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By Mrs. Azduwin Khasri 23rd October 2012

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1 By Mrs. Azduwin Khasri 23rd October 2012
ALKYLHALIDE By Mrs. Azduwin Khasri 23rd October 2012


3 ELIMINATION REACTION Elimination reactions involve the loss of elements from the starting material to form a new  bond in the product. The product of elimination reaction is an Alkene

4 E1 AND E2 REACTION E2 E1 E-Elimination E-Elimination 2-Bimolecular

5 E2 REACTION Hydroxide cannot act as a nucleophile in this reaction because of the bulky tertiary halide. Rather, hydroxide acts as a base and abstracts a proton.

The removal of a proton and a halide ion is called dehydrohalogenation

Carbon attached to halogen Carbon adjacent to α-carbon An E2 reaction is also called a b-elimination or a 1,2-elimination reaction

8 The weaker the base, the better it is as a leaving group

9 2 structurally β-carbon
The Regioselectivity of the E2 Reaction 2 product produce 2 structurally β-carbon base The major product of an E2 reaction is the most stable alkene

10 Reaction coordinate diagram for the E2 reaction
2-Butene formed faster than 1-butene

11 The Zaitsev Rule The more substituted alkene product is obtained when a proton is removed from the b-carbon that is bonded to the fewest hydrogens 2-β-Hydrogens 3-β-Hydrogens Monosubstituted Disubstituted

12 The more substituted alkene is not always the most stable

13 Conjugated alkene products are preferred over the
more substituted alkene product: Do not use Zaitsev’s rule to predict the major product If the alkylhalide has a double bond or a benzene ring.

14 The anti-Zaitsev Rule (Hoffmann Product)
Bulky Alkyl halide Zaitsev product Hofmann product Bulky bases affect the product distribution resulting in the Hofmann product, the least substituted alkene Less sterically hindered Major product- Most stable product


16 Another exception to Zaitsev’s rule
When the halogen is fluorine-The major product of the E2 reaction of alkyl Fluoride is the less substituted alkene. STRONGEST BASE-POOREST LEAVING GROUP The Fluoride ion does not have as strong a propensity to leave as another halide ion.

17 Consider the elimination of 2-fluoropentane…
Major product-Less substituted alkene A carbanion-like transition state

18 Carbocation vs Carbanion
Carbocation stability 3° > 2° > 1° Carbanion stability 1° > 2° > 3°

19 CLASS EXERCISE 1-(E2) Which of the alkyl halides is more reactive in an E2 reaction? Producing more stable alkene Br better leaving group Producing more stable alkene

Give the major elimination product obtained from an E2 reaction of the following alkyl halides with hydroxide ion: ANSWER:

21 E1 REACTION First order elimination reaction
Must have at least two step

22 Mechanism of E1 reaction
Alkylhalide dissociates,forming carbocation A Base removes a proton from β-carbon

23 How does a weak base like water remove a proton from
an sp3 carbon? 1) The presence of a positive charge greatly reduces the pKa 2) Hyperconjugation weakens the C-H bond by electron density

24 The Regioselectivity of the E1 Reaction
Fewest β-Hydrogen (According to Zaitsev rule) The major product in an E1 reaction is generally the more substituted alkene

25 Reaction coordinate diagram for the E1 reaction

26 Because the first step is the rate-determining step, the rate of an
E1 reaction depends both on the ease with which the carbocation is formed and how readily the leaving group leaves

27 E1 AND E2 REACTION Major product generally the most stable alkene
Tertiary alkyl halides are the most reactive and the primary alkyl halides are the least reactive. For alkyl halide with the same alkyl group, alkyl iodide are the most reactive and alkyl fluoride are the least reactive.

28 Because the E1 reaction forms a carbocation
intermediate, we need to consider carbocation rearrangement

29 1° Carbocation are too unstable to be formed in E1 reaction
Competition Between E2 and E1 Reactions 1° Carbocation are too unstable to be formed in E1 reaction An E2 is favored by a high concentration of strong base and an aprotic polar solvent An E1 is favored by a weak base and a protic polar solvent

30 Parallel on the opposite side Parallel on the same side
Stereochemistry of the E2 Reaction The bonds to the eliminated groups (H and X) must be in the same plane Parallel on the opposite side Parallel on the same side Syn elimination Anti elimination The anti elimination is favored over the syn elimination

31 Consider the stereoselectivity of the E2 reaction
(E)- : the higher priority groups are on opposite sides of the double bond. (Z)- : the higher priority groups are on the same side of the double bond. Syn-elimination Anti-elimination The alkene with the bulkiest groups on opposite sides of the double bond will be formed in greater yield, because it is the more stable alkene

32 Syn-elimination Anti-elimination

33 When only one hydrogen is bonded to the b-carbon, the
major product of an E2 reaction depends on the structure of the alkene

34 Stereochemistry of the E1 Reaction
The major stereoisomer obtained from an E1 reaction is the alkene in which the bulkiest substituents are on opposite sides of the double bond Both syn and anti elimination can occur in an E1 reaction, both E and Z formed In contrast,E2 forms both E and Z only if the β-carbon is bonded to 2 Hydrogen. If β-carbon bonded to only 1 Hydrogen,E2 form only 1 product because anti elimination favored.

35 Substitution and Elimination
Competition Between Substitution and Elimination Alkyl halides can undergo SN2, SN1, E2, and E1 1) HO- is called nu in a substitution reaction (Attack carbon) and a base in elimination reaction (removes proton). 2) Decide whether the reaction conditions favor SN2/E2 or SN1/E1 SN2/E2 reactions are favored by a high concentration of a good nucleophile/strong base SN1/E1 reactions are favored by a poor nucleophile/weak base 3) Decide how much of the product will be the substitution product and how much of the product will be the elimination product

36 SN2/E2 conditions

37 A bulky alkyl halide or a sterically hindered nucleophile
encourages elimination over substitution

38 A strong or a bulky base encourages elimination over

39 High temperature favors elimination over substitution:
Why? Because elimination is entropically favorable.

40 Tertiary alkyl halides undergo only elimination under
SN2/E2 conditions:

41 SN1/E1 conditions The elimination reaction favored at higher temperatures. Primary alkyl halides do not form carbocations; therefore they cannot undergo SN1 and E1 reactions.

Submit on 6th November (Tuesday)

43 -The End-

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