CH203 Fall 2014 Lecture 12 September 29 slides.pdf

8/11/2019 CH203 Fall 2014 Lecture 12 September 29 slides.pdf

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Exam 1

Our overflow room is CAS224, 725Commonwealth Avenue

Students whose last names begin with A throughM will take the exam in SCI109 (here)

Students whose last names begin with N throughZ will take the exam in CAS224.

Do not be late.

No make-up lecture exams are given for anyreason. Do not make travel plans that conflict with

the lecture exams. If you are sick, have a death inthe family, religious conflict, etc. please contact Dr.Pounds as soon as possible.


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Exam 1

!"#$% '() *+,-./ $0+1% "#$ %&'() '&& *)+,-.'& /)&-.01.0, ")2023 /'(4*'(4,3 (&-561.0 56'5 7-8 91&& .-5

/) 9)'+1.0$ '5 56) :+-.5 -: 56) +--;2 1/+'5-+ :8.(B-.2 ) &'5)3 7-8 '+) .-5 01>). '==1B-.'& B;) 5- (-;*&)5) 56))F';2 J: 7-8 K.1,6 56) )F'; )'+&73 6'.= 15 1. '.= &)'>) 56) )F';1.'B-. 6'&&9156 56) ;1.1;8; =1,58+/'.()2 ?- .-5 (-.0+)0'5) 1. 56) (-;;-. '+)'1;;)=1'5)&7 -85,1=) 56) )F';1.'B-. 6'&& 9'1B.0 :-+ 7-8+ :+1).=,2


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IUPAC Nomenclature of alkenes

1. Number the longest chain of carbon atoms that contains the doublebond in the direction that gives the carbons of the double bond thelowest numbers.

2. Locate the double bond by the number of its first carbon.

3. Name substituents.

4. Number the carbons, locate and name substituents, locate thedouble bond, and name the main chain.

1-Hexene 4-Methyl-1-hexene 2-Ethyl-4-methyl-1-pentene

3

4

5

6

35

6

3

5


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Dienes, Trienes, and Polyenes

For alkenes containing two or more double bonds, change the infix -en-to -adien-, -atrien-, etc.

Those containing several double bonds are often referred to moregenerally as polyenes.

Following are three dienes:

1,4-Pentadiene 2-Methyl-1,3-butadiene (Isoprene)

1,3-Cyclopentadiene

12

34

51

23

4 1

2 3

4

5


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Stereocenters and Chiral Centers

Stereocenter: An atom about which exchange of two groups produces astereoisomer.

Chiral center: A tetrahedral atom, most commonly carbon, that is bondedto four different groups.

All chiral centers are stereocenters, but not all stereocenters are chiralcenters.


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Double bonds as stereocenters

When an alkene can exist as cis and trans(or E and Z) stereoisomers, bothcarbons of the double bond are stereocenters.

Stereocenter - an atom at which the interchange of two groups gives a

stereoisomer. At a chiral carbon, such a swap changes R to S or S to R.

different molecules(stereoisomers)

interchange

circled groups

$ $

$ $

H CH2CH3

H3C CH3

C AC

$ $

$ $H

H3C

CH3

C AC

CH2CH3

identical molecules

interchange

circled groups$

$

$ $

CH3

H3C

H3C

H

C AC$ $

$ $

CH3H3C

H3C H

C AC

These carbonsare stereocenters

$ $

$$

H

CH2CH3H3C

CH3

C AC

stereoisomers

These carbonsare stereocenters

$$

$H CH2CH3

H3C $CH3

C AC

"


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Double bond acts as aDouble bond acts as a stereocenterstereocenter when E,Z isomers arewhen E,Z isomers are possiblepossible

(E)(S)

Cl

(E)(R)

Cl(Z)

(S)

Cl

(Z)(R)

Cl

22

= 4

(4S)-(2E)-4-chloro-2-hexene

(4R)-(2E)-4-chloro-2-hexene (4S)-(2Z)-4-chloro-2-hexene (4R)-(2Z)-4-chloro-2-hexene

O

OH

O HO

How many stereoisomers are possible? 24

= 16


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For alkenes with ndouble bonds, each of which canshow cis,transisomerism, 2nstereoisomers are

possible.

Example: 22= 4 cis,transisomers are possible for 2,4-heptadiene.

(2E,4E)-2,4-Heptadiene (2E,4Z)-2,4-Heptadiene

12 3

4 5

6 7

2

4

2

4

2

4

(2Z,4E)-2,4-Heptadiene (2Z,4Z)-2,4-Heptadiene

C

C

3

C

4

C

5

trans trans

trans ciscis trans

cis cis

Double bond


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Vitamin A is a biologically important compound with five doublebonds.

cis,transisomers are possible at four of these double bonds,

for 24= 16 stereoisomers.


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IUPAC Nomenclature of alkenes - more

1. Name the parent hydrocarbon by locating the longest carbon chain that contains

the double bond and name it according to the number of carbons with the suffix -ene.

2. a. Number the carbons of the parent chain so the double bond carbons have the

lowest possible numbers.b. If the double bond is equidistant from each end, number so the first substituent

has the lowest number.

3. Write out the full name, numbering the substituents according to their position in

the chain and list them in alphabetical order.

4. Indicate the double bond by the number of the first alkene carbon.

5. If more than one double bond is present, indicate their position by using the

number of the first carbon of each double bond and use the suffix -diene (for 2

double bonds), -triene (for 3 double bonds), -tetraene (for 4 double bonds), etc.

6. a. Cycloalkenes are named in a similar way. Number the cycloalkene so thedouble bond carbons get numbers 1 and 2, and the first substituent is the lowest

possible number.

b. If there is a substituent on one of the double bond carbons, it gets number 1.


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2. b. If the double bond is equidistant from each end, number so the first substituent

has the lowest number.

5. If more than one double bond is present, indicate their position by using thenumber of the first carbon of each double bond and use the suffix -diene (for 2

double bonds), -triene (for 3 double bonds), -tetraene (for 4 double bonds), etc.

H3C CH CH CH CH2 CH3

1 2 3 4 5 6

2-methyl-3-hexene

CH3

H2

C CH CH2

C H CH2

1 2 3 4 5

1,4-pentadiene

H2C CH CH CH CH3

1 2 3 4 5

1,3-pentadiene


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Name to structure


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LM@%GNO#H

PQLRN@OS@TPNGJL

ONAUP@PLNA%NLGOUALU%


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NMR of ethanol, 1951

NMR of ethanol, 2014


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P8(&)'+ A*1., 1. VW

X108+) #H2# :-+ #M '.= #HL3 -.&7 59- -+1).5'B-., '+) '&&-9)=2


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P8(&)'+ A*1. 1. VW

X108+) #H2D G6) ).)+07 =1Y)+).() /)59)). '&&-9)= .8(&)'+,*1. ,5'5), 1.(+)',), &1.)'+&7 9156 '**&1)= K)&= ,5+).0562

Z'&8), ,6-9. 6)+) '+) :-+ #M .8(&)12


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P8(&)'+ S'0.)B( O),-.'.()

X108+) #H2H G6) -+101. -: .8(&)'+ ;'0.)B(

+),-.'.()

2 "'$%+)(),,1-. /):-+) '.= "/$ '[)+ '/,-+*B-. -:

)&)(5+-;'0.)B( +'=1'B-.2


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PSO A*)(5+-;)5)+

X108+) #H2I A(6);'B( =1'0+'; -: ' .8(&)'+ ;'0.)B(+),-.'.() ,*)(5+-;)5)+2


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PSO A*)(5+8;

X108+) #H2\#M]PSO ,*)(5+8; -: ;)567& '()5'5)2

M106 :+)C8).(7^G6) ,61[ -: '. PSO ,10.'& 5- 56) &)[2

R-9 :+)C8).(7^G6) ,61[ -: '. PSO ,10.'& 5- 56) +10652


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NC81>'&).5 M7=+-0).,

@ ;-&)(8&) 9156 59- -+ ;-+) ,)5, -: )C81>'&).5 67=+-0).,01>), ' =1Y)+).5 PSO ,10.'& :-+ )'(6 ,)52


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A10.'& @+)',

O)&'B>) '+)', -: ,10.'&, '+) *+-*-+B-.'& 5- 56) .8;/)+ -: M01>1.0 +1,) 5- )'(6 ,10.'&2 S-=)+. PSO ,*)(5+-;)5)+,

)&)(5+-.1('&&7 1.5)0+'5) '.= +)(-+= 56) +)&'B>) '+)' -: )'(6

,10.'&2

X108+) #H2_ #M]PSO ,*)(5+8; -: !"#!]/857& '()5'5)2


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L6);1('& A61[ ] #M]PSO

X108+) #H2` @>)+'0) >'&8), -: (6);1('& ,61[, -:+)*+),).5'B>) 57*), -: 67=+-0).,2

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CH203 Fall 2014 Lecture 12 September 29 slides.pdf

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