Download 1 Organic Chemistry 101 Problems #13 Answers 1. 2. 3. Each of the

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Asymmetric induction wikipedia , lookup

Physical organic chemistry wikipedia , lookup

Alcohol wikipedia , lookup

Transcript
Organic Chemistry 101
1.
Problems #13
Name each of the following compounds according to the IUPAC rules, including sterereochemistry when
shown.
HO
CH3
trans-2-methoxycyclohexanol
O
d) HOCH CH CH OCH
b)
a)
c)
2
2
2
3
CH3
OCH3
OH
2,2-dimethyltetrahydrofuran
2.
Answers
H3C
O
(1S,2R,3R)-2,3-epoxy-1-methylcyclohexanol
Draw structures for the following compounds. Show stereochemistry clearly.
a) 1-chloro-1-ethoxyethene
c) allyl vinyl ether
b) (E)-3-propoxy-2-hexene
Cl
C
O
CH2
CH3CH2-O
3.
3-methoxy-1-propanol
OCH2CH2CH3
For each set of experimental facts presented below, give reasonable explanations for the observed trend,
using the relationships between structure and physical properties.
CH 3
Each of the alcohols can
hydrogen bond with the water
OH
to help it dissolve. The more
solubility
infinitely soluble
8.3g/100 mL H O
26.0g/100 mL H O
in H O
in water
Explain
polar of the three is 1-butanol,
both
boiling
118°C
99.5°C
82.5°C
separately which would seem to indicate
point
it should be the one that is
most soluble in water. Therefore there must be another explanation. The
intermolecular interactions between the 2° and 3° alcohols are less due to smaller
surface areas and poorer intermolecular hydrogen bonding. This makes it easier for the
water molecules to separate the alcohols. The other reason is that the 2° and 3° alcohols
are more compact and disrupt the water molecules less.
Boiling Point: This is just what one would expect based on intermolecular interactions.
The 1° alcohol would have more effective H-bonding, as well as more effective London
Dispersion forces (more surface area). We learned earlier that branching reduces
boiling point due to a reduction in the dispersion forces. But also here, the hydrogen
bonding is made less effective with an increase in the steric size of the alkyl group near
the oxygen.
CH3CH2CH2CH 2OH
a)
CH3 CH 2 CH CH3
OH
2
H3C
C
CH 3
}
2
2
b)
CH3CH2CH2CH2OH CH3CH2CH2CH2CH2CH2OH
CH3 CH2 CH CH CH2 CH3
OH OH
solubility
in water
boiling
point
7.9 g/100mL H2O
118°C
0.59 g/100mL H2O
157°C
infinitely soluble
in H2O
207°C
}
Explain
both
separately
Solubility: the number of
hydroxyl groups will
increase the solubility in
water due to increased Hbonding interactions.
Butanol has fewer non-polar
methylene groups than hexanol making it more soluble.
Boiling point: both butanol and hexanol are primary alcohols but hexanol has a higher
molecular weight, therefore increasing intermolecular London forces and the boiling
point. The hexanediol, with the two OH groups, has extra intermolecular interactions
including additional H-bonding, that will raise the boiling point.
1
4.
Write balanced equations for the reactions of 1-propanol with each of the following reagents under the
conditions shown.
a)
OH
Br
HBr
+
H2O
b)
OH + 1/3 PBr3
Br +
1/3 H3PO3
c)
OH + SOCl2
Cl
SO2
d)
OH + NaH
O Na
OH +
f) 2
+
+
+
H2SO4
OH
e)
5.
+
+
2
2 Na
+
H2
+
H2
HCl
H2O
+
O Na
Predict the major organic products expected in the following reactions. Pay particular attention to
stereochemistry in your answers. When mixtures are expected, indicate the relative amounts of each.
OH
Br
HBr
a)
HCl/ZnCl2
b)
Cl
OH
CH3
Br
HBr
c)
+
CH3
OH
PBr3
"
d)
Br
1. Ts-Cl/py
e)
OH
CH 2
2. KO(t-Bu)/HO(t-Bu)
1. Ts-Cl/py
f)
H
OH
2. NaI/acetone
1. BH3:THF
g)
I
H
2. H2O 2/NaOH
OH
h)
Br
CH3
"
1. Hg(OAc) 2/THF/H2O
H
CH3
+ enantiomer
H
OH
2. NaBH4/NaOH
2
6.
Draw the major organic product (or products) expected from each of the following reactions. Show
stereochemistry where important.
CH3
1. Hg(OAc)2/CH3OH
a)
CH3
O CH3
2. NaBH4/NaOH/H2O
CH3
+
b)
H / HOCH3
"
c)
O CH3
CH3
C6H5CO3H
"
1. C6H5CO3H
d)
"
2. H+/H2O
O H
H
1. Cl2/H2O
e)
"
H
C
g) H3C C
H2
CH
H 3
CH 2
C
H
C
C C
H2
C
HO
HOCH3
O
3. H 2O
HBr
C C
H2
O-CH 3
C
H2
SCH2CH3
C
CH2CH2OH
+
HOCH 2CH3
CH 3
H 3C
C
Br
CH 3
CH3
7.
C
H2
OH
H3C
+ enant.
CH3
H
H3C C C
H2
2.
O-CH 2CH3
O CH3
C
OH
1. NaNH 2
CH
C
H
H3C C C
H2
2. H2O
O
H
H
1. NaSCH2CH3
CH3
j) H3C
H3C
NaOCH 3
CH2
+ enantiomer
O
2. NaOCH3
O
h) H3C C
H2
i) H3C
1. C6H5CO3H
C
H
CH3
2. NaOH
H3C
f)
O
CH3
HO
+ enantiomer
Suggest methods for carrying out the following conversions, using any reagents you
wish. More than one step may be required in each case.
Cl
a)
1. NaOEt/HOEt
2. PhCO3 H
OH
3. H+/H2O
Cl
b)
1. NaOEt/HOEt
OH
2. OsO4
OH
3. Na2SO3/H2O
OH
1. HBr/peroxides
c)
CH2
2. NaC≡CH/NH3
3. H2O
CH2
C
CH
3