Chemistry 1152
Review for Exam III — Answers
1. a.
c. H3C
CH2
CH
O
CH3
d.
O
CH2
C
H
O
C
g.
CH3
H3C CH C
O
h.
O
H3C C CH2 CH3
H
O
k.
2.
C CH2 CH3
b) benzaldehyde
c) diphenyl ketone (benzophenone)
f) 4-methyl-3-hexanone
g) 5-chloro-4-ethyl-6-methylheptanal
i) 2-pentanone
k) propanone (acetone)
5. a) amylopectin—common form of starch. It is heavily branched with α1-6
connections.
b) amylose—less common form of starch. No α1-6 branches as seen in amylopectin.
d) aldose—aldehyde sugar. This term is often used for the aldehyde form of glucose.
f) hemiacetyl—functional group formed from aldehyde and alcohol. Four criteria
need to be met: 1) an R—group 2) a Hydrogen 3) an alcohol and 4) an ether all
bonded to the same carbon.
j) monosaccharide—simplest of all carbohydrates. It cannot be broken down further
by hydrolysis.
k) acetal—functional group formed when a hemiacetyl reacts with an alcohol. Four
criteria need to be met: 1) an R—group 2) a Hydrogen 3) an ether 4) another
ether all bonded to the same carbon.
l) glycoside—a carbohydrate acetal formed by reacting a glycose with an alcohol.
m) hydrolysis—reaction with water. Usually applies to the reverse of condensation
reaction.
n) hyperglycemia—a medical condition where the level of glucose in the blood is
above normal.
7. I would use Tollen’s Reagent. A positive test will indicate aldehyde, a negative test
will indicate the ketone.
9.
Alpha
H2C OH
O
H
HO
OH
H
H2C OH
H
HO
OH
H
O
H
OH
Aldose
OH
H
O
H
H
HO
OH
H
Beta
H2C OH
O
OH
H
OH
H
OH
H
10. Amylose, amylopectin, glycogen and dextrins are all starches. Amylose is the
straight chain type, amylopectin is branched, glycogen is heavily branched and is
the human form of starch. Dextrins are branched oligosaccharides formed from
the partial hydrolysis of amylopectin. Cellulose is the β1-4 glucose polymer that
cannot be digested by humans.
11.
O
H3C C
more
OH
+ H3C CH2 OH
H
H3C C O CH2 CH3
H
Hemiacetal
12.
H3C
OH
CH
KMnO4
CH3
H3C
CH 3CH 2OH
O CH2 CH3
H3C C O CH2 CH3
H
acetal
O
C
CH3
One cannot make propanal from an alcohol as the primary alcohol would further
oxidize to a carboxylic acid. It can however, be made from a reaction of ozone
with an alkene…
H2C OH
13.
O
H
HO
H2C OH
OH
H
+ CH3OH
H
HO
OH
OH
H
O
H
OH
H
H
O
CH3
OH
H
glucoside = glycoside of glucose
14. This is the β1-4 polymer of glucose, which is cellulose.
20. Hot tea will hydrolyze the sucrose into glucose and fructose. The sum of sweetness
from glucose and fructose is greater than the sweetness from sucrose alone.
21. Glucose is sweet, it is a monosaccharide. Starch is not sweet, neither is cellulose as
the polymers are too large. Of course, the instructor is sweet, very sweet.
22.
a.
O
H3C
OH
C
CH2
CH2
CH3
+ H2
Ni
H3C
CH
CH2
CH2
CH3
O
b.
+ H2
H3C C
Pt
H3C CH2 OH
H
O
c.
H2C
CH
OH
H3C CH2
CH2 C CH3
+ Excess H2
CH2 CH CH3
Ni
HO
HO
H2C OH
C
O
H
d. R
C
No Reaction
KMnO4
Ar
R-CH2OH KMnO4
R
O
O
R
C H KMnO4
R
O
e.
C OH
R
C OH
H+
Δ
O
H+
f.
+ CH3OH Δ
H3C CH2 C
KMnO4
OH
O
Starch + H2O
R'
CH
R
C
R'
O
glucose
CH2 CH3
O
OH
CH3CH2OH
H3C CH2 C O CH3
H3C CH2 C O CH3
+
H
H
H
H
O
+ Tollen’s Reagent
C
q.
Positive Test
(Should give an acid)
H
HO
O
r.
H3C
+ CH3OH
C
CH2
H+
H3C
O CH3
C
CH2
H
H
O
H+
s. Product of (r) + CH3CH2CH2CH2OH
O
u) H3C CH2 C CH CH
H3C
H2, Pt
C
CH2
CH2 CH2 CH2 CH3
O CH3
H
OH
H3C CH2 CH CH2 CH2
H2C OH
H2C OH
x.
O
H
HO
OH
C
OH
H
H
OH
OH
H
β formation
HO
H2C OH
OH
H
H
α formation
C
HO
O
H
OH
H
O
H
H
C
OH
H
H
OH
OH
O
z.
H3C CH CH2 CH2 C + Benedict’s Solution
CH3
H
Positive Test
CH3
23. H3C C CH3
a.
O
+ CH3CH2OH
H3C
H+
C
O
CH2 CH3
OH
CH3
b.
Product of (a) + CH3CH2CH2OH
H3C
H+
C
O
O
CH2 CH2 CH3
CH2 CH3
CH3
c.
H+
Δ
Product of (b) + H2O
H3C
C
O
CH2 CH3 + CH CH CH OH
3
2
2
OH
Pt
d.
OH
O + H2
O
e.
H3C CH CH2 CH2 C
CH3
+ Benedict’s Solution
CH3
Negative Result
k. Glucose + Benedict’s Solution
Positive Test
l. Starch + Benedict’s Solution
Negative Test
m. Cellulose + Benedict’s Solution
Negative Test
n.
H3C
CH2
CH2
CH
CH3
CH2
KMnO4
CH2
C
H3C
CH3
O
OH
o.
KMnO4
O
No Reaction
p.
O
+ Benedict’s Solution
Negative Test
O
q.
O CH2 CH3
CH C
H2C
LiAlH4
O
CH2 CH3
CH3
CH C CH2 OH
CH3
C
CH3
CH3
H3C CH CH2 C
H
CH3
H3C CH2 C CH2 OH
CH3
H2, Pt
O
KMnO4
CH3
H3C CH CH2 C
O
CH3
H3C CH CH2 C CH3
O
OH
H
u.
HO
H
CH3
H2C
+
H3C C
H3C C H
OH
r.
t.
Dissociation (?)
+ Tollen’s Reagent
Negative Test
H2C OH
H2C OH
v.
O
H
OH
HO
+
H
H
+ CH3CH2OH
OH
H+
Δ
H2C OH
OH
H
OH
H
+
HO
OH
H
OH
H2C OH
O
H
Galactose
OH
H
H
β1-4 Connection
H
OH
H2C OH
O
HO
O
H
OH
O
H
H
OH
H
OH
H
O CH2 CH3
The reactants of (v)
H2C OH
O
HO
H
H
H
w. Product of (v) + H2O
H
OH
HO
OH
H
z.
H
O
H
Glucose
OH
H
Lactose
24. Even more reactions…
OH
O CH2 CH3
b. H C CH
3
O
Δ
+ H2O
H3C CH
O
CH2 CH3
+ CH3CH2OH
CH2 CH3
CH3
g.
OH
O
Intramolecular
H3C CH CH2 CH2 CH2 C
H
O
OH
H
H
OH
H
OH
25. Heat can be used as a denaturing agent. The elevated heat denatures the protein
enzyme and it stops working.
29.
H2C OH
O
HO
Notice that the –OH group
is in a different orientation
here than it is in glucose.
H
H2C OH
OH
H
H
OH
galactose
H
or
OH
OH
HO
H
OH
H
H
C
O
H
OH
galactose in the
aldose form