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Unformatted text preview: r 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...
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