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Bromobenzene undergoes no reaction for both SN1 and SN2. This is because bromobenzene is very stable, and contains allylic and vinyllic bromine, which is also very stable, and cannot be a good nucleophile.
Bromocyclopentane reacts under SN1 and SN2, but it shows a faster reaction in AgNO3/ethanol reagent, that is SN1. This is because bromocyclopentane is secondary bromine, and have bigger steric strain, since it is a cyclic compound. The bigger steric in a molecule, the harder it is for the nucleophile to attack the leaving group (-Br) from the opposite sides, therefore, SN2 reaction is slower than SN1 for bromocyclopentane.
From this reaction, precipitation occurs to give out AgBr in ethanol and NaBr in acetone.
Bromocyclohexane, on the other hand, shows no reaction in SN2 reagent, but almost an immediate reaction in SN1 reagent. The reasons are very much the same with bromocyclopentane above - steric strain. As bromocyclohexane is bigger than that of bromocyclopentane, the steric in bromocyclohexane is too much for the nucleophillic attack from the opposite side of the leaving Bromine.
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In the end, only SN1 reaction could pass this nucleophillic substitution.
2-bromobutane undergoes a slow SN2 reaction, but shows an immediate result in SN1 reaction. As 2-bromobutane consists of secondary bromine, both SN1 and SN2 could happen. The precipitate for this reaction should be NaBr and AgBr. On the other hand, 2-chlorobutane undergoes an even slower SN1 and SN2 reactions. This happens because bromine is a better leaving group than chlorine. Bromine is larger in diameter, and less electronegative, too; therefore making it easier to leave the alkyl group.
The precipitate for this reaction is NaCl and AgCl.
The reaction of chloroacetone, by comparison is mainly SN2, because the reaction in SN1 is quite slow. As chloroacetone consists of primary chlorine, SN2 reaction is very favorable. Moreover, the stereochemistry of this compound helps the reaction go faster (nucleophile came through from the opposite of leaving group). The product for this reaction is the precipitate of NaCl.
1-chlorobutane is slow in reaction with both, either SN1 or SN2. 1-chlorobutane has primary chlorine; theoretically it is available only for SN2 reaction. But it also reacted in SN1 - only even slower. The reasoning is the same as the above; chlorine makes a hard-to-leave leaving group. The precipitates for 1-chlorobutane in both reagents are NaCl and AgCl.
For t-butyl chloride, no reaction occurred in SN2 reaction, but instant result appears in SN1 reaction. This is mainly because the leaving group, -Cl is a tertiary atom, making it only available for SN1 substitution. Furthermore, there is no carbocation rearrangement. The precipitate is AgCl.
Benzyl chloride reacts instantaneously with SN2 reagent and around several minutes in the other reagent, producing NaCl in the acetone solution and AgCl in the ethanol solution. Different from bromobenzene, benzyl chloride is consisted of a carbon in between the leaving group and the arene. This creates a partial charge, ( on the carbon next to the leaving group that has partial charge. With the attack of nucleophile, the leaving group leaves more easily than the one on bromobenzene. As -Cl in benzyl chloride is primary halide, SN2 substitution is more favorable, and cannot undergo SN1 reaction.
Title: the Diels-Alder reaction of Maleic Anhydride and Anthracene
Introduction: The Diels-Alder reaction is a reaction of creating a six-member ring from an open chain. The two reactants are called Diene and Dienophile ("Diene-seeker"). The goal of this lab was to produce a pure cyclic product from the reaction of maleic anhydride and anthacene.
Procedure: The procedure is followed according to the lab manual.
Result: _refer to the datasheet_.
% recovery of recrystallization:
Mass of crystal before recrystallization: 0.5822 g
Mass of crystal after recrystallization: 0.549 g
% recovery = 94.3 %
Melting point = 262˚C - 266˚C
Conclusion
The reaction of maleic anhydride and anthracene produces 10-dihydroanthracene-9,10-,-succinic acid anhydride. The use of xylene as a solvent was to enable recrystallization as xylene dissolves the two compounds on high temperature only. The IR result shows three significant troughs, that is peaked at 3449.67 cm-1, 3075.19 cm-1 and 2968.73 cm-1. The first trough was the sp and sp2 stretches, where here, I conclude the double in the product.
The second and third trough was probably the C=O bond stretch, where it happens to look like a very broad 'swelling', according to Zubrick, page 287.
Questions
Reaction of water with anhydride
By adding more than the suggested xylene could disable it from recrystallizing, and dissolve the product forever.
The carbonic region in the starting material - maleic anhydride has a very sharp, deep, 'V' shaped trough in the frequency of ~3500 cm-1 and 3100 cm-1. This shows the appearance of the -OCH group in maleic anhydride. On the other hand, the troughs in the product shows a very shallow one compared to the starting material. The stretches are also more to be like, 'U' shaped rather than 'V'.
The specific band in the IR is sp3 in between carbon and carbon, which takes the region around ~3500 cm-1 in the IR. This happens because the double bond in maleic anhydride is opened to make a cyclic compound.
Self-dimerization of cyclopentadine
Title: Nitration of Toluene/Electrophillic Aromatic Substitution
Introduction: The goal for this experiment was to determine the regioselectivity of nitration of toluene. The method used in this lab was reflux, and the products were determined by CG.
Procedure: According to the lab manual.
Results:
_See CG result in attached sheet._
Area of peak 1 (o-nitrotoluene): 11.933
Area of peak 2 - estimation (m-nitrotoluene): 1.5
Area of peak 3 (p-nitrotoluene): 5.833
Ratio of o:m:p = 11.933:1.5:5.833 ~~ 8:1:4
Conclusion
The products of nitration of toluene yield three products, which are o-nitrotoluene, m-nitrotoluene, and p-nitrotoluene. As toluene has a substituent -CH3, which is a donating group in an arene, the product tends to be ortho and para directing. Therefore, o-nitrotoluene and p-nitrotoluene should be the two major product, and m-nitrotoluene should be the least. The percentage for all these compounds is 62%, 8% and 30%, according to ortho, meta, and para respectively.
This is proven in the result of my CG analysis. The area of peak 1 and peak 3 is far larger than the area of peak 2.
Questions
The ration of my products, o:m:p is about 8:1:4, or the percentage, calculated by the amount of area; 62% for ortho-nitrotoluene, 8% for meta-toluene, and 30% for para-nitrotoluene. This ration is definitely what I expected, as ortho-para have far more products than the meta configuration.
Nucleophillic Substitution Reaction. (2016, Aug 03). Retrieved from https://studymoose.com/nucleophillic-substitution-reaction-essay
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