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Study Guide for Exam 2

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by: Hayley Lecker

Study Guide for Exam 2 CHEM 2325 - 001

Hayley Lecker
GPA 3.42

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Covers topics from practice exam.
Organic Chemistry - 12551
James Salvador
Study Guide
Organic Chemistry, University of Texas at El Paso, Ochem
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This 7 page Study Guide was uploaded by Hayley Lecker on Friday March 18, 2016. The Study Guide belongs to CHEM 2325 - 001 at University of Texas at El Paso taught by James Salvador in Fall 2015. Since its upload, it has received 155 views.


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Date Created: 03/18/16
OrganicChemistry 2 Exam2Study Guide Reactions – From TEST 2014 with explanation after the schemes The major products of the reaction would be, 20. Product : c. The reaction is an electrophilic aromatic substitution with bromine. 21. Product : a. Sulfonation at para position, followed by bromination at ortho position of iso-propyl group. Finally hydrolysis of SO3H to yield product a. 22. Product : b. Nitration at para position, reduction to amine. Bromination at ortho to amine group. Diazotization and substitution with H to form product b. 23. Product : a. Nitration at para posiiton, bromination and meta-to nitro group. Reduction and dizotization followed by substitution with H to give product a. Major Intermediates with Explanations (Exam 2014) 1) NaNH2 are strong bases, very efficient for syn eliminations of organic halides, useful in aromatic, heteroaromatic, and cycloalkenyl elimination-addition reactions; initiate anionic polymerizations. So, for (15). Br elimination takes place and that electron forms arynes as intermediate thus and intermediate is (b) for 15. 2)Treatment of an aqueous solution of diazonium salt with fluoroboric acid under cold conditions gives diazonium fluoroborate as precipitate, which could be dried and gently heated to afford the flurobenzene by decomposition. The reaction involves SN1 mechanism, So, answer for (16) is (c) 3) Sandmeyer Reaction: This method provides an effective route for the preparation of aromatic bromides and chlorides. Addition of cold aqueous solution of diazonium chloride to a solution of CuBr gives a sparingly soluble complex which is separated and heated to give aryl bromide by decomposition. So, answer for 17 is (e) where X is Br- or Cl- or CN 4) Allowing bromine to react with iron metal first generates FeBr3, which then interacts with the remaining Br2 to form a highly polarized system. It is this highly polarized bromine that becomes a source of “Br+.” The reaction proceeds by the mechanism shown above to give brominated benzene, So (18) answer is (d) 5) The presence of electron-withdrawing groups (such as nitro) ortho and para to the bromine substantially enhance the rate of substitution. mechanism is characterized by initial addition of the nucleophile (hydroxide ion or water) to the aromatic ring, followed by loss of a halide anion from the negatively charged intermediate. So 19 answer is (a) where W is NO2, Instead of Cl write Br, Nu: is OH- (Nucleophile) Nitration Problem14 on Exam 2014 12. Which position will be substituted in the major product of the reaction of HNO /3 S2 an4 O c b Br a d e ? b In the presence of a strong electron withdrawing group, the incoming nucleophile aka NO2+ is substituted at the meta position. In the given substrate one of the meta positions is bonded with the Br atom, therefore it has to go to the other meta position. Reactions with Br2/FeBr3 Bromine in presence of lewis acid such as FeBr3 acts as brominating agent. Aromatic compounds containing benzene ring reacts with Br2/FeBr3 to form corressponding bromo derivatives. The reaction is an example of aromatic electrophilic substitution. Presence of electron donating group on benzene ring makes it more electrophilic, thus, increases the rate of reaction. Out of the five given compounds, only compound e has electron donating dimethyl amino group, which makes the aromatic ring more electron rich resulting in increased rate of bromination. Compound a has trimethyl ammonium group carrying positive charge, which makes aromatic ring electron deficient. Compound b, c and d has electron withdrawing groups, thus, they deactivate the aromatic ring towards aromatic electrophilic substitution. Bond Lengths – Exam 2014 b c e a d Longest bond of styrene is d. The other bonds a,b and c are between sp3 and sp2 bond lengths. Bond e is between two sp2 carbons with a pi-bond. Bond d has more single bond character among all and thus is the longest. Also remember in an aromatic ring the pi system is moving so A and C could have c-c double bond properties, d is the only true single bond in this molecule. Aromatic, Antiaromatic, Non-aromatic Based on the properties of aromatic compounds, there are FOUR criteria about the π system that need to be met inorder for the "special" aromatic stabilisation to be observed: 1. Conjugated : there needs to one "p" orbital from each atom in the ring, so each atom must be either sp or sp hybridised. 2. Cyclic : linear systems are not aromatic, all atoms in the ring must be involved in the π system (i.e. no sp atoms) 3. Planar : if the ring is planar flat then this means there is good overlap / interaction between the "p" orbitals....not always easy to consider. 4. The Huckel Rule..... 4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.) This is equivalent to an odd number of π-electrons pairs). N this is an example of a nonaromatic compound, it has 10 pi electrons so it is a 4n+2 compound, however since the ring is bigger than 5/6 carbons it won’t be flat. Diels-Alder Reactions The Diels Alder reaction converts a diene and an alkene (usually electron-poor, called a “dienophile”) into a six-membered ring containing an alkene (cyclohexene). Mechanism: Example Problem 5 on Exam 2014 This reaction is a 1,4 type. Br is an electronegative atom so they would not want to be near each other. Also the double bond here is tri-substituted. HOMO LUMO LUMO lowest  LUMO receives electrons unoccupied  lowest energy orbital available molecular  characteristic for electrophilic component orbital  electrons from the HOMO are donated HOMO highest occupied  most available for bonding  most weakly held electrons molecular  characteristic for nucleophilic component orbital For Exam 2014 the reason it was E because of the isolation, the further away pi bonds are away from each other the more isolation there is and the higher the gap. Adjacency Matrix To understand the number of 1’s you have to look at the pi system and how the pi bonds can move around the ring. An example is pyrrole, the double bond can move around the ring to the nitrogen, then around to ring to different carbons so 5 places then times by 2 to get the magic number of 10 for 1’s in the matrix. N H


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