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Draw the product formed when \(1,3,5\) -cycloheptatriene \(\left(\mathrm{p} K_{\mathrm{a}}=39\right)\) is treated with a strong base. Why is its \(\mathrm{pK}_{\mathrm{a}}\) so much higher than the \(\mathrm{p} K_{\mathrm{a}}\) of cyclopentadiene?

Short Answer

Expert verified
Aromatic cycloheptatrienyl anion; pK鈧 higher due to initial instability of cycloheptatriene.

Step by step solution

01

Recognize the Reactant

The reactant is 1,3,5-cycloheptatriene, a seven-membered ring with three alternating double bonds.
02

Identify Strong Base Reaction

When treated with a strong base, a hydrogen atom is abstracted from cycloheptatriene, as strong bases deprotonate compounds to form anions.
03

Analyze Anion Formation

Deprotonation occurs at a hydrogen atom adjacent to a double bond, creating a negatively charged carbon and converting the system to a more stable aromatic anion.
04

Consider Aromaticity

The product formed is a cycloheptatrienyl anion, which is stabilized by resonance, making it aromatic by fulfilling the (4n+2) rule for 6 蟺-electrons, allowing for electron delocalization.
05

Compare with Cyclopentadiene

Cyclopentadiene, with a pK鈧 of approximately 16, is much more acidic because the resulting anion, cyclopentadienyl anion, acquires aromatic stability with 6 蟺-electrons, which is much more favorable compared to cycloheptatriene's initial state.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aromaticity in Cycloheptatriene
Aromaticity is a concept in chemistry where a compound exhibits extra stability due to the delocalization of electrons across a cyclic, planar structure. Cycloheptatriene, by itself, doesn't fall under the traditional definition of an aromatic compound. This is because its seven-membered carbon ring with three alternating double bonds does not complete the (4n + 2) 蟺-electron rule, also known as H眉ckel's Rule, necessary for aromaticity. Each double bond contributes two 蟺-electrons, and therefore, cycloheptatriene originally has six 蟺-electrons. However, when a strong base removes a proton (deprotonation), the resulting cycloheptatrienyl anion achieves aromaticity. This is because the anion allows for the delocalization of the sixth electron, satisfying the H眉ckel's Rule and stabilizing the molecule via resonance.
Understanding pKa Comparison
The pKa value is a measure of the acidity of a substance. It tells us how easily a substance can give up a hydrogen ion. Cycloheptatriene's pKa of 39 indicates a much lesser tendency to lose a proton compared to cyclopentadiene, whose pKa is around 16. This stark difference can be attributed to the stability of the anions formed after deprotonation. Cyclopentadiene forms an aromatic cyclopentadienyl anion, which is very stable due to its 6 蟺-electron aromatic system. In contrast, cycloheptatriene, before deprotonation, isn't naturally fulfilling the aromaticity conditions, thus less eager to lose a proton. It only achieves aromatic stabilization upon forming the cycloheptatrienyl anion, which happens with a greater push, thus explaining the higher pKa.
Anion Formation and Stability
When cycloheptatriene is treated with a strong base, a hydrogen is removed, allowing one of the carbons in the ring to bear a negative charge, forming the cycloheptatrienyl anion. This transition is significant because the anion becomes more stable than the original neutral molecule. By achieving aromaticity, the cycloheptatrienyl anion stabilizes through resonance, redistributing the negative charge over the entire structure. This stabilization makes the resulting anion more favorable energetically, as the energy is lowered by the uniform distribution of electrons.
Resonance Stabilization in Anions
The concept of resonance stabilization explains why certain anions are more stable than others. In the cycloheptatrienyl anion, the removal of the hydrogen allows the electrons to spread out or delocalize across the entire ring structure. This spread of electrons occurs over all the atoms in the ring, reducing energy and increasing stability. It leads to a scenario where the electrons don't just sit as static charges on individual atoms, but they move around, effectively reducing the potential energy of the molecule. Thus, resonance dramatically enhances the anion's stability, as the structure resonates between different possible forms, distributing electron density more uniformly throughout the molecule.

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Most popular questions from this chapter

Early structural studies on benzene had to explain the following experimental evidence. When benzene was treated with \(\mathrm{Br}_{2}\) (plus a Lewis acid), a single substitution product of molecular formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\) was formed. When this product was treated with another equivalent of \(\mathrm{Br}_{2}\), three different compounds of molecular formula \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Br}_{2}\) were formed. a. Explain why a single Kekul茅 structure is consistent with the first result, but does not explain the second result. b. Then explain why a resonance description of benzene is consistent with the results of both reactions.

Propose a structure consistent with each set of data. a. Compound \(\mathbf{A}\) : Molecular formula: \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}\) IR absorption at \(3150-2850 \mathrm{~cm}^{-1}\) \({ }^{1} \mathrm{H}\) NMR data: \(1.4\) (triplet, \(3 \mathrm{H}\) ), \(3.95\) (quartet, \(2 \mathrm{H}\) ), and \(6.8-7.3\) (multiplet, \(5 \mathrm{H}\), ppm b. Compound \(\mathrm{B}\) : Molecular formula: \(\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}_{2}\) IR absorption at \(1669 \mathrm{~cm}^{-1}\) \({ }^{1}\) H NMR data: \(2.5\) (singlet, 3 H), \(3.8\) (singlet, \(3 \mathrm{H}\) ), \(6.9\) (doublet, \(2 \mathrm{H}\) ), and \(7.9\) (doublet, \(2 \mathrm{H}\) ) ppm

How many \({ }^{13} \mathrm{C}\) NMR signals does \(\mathrm{C}_{60}\) exhibit?

Draw a structure corresponding to each name. a. \(p\) -dichlorobenzene e. 2,6 -dimethoxytoluene b. \(m\) -chlorophenol f. 2-phenyl-1-butene c. \(p\) -iodoaniline g. 2-phenyl-2-propen-1-ol d. o-bromonitrobenzene h. trans-1-benzyl-3-phenylcyclopentane

Draw five resonance structures for the cyclopentadienyl cation.

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