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Which compound in the following pairs would react faster under the reaction conditions? Draw the structures of the major products expected. a. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) or \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)(\mathrm{Br}) \mathrm{CH}_{3}\) in ethanol-water solution. b. Same as in Part a, but with potassium iodide in acetone. c. Same as in Part a, but with potassium hydroxide in ethanol. d. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{BF}_{4}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \stackrel{\oplus}{\mathrm{N}}\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\ominus}{\mathrm{O} \mathrm{CH}_{3}}\) on heating in methanol solution.

Short Answer

Expert verified
a. \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)(\mathrm{Br}) \mathrm{CH}_{3} \) reacts faster. b. \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \) reacts faster. c. \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)(\mathrm{Br}) \mathrm{CH}_{3} \) reacts faster. d. \( \mathrm{CH}_{3} \mathrm{CH}_{2} \stackrel{\oplus}{\mathrm{N}}\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\ominus}{\mathrm{O}\mathrm{CH}_{3}} \) reacts faster.

Step by step solution

01

Analyze First Reaction Conditions

In an ethanol-water solution, we expect an SN1 reaction due to the polar protic nature of the solvent, which favors carbocation formation. Compare the two substrates: \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \) and \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)(\mathrm{Br}) \mathrm{CH}_{3} \). The latter can form a more stable tertiary carbocation after leaving group departure due to the resonance and hyperconjugation effect.
02

Determine Faster Reacting Compound in Part a

The compound \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)(\mathrm{Br}) \mathrm{CH}_{3} \) will react faster in ethanol-water because it can form a more stable tertiary carbocation than the primary carbocation from \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \).
03

Analyze Second Reaction Conditions

In acetone with potassium iodide, an SN2 mechanism is favored due to the polar aprotic nature of the solvent that promotes bimolecular nucleophilic substitution. This favors less hindered substrates for backside attack.
04

Determine Faster Reacting Compound in Part b

The compound \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \) will react faster in acetone with potassium iodide via an SN2 reaction, as it is a primary halide, which experiences less steric hindrance.
05

Analyze Third Reaction Conditions

In ethanol with potassium hydroxide, we expect an E2 elimination reaction, particularly favoring a tertiary substrate providing better stabilization of the developing anion and the formation of alkenes.
06

Determine Faster Reacting Compound in Part c

The compound \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)(\mathrm{Br}) \mathrm{CH}_{3} \) will react faster due to its tertiary structure, favoring E2 elimination, leading to the formation of a more substituted double bond.
07

Analyze Fourth Reaction Condition

Upon heating in methanol, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{BF}_{4} \) and \( \mathrm{CH}_{3} \mathrm{CH}_{2} \stackrel{\oplus}{\mathrm{N}}\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\ominus}{\mathrm{O}\mathrm{CH}_{3}} \) are examined. This condition favors solvolysis or elimination. Consider stability of the intermediates or products formed under heating.
08

Determine Faster Reacting Compound in Part d

The compound \( \mathrm{CH}_{3} \mathrm{CH}_{2} \stackrel{\oplus}{\mathrm{N}}\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\ominus}{\mathrm{O}\mathrm{CH}_{3}} \) reacts faster under heating in methanol due to the internal salt nature, leading to easier solvolysis and rearrangement of ions.

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Key Concepts

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

Carbocation Stability
When exploring organic reactions such as SN1 and SN2, understanding carbocation stability is essential. Carbocations are positively charged carbon ions and play a key role in reaction mechanisms.
The stability of these ions greatly influences how fast a reaction proceeds. In an SN1 reaction, the formation of a stable carbocation is critical. It typically occurs in two main steps. First, the leaving group departs, forming a carbocation. Then, the nucleophile attacks. The more stable the carbocation, the faster and smoother the reaction will be.
Tertiary carbocations are more stable than secondary, which are more stable than primary carbocations. This is due to two main reasons:
  • **Hyperconjugation**: More alkyl groups allow for electron donation to the positively charged carbon, stabilizing it.
  • **Resonance effects**: The delocalization of charge over a larger structure adds stability, especially if the carbocation can be part of an aromatic ring.
These factors explain why in a polar protic solvent like ethanol, a compound that can form a tertiary carbocation often reacts faster under SN1 conditions.
Reaction Mechanisms
Understanding reaction mechanisms helps explain how reactions occur at a molecular level. Two important mechanisms are SN1 and SN2, which dictate how substitution reactions proceed. **SN1 Mechanism**: This is a two-step process:
  • The first step is unimolecular, where the leaving group departs, forming a carbocation. This step is slow and determines the rate of the reaction.
  • The second step involves a nucleophile attacking the carbocation, leading to product formation.
SN1 reactions typically happen in polar protic solvents that stabilize the carbocation intermediate. **SN2 Mechanism**: This is a single-step bimolecular process where the nucleophile attacks the substrate from the opposite side of the leaving group. This process involves a concerted transition state where bonds are being formed and broken simultaneously.
SN2 reactions favor polar aprotic solvents, enhancing nucleophilic strength and promoting faster reactions.
They are also highly sensitive to steric hindrance, which affects the approach of the nucleophile. Primary halides, with less crowding around the carbon center, are most conducive for SN2 reactions.
Tertiary and Primary Halides
Tertiary and primary halides have distinct reactivities based on their carbon structure and the types of reactions they undergo. Here's how they differ: **Tertiary Halides**: These compounds have a halogen atom attached to a carbon that is connected to three other carbon atoms. They are highly favored in SN1 reactions due to their ability to form stable tertiary carbocations. This stability allows for efficient carbocation formation needed in SN1 processes. Tertiary halides are also suitable for E2 reactions, where a strong base induces elimination to form alkenes. **Primary Halides**: In contrast, primary halides have the halogen attached to a carbon bonded to only one other carbon. They undergo SN2 reactions effectively because the steric hindrance is minimal, allowing the nucleophile easy access to perform a backside attack. The type of halide thus determines the preferred mechanism under given conditions:
  • **Tertiary halides**: Ideal for SN1 and E2 mechanisms.
  • **Primary halides**: Suit SN2 reactions best due to minimal steric interference.
Knowing these distinctions helps predict reaction pathways and expected products in different scenarios.

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Most popular questions from this chapter

Write equations and mechanisms for all the products that might reasonably be expected from the reaction of 2 chlorobutane with a solution of potassium hydroxide in ethanol.

Which compound in each of the following pairs would you expect to react more readily with (A) potassium iodide in 2 -propanone, (B) concentrated sodium hydroxide in ethanol, and (C) silver nitrate in aqueous ethanol? Write equations for all the reactions involved and give your reasoning with respect to the predicted orders of reactivity. a. methyl chloride and isobutyl chloride with \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) b. methyl chloride and tert-butyl chloride with \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) c. tert-butyl chloride and 1 -fluoro-2-chloro-2-methylpropane with \(\mathrm{B}\) and \(\mathrm{C}\) d. 1 -chloro-2-butene and 4 -chloro-1-butene with \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\)

Write Lewis structures for each of the following reagents and classify them as either electrophilic, nucleophilic, both, or neither by evaluating whether they will react appreciably with hydroxide ion, \(\mathrm{HO}^{\ominus}\), or hydronium ion, \(\mathrm{H}_{3} \mathrm{O}^{\oplus}\). Write equations for each of the reactions involved. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{NH}_{2}^{-}\) c. \(\mathrm{Na}^{\oplus}\) d. \(\mathrm{Cl}^{\ominus}\) e. \(\mathrm{Cl}_{2}\) f. \(\mathrm{CH}_{4}\) g. CN h. \(\mathrm{CH}_{3} \mathrm{OH}\) i. \(\mathrm{CH}_{3} \mathrm{O} \mathrm{H}_{2}\) j. \(\mathrm{BF}_{4}^{-}\) k. \(\mathrm{HBr}\) l. \(\mathrm{HC} \equiv \mathrm{C}:^{\ominus}\) \(\mathrm{m}\). : \(\mathrm{CH}_{2}\) n. \(\mathrm{FSO}_{3} \mathrm{H}\) o. \(\mathrm{SO}_{3}\)

Give a plausible explanation for each of the following observations: a. Aqueous sodium chloride will not convert tert-butyl alcohol to tert-butyl chloride but concentrated hydrochloric acid will. b. Better yields are obtained in the synthesis of isopropyl methyl ether starting with methyl iodide rather than sodium methoxide: $$ \begin{array}{l} \mathrm{CH}_{3} \mathrm{I}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHO}^{\ominus} \mathrm{Na}^{\oplus} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOCH}_{3}+\mathrm{Na}^{\oplus} \mathrm{I}^{\ominus} \\\ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHI}+\mathrm{CH}_{3} \mathrm{O}^{\ominus} \mathrm{Na}^{\oplus} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOCH}_{3}+\mathrm{Na}^{\oplus} \mathrm{I}^{\ominus} \end{array} $$ c. The following reaction proceeds only if an equivalent amount of silver fluoroborate, \(\mathrm{Ag}^{\oplus} \mathrm{BF}_{4} \ominus\), is added to the reaction mixture: d. 1 -Bromo- 2 -butene reacts with water to give a mixture of 2 -buten- 1 -ol, 3 -buten-2-ol, and some 1,3 -butadiene.

Explain the following observations: a. The tertiary chloride, apocamphyl chloride, is unreactive in either \(S_{N} 1\) or \(S_{N} 2\) reactions. For example, no reaction occurs when its solution in aqueous ethanol containing \(30 \%\) potassium hydroxide is refluxed for 20 hours. CC1(C)CCCCC1(C)C chlonide b. Chloromethyl alkyl (or aryl) ethers, \(\mathrm{ROCH}_{2} \mathrm{Cl}\), are very reactive in \(S_{\mathrm{N}} 1\) solvolysis reactions. Compared to chloromethane, the rate of hydrolysis of chloromethyl phenyl ether is about \(10^{14}\). Also, the rate of hydrolysis is retarded significantly by lithium chloride.

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