CHAPTER SOLUTIONS > CH 4. University of North Texas CHEM 3210__Organic Chemistry_Chapter 4: Alkanes

Organic Chemistry I Prof. Daniel Cruz

Chapter 4: Alkanes and Cycloalkanes -a.j

CHAPTER 4 alkanes G cyclo alkanes Then we 'll learn how Alkanes and Cycloalkanes to asses the stability 4 09.21.2020 energies of different first we're going to learn how to name .

alkanes 4 cyclo alkanes

-

=

composed

Hydrocarbon It

C

-

l

c

H

-

C

-

l

H

H

c

but they still have no pl bonds

Alkenes : c

saturated hydrocarbons

C IC

-

-

,

= #

slatvrated

C

hydrocarbons

unsaturated

Benzene, Getty

'

Pi bonds or

or

rings

both

unsaturated

Or

-

ll

X

Acetylene , GHz

Ethylene city

l

H

H

H

r

Hydrocarbons are

H

l

H

Ethane , Cz Hce

'

=

,

.

-

they have a ring

/ nyaroge;a#÷i"inim÷!!÷

type of hydrocarbon

"

'

unsaturated b/c

using a series of approaches

0:43

4.1 Alkanes

(

-

H

Rings cyclic structures O

D

O

O

carbons have as many H 's as possible '

NO

IT bonds or

rings

04:12

IUPAC Nomenclature - Alkanes The

•

IUPAC system

-

( Alkanes

=

C

-

c

single bonds)

systematic naming of compounds

.

IUPAC name includes :

'

-

-

-

( longest carbon chain )

parent name Names

of substituents

Location of substituents

4.2 Selecting the Parent Chain 7

.

Identify the parent chain

-

06:09

the

.

start by selecting parent name of compound

by selecting the parent chain in compound

longest consecutive chain of carbons

-

•

identify the

longest consecutive chain of carbon

•

⑥

•

You just need to

.

•

look at different directions,

•

horizontal , vertical , whatever

•

a

once

'

•

you find the longest consecutive

chain , that will

4.2 Selecting the Parent Chain 7 -

.

Identify the parent chain if there

is more

-

name

give you parent

08:08

the

longest consecutive chain of carbons

than one possible parent chain, choose the one with more substituents attached

.

-

-

•

•

•

•

•

•

•

• •

•

or

•

substituents : groups that sustltvte a

hydrogen in a parent chain

• •

.

*

Remember : we only check the # of substituents to decide the parent chain If YOU end up with a tie -

.

If only 1 Chain that is the longest one , thats the parent chain , it doesnt matter how many substituents It has

-

Only check if there is a tie

.

.

4.2 Selecting the Parent Chain

MEMORIZE UP TO 12 !

10:10

.

Once you know the amounts of carbons in chain

the longest

②

sketches

-

-

,

you will know the parent

name b/c YOU NEED TO MEMORIZE THIS

-

/

TABLE ( ONLY UNTIL 127 !

s.int/Xi: n: : :n: : :n:.no:: owe::ei: .n.rr o

O

-

(

..

first 4 acronym : me estoy poniendo buena methane

# ethane Tspropane

Jane

SMEPB

13:18

4.2 Selecting the Parent Chain 7

.

Identify the parent chain -

if the parent

-

longest consecutive chain of carbons

the

cyclic

chain is

"

add the prefix cyclo

,

4 .

.

B

M

-

f

.

cyclic

.

O

n

e

o

just in a ring so ,they are

O

} &

still alkanes bk

they have c c single bonds ,

"

.

:

-

•

S

.

cyclo alkanes

" -

M cyclohexane

C

y

r

4.2 Selecting the Parent Chain 14:39 Practice Identify and name the parent chain in each of the following compounds.

3pubstHve :

]

.

o

g

p t

G

.

①

.

.

o

•

a

z

hexane

.

*

chain

.

know

-

4.2 Naming Substituents 17:40 Identify { name substituents

longest

I

.

substituting H &

•

⑨

•

•

q

D

a

group

,

H a substituent bk It sort of

substitutes a

0

a •

co

11

2

defiant

D

checking

Carbons

If a structure has substituents,

point of systematic IUPAC name

ly

is for each

compound

3. Since It has

substituents , name isn't just .

.

substituents b/c the whole

a row

decane We need to name them

.

hydrogen

we also need usys.to name

mm

how many carbon atoms the substituent has

=

group of atoms but we call

10 in

4. we name substituents by

Again , a substituent

It could be an atom or a

the

-

.

cyclo heptane

substituent ;

heptane

2.

team:S:b:ri:i ::* :::: ::c: me

name .

unique

to have a Unique

.

*

the same amount of carbons That is how -

-

we

prefixes as the names of the parent chains

know If we're talking about a parent chain or a substituent

Parent

chain t

substituent prefix

parent

chain

suffix

4.2 Naming Substituents .

same

However also notice that the name Of the substituents end with the

.

=

=

with

suffix

yl

-

.

.

same

ane

-

substituent chain suffix

-

2

have the

Notice that the names of the substituents

-

-

.

yl

22:13

Identify 4 name substituents • a

-

parent :

new @

•

•

O -

o

-

substituents in

yl instead of

4.2 Naming Substituents 2

Delane

ane

.

23:00

Identify 4 name substituents

.

•

A

ring

can be either

Since the reg

•

a

•

chain or a substituent depending on the # Of carbons

a parent

•

•

•

has more

tphaererlnn.gl

.

•

•

.

linear chain

carbon than

't becomes the

I •

It comes to comparing

have

ring Which one will be parent 4 Which

•

-

•

•

.

• -

2

.

④ •

parent

:

.

.

.

Methyl

.

.

.

.

# the longest carbon

.

2 3

.

.

a °

chain

Name the substituent Name -

-

The

whichever one has the

largest amount

coming soon) But , for the substituents that .

assigning meaning to It (this Sec ) .

.

common

WITHIN the case

,

.

-

-

methyl butyl)

substituents first 4th en the parent name

2- methyl but ywnde cane

attached to the parent chain

( in this case 2 methyl)

side group

a molecule

substituent Start with the carbon

butyl)

of the substituent is (2

when naming

(

name )

(in this

9 # the substituent 's name

Depends # of carbons

already have °

Un decane

1

substituent?

don't have a common name , you need a systematic way of

that don't

@

.

have to MEMORIZE

I

I

-

a

of

some substituents are going to have common names you will

for substituents with complex branches

butyl

,

14:52

Identify 4 name substituents -

a

consecutive carbons will be the parent chain

-

4.2 Naming Substituents

structures that

when

one is going to be

.

.

( single word tho) :

.

.

Remember : 27.52 (

Memorize !

methyl

⑧

(

substituent

4.2 Naming Substituents

28 : 53

-

-

some branches have Two

2 carbon

common names

.

IO O

① .

•

•

-

.

-

some branches have

3 types of

Enadrbon

car ⑥

( butyl

.g

4.2 Naming Substituents

Middle carbon

-

.

33 : 06

.

.

font:

. ° .

o

K

-

MEMORIZE

Identify 4 name substituents some branches have

2 types of

-

vs

.

4 carbons)

-

.

-

.

carbon

MEMORIZE

common names

butyl groups

end

2

, ,

Identify 4 name substituents -

-

isopropyl cyclohexane

II

end

31 :O S

vs

propyl cyclohexane

f

f

-

2

④ O

.

.

4.2 Naming Substituents

long

3 carbon = propyl

types of propyl groups

•

Substituent

MEMORIZE

Identify 4 name substituents

2.

←

Parent

carbon

parent long 1

ethyl

common names

pentyl groups

(

pentyl

-

-

.

S carbons)

-

-

Assembling the IUPAC Name °

carbons

in the

parent chain have to be number

① -

•

2.methyl

33 : 34 o

1st thing

after

Identifying parent chain is you need to

decide at which end of the parent chain

①

.

Need to #

to start

# Ing

you can say where on the parent chain the substituent is ( less ambiguous) so

.

I

pentane means there

is a

methyl group on carbon # 2 Of the pentane chain

•

•

.

Assembling the IUPAC Name 36 : is -

Guidelines to follow when # ing the parent chain 1

if

.

ONE

sub St vent is

present , # the parent chain

so that the substituent has the lowest #

possible

Assembling the IUPAC Name 37 : Ss -

Guidelines to follow when # ing the parent chain 2

When

.

multiple substituents are present , # the parent chain to give the

0

2 substituents on

#

1st substituent the lowest # possible

same carbon,

listed by Itself each must be

Assembling the IUPAC Name 38 : 40 -

Guidelines to follow when # ing the parent chain 3

1st

.

If there

is a tie ,then # the

regarding ✓ substituent

parent

chain so that the

second 10 cant gets the lowest # possible

Guiding principle : ALWAYS go by the lowest # possible

Assembling the IUPAC Name -

39 : 03

Guidelines to follow when # ing the parent chain 4

.

If there

is

no other tie breaker , then assign the lowest #

0

O.O .

.

Both are

1,S

alphabetically

0

Assembling the IUPAC Name -

Guidelines to follow when # ing the parent chain -

The same rules

apply for cyclo alkanes # Ing

can be

clockwise or counterclockwise But still want substituents

to be

lowest

possible # Is )

.

.

Assembling the IUPAC Name To assemble the

'

1

complete name :

put the # da name of each substituent before the parent chain name in alphabetical order .

A prefix

is

used

-

Note :

-

"

di

"

( di , tri tetra penta etc ) If multiple substituents are Identical ,

,

or

"

th

"

is

4.2 IUPAC Rules - Summary 1 Identify the parent .

2 3

.

,

.

2

41 : 48

,

.

.

-

ignored when alphabetizing the substituents

.

42 :3 S

chain

Identify 9 name the substituents

.

Number the parent chain ; assign a 10 cant to each substituent

.

list the Numbered substituents before the

4.

parent

-

Remember substituent needs lowest #

alphabetical

name in

possible

order

.

4.2 IUPAC Rules - Practice 43: Os Following the rules, we can name the following compound: 1

•

,n¥,qg Z

•

o

Parent name : octane

methyl

substituents :3

,

methyl g 4 ethyl -

-

•

e •

④

⑨

name , y

, up , ,

e, ny ,

,

,

3. me, my ,

ane

0

4.2 IUPAC Rules - Practice 47 : 45 Following the rules, we can name the following compound: Parent name : cyclohexane

go by Bbk space

4

b substituents : l

Z 4

④ 3

methyl, I methyl 3 ethyl -

-

,

group :

3

IUPAC Name : 4

z

-

I, I

tert-butyl

-

-

3

-

ethyl

,

4

-

tert-butyl -

( but when listing in IUPAC still Use m)

dimethyl -

l , I dimethyl cyclohexane

I

,

2

-

#

I

4

-

a-reentl.03.IE lowest #

red S 1,02, 4,4 Us

.

↳ s

to

group constituents with identical names they ,

don't need to be on same carbon

.

like , If we had 3

methyl scythe 3rd One was on carbon 4 we would ,

just call it 1,1 , 4

Practice 53 : 18 What is the systematic IUPAC name for the following compound?

÷÷.;÷÷.s,÷!#

10

6

8

1%917%7

-

tri methyl

MEPB

-

-

" " ane : 3-

methyl 47 ethyl -

IUPAC Name : 7. ethyl 3 methyl decane -

-

•

?

44 8

us

3④ * time is

best spent practicing but memorize prefixes

.

4.6 Drawing Newman Projections S7 : 25 '

single bonds rotate resulting in 3D shapes , called conformations ,

°

There are various ways to represent the 3D

shape of a compound

.

.

O

o

-

•

Newman projections

=

Ideal for comparing the relative stability of possible conformations resulting from single bond F "

rotation

good for alkanes G cyclohexane s

÷: ::: : :::aan" ::a%r::: :c:: wot's:

iii. omninneesamaenkanwenunweuseare.ua

"

n

.

.

e

.

§µ H

which we Use

H

This

"

ethane HOW

1 2

.

.

is

a

dashed line

Draw the bond line I wedge

only rel a vent if the 4 groups bonded to the carbon

dash

Into

a new man

In what we

?

call a sawhorse projection

Now for sawhorse , what you need to do is look at the bond on the to

compare

to the bond on the plane on the 2nd

it

pointing in opposite directions

bonds need to be

,

.

do we translate a bond line -

wedged line, 4 one bond to the back

.

often not shown explicitly b/c

different

are •

bond to the front , which we use a

line for ; one

.

'

plane on the first carbon and you want

carbon For it to be considered saw horse , those two .

.

4.6 Drawing Newman Projections 01:03:56 '

'

A new man

projection

show the front

is the

perspective of looking straight down at a particular

carbon as a

OO

point 4 the back carbon as a large circle behind it

front

-

tho

O

08

①

O rep

O

I

.

0

C

-

C

bond

.

bond we're

looking thr

Bonds to the carbon in the back are only drawn

0

until

the circumference of the circle -

4.6 Drawing Newman Projections '

Another example

-

-

jojflheaeaf

④

q

01:07:47

be

to

-

vertical

⑨ ⑧

specify which C C bond youre looking through Typical exam question will have structure 4 eye On left He will

-

first carbon eye is at

=

front

.

.

.

④0

-

.

.

④1

t

④

you draw all dashes da wedges Order doesn't

matter

.

④

Here come nd

§

CH3

-

when

translating a bond line to a new man drawing your ,

dashes 4 wedges 4 make sure in sawhorse form make sure

Its in

sawhorse form? look at the

.

How do you

bonds

on the

Plane 4 make sure they are pointing in opposite directions

.

4.6 Drawing Newman Projections 01:16:33 -

-

Oi

<

a.

"

Dash end up on same side

wedge

end

up

on

same side

You just gotta use mind eye to figure out which side

CH

.IE

H

,

Cl

.

.

Truman Roland

Help : atoms with the

-

CHAPTER 4 If you're looking to the right dashs goes on left Alkanes and Cycloalkanes 4wedges goes on the right If you're looking to the left the dashes goes on 09.23.2020

wedge line need to be on

,

same side 4 same with dashed atoms -

'

atom on plane attached to front atom is

if

-

.

gets lines in circle

front C

Help :

'

pointing up, Vertical line goes

right 4 wedges go on left

.

Practice on top of circle ; If pointing down vertical line goes on bottom 1. Draw a Newman projection by looking through the C3-C4 with the C3 in the front. 00:28 2. Draw a Newman projection by looking through the C3-C4 with the C4 in the front. .

,

'

Ph

Hzc ,

(

,

"

µ

%

12

eye may be on left but its

1.

.

I

=

\

'

2

(turn head in direction eye is facing

,

facing right

,

l

.

,

.

I

doonwnbos.Eo9moes@Xlp.r#

4¥ .sn

Hsc

/

.

-

Ph

Ph

'

-

" needs

f.gifting !Pop

H

Br

f

Cz in front (y

in

back

helps me)

-

Hsc

it

'

f 1

Cy In

/

Cz

front back

in

4.7 Conformational Analysis 07:48 The

-

angle between

molecule below

°

HSI

.

H

H

/

⑦ \

-

TH

Dihedral

.

Newman structures are used for

confirmational

.

analysis You will need to understand what ,

.

H

gdnmeeddirnae.la/aY'aenlsgie

is the

groups in the Newman '

angle =6O°

angle that separates 2

.

Newman has like circle , Circle '

3600 , ÷ 0 in 6 '

.

parts , you get dihedral angles

Staggered dihedral angle 600 ; eclipsed -

-

-

-

a

1600 )

Oo

staggered conformations are more stable (lower in energy ) than eclipsed conformations

rotation causes the front

④

iganrgbanngptfobn

°O° ④t

fX

the ,pb%dk= ,

④

00 '

.

Groups connected to back carbon stayed in place BUT

gyu: oenneotedto °

④

⑦ ④ Eclipsed conformation

Highest in energy

staggered then do eclipsed

12 •

I

Newmans are used for

The

rotate

conformational

.

Rot citation is either clockwise or

counterclockwise ( be consistent) ht he carbon you rotate

torsional strain

O bond

confirmationat analysis 4What that means

as the O bond rotates

.

.

Here , the difference

in energy

is

is that we

use it to asses

the energy change or

.

we're talking about is the

thats rotating in a conformational -

to do

Eclipsed conformation - higher In energy than staggered

stability change '

rotating a 0 bond

can either be the front or the back one ( Be consistent)

.

The difference in energy between these conformations is due to " IMO

When we are

front carbon rotated 600

analysis, one of the 2 carbons needs to have 3 groups stay where they are { the other carbons groups

,

lowest in energy start c. analysis with

O

O ④

staggered conformation

'

torsional angle His 600 In the

analysis But in order to understand confirmation al

H

4.7 Conformational Analysis - Ethane 8:57 -

dihedral angle or

atoms on adjacent carbons Is called a

c C O bond -

we're looking through That is the O bond, the C C bond , -

.

analysis problem we monster energy 9 Stability changes as that O bond rotates energy changes b/c the conformations will alternate between staggered 9 eclipse .

.

.

.

4.7 Conformational Analysis - Ethane

14:29 *

eclipse

=

Hoon

able to conformational analysis of

is to be

any alkanes

a

•

•

Goal

:#

staggered

•

a

•

l

'-

l

o

1-1

I

-

-

4.7 Conformational Analysis - Ethane 16:26 It

'

lot or Matton is 12141 MOI less stable b/c of

is possible that the eclipsed

14 Ks IM 01 )

at

pair repost on between the eclipsing bonds

The reason why the eclipse conformation will Always be too close

so collide

higher in energy is b/c the groups are elapsing each other

,

=

This means the e

'

destabilize due to repulsion

with a

-

for each eclipse interaction)

0 0

•

e

difference of

Clouds are able to destabilize the structure

clouds repel each Other

staggered conformation But, on avg ,they will usually

12 Kj Im 01 In stability , at room temp , 99% Of the molecules will be in .

any given point in time

spend more time

bio e

-

in staggered

o bonds are

.

in

constant rotation at room temp for ethane

conformation bio more stable

.

.

.

4.7 Conformational Analysis - Propane 18:03 '

The analysis Of torsional strain for propane ( below) Is similar to ethane

①①

mo. .

::÷

H

"

→

pH

①

9 H ,

①

The larger the group is

l

A Its

H

" "

for

H

=

'

•

21:16

l

H

more opportunities for

larger the energy change will

4.8 Conformational Analysis - Butane

H

be

repulsion

=

4.8 Conformational Analysis - Butane 22:09

ghbqhffemtf.ly

more variety blcmore possible combinations

-

are

in order

-

close

-

,

-

-

But Mthe other 2 , chance of collision

£0

staggered

energy will be

q

Og

C

O

↳f8pYI9%ctiy

-

-

.

.

Remember, the less stable something Is , the higher the

-

•

youre done conformational

analysis completely you need to end up with 6 newmans Manti conformation , stable tgnobhaha to collide

-

eclipse

to make sure

*

.

You need to be able to draw a conformational

analysis for any molecule that includes all 6

cap

.

only a rotations

newmans

9 YOU need to be able their Stabllltles

among them Keep in mind that all the staggered .

.

ones will be lower in energy than eclipse

opposite

among the the

.

But

standard ones , we can still rank

them depending on whether they are untidy

gvache 9 Among the Eclipse ones we can further -

depending on whether the larger groups are elapsing each

rank them

other or not

4.8 Conformational Analysis - Butane 2%45 "

when the

methyl groups are

they are

anti to one

gauche to one another there is steric strain ha higher energy conformation than whey ,

,

another

I

-

The highest energy conformation for butane results when the methyl groups eclipse one another

4.8 Practice: Conformational Analysis 21:58 Identify the highest and lowest energy conformations of the molecule below. weak

LOOKING R

"

"

"

"

E

H H

strong

S

H

II

D

ft

H

E

H S

¥

S

"

H

H

C

B

A

Al

H

H

eclipse

) ,e)dBiDif me

staggered

"

ti

E

A f

JH

ALCSECBSFCD E

Ring strain

4.9 Cycloalkanes 40.09 cyclo alkanes don't have the

'

same stability due to ring strain

-

:

puts burden

strain

=

destabilize

2 kinds ; both destabilize

Ring strain comprises :

'

-

Angle strain : result of deviation from ideal bond angles caused by inherent structural constraints -

-

Torsional strain :

"

result of

4.9 Cycloalkanes -

can 't be

relieved due to restricted

conformational mobility

angles for sp hybridized carbon is 109.50 Due to carbon atoms being locked in ring the bond angles must deviate from the ideal Value 3

,

'

Furthermore , most

cyclohexanes will exhibit some degree Of torsional strain

All

O

60

Alot of angle q H 's are

torsional

eclipsed

cyclopropane

-

"

strain

I

"

•

It

Internal bond

.

.

4.9 Cyclopropane 41:so

'

.

clouds Coll de

-

skipped

Ideal bond

'

having their e

eclipsed groups

.

dispersion forces that

angle ( O)

-

VH

tort what strain 's

¥

sp

109.90

H

600 649.50 www.fnmthe#eatetrahedraange- angle strain

forcing It to be 600

in Steed

Of 49 so .

4.9 Cyclobutane 44: 37 bond angles of 880

1.

Angle strain

2.

Slight torsional strain blc adjacent C H bonds are neither fully eclipsed nor full staggered -

7%9: :ndan%YaY%%;Ymna%gs

g

less torsional

so

bonds

is 880

some Angle 4 tort what Stram

4.9 Cyclopentane 46 : 08 7. very little

angle strain

1. Slight torsional strain

-

-

bond angles

nearly 109.50

adopts an envelope conformation to avoid most of it

①

NO angle strain 4 some tort tonal strain

4.10 Conformations of Cyclohexane cyclohexane

>

can

adopt

4729

most import at

conformations , but

a variety of

O

4 interesting one

Its the chair conformation that is the most stable

O

4.10 Conformations of Cyclohexane 48 : 12 cyclohexane has no ring strain in a

-

I 2

No angle strain

.

.

bond angles are

-

No torsional strain

-

all

chair conformation

109.90

adjacent C H bonds are staggered -

6 member b/c

"

The

other

most stable

possible conformations of cyclohexane have some amount Of angle 410 r torsional strain lie ring strain) .

4.11 Drawing Chair Conformations (Need to know ! ) '

rings most common

50 : ol

Drawing a chair conformation Draw

a

wide v

Draw a line going down at a 600 angle, ending just before the center of the V

.

Draw a line parallel to the left side of the V, ending just before the left side of the V

Draw a line parallel to the line from step 2, going down exactly as low as that

connect the dots

.

line

.

4.11 Drawing Chair Conformations Sl : 07 '

'

if drawn

correctly the

Each carbon

,

in the

chair should contain 3 sets

of parallel lines

ring has 2 substituents : one is in axial position { the other

in a

equatorial position

.

4.11 Drawing Chair Conformations Adding the

-

The

.

axial substituents

is

S2 : 14

easy , as they point straight up da down alternating around the ring

equatorial Groups are drawn off the ring , 4 they run parallel to the lines in the chair ←

4.12 Monosubstituted Cyclohexane when cyclohexane has one

'

.

toeing!nates

lines can exist

Ss :B

substituent there are 2 possible chair conformations, ,

chairs can flip da are in constant motion s

6

# 24

4

s

3

4

Ho z

z

-

when

chair flips , the positioning of the

Substituents changes

£

equatorial UP

axial , up

,

Ring flipping occurs by rotation of all the c

-

6

s

-

C bonds in the

ring Axial substituents become equatorial d. Vice versa .

-

whatever was axial becomes

equatorial 4 whatever was equatorial becomes axial But , groups still point in same .

direction that they were pointing

.

4.12 Monosubstituted Cyclohexane 01:02:10 consider

-

methyl cyclohexane The chair conformation where the methyl group is equatorial

is the

more stable

LN

.

chair

why ? ble steric interactions l

-

clouds cold ing with each

other

At room temp

"

in the axial

-

.

Methyl cyclohexane will be in the more stable chair position

position the methyl group causes ,

Sterk Interactions that destabilizes the conformation

.

4.12 Monosubstituted Cyclohexane 01:03:13 '

The steric strain

from a substituent being In axial POSITION

IS the result

Collison

of 1,3 di axial interactions -

point away so b chance of COLLISION

'

The

I

1,3 diaxal interactions are actually gauche interactions , which aren't present when the methyl groups are -

equatorial

.

.

4.12 Monosubstituted Cyclohexane

01:04:39

larger groups will cause more steric crowding in axial position

-

{

gets

bigger

4.12 Monosubstituted Cyclohexane 01:05:10 Practice: Draw the t wo chair conformations of isopropylcyclohexane and determine the more stable conformation.

*

4.13 Disubstituted Cyclohexane '

with

iii.

ma ,

01:07:58

multiple substituents solid or dashed wedges are used to show positioning of the groups on the rings chair wedge dash bond line ,

-

-

-

Realize the Cl group Is -

-

up

in

both possible conformations , 4the methyl group

cyclohexane is both equatorial dy axial b/c constantly flipping make

sure placement

.

.

up

larger group being equatorial more stable

Axial

l

l

equatorial

/ axial

-

y

equatorial

wedge sup =

down

In both possible chair conformations

4.13 Disubstituted Cyclohexane 01:11:17 Practice: Draw both chair conformations for the following molecule

down

-

substituents dash

down

is

of substituents agrees with OG dashes { wedges

-

wedge 4 dach dont impact axial / equatorial

'

-

substituent

:

.

one up

or

both up n both down ll

4.14 cis-trans Stereoisomerism 01:14:48 when naming

r

a

di substituted cyclo alkane , use the prefix Cls when

opposite sides of ring

2 substituents are on

they

Gone down n

H

are 2 groups on same side of

ring da trans when ,

.

These 2 compounds are stereoisomers since they are 6 Member rings they are best represented as Chair conformations

-

.

,

4.14 cis-trans Stereoisomerism 01:16:18 Each compound exists as 2 equilibrating chairs , spending more time in the more stable chair conformation

-

p

l

-

s

-

Practice 01:18:24 O Cis-1-tert-butyl-4-methylcyclohexane -

-

I b

l

s

4

r

'

U

f

z

more stable b/c

large group is equatorial

Practice 01:20:48 Which represents the ring flip of the following structure?

l l

l O

l CHAPTER 4 FIN.