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Date:_____________________________
Exam I
Practice Exam
Chem. 212
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1 of 6
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 6Br
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-methylpropanol
B. 1-bromo-2-methylbutane
C. 3-bromo-1-chloro-2,3-dimethylbutane
D. 3-bromo-1-chloro-2,2-dimethylbutane
E. 5-bromopentanol
(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 6
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 6
<-,$)(+&).$"9$%-#,,$1231%&%2%,*$3,)4,),$101%,+1$*"$)"%$9"//"8$
>?@AB$)"+,)6/(%2#,$&)$%-,$!"#$%&'()*+,-.'&-, &)*,:&).$
=#,9,#,)6,1;$$=-,)"/5$3,)4(/*,-0*,5$()*$3,)4"&6$(6&*C
F,7,#(/$"%-,#$6"++"
&)6/2*&).;
D&).$1231%&%2%,*$6"+="2)*1$"9$%-,1,$1231%()6,1$(#,$)(+,*$30$
Q) Samples for UV and visible spectroscopy are usually dissolved in what type of solvent?
)2+3,#&).$%-,$#&).$="1&%&")1$"#$21&).$%-,$=#,9&:,15$"'5$='5$()*$+'C
A) solvents with no absorption peaks above 200nm. Generally, solvents of choice should not absorb
<-,$6(#3")$6(##0&).$%-,$1231%&%2%,)%$.&7&).$%-,$6"+="2)*$&%1$
radiation in the region of interest no matter what type of spectroscopy you are doing.
3(1,$)(+,$&1$.&7,)$%-,$)2+3,#$EC
CH3
!
!
CH3
!Br
Q) Provide an acceptable name for this compound:
A) 1-bromo-3-ethenylbenzene (IUPAC) or m-bromostyrene (common). You are much more likely to be
TMEDA
asked to provide
CH3IUPAC names.
CH3
Cl
Br
!Cl
PBr3
NO2
aH3signal
C CH3 at
CH3
Q) A 1H
CH2CH3
!
CH3
!
O
!
!
Br!
!
NMR spectrum at 300MHz records
617 Hz downfield from tetramethylsilane. What
is this shift in δ (ppm) units?
HOCH2CHCHCH2OH
NO2
CH=CH2
CH3
CH3A) 2.06ppm
CH3
CH2Br
CH2Cl
H3C CH3
BrCH2CHCHCH2Br
Cl!
Br
NO2 an odd mass HCΞC
Q) Explain whether C23H30N2O should
have
number for the molecular ion.
A) The mass number should be even because the molecule has an even number of nitrogen atoms
OCH3
HOCH2CHCH=CH2
+
•
!
HOCH2CH=CHCH2OCH3
A$1231%&%2%,*$3,)4,),$&1$6(//,*$()$(#,),C+
Q) Predict the product(s) of this reaction:
•
+ HCl →
!
A)$(#,),5$8-,)$21,*$(1$($1231%&%2,)%5$&1$6(//,*$()$!"#$%&"'()
A) There are no products because no reaction occurs without a catalyst.
GA#HC
<-,$=(#,)%$(#0/$1231%&%2,)%$&1$)*+,#$5$BIJK'C$$<-,$.#"2=5$
BIJKBJL' &1$6(//,*$)*+,#$-+.*#$
+
•G/+,0#$HC
!
15-2 Structure an
Benzen
A%$#""+$%,+=,#(%2#,
NO)PQ5$#,(.,)%1$%-(
<-,$606/&6$I',/,6%#")
1=,6&(/$1%(3&/&%0$&)$%-
Q) Explain whether this molecule is aromatic:
A) It is cyclic with 6 electrons, but it is not aromatic because the p-orbital overlap is interrupted by the
single bond with sp3 hybridization.
Q) Explain why this compound below, pentalene, is not stable.
: !
A) The compound is antiaromatic, 8 electrons, based on the Huckel 4n+2 rule for cyclic compounds.
Q) The diagram below is the reaction coordinate for electrophilic aromatic substitution of benzene. Use
E as your electrophile. Label the axes of the diagram, then write the species that is present at each of the
5 labeled points.
5 of 6
Cl
CH3
Br
TMEDA
CH3
Y2
Y3
0.315
3
0.1
0.01
6
2.5
BrCH2CHCHCH2Br
4
!"#$%&#'())$'#(*+,%-$,.$#/%+"#'0,*$1#*(2.#$+"#$1%-3.$4%'0#3$(
.+'%-5#'$+"(-$+"#$1%-3.$1'%6#-7
3
H3
H=CH2
O2
0.5
3.8
0.05
1
6
HOCH2CHCHCH
2OH
-10
H3C CH3
2
!
!
CH2Cl
O
H3C0 CH3
1.3
2
3
CH2Br
PBr3
HOCH2CH=CHCH2OCH3
/%+"#'0,*$1#*(2.#$+"#$1%-3.$4%'0#3$('#$
1
$1'%6#-7
!"#$%&#'())$'#(*+,%-$,.$#/%+"#'0,*$1#*(2.#$+"#$1%-3.$4%'0#3$('#$
15-9
5.+'%-5#'$+"(-$+"#$1%-3.$1'%6#-7
Halogenation of Benzene needs a Catalyst
15-9
H
8#-9#-
+ E+. At point 2,
A) At point 1,
NO2
8#-9#-#$,.$-%'0()):$2-'#(*+,&#$+%$"()%5#-.$1#*(2.#$+"#:$('#$-%+$
#)#*+'%
#)#*+'%-#5(+,&#$#-%25"$+%$3,.'2;+$,+.$('%0(+,*,+:7
. At point 3,
CH2CH3
E
. AtCH=CH
point 5,
2
NO2
NO2
!
•
!
<()%5#
+ H+.
HCΞC
•
+
. At point 4,
(.$4#''
<()%5#-.$*(-$1#$(*+,&(+#3$1:$=#>,.$(*,3$*(+():.+.$"%>#&#'?$.2*"$
(.$4#'',*$"(),3#.$@A#BCD$%'$()20,-20$"(),3#.$@E)BCD?$+%$1#*%0#$ 02*"$0
02*"$0%'#$;%>#'42)$#)#*+'%;",)#.7
Q) Provide the IUPAC name for this compound
A) benzenol (if there were 2 -OH groups, the compound would be a benzendiol where the numerical
prefixes would indicate the attachment positions of the -OH groups)
Q) !Which species has a molecular ion at 30.026?
A. C2H4O
B. CH2O
C. CH4N
D. Si
E. NO2
!",.$(*+,&(+#3$1'%0,-#$*%0;)#/$*(-$(++(*6$+"#$1#-9#-#$0%)#*2)#$
())%>,-5$+"#$%+"#'$1'%0,-#$(+%0$+%$3#;('+$>,+"$+"#$5%%3$)#(&,-5$
5'%2;$A#8'FG
H2C=CH2
8%-3$#
%4$1#-9
(Answer B, calculate the exact mass of the molecule using the periodic table)
!CH3
!CH3
%0;)#/$*(-$(++(*6$+"#$1#-9#-#$0%)#*2)#$
-#$(+%0$+%$3#;('+$>,+"$+"#$5%%3$)#(&,-5$
!CH3
8%-3$#-#'5:$*()*2)(+,%-.$."%>$+"(+$+"#$#)#*+'%;",),*$1'%0,-(+,%CH3CH=CH2
%4$1#-9#-#$,.$#/%+"#'0,*H
1
I"#
8'G
Q) Assigning values in the order that the hydrogens appear in the molecule, which one gives H NMR
!",.$(*+,&(+#3$1'%0,-#$*%0;)#/$*(-$(++(*6$+"#$1#-9#-#$0%)#*2
I"#
CH3
H
H
H
CH3CH2CH=CH
2
())%>,-5$+"#$%+"#'$1'%0,-#$(+%0$+%$3#;('+$>,+"$+"#$5%%3$)#(&,C=C
C=C GK
<G8
I"#-:)G<
GGGGGG
H JKKL$6*()$0%)
CH3
chemical shifts in ppm for the hydrogens
in CH3 FG
?CH3
5'%2;$A#8'
GK
CH3CH2CH2CH=CH2
8'G8'
JFM$6*()$0%)
G
Q#(
!"#$A#8'
2
F -#/+$(1.+'(*+.$($;'%+%-$4'%0$+"#$*:*)%"#/(3,#-:)$
A. 5.55,
1.58, 5.55, 1.58(CH3)3CCH=CH
(CH3)2CHCH=CH2
GK
*(+,%-$,-+#'0#3,(+#?$(-3$,-$+"#$;'%*#..$'#5#-#'(+#.$+"#$%',5,-()
I"#-:)G8'
GNK$6*()$0%)
CH3
CH3
H
CH3
H
H
B. 5.55, 1.58, 1.58, 5.55
A#8'C *(+():.+7
CH3CH2
C=C
C=C
C=C GK
<G8'
GNO7P$6*()$0%)
C. 1.58, 5.55,C=CH
5.55,
CH3
2 1.58
H CH3CH2 GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG
H CH3CH2
CH3
A)2%',CH
3
D. 1.58, 5.55, 1.58, 5.55
GK
S")%',Q#(*+,%-$$$$$$$$$$$$$GKR7P$6*()$0%)
.$($;'%+%-$4'%0$+"#$*:*)%"#/(3,#-:)$
E.CH1.58,
5.55,
1.58
(CH
)
CHCH
CH=CH
(CH
)
CCH=CH
3
2
2
2
3
3
2
$,-$+"#$;'%*#..$'#5#-#'(+#.$+"#$%',5,-()
3CH2CH2CH2CH=CH2
CH3
H
H
C=C
H
U%3,-(+
C=C your way across the carbon chain to the right. You
(Answer
work
(CHin
CH 3 comes first, then
(CH3)2CH
3)2CH
CH3
CH3CH2 C. The H(CH
3)2CHA)2%',-(+,%-$%4$1#-9#-#$,.$&#':$#/%+"#'0,*$@#/;)%.,&#D7
H
C=CH
2 !"#$A#8' G -#/+$(1.+'(*+.$($;'%+%-$4'%0$+"#$*:*)%"#/(3,#-:)$
C=CH
2
need to remember thatCH
a 3H- attachedHto a carbonFCHcontaining
to
a
double
bond
is further downfield than
H
2CH3
H
CH3CH2
S")%',-(+,%-$(-3$8'%0,-(+,%-$'#T2,'#$(-$(*+,&(+,-5$*(+():.+7
*(+,%-$,-+#'0#3,(+#?$(-3$,-$+"#$;'%*#..$'#5#-#'(+#.$+"#$%',5,-(
C=C
C=C
one that is not. The exact magnitude (i.e. the fact that it is 5.55) of the downfield shift is less important.)
CH3CH2CH2
CH3CHA#8'
CH3CH2
2
H
CH2CH3
*(+():.+7
C U%3,-(+,%-$,.$#-3%+"#'0,*$(-3$3%#.$-%+$%**2'7
C=CH2
CH3
CH3
CH3
C=C
CH3
CH3
H
CH3CH2CH2
C=C
CH3
H
H 6CH3CH2CH2
6 of
C=C
H
CH3
15-10 Nitration and Sulfonation of Benzene
!"#$;2'
#)#*+'%;