ChemistryChemistry 205205 “Introduction“Introduction toto OrganicOrganic Chemistry”Chemistry” FallFall SemesterSemester 20032003 Dr.Dr. RainerRainer GlaserGlaser
Examination #2 “Making & Breaking Multiple Bonds: Elimination & Addition. Aromatic Compounds and Electrophilic Substitution.”
Wednesday, October 22, 2003, 11 – 11:50 am
Name: Answer Key
Question 1. Multiple Bonds & Aromaticity. 25 Question 2. Making Multiple Bonds: Eliminations. 25 Question 3. Breaking Multiple Bonds: Additions. 25 Question 4. Breaking & Remaking Double Bonds: Aromatic Subst. 25 Total 100
— 1 — Question 1. Multiple Bonds and Aromaticity. (25 points)
(a) For each of the following molecules, draw the complete molecular structure and indicate whether the multiple bonds are isolated, conjugated, or cumulated. Draw the structures nicely, that is, try to draw bond angles that mimic hybridization. (9 points) 1,2-pentadiene Cyclopentadiene 1,4-Pentadiyne
H2C CH3 H2 HC CH H C C C C C C HC CH C H H2 H HC CH isolated cumulated conjugated
(b) “Being aromatic” implies that an unsaturated molecule is planar, fully __conjugated___ and (poly)__cyclic____. In addition, the number of pi-electrons has to be ___4n+2____ (provide the formula) according to the ___Hueckel___-rule (name the rule’s inventor). It is the number of pi- electrons that matters and the number of C-atoms in the ring ______(does, does not) matter and the overall charge of the system ______(does, does not) matter. (6 points)
(c) A C-C single bond is _1.54 Å_ long. A C=C double bond is _1.34 Å_ long. A CC bond in benzene is _1.40 Å_ long; this makes sense because the pi-bond order of each CC bond in benzene is ____ (0, 0.5, 1, 1.5, 2). (Provide bond length values with units, 4 points.)
(d) The sigma framework is shown for benzo[a]pyrene, an important example of a __polycyclic__ __aromatic__ __hydrocarbon___ (abbreviated PAH) well known to be a carcinogen. Draw two resonance forms of benzo[a]pyrene by adding pi-bonds as you see fit. The number of pi-electrons in benzo[a]pyrene is _20_. (6 points)
— 2 — Question 2. Making Multiple Bonds: Eliminations. (25 p.)
(a) HBr elimination from 2-bromo-2-methylbutane with potassium ethoxide, KOCH2CH3, in the solvent ethanol at 70°C produces two products (major ca. 70%, minor ca. 30%). Draw the structures of both products in the correct box! Provide the names of both products. These two products are ___structure___ isomers (2 points). Your choice of the major product was informed by the ____Zaitsev____ rule (2 p.). In the box below, state as briefly as possible what that rule says. MAJOR Product (3 p.) MINOR Product (3 p.)
H
H H 2-methyl-2-butene 2-methyl-1-butene (tri-substituted double bond) (di-substituted double bond) Define “the rule” (no need to explain the reason for the rule, 2 points): Make the product with the more substituted double bond!
(b) 2-Bromo-2-methylbutane is a ______(primary, secondary, tertiary) alkyl bromide. Eliminations from such substrates usually are __uni__-molecular and we denote this kind of reaction mechanism as __E1__. (3 points)
(c) HBr elimination from substrate X with potassium ethoxide, KOCH2CH3, in ethanol produced the two products shown. These two products (are, are not) ______structure isomers, (are, are not) ______geometrical isomers, (are, are not) ______stereoisomers, (are, are not) ______enantiomers, (are, are not) ______diastereoisomers. (5 p.) Provide the names of both products including any descriptor of stereochemistry if needed. Draw the structure of the substrate X (and if there are several suitable substrates, drawing one suffices). Substrate X (3 p.) Product 1 (2 p.) Product 2 (2 p.)
Br H
H E-3-methyl-2-pentene Z-3-methyl-2-pentene
— 3 — Question 3. Breaking Multiple Bonds: Additions. (25 p.)
(a) The structures of the products of major product minor product HBr addition to 1-butene. Circle Br the major product. Your choice of Br the major product was guided by the __Markownikov_ rule. (8 p.)
(b) Consider the “dissolved metal” Reagents Product reduction of 2-hexyne. In the first trans or E-2-hexene box, state what metal is used and what solvent the metal is dissolved Sodium, Na H in. In the other box, draw the structure of the reduction product Ammonia, NH3 and name it (include stereochemical H information if needed). (6 p.)
(c) Bromination of trans-2-butene. This reaction involves Br-Br Perspective Drawing bond _heterolysis, The Br-cation is the ______of Product (electrophilic, nucleophilic, hydraphilic) species that first adds to Me the double bond. The bromide anion adds second from the ______Br H (same, opposite) side. Draw the structure of the product in the box R on the right (3 p.). Name each chiral C-atom as “R” or “S” (1 p. S Me each), and state whether the product is chiral (1 p.). (9 points) H Br not chiral
(d) Dibromides 1 & 2 are ______(identical, enantiomers, diastereoisomers). (2 points) 1 2
— 4 — Question 4. Breaking & Remaking Double Bonds: Aromatic Substitution. (25 points)
(a) Addition/Elimination versus Addition. In the box to the right, draw Br the structure of the product of the FeBr3 catalyzed bromination of benzene. (Consider only monobromination!) This product is formed by the mechanism of _electrophilic aromatic _substitution_. (6 p.)
To the right, draw the structure of the product that would be obtained if
Br2 would add to benzene in the same way Br2 adds to alkenes. (3 p.) Br H H Br Below, explain why the outcomes of the brominations of aromatic compounds and of alkenes differ in this fundamental fashion. (4 p.)
The substitution product stays aromatic. The addition product no longer is aromatic. Loss of benzene resonance energy is avoided.
(b) Bromination of benzene requires a catalyst, such as FeBr3 or AlBr3. This catalyst is a Br Br _Lewis_ acid and it is electron-______(poor, Br Br Fe Br Br Br Fe Br rich). This catalyst helps in the __heterolysis of Br Br the Br-Br bond. Using curved arrows, show the reaction between Br2 and FeBr3. (6 points)
(c) The bromination of nitrobenzene could give 3 products depending as to whether the NO2 bromination occurs in the ortho, meta, or para position. Draw the major product(s) formed. The bromination of nitrobenzene is ______(slower, faster) than the bromination of benzene itself, because the nitro group is ______Br mostly meta (activating, deactivating). (6 points)
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