Write your name and date on the cover page Do not open exam until instructed to do so

Name:____________________________

Date:_____________________________

Exam III

Chem. 210

!

Do not open exam until told to do so.

Get out your pencil, eraser, and scientific nongraphing calculator. Put everything else under the desk or on the floor. Turn off or silence then stow all electronic devices.

You may leave the exam room after turning in your exam. You may not return to the exam room after leaving (until the exam is over).

Provide the best answers as requested. You must show work for credit. Label your work should you use the back side of the paper. If you have any questions during the exam, write them on the exam. When told to do so, check that exam has all of its pages.

! of !1 12

1) How does M+• form in mass spectrometry?

2) Draw the product(s) expected from this reaction. + HNO3

3) Which species gives a 1:1 ratio for the M+ to M+2 peak in the mass spectrum?

4) Where would the M+2 peak occur for bromoethane?

5) Describe the -OH absorption band in infrared spectroscopy.

6) A 1H NMR spectrum at 300MHz records a signal at 307 Hz downfield from tetramethylsilane. What would be the position in hertz (Hz) on a 90MHz instrument?

7) How many degrees of unsaturation are in this compound: C6H14

8) The sharp peak at 1700cm-1 in this infrared spectrum corresponds to what type or class of compound?

!

9) Ignoring spin-spin splitting, how many types of 1H NMR signals should appear from ? The answer is the same as the number of chemically equivalent hydrogens.

10) Draw a structure having the formula C5H11Cl that is consistent with this 300MHz 1H NMR spectrum.

!Br !Cl

!CH3 !CH3 CH3 CH3Cl Br

!CH3

Br! Cl!

!CH3 !CH3 !CH2Br !CH2Cl

O

HOCH2CHCH=CH2

OCH3

HOCH2CH=CHCH2OCH3

TMEDA

HOCH2CHCHCH2OH

CH3H3C

BrCH2CHCHCH2Br

CH3H3C

PBr3

OH

CH3HO

O

HHO

OH

CH3(CH3)3CO

OH

CH3(CH3)3CO

O

H(CH3)3CO

CH3

CH3

O

CH3OCH2C(CH3)2

OH

CH3C(CH3)2

OH

CH3CHCH3

OH

!CH3

O

O

CH3

H

OH

CH3

D

H

H2SO4

CH3

H

O

Br

CH3

HBr

HCH3 CH3

Br

BrH

HCH3 Br

H

HBr

CH3CH3

Cl

CH3

HCl

HCH3 CH3

Cl

ClH

HCH3 Cl

H

HCl

CH3CH3

OH

Br

OH

Br

OH

Cl

OH

Cl

OCH3

Br

OCH3

Br

OCH3

Cl

OCH3

Cl

H

O

H

H

O

H

CHCl3

CH2Cl2

CHCl3

CH3COOH

O

CH3CH

O

HCH

O

CH3CCH3

O

CH3CCH2CH3

OHCCH2CH3

O

HCCH2CH2CH3

O

CH3CCH2CH2CH3

O

CH3CH2CCH2CH3

O

! of !2 12

!

11) Which compound gives an integrated 1H NMR signal with a 1:1:2:2:3 ratio? Ignore spin-spin splitting. A. 3-chloro-2-methyl-1-propanol B. 1-bromo-2-methylbutane C. 3-bromo-1-chloro-2,3-dimethylbutane D. 3-bromo-1-chloro-2,2-dimethylbutane E. 5-bromo-1-pentanol

12) Using Planck’s constant as 6.626x10-34Js and the speed of light as 3.00x108m/s, convert a bond vibration of 2900cm−1 into kJ/mol.

13) How many peaks would be in the proton-decoupled 13C NMR spectrum of ?

14) Given that the UV range is from 200-400nm, Planck’s constant is 6.63x10-34Js, the speed of light is 3.00x108m/s, and Avogadro’s number is 6.02x1023, what is the approximate range of photon energies in J for UV light?

15) A spectrometer indicates that a compound absorbs UV-visible radiation at 325nm. Given that Planck’s constant is 6.63x10-34Js, the speed of light is 3.00x108m/s, and Avogadro’s number is 6.02x1023, what is the approximate frequency in Hz (s-1) required for this electronic transition?

16) Given that Planck’s constant is 6.63x10-34Js, the speed of light is 3.00x108m/s, and Avogadro’s number is 6.02x1023, calculate the energy in kJ/mol for the transition that occurs at λmax?

Y1 Y2 Y3 Y40.005 1 6 0 1

-4.31 -1.88 -0.799999 0.000001

0.06 0.06 0.06 0.06

0.1 0.1 0.1 0.1

7 2 2 1

1/pi 1/2 W

(x-M)2 + (1/2W)2(x-M)2 + (1/2W)2 View Page 66

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

δ, ppm

2-iodopropane

6H

1Hseptet

Y1 Y2 Y3 Y40.005 4 2 0 1

-3.53 -2.34 -0.799999 0.0000010.06 0.06 0.06 0.06

0.1 0.1 0.1 0.13 5 2 1

View Page 66

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

δ, ppm

1,3-dibromopropane

2H

4H

Y1 Y2 Y3 Y4 Y50.005 3 6 2 1 1

-0.8 -1.20 -1.6 -2.1 00.06 0.06 0.06 0.2 0.060.1 0.1 0.1 0.1 0.1

3 1 4 1 1

View Page 66

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

δ, ppm

2-methyl-2-butanol

1H2H

6H

3H

1-chloro-2,2-dimethylpropane

Y1 Y2 Y3 Y4 Y50.005 9 2 0 0 1

-1.1 -3.3 -1.6 -2.1 00.06 0.06 0.06 0.2 0.06

0.1 0.1 0.1 0.1 0.11 1 4 1 1

View Page 66

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

δ, ppm

2H

9H

(CH3)4Si

! of !3 12

!

17) What type of spectrum is this?

!

18) Predict the primary product of this reaction sequence: ! 1) ! 2) ! 3) !

19) What is a typical minimum number of conjugated double bonds for an organic compound to absorb in the visible region?

20) What is λmax in UV-visible spectroscopy?

21) Samples for UV and visible spectroscopy are usually dissolved in what type of solvent?

22) Provide an acceptable name for this compound: !

23) A 1H NMR spectrum at 300MHz records a signal at 617 Hz downfield from tetramethylsilane. What is this shift in δ (ppm) units?

24) Explain whether C23H30N2O should have an odd mass number for the molecular ion.

Y2 Y3 O O210 0.15 0.1 0 0

450 525 0 0

20 25 800 6

200 2.5

200 300 400 500 600 700 800

Ab

sorb

ance

Wavelength (nm)

CH3

CH3

Br

NO2

NO2

NO2

CH2CH3

CH=CH2

HCΞC

+ • !

+ • !

!•+

++ !!

E

F

Br

Cl

OH

=O

PCC

PyH+, CrO3Cl-, CH2Cl2

Pyridinium chlorochromate

HO

CH3

NaBH4

LiAlH4 LiAlH4, (CH3CH2)2O

NaBH4, CH3CH2OH

Na2Cr2O7, H2SO4, H2O

!H

CrO3Cl!, CH2Cl2

+ N

H3O+

H+, H2O

O

= =O

=

O

H

O

CH3

=

=

O

HHO

=

O

H

LiAlD4 LiAlD4, (CH3CH2)2OD+, D2O

Mg, (CH3CH2)2O

Li

FeBr3 Acid BaseFeCl3 AlBr3 AlCl3

H2SO4, 100°C

SbCl5 BF3

HF 0°C

=O

HCCH3

=O

HCCH2CH3

=O

HCH

=O

CH3CCH3

OH

=O

PCC

PyH+, CrO3Cl-, CH2Cl2

Pyridinium chlorochromate

HO

CH3

NaBH4

LiAlH4 LiAlH4, (CH3CH2)2O

NaBH4, CH3CH2OH

Na2Cr2O7, H2SO4, H2O

!H

CrO3Cl!, CH2Cl2

+ N

H3O+

H+, H2O

O

= =O

=

O

HHO

Structure and Resonance Energy of Benzene: A First Look at Aromaticity

15-2

! of !4 12

25) Which species has a molecular ion at 30.026? A. C2H4O B. CH2O C. CH4N D. Si E. NO2

26) Assigning values in the order that the hydrogens appear in the molecule, which one gives 1H NMR

chemical shifts in ppm for the hydrogens in ! ? A. 5.55, 1.58, 5.55, 1.58 B. 5.55, 1.58, 1.58, 5.55 C. 1.58, 5.55, 5.55, 1.58 D. 1.58, 5.55, 1.58, 5.55 E. 1.58, 5.55, 1.58

27) Draw (R)-2-bromobutane

28) Predict the product and dominant mechanism of this reaction. CH3CH2CH2CH2Br + (CH3)3COK

29) Provide the nucleophilic substitution product(s) for the reaction below CH3Cl + CH3CH2O− →

Bond DH° (kJ/mol) Bond DH° (kJ/mol) Bond DH° (kJ/mol) Br−Br 192 CH2=CH2 272 C−Br 292 C−Cl 351 C−F 460 C−H 423 C−I 234 C−OH 393 Cl−Cl 243 F−F 159 H−Cl 431 H−H 435 H−OH 498 HO−OH 205 I−I 151

30) Using the chart of DH° values, calculate ΔH° (in kJ) for the reaction below. CH2CH2 + HCl → CH3CH2Cl

31) Show the stereochemistry of starting material needed to get the product shown by SN2.

+ NH3 + Br−

C=CH2CH3

CH3CH2

H2C=CH2

C=CCH3

CH3 CH3

CH3

C=CCH3

CH3 CH3

H

C=CCH3

H CH3

HC=C

CH3

H H

CH3

C=CCH3CH2

H CH3

H

!CH3 !CH3 !CH3

C=CCH3CH2

H H

CH3

C=CCH3CH2

H H

CH2CH3C=C

CH3CH2

H CH2CH3

H

C=CCH3CH2CH2

H H

CH3C=C

CH3CH2CH2

H CH3

H

CH3CH=CH2

CH3CH2CH=CH2

(CH3)2CHCH=CH2

CH3CH2CH2CH2CH=CH2

C=CH2CH3CH2

CH3CH2

C=CH2CH3

CH3CH2CH2

C=C(CH3)2CH

H H

CH3C=C

(CH3)2CH

H CH3

H

CH3CH2CH2CH=CH2 (CH3)3CCH=CH2

C=CH2CH3

(CH3)2CH

(CH3)2CHCH2CH=CH2 (CH3)3CCH=CH2

H2OEthanol

Acetone

Acetonitrile

DMF

DMSO

HMPA

Nitromethane

Methanol

Isopropanol

Propanol

Formic acid

Acetic acid

Formamide

N-methylformamide

N,N-dimethylformamide

Water

CH3CH2OH

CH3OH

(CH3)2CHOH

CH3CH2CH2OH

HCOOH

CH3COOH

HCONH2

HCONHCH3

HCON(CH3)2

CH3NO2

((CH3)2N)3PO

Hexamethylphosphoric triamide

(CH3)2SO

HCON(CH3)2

CH3CN

CH3COCH3

Dimethyl sulfoxide

Dimethyl formamide

Diethyl ether (CH3CH2)2O

NH3Ammonia

BMIM hexafluorophosphate

Solvent Fast Slow

N NCH2CH2CH2CH3CH3 +

PF6!

(CH3)3COHtert-butyl alcohol

Cl! + (CH3)3C+

HO! + H!CH2!CH2+

(CH3)3CCl

H2O + H2C=CH2

H2O

Ethanol

Acetone

CCH3CH2

O2CCH3

CH2CH2CH2CH2CH3

H

CCH3CH2

H

CH2CH2CH2CH2CH3

O2CCH3C

CH3CH2I

CH2CH2CH2CH2CH3

H

CCH3CH2

H

CH2CH2CH2CH2CH3

Br

CCH3

H

NH3

CO2!

+

CCH3

Br

H

CO2!

! of !5 12

32) Which methylcyclohexane conformer is more stable? Explain why.

! !

33) Predict the major product of this reaction.

! !

34) Given the following energy diagram for a hypothetical reaction, which statements would be true of the reaction? 


! 
A. Product B will be formed faster, but product A would predominate at equilibrium if both reactions are reversible. 
B. Product A will be formed faster, and product A would predominate at equilibrium if both reactions are reversible. 
C. Product B will be formed faster, and product B would predominate at equilibrium if both reactions are reversible. 
D. Product A will be formed faster, but product B would predominate at equilibrium if both reactions are reversible. 
E. Product B will predominate whether or not the reactions are reversible.

35) Which one is the bond line drawing for this condensed structure? ClCH2CH(CH3)CH2CH3

A) !

CH3

CH3H

H

H

HH H

HH

HH

H H

HH

H

H

H H

H

H

H

HCH3

CH3H

H

H

HH H

HH

HH

H H

HH

H

H

H H

H

H

H

H

CH3CH2CCH2CH2CH3

Br

CH3CCH2CH2CH2CH3

Br

CH3 CH3

CH3CHCHCH2CH2CH3

Br

CH3

CH3CH2CCH2CH2CH3

OH

CH3CCH2CH2CH2CH3

OH

CH3 CH3

CH3CHCHCH2CH2CH3

OH

CH3

CH3CH2CHCHCH2CH3

Br

CH3

CH3CH2CHCHCH2CH3

OH

CH3

CH3CHCHCH2CH2CH3

BH2

CH3

CH3CH2CHCHCH2CH3

BH2

CH3

CH3CH2CHCHCH2CH3

HgC2H3O2

CH3

CH3C≡CCHCH2CH3

Br

CH3CC≡CH

CH3

CH3

C≡CH

C≡CH

!C≡CH

!CH2C≡CH

HC≡CCH2CH2CHCH3

OH

HC≡CCH2CH2CH2CH3CH3C≡CCH2CH2CH3

CH3CH2C≡CCH2CH3

HC≡CCH2CHCH3

CH3

HC≡CCHCH2CH3

CH3

CH3C≡CCHCH3

CH3

HC≡CCCH3

CH3

CH3

HC≡CCH2CH2OH

HC≡CCHCH3

OH

CH3C≡CCH2OH

CH3CH2C≡COH

CH3CH2CH2CH2CHCH2Br

Br

CH3CHCHCH2CH2CH3

Br BrCH3CH2CHCHCH2CH3

BrBrO O

OsOO

OO

O OH

OH

H

OH

HHO

CH3H3C

H

CH3

HHO

HOCH3

OH

OH

OH

OH

(CH3)2S

Zn, HC2H3O2

O3, CH2Cl2

H2, Lindlar catalyst

Na, liquid NH3

Na, NaNH2, NH3

C=CC=C

H CH2CH3

H

CH3 H

H

C=CCH3

H CH3

BrC=C

CH3

I H

H

C=CCH3CH2

Br CH2CH3

Br

C=CCH3

Cl CH3

Cl

C=CCH3CH2

Cl H

Cl

H2O, H+, HgSO4

CH3C=CHCH2CH3

OH

CH3CH=CCH2CH3

OH

CH3C=CHCH2CH2CH3

OH

CH3CH=CCH2CH2CH3

OH

CH3CH2C=CHCH2CH3

OH

CH3CHC≡CCH2CH3

CH3

Ni(C2H3O2)2

Pd(C2H3O2)2

NaOCH3

NaOCH2CH3

! of !6 12

B) ! 


C) ! 


D) ! 


E) !

36) Given that free-radical bromination proceeds with a selectivity ratio of 1700:80:1:0.002 (tertiary:secondary:primary:methyl), predict the major product of the following reaction.

+ Br2

A.! B.! C.!

D.! E. !

37) Provide the product(s) for the free-radical chlorination of the following compound at C5.

! {(S)-1,3-dichlorobutane

38) Calculate the formal charges on each atom in CN−.

39) Predict the product(s) of this reaction.

! + HBr →

40) Explain the aspects of the mechanism that leads to the results shown below.

+ NaOCH3 + CH3OH The rate of reaction depends on two reactants.

hνF

F

F

F

F

Cl

Cl

Cl

Cl

Cl

light

Δ

I

I

I

I

I

hν FF

F

F

Cl

Cl

Cl

Cl

light

Δ

I

I

I

I

Br

Br

Br

Br

hν FF

F

F

F

Cl

Cl

Cl

Cl

Cl

light

Δ

I

I

I

I

I

Br

Br

Br

Br

Br

hν FF

F

F

F

Cl

Cl

Cl

Cl

Cl

light

Δ

I

I

I

I

I

Br

Br

Br

Br

Br

hν FF

F

F

F

Cl

Cl

Cl

Cl

Cl

light

Δ

I

I

I

I

I

Br

Br

Br

Br

Br

hν FF

F

F

F

Cl

Cl

Cl

Cl

Cl

light

Δ

I

I

I

I

I

Br

Br

Br

Br

Br

hν FF

F

F

F

Cl

Cl

Cl

Cl

Cl

light

Δ

I

I

I

I

I

Br

Br

Br

Br

Br

CCH2CH2ClH

Cl

CH2CH3

CC —CH2ClH

Cl

CH2CH3

H Cl

CC —CH2ClH

Cl

CH2CH3

Cl H

CCH2CH2ClCl

Cl

CH2CH3

CCH2CH2ClH

Cl

C — CH3

ClH

CCH2CH2ClH

Cl

C — CH3

HCl

!OH

!

CH3

Br

CH3

I

CH3

!OH + H+ !OH2+

!

CH3

!

CH3

+

!

CH3

+

HCH3

!H

!

CH3 +

+ Br!Br

(CH3)3CH

H

Br

(CH3)3CH

H

HH

HH

(CH3)3C

BMIM hexafluorophosphate

N NCH2CH2CH2CH3CH3 +

PF6!

Solvent

N≡N

F!Cl

CH3!I

N≡C CH3!I!O=O CH2=CH2

O≡C CH3!I

Fast Slow

Br

(CH3)3CH

H

Br

(CH3)3CH

H

HH

HH

(CH3)3C

! of !7 12

+ NaOCH3 + CH3OH The rate of reaction depends on one reactant.

41) Rank these alcohols in the order of increasing acidity, starting with the least acidic first: A B C D

! ,! ,! ,!

42) The following structure would be classified as containing what functional group?

! A. ester B. aldehyde C. alcohol D. carboxylic acid E. ketone F. amide

43) Predict the product(s) of this reaction sequence.

CH3CH2MgBr + 1) 2)

44) Calculate the enthalpy of reaction for the free radical fluorination of methane.

Table of bond-dissociation energies. Entries are kJ/mol −H −F −Cl −Br −I −C H− 435 569 431 364 297 CH3− 439 460 356 293 238 CH3(CH2)n− 423 464 352 293 234 370 (CH3)2CH− 412 464 352 297 234 (CH3)3C− 404 460 356 297 230 F− 569 159 Cl− 431 243 Br− 364 192 I− 297 151

Br

(CH3)3CH

H

Br

(CH3)3CH

H

HH

HH

(CH3)3C

BMIM hexafluorophosphate

N NCH2CH2CH2CH3CH3 +

PF6!

Solvent

N≡N

F!Cl

CH3!I

N≡C CH3!I!O=O CH2=CH2

O≡C CH3!I

Fast Slow

Br

(CH3)3CH

H

Br

(CH3)3CH

H

HH

HH

(CH3)3C

OH

OH

Cl

OH

Cl

OHHO

OH

OH

Cl

OH

Cl

OHHO

OH

OH

Cl

OH

Cl

OHCl

OH

OH

Cl

OH

Cl

OHHO

!CH2OH =OCH3CHCH2MgCl

CH3

CH3CH

=O!

CH3CHCH2CHCH3

CH3!

OH

!CH(CH3)2

OH

!CHCH(CH2CH3)2

OH

!

CH3CH2CHCHCH3

O=CH

!O

=

O

=

O

=

O

=

O

=

O

=

O =

OH

!

OH

!

OH

!

OH

!

OH!

OH

!

OH

!

Cl! + (CH3)3C+

HO! + H!CH2!CH2+

(CH3)3CCl

H2O + H2C=CH2

H2OEthanol

Acetone

Acetonitrile

DMF

DMSO

HMPA

Nitromethane

Methanol

Isopropanol

Propanol

Formic acid

Acetic acid

Formamide

N-methylformamide

N,N-dimethylformamide

Water

CH3CH2OH

CH3OH

(CH3)2CHOH

CH3CH2CH2OH

HCOOH

CH3COOH

HCONH2

HCONHCH3

HCON(CH3)2

CH3NO2

((CH3)2N)3PO

Hexamethylphosphoric triamide

(CH3)2SO

HCON(CH3)2

CH3CN

CH3COCH3

Dimethyl sulfoxide

Dimethyl formamide

Diethyl ether (CH3CH2)2O

OH

=O

PCC

PyH+, CrO3Cl-, CH2Cl2

Pyridinium chlorochromate

HO

CH3

NaBH4

LiAlH4 LiAlH4, (CH3CH2)2O

NaBH4, CH3CH2OH

Na2Cr2O7, H2SO4, H2O

!H

CrO3Cl!, CH2Cl2

+ N

H3O+

H+, H2O

O

= =O

=

O

HHO

! of !8 12

45) Explain the mechanism(s) producing these products: (CH3)2CHBr + NaI (CH3)2CHI + NaBr 100%

46) Terpenes are made from which unit?

A.!

B.!

C.! D.!

E. !

Cl! + (CH3)3C+

HO! + H!CH2!CH2+

(CH3)3CCl

H2O + H2C=CH2

H2OEthanol

Acetone

Acetonitrile

DMF

DMSO

HMPA

Nitromethane

Methanol

Isopropanol

Propanol

Formic acid

Acetic acid

Formamide

N-methylformamide

N,N-dimethylformamide

Water

CH3CH2OH

CH3OH

(CH3)2CHOH

CH3CH2CH2OH

HCOOH

CH3COOH

HCONH2

HCONHCH3

HCON(CH3)2

CH3NO2

((CH3)2N)3PO

Hexamethylphosphoric triamide

(CH3)2SO

HCON(CH3)2

CH3CN

CH3COCH3

! of !9 12

Answers 1) Loss of an electron from a molecule (ionizing)

2) ! 3) bromine 4) Use 81Br to calculate the molecular mass as 110. 5) a broad peak at 3300cm-1 6) 92Hz 7) none or zero 8) The sharp peak at 1700cm-1 occurs with a carbonyl functionality. It could be a ketone, aldehyde, ester, amide, anhydride, or carboxylic acid. (In this case the spectrum is for butanone, a ketone.) 9) 3 10) 1-chloro-2,2-dimethylpropane 11) A. It is best to draw each of the molecules out to determine the numbers of equivalent protons. 12) 34.7kJ/mol 13) 3, the number of chemically different carbon atom environments based on symmetry. 14) You need to know that UV light has a wavelength of 200nm to 400nm to get 5.0x10-19J to 9.9x10-19J 15) 9.2x1014Hz 16) Using λmax of 450nm from the spectrum gives an energy of 270kJ 17) It is a UV-visible spectrum (you can tell by the wavelength axis), with a λmax of 631nm.

18) ! 19) 8 20) The wavelength with maximum absorbance 21) Solvents with no absorption peaks above 200nm. It is better if solvents do not absorb radiation in the region of interest. 22) 1-bromo-3-ethenylbenzene (IUPAC) or m-bromostyrene (common). 23) 2.06ppm 24) The mass number should be even because the molecule has an even number of nitrogen atoms 25) Answer B, calculate the exact mass of the molecule using the periodic table provided 26) C. The H- in CH3 comes first, then work your way across the carbon chain to the right. Remember that a H- attached to a carbon containing to a double bond is further downfield than one that is not. The exact magnitude (i.e. the fact that it is 5.55) of the downfield shift is less important.

27) ! 28) CH3CH2CH=CH2 + (CH3)3COH + KBr The mechanism is E2 29) CH3CH2OCH3 + Cl−

CH3

CH3

Br

NO2

NO2

NO2

CH2CH3

CH=CH2

HCΞC

+ • !

+ • !

!•+

++ !!

E

F

Br

ClSO3H

NO2

CH2CHCH3

CH3

CH2CH2CH2CH3

CHCH3

CH3

CHCH2CH3

CH3

CHCH3

CH3

CH3

CCH3

=O

CCH2CH3

=O

CH2CH2OH

MgBr

MgCl

CHCH3

OH

carvonecaraway spearmint

alanine lactic acid2-methyl-5-(1-methylethenyl)-2-cyclohexenone 2-aminopropanoic acid 2-hydroxypropanoic acid

C

C

CH3

H3C

HBr

H

H

C

C

H3C

CH3

HBr

H

H

D(+)-2-bromobutane L(!)-2-bromobutane

! of !10 12

30) [272+431]-[423+351]= -71kJ

31) ! R starting material gives S product by SN2 inversion of configuration. 32) The second one is more stable, because the first has 1,3 diaxial interactions with the methyl group. The second form does not. 33) ! 34) D 35) D 36) E

37) at C5 ! the product, 1,3,5-trichlorobutane is not optically active. 38) C has a -1 charge=4-2-0.5*(6), and N has a 0 charge=5-2-0.5(6) in :C≡N:

39) + H2O (minor amounts of 1,2 product) 40) The cis isomer, cis-1-bromo-4-(1,1-dimethylethyl)cyclohexane, has the correct anti orientation of Br and H to occur quickly by E2. The trans isomer does not. The conformer with the tert-butyl, Br and H groups in the axial position is energetically not favorable, so that reaction is slow by that mechanism. The observed product probably occurs by E1 starting with dissociation of Br.

41) ( !

43) ! 44) Calculate the energy for the two progagation steps Initiation: F2 → 2F⋅

Propagation 1: F⋅ + CH4 → ⋅CH3 + HF Propagation 2: ⋅CH3 + F2 → CH3F + F⋅

Termination steps: 2F⋅ → F2 ; 2 ⋅CH3 → C2H6 ⋅CH3 + F⋅ → CH3F

CCH3CH2

O2CCH3

CH2CH2CH2CH2CH3

H

CCH3CH2

H

CH2CH2CH2CH2CH3

O2CCH3C

CH3CH2I

CH2CH2CH2CH2CH3

H

CCH3CH2

H

CH2CH2CH2CH2CH3

Br

CCH3

H

NH3

CO2!

+

CCH3

Br

H

CO2!

CH3CH2CCH2CH2CH3

Br

CH3CCH2CH2CH2CH3

Br

CH3 CH3

CH3CHCHCH2CH2CH3

Br

CH3

CH3CH2CCH2CH2CH3

OH

CH3CCH2CH2CH2CH3

OH

CH3 CH3

CH3CHCHCH2CH2CH3

OH

CH3

CH3CH2CHCHCH2CH3

Br

CH3

CH3CH2CHCHCH2CH3

OH

CH3

CH3CHCHCH2CH2CH3

BH2

CH3

CH3CH2CHCHCH2CH3

BH2

CH3

CH3CH2CHCHCH2CH3

HgC2H3O2

CH3

CH3C≡CCHCH2CH3

Br

CH3CC≡CH

CH3

CH3

C≡CH

C≡CH

!C≡CH

!CH2C≡CH

HC≡CCH2CH2CHCH3

OH

HC≡CCH2CH2CH2CH3CH3C≡CCH2CH2CH3

CH3CH2C≡CCH2CH3

HC≡CCH2CHCH3

CH3

HC≡CCHCH2CH3

CH3

CH3C≡CCHCH3

CH3

HC≡CCCH3

CH3

CH3

HC≡CCH2CH2OH

HC≡CCHCH3

OH

CH3C≡CCH2OH

CH3CH2C≡COH

CCH2CH2ClH

Cl

CH2CH3

CC —CH2ClH

Cl

CH2CH3

H Cl

CC —CH2ClH

Cl

CH2CH3

Cl H

CCH2CH2ClCl

Cl

CH2CH3

CCH2CH2ClH

Cl

C — CH3

ClH

CCH2CH2ClH

Cl

C — CH3

HCl

CCH2CH2ClH

Cl

CH2CH2Cl

!OH

!

CH3

Br

CH3

I

CH3

!OH + H+ !OH2+

!

CH3

!

CH3

+

!

CH3

+

HCH3

!H

!

CH3 +

+ Br!OH

OH

Cl

OH

Cl

OHHO

OH

OH

Cl

OH

Cl

OHCl

OH

OH

Cl

OH

Cl

OHHO

OH

OH

Cl

OH

Cl

OHHO

!CH2OH =OCH3CHCH2MgCl

CH3

CH3CH

=

O

!

CH3CHCH2CHCH3

CH3!

OH

!

CH(CH3)2

OH

!CHCH(CH2CH3)2

OH

!

CH3CH2CHCHCH3

O=CH

!

O

=

O

=

O

=

O

=

O

=

O

=

O

=

OH

!

OH

!

OH

!

OH

!

OH

!

OH

!

OH

!

! of !11 12

Fluorination: ΔH°= -431kJ/mol Propagation step 1: C−H (439) − H−F (569) = -130kJ/mol Propagation step 2: F−F (159) − C−F (460) = -301kJ/mol

45) SN2 reactions are favored by polar, aprotic solvent, a good nucleophile that is a weaker base than OH−, a good leaving group, and a secondary substrate. NaBr precipitates from acetone, helping to drive the reaction to form products. 46) E

! of !12 12