Chemistry 2.222 Organic Chemistry II: Reactivity and ...hultin/chem2220/Archive/2002/final... · Name: Student No: Page 1 of 12 Chemistry 2.222 Organic Chemistry II: Reactivity and

Name: Student No: Page 1 of 12

Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis

FINAL EXAM – Winter 2002 Paper Number 332

Friday April 19, 2002 6:00 – 9:00 pm

Frank Kennedy Gold Gym Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper, with any HANDWRITTEN notes they wish (both sides). Molecular model kits are also permitted but no other aids may be used. The exam is in two parts. Answers to PART I are to be entered in the indicated spaces on the exam paper itself. An answer to ONE of the “Challenge Questions” in PART II is to be written in the exam booklet.

PART I: Question 1 (20 Marks)

Question 2 (4 Marks)





PART II: (8 Marks)

TOTAL: (70 Marks)


PART I: DO ALL QUESTIONS – THERE IS CHOICE IN QUESTION 1 ONLY. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions

to correctly complete TEN (10) of the following. All reactions do in fact lead to a new product. Be sure to indicate clearly which ones you want marked.

(a)

(C6H5)3P

Ethanol

n-BuLiTHF

then addO

H

Br

(b)

OH

heat

(c)

Br2, CH2Cl2

(d)

O

CN

1) NaOEt/EtOH

EtO OEt

O

2) NaOEt/EtOH CH3I

C19H17NO3

(e)

O

OCH31) CH3OH, H2SO4 (cat.)

O

2) 2 equiv. CH3MgBrEther (H2O workup)

(f)

O

H2 (g), Pd/CAcetic acid (solvent)

O


(g)

NO2

AlCl3, CH2Cl2

(h)

NaOEt, EtOHheat

OO

C12H18O

(i)

NH

O

NH2

O

OCF3CO2H

(j)

HCO2H, H2O2H2SO4 (cat.)

O

(k)

O

H

NH

toluene, heat

(l)

O

O

(m)

OO

OCH3

1) 2 NaOCH3CH3OH

2) H2SO4, H2OheatO

BrBr+

(n) (Two Steps)

OO


(o)

LiAlH4THF

followed byH2O workup

O

N

(p)

Hg(OAc)2H2O, THF

followed byNaBH4

CH2

(q)

+ heatNO2

O

NO2

(r)

S SO O

OH

O

OH

OO

2. (4 MARKS) Four steps of a synthesis of the plant terpene juvabione are shown below, with some items missing. All chemistry in this sequence has been covered in this course. Fill in the blanks. Keep your eyes on what is changing, don’t be distracted by what stays the same!

O

H

H3C HOCH3

O O

OCH3H3C

NaH, THF

OH

H

H3C H

OCH3

O

OH3C

H3C S Cl

O

OEt3N, CH2Cl2

H

H3C H

OCH3

O

OH3C

C13H20O4

C14H24O6S


3. (10 MARKS) Provide detailed stepwise mechanisms for both reactions in the following two-step synthetic transformation.

O

OHCl

Cl

O

OCH2Cl2

O

Cl

( + CO2 + CO + HCl )

CH3CH2SH

Et3N, CH2Cl2

O

S


4. (a) (1 MARK) The presence of nitrogen in an organic compound can be demonstrated by carrying out a sodium fusion. After dissolving the fusion product in water, the presence of nitrogen in the original compound is confirmed by adding FeSO4 and observing the formation of “Prussian Blue”, Fe4[Fe(CN)6].

On the basis of this information, what molecular or ionic nitrogen-containing species is formed when a nitrogen-containing compound is fused with sodium?

(b) (2 MARKS) Silver halides (AgX; X = Cl, Br or I), formed when testing a sodium fusion residue for halogens, are distinguished from one another in two ways:

• Physical appearance • Solubility in aqueous ammonium hydroxide.

Briefly describe how these two tests would distinguish AgCl from AgI.

(c) (2 MARKS) A student measured the boiling temperature of a liquid by distilling a sample. A value of 157 oC was obtained, but the literature value was found to be 138 oC. Indicate two experimental errors that might have led to the incorrect boiling temperature being measured.

(d) (3 MARKS) An unknown compound (X) was dissolved in cold hydrochloric acid and the solution was treated with a cold aqueous solution of sodium nitrite. The mixture was then carefully added to a cold alkaline solution of β-naphthol. A red solid (Y) was formed.

i) What type of compound was X? ii) Draw a reasonable structure for Y.

HO

β-naphthol


(e) (2 MARKS) A solution of sodium hydroxide in aqueous ethanol was treated with a 1 : 1 mixture of phenylethanal (C6H5CH2CHO) and propanone. The mixture was shaken for about 20 minutes. A yellowish solid precipitated and was filtered off. The solid was shaken with water and this mixture was then treated with dilute hydrochloric acid until it was weakly acidic (pH 4 - 7). The product of the reaction was “worked up”. Draw the structure of this product.

(f) (2 MARKS) A student is given a sample of aniline and told to use it to prepare a sample of p-nitroaniline. Indicate the steps that would be required to ensure that the synthesis would work and that little or no m-nitroaniline by-product would be formed.


4. (8 MARKS) The Claisen Rearrangement is a common example of a sigmatropic process. Two very similar molecules shown below were subjected to identical conditions expected to lead to Claisen Rearrangement. The outcomes of the two reactions were, however, rather different. Using your knowledge of organic structures and reaction mechanisms, EXPLAIN why the product of reaction B does not appear to follow the standard Claisen pattern.

O

toluene, heat

OH

O

toluene, heat

OH

H3CH3C

A

B


5. (8 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the molecular formula C7H13NO are shown below.

4000 3000 2000 1500 1000 500

br s

(a) Draw the structure of the compound in the

box at right (4 Marks).

(b) Clearly indicate which H atom(s) in your structure is(are) responsible for each of the signals in the 1H NMR spectrum (4 Marks).


PART II: CHALLENGE PROBLEM. DO ONE PROBLEM ONLY. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Part II is worth 8 Marks.

A. SYNTHESIS The simple alkaloid (R)-δ-coniceine has been synthesized from the intermediate A shown. The synthesis used reactions that have been discussed in this course. Construct a route to coniceine from intermediate A in 4 or 5 steps. The first reaction should form the bicyclic ring system found in the product.

B. MECHANISM The Benzoin Condensation, shown below, may appear surprising at first but its mechanism is actually quite simple. The process is catalyzed by cyanide ion – stoichiometric KCN is not needed. However, the catalyst must be cyanide (or one of a very few other substances). Hydroxide or similar basic reagents will NOT work.

O

HKCN

EtOH/H2Oheat

O

HO H

Provide a mechanism for the Benzoin condensation and explain why cyanide is essential. Can you suggest what other kinds of substances besides cyanide might catalyze the Benzoin condensation?

C. SPECTROSCOPY

You have been presented with an unknown organic compound to analyze. It is a clear liquid with a spicy odour similar to nutmeg. You perform several chemical tests, with the following results:

• Insoluble in water or dilute acid, soluble in aqueous NaOH. Also soluble in CHCl3, ether, alcohol.

• Decolourises bromine water fairly rapidly.

• Forms a blue solution when treated with FeCl3.

A mass spectrum shows the molecular ion at m/z 164, and a significant fragment at m/z 149, along with numerous other fragments. You then obtain IR, 1H and 13C NMR spectra, which are shown on the next page.

1. What is the structure of the unknown material?

2. Explain HOW the chemical and spectroscopic evidence supports your proposed structure.

N

(R)-δ-coniceineNH

O

COOEt

4 or 5 steps

A


There are 2 peaks inthis signal

multiplet singlet singlet

Double doublet

Doublet of quartets

Doublet of quartets

Spectroscopy “Crib Sheet” for 2.222 – Introductory Organic Chemistry II 1H NMR – Typical Chemical Shift Ranges

Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)

C CH3 0.7 – 1.3 C C H 2.5 – 3.1

C CH2 C 1.2 – 1.4 O

H 9.5 – 10.0

C H

C

C

C

1.4 – 1.7 O

OH

10.0 – 12.0 (solvent dependent)

C H

1.5 – 2.5 C OH


O

H

2.1 – 2.6 CO H

3.3 – 4.0

Aryl C H 2.2 – 2.7 CCl H

3.0 – 4.0

H 4.5 – 6.5 CBr H

2.5 – 4.0

Aryl H 6.0 – 9.0 CI H

2.0 – 4.0

13C NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands Group Frequency

(cm-1) Intensity Group Frequency (cm-1) Intensity

C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium Aryl C=C 1600, 1500 Strong RNO2 1540 Strong

12 11 10 9 8 7 6 5 4 3 2 1 0

←δ

R3C–H Aliphatic, alicyclic

X–C–H X = O, N, S, halide

Y

HH

Aromatic, heteroaromatic

Y

H

RCO2H

Y = O, NR, S Y = O, NR, S

“Low Field” “High Field”

220 200 180 160 140 120 100 80 60 40 20 0

←δ

CH3-CR3 CHx-C=O

CR3-CH2-CR3

CHx-Y Y = O, N Alkene

Aryl

AmideEster

Ketone, Aldehyde

Carbox. Acid

ANSWER KEY Page 1 of 12

Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis

FINAL EXAM – Winter 2002

Paper Number 332

Friday April 19, 2002 6:00 – 9:00 pm

Frank Kennedy Gold Gym Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper, with any HANDWRITTEN notes they wish (both sides). Molecular model kits are also permitted but no other aids may be used. The exam is in two parts. Answers to PART I are to be entered in the indicated spaces on the exam paper itself. An answer to ONE of the “Challenge Questions” in PART II is to be written in the exam booklet.

PART I: Question 1 (20 Marks)






PART II: (8 Marks)

TOTAL: (70 Marks)


PART I: DO ALL QUESTIONS – THERE IS CHOICE IN QUESTION 1 ONLY. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions

to correctly complete TEN (10) of the following. All reactions do in fact lead to a new product. Be sure to indicate clearly which ones you want marked.

(a)

(C6H5)3P

Ethanol

n-BuLiTHF

then addO

H

Br

H

(b)

OH

heat

O

(c)

Br2, CH2Cl2

Br

Br

(d)

O

CN

1) NaOEt/EtOH

EtO OEt

O

2) NaOEt/EtOH CH3I

C19H17NO3

O

CN

O OEt

CH3

(e)

O

OCH31) CH3OH, H2SO4 (cat.)

O

2) 2 equiv. CH3MgBrEther (H2O workup)

OH

CH3

OCH3H3CO

CH3

(f)

O

H2 (g), Pd/CAcetic acid (solvent)

O O


(g)

NO2

AlCl3, CH2Cl2 O2N

(h)

NaOEt, EtOHheat

OO

C12H18O O

(i)

NH

O

NH2

O

OCF3CO2H

NH2

O

NH2

(j)

HCO2H, H2O2H2SO4 (cat.)

O

O

O

(k)

O

H

NH

toluene, heat

H

N

(l)

O

O(CH3)2CuLi or CH3MgBr, CuI (cat.)THF or Ether

(m)

OO

OCH3

1) 2 NaOCH3CH3OH

2) H2SO4, H2OheatO

BrBr+

O

O

(n) (Two Steps)

OO

O3, CH2Cl2thenZn/Acetic Acid


(o)

LiAlH4THF

followed byH2O workup

O

N N

(p)

Hg(OAc)2H2O, THF

followed byNaBH4

CH2OH

(q)

+ heatNO2

O

NO2

O

(r)

S SO O

OH

O

OH

OO

Cl

O

AlCl3, CH2Cl2

2. (4 MARKS) Four steps of a synthesis of the plant terpene juvabione are shown below, with some items missing. All chemistry in this sequence has been covered in this course. Fill in the blanks. Keep your eyes on what is changing, don’t be distracted by what stays the same!

O

H

H3C HOCH3

O O

OCH3H3C

NaH, THF

OH

H

H3C H

OCH3

O

OH3C

H3C S Cl

O

OEt3N, CH2Cl2

H

H3C H

OCH3

O

OH3C

C13H20O4

C14H24O6S

O

H

H3C HOCH3

OOH3C

NaBH4, CH3OH or H2O

O

H

H3C HOCH3

OO

H3C

SCH3

O O

NaOCH3, CH3OH


3. (10 MARKS) Provide detailed stepwise mechanisms for both reactions in the following two-step synthetic transformation.

O

OHCl

Cl

O

OCH2Cl2

O

Cl

( + CO2 + CO + HCl )

CH3CH2SH

Et3N, CH2Cl2

O

S

O

OH

O Cl

O

Cl - H+O

O

O Cl

O

Cl

O

O

O

O

Cl

Cl

O

OO

OCl

ClO

Cl + CO2 + CO + Cl-

CH3CH2SH

O

ClSH

H3CH2C

NEt3O

ClS

H3CH2C

O

SCH2CH3- Cl-

- Et3NH+


4. (a) (1 MARK) The presence of nitrogen in an organic compound can be demonstrated by carrying out a sodium fusion. After dissolving the fusion product in water, the presence of nitrogen in the original compound is confirmed by adding FeSO4 and observing the formation of “Prussian Blue”, Fe4[Fe(CN)6].

On the basis of this information, what molecular or ionic nitrogen-containing species is formed when a nitrogen-containing compound is fused with sodium?

(b) (2 MARKS) Silver halides (AgX; X = Cl, Br or I), formed when testing a sodium fusion residue for halogens, are distinguished from one another in two ways:

• Physical appearance • Solubility in aqueous ammonium hydroxide.

Briefly describe how these two tests would distinguish AgCl from AgI.

(c) (2 MARKS) A student measured the boiling temperature of a liquid by distilling a sample. A value of 157 oC was obtained, but the literature value was found to be 138 oC. Indicate two experimental errors that might have led to the incorrect boiling temperature being measured.

(d) (3 MARKS) An unknown compound (X) was dissolved in cold hydrochloric acid and the solution was treated with a cold aqueous solution of sodium nitrite. The mixture was then carefully added to a cold alkaline solution of β-naphthol. A red solid (Y) was formed.

i) What type of compound was X? ii) Draw a reasonable structure for Y.

HO

β-naphthol

The only nitrogen in Prussian Blue is in the ion [Fe(CN)6]4- where it occurs as CN–. Therefore cyanide ion (CN–) is formed during sodium fusion.

Silver chloride (AgCl) is white and soluble in aqueous NH4OH. Silver iodide (AgI) is yellow and insoluble in aqueous NH4OH.

1) The thermometer bulb may have been immersed in the liquid rather than just the vapour.

2) Rapid boiling of the liquid may have caused heating of the vapour above the true boiling point.

Other reasonable suggestions will be considered on their merits.

Compound X was a primary aromatic amine, Ar-NH2.

N N

HO

Y =

this is a minimal structure. Any modified aryl ring is acceptable


(e) (2 MARKS) A solution of sodium hydroxide in aqueous ethanol was treated with a 1 : 1 mixture of phenylethanal (C6H5CH2CHO) and propanone. The mixture was shaken for about 20 minutes. A yellowish solid precipitated and was filtered off. The solid was shaken with water and this mixture was then treated with dilute hydrochloric acid until it was weakly acidic (pH 4 - 7). The product of the reaction was “worked up”. Draw the structure of this product.

(f) (2 MARKS) A student is given a sample of aniline and told to use it to prepare a sample of p-nitroaniline. Indicate the steps that would be required to ensure that the synthesis would work and that little or no m-nitroaniline by-product would be formed.

In lab, you did the Claisen-Schmidt reaction between acetone and benzaldehyde. Note that this is a crossed aldol, which is only feasible because benzaldehyde is NOT enolizable. In this question, the aldehyde IS enolizable, and thus the major product will not be derived from acetone – it will be a self-condensation of the phenylethanal.

O

H

1) React aniline with acetic anhydride or acetyl chloride to form acetanilide.

NH2

(CH3CO)2Oor CH3COCl

HN

O

2) React acetanilide with cold HNO3/H2SO4.

HN

O

HNO3/H2SO4

ice bath temperature

HN

O

NO2

3) Hydrolyse with aqueous NaOH.

HN

O

NO2

NaOH/H2O

NH2

NO2

+ CH3COO


5. (8 MARKS) The Claisen Rearrangement is a common example of a sigmatropic process. Two very similar molecules shown below were subjected to identical conditions expected to lead to Claisen Rearrangement. The outcomes of the two reactions were, however, rather different. Using your knowledge of organic structures and reaction mechanisms, EXPLAIN why the product of reaction B does not appear to follow the standard Claisen pattern.

O

toluene, heat

OH

O

toluene, heat

OH

H3CH3C

A

B

The key difference between these two reactions is that the substrate in reaction B has an alkyl group at both ortho positions, which means that it is impossible for it to re-aromatize by loss of a hydrogen if the rearrangement stops at this position.

O

H

OH

OCH3

cannot reform aromatic ring because CH3 group cannotbe lost in a tautomeric equilibrium processHOWEVER, this intermediate product is also a 1,5-diene and thus can undergo a second Cope-type rearrangement!

O

H3C

H

OH

H3C


6. (8 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the molecular formula C7H13NO are shown below.

4000 3000 2000 1500 1000 500

br s

(a) Draw the structure of the compound in the

box at right (4 Marks).

(b) Clearly indicate which H atom(s) in your structure is(are) responsible for each of the signals in the 1H NMR spectrum (4 Marks).

O

NH

CH3

CH3

CH3

H

H

Hor

HN

CH3

CH3

CH3

H

H

H

O

Full marks will be given for either of these structures, although the spectra are actually of the one on the left – t-Butyl Acrylamide. Part marks will be given for reasonable fragments.

t-Butyl group CH3 hydrogens

Vinylic (C=C-H) hydrogens

Amide N-H


PART II: CHALLENGE PROBLEM. DO ONE PROBLEM ONLY. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Part II is worth 8 Marks.

A. SYNTHESIS The simple alkaloid (R)-δ-coniceine has been synthesized from the intermediate A shown. The synthesis used reactions that have been discussed in this course. Construct a route to coniceine from intermediate A in 4 or 5 steps. The first reaction should form the bicyclic ring system found in the product.

B. MECHANISM The Benzoin Condensation, shown below, may appear surprising at first but its mechanism is actually quite simple. The process is catalyzed by cyanide ion – stoichiometric KCN is not needed. However, the catalyst must be cyanide (or one of a very few other substances). Hydroxide or similar basic reagents will NOT work.

O

HKCN

EtOH/H2Oheat

O

HO H

Provide a mechanism for the Benzoin condensation and explain why cyanide is essential. Can you suggest what other kinds of substances besides cyanide might catalyze the Benzoin condensation?

C. SPECTROSCOPY

You have been presented with an unknown organic compound to analyze. It is a clear liquid with a spicy odour similar to nutmeg. You perform several chemical tests, with the following results:

• Insoluble in water or dilute acid, soluble in aqueous NaOH. Also soluble in CHCl3, ether, alcohol.

• Decolourises bromine water fairly rapidly.

• Forms a blue solution when treated with FeCl3.

A mass spectrum shows the molecular ion at m/z 164, and a significant fragment at m/z 149, along with numerous other fragments. You then obtain IR, 1H and 13C NMR spectra, which are shown on the next page.

1. What is the structure of the unknown material?

2. Explain HOW the chemical and spectroscopic evidence supports your proposed structure.

N

(R)-δ-coniceineNH

O

COOEt

4 or 5 steps

A


There are 2 peaks inthis signal

multiplet singlet singlet

Double doublet

Doublet of quartets

Doublet of quartets

Answer to Synthesis Challenge Problem You know that the conversion of A to coniceine can be achieved in 4 or 5 steps, and that you should start by forming the bicyclic ring system. This means making a C-N bond, and the obvious way to do this is to make the nitrogen attack the electrophilic ester carbonyl. This is easily achieved using a base. Notice that because of conjugation with the enone, this amine will be quite acidic.

Now all that has to be done is to remove excess functional groups – an alkene, a ketone carbonyl and an amide carbonyl. The order in which you do this matters to some extent however. The only way we know to remove an amide carbonyl is by reduction with LiAlH4 in THF or ether, but this will also reduce a ketone to a secondary alcohol. If the ketone were reduced, we would have to re-oxidize it back to the ketone because the only obvious way we know to completely remove an oxygen from an aliphatic molecule is the Wolff-Kishner reduction, which doesn’t work with alcohols. The best route is thus to hydrogenate the alkene, then use Wolff-Kishner to remove the ketone, and finally LiAlH4 to reduce the amide to an amine. Now, you could have treated the keto-amide with LiAlH4 and proceeded from the alcohol, if you had remembered from Chapter 8 that alcohols, when converted to sulfonate esters, can be attacked by nucleophiles. Thus, a possible 5-step synthesis would have involved the following steps:

There is considerable leeway in the details of these routes, and there are other alternatives that you may have come up with as well. Credit will be given for any reasonable approach or portion thereof.

N

(R)-δ-coniceineNH

O

COOEt

4 or 5 steps

A

NH

O

COOEt

NaOEt, EtOH

or NaH, THFN

O

O

H2 (g)Pd catalystEthanol

N

O

O

KOH, H2NNH2

N

O

N

LiAlH4THF

N

O

O

LiAlH4THF

N

OH

S Cl

O

O

H3C

Et3N or pyridine N

OS

O

O

CH3

LiAlH4THF

N

NH

O

COOEt

Base

N

O

COOEt

N

O

COOEt

N

O

COOEt

Answer to Mechanism Challenge Problem

Looking at the reaction to find the starting material in the product, it should be evident that the product is a dimer of the starting aldehyde. You also can see that the cyanide ion is not incorporated into the product, but that should not be a surprise since you are told that it functions as a catalyst. The product has a C-C bond between the carbons that were the aldehyde carbonyls in the starting material. Notice that one of the carbonyls has been converted to an alcohol in a pattern that suggests a possible nucleophilic addition. The other carbonyl is still present but has lost its terminal hydrogen, suggesting that somehow it has been deprotonated and is acting as a nucleophile. The problem is, aldehyde C-H groups are NOT acidic. So, what is the cyanide doing to change this fact? Recall that cyano groups are strongly electron withdrawing. Moreover, we have seen (in Chapter 15) that this makes the C-H bond adjacent to a CN group quite acidic. We also know that cyanide reacts with aldehydes to make cyanohydrins. These facts provide the solution to the problem.

You can see that the cyanide is important because it stabilizes the nucleophilic anion by resonance. This is why hydroxide etc. will not work in this reaction. Any other nucleophile that can add to a carbonyl AND can stabilize an adjacent anion by resonance would also work.

O

HKCN

EtOH/H2Oheat

O

HO H

O

HCN

O

HO H

O

H

CNHOEt

HO

H

CN

EtO

OH

CN

OH

CN

O

H

HO

O H

CN

EtOH

HO

HO H

CN

EtO

O

HO H

CN

Answer to Challenge Spectroscopy Problem The mystery compound is trans-isoeugenol, shown at right. However, from the spectra given it is actually very difficult to ascertain the positions of the OH and OCH3 groups absolutely. This is because the 3 aromatic hydrogen signals all happen to overlap, so the pattern of a singlet and a pair of doublets that the structure would suggest is not visible. Thus, full marks will be given for isomers of trans-isoeugenol that differ in the positions of these groups. This structure is indicated by the following facts:

1) Solubility suggests a phenol (sol. in aqueous NaOH but not in other aqueous solvents).

2) Decolourises bromine water – suggests electron-rich aromatic ring (such as a phenol) and/or an alkene.

3) Blue colour with FeCl3 is a clear indication of a phenol. 4) Mass spectrum fragment at 149 indicates loss of a CH3 group from the parent ion

at 164. 5) IR spectrum shows OH but NO C=O. If you looked further you might have

pointed out the presence of aromatic C=C stretches at 1600 and 1500 cm-1. 6) The 13C NMR shows 10 separate signals. Only two of these signals are above 60

ppm, and the signal at 55 ppm strongly suggests an O-CH3 grouping. 7) The 1H NMR shows 6 groups of signals in a ratio of 3:1:1:1:3:3. A 3H singlet at

about 3.7 ppm definitely says OCH3, while the 3H double doublet at about 1.9 ppm suggests another CH3 group, probably adjacent to an alkene, and coupled to two different hydrogens. There are only 3 aromatic protons, evidence for a trisubstituted ring. The 1-proton doublets of quartets are at the right chemical shift for vinylic hydrogens, and the splitting pattern suggests that they are coupled to each other (doublet) and to a CH3 group (quartet). The remaining 1H signal must be the phenolic OH.

At this point it is possible to check for a formula. We have evidence for 10 carbons, at least 2 oxygens, and 12 hydrogens. This does in fact add to 164, so this is probably the formula. This formula has 5 degrees of unsaturation, consistent with having both an aromatic ring and an alkene. These deductions give us the following fragments:

Since the aromatic hydrogen signals don’t offer much help, we really have to guess about the correct placement of the OCH3 and alkenic groups.

OH

O

isoeugenol

C10H12O2Mol. Wt.: 164.20

OH

with 2 other substituents

OCH3CH3

H

H

Spectroscopy “Crib Sheet” for 2.222 – Introductory Organic Chemistry II 1H NMR – Typical Chemical Shift Ranges

Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)

C CH3 0.7 – 1.3 C C H 2.5 – 3.1

C CH2 C 1.2 – 1.4 O

H 9.5 – 10.0

C H

C

C

C

1.4 – 1.7 O

OH


C H

1.5 – 2.5 C OH


O

H

2.1 – 2.6 CO H

3.3 – 4.0

Aryl C H 2.2 – 2.7 CCl H

3.0 – 4.0

H 4.5 – 6.5 CBr H

2.5 – 4.0

Aryl H 6.0 – 9.0 CI H

2.0 – 4.0

13C NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands Group Frequency

(cm-1) Intensity Group Frequency (cm-1) Intensity

C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium Aryl C=C 1600, 1500 Strong RNO2 1540 Strong

12 11 10 9 8 7 6 5 4 3 2 1 0

←δ

R3C–H Aliphatic, alicyclic

X–C–H X = O, N, S, halide

Y

HH

Aromatic, heteroaromatic

Y

H

RCO2H

Y = O, NR, S Y = O, NR, S

“Low Field” “High Field”

220 200 180 160 140 120 100 80 60 40 20 0

←δ

CH3-CR3 CHx-C=O

CR3-CH2-CR3

CHx-Y Y = O, N Alkene

Aryl

AmideEster

Ketone, Aldehyde

Carbox. Acid

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