Review for Exam III answers




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Chemistry 1152

Review for Exam III — Answers




1. a. c.




d.


g. h.


k.
2. 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

Aldose

Beta
.

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.


KMnO4

12.


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…




13.



glucoside = glycoside of glucose
+ CH3OH

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.



Ni

+ H2

b
Pt


.

+ H2


c.


Ni

+ Excess H2

No Reaction



d
KMnO4



KMnO4
. R-CH2OH

KMnO4



KMnO4





H+

Δ


glucose

e. Starch + H2O



H+

Δ

CH3CH2OH

f. + CH3OH

H+





q. + Tollen’s Reagent Positive Test

(Should give an acid)





H+

r. + CH3OH

H+


s. Product of (r) + CH3CH2CH2CH2OH




H2, Pt


u)


x
α formation



β formation
.


z.


+ Benedict’s Solution Positive Test




2
H+


3.

a. + CH3CH2OH



H+

b. Product of (a) + CH3CH2CH2OH


H+

Δ

c. Product of (b) + H2O + CH3CH2CH2OH



Pt

d. + H2




e. + Benedict’s Solution Negative Result


k. Glucose + Benedict’s Solution Positive Test

l. Starch + Benedict’s Solution Negative Test

m. Cellulose + Benedict’s Solution Negative Test

KMnO4

n.


o
KMnO4


.

No Reaction

p.


+ Benedict’s Solution Negative Test
q
Dissociation (?)
.


LiAlH4



r
H2, Pt


.






t
KMnO4


.

u.


+ Tollen’s Reagent Negative Test





v.


H+

+ CH3CH2OH


H+

Δ

w. Product of (v) + H2O The reactants of (v)





β1-4 Connection

z. +

Galactose



Glucose

Lactose

24. Even more reactions…





Δ

b. + H2O






g
Intramolecular


.

25. Heat can be used as a denaturing agent. The elevated heat denatures the protein enzyme and it stops working.

29.


Notice that the –OH group is in a different orientation here than it is in glucose.




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