H

-OHO

H

7) (This one requires you to stretch your thinking a bit, but you can do it.)

Treatment of an αααα,ββββ-unsaturated ketone with basic aqueous hydrogen peroxide

yields an epoxy ketone. The reaction is specific to unsaturated ketones, isolated

alkene double bonds do not react. Propose a mechanism.

H2O2

NaOH, H2O

OO

O

Given that this reaction only occurs with α,β-unsaturated ketones, we know that

the carbonyl must be a vital part of the mechanism. This tells us that we are not looking

at a simple epoxidation. Another important part is to realize that we are working in basic

conditions—why will become more apparent as we go through the arrow formalism.

Here is how I saw it:

O

O

O

HO O H-OH

HO O

O

OHO

Note we lost OH- in the last step; in basic solution, this otherwise poor leaving group can

become an acceptable leaving group.