Download Chapter 4 Study Guide

1
Chapter 4
Definitions:
Alkanes
Alkenes
Alkynes
Cycloalkanes
Cracking(thermal and catalytic)
Polymerization
Constitutional Isomers
Haloalkanes
Alcohols
Cyclics
Bicyclo
Homologs/homologous series
Solubility
Acetylenic hydrogen/alkynide ion/acetylide ion
Conformations/Conformational Analysis
Newman Projections
Eclipsed/Staggered/Gauche/Anti
Chirality
Stereochemistry
Ring Strain
Angle/Torsional Strain
Chair/Boat
Catenanes
Pheromones
Paraffins
Alkylation
Retrosynthesis
1. Oil Refining/Fractional Distillation
Petroleum
source of alkanes
types of alkanes
Petroleum refining
fractional distillation
catalytic/thermal crackinGg
polymerization
2
2. Alkane structure
molecular formula/structural formula/condensed formula
line formula(zig-zag)
geometry
alkyl substituents
isopropyl
isobutyl
sec-butyl
tert-butyl
neo-pentyl
3. Nomenclature
A. Alkanes--chart p. 136(table 4.4)
B. IUPAC rules/common names
1. Locate longest continuous chain. Name it according to p. 136.
a. if 2 chains have equal length. The chain with most substituents
becomes the parent chain.
2. Number the chain
a. if only one substituent, number to get to that substituent first
b. if two substituents, number from side that first gets to a
substituent
c. if both substituents are equal distant from both sides and
both are alkyl, use alphabet to determine the numbering
d. if both substituents are equal go to next carbon with a
substituent. Number from side that gives smaller total.
I.E. 2,4 is better than 2,5
e. halides have same priority as alkyls. Use alphabet to determine
priority.
3
f. alcohols have highest priority. If an alcohol is present number
the chain to get to the alcohol first regarldless of alkyl and halide
groups. Alcohols also take priority over alkenes and alkyenes.
g. ethers have same priority as alkyls. Use alphabet to determine
priority
h. alkenes and alkynes are used to determine priority if there are
no alcohols present in a molecule
3. Give each substituent a number and a name(3-bromo, 2-methyl…)
a. if two groups are on the same carbon it must be numbered twice
i.e 2,2-dimethyl
4. Common substituents can be grouped together(2-methyl, 3-methyl can
be called 2,3-dimethyl).
5. Numbers are separated by commas. Numbers and letters are
separated by dashes.
6. Prefixes
a. alkyl substituents are named alphabetically in the front(di, tri,
sec,…do not count for alphabetical, dimethyl comes after ethyl)
b. Halides are named alphabetically in the front
1.
2.
3.
4.
I—iodo
Br—bromo
Cl—chloro
F—Fluoro
c. ethers are named alphabetically in the front
1. methyl ether—methoxy
2. ethyl ether—ethoxy
3. propyl ether—propoxy…and so on
4
d. alkenyl side chains are named alphabetically in front
1. vinyl
2. allyl
e. stereochemical designations come in front as prefixes
1. alkenes—cis/trans/E/Z
2. alkanes—R/S
7. Suffixes
a. alcohols are named by dropping e and adding ol
b. alkenes are named by changing ending to ene. the location
of the first carbon is placed in front of alkane name
1. butane to butene
propane to propene
if double bond is at 1 st carbon—1-octene
if double bond is at 2 nd carbon—2-octene
if double bond is at 3 rd carbon—3-octene
c. if two double bonds—diene, three double bonds—triene…
1. 8 carbon unit with double bonds at 1 st and 5th would be
1,5-octadiene
2. 9 carbon unit with double bonds at 1 st, 4th and 7th
1,4,7-octatriene
d. alkynes use the same rules as alkenes but change ending to yne.
8. Cyclic Alkanes/Alkenes
a. use all previous rules add cyclo to root name(i.e octane
becomes cyclooctane; 2-hexene becomes 2-cyclohexene)
b. if a double bond is present it will be numbered 1 and 2.
Numbering will precede to get to the substituent with lowest
number.
5
1. if there is more than one substituent present use rules
for alkanes(I.E. 2,2,3 is better than 2,2,5. if two equal
substituents can be reached use alphabetical to
determine which direction to precede.
2. if an alcohol is present it will be the 1st carbon and then
precede to the next closest substituent in either
clockwise or anticlockwise manner.
9. Other alkyl/alkene rules
a. when there is a cyclic chain attached to an alkyl group, the
larger group becomes the base name.
1. cyclohexane ring with a pentyl group will be called
1-pentylcyclohexane
2. cyclopentyl ring with a hexane group will be called
1-cyclopentylhexane
b. if an alkyl substituent is complex, use nomenclature rules to
name it as its own molecule then change ending to yl.
1. 2,3-dimethy-4-propyloctane becomes 2,3-dimethyl4-propyloctyl
10. Bicyclics
a. the total number of carbons in the ring becomes the parent
name with the word bicyclo in front.
b. the carbons where the rings are fused are called bridgehead
carbons.
c. the carbons between the bridgeheads are called bridges
d. between bicyclo and the parent name is inserted in brackets
the number of carbons of each bridge in order of decreasing
length(bicyclo[3.2.1]octane). Periods are used to separate
the numbers.
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More information on bicyclics...
On bicyclics, the longest chain is always numbered 1st then the next longest...
In naming bicyclics, the name is bicyclo[x.y.z]alkane, for example. Number to
get to the substituent quickest in the longest chain. Substituents in shorter chains
DO NOT take precedence over any substituent in a longer chain.
11
10
6
4
5
3
7
1
2
9
8
11
10
Br
10
11
6
4
5
3
2
8
Br
2
4
9
11
1
3
7
1
10
Br
5
9
6
1
3
2
4
7
5
9
6
8
7
OH
8
11. Common names
a. alkyl halides are often named by a common name(pentyl
chloride instead of 1-chloropentane).
b. alcohols are often named by common name(ethyl alcohol
instead of 1-ethanol).
c. other common names include acetic acid, formic acid,
ethyl alcohol, ethylene glycol, glycerol, isopropyl alcohol,
acetylene, dichloromethane, chloroform
12. Hydrogens
a.
b.
c.
d.
Carbon with 3 H’s—primary hydrogens/carbon
Carbon with 2 H’s—secondary hydrogens/carbon
Carbon with 1 H—tertiary hydrogen/carbon
Carbon with 0 H—quarternary carbon
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RULES FOR NOMENCLATURE
Select the longest chain and name it methane, ethane,….
If cyclic add prefix cyclo.
If contains double bond change ending from -ane to –ene.
If there is an alcohol, number the chain to get to the alcohol first.
If there is no alcohol but there is a triple bond, number the chain to get to the triple bond first.
If there is no alcohol or a triple bond but there is a double bond, number the chain to get to the
double bond first.
If no alcohol or double or triple bonds, number the chain starting with the end with a
functional group.
If there is not a functional group on the first carbon go to the second…
If there are functional groups on both sides, priority is determined by alphabet. I. E. A
bromine would get preference over a chlorine.
If there are substituents hanging off the base chain name them.
I—iodo, Br—bromo, Cl—chloro, F—Fluoro
If there are more than one substituent you can group them together with prefixes.
2—di, 3—tri, 4—tetra, 5—penta, 6—hexa…
Put all substituents in alphabetical order with a number to indicate their position on the base
chain.
Separate numbers with commas. Separate numbers from letters with dashes.
If there are two substituents on the same carbon you must use the number twice.
The position of double and triple bonds are indicated by a number in front of the base name.
If there are more than one double or triple bond you use the prefixes as above, i.e.
di, tri, tetra…For example, hexane with two double bonds is 1,3-hexadiene.
Cyclic compound are numbered with largest group being number 1. The numbering then
proceeds in whichever direction gets to the next group quickest.
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EXAMPLES
First, the molecule is 10 carbons in length so it is a decane. Since there are double bonds
it becomes decadiene. Since there are no alcohols, alkenes become highest priority. The
two methyls are combined into one name. Therefore, the name of this molecule is
6,8-dimethyl-1,3-decadiene
First, the molecule contains 6 carbons in a circle with a double bond. Thus, the base
name is cyclohexene. A double bond in a carbon will always be the first carbon. We
then head towards the bromine since bromine is alphabetically before methyl.
Br
3-bromo-6-methyl-1-cyclohexene
First the base chain is heptane. We could number it from either side and get to a group
at carbon #2. Since Br comes first alphabetically before Cl we number from the left to
right. Therefore,
Br
Cl
2-bromo-6-chloro-4,4-dimethylheptane
First the base chain is 9 carbons long with a double bond so it is nonene. If we number
from the right we start with chlorine. From the left we start with double bond. Since
double bonds take precedence over all non alcohol compounds we number from the left.
Cl
9-chloro-4,4,5,6,6,7-hexamethyl-1-nonene
9
OH
I
Alcohols always are numbered to be as low as possible. We would name this
compound 4-iodo-2-buten-1-ol.
Br
OH
F
OH
We would name this compound 2-bromo-3-fluoro-2,4-octadiene-1,5-diol
OH
Here the longest chain is 11 carbons long. See if you can find it. In either
direction alcohol comes on the 6 carbon. If you number from the left you get to
the double bond at the third carbon. From the right there is a methyl on the
second carbon. Therefore, this is 2-methyl-4,6,8-undecatrien-6-ol.
At first glance you may try to number this from the left. There is an ethyl on the
third carbon from the left and a methyl on the third carbon from the right. By
rule ethyl comes alphabetically before methyl so you would assume that you
would number from the left. But actually you number from the right because
double bonds have the highest priority.
7-ethyl-3-methyl-4-nonene is the correct name.
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I. For the following structures give the name.
Br
Br
Cl
II. For the following names give the structure.
1. Octene
2. Cycloheptane
3. 4-methylnonane
4. 1,3-cyclooctadiene
5. 2-bromo-3-chlorobutane
6. 1-ethyl-2-methyl-3-pentylcyclopropane
7. 1,3,5-cyclohexatriene
8. 3-ethyl-4-fluoro-5,6,7-trimethyldecane
9. 3-bromo-1-cyclopentene
10. 7-methyl-5-propyl-3-octyne
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I. For the following structures give the name.
4-ethyl-6-methylnonane
Br
3-bromo-heptene
6,7,8-trimethyl-3-decene
4-propylcyclohexene
3-bromo-5-ethyl-1,5-octadiene
Br
1,2,3-trimethylcyclopropane
4-methyl-1-pentyne
Cl
1-chlorocyclopentene
1,3,5,7-cyclooctatetraene
7-methyl-3,5-nonadiene
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II. For the following names give the structure.
1. Octene
2. Cycloheptane
3. 4-methylnonane
4. 1,3-cyclooctadiene
5. 2-bromo-3-chlorobutane
Cl
Br
13
6. 1-ethyl-2-methyl-3-pentylcyclopropane
Technically, the name I gave is wrong. This
should be named based on pentyl side chain. It
should be called
1-(2-ethyl-3-methylcyclopropanyl)pentane
7. 1,3,5-cyclohexatriene
8. 3-ethyl-4-fluoro-5,6,7-trimethyldecane
F
9. 3-bromo-1-cyclopentene
Br
10. 7-methyl-5-propyl-3-octyne
14
Practice Problems—Chapter 4—Nomenclature
OH
A
F
C
B
Br
Cl
Br
E
D
OH
OH
OH
F
Br
F
F
G
Cl
Br
OH
H
Br
I
Br
J
F
15
Draw the structures of the following:
1. 5-methyl-4-propyl-2,4,6,8,10-pentadecapenten-1-ol
2. Bicyclo[4.2.1]nonane
3. 3-chloro-4,5,6-trimethyl-1,4-cycloheptadien-1-ol
4. 5-sec-butyl-2-chloro-3-ethyl-8,9,9,11,12,13-hexamethyl-1,3,5-octadecatriene
5. 1-cyclopentyl-2-bromo-4-penten-2-ol
6. 1-bromo-2-chloro-3-ethyl-4-isobutyl-5- isopropyl-6- methyl-7-pentyl-1,3,5cycloheptatriene
7. 2-chloro-4,5-diethyl-7-iodo-3-vinyl-1,3-octadiene
8. 6-allyl-2-tert-butyl-3-chloro-5-ethyl-7,8,9-trimethyl-8-decen-6-ol
(after drawing this molecule rename it from the picture. You will come up with a
different name because the name above has a mistake. What is the mistake?)
9. 1-bromo-3-chloro-4-(2-cyclopropyl-3-ethyl -4,5-diiodo-4-octenyl)-2-nonene
(after drawing this molecule rename it from the picture. You will come up with a
different name because the name above has a mistake. What is the mistake?)
10. 2,3-dichloro-4,6-diethylbicyclo[4.3.2]undecane
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Answer Key to Practice Problems
A. 4-allyl-7-bromo-6-chloro-2,8-cyclononadien-1-ol
B. 4-ethyl-3,7,8-trimethyl-5-decene
C. 4-bromo-8-fluorobicyclo[10.2.2]hexadecane
D. 5-ethyl-3,4,6,7-tetramethyl-2,4,6-nonatrien-1,9-diol
E. 5-ethyl-6-(1-fluorobutanyl)-4-isopropyl-3,8-tridecadien-2-ol
F. 8-bromo-5-sec-butyl-7-chloro-4-fluoro-6-methyl-2-nonene
G. 5-(2-bromo-6,6-dimethyl-2,4-cyclohexadien-3-olyl)-4,4,6,8,9,9,10,10octamethyl-1,5,7-undecatriene
H. 4-bromo-1-(3-ethyl-2,4-dimethylcyclobutanyl)-2-pentene
I. bicyclo[5.2.2]undecane
J. 2-bromo-8-butyl-11-tert-butyl-17-fluoro-6-isopropyl-3,4,16-trimethyl4,8,14-octadecatriene
1.
OH
2.
17
3.
OH
Cl
4.
Cl
5.
HO
Br
18
6.
Br
Cl
7.
Cl
I
8.
HO
Cl
The correct name is 5-allyl-8-chloro-6-ethyl-2,3,4,9,10,10-hexamethyl2-undecen-5-ol
19
9.
I
I
Br
Cl
The correct name is 9-(3-bromo-1-chloro-1-propenyl)-7-cyclopropyl-6-ethyl4,5-diiodo-4-tetradecene
10.
Cl
Cl
20
Br
Br
Br
Br
Cl
Cl
Cl
Br
OH
Br
OH
Br
21
2
4
1
6
3
5
3-bromohexane
Br
Br
2
4
1
5
2
7
4
1
2-bromo-5-methylheptane
6
3
6
3
8
5
7
Br
Cl
Br
Cl
Cl
7
5
3
8
3-bromo-7-chlorononane
9
6
2,7-dibromo-3,5-dichlorooctane
1
4
2
Br
OH
2
4
1
6
3
8
5
7-bromooctan-4-ol
7
Br
2
4
1
6
3
8
5
7
Br
OH
2
1
4
3
6
5
9-bromo-7-methylnon-2-en-5-ol
9
8
7
9-bromo-7-methyl-2-nonen-5-ol
10
9
Br
9-bromodec-4-ene
9-bromo-4-decene
22
4. Physical Properties
mp/bp
C 1-4—gases; 5-17—liquids; 18+--solids
as molecular weight increase—bp/mp increases
as branching increases—bp decreases
symmetry increases mp
Density
very low; halogens increase density considerably
Solubility(alkanes)
very low in water
high solubility in low polar solvents
5. Conformational analysis
a. Newman projections
1. eclipsed/staggered/anti/gauche
anti staggered most favorable
eclipsed least favorable
gauche in between eclipsed and staggered
gauche is the non favored staggered positions
anti is the most favored staggered positions
eclipsed have favored and non favored positions also
23
Draw the Newman conformation for the following molecule exactly as drawn sighting down
the indicated carbons. Then draw the other 5 conformations in order by rotating the front
carbon clockwise.
H
sight down Carbons
2-3 with 2 in
Cl
front
tert-butyl is largest group, then phenyl,
then pentyl chain, then Cl, then H.
24
Draw the Newman conformation for the following molecule exactly as drawn sighting down
the indicated carbons. Then draw the other 5 conformations in order by rotating the front
carbon clockwise.
H
sight down Carbons
2-3 with 2 in
Cl
front
C H3
H
Ph
Cl
C 4H 9
C 5H 12
C 4 H9
H
Ph
tert-butyl is largest group, then phenyl,
then pentyl chain, then Cl, then H.
Cl
H
Ph
C H3
C4 H9 C H
5 12
Ph
Cl
H3 C C 5 H1 2
H3 C
H
P
h
C 4H 9
C
H
C 5 H 123
C4 H 9
H
Cl
H 3C
H
P
h
C
4
Cl
C 5 H 12
Cl
H
C5 H1 2
9
25
Draw the Newman projections for the following molecule. In the first picture draw the
molecule exactly as shown. Holding the first carbon fixed, rotate the back carbon to
show all 6 conformations of this picture. Identify the gauche, anti and eclipsed pictures.
Ph
H
F
OH
Size Ph > Hexyl > F and CH3 > OH > H
Ph
H
CH3
Ph
F
H
OH
Hexyl
F
OH
Size Ph > hexyl > F and CH3 > OH > H
H
F
Hexyl
F
OH
H
Hexyl
GAUCHE
Hexyl CH3
CH3
Ph CH 3
ECLIPSED
ECLIPSED
F
H
OH
F
GAUCHE
F CH3
CH 3
Ph
OH
Ph
Hexyl
H
OH
Ph
ANTI
H
Ph
He
xyl
OH
ECLIPSED
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6. Cyclic Alkanes/Ring Strain
a. ring strain
1. heats of combustion have been used to determine ring strain
2. table 4.6 page 155
cyclopropane has greatest ring strain. cyclononane,
cyclodecane and cyclooctane come next in ring strain.
cyclopentane and heptane have very little ring strain
cyclopentadecane and cyclohexane have almost no ring strain
3. angle strain makes up most of ring strain
due to angles being less than the tetrahedral angles that
are preferred
cyclopropane is 60 degrees instead of 109.5
because of ring strain orbital overlap is low which
weakens bonds.
4. Torsional strain due to locked ring structures leads to eclipsed
newman projections
cyclobutane actually is not planar(like a square) but bent.
in bent conformation angles are 88 degrees which is even
further from 109.5 than the 90 degrees of a square. However,
the bent structure provides much less torsional strain which
makes it a more favorable structure than the planar form.
cyclopentane has angle of 108 vs 109.5 if planar. Therefore
little angle strain. However it doesn’t take this form due to
intense torsional strain. Almost all of the ring strain is due to
torsional strain. Therefore it bends out of plane increasing
its angle strain but relieving its torsional strain
cyclohexane has a form that has almost no angle or torsional
strain.
27
7. Cyclohexane
a. boat vs. chair
1. cyclohexane actually has 7 conformations it can take.
However, the boat and chair forms predominate. The chair
form is free of ring and torsional strain. The boat form is
free of ring strain but has torsional strain. Based on thermodynamical data, 99% of cyclohexane molecules exist in the
chair form.
2. Boat has torsional strain(due to eclipsed hydrogens—2 and 3
eclipsed, 5 and 6 eclipsed).
Boat has flagpole interactions(due to van der Waal
interactions between H-1 and H-4).
3. half-twist forms has less torsional strain and less flagpole
interactions
4. cycloheptane, cyclooctane, cyclodecane…have small to no
ring strain. Most strain is due to eclipsed
hydrogens(torsional strain) and due to van der waal
interactions also called transannular strain.
b. axial vs equatorial
1. there are 6 equatorial and 6 axial positions on chair form
2. the equatorial positions are best for larger group as
the equatorial positions avoid van der waal interactions.
3. As cyclohexane flips through all the axial positions become
equatorial and vice versa
4. In cyclohexane these interactions are called 1,3 diaxial
interactions. They are akin to gauche interactions in
butane newman conformations.
5. as stated in #2, larger groups prefer equatorial position
because larger groups will have larger diaxial interactions.
28
c. cis vs trans in disubstituted alkanes
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12,
1,2 cis are e,a and a,e
1,2 trans are e,e and a,a
1,3 cis are e,e and a,a
1,3 trans are e,a and a,e
1,4 cis are e,a and a,e
1,4 trans are e,e and a,a
1,2 cis is meso; 1,2 trans is pair of enantiomers
1,3 cis is meso; 1,3 trans is pair of enantionmers
1,4 cis is meso and 1,4 trans is meso.
1,4 cis is a diastereomer of 1,4 trans
1,3 cis is a diastereomer of 1,3 trans
1,2 cis is a diastereomer of 1,2 trans
8. Decalin/other bicyclics
a. Know structures for decalin(cis and trans)
b. Know adamantine, diamond, cubane, prismane, and
bicyclo[1.1.0]butane, dodecahedrane
9. Pheremones
a. undecane, 2-methylheptadecane, muscalure
10. Synthesis/Retrosynthesis
a. hydrogenation of alkenes/alkynes
b. reduction of alkyl halides
c. deprotonation of alkynes/alkylation of alkynide ions
d. retrosynthesis/identity of precursors/E. J. Corey
29
RING FLIP OF CHAIR
CHAIR
A
A
B
B
A
B
B
5
1
6
B
B
6
A
B
3
2
B
A
A
4
A
B
A
A
A
2
1
A
B
4
5
3
A
B
B
CIS 1,2 and ring flip
5
1
6
6
B
4
5
3
4
2
B
A
2
1
3
A
CIS 1,3 and ring flip
5
1
6
6
B
4
5
3
4
2
B
2
1
3
B
B
CIS 1,4 and ring flip
5
1
6
6
B
3
2
4
5
4
1
A
B
A
2
3
30
CHAIR
A
A
RING FLIP OF CHAIR
B
B
A
B
5
1
B
6
B
B
6
A
B
3
2
B
A
4
5
A
4
A
A
2
1
B
3
B
B
A
A
A
B
B
TRANS 1,2 and ring flip
A
5
1
6
6
A
3
4
2
4
5
2
1
3
A
A
TRANS 1,3 and ring flip
A
5
1
6
6
A
3
4
2
4
5
2
1
3
B
B
TRANS 1,4 and ring flip
A
5
1
6
6
3
2
A
4
A
4
5
1
A
2
3
31
Draw the following cyclohexane in chair form. Draw the other chair form by
flipping the chair. Pick which one is the more favored product.
Phenyl > tert-butyl > pentyl > OH
OH
up
4
up
3
d ow n
2
5
6
1
up
OH
OH
4
2
3
5
6
3
Ph
1
OH
be tte r pi cture all grou ps equ ato rial
4
5
1
2
6
32
Homework #2—Give the name for the structures below.
OH
Br
F
Br
F
OH
OH
Br
33
Redraw the following molecule in Newman Configuration in the first box. Put carbon 4 in front
and carbon 5 behind. Then draw the other 5 main Newman Configurations by rotating the
BACK carbon counterclockwise. Label the configurations, gauche, anti, eclipsed…
Cl
1
3
2
5
4
Br
Ph
7
6
H
9
8
10
On carbon 4--Br is largest, next is propyl, and
hydrogen is the smallest.
On carbon 5--phenyl is largest, next is pentyl,
and hydrogen is the smallest.
34
Draw the following two molecules in chair configuration, then draw the ring flip of each.
Br
Cl
35
Homework #2—Give the name for the structures below.
OH
Br
F
11-bromo-6-ethyl-13-fluoro-9-methyl-8-vinyl-2,6-tridecadien-5-ol
Br
F
7-bromo-2-fluorobicyclo[4.2.2]decane
OH
7,10-dimethylbicyclo[4.4.0]decan-3-ol
OH
Br
8-(3-bromo-2-methyl-2-pentenyl)-11-ethyl-10-isobutyl-4-isopropyl-3,7,11,13tetradecatetraen-7-ol
36
Redraw the following molecule in Newman Configuration in the first box. Put carbon 4 in front
and carbon 5 behind. Then draw the other 5 main Newman Configurations by rotating the
BACK carbon counterclockwise. Label the configurations, gauche, anti, eclipsed…
Cl
1
3
Ph
5
2
4
Br
7
6
On carbon 4--Br is largest, next is propyl, and
hydrogen is the smallest.
On carbon 5--phenyl is largest, next is pentyl,
and hydrogen is the smallest.
9
8
10
H
ANTI
ECLIPSED
P
P Ph
Cl
Ph
Br
H
Cl
Br
GAUCHE
P Pentyl
H
Cl
Ph
Pentyl
Br
H
Cl
ECLIPSED
P Cl
P
Pentyl
Ph
Br
P
H
Pentyl
Pentyl
ECLIPSED
GAUCHE
Cl
Br
H
PH
Pentyl
Br
Ph
37
Draw the following two molecules in chair configuration, then draw the ring flip of each.
4
5
3
Br
6
Br
6
2
5
1
5
4
4
1
2
2
6
3
1
3
Br
6
4
3
5
2
6
1
Cl
5
1
Cl
4
2
3
3
5
6
4
1
2
Cl