Ether Synthesis (Williamson) and Cleavage - Organic Chemistry 2

Which substrate is best for Williamson with sodium ethoxide to form ethyl benzyl ether efficiently?
Answer: benzyl bromide This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, using benzyl bromide for efficient SN2. A common distractor fails because it suggests aryl halides resistant to SN2. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
Which balanced equation correctly represents Williamson: NaOEt + CH3I → ?
Answer: CH3OCH2CH3 + NaI This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, producing the ether and NaI via SN2. A common distractor fails because it suggests hydrolysis or elimination products. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
Which of the following ethers would be cleaved most slowly when heated with concentrated HBr?
Answer: Diphenyl ether When you encounter ether cleavage reactions, you need to consider how readily the ether can be protonated and how stable the resulting carbocation intermediate will be. Ether cleavage with HBr follows an SN1 or SN2 mechanism depending on the substrate. Diphenyl ether (D) cleaves most slowly because the oxygen atom's lone pairs are extensively delocalized into the aromatic π system through resonance. This delocalization makes the oxygen much less basic and nucleophilic, so it resists protonation by HBr. Additionally, even if protonation occurred, the resulting carbocation would be highly unstable since it would disrupt the aromatic system. Looking at the other options: Tetrahydrofuran (A) is a cyclic ether that cleaves readily because ring strain is relieved upon opening, and the oxygen is quite basic. Di-tert-butyl ether (B) cleaves very rapidly via an SN1 mechanism because it can form highly stable tertiary carbocations after protonation and C-O bond cleavage. Diethyl ether (C) also cleaves relatively easily through either SN1 or SN2 pathways, forming moderately stable secondary carbocations or undergoing direct displacement. The key pattern to remember is that aromatic ethers are much more resistant to cleavage than aliphatic ethers because resonance stabilization reduces the basicity of the ether oxygen. When ranking ether reactivity toward acid-catalyzed cleavage, always consider: can the oxygen be easily protonated, and how stable will any carbocation intermediates be?
An efficient synthesis of 2-ethoxy-2-methylpropane (ETBE) is desired. Which of the following conditions represents the best application of the Williamson ether synthesis for this target?
Answer: Sodium 2-methylpropan-2-oxide (tert-butoxide) reacting with bromoethane To synthesize an ether R-O-R', the Williamson synthesis requires one part to be an alkoxide and the other an alkyl halide. For the SN2 reaction to be efficient, the alkyl halide should be primary or methyl. For the target ether, there are two disconnections: (1) ethoxide + tert-butyl halide or (2) tert-butoxide + ethyl halide. Route (1) fails because the tertiary halide undergoes elimination. Route (2) succeeds because it uses a primary halide (bromoethane), which readily undergoes SN2. Therefore, reacting sodium tert-butoxide with bromoethane is the correct choice.
Given cyclohexyl methyl ether + HBr, which major products expected after cleavage and workup?
Answer: cyclohexanol + methyl bromide This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, cleaving at the methyl group via SN2. A common distractor fails because it suggests cleavage at the secondary cyclohexyl group or elimination. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
Consider the intramolecular Williamson ether synthesis of 3-chloro-1-propanol to form oxetane. Which statement best describes a key feature of the mechanism?
Answer: The alkoxide formed attacks the electrophilic carbon in an intramolecular SN2 displacement, leading to ring formation. The intramolecular Williamson ether synthesis follows the same general mechanism as the intermolecular version. A strong base (like NaH) deprotonates the alcohol to form an alkoxide nucleophile. This nucleophile then attacks the carbon atom bonded to the leaving group (chlorine) in the same molecule. This step is an intramolecular SN2 reaction, which results in the formation of the cyclic ether (oxetane) and the expulsion of the chloride ion.
What is major product when tert-butyl methyl ether reacts with excess HI and heat?
Answer: (CH3)3CI + CH3OH This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, favoring SN1 cleavage at the tertiary carbon with excess HI. A common distractor fails because it suggests SN2 at the methyl group or no reaction. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
Which side reaction most competes with Williamson when using secondary alkyl halides and strong alkoxides?
Answer: E2 elimination to give an alkene This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, identifying E2 as the main competitor for secondary halides. A common distractor fails because it suggests unrelated reactions like oxidation. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
Given CH3OCH2CH3 + 1 equiv HBr, predict major products under typical cleavage conditions.
Answer: CH3Br + CH3CH2OH This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, preferring SN2 cleavage at the less hindered methyl group. A common distractor fails because it suggests cleavage at the ethyl group or symmetric products. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
Given PhCH2OCH3 + excess HBr, what major products form after cleavage?
Answer: PhCH2Br + CH3OH This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, cleaving to benzyl bromide via benzylic stabilization. A common distractor fails because it suggests aryl-like cleavage or unrelated products. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
In mixed ether synthesis, which pairing minimizes byproducts: choose nucleophile and electrophile roles correctly.
Answer: EtO⁻ as nucleophile with methyl iodide as electrophile This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, pairing unhindered components to minimize byproducts. A common distractor fails because it suggests hindered pairings that promote elimination. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
Which of the following statements regarding ethers is FALSE?
Answer: Aryl ethers are readily cleaved by concentrated HI to produce an aryl iodide and an alcohol. Statement (C) is false. The carbon-oxygen bond in aryl ethers, where the oxygen is directly attached to the benzene ring, is very strong due to partial double-bond character from resonance and the sp² hybridization of the carbon. This bond is not susceptible to cleavage by HI or HBr. Instead, cleavage occurs at the alkyl-oxygen bond to produce a phenol and an alkyl iodide.
The following ethers are each treated with one equivalent of HI under identical conditions. Which ether is expected to undergo acid-catalyzed cleavage at the fastest rate?
Answer: tert-Butyl methyl ether The rate of ether cleavage depends on the reaction mechanism. Ethers that can react via an SN1 mechanism are generally cleaved faster than those restricted to an SN2 mechanism. Tert-butyl methyl ether is unique among the choices because it can form a highly stable tertiary carbocation upon protonation and loss of methanol. This low-activation energy SN1 pathway makes its cleavage significantly faster than the SN2 cleavage of primary ethers (A), secondary ethers (B), or aryl ethers (D).
What is major product when di-n-propyl ether is cleaved with 1 equiv HBr (no excess)?
Answer: C3H7Br + C3H7OH This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, yielding one halide and one alcohol with limited HBr on symmetric ether. A common distractor fails because it suggests products from excess reagent or elimination. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
The chiral ether (R)-2-methoxybutane is treated with concentrated HI. Assuming the reaction proceeds exclusively via an SN2 mechanism, what are the expected major products?
Answer: (R)-Butan-2-ol and iodomethane Cleavage of this ether involves an SN2 attack by iodide after the ether oxygen is protonated. The two possible sites of attack are the methyl carbon and the secondary carbon at C2 of the butyl group. The SN2 attack occurs at the less sterically hindered methyl carbon. Since the reaction does not occur at the chiral center (C2), its configuration is retained. The products are therefore (R)-butan-2-ol and iodomethane.
The compound 1-methoxy-1-phenylethane is heated with one equivalent of concentrated HBr. What are the major organic products?
Answer: Methanol and 1-bromo-1-phenylethane When you encounter ethers being heated with HBr, you're looking at an ether cleavage reaction. The key is understanding that ethers break apart under acidic conditions, and the cleavage follows specific patterns based on carbocation stability. In 1-methoxy-1-phenylethane, you have a methoxy group ( $-OCH_3$ ) attached to a secondary carbon that's also bonded to a phenyl ring. When heated with HBr, the ether oxygen gets protonated first, making it a better leaving group. The C-O bond then breaks, and since the secondary benzylic position can stabilize a carbocation through resonance with the phenyl ring, the cleavage occurs at this position. This produces methanol (from the $-OCH_3$ group) and a secondary benzylic carbocation, which immediately captures the bromide ion to form 1-bromo-1-phenylethane. Answer D correctly identifies these products. Answer A is wrong because ethylbenzene would require reduction, not substitution. Answer B suggests 1-phenylethanol forms, but alcohols don't typically form directly from ether cleavage with HBr - you get halide substitution instead. Answer C proposes styrene formation, which would require elimination rather than substitution, and elimination isn't favored under these specific conditions with concentrated HBr. Remember this pattern: ether cleavage with HBr breaks the C-O bond at the position that can best stabilize a carbocation (tertiary > secondary > primary), and benzylic positions are particularly favored due to resonance stabilization. The alcohol component becomes the corresponding halide.
An unknown ether with the molecular formula C₆H₁₄O is heated with excess concentrated HBr. The only organic product isolated after the reaction is 2-bromopropane. Based on this information, what was the structure of the original ether?
Answer: Diisopropyl ether The reaction is an ether cleavage with excess HBr. The fact that only one organic product, 2-bromopropane, is formed implies that both alkyl groups attached to the ether oxygen must give rise to this product. This can only happen if the starting ether is symmetrical and composed of two isopropyl groups. Diisopropyl ether, (CH₃)₂CH-O-CH(CH₃)₂, fits this description and has the correct molecular formula C₆H₁₄O.
Which leaving group generally gives fastest Williamson SN2 with RO⁻: compare R-I, R-Br, R-Cl, R-F?
Answer: alkyl iodide (R-I) This question tests the ability to understand and apply knowledge of Williamson ether synthesis and ether cleavage reactions in Organic Chemistry 2. The Williamson ether synthesis involves an SN2 reaction where an alkoxide ion reacts with a primary alkyl halide, favored by conditions that prevent elimination. For ether cleavage, strong acids like HI or HBr are used to break the C-O bond, forming alkyl halides and alcohols. The correct answer works because it accurately reflects the reaction conditions and mechanistic pathway taught in the course, with iodide as the best leaving group for SN2. A common distractor fails because it suggests poorer leaving groups like fluoride. Teaching strategies include emphasizing the importance of substrate choice and reaction conditions in predicting the outcome of ether synthesis and cleavage, as well as practicing with a variety of examples to understand regioselectivity and mechanism logic.
To maximize the rate of the SN2 reaction between sodium methoxide and 1-bromopropane, which solvent is the most suitable?
Answer: Acetone The Williamson ether synthesis is an SN2 reaction. SN2 reactions are fastest in polar aprotic solvents. These solvents can dissolve the ionic alkoxide but do not strongly solvate the nucleophilic anion, leaving it 'naked' and highly reactive. Acetone is a polar aprotic solvent. Water and methanol are polar protic solvents, which would solvate the methoxide and reduce its nucleophilicity. Toluene is nonpolar and would not effectively dissolve the sodium methoxide.
Treatment of anisole (methyl phenyl ether) with excess concentrated HBr at high temperature yields which final products?
Answer: Phenol and bromomethane In the cleavage of aryl alkyl ethers, the C(sp²)-O bond to the aromatic ring is very strong and is not cleaved. The reaction occurs at the alkyl-oxygen bond. The bromide ion attacks the methyl group via an SN2 mechanism, displacing phenol. Phenols are not converted to aryl halides under these conditions, so even with excess HBr, the final products are phenol and bromomethane.

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