Write your name and date on the cover page Do not open ...drlchem.com/CHEM210/Lecture 15 - Exam III...
Transcript of Write your name and date on the cover page Do not open ...drlchem.com/CHEM210/Lecture 15 - Exam III...
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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
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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
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!
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
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!
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
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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
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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
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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
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+ 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