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 Structure, nomenclature of alkenes and naming alkenes using the E,Z system:
DEFINE and ILLUSTRATE the principle in naming alkenes on few examples

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
Hydrocarbons that contain a carbon-carbon double bond.
Important roles in biology, e.g ethene- plant hormoneeffect seed germination, flower germination/fruit ripening.

 The general formula of alkenes: C n
H
2n
– as a result of double bond, alkene has 2 fewer
H than alkane.
 The general formula for cyclic alkene:
C n
H
2n-2
.
 Double bond is the functional group of an alkene, its presence is denoted by the suffix “ene”.
Eg ethene, propene.

1. Longest continuous chain containing the functional group:
- The functional group is numbered in the direction that gives the suffix the lowest possible no.

2. For compound with 2 double bonds, the suffix is ‘diene’
3. Name of substituent is stated before the name of the longest continuous chain that contains the functional group, together with a no to designate the carbon to which the substituent is attached.
Note: When there are both a functional group and a substituent, the functional group gets the lowest number

4. If a chain has more than 1 substituent , the substituents are stated in alphabetical order . Appropriate no is assigned to each substituent.
5. If counting in either direction results in the same no for the alkene functional group, Name with the lowest functional group number and then the lowest substituent numbers :

6. A number is not needed to denote the position of the double bond in a cyclic alkene- the ring is always numbered so that the double bond is between C1 and C2.
To assign number to substituents, count around the ring in the direction that put the lowest no into the name.

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
The sp 2 carbons of alkene called vinylic carbons sp 3 carbon that is adjacent to vinylic carbon called an allylic carbon.
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Two groups containing a C-C double bond are used in common names- the vinyl group and allyl group.
Vinyl group- smallest group containing vinylic carbon
Allyl group- smallest group containing allyllic C.

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When ‘vinyl’ or ‘allyl’ is used, the substituent must be attached to the vinylic or allylic C.
A hydrogen bonded to a vinylic C- vinylic H
A hydrogen bonded to a allylic C- allylic H

 What is the IUPAC name for the following compound?
A) 5-methylcyclohexene
B) 4-methylcyclohexene
C) 1-methyl-3-cyclohexene
D) 1-methyl-4-cyclohexene
E) methylcyclohexene

 Which of the following is an allylic alcohol?
A) CH
2
=CHCH
2
OCH
3
B) CH
2
=CHCH
2
CH
3
C) HOCH=CHCH
2
CH
3
D) CH
3
CH=CHCH
2
OH
E) CH
2
=CHCH
2
CH
2
OH

• Each double bond of an alkene has 3 sp 2 orbitals overlaps an orbital of another atom to form a bond.
Thus one of the C-C bonds in a double bond is a σ bond.
• The 2 nd C-C bond in the double bond (the π bond) is formed from side-to-side overlap of the remaining p orbital of 1 C with sp 2 C with the remaining p orbital of the other sp 2 C.

 Since 3 points determine a plane, each sp 2 C and the two atoms singly bonded to it lie in a plane.
(a plane-defined as sp 2 hybridized carbon is bonded to 3 atoms)
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
In order to achieve max orbital-orbital overlap, the
2 p orbitals must be parallel.
Therefore, all 6 atoms of the double bond system are in the same plane.

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
To achieve max overlap- 2 p orbitals forming the π bond must be parallel.
Therefore rotation about double bond does not readily occur. If rotation to occur, the 2 p orbitals would cease to overlap and the π bond would break. the energy barrier to rotation about a double bond is high.

 Because of the high energy barrier to rotation about C-C double bond, an alkene such as 2-butene can exist in 2 distinct form. The H bonded to the sp 2 C can be on the same side of double bond or on opposite side of double bond.
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 Compound with Hs on the same side of double bondcis isomer
Hs on the opposite side of double bondtrans isomer
Cis isomer- substituents on the same side of the ring
Trans isomer- substituents on opposite sides of the ring.

 If one of the sp 2 Cs of the double bond is attached to 2 identical substituents, there is only 1 possible structure for the alkenes.
 Eg: CH
3
CH
2
C=CHCH
3
CH
2
CH
3

 Because of the energy barriercis and trans isomers of alkenes cannot interconvert
(except under extreme conditions that can overcome the energy barrier)
 They can be separated from each other
 The 2 isomers are different compounds with different physical properties ie. bp and different dipole moments

 as long as each of the sp 2 carbon of an alkene is bonded to only 1 substituent, we can use the term cis and trans to designate the structure of alkenes
How to designate the isomers such as
1-bromo-2-chloropropene?

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
The E,Z system of nomenclature for alkenes with 2 substituents on 1 or both of the sp 2 carbons.
To name an isomer by E, Z system- determine the relative priorities of the 2 groups bonded to 1 sp 2 C
Then, the relative priorities of the 2 groups bonded to the other sp 2 carbon.
E,Z is from German:
Z, Zusammen (together) E, Entgegen (opposite
)

 The relative priorities of the 2 groups bonded to a sp 2 carbon are determined using the following rules:
1. Relative priorities of the 2 groups depend on the atomic number of the atoms bonded directly to the sp 2 C.
the greater the no, the higher priority.

2. If the 2 groups bonded to sp 2 C start with the same atom , then move outward from the point of attachment and consider the atomic no of the atoms that are attached to the ‘tied’ atoms.
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The C of the CH
2
Cl group is bonded to Cl, H, H and the C of the
CH
2
CH
2
Cl group is bonded to C, H, H.
Cl has a greater atomic number than C , so CH
2
Cl group has higher priority.
In CH
2
OH group and CH(CH
3
)
2 group, both atoms are bonded C.
The C of CH
2
OH group is bonded to O, H and H and the C for
CH(CH
3
)
2 group is bonded to C, C and H.
Of these 6 atoms, O has the greatest atomic number so CH
2
OH has a higher priority.

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3. Multiple bonds are treated as attachment of multiple single bonds using “divide-duplicate.”
If an atom is doubly bonded to another atom, the priority system treats it as if it were singly bonded to two of those atoms.
If an atom is triply bonded to another atom, the priority system treats it as if it were singly bonded to 3 of those atoms
-
-
because the Cs immediately bonded to the sp 2 carbon on the left are both bonded to C, H and H- ignore them-look at the groups that attached to them.
1 of the group is CH
2
OH and the other is C CH.
1 C is bonded to H, H and O; the triple bonded C is considered to be bonded to C, C and C. of the 6 atoms, O has the greatest atomic number , so CH
2
OH has higher priority .

 At the other sp 2 C- both atoms are Cs.
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The 1 st carbon of the CH
2
CH
3
H and H.
group is bonded to C,
1 st C of the CH=CH
2 is bonded to an H and doubly bonded to a C, so it is considered to be bonded to
H, C and C.
1 C cancels in each groups (cancel atoms that are identical in the 2 groups), leaving H and H in
CH
2
CH
3 group and H and C in the CH=CH
2
. C has greater atomic no than H, so CH=CH priority.
2 has greater

4. If 2 isotopes (atoms with same atomic number but different mass no) are compared, mass number is used to determine priorities.
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Deuterium (D) and H have same atomic number but different mass , D has higher priority than H.
the C that are bonded to the other sp 2 C are both bonded to C, C and H so the next atoms are looked to.

 Provide the proper IUPAC name for the alkene shown below.
a)
(E)-3,4-dimethyl-3-heptene b) (Z)-4-ethyl-3-methylheptene

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Alkenes are more reactive than alkanes due to the presence of
π bond.
The bond has high electron density or is electron rich site and susceptible to be attacked by electrophiles (electron deficient species/low electron density).
Alkenes undergo ADDITION reaction which means the C=C are broken to form C-C bonds.

 If the electrophilic reagent that adds to an alkene is a hydrogen halide  the product of the reaction will be an alkyl halide :
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Alkenes in these reactions have the same substituents on both sp 2 carbons, it is easy to predict the product of the reaction
The electrophile (H + nucleophile (X -
) adds to 1 of the sp
) adds to the other sp 2
2 carbons, and the carbondoesn’t matter to which C it will attach to- same product.

- Mechanism of reaction.
• Arrow shows- 2 electrons of the π bond of the alkene are attracted to the partially charged H of HBr.
• π electrons of the alkene move toward the H, the H-Br bond breaks, with Br keeping the bonding electrons

• Notice that π electrons are pulled away from 1 C, but remain attached to the other.
• Thus, the 2 electrons that originally formed the π bond – form a new σ bond between C and the H from HBr.
• The product is +vely charged since the sp 2 C that did not form a bond with H has lost a share in an electron pair.
• In 2 nd step of reaction: a lone pair on the –vely charged bromide ion forms a bond with the +vely charged C of the carbocation.

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1 st step of reaction: the addition of H + to a sp 2 carbon to form either tert-butyl cation or isobutyl cation.
Carbocation formation- rate-determining step
If there is any difference in the formation of these carbocations- the 1 that formed faster will be the predominant product of the first step.
Since carbocation formation-rate determining step, carbocation that is formed in 1 st step, determines the final product of reaction.

Since the only product formed is tert-butyl chloride- tert-butyl cation is formed faster than isobutyl cation.

tert
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 Factors that affect the stability of carbocationsdepends on the number of alkyl groups attached to the +vely charged carbon.
Carbocations are classified according to the carbon that carries the +ve charge.
Primary carbocation- +ve charge on primary C
Secondary carbocation+ve charge on secondary C
Tertiary carbocation- +ve charge on tertiary C

 Thus, the stability of carbocations increases as the no of alkyl substituent attached to +vely charged carbon increases.
Alkyl groups decrease the concentration of positive
Charge- makes the carbocation more stable!

• In ethyl cation: the orbital of an adjacent C-H σ bond can overlap the empty p orbital (empty p orbital: positive charge on a carbon)
• Note: no such overlap is impossible for methyl cation.
• Movement of electrons from the σ bond orbital toward the vacant p orbital decreases the charge on the sp 2 carbon- causes a partial positive charge to develop on the atoms bonded by the σ bond
• Thus, the +ve charge is no longer concentrated on 1 atom but is delocalized (spreading out).
• The dispersion of positive charge stabilizes the carbocation because a charged species is more stable if its charge is spread out.
• Delocalization of electrons by overlap of a σ bond orbital with empty p orbital on an adjacent carbonhyperconjugation .

 Which of the following is the most stable carbocation?

 When alkene has different substituents on its sp 2 carbons, undergoes electrophilic addition reaction - the electrophile can add to
2 different sp 2 carbonsresult in the formation of more stable carbocation .

• In both cases, the major product- that results from forming the more stable tertiary carbocation- it is formed more rapidly.
• The 2 nd products known as constitutional isomers – same molecular formula, differ in how their atoms are connected.
• A reaction in which 2 or more constitutional isomers could be obtained as products but 1 of them are predominatesregioselective reaction.
• 3 degrees of regioselective:
Moderately regioselective
highly regioselective
completely regioselective

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Completely regioselective:
1 of the possible products is not formed at all
For E.g: addition of a hydrogen halide to 2-methylpropane- 2 possible carbocations are tertiary and primary.
Addition of a hydrogen halide to 2-methyl-2-butene- 2 possible carbocations are tertiary and secondary- closer in stability.
Addition of HBr to 2-pentene- not regioselective. Because the addition oh H + to either of the sp 2 carbons produces a secondary carbocation- same stability so both are formed with equal ease.
CH
3
CH=CHCH
2
CH
3
+ HBr  CH
3
CHCH
2
CH
2
CH
3
+ CH
3
CH
2
CHCH
2
CH
3
Br
2-bromopentene
50%
Br
3-bromopentene
50%


Markovnikov’s rule : the electrophile adds to the sp 2 carbon that is bonded to the greater number of hydrogens.
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In the above reaction, the electrophile (H + ) adds preferentially to C-1 because C-1- bonded to 2 H.
C-2 is not bonded to H.
Or we can say that: H + adds to C-1 bacause it results in the formation of secondary carbocation, which is more stable than primary carbocation- would be formed if H + added to C-2.

 What alkene should be used to synthesize 3-bromohexane?
? + H-Br  CH
3
CH
2
CHCH
2
CH
2
CH
3
Br
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Solution:
1.
List the potential alkenes that can be used to produce 3-bromohexane.
Potential alkenes  2-hexene and 3-hexene
2. Since there are 2 possibilities- deciding whether there is any advantage of using 1 over the other

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The addition of H + to 2-hexene- form 2 different carbocations- both secondary, same stability- equal amounts of each will be formed.
½ 3-bromohexane and ½ 2bromohexane.
The addition of H + to either of the sp 2 carbons of 3-hexene- forms the same carbocation because the alkene is symmetrical. Thus, all product will be 3bromohexane.
Therefore, 3-hexene is the best alkene to use to prepare 3-bromohexane.

 What alkene should be used to synthesize 2-bromopentane?

 Sometimes, in the electrophilic addition reactions, the products obtained are not as expected.
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 For eg: the addition of HBr to 3-methyl-1butene forms 2 products.
2-bromo-3-methyl butane (minor product)predicted
2-bromo-2-methylbutaneproduct- major product unexpected

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
F.C. Whitmore- 1 st to suggest that the unexpected products results from a rearrangement of the carbocation intermediate.
Carbocations rearrange if they become more stable as a result of the rearrangement.

 Result of the carbocation rearrangement2 alkyl halides are formed
 1 st from the addition of the nucleophile to the unrearrange carbocation and
 1 st from the addition of the nucleophile to the rearranged carbocation- major product.
 Because it entails the shifting of H with its pair of electrons- the rearrangement is called a hydride shift (1,2-hydride shift). The hydride ion moves from 1 carbon to an adjacent C.

• In this reaction, after 3,3-dimethyl-1-butene acquires an electrophile to form a secondary carbocation, one of the methyl groups, with its pair of electrons shifts to the adjacent +vely charged C to form a stable tertiary carbocation.
• 1,2-methyl shift
• Major product- is the most stable carbocation.

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If a rearrangement does not lead to a more stable carbocation, then the rearrangement does not occur.
For eg: when a proton adds to 4-methyl-1-pentene, a secondary carbocation is formed.
A 1,2-hydride shift would form a different secondary carbocation- but since both carbocations are equally stableno advantage to the shift. Rearrangement does not occur.

 Carbocation rearrangements also can occur by ring expansion- another type of 1,2-shift.
 Ring expansion produces a carbocation that is more stable because it is tertiary rather than secondary - five-membered ring has less angle strain than 4-membered ring.

 Which of the following carbocations would be expected to rearrange?

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The halogens Br
2 and Cl
2 add to alkenes.
It is not immediate apparent- electrophile
Electrophile- necessary to start electrophilic addition reaction
Reaction is possible- the bond joining the 2 halogen atoms is relatively weak- easily broken.

• As the electrons of the alkene approach a molecule of Br
2
, 1 st of the Br atoms accepts those electrons and releases the electrons of the Br-Br bond to the other Br atom
• Br atom acts as nucleophile and electrophileadds to the double bond in a single step.
• The intermediate- unstable because there is considerable +ve charge on the previously sp 2 carbon.
• Thus, the cyclic brominium ion reacts with a nucleophile, the bromide ion
• product is vicinal dibromide. Vicinus: near

• The product for 1 st step: cyclic bromonium ion. NOT carbocation- Br electron cloud is close to the other sp 2 carbonform a bond.
• Bromonium ion more stable- its atom have complete octets.
• Positively charged carbon of carbocation – does not have complete octet.

 Cl
2 adds to an alkene- a cyclic chloronium ion is formed.
 Final product- vicinal dichloride
The same reaction with chlorine affords a chloronium ion:
 If H
2
O rather than CH
2
Cl
2 is used as solventmajor product will be a vicinal halohydrin
 Halohydrin- organic molecule that contains both halogen and an OH group.

• Mechanism for halohydrin formation- 3 steps.
• 1 st step: a cyclic bromonium ion/chloronium is formed in the 1 st step because Br+/Cl+ the only electrophile in the reaction mixture
• 2 nd step: the unstable cyclic brominium ion rapidly reacts with any nucleophile it bumps into. 2 nucleophiles present: H
2
O and Br , but because H
2
O is the solvent, its conc > Br-. Tendency to collide with
H
2
O is more.
• The protonated halohydrin is strong acid- so it loses proton.

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Notice that in the preceding reaction, the electrophile
(Br + ) end up on the sp 2 carbon bonded to the greater no of H. why?
In the 2 nd step of reaction: the C-Br bond has broken to a greater extent than the C-O bond has formed.
As a result, there is a partial positive charge on the carbon that is attacked by the nucleophile.

• Therefore, the more stable transition state is the 1 achieved by adding the nucleophile to the more substituted sp 2 carbon- carbon bonded to fewer H.
• because, in this case the partial +ve charge is on a secondary carbon rather than on a primary carbon.
• thus, this reaction too follows the general rule for electrophilic addition reaction: the electrophile (Br + ) adds to the sp 2 carbon that is bonded to the greater no of H.

 When nucleophiles other than H
2
O are added to the reaction mixture- change the product of reaction, from vicinal dibromide to vicinal bromohydrin
 Because, the concentration of the added nucleophile will be greater than the conc of halide ion (Br
2
/Cl
2
)- the added nucleophile most likely to participate in the 2 nd step reaction.

 Complete the following reaction and provide a detailed, step-by-step mechanism for the process.
 Answer: