Examination%20_1%20Spr%202008

Examination%20_1%20Spr%202008 - CHEM 235 – Organic...

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Unformatted text preview: CHEM 235 – Organic Chemistry II Spring 2008 Mid‐Term Examination #1 Feb. 8th, 2008 This examination is worth 100 points out of a possible total of 700 points for this course, and consists of 8 questions. You are required to answer all questions, but are not necessarily required to answer all parts of a given question, so read the questions carefully. In cases where you are not required to answer all parts, indicate clearly which parts you wish to have graded. Otherwise, I will grade the fist answers I see. Question 1 (15 points) Provide systematic names for three of the following five compounds. O2N Br Cl CH3 Br Br OH NH2 Cl Question 2 (12 points). Predict, with reference to the relevant reaction mechanism, the outcome of the following reaction (including stereo‐ and regiochemical outcomes). You must include the detailed reaction mechanism in your answer, and may include other structures or diagrams to explain your answer. NBS, hν CCl4 Question 3 (10 points). Explain why 1,3‐cyclopentadiene has a pKa of only 16, whereas 1,3,5‐ cycloheptatriene has a pKa of above 40. H H H pKa > 40 pKa = 16 H Question 4 (8 points). Indicate which of the following compounds are aromatic. N N NH O O N O O BF4 Question 5 (12 points). Predict, with reference to the relevant reaction mechanism, the outcome of the following reaction (including stereo‐ and regiochemical outcomes). You must include the detailed reaction mechanism in your answer, and may include other structures or diagrams to explain your answer. C N Cl2 FeCl3 Question 6 (12 points). Predict, with reference to the relevant reaction mechanism, the outcome of the following reaction (including stereo‐ and regiochemical outcomes). You must include the detailed reaction mechanism in your answer, and may include other structures or diagrams to explain your answer. OCH3 HNO3/H2SO4 ∆ Question 7 (21 points). Predict the outcome of seven of the following ten reactions, including any stereochemical or regiochemical outcomes. Br2, hν NBS, AIBN CCl4, ∆ 1) NaNH2/NH3 2) I NO2 HNO3/H2SO4 ∆ O + Cl AlCl3 SO3/H2SO4 ∆ OCH3 NO2 CH3 OCH3 CH3 SnCl2/HCl Cl2 FeCl3 KOH + CHCl3 OCH3 + Br AlCl3 Question 8 (10 points). Provide structures for two of the following three names p‐chloronitrobenzene 1,3,5‐trimethyl‐2‐chlorobenzene m‐nitrotoluene BONUS QUESTION (10 points). Design a rational synthesis for the following: and starting from O2N Cl You may use any reagents you deem appropriate, and you may use the back of the page. CHEM 235 – Organic Chemistry II Spring 2008 Mid‐Term Examination #1 – ANSWER KEY Feb. 8th, 2008 This examination is worth 100 points out of a possible total of 700 points for this course, and consists of 8 questions. You are required to answer all questions, but are not necessarily required to answer all parts of a given question, so read the questions carefully. In cases where you are not required to answer all parts, indicate clearly which parts you wish to have graded. Otherwise, I will grade the fist answers I see. Question 1 (15 points) Provide systematic names for three of the following five compounds. O2N 1‐isobutyl‐2‐methyl‐4‐nitrobenzene 1‐(2‐methylpropyl)‐2‐methyl‐4‐nitrobenzene Br Cl CH3 2‐bromo‐1‐chloro‐4‐methylbenzene 3‐bromo‐4‐chlorotoluene Br p‐dibromobenzene 1,4‐dibromobenzene Br OH 2,4,6‐trimethylphenol NH2 Cl 2‐chloroaniline o‐chloroaniline Question 2 (12 points). Predict, with reference to the relevant reaction mechanism, the outcome of the following reaction (including stereo‐ and regiochemical outcomes). You must include the detailed reaction mechanism in your answer, and may include other structures or diagrams to explain your answer. NBS, hν NBS generates small amounts of Br2 in situ on reaction with HBr. CCl4 hν Br-Br CCl4 Br* is highly selective due to the endothermicity of the hydrogen abstraction step (the rate limiting step). Due to the symmetry of the molecule, abstraction at any of the secondary allylic sites will generate the same product distribution. HH Br The allylic radiacl is planar. Thus, bromine may attack from either face of the molecule in the second step, leading to pairs of enantiomers. 2 Br Hydrogen atom abstraction occurs at the secondary allylic site to generate a resonance stabilized allylic radical. The free electron is delocalized with partial spin density on secondary and tertiary sites. Reaction at both sites is possible. + H HBr + Br Br Br Br + Br Br Br Br + Br + Br H Question 3 (10 points). Explain why 1,3‐cyclopentadiene has a pKa of only 16, whereas 1,3,5‐ cycloheptatriene has a pKa of above 40. H H H pKa > 40 pKa = 16 H pKa is a measure of acidity; ie. the ease of deprotonation. Any factor that stabilizes a conjugate base (makes its formation more thermochemically favorable) will increase the acidity (lower the pKa). H H (A) H H +H H H +H (B) The cycloheptatrienylide anion (A) has 8π electrons in the cyclic π‐system, including the anion electron pair. It is anti‐aromatic, and its formation is not favored any more than any other allyl anion (or bisallyl anion). On the other hand, the cyclopentadienylide anion (B) has 6 electrons in the cyclic p‐system, which satisfies the Hückel Rule for aromaticity. This anion is greatly stabilized by aromaticity, and is readily formed. The acidity of the parent acid is correspondingly greater than cycloheptatriene. Question 4 (8 points). Indicate which of the following compounds are aromatic. N N NH O O N O O BF4 Non aromatic Antiaromatic Aromatic Aromatic Aromatic Question 5 (12 points). Predict, with reference to the relevant reaction mechanism, the outcome of the following reaction (including stereo‐ and regiochemical outcomes). You must include the detailed reaction mechanism in your answer, and may include other structures or diagrams to explain your answer. C N Cl2 FeCl3 δδ+ The -CN group is resonance electron-withdrawing: it is ring deactivating and m-directing. CN CN Cl2 + FeCl3 Cl3Fe Cl Cl = FeCl4 ClH2 C N MINOR δ+ δ+ δ+ C N Cl H N Cl ortho-substitution intermediate is destabilized due to the proximity of partial carbocation charge to the electron withdrawing group. A resonance contributor with a repulsive interaction between adjacent carbon atoms exists. Formation of this intermediate is disfavored. The intermediate for para-substitution is similarly destabilized. MINOR δ+ δ+ C Cl H MAJOR Cl H Cl3Fe Cl δ+ δ+ δ+ N C δ+ The intermediate for meta-substitution is not destabilized in this manner, and is therefore the most readily formed. The m-isomer is the dominant product Cl CN + HCl + FeCl3 Question 6 (12 points). Predict, with reference to the relevant reaction mechanism, the outcome of the following reaction (including stereo‐ and regiochemical outcomes). You must include the detailed reaction mechanism in your answer, and may include other structures or diagrams to explain your answer. OCH3 HNO3/H2SO4 ∆ H HNO3 + H2SO4 O ON H O + HSO4 δ+ δ+ O -OCH3 is resonance electron donating. It is a ring activating group that directs ortho- and para-. ONO + H2 O OCH3 MAJOR δ+ CH3 Positive charge may be delocalized onto O, due to resonance electron donationg character of -OCH3 group. The additional delocalization leads to stabilization of the ortho-intermediate; the orthoproductis favored. OCH3 NO2 OCH3 +H O2N δ+ NO2 H ONO MAJOR δ+ δ+ +H O δ+ CH3 H O2 N MINOR δ+ The para-intermediate is stabilized in a similar manner by the -OCH3 group, and so the p-isomer is also favored. δ+ δ+ O δ+ CH3 H NO2 The meta-intermediate is not stabilized by the methoxy group, and so this intermediate is the least thermodynamically favored of the three. The m-isomer will be a minor product. Question 7 (21 points). Predict the outcome of seven of the following ten reactions, including any stereochemical or regiochemical outcomes. Br2, hν Br Br + NBS, AIBN CCl4, ∆ Br + Br 1) NaNH2/NH3 2) I O2N NO2 NO2 HNO3/H2SO4 ∆ O O AlCl3 SO3/H2SO4 ∆ OCH3 + SO3H NH2 CH3 Cl OCH3 + Cl CH3 HO3S OCH3 + Cl OCH3 NO2 CH3 OCH3 CH3 SnCl2/HCl Cl2 FeCl3 OCH3 CH3 KOH + CHCl3 CCl2 + CCl2 OCH3 OCH3 + Br AlCl3 OCH3 + OCH3 + OCH3 Question 8 (10 points). Provide structures for two of the following three names p‐chloronitrobenzene Cl NO2 1,3,5‐trimethyl‐2‐chlorobenzene Cl m‐nitrotoluene BONUS QUESTION (10 points). Design a rational synthesis for the following: and NO2 starting from O2N Cl You may use any reagents you deem appropriate, and you may use the back of the page. H2 1) NaNH2/NH3 2) CH3-I NH3 HBr/ether HBr/ether (hydride shift) Pd/BaSO4 Li H2SO4 H2SO4 HBr addition from any of the alkene species will yield the desired 3° alkyl bromide, but the disubstituted alkenes will yield mixtures. The trisubstituted alkene (the most stable fromed by equilibration with H2SO4 or H3PO4 will give one alkyl halide. See over for remainder of synthesis. Br Br + AlCl3 HNO3/H2SO4 + dialkylated product TARGET Cl2 O2N Cl FeCl3 O2N + o-isomer. Chromatography or recrystallization required. ...
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