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There is a smooth transition between the reactants and products. Analogous to the S N 2 size 12{S rSub { size 8{N} } 2} {} reaction, no intermediate has been isolated or detected. Furthermore, no rearrangements occur under E2 conditions. This situation is in marked contrast to E1 elimination reactions, where carbocation intermediates are generated and rearrangements are frequently observed.

The alkyl halide adopts an anti-periplanar conformation in the transition state, and experimental evidence demonstrates that if the size of the base is increased, then it becomes more difficult for the large base to abstract an internal β size 12{β} {} -hydrogen atom. In such cases, the base removes a less hindered β size 12{β} {} -hydrogen, leading to the predominance of the thermodynamically less stable (terminal) alkene in the product mixture. This type of result is often referred to as anti-Zaitsev or Hofmann elimination. Thus, in the reaction of 2-bromo-2,3-dimethylbutane given above, 2,3-dimethyl-1-butene would be the major product (anti-Zaitsev) if the conditions involved use of a bulkier base. The anti-periplanar conformations are illustrated in the Newman projections below:

Dehydrohalogenation of alkyl halides in the presence of strong base (E2) is often accompanied by the formation of substitution ( S N 2 size 12{S rSub { size 8{N} } 2} {} ) products. The extent of competitive substitution depends on the structure of the alkyl halide. Primary alkyl halides give predominantly substitution products (the corresponding ether), secondary alkyl halides give predominantly elimination products, and tertiary alkyl halides give exclusively elimination products. For example, the reaction of 2-bromopropane with sodium ethoxide proceeds as follows:

In general, for the reaction of alkyl halides with strong base:

Physical properties of reactants: (table 1.1)

Compound MW Amount mmol bp (°C) D np

2-Bromobutane 137.03 100 mL 0.92 91.2 1.26 1.4366

Methanol 32.04 3.5 mL 64.9 0.791 1.3288

2-Propanol 60.09 3.5 mL 82.4 0.785 1.3776

2-Methyl-2-propanol 74.12 3.5mL 82-83 0.786 1.3838

(tertbutanol)

3- Ethyl-3-pentanol 116.20 3.5 mL 140-142 0.839 1.4266

Sodium 22.98 60 mg 2.6 883 0.97

Potassium 39.10 60 mg 1.5 760 0.86

Reagent combinations: (table 1.2)

Alcohol Solvent Metal Alkoxide Base Produced

Methanol Sodium Sodium methoxide

2-Propanol Potassium Potassium 2-propoxide

2-Methyl-2-propanol Potassium Potassium 2-methyl-1-2-

(tertbutanol) propoxide (potassium

tert- butoxide)

3-Ethyl-3-pentanol Potassium Potassium 3-ethyl-3-

pentoxide

Reagents and equipment:

The combinations of reagents in Table 1.2 may be used to prepare the alkoxide base. Students should compare results to observe a total picture of the effect.

Reaction conditions: (table 1.3)

Temperature Conditions Base Temperature (°C)

NaOCH 3 size 12{ ital "NaOCH" rSub { size 8{3} } } {} 100-110

KOCH ( CH 3 ) 2 size 12{ ital "KOCH" \( ital "CH" rSub { size 8{3} } \) rSub { size 8{2} } } {} 130-140

KOC ( CH 3 ) 3 size 12{ ital "KOC" \( ital "CH" rSub { size 8{3} } \) rSub { size 8{3} } } {} 140-150

KOC ( CH 2 CH 3 ) 3 size 12{ ital "KOC" \( ital "CH" rSub { size 8{2} } ital "CH" rSub { size 8{3} } \) rSub { size 8{3} } } {} 175-180

Important table to memorize:

Prelab: the e2 elimination (total 10 points)

On my honor, in preparing this report, I know that I am free to use references and consult with others. However, I cannot copy from other students’work or misrepresent my own data.

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Questions & Answers

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SUYASH
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s. Reply
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Ebrahim
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Ebrahim
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s.
Graphene has a hexagonal structure
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Cied
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Porter
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Chem 215 spring08. OpenStax CNX. Mar 21, 2008 Download for free at http://cnx.org/content/col10496/1.8
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