a lke ne reactions

7/17/2019 A Lke Ne Reactions

http://slidepdf.com/reader/full/a-lke-ne-reactions 1/95

Surviving Alkene ReactionsA logical approach



Surviving Reactions

• Seems like “surviving” is always the word that bestdescribes what has to happen in this class… but Ithink with a little practice you’ll find this isn’t all

that difficult…• es! its time for that memori"ation part so many

people dread# $rganic chemistry has this awfulreputation and I think %&IS is where it all sprung

from#• es! you’ll need to remember some facts… but even

worse! you’ll need to A''( these in order to draw your product#



)here to begin…*

• %here are several pieces that have tocome together when drawing the

product of a reaction#• ou have to maintain a specific carbon

skeleton for the molecule that contains

the alkene#• ou have to know what the reagents will

transform the alkene into#



• ou have to know the details of thepieces that need to be attached#

•

Is the reaction “+arkovnikov” or “,on-+arkovnikov”*

• Is the reaction going to proceed “Syn”

or “Anti”*



•

If regiochemistry is the concern! you’llneed to determine which end of thealkene is more substituted! versus less

substituted#

• If stereochemistry is the concern! you’ll

have to use wedges and dashes tocommunicate if the pieces are attachingon the same side or opposite sides#



• And! of course! when all is said and

done! you have to draw the actualproduct…

• &opefully this will help walk you throughthe trials and tribulations of your firstreaction chapter…

• (et’s begin…



.nderstand your alkene

• Alkenes come in all shapes and si"es#• )hen you want to do a reaction on an

alkene! you first have to identify with thetwo carbons in the skeleton thatconstitute the alkene#

• /epending on the reaction! you may needto determine which end is the moresubstituted end of the alkene#

• (et’s start here…



• (et’s look at an alkene so you can see

what needs to be seen#

• Identify how many carbons are

attached to each sp0 carbon of thealkene#



• 1irst! 2ust identify the alkene carbons3shown in blue4#

• %hen count carbons attached to each#

C

C

C

C

C

C

C



• %he right-hand alkene carbon has two carbons 3inred4 attached#

•

%he left-hand alkene carbon has only one carbon 3ingreen4 attached#

C

C

C

C

C

$ , 5

c a r b o n

a t t a c h e d

% ) $

c a r b o n s

a t t a c h e d



• )ith this in mind! you should be ableto see that the right carbon is “moresubstituted” with more alkyl groups#

• %he left carbon is “less substituted”with only one alkyl group attached#

C

C

C

C

C

$ , 5

c a r b o n

a t t a c h e d

% ) $

c a r b o n s

a t t a c h e d



• )hat about the ne6t alkene* )hichend is more substituted*



• %he right side is more substitutedwith two carbon groups# %he left sidehas only one carbon group#

• %ry another…

C

CC



• )hat about this alkene* )hich end ismore substituted*



• %he left side is more substituted with

two carbon groups# %he right sidedoesn’t have any carbon groups#

•

7ot the idea*

C

C



• Remember that any reaction that hasregiochemistry 3i#e# that reacts

+arkovnikov or ,on-+arkovnikov4 willneed you to determine which end ismore 3or less4 substituted#



Reagents…

• 1or each reaction that is covered! thereare many with mechanistic informationthat you need to remember 3the “why isit this or that*”4#

• 1or each reaction! there are also some8uick and easy pieces to remember

3which will take some time to getembedded in yours brains4…



Reagents…

• %he &alogenation Reaction uses &9 toadd “&” and “9” to the alkene in a+arkovnikov manner#

•

&9 can be &:l! &;r or &<'$= and >I#• “9” will attach to the more substituted

side#

& : l



Reagents…

• Answer?

• ,eed to remember?

& 9 +ark

&:l

:l

&



Reagents…

• Addition of 90 will add “9” and “9” tothe alkene in an A,%I manner#

•

90 can be ;r0 or :l0#

:l0



Reagents…

• Answer?


9 9 A,%I

:l0

:l

:l



Reagents…

• Addition of 90 and &0$ will add “9” and“$&” to the alkene in a +arkovnikov!A,%I manner#

• 90 can be ;r0 or :l0#

:l0

&0$



Reagents…

• Answer?


9 $& +ark A,%I

:l0:l

&0$

$&



Reagents…

• @ariation? Addition of 90 and R$& willadd “9” and “$R” to the alkene in a+arkovnikov! A,%I manner#

• 90 is :l0 or ;r0 and R$& can have anyalkyl group 3R4#

;r0

:&<$&



Reagents…

• Answer?


9 $R +ark A,%I

;r0

;r

:&<$&

:&<$



Reagents…

• %here are three reactions to makealcohols from alkenes#

&$ &



Reagents…

• Addition of &0$ and & will add “&”and “$&” to the alkene in a+arkovnikov manner#

•

&

can come from any acid like &0S$= or &<'$=#

&0$

&0S$=



Reagents…

• Answer?


& $& +ark

&0$&

&0S$=

&$



Reagents…

• Addition of B# &g3$Ac40! &0$ and 0#,a;&= will add “&” and “$&” to thealkene in a +arkovnikov manner#

B# &g3$Ac40! &0$

0# ,a;&=



Reagents…

• Answer?


& $& +ark

B# &g3$Ac40! &0$

0# ,a;&=

&&$



Reagents…

• Addition of B# ;&< and 0# ,a$&! &0$0 will add “&” and “$&” to the alkene ina ,$,-+arkovnikov and S, manner#

B# ;&<

0# ,a$&! &0$0



Reagents…

• Answer?


& $& ,on-+ark S,

B# ;&<

0# ,a$&! &0$0

$&&



Reagents…

• &ydrogenation by the addition of &0!'dC: will add “&” and “&” to the alkenein a S, manner#

&0

'dC:



Reagents…

• Answer?


& & S,

&0

'dC:

&

&



Reagents…

• /ihydro6ylation will add “$&” and“$&” to the alkene in a S, manner#

•

:an use either the addition of >+n$=!,a$&! &0$ $R

• B# $s$= and 0# ,a&S$<! &0$

B# $s$=

0# ,a&S$<! &0$



Reagents…

• Answer?


$& $& S,

$&

$&

B# $s$=

0# ,a&S$<! &0$



Reagents…

• %here are three cyclopropanationreactions#



Reagents…

• Addition of :&0,0 will form a S,cyclopropane ring! with two hydrogenson its ape6#

:&0,0



Reagents…

• Answer?

• S, addition for ring formation#

:&0,0

&&



Reagents…

• Addition of >$&! :&:l< will form aS, cyclopropane ring! with twochlorines on its ape6#

:&:l<

>$&



Reagents…

• Answer?

• S, addition for ring formation

:&:l<:l

:l

>$&



Reagents…

• Addition of :&0I0! Dn3:u4 will form aS, cyclopropane ring! with twohydrogens on its ape6# %he hydrogens

can be replaced by one or two alkylgroups# :&0I0

Dn3:u4

:&3:&<4I0

Dn3:u4



Reagents…

• Answer?

• S, addition for ring formation

:&0I0&

&

Dn3:u4

:&3:&<4I0 &

:&<

Dn3:u4



Reagents…

• Addition of either of these last tworeagents will do an o6idative cleavagereaction! splitting the molecule into

pieces# 5very double bond must be split#

• $"onolysis! B# $< and 0# Dn! &<$! willform aldehydes and ketones#

B # $ <

0 # D n ! & < $ A$

$

&



Reagents…

• In either of these last two reactions!an o6idative cleavage reaction willoccur! splitting the molecule into

pieces# 5very double bond must besplit#

$ $



Reagents…

• Addition of >+n$=! &0$ or &<$ willform carbo6ylic acids! ketones or :$0 3&0:$<4#

> + n $ =

& 0 $ o r & < $ A$

$

$ &

> + n $ =

& 0 $ o r & < $ A& $ $

$

$ &



>now your reagents…

• +ake a single list of all of your reagentsand their important “must know” tidbits…

• )riting this list out will help you

remember all those little facts#• )riting the list multiple times! over and

over again will R5A(( ingrain it in your

memory#• Start working on this early and the night

before the e6am won’t be so painfulE



,ow! put it all together…

• Identify what the reagent is doing tothe alkene

• %hen identify the carbons in your alkene

carbon skeleton• 1ind the two carbons of the alkene in your product skeleton

• /raw in the appropriate parts! withcorrect regiochemistry andstereochemistry#



• (et’s start with a simple e6ample usingthe reagent! &:l# &:l! like all &9reactions! adds one & and one 9 3in

this case! :l4 to the ends of youralkene in a +arkovnikov fashion#

& : l



• Start by identifying the carbons in your alkene carbon skeleton# Attachall of the original carbons in a similar

"ig-"ag format 3unless of course! itsan o6idative cleavage reaction4#



• Identify the two carbons that were inthe original alkene# our reaction willoccur between these two carbons#



• In the e6ample here! the moresubstituted end is on the right?

$ , 5

c a r b o n

a t t a c h e d

% ) $

c a r b o n s

a t t a c h e d



• 1or the addition of &:l! the reactionoccurs +arkovnikov-style with :lattaching to more substituted end#

& : l

: l&



• %his would give you the final productshown below?

• $r simply?

C l

HC lH

C lH

& : l



(et’s try a few more…

• %he addition of 90 adds 9 and 9 in anA,%I fashion#

• %ry this problem?

; r 0



• 1ind the two carbons of the alkene?

; r 0



• (ine up the pieces to be attached#

,otice that if the same piece is beingadded to both ends! you don’t have toworry about regiochemistry#

; r 0

B r

B r



• Attach the pieces?

; r 0

B r

B r



•

And finish with your A,%Istereochemistry?

; r 0

B r

B r



(et’s do another…

• %he addition of 90 and &0$ adds 9 and$&! +arkovnikov style and in an A,%Ifashion#


: l 0 ! & 0 $




: l 0 ! & 0 $



• (ine up the pieces to be attached#+arkovnikov addition places the $& onthe more substituted end! 9 on theless substituted end#

: l 0 ! & 0 $

C l

O H




: l 0 ! & 0 $

O HC l



• And finish with your A,%Istereochemistry?

: l 0 ! & 0 $

O HC l



And another…

• %he addition of &0 with 'dC: adds &and & in a S, fashion#


& 0 ! ' d C :




& 0 ! ' d C :



• (ine up the pieces to be attached# ,oregiochemistry since you are addingthe same thing to both ends#

H

& 0 ! ' d C :

H




& 0 ! ' d C :

H

H



• And finish with your S,stereochemistry?

& 0 ! ' d C :

H

H



A different kind of reaction…

• %he formation of cyclopropane ringsfrom alkenes occurs in a S, fashion#


: & 0 , 0




: & 0 , 0



• 1orm the cyclopropane 3with &’s atthe ape64 across the two carbons ofthe alkene?

: & 0 , 0H

H



• And finish with your S,stereochemistry?

: & 0 , 0H

H



• .sing :&:l< and >$& forms a

cyclopropane with :l’s on the ape6#• .sing :&0I0 and Dn3:u4 forms a

cyclopropane with &’s on the ape6#

Alternatively! you might have alkylgroups! instead of &’s# 1or e6ample! you might use :&3:&<4I0 and form a

cyclopropane that has one & and one:&< group on the ape6# +ore1le6ibilityE



%hose last reactions…

• %he o6idative cleavage reactions splitan alkene into two pieces#

• )hen using the o"onolysis reaction!

the alkene splits to form ketonesandCor aldehydes?

a n d

R

R

H

R

O

R

RO

H

R



• $6idatively cleave the following?

B # $ <

0 # D n ! & < $ A



• It may help to draw in the &’s

attached to the alkene#• %hen S%R5%:& that alkene! and insert

the o6ygen atoms that will form the

aldehydes or ketones# 3%his reactiondoesn’t happen this way but it mayhelp you draw the products#4

O

H H

O



• 1inish the cleavage of both the pi and

sigma bonds#• %he final products should be?

B # $<

0 # D n ! & < $ A

H H

O O

%h fi l f i



%he final set of reactions…

• )hen using the >+n$= reaction! witheither &0$ or &<$! the alkene splitsto form ketones andCor carbo6ylic

acids?a n d

R

R

O H

R

O

R

R

O

O H

R



• $6idatively cleave the following?

> + n $ =

& 0 $ o r & < $ A

H



• It may help to draw in the &’sattached to the alkene#

• %hen S%R5%:& that alkene! and insertthe o6ygen atoms that will form thecarbo6ylic acids or ketones# Any

aldehydes that typically form foro"onolysis are converted to carbo6ylicacids with >+n$=#

O

H O H

O



• %he final products should be?

> + n $ =

& 0 $ o r & < $ A

H O H

OO

$th thi t id



$ther things to consider…

• As you get started! you will noticethat you spend a lot of your timelearning how to draw the products#

• ou should also be able to look at areaction! given the starting material

and product! and determine whatreagent did the transformation shown#



• :onsider this reaction…

•

)hat reagent was used* )hatchanged when the starting moleculeturned into the product molecule*

; r



• @iew the alkene carbons F from both

the perspective of the starting alkeneand the product#• Identify those carbons•

%hen identify what changed on thosecarbons#

; r



• ,otice that a ;r is now attached to the

left-hand side# An e6tra & is nowattached to the right-hand carbon#• Ask yourself what reagent adds an &

and a ;r and does so in such a way thatthe ;r is on the more substituted end*

• &;r! of course…

; r



• &ere’s another e6ample F whatreagent is causing %&IStransformation*

$ &

: l



• Identify the carbons of the alkenes Fwhat pieces are adding to the alkene*

$ &

: l



• %he more substituted carbon 3bottom4

now has an $& group! while the lesssubstituted carbon 3top4 has a :l nowattached#

• %he reagent that was used was :l0 and&0$#

$ &

: l

% fi l ti t thi k b t



%wo final reactions to think about…

• /on’t forget that you need to knowhow to +A>5 an alkene#

• An acid like sulfuric acid or phosphoric

acid is used to dehydrate an alcohol#• A strong base like sodium hydro6ide

or potassium tert-buto6ide! >$t;u! is

used to dehydrohalogenate an alkylhalide#

S th sis



Synthesis…

• Synthesis problems re8uire thestringing together of a series ofreactions as one molecule is turned

into another molecule#• :onsider the following:

; r

; r; r

;;



• Synthesis problems need to be workedfrom both directions! forwards andreverse#

• Identify what’s changing#• ou started with a single ;r# %he product

has two ;r! one on the original position

3note that I didn’t say the SA+5 ;r4 andone on the carbon ne6t door#

; r

; r; r

;;



• ou do ,$% know how to randomly add;r’s to 2ust any carbon#

•

ou actually only know $,5 reaction thatalkyl halides do - dehydrohalogenate#• ou only know $,5 reaction that adds

%)$ halides 3A,%IE4 on two consecutivecarbons#

• 'ut the se8uence together…

; r

; r; r



• 1irst add base 3like >$t;u or ,a$&4

to form an alkene#• %hen add ;r0 to add two ;r’s in an

A,%I fashion#

; r 0

; r

; r; r> $ t ; u



• Alkene reactions will go easier for you

if you start sooner not later forstudying all those bits and pieces youneed to know# %his will also make

practicing problems easier! so you canbetter learn to draw the parts and newmolecules#

•

As always! if this has been helpful!drop me a line at 2discordGtowson#edu

and let me know#

mailto:[email protected]

mailto:[email protected]

a lke ne reactions

Documents

alkene carbons

righthand alkene carbon

lefthand alkene carbon

substituted end

carbon groups

reaction markovnikov

right carbon

left carbon