CHM 112
1.
PRACTICE SHEET FOR MID-TERM EXAM ANSWERS
GAGE
Write the IUPAC name for each of the following:
CH 3
H3C
Cl
H2
C
CH
C
CH 2CH 2CH 2CH 2CHClCH 3
CH 2CH 3
Cl
C
H2
C
H2
H2
C
O
H3C
C
OH
5-chloro-1-phenyl hexane
CH 3 H
2
C
C
C
CH 3
O
H2
C
CH
C
H2
CH 3
4, 6-dichloro-6-methyl heptanoic acid
6-ethyl-3, 3-dimethy5-nonanone
H
H2
C
H2
C
H3C
C
H2
CH 3
C
CH
O
CH
H2
C
H3C
C
C
H2
H3C
OH
CH 3
CH 3
C
HO
CH
Br
H3C
2-butyl hexanal
2-bromophenol
2, 3, 4-trimethyl-2-pentene 1-hydroxy-1o-bromophenol
methyl cyclohexane
1-hydroxy-2-bromobenzene
H3C
CH 3
CH 3
H2
C
CH
H3C
O
C
H2
CH 3
H3C
C
C
H2
H3C
H2
C
C
C
H2
C
H2
C
C
H2
4-nonyne
CH 3
H
CH
C
CH
O
CH3
H3C
C
HO
benzoic acid
CH 3
C
H2
H2
C
O
C
O
butyl propanamide
C
H2
3-isopropyl heptanal
C
H2
O
H2
C
C
H2
H2
C
H2
C
H
N
H2
C
H2
C
isopropoxy butane
butyl isopropyl ether
H2
C
CH 3
CH 3
C
H2
O
propyl propanoate
CH 3
2.
Write the full or condensed structural formula for each of the following:
3,5-dimethyl-2-heptene
H2
C
H
C
H3C
m-ethyl toluene
O
CH
C
CH 3
4-methyl pentanoic acid
CH 3
H2
C
CH 3
H2
C
C
HO
CH 3
CH 3
CH
C
H2
CH 2CH 3
CH 3
cyclohexanone
3,3,4,5-tetramethyl-2-nonanol
OH
phenyl butanoate
CH 3
CH 3
CH
H2
C
CH
H3C
CH
C
O
CH 3
O
H2
C
C
H2
H2
C
C
CH 3
O
C
H2
CH 3
CH 3
methyl cyclopentyl ether
3-chloro cyclohexene
3-isopropyl pentanal
H3C
O
CH
C
CH 3
H
Cl
3,4-dimethyl-4-ethyl decane
CH 3
CH
O
C
H2
benzaldehyde
CH 3
H2
C
H2
C
C
CH
H3C
C
H2
H2C
CH 3
H2
C
C
H2
CH 3
O
C
H2
C
CH 3
H
N-butyl-N-methyl-2-iodo hexanamide
2-methyl-3-heptyne
O
H2
C
H3C
H2
C
C
H2
H2
C
C
CH
I
N
CH 3
C
H2
H2
C
H3C
H2
C
CH 3
C
H2
C
C
CH 3
CH
CH 3
CH 3
C
H2
3. a. For each pair of compounds below, decide which will have the higher boiling point and
explain why.
octane
and
2,2,3-trimethyl pentane
octane - it is straight-chained so there is more surface for interaction (dispersion forces) so more
energy will be required to separate molecules.
benzene
and
o-diethyl benzene
o-diethyl benzene - it has a higher molar mass so more energy will be required to convert to gas
state
butanol
and
diethyl ether
butanol – this molecule can hydrogen bond with other butanol molecules so greater energy will
be required to separate them
pentanoic acid and
2-methyl pentanal
pentanoic acid – can hydrogen bond to form dimers, effectively increasing molar mass and
requiring more energy to separate and vaporize
b.
Determine which of the following pairs will be more soluble in water and why.
pentanal
and
pentanol
pentanol – both a H and O that can hydrogen bond with water; pentanal has only O to form
bonds
benzene
and
benzoic acid
benzoic acid – can hydrogen bond with water while benzene cannot
hexanoic acid and
ethyl butanoate
hexanoic acid – 2 O’s and an H that can hydrogen bond with water
N,N-dimethyl butanamide
and
hexanamide
hexamide – this has a N and two available H’s to bond with water while the butanamide only has
N
4.
a.
Why are alkanes so unreactive?
The C-C and C-H bonds are very stable and relatively low energy so there is little reason
for the carbons to react.
b.
What is required in a compound so that it can undergo a substitution reaction?
Explain and give an example.
There should be a group that can easily be removed (not a hydrogen or an alkyl group) like an
alcohol group.
CH3CH2OH + HCl ---> CH3CH2Cl + H2O
c.
What test(s) would you use to determine which compound you had in each of the pairs
below? What would you expect to observe in each case?
pentanal
and
pentanol
Use Fehing’s or Tollen’s reagent which will react with pentanal but not pentanol; will observe a
red ppt or silver mirror
benzoic acid
and
methyl benzoate
pH – acid will be low value and ester will be neutral
2-pentene
and
pentane
Add purple KMnO4 – alkene will react and produce brown MnO2
cyclopentanone
and
pentanal
Use Fehing’s or Tollen’s reagent which will react with pentanal but not pentanone; will observe
a red ppt or silver mirror
2-methyl-2-pentanol and
1-hexanol
Use HCl – pentanol will react very fast and produce cloudy solution because it is a 3o alcohol
d.
Benzene has three double bonds but it does not behave like an alkene when you
try to do addition reactions. Explain why.
The electrons in the three double bonds are “delocalized” which means that they circulate
between the 6 carbon atoms. Therefore, they are not easily available to be used to add
something to the ring
e.
Why is a double bond higher energy than a single bond?
There are two pairs of negative electrons that are repelling so more energy is required to
maintain the double bond
f.
H
Determine the bond angles for the labeled
molecule here.
O
C
C
A = B = D = 120oC
C = E = 109.5oC
B
C
A
O
H
C
H
HO
H
C
C
H
E
C
C
D
C
H
H
H
H
H
H
5.
Complete the following reactions supplying the missing reactant or product:
Cu2+
--------------->
O
C
O
C
H
H2
C
O
H3C
C
HO
H2
C
C
H2
H2
C
NaOH
----------------------->
CH 3
ONa+
H3C
C
O
C
H2
CH 3
C
HBr
--------------------->
CH 3
H3C
C
Br
CH
CH
H3C
C
or
C
CH
H3C
H2
C
C
C
H2
CH 3
H3C
CH 3
OH
2 H2O
----------------->
H2
C
CH 3
H3C
CH
CH 3
C
H2
CH 3
Br
H3C
CH3
CH 3
C
CH
H3C
H2
C
HO
O
CH 3
H2
C
+
H2
C
H3C
CH 3
H2
C
CH
CH
C
H2
H2
C
C
H2
CH 3
OH
H
H
H2
C
H3C
H2
C
C
H2
LiBH4
--------------------->
C
CH
O
CH
H2
C
H3C
C
H2
H3C
O
OH
CH
H2
C
C
H2
CH 3
CH 2CH 3
C
H2
K2Cr2O7/H
---------------->
CH 3
H2
C
CH
H3C
C
C
H2
CH 3
O
H2
C
H
N
C
CH
H3C
OH
H2
C
CH
C
H2
+
H2
C
CH
CH
H3C
H2
C
CH 3
CH 2CH 3
H3C
H2
C
C
H2
CH3
C
H2
H2O/H+
------------------------>
H2
C
H3C
OH
C
O
+
H2
C
H2N
C
H2
CH3
C
H2
O
H+
+ CH3CH2CH2CH2OH ---------->
C
H2
C
H3C
O
O
C
H2
C
C
H2
HO
H3C
CH
C
CH
O
H3C
CH 3
H
H2
C
C
H2
CH 3
C
H2
CH
H3C
C
H2
H2
C
H2
C
C
H2
C
H
H3C
C
H2
C
C
H2
C
H2
CH 3
C
H2
CH 2CH 2CH 2CH 2CHClCH 3
------------------->
Ag+/NH3
-------------------->
H2
C
C
H2
CH
CH3
C
H2
H
N
CH
CH 2
OH
H2
C
CH
HCl
H2
C
C
H2
CH 3
H
HO
C
H2
H2
C
H2
C
H2/Pt
------------------>
CH3
C
H2
No reaction
KMnO4
----------------->
No reaction
O
H2
C
+NaO
H2
C
HO
C
CH 3
C
H2
C
NaOH
------------------------>
O
O
CH3
CH 3
C
H2
and H2O