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Date:_____________________________
Exam III
Chem. 210
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1 of 10
CH2CH3
Br!
NO2
BrCH2CHCHCH2Br
Cl!
E
CH=CH2
O
OCH3
HOCH2CHCH=CH2
HCΞC
F
HOCH2CH=CHCH2OCH3
Q) How does M+• form in mass spectrometry?
A) Loss of an electron from a molecule
(ionizing)
Br
!
HO
•
!
!
NO2
!
O Cl
SO3H
OH
(CH3)3CO
Q) Draw the product(s) expected
from this reaction.
H
CH3
+ HNO3
H2SO4
NO2
A)
O
O
(CH3)3CO
!CH3
H
Q) Which species gives a 1:1 ratio for the M+ to M+2 peak in the mass spectrum?
A) bromine
O
CH3
OH
OH
OH
CH
CH3C(CH3)2
3OCH2C(CH3)2
Q) Where would the
peak
occur for bromoethane?
CHM+2
3
A) Use 81Br to calculate the molecular mass as 110.
CH3CHCH3
OH
O
CH3
Q) Describe the -OH absorption band in infrared spectroscopy.
CH3
H
A) a broad peak at 3300cm-1
H
OH
(CH3)3CO
D
CH3
Q) A 1H NMR spectrumOat 300MHz
CH3 records a signal at 307 Hz downfield from tetramethylsilane. What
OH
would be the position in hertz (Hz) on a 90MHz instrument? HO
H
A) 92Hz
CH3
Q) How many degrees of unsaturation are in this compound: C6H14
A) none or zero
Q) The sharp peak at 1700cm-1 in this infrared spectrum corresponds to what type or class of compound?
O
O
H
H
Br
H
A) The sharp peak at 1700cm-1
occurs withHa carbonyl functionality.
It could be a ketone,HCH
aldehyde,
Br
Br
CH3CH
H
H
CH3
ester, amide, anhydride, or carboxylic acid. (In this case the spectrum is for butanone,Oa ketone.)
O
CH3
Br CH3
CH3 CH3
Br
CH3CCH3
CH3
Br
H
H
O
H
Cl 1
H
H
Cl
Cl signals should appear from HCCH2CH3? The
Q) Ignoring spin-spin splitting,
how many Htypes of H NMR
H
H
CH3
O
answer is the same as theCHnumber
of chemically
equivalent
hydrogens.
CH
CH
3
3
3
O
Cl
CH3
Cl
CH3CCH2CH3
O
CH3
Cl
H
H
HCCH2CH2CH3
Br
Br
Br
2 of 10Br
O
OH
OH
OCH3
OCH3
CH3CCH2CH2CH3
O
0.005
-1
0
-10
9
-1.1
0.06
0.1
-9 1
-8
2
-3.3
0.06
0.1
-71
-6
0
-1.6
0.06
0.1
4-5
0
-2.1
0.2
0.1
-4
1 -3
1
0
0.06
0.1
-2 1
View Page 66
ol
0
View Page 66
A) 3
1-chloro-2,2-dimethylpropane
Q) Draw a structure having the formula C H Cl that is consistent with this 300MHz H NMR spectrum.
5
1
11
9H
6H
H
-1
δ, ppm
3H
2H
(CH3)4Si
-1
0
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
δ, ppm
A) 1-chloro-2,2-dimethylpropane
Q) 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
(Answer A. It is best to draw each of the molecules out to determine the numbers of equivalent protons.)
Q) 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.
A) 34.7kJ/mol
Q) How many peaks would be in the proton-decoupled 13C NMR spectrum of
A) 3, the number of chemically different carbon atom environments based on symmetry.
?
Q) 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?
A) You need to know that UV light has a wavelength of 200nm to 400nm to get 5.0x10-19J to 9.9x10-19J
3 of 10
Y3
0.15
450
20
O
0.1
525
25
O2
0
0
800
200
0
0
6
2.5
Q) 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?
Absorbance
200
300
400
500
600
700
800
Wavelength (nm)
A) Using λmax of 450nm from the spectrum gives an energy of 270kJ
Q) What type of spectrum is this?
PCC
OH
CH3
CH2CHCH3
CH3
CH3
CHCH3
CHCH3
OH
+ N !H
CH3
NO2
NO2
NaBH4
=O
NaBH4, C
+ N !H
E
CH=CH2
=O
!
3Cl , CH2Cl2
PyridiniumCrO
chlorochromate
HO
CH2CH3
CHCH2CH3
CH3
PCC
PyH+, CrO3Cl-, CH2Cl2
CH2CH2CH2CH3
CH3 CH3
Pyridinium chloro
PyH+, CrO3Cl-, CH2Cl2
=O
CH3
=O
4
CrO3LiAlH
Cl!, CH
2Cl2
CH3
O
LiAlH4, (C
of this reaction sequence:
1)
=O
Br
=O
HO
=O
CH2CH2OH
+
• product
!
the primary
O
HF 0°C
NaBH4, CH3CH2OH
H2SO4, 100°C NaBH
4
+
H
H
3O
MgCl
O
HO
=O
HCΞC
=
Q) Predict
NO2
=
Br
=
Mg, (CH
3CH
2)72, O
NaAlCl
HAcid
O
A) It is a UV-visible
(you
can
tell by the wavelength axis), with
a λ3MgBr
of 631nm.
CCH3 spectrumCCH
2Cr
23O
2SO4, H2Base
maxFeCl
2CH
3
F FeBr
AlBr
3
3
+
HCH
Li O HCCH
LiAlH
O 3 4,CH
(CH3CCH
3 4 HCCH
2LiAlH
3CH23)2O
2)
3) H , H2CH
+
•
!
O
Cl
Na2Cr2O7, H2SO4, H2O
O
H
H
=
A)
=
CHCH3
=
OH
=
Y2
10
Q) 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?
A) 9.2x1014Hz
H3O+
HO
Q) What is the minimum number of conjugated double bonds
does it O
take for an organic
compound to
H+, H2O
O
absorb in the visible region?
+
•
!
LiAlD4
LiAlD4, (CH3CH2)2O
A)
8
D+, D2O
Q) What is λmax in UV-visible spectroscopy?
!
++ wavelength
A) The
with maximum absorbance
!
4 of 10
<-,$)(+&).$"9$%-#,,$1231%&%2%,*$3,)4,),$101%,+1$*"$)"%$9"//"8$
>?@AB$)"+,)6/(%2#,$&)$%-,$!"#$%&'()*+,-.'&-, &)*,:&).$
=#,9,#,)6,1;$$=-,)"/5$3,)4(/*,-0*,5$()*$3,)4"&6$(6&*C
F,7,#(/$"%-,#$6"++"
&)6/2*&).;
D&).$1231%&%2%,*$6"+="2)*1$"9$%-,1,$1231%()6,1$(#,$)(+,*$30$
Q) Samples for UV and visible spectroscopy are usually dissolved in what type of solvent?
)2+3,#&).$%-,$#&).$="1&%&")1$"#$21&).$%-,$=#,9&:,15$"'5$='5$()*$+'C
A) solvents with no absorption peaks above 200nm. Generally, solvents of choice should not absorb
<-,$6(#3")$6(##0&).$%-,$1231%&%2%,)%$.&7&).$%-,$6"+="2)*$&%1$
radiation in the region of interest no matter what type of spectroscopy you are doing.
3(1,$)(+,$&1$.&7,)$%-,$)2+3,#$EC
Q) Provide an acceptable name for this compound:
A) 1-bromo-3-ethenylbenzene (IUPAC) or m-bromostyrene (common). You are much more likely to be
asked to provide IUPAC names.
Q) A 1H NMR spectrum at 300MHz records a signal at 617 Hz downfield from tetramethylsilane. What
is this shift in δ (ppm) units?
A) 2.06ppm
Q) Explain whether C23H30N2O should have an odd mass number for the molecular ion.
A) The mass number should be even because the molecule has an even number of nitrogen atoms
Q) Which species has a molecular ion at 30.026?
A. C2H4O
B. CH2O
A$1231%&%2%,*$3,)4,),$&1$6(//,*$()$(#,),C
C. CH4N
A)$(#,),5$8-,)$21,*$(1$($1231%&%2,)%5$&1$6(//,*$()$!"#$%&"'()
D. Si
GA#HC
E. NO2
15-2 Structure an
Benzen
A%$#""+$%,+=,#(%2#,
NO)PQ5$#,(.,)%1$%-(
<-,$=(#,)%$(#0/$1231%&%2,)%$&1$)*+,#$5$BIJK'C$$<-,$.#"2=5$
BIJKBJL' &1$6(//,*$)*+,#$-+.*#$ G/+,0#$HC
H2C=CH2
(Answer B, calculate the exact mass of the molecule using the periodic
table)
!CH3
!CH3
!CH3
<-,$606/&6$I',/,6%#")
1=,6&(/$1%(3&/&%0$&)$%-
CH3CH=CH2
Q) Assigning values in the order that the hydrogens appear in the molecule, which one gives 1H NMR
H
CH3CH2CH=CH2
C=C
CH3
H
H ?CH3
chemical shifts in ppm for the hydrogens in CH3
CH3CH2CH2CH=CH2
A. 5.55,
1.58, 5.55, 1.58(CH3)3CCH=CH2
(CH3)2CHCH=CH2
CH3
CH3
CH3
H
H
B. 5.55, 1.58, 1.58, 5.55
CH3CH2
C=C
C=C
C. 1.58, 5.55,C=CH
5.55,
CH3
2 1.58
H CH3CH2
H CH3CH2
CH
3
D. 1.58, 5.55, 1.58, 5.55
E.CH1.58,
5.55, 1.58
(CH3)2CHCH2CH=CH2 (CH3)3CCH=CH2
3CH2CH2CH2CH=CH2
C=C
C=C
H
CH3
H
CH3
H
H
CH3
C=C
C=C
(Answer
(CHin
CH 3 comes first, then work your way
(CH3across
)2CH the carbon
3)2CH
CH3 chain to the right. You
CH3CH2 C. The H(CH3)2CH
H
C=CH
2
C=CH
2
need to remember thatCH
a 3H- attachedHto a carbonCHcontaining
to
bond is further downfield than
2CH3
H a double H
CH3CH2
C=C
C=C
one that is not. The exact magnitude (i.e. the fact that it is 5.55) of the downfield shift is less important.)
CH3CH2CH2
CH3CH2
CH3CH2
H
CH2CH3
C=CH2
CH3
CH3
H
H
H
CH3
CH3
C=C
C=C
C=C
CH3CH2CH2
H CH3CH2CH2
CH3
CH3
CH3
H
5 of 10
PF6!
(CH3CH2)2O
BMIM hexafluorophosphate
spearmint
caraway CH
3COCH3
Ammonia
NH
lactic
acid
alanine
Water
carvone
CH CN
3
-methyl-5-(1-methylethenyl)-2-cyclohexenone
3
H2O
2-hydroxypropanoic acid
CH3CH2OH
2-aminopropanoic acid
Ethanol
2)3)2Draw (R)-2-bromobutane
HCON(CH
H
H
Methanol
l formamide
CH3 H3C
Br
Br
C
Isopropanol
C
(CH3)2SO
Propanol
C
yl sulfoxide
C
H
H
CH3 Formic acid
H3C
((CH3)2N)3PO
Acetic acid
H
H
sphoric
triamide
D(+)-2-bromobutane
ne
ormamide
(CH3)2CHOH
CH3CH2CH2OH
HCOOH
CH3COOH
Formamide
L(!)-2-bromobutane
CH33)
NOPredict
2
CH3OH
HCONH2
theN-methylformamide
product(s) and
HCONHCH
3
mechanism
of this
tert-butyl alcohol
CH3CH
HCON(CH
3)2 2CH2CH2Br
+ (CH3)3COK
The mechanism is E2
(CH3)3COH
reaction.
CH3CH2CH=CH2 + (CH3)3COH + KBr
6) Provide the nucleophilic substitution product(s) for the reaction below
CH3Cl + CH3CH2O− → CH3CH2OCH3 + Cl−
Bond
Br−Br
C−Cl
C−I
F−F
H−OH
CH3CH2
DH° (kJ/mol) Bond
DH° (kJ/mol) Bond
192
CH2=CH2
272
C−Br
351
C−F
460
C−H
CH2CH2234
CH2CH
CH22CH
CH32CH2C−OH
CH2CH3 CH CH CH393
Cl−Cl
CH22CH
CH32CH2CH2CH3
2
2
2CH
159
H−Cl
431
H−H
C
C
H
H
C
C
O2CCH3 I−I
498
HO−OH O2CCH205
3
CH3CH2
I
I
CH3CH2
CH3CH2
H
DH° (kJ/mol)
292
423
243
435
151
H
7) UsingCH
the
chart
of 22DH°
values,
ΔH°
for the reaction below.
CH
2CH
2CH2CH
CH
CH32CH
2CH2CHcalculate
3 CH2CH2CH
2CH
22CH
32CH2CH2CH3
CH(in
CHkJ)
CH2CH2 + HCl → CH3CH2Cl
C
C
! + (CH C) CH+
C
Br
Br-71kJ)
([272+431]-[423+351]=
H
Cl
(CH3)3CCl
3 3
CH3CH2
CH3CH2
H
H
CH3CH2
CH3CH2
O2CCH3
O2CCH3
8) Show the stereochemistry of starting material needed to get the product shown by SN2.
H
H
+
+
NH
HO! +
H!CHNH2!CH2+
3
3
H2O + H2C=CH2
C
C
CO2! C
CO2! C
CO2!
CO2!
CH3
CH3
CH3
H2O
Br
H
Br3
+ NH
+ Br−
H
Ethanol
starting material gives S product by SN2 inversion of
CH3
(R
configuration.)
Acetone
9) Which methylcyclohexane conformer is more stable? Explain why.
HH
HH
HH
HH
HH
HH
HH
CH
HH
33
CH
HH
HH HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
CH
33
CH
HH
(The second one is more stable, because the first has 1,3 diaxial interactions with the methyl group. The
second form does not.)
6 of 10
H3
HC≡CCH2CH2OH
Br
OH
HCH2CH3
CH3
HC≡CCHCH3
C≡CCHCH3 CH3C≡CCH2OH
Br
Br
Br
CH3CH2CCH2CH2CH3 CH3CCH2CH2CH2CH3 CH3CHCHCH2CH2CH3 CH3CH2CHCHCH2CH3
CH3
CH3
CH3
CH3
(CH3)2S
CH3CH2C≡COH
OH2Cl2
O3, CH
OH
OH
OH
OH
O
Br
Br
Ni(C
H
O
)
2
3
2
2
H3
Zn, HC2H3CH
O2 3CH2CCH2CH2CH3 CH3CCH2CH2CH2CH3 CH3CHCHCH2CH2CH3 CH3CH2CHCHCH2CH3
Pd(C2H3O2)2
CH3CH
3
2CHCHCH
2CH3
10)
the product(s)
of this reaction.
CH3
CH3
CH3
CH3
HPredict
2, Lindlar catalyst
NaOCH3
Br Br
Br
HgC2H3O2 HC≡CCH2CH2CH2CH3
BH2
BH2
Na, liquid NH3
NaOCH2CH3 CH C≡CCH CH CH
CH
CHCHCH
CH
CH
CH2CH2CHCH
Br
3
2
2
3
2
3
2
2
3
OH
Na, NaNH2, NH3
3
CH3CH2CHCHCH2CH3
H
H
H3
H
C=C
CH3
C=C
3
CH3CH
C≡CCHCH
3
2CH3Cl
H
I
C=C
Br CH3
C=C
CH3CHCHCH2CH2CH3 CH3CH2CHCHCH2CH3
CH3
CH3
2CH3
11)
GivenCHthe
following energy diagramCH
for
3 a hypothetical reaction, which statements would be true of
C=C
CH3
CH3
Cl Br
HC≡CCH2CH2OH
H
the reaction?
C=C
HC≡CCH2CHCH3
CH3
Br
H3CH2
CH3
CH3CH2C≡CCH2CH3
CH2CH3
C≡CH
H
CH3CH2
C≡CH
C=C
CH3
Br
OH
H
Cl
OH
CH3CC≡CH
CH3CH=CCH2CH3 CH
CH33C=CHCH2CH3
OH
OH
HC≡CCHCH3
CH3C≡CCHCH3 CH3C≡CCH2OH
CH3
!C≡CH
OH
HC≡CCHCH2CH3
CH3
HC≡CCH2CH2CHCH3
H
Cl
OH
CH3CH2C≡COH
HC≡CCCH3
CH3
!CH2C≡CH
CH2CH3 CH3CH=CCH2CH2CH3 CH3C=CHCH2CH2CH3
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.
12) Which one is the bond line drawing for this condensed structure?
ClCH2CH(CH3)CH2CH3
A)
B)
C)
D)
E)
7 of 10
F
Cl
F
F
F 14)
I
Br
Br
+
Br
F
A.
light F
Δ
D.
Δ
Cl
C F
Br
CH2CH2Cl
H
E.
ClC
HF F
Cl
Cl
Cl
Br
Δ
I
I
I
I
Cl
I
I
Cl I
H
Br
Br
C —CH2Cl Cl
I
C — CH3
Br
F
Br
Cl
C
I
Cl
I
I
I Cl
H
CH
CH2Cl
15) Provide the product(s) for
free-radical chlorination
of 2the
following compound
at C5.
H the
Br
CH CH3 Br
CH22CH3Br
Cl Cl
ClH
Cl
F
CC
H
H
CH2CH
CH2Cl
Cl
CH
2
2
Cl
Br
Cl F
CH
CH22CH
CH33
Cl F
H
Cl
CC I
Cl
CC—CH
—CH22Cl
Cl
Br Cl
Cl
Cl
Cl
Cl
F H
Cl H
{(S)-1,3-dichlorobutane CH2CH3
CH2CH3CH CH Cl
2
2
Cl Cl
Br
C—
H CH3
I
Cl
I
I
CC — CH3
I
CH2CH2Cl
I
Cl C
H
CH2CH2Cl
Cl H
Cl
C — CH3
H
F
C
Cl C
Br
F
C
H
C —CH2Cl
CH
CH
Cl
2
2
H
CH2CH2Cl
Cl
Cl
H has a
Cl
at C5
the Cl
product
Cl
I
I
Br
C.
Br
Br
Br
I
Cl
F
Cl
BrF
Cl
CHF2CH3
I
Cl I
Cl
light
I
B.
BrCl
hν
I
Br
2
F
CH2CH3
F
Cl
I
I
Δ
hν
Cl
Cl
light
Cl
I
Br
F
I
Br
hν
Given that Cl
free-radical bromination
proceeds with a selectivity ratio of 1700:80:1:0.002
Br
light
F
F
Cl
Br
I
Cl
Cl Br
I predict the major product
(tertiary:secondary:primary:methyl),
Br of the following reaction.
F
Cl
F
Cl
hν
I
Br
mirror plane and is not optically active even though it has a
C
stereocenter. As a result, the R and S designations Hare not meaningful.
Name is 1,3,5-trichlorobutane}
CH2CH2Cl
Cl
16) Calculate the formal charges on each atom in CN−.
(C has a -1 charge=4-2-0.5*(6), and N has a 0 charge=5-2-0.5(6) in :C≡N:)
17) Predict the product(s) of this reaction.
!
+ HBr →
(CH
C 3
CH33)3CH
!
!
!
CH3 CH3
I
I
CH3 CH3
+ H2O
(minor amounts of 1,2 product)
H
CH3 CH3
H H
The rate+ of
(CH3)3C
H
H
Br
H
2
!OH!OH
2
(CH3)3C
H
to the results shown below.
H
+H +hexafluorophosphate
BMIM
Br H H
N≡N
CH !I
!
CH3!I
N≡C
N≡N
CH3!I
!
F!Cl
CH
3!I
N≡C
8 of 10
H
H
H
H
O=O
O≡C
CH3
N
Br
+ CH2CH2CH2C
N
PF6!
H
!
The rate of reaction depends on one reactant.
(CH3)3C
(CH3)3C
H Solvent
H 3)3CBr (CH3)3C
(CH
3 3)3C
H+ CH+3 CH
(CH
HFast
Slow
CH3
Br + NaOCH3 + CH3OH
H
!+ Br!
+
Br
!H
!H
+
reaction
depends
on twohexafluorophosphate
reactants. H
(CH3BMIM
)3C
H
Br
H
H
Solvent
3
(CH3)3C
Br
Fast + CH OH Slow
H
+ NaOCH
3
3
!
(CH3)3C
+ + H+
+ HH
!OH!OH
!
18) Explain the aspects of the +mechanism
+
CH3 CH
H 3
H leads
that
!
Br Br
CH3 CH3
!OH!OH
!
!
CH3 CH3
N
H
H
+ CH2CH2CH2CH3
N
PF
Br6!
H
O=O
O≡C
CH2=CH2
CH3!I
CH2=CH2
CH3!I
OH
(The cis isomer, cis-1-bromo-4-(1,1-dimethylethyl)cyclohexane, has OH
the correct anti orientation of Br
and H to occur quickly by E2. The trans isomerOH
does not. The ring flip to get 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 probablyOH
occurs by E1
Clstarting with dissociation of Br.)
OH
OHalcohols in the order of increasing acidity, starting with the least acidic first:
19) Rank these
OH OH
A
B
C
D Cl
OH
Cl
Cl
OH Cl
,
OH
OH
Cl
,
OH
Cl
OH
Cl
OH Cl Cl
OH
(
<
HO
( B < C < D < A)OH
OH
OH
OH
, Cl
Cl
HO
OH
< Cl
OH
OH
OH
Cl
<
)
OH
HO
OH
Pyridinium chlorochromate
PCC
OH
OH
OH
20) The following structure would be classified
containing
PyH , as
CrO
Cl , CH Cl what functional group?
Cl
Cl
Cl
=O
+ N !H
HO
OH
HO
OH
CrO Cl , CH Cl
OH
A. ester
!
H O + H C=CH
HCON(CH3)2
O
Methanol OH
!
Formamide
OH
N-methylformamide
!
O
OH
=
!
O
!
O
CH3CH2CH2OH
HCOOH
OH
!
=
−I
−C
Isopropanol
297
Propanol
238
Formic acid
234
370
Acetic acid
234
9 of 10
CH3OH
OH
(CH3)2CHOH
!
=
−Br
364
293
293
O
297
=
!
!
=
Hexamethylphosphoric triamide
O
CH3NO
Nitromethane
OH2
O
H2O
CH3CH2OH
=
Ethanol
HCON(CH3)2
!
!
=
!
=
=
!
!
=
!
OH
O
Table of bond-dissociation energies.
Dimethyl formamide
OH
Entries are kJ/mol
−H O −F
−Cl
DMSO
(CH3)2SO
H−
435
569
431
Dimethyl439
sulfoxide 460
CH3−
356
CH3(CH2)n− HMPA
423 ((CH
4643)2N)3352
PO
(CH3)2CH−
412
464
352
N,N-dimethylformamide
O=CH
=
CH3COCH3
O
CH3CHCH2CHCH3
OH
!CHCH(CH2CH3)2 CH3CH2CHCH
O=CH
OH
CH(CH3)2
Na2Cr2O7, 3H23
SO4, H2O
!CHCH(CH2CH3)2 CH3CH2CHCHCH3
CH(CH3)2 OH
OH
O
O
H 3O +
2
2
2
OH
O
H+, H2O
22) Calculate
the enthalpy of reaction
the free radical
fluorination of methane.
Acetonitrile
CH3CN for
Water
OH
DMF
OH
! !
=
!
!
=
!
O 3CH2)2O 2)
1) (CH
(CH ) CCl
CH3
=
=
!
=
2
!
=
=
O
! + H!CH !CH
H 2+
HO
HO
2 OH
O
Acetone
2
!
!CH2OH
21) Predict the product(s) of this reaction.
Diethyl
CH3CH2MgBr
+ ether
2
O
NaBH4
NaBH4, CH3CH2OH CH3
!CH2OH
=O
OH CH3CH
CH3
CH3 2MgCl
O CH3CHCH
=O
CH
CH
CH
CHCH
CH
CHCH
MgCl
CHCH3
3
3
3
2
2
LiAlH4
LiAlH4, (CH3CH2)2O
OH
OH
Cl! + (CH3)3C+
O
2
!
3
CH3
=
B. aldehyde
C.
Clalcohol
D. carboxylicHO
acid
E. ketone
HO
OH
F. amide
-
3
!
+
CH3COOH OH
HCONH2
HCONHCH3
(CH3)3C−
F−
Cl−
Br−
I−
404
569
431
364
297
460
159
356
297
230
243
192
151
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
Cl! + (CH3)3C+
(CH3)3CCl
Fluorination: ΔH°= -431kJ/mol
Propagation step 1: C−H (439) − H−F (569) = -130kJ/mol
! F−F
+ = -301kJ/mol
Propagation HO
step 2:
(159) −2C−F
+ H!CH
!CH(460)
H2O
2
+ H2C=CH2
23) Explain the mechanism(s) producing these products:
Water
H2O
Acetone
CH3COCH
3
(CH3)2CHBr + NaI
(CH3)2CHI
+ NaBr
Ethanol
CH3CH2OH
Acetonitrile
100% CH3CN
Methanol
CH3OH
(This is an SN2 reactionDMF
favored by polar,
aprotic
solvent,
a
good
nucleophile that is a weaker base than
HCON(CH3)2
Isopropanol
(CH3)2CHOH
OH−, a good leaving group,
substrate. NaBr precipitates
from acetone, helping to drive
DMSO and a secondary
Propanol
(CH3)2SO
CH3CH2CH2OH
the reaction forward.)
Formic acid
HMPA
Acetic acid
Hexamethylphosphoric
triamide
24) Terpenes are made
from which unit?
A.
B.
Nitromethane
N,N-dimethylformamide
HCOOH
((CH3)2N)3PO
CH3NO2
HCON(CH3)2
C.
D.
E.
10 of 10
CH3COOH
Formamide
HCONH2
N-methylformamide
HCONHCH3