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Chemistry 303 Fall, 2011

FINAL EXAMINATION KEY

1:30 PM, January 18TH, 2012

Duration: 20 minutes reading and then 3.0 hr to write Name_____KEY______________________________________________________ (Official Name) This is an “open book” examination; you may use anything that is not alive or connected to the Web.

Note: if you do not know the complete or specific answer, give a partial or general answer—

We love to give partial credit.

If there seems to be more than one good answer, explain your thinking.

If you invoke resonance delocalization as part of your answer, draw the relevant resonance structures.

If you draw a chair cyclohexane, be sure to orient the bonds carefully.

If you do not know a structure and need to write a mechanism, write a general mechanism for partial credit.

You need not draw transition states as part of a mechanism unless expressly instructed to do so.

USE THE ARROW FORMALISM CAREFULLY FOR ALL MECHANISMS. SHOW ALL INTERMEDIATES.

BE SURE TO INCLUDE ALL FORMAL CHARGES.

Write only in the space provided for each question. Score: p2___________/ 10 p3___________/ 10 p4___________/ 18 p5___________/ 18 p6___________/ 15 p7___________/ 12 Lab question _________/26 p8___________/ 6 p9___________/ 14 p10__________/ 10 p11__________/ 12 p12__________/ 15 p13__________/ 14 p14__________/ 30 p15__________/ 16 Lecture Total: /200 There are 19 pages in this exam; please check now to be sure you have a complete set. The last page is Table 3.2 from the text related cyclohexane conformational data, and a glossary of definitions. Pledge:_________________________________________________________________________________

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1. Fumaric acid is an intermediate in the Krebs cycle and is involved in the production of energy in the form of ATP from the breakdown of carbohydrates, fats, and proteins. Fumaric acid and its closely related isomer maleic acid have several unique properties that significantly alter their relative properties in living organisms and in the laboratory. For example, consider the aqueous pKa values for the two compounds.

(a) (5 pts) Provide the single best reason why the most acidic proton (pKa1) on maleic acid is approximately 12.5 times more acidic than the corresponding proton on fumaric acid.

1 pt for drawing conjugate base of maleic acid 3 pts for identifying intramolecular H-bond 1 pt for identifying that intramolecular H-bond not possible for fumaric acid

(b) (5 pts) Interestingly, the second proton to be deprotonated on maleic acid (pKa2) is approximately 70 times less acidic than the corresponding second proton on fumaric acid. Provide the single best reason to explain this data.

maleic acid: 1 pt for stabilization of acid by H-bond; 3 pts for dipole-dipole destabilization of conjugate base fumaric acid: 1 pt for dipole-dipole stabilization of conjugate base

CO2HHO2C HO2CCO2H

fumaric acid

pKa1: 3.02pKa2: 4.38

maleic acid

pKa1: 1.92pKa2: 6.23

CO2HHO2C CO2

HO2C

HO2CCO2H

OH O

O

O

CO2HO2C

OH O

O

O

CO2O2C

OO

O

O

3

2. (5 pts) (a) Label the following molecules in order of basicity (most, least, intermediate), and defend your choice. You must draw pictures to receive credit.

2 pts for ordering 1 pt for accurate resonance structures showing delocalization of nitrogen lone pair into ring 1 pt for effect of nitro group 1 pt for effect of methoxy group

(b) (5 pts) Now consider molecules A and B. Contrary to what you might expect at first glance, the basicity of A is approximately the same as the basicity of B. Provide the single best reason why. Explain your choice with carefully drawn pictures.

3 pts for non-planarity of ring system 2 pts for no delocalization of nitrogen lone pair into the bottom ring

O2N

NMe2

Me

NMe2

MeO

NMe2

least mostintermediate

N

NMe2

O

O N

NMe2

O

ON

NMe2

O

O

Me

NMe2

Me

NMe2

MeO

NMe2

MeO

NMe2

destabilized due to electron-electron repulsion

O2N

t-BuNMe2

O2N

t-BuNMe2

A B

O2N

O2N

t-BuNMe2

A

O2N

The two aromatic rings cannot be in the same plane due to steric interactions between the amine and tert-butyl group. Thus, there is no resonance delocalization of the nitrogen lone pair into the bottom ring.

x

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3. (18 pts) For each of the following pairs of reactions, (i) draw the product(s) of each reaction and (ii) circle which member of the pair would be FASTER. (iii) Explain briefly the most important reason for your choice.

1 pt for each correct product (2 pts total) 1 pt for correct choice of faster reaction 1 pt for SN1 (cation formation is rds) 1 pt for stabilization of allylic cation 1 pt for destabilization of !-carbonyl cation

1 pt for each correct product (2 pts total) 1 pt for correct choice of faster reaction 1 pt for SN2 mechanism (rds is dependent on electrophile and nucleophile identity) 2 pt for starting material destabilization (transition states similar energies) -or- (1 pt for nucleophile attack on opposite face of t-Bu group (steric argument))

1 pt for each correct product (2 pts total) 1 pt for correct choice of faster reaction 1 pt for SN2 mechanism (rds is dependent on electrophile and nucleophile identity) 2 pts for intramolecular faster than intermolecular

O

MeMe

Br

MeMe

Br

Me

(a) AgNO3

EtOH

AgNO3

EtOH

O

MeMe

OEt

MeMe

OEt

Me

O

MeMe

MeMe

Me

MeMe

Me

O

MeMe

t-Bu

t-Bu Cl

Cl

NaI

acetone

NaI

acetone

(b)

t-Bu

t-Bu I

I

t-Bu

Cl

t-Bu Cl

t-BuHCl

I

t-BuHI

Cl

!

!

!

!

!

!

(c) Br

Br

OH

NaH

Et2O(0.1M)

NaH

Et2O(0.1M)

HO

Me

Me

O

OMe Me

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4. Consider the following transformations of compound C.

(a) (6 pts) Predict the structure of D and draw the best mechanism for step (1) of its formation. Include a carefully rendered drawing of the rate-determining transition state of the reaction.

1 pt for correct structure of D (anti-Markovnikov) 1 pt for concerted addition of B and H 2 pts for correct direction of two arrows 1 pt for correct position of B and H relative to alcohol product 1 pt for 4-membered transition state (no penalty for not showing di- and trihydroboration)

(b) (8 pts) Predict the structure of E that is consistent with the IR data provided, and indicate the single best mechanism for its formation.

3 pts for correct structure of E (1 pt for wrong structure but consistent with IR data) 1 pt for Markovnikoff protonation 1 pt for cyclopropane ring opening 1 pt for cleavage of correct cyclopropane bond 1 pt for water addition to cation 1 pt for deprotonation of oxonium (no penalty for showing E or Z alkene isomer)

Ph H2SO4

H2O

1. BH3, Et2O

2. H2O2, NaOHD: C11H14O

E: C11H14O

C

select IR data: 3432, 3083, 1644, 1610 (phenyl) cm–1

Ph BH3 PhHH2BPh

HH2BPh

HH2B

PhHB

Ph

PhH2O2/NaOHPh

D

HO

!

!

PhH+

PhH H Ph H2O

H Ph

OH HH2O

H Ph

OHE

H+

formation of stabilized benzylic cation is favorable even though it is still secondary

6

(c) (4 pts) Compounds D and E were both treated with MnO2, but only one of the two underwent reaction. Predict which compound reacted with MnO2; draw the structure of the resulting product; and briefly rationalize your choice.

Compound E will react with MnO2: MnO2 only oxidizes allylic or benzylic alcohols because it is a mild oxidant. E contains a benzylic alcohol whereas D contains an aliphatic alcohol.

2 pts for correct alcohol substrate 1 pt for correct product 1 pt for explanation

H Ph

O

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5. Consider the following reaction in the gas phase and the provided data:

(a) (2 pts) Draw the product G in the box provided above. 2 pts all or nothing (b) (2 pts) Identify the nucleophile and the electrophile of the reaction. Alkyl chloride F is the electrophile; trimethylamine is the nucleophile 1 pt for each (c) (4 pts) Carefully draw the orbitals corresponding to the HOMO of the nucleophile and LUMO of the electrophile.

1 pt for CCl antibonding as LUMO; 1 pt for correct picture 1 pt for lone pair as HOMO; 1 pt for correct picture (d) (3 pts) Which of the thermodynamic terms above favors this reaction going toward G? The entropy term is the only one that is favorable for the forward direction because it is positive (the products are more disordered than the starting materials). The enthalpy, the free energy, and the Keq all favor the starting material over the products. 3 pts for correct answer. No partial credit. (e) (4 pts) The equilibrium constant for this reaction in aqueous solution is much larger than the one provided for the gas phase. How can you account for this fact? Relative to the gas phase, both the starting materials and the products will be stabilized. In aqueous solution, the charged products will be highly stabilized by H-bonding and dipole-dipole interactions with water. The starting materials will only be minimally stabilized in water compared to the products due to the fact that they are neutral molecules. Thus, the Keq value will increase. 2 pts for stabilization of products

2 pts for specific molecular associations that will stabilize products (H-bonding, dipole-dipole interactions)

!H0 = +6.36 Kcal/mol!S0= +0.00115 Kcal/K*mol

T = 25°C!G0 = +6.02Kcal/mol

Keq = 3.85*10–5

Cl NMe3 ClMe Me

F G

NMeMe

Me

ClMe

F

LUMO: !* C–Cl

NMe

MeMe

HOMO: sp3 lone pair on Nitrogen

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(f) (12 pts) On one plot, draw reaction coordinate diagrams for the gas phase and aqueous reactions. Be sure that your diagrams account for the relative energies of the starting materials, products, and activation energies between the two conditions. Label "Go

gas, "G‡gas, "Go

aq and "G‡aq.

1 pt for single TS (SN2) for both reactions 1 pt for correct identification of activation energy ("G‡) 1 pt for correct identification of free energy ("Go) 1 pt for aq starting materials lower energy than gas phase starting materials 1 pt for aq products lower energy than gas phase products 1 pt for aq transition state lower energy than gas phase transition state 2 pts for "Go

gas > 0 as given 2 pts for "Go

aq less than "Gogas (probably negative - part (e))

2 pts for "G‡aq < "G‡

gas – transition state stabilization more significant than starting material stabilization.

reaction coordinate

pote

ntia

l ene

rgy

F + NMe3(H2O)

F + NMe3(gas)

TSgas

TSaq

G + Cl–(H2O)

G + Cl–(gas)!Gogas

!G gas

!Goaq < 0

>0

!G aq

!G aq !G gas<

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6. Consider the following data regarding the reactions depicted below:

(a) (2 pts) What type of isomers are H and I? Diastereomers 2 pts for diastereomers (or 1 pt for stereoisomer) (b) (4 pts) What elimination mechanism is most likely for reactions (i) and (ii)? Draw the curved arrow mechanism that accounts for the product generated in (i). Note: you do not need to draw a 3-dimensional picture of H for your mechanism. E2 mechanism

1 pt each for correct arrows (concerted) (3 pts total)

1 pt for E2

MeMe

Br

MeMe MeMe

MeMe

Br

MeMe MeMe

Me Me Me

100% 0%

Me Me Me

25% 75%

NaOEt

EtOH

NaOEt

EtOH(i)

(ii)

H

I

MeMe

Br

MeMe

Me MeNaOEt

EtOH

HOEt

H

+ HOEt + NaBr

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(c) (12 pts) Draw the chair conformations of H and of I (4 total). Circle the lowest energy chair conformation for each molecule and calculate how much lower in energy each is relative to the higher energy form (note: the back page of the exam has useful data for doing this). Show your work and if necessary, identify any ambiguities in your estimate.

1 pt each for well-drawn/correct chairs (2 pts total) 1 pt for correct low energy structure 2 pts for correct value of energy difference 1 pt for ambiguities

1 pt each for well-drawn/correct chairs (2 pts total) 1 pt for correct low energy structure 2 pts for correct value of energy difference 1 pt for no ambiguities (d) (2 pts) Why does reaction (i) give only one product? Explain your answer using 3-dimensional pictures.

There is only one available antiperiplanar beta hydrogen for E2 elimination. 1pt for E2 from correct chair conformation 1 pt for antiperiplanar beta-H

i-PrBr

Me

i-Pr

Br Me

axial i-Pr: + 2.1 kcal/molaxial Br: + 0.5 kcal/molaxial Me: + 1.7 kcal/mol1,3 diaxial Br/Me interaction + 4.3 kcal/mol+ 1,3 diaxial Br/Me interaction– i-Pr/Br gauche interaction

ambiguities:

i-Pr/Br gauche interaction

i-Pr Me

i-Pr

Me

axial i-Pr: + 2.1 kcal/molaxial Me: + 1.7 kcal/molgauche i-Pr/Br

BrBr

axial Br: +0.5 kcal/molgauche i-Pr/Br – 0.5 kcal/mol

– gauche i-Pr/Br

+3.3 kcal/mol

H

H

Br

Hi-Pr

BrMe

i-Pr

Br Me

HH

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(e) (4 pts) Why does reaction (ii) favor the indicated product? Explain your answer using 3-dimensional pictures.

Elimination will occur from the low energy chair conformation. Two axial C-H bonds are aligned antiperiplanar to the C-Br bond and could undergo elimination. Elimination to favor the more stable/more-substituted double bond is faster because the transition state leading to its formation is lower in energy.

1 pt for E2 from correct chair conformation 2 pts for identifying both antiperiplanar beta C-H bonds 1 pt for more substituted double bond, more stable = lower energy transition state

(f) (6 pts) Assign the 1H NMR spectra below to the two products of reactions (i) and (ii). Describe one diagnostic feature of the spectra that justifies your answer.

possible answers: vinyl C-H bond (Ha) is a triplet in the bottom spectra, consistent with the trisubstituted

alkene product and not the disubstituted alkene (Hb/Hc).

Hd is more downfield in the bottom structure, consistent with it being allylic whereas He is not.

3 pts for structure matching 3 pts for consistent justification

HH

Br

Me

Me

i-Pr MeBr

H H

01234567PPM

0123456PPM

MecMec

Med

MeaMea

Meb

Ha

Hb

Hc

Hb/Hc

Ha

Mea

Meb

Mec

Med

Hd Hd

HeHe

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7. In 1990 the Sharpless lab at MIT devised an elegant synthesis of L-hexose derivatives beginning from the allylic alcohol J shown below. (a) (4 pts) Draw the product (K) of epoxidation of J with mCPBA and indicate how many stereoisomers are possible for the product.

2 pts for correct structure 2 pts for 4 diastereomers (all or nothing). No need to draw all four isomers so long as a structure is drawn that clearly indicates the trans geometry of the epoxide.

(b) (8 pts) If K is subjected to a reaction containing a sulfur nucleophile, what's known as a Payne reaction can occur. The product of the Payne reaction is depicted below. Provide a mechanism for the formation of product L.

1 pt for deprotonation of alcohol

3 pts for formation of terminal epoxide by SN2 process 1 pt for protonation of alkoxide 1 2 pts for nucleophilic attack of thiolate as SN2 process 1 pt for protonation of alkoxide 2

OO

MeMe

OH

(+/–)-J

OO

MeMe

OH

K

OmCPBA OO

MeMe

OHO

OO

MeMe

OHO OO

MeMe

OHO

OO

MeMe

SPh

NaSPhNaOH

H2O, t-BuOHOH

OHK

L

OO

MeMe

OO OO

MeMe

OO

OHO

OMeMe

O

O

OO

MeMe

SPh

OH

O

KH

OO

MeMeO

OH

SPhHO H

OHHOO

MeMe

SPh

OH

OH

L

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8. The following reaction sequence can be used to provide P from M.

(a) (4 pts) Draw the structure of N and describe the two most diagnostic features of N that would show up in an IR spectrum (not the fingerprint region).

2 pts for correct structure (1 pt for wrong structure consistent with molecular formula or wrong stereochem) 2 pts for IR peaks (1 each) (consistent with proposed structure) (b) (4 pts) Draw the structure of O. About where would you expect the most downfield (highest ppm) signal to be in the 1H NMR spectrum of O?

2 pts for correct structure (1 pt for aldehyde) 2 pts for NMR signal (consistent with proposed structure) (c) (2 pts) Label all of the stereocenters in P as (R) or (S)

1 pt for each stereocenter (d) (5 pts) Draw the structure of P in its lowest energy chair form (make sure that you draw the correct enantiomer of P).

3 pts for trans decalin; 2 pts for correct enantiomer

I

NaOH 1. O3

2. H2O2

O OH

H

M

N O

P

one step

(next semester)

formula: C10H18Oacetone

OH

SN2

C=C stretch: 1500-1650 cm-1

O-H stretch: broad peak at ~3200-3500 cm-1

=C-H stretch: 3150-3100 cm-1

OH

O

OHozonolysis followed by oxidative workup

for a carboxylic acid, a signal at ! 10-12 ppm

(R)(R)(S)(S)

O OH

H

P

HO O

H

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9. Labeling molecules with unnatural isotopes can change their spectral properties. Consider the following pair of molecules, one of which has ONLY 12C carbons (Q), and the other of which has TWO 13C labeled carbons (R) and ONE 12C carbon.

(a) (6 pts) If the molecular ion of Q is set to 100% in the mass spectrum, what percentage is M+1 for Q? M+2? Would the mass spectrum be different in a predictable way in R? If so, how?

1 pt for no M+1 for Q 1 pt for M+2 at (100%) for Q 1 pt for answer of YES For R: 1 pt for M at +2 units relative to Q For R: 1 pt for M+2 at (100%) for R 1 pt for no M+1 for R (b) (4 pts) Q has an absorption peak at 1637 cm–1 in the IR. What functional group does this correspond to? Would this stretching frequency shift in a predictable way for R? If so, how? Briefly explain your answer.

1 pt for C=C functional group 1 pt for answer of YES 2 pts for decrease in frequency based on reduced mass

For part (c), use the labeling scheme Cx-Cz shown below:

(c) (4 pts) How many signals would you observe in the proton-decoupled 13C NMR spectrum for Q? What splitting patterns would they have and why? How many signals would you observe in the proton-decoupled 13C NMR spectrum for R? What splitting patterns would they have and why?

1 pt for no signals for Q 2 pts for two signals for R 1 pt for doublets

H312C12C

12C

Br

H

H

H312C13C

13C

Br

H

HQ R

QM = 120 (100%)

M+1 (0%)M+2 (100%)

RM = 122 (100%)

M+1 (0%)M+2 (100%)

Because we have defined the carbon isotopes exactly, there is no 1.1% chance of a 12C being replaced with a 13C in either molecule. M+2 is from 81Br.

QC=C stretch (1500-1650 cm-1)

RBecause frequency is proportional (Mr)-1/2, where Mr is reduced

mass, C=C stretch will decrease in frequency.

CxH3Cy

Cz

Br

H

H

QNone! There are no 13C's

RTwo. Cy and Cz are both doublets (they couple to each other), and CX is 12C

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10. (30 pts) Provide the necessary reagents to convert 1-methylcyclohexene (S) into the target compounds T–Y. (Hint: More than one step may be required in some of the conversions.) 5 pts each… 2/5 points for correct connectivity and wrong stereochemistry 3 points for correct first step and wrong second step

Me Br

T (racemic)

Me OHBr2/H2O

(a)

S

OH

U (racemic)

Me Br1. m-CPBA

2. HBr(b) S

MeOCH3

V (racemic)

1. BH32. H2O2/NaOH

3. NaH4. CH3I

(c) S

OH

W (racemic)

Me OH1. m-CPBA

2.NaOH/H2O(d) S

OH

X (racemic)

Me OH1. OsO4

2. NaHSO3/H2O(e) S

O

Y (racemic)

(f) S

Me1. BH32. H2O2/NaOH

3. PCC or Collins Reagent

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11. Caryophyllene is obtained from the oil of cloves. When it is treated with aqueous acid, two products are obtained, caryolanol and clovene.

(a) (8 pts) Draw the best arrow-pushing mechanism for the formation of caryolanol (you do not need to account for the stereochemical outcome of the reaction).

2 pts for protonation of correct alkene 1 pt for formation of correct cation (more substituted) 2 pts for alkene attack of cation 1 pt for correct cation formation (more substituted) 1 pt for water attack 1 pt for deprotonation (b) (8 pts) Draw the best arrow-pushing mechanism for the formation of clovene (you do not need to account for the stereochemical outcome of the reaction).

3 pts for common intermediate (or redoing mechanism) 2 pts for alkyl shift 1 pt for breaking correct C–C bond 2 pts for elimination of C–H

H

H

MeMe Me

HH2SO4

H

H

MeMe Me

HO

Me

MeMe H

clovenecaryolanolcaryophyllene

H2O

H

H

MeMe Me

H

H+

H

H

MeMe Me

HH

H

H

MeMe Me

H2O

H

H

MeMe Me

HOH

OH2

H

H

MeMe Me

HO

caryolanol

+ H+

Intermediate

H

H

MeMe Me

Intermediate

Me

MeMe H

HH2O

Me

MeMe H

clovene

+ H+

although this is also a secondary cation, it is more stable than the intermediate because the ring strain in the 4-membered ring has been relieved

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Lab-Related Question (26 pts)

Bromination of “butene” isomers (C4H8) 1 and 2 gives dibromides 3 and 4, respectively.

Proton-decoupled 13C NMR spectra of “butene” isomers 1 and 2 and the 1H NMR spectra of dibromides 3 and 4 are summarized below:

“Butene” 1 proton-decoupled 13C NMR (CDCl3): d 25.8, 112.1, 141.8 “Butene” 2 proton-decoupled 13C NMR (CDCl3): d 14.0, 26.7, 116.4, 134.3 Dibromide 3 1H NMR (CDCl3): d 1.87 (s, 3H), 3.86 (s, 1H) Dibromide 4 1H NMR (CDCl3): d 1.08 (t, J = 7 Hz, 3H), 1.75-1.91 (m, 1H), 2.12-2.26 (m, 1H), 3.63 (t, J = 10 Hz, 1H), 3.84 (dd, J = 10 Hz, J = 4.5 Hz, 1H), 4.10-4.20 (m, 1H) (a) (4 pts) Deduce the structures of “butene” isomers 1 and 2. Indicate how the proton-decoupled 13C NMR spectra of “butenes” 1 and 2 support your structural assignments. (Hint: There are only four possible “butene” isomers. Draw the four possible isomers.)

There are only four possible “butene” isomers, A-D. We can use the proton-decoupled 13C NMR spectral data to differentiate these four isomers. From symmetry considerations alone, we can differentiate 1-butene (A) [4 unique carbons] and isobutylene (D) [3 unique carbons] from the cis- and trans-2-butenes (B) and (C) [each with 2 unique carbons]. Thus, “butene” 1 must be isobutylene (D), which shows three unique carbons in its proton-decoupled 13C NMR spectrum. Similarly, “butene” 2 must be 1-butene (A), which shows four unique carbons in its proton-decoupled 13C NMR spectrum. (Chemical shift assignments are not required for full credit.)

1PyHBr3

CH2Cl2 0 °C

3

2PyHBr3

CH2Cl2 0 °C

4

A B C D

19

(b) (3 pts) Deduce the structure of dibromide 3. Draw your proposed structure for dibromide 3, and label the hydrogens on your proposed structure. Support your structural assignment by assigning the 1H NMR signals to the appropriate hydrogens.

Note: The “relative” integrals need to be multiplied by a factor of two to give the “absolute” integrals. (c) You should be able to deduce the structure of dibromide 4, based on your proposed structure of “butene” 2. (There are no rearrangements here – THINK SIMPLE!) Of course, we eventually want you to assign all of the 1H NMR signals for dibromide 4 and to explain the splitting patterns. However, we suspect that you may find the 1H NMR spectrum of dibromide 4 more complicated than expected. So, a little guidance is in order. (Hint: Chemical shift calculations are not required, but may prove helpful for your spectral analyses.) (1) (1 pt) Draw the structure for dibromide 4, based on your proposed structure for “butene” 2.

This bromination reaction produces a racemic mixture of 1,2-dibromobutane (4). The 1H NMR spectral “complications” are a result of the newly created stereogenic center at C-2. The chemical shift calculations (not required) are summarized below: # (H4) = 0.90 ppm # (H3) = 1.20 + 0.60 = 1.80 ppm # (H2) = 1.55 + 2.20 + 0.25 = 4.00 ppm # (H1) = 1.20 + 2.15 + 0.60 = 3.95 ppm (2) (10 pts) Draw the lowest energy staggered Newman projection looking down the carbon-carbon bond, connecting

the two bromine-substituted carbon atoms of dibromide 4. Label the hydrogens on these two carbon atoms. Now, assign the 1H NMR signals to these labeled hydrogens and explain the observed splitting patterns. Does the magnitude of the observed coupling constants support your lowest energy conformational assignment? Explain.

1 (= D)

112.1

25.8

25.8

141.8

2 (= A)

14.0

26.7

134.3

116.4

1

PyHBr3

CH2Cl2 0 °C 3Br

Br

b

a

a

! 1.87 (s, 3H) Ha 3.86 (s, 1H) Hb

4Br

Br

123

4

Hb Ha

Br Et

Br

Hc

4

! 3.63! 3.84

! 4.10-4.20

20

The methylene protons at C-1 (Ha and Hb) are diastereotopic. Even though there is free rotation about the C-1/C-2 bond, these methylene protons are not equivalent, as nicely illustrated by the Newman projection of the most populated (lowest energy) staggered conformation (looking down the C-1/C-2 bond), shown above, for one enantiomer of 4. As you learned in lab, the lowest energy conformation here keeps the two bromine atoms as far apart as possible to minimize any Br/Br dipole destabilization. The 1H NMR signal at # 3.63 ppm is due to Ha. This signal appears at as a triplet because the two coupling constants are approximately equal (Jab ~ Jac = 10 Hz). Jab is large, of course, because the two protons are geminally coupled. The large vicinal coupling constant (Jac = 10 Hz) is consistent with a large dihedral angle (~180°) for Ha-C-C-Hc. The 1H NMR signal at # 3.84 ppm is due to Hb. This signal appears as a doublet of doublets with Jab = 10 Hz and Jbc = 4.5 Hz. Again, Jab is large because it is a geminal coupling constant. The smaller vicinal coupling constant (Jbc = 4.5 Hz) is consistent with a smaller dihedral angle (~60°) for Hb-C-C-Hc. Finally, the multiplet at # 4.10-4.20 ppm is due to Hc. Thus far, Hc should be a dd with Jac = 10 Hz and Jbc = 4.5 Hz. However, as we shall see, Hc is split by two additional protons and will be more complicated than a dd. (3) (4 pts) Please explain why the multiplets at d 1.75-1.91 and 2.12-2.26 ppm appear as separate 1H multiplets,

rather than as one 2H multiplet. Which two protons in dibromide 4 give rise to these two multiplets? The 1H multiplets at # 1.75-1.91 and 2.12-2.26 ppm are due to the C-3 methylene protons (Hd and He). These methylene protons are also diastereotopic due to the C-2 stereogenic center. This situation can nicely be illustrated by examining the lowest energy staggered Newman projection, looking down the C-3/C-2 bond, shown below.

Alternately, you can demonstrate the diastereotopicity of Hd and He by performing the “X substitution” test; i.e.,

You don’t have enough information to assign which proton is which here, as both signals were only reported as multiplets. Hd should be a ddq (Jde, Jcd, and Jdf), which could be reduced to a tq, if Jde ~ Jcd. He should be a dqd (Jde, Jdf, and Jce). (4) (2 pts) Please assign the remaining signal in the 1H NMR spectrum of dibromide 4, and explain the observed

splitting pattern. Don’t forget the label(s). The remaining signal in the 1H NMR spectrum of dibromide 4 is the 3H triplet at # 1.08 ppm. This signal must be due to the C-4 methyl group (Hf). The signal appears as a triplet because the methyl protons are coupled to Hd and

He Hd

CH3Br

CH2Br

Hc

4

! 1.75-1.91 or 2.12-2.26

! 4.10-4.20

! 1.75-1.91 or 2.12-2.26

CH2Br

H3C

HeHd

HcBr

4

"X substn."

"X substn."

CH2Br

H3C

HeX

HcBr

CH2Br

H3C

XHd

HcBr

Diastereomers!

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He with Jdf ~ Jef ~ 7 Hz (due to rotational averaging). Note that coupling constant for the methyl hydrogens (Hf) with Hd and He must both be 7 Hz; otherwise, the methyl signal would appear as a dd. (5) (2 pts) The signal at # 4.10-4.20 ppm is reported simply as a multiplet (“m”), although perhaps “mess” might be a

more apt description. The proton, giving rise to this mutiplet, should already be labeled in your Newman Projection from Part 2 above. What should be the actual multiplicity of this signal (e.g., dd, td, etc.)? Explain.

The signal at # 4.10-4.20 ppm has presumably already been assigned to the C-2 methine proton (Hc). This proton is coupled to four different protons (Ha, Hb, Hd, and He) with four different coupling constants (Jac, Jbc, Jcd, and Jce). Thus, Hc should actually appear as a doublet of doublets of doublets of doublets (dddd) with 16 lines (assuming none of the lines overlap, which is unlikely).

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Typical gauche interactions: 0.9 kcal/mol Typical 1,3-diaxial interactions: approx 2.0 kcal/mol Glossary: Me = methyl Et = ethyl Ph = phenyl t-Bu = tert-butyl OMe = methoxy

O THF

LDA = lithium disopropylamide N

MeMe

MeMe

Li

Ts = tosylate =MeS

O

O

Me SO

MeDMSO DMF

O

H NMe2