- A retrosynthetic plan for 4-bromo-3-nitrobenzoic acid requires careful consideration of directing effects. Which disconnection corresponds to the most viable final step in the forward synthesis?
Answer: Disconnection of the nitro group, implying nitration of 4-bromobenzoic acid.
The target is 4-bromo-3-nitrobenzoic acid. We must consider the directing effects of the substituents. -COOH and -NO2 are meta-directors and deactivating. -Br is an ortho/para-director and deactivating. In choice A, the precursor is 4-bromobenzoic acid. The -Br group (at C4) directs ortho to itself (to C3 and C5). The -COOH group (at C1) directs meta to itself (to C3 and C5). Since both groups direct to C3, nitration will successfully install the nitro group at C3. In choice B, the precursor is 3-nitrobenzoic acid. Both -NO2 and -COOH are meta-directors. They will direct an incoming electrophile to C5, not C4. In choice C, forming a Grignard reagent from 1-bromo-2-nitrobenzene is not possible because the nitro group would react with the Grignard. In choice D, starting with 3-bromobenzoic acid, the -COOH directs meta (to C5) and the -Br directs ortho/para (to C2, C4, C6). The result would be a mixture of products, not cleanly the desired isomer.
- A key retrosynthetic disconnection for a tertiary alcohol containing a carboxylic acid, such as 4-(2-hydroxypropan-2-yl)benzoic acid, must account for chemoselectivity. What strategic modification is necessary in the forward synthesis to accommodate a Grignard reaction?
Answer: The carboxylic acid must be protected as an ester before reacting with the Grignard reagent.
The target molecule has an acidic proton on the carboxylic acid group. Grignard reagents are incompatible with acidic protons, as they are strong bases and will be quenched by an acid-base reaction. Therefore, a direct Grignard addition to a molecule containing a carboxylic acid is not feasible. The retrosynthesis must include a step that accounts for this. The standard strategy is to protect the carboxylic acid by converting it into a functional group that does not have acidic protons and is also reactive toward the Grignard reagent in a productive way. An ester is the perfect choice. The Grignard reagent (e.g., MeMgBr) can then react with the ester (in this case, a benzoate ester) to form the tertiary alcohol. The ester serves as both a protecting group for the acid and the electrophile for the C-C bond formation.
- The synthesis of 5-hydroxy-5-phenyl-2-pentanone requires a C-C bond formation. A direct retrosynthetic disconnection of the C4-C5 bond suggests reacting 4-oxopentanal with phenylmagnesium bromide. Why is this proposed forward synthesis fatally flawed?
Answer: The Grignard reagent is a strong base and will deprotonate the acidic α-protons of the ketone and aldehyde, quenching the reagent.
The proposed precursor, 4-oxopentanal, has two carbonyl groups and, more critically, acidic α-protons at C3. Grignard reagents are extremely strong bases (pKa of conjugate acid is ~50). They will react irreversibly with any available acidic proton (pKa < ~25) before acting as a nucleophile. The α-protons of ketones/aldehydes have a pKa of ~19-20. Therefore, phenylmagnesium bromide would be consumed by an acid-base reaction, preventing the desired nucleophilic attack. A successful synthesis would require protecting the ketone (e.g., as an acetal) before the Grignard addition step.
- The synthesis of 2-heptanone from ethyl acetoacetate requires an alkylation step. A retrosynthetic analysis based on this pathway would identify which alkyl halide as the necessary electrophile?
Answer: 1-Bromobutane
The acetoacetic ester synthesis produces a methyl ketone. The general structure of the product is CH₃-CO-CH₂-R. The R group is introduced by alkylating the enolate of ethyl acetoacetate with an alkyl halide R-X. The target molecule is 2-heptanone, which has the structure CH₃-CO-(CH₂)₄-CH₃. Comparing this to the general product, the group attached to the CH₂ next to the carbonyl is a butyl group (-(CH₂)₃CH₃). Therefore, the required alkyl halide is 1-bromobutane. After alkylation, the synthesis is completed by hydrolysis and decarboxylation.
- A retrosynthetic analysis for a secondary amine, such as N-benzylethanamine, using reductive amination would involve disconnecting a C-N bond. Which pair of starting materials corresponds to this strategy?
Answer: Benzaldehyde and ethanamine.
Reductive amination is a method to form amines by reacting a carbonyl compound with an amine (or ammonia) to form an imine (or enamine) intermediate, which is then reduced in situ (e.g., with NaBH₃CN). Disconnecting a C-N bond of the target amine, N-benzylethanamine (C₆H₅CH₂-NH-CH₂CH₃), reveals two possibilities. (1) Disconnecting the benzyl-N bond gives benzaldehyde (the carbonyl precursor) and ethanamine. (2) Disconnecting the ethyl-N bond gives ethanal (acetaldehyde) and benzylamine. Both are valid. Choice A presents the first option. Choice B describes direct alkylation, which often leads to over-alkylation and is less controlled. Choice C would form an amide, which would then require a strong reducing agent like LiAlH₄ to become an amine. Choice D is not a standard amine synthesis pathway.
- Retrosynthetic analysis of a β-keto ester like ethyl 2-oxocyclopentanecarboxylate points to an intramolecular Dieckmann condensation. This disconnection implies that the starting material is what type of molecule?
Answer: A linear diester, specifically diethyl hexanedioate (diethyl adipate).
The Dieckmann condensation is an intramolecular version of the Claisen condensation, used to form five- or six-membered rings. The product is a cyclic β-keto ester. A retrosynthetic disconnection breaks the α-β bond within the ring that was formed during the condensation. This opens the ring to reveal a linear precursor. Since the Claisen condensation involves the reaction of two ester functional groups, the linear precursor must be a diester. For the target, ethyl 2-oxocyclopentanecarboxylate (a 5-membered ring), the precursor must be a 6-carbon chain with esters at both ends: diethyl hexanedioate, also known as diethyl adipate.
- A retrosynthesis of 3-phenylpropanoic acid is devised using a Grignard reagent to form the C2-C3 bond. Which pair of starting materials does this disconnection imply?
Answer: Phenylmagnesium bromide and ethylene oxide, followed by oxidation.
The target is 3-phenylpropanoic acid (C6H5-CH2-CH2-COOH). Disconnecting the C2-C3 bond means breaking the bond between the second and third carbons from the carboxyl group. This suggests adding a two-carbon chain to a phenyl-containing nucleophile. Choice D describes reacting phenylmagnesium bromide (a C6H5 nucleophile) with ethylene oxide (a two-carbon electrophile) to form 2-phenylethanol (C6H5-CH2-CH2-OH). A subsequent oxidation step would convert the primary alcohol to the target carboxylic acid. Choice A would form phenylacetic acid (C6H5-CH2-COOH), breaking the C1-C2 bond. Choice B involves a conjugate addition and would also form the target, but it corresponds to a C3-Cβ disconnection relative to the ester, not a simple C2-C3 bond break with a Grignard. Choice D is the most direct interpretation of a Grignard reaction forming the C2-C3 bond to eventually make the target acid.
- Retrosynthetic analysis of (E)-1-phenylprop-1-ene via a Wittig reaction suggests two possible routes. Which pair of reagents provides the most reliable and stereoselective synthesis of the E-isomer?
Answer: Acetaldehyde and benzyltriphenylphosphonium bromide, which forms a resonance-stabilized ylide.
The stereochemical outcome of the Wittig reaction depends on the stability of the phosphonium ylide. Non-stabilized ylides (e.g., from simple alkyl halides like ethyl bromide) typically react quickly and irreversibly to form a cis- or Z-alkene as the major product. Resonance-stabilized ylides (e.g., from benzyl or allyl halides, or halides alpha to a carbonyl) react more slowly and reversibly, leading to the thermodynamically more stable trans- or E-alkene. The target is the E-isomer. The route in choice B uses acetaldehyde and a benzyl ylide. The benzyl ylide is stabilized by resonance with the phenyl ring, and therefore this route will selectively produce the E-alkene. The route in choice A uses a non-stabilized ethyl ylide and would primarily yield the Z-alkene.
- Considering the synthesis of 4-methoxyacetophenone, a retrosynthetic disconnection of the C-C bond between the acetyl group and the aromatic ring suggests a Friedel-Crafts acylation. What would be the starting aromatic compound?
Answer: Anisole (methoxybenzene)
The disconnection suggests a Friedel-Crafts acylation, where an acyl group (in this case, an acetyl group, CH₃CO-) is added to an aromatic ring. The target molecule is 4-methoxyacetophenone, which is an aromatic ring substituted with a methoxy group (-OCH₃) and an acetyl group. The methoxy group is a strong activating group and an ortho/para-director. Therefore, starting with anisole (methoxybenzene) and reacting it with an acylating agent like acetyl chloride in the presence of a Lewis acid (e.g., AlCl₃) will readily produce the desired para-substituted product as the major isomer due to sterics. Starting with phenol is problematic as the -OH group coordinates with the Lewis acid catalyst.
- A Diels-Alder reaction is planned to synthesize 4-methylcyclohex-4-ene-1,2-dicarboxylic anhydride. What diene and dienophile are required for this [4+2] cycloaddition?
Answer: Isoprene (2-methyl-1,3-butadiene) and maleic anhydride.
A retrosynthetic disconnection of a cyclohexene derivative via the Diels-Alder reaction involves breaking the two sigma bonds that were formed and reforming a pi bond in the six-membered ring. This reveals the diene and dienophile. The target has a six-membered ring with a double bond. The double bond originated from the diene. The anhydride part of the molecule must have come from the dienophile. Maleic anhydride is a classic, highly reactive dienophile. The diene must be a conjugated four-carbon system. The methyl group at C4 of the product ring comes from the diene. Therefore, the diene is isoprene (2-methyl-1,3-butadiene). The reaction between isoprene and maleic anhydride gives the desired product.
- A proposed synthesis of cyclohexylacetic acid involves the malonic ester synthesis. A retrosynthetic analysis based on this method would disconnect which bonds to identify the required starting materials?
Answer: The Cα-Cβ bond and one Cα-COOH bond, identifying cyclohexyl bromide and diethyl malonate.
The malonic ester synthesis is used to create carboxylic acids with substituents on the alpha-carbon. The target is cyclohexylacetic acid. The 'acetic acid' part comes from the two-carbon core of malonic ester. The 'cyclohexyl' part is the substituent added to the α-carbon. The retrosynthesis thus involves two disconnections: (1) removing the added substituent (the cyclohexyl group) from the α-carbon, which corresponds to breaking the bond between the ring and the side chain (the Cα-Cβ bond), and (2) recognizing that one of the carboxyl groups in malonic ester is removed via decarboxylation. This process reveals the two starting materials: diethyl malonate and an electrophile to add the cyclohexyl group, which is cyclohexyl bromide.
- Retrosynthetic analysis of a 1,5-dicarbonyl compound, such as 2,6-heptanedione, points to a Michael addition as the key C-C bond-forming step. What are the synthons that correspond to this disconnection?
Answer: An α,β-unsaturated carbonyl (Michael acceptor) and an enolate (Michael donor).
The Robinson annulation and other syntheses build 1,5-dicarbonyl compounds using a Michael (or 1,4-conjugate) addition. The key disconnection is between the β- and γ-carbons relative to one of the carbonyl groups. This disconnection reveals a Michael acceptor (an α,β-unsaturated carbonyl compound) and a Michael donor (an enolate). For 2,6-heptanedione, this would be methyl vinyl ketone (MVK) and the enolate of acetone. An aldol reaction produces a 1,3-relationship, and a Claisen condensation produces a β-keto ester (a 1,3-dicarbonyl relationship).
- A retrosynthetic analysis of 3-hydroxy-3-phenylbutanal involves an aldol addition disconnection. Which pair of molecules represents the required carbonyl precursors for the forward synthesis?
Answer: Acetophenone and ethanal (acetaldehyde)
The target molecule is a β-hydroxy aldehyde. An aldol disconnection breaks the bond between the α- and β-carbons. In 3-hydroxy-3-phenylbutanal, this is the C2-C3 bond. The β-carbon (C3) and its substituents (the hydroxyl, phenyl, and methyl groups) originate from the electrophilic carbonyl component. This corresponds to acetophenone (a phenyl methyl ketone). The α-carbon (C2) and its substituent (the aldehyde group) originate from the enolate component, which is formed from ethanal (acetaldehyde). Therefore, the precursors are acetophenone and ethanal.
- A retrosynthetic disconnection for 3-cyclohexylpropan-1-ol breaks the C1-C2 bond. This strategy implies the formation of the carbon skeleton via which type of reaction and precursors?
Answer: Nucleophilic ring opening of an epoxide, using cyclohexylmagnesium bromide and ethylene oxide.
The target molecule is a primary alcohol with a cyclohexyl group on C3. Disconnecting the C1-C2 bond means that the two-carbon unit (C1-C2 with the OH on C1) was the electrophile, and the cyclohexyl group was the nucleophile. This pattern perfectly matches the reaction of a Grignard reagent (cyclohexylmagnesium bromide) with an epoxide (ethylene oxide). The Grignard reagent attacks one of the carbons of the epoxide, opening the ring and forming the C1-C2 bond. Aqueous workup then protonates the resulting alkoxide to give the primary alcohol. The other options either form the wrong C-C bond (aldol, choice B) or do not correspond to a C1-C2 disconnection (ester reduction, choice C).
- Two retrosynthetic disconnections are proposed for benzyl sec-butyl ether via the Williamson ether synthesis. Which statement accurately compares the viability of the two corresponding forward syntheses?
Answer: Route 1 (benzyl bromide + sodium sec-butoxide) is superior as SN2 on a primary halide is fast, minimizing elimination from the bulky secondary alkoxide.
The Williamson ether synthesis is an SN2 reaction between an alkoxide and an alkyl halide. This reaction is sensitive to steric hindrance on the alkyl halide, as hindrance favors the competing E2 elimination pathway. Route 1 uses a primary alkyl halide (benzyl bromide) and a secondary alkoxide (sodium sec-butoxide). SN2 on a primary halide is efficient. While the alkoxide is bulky and basic, the lack of hindrance at the electrophile allows SN2 to dominate. Route 2 uses a secondary alkyl halide (sec-butyl bromide) and a primary alkoxide (sodium benzoxide). SN2 on a secondary halide is slow, and when a strong base like an alkoxide is used, E2 elimination becomes the major pathway. Therefore, Route 1 is far superior.
- The retrosynthesis of N,N-diethyl-m-toluamide (DEET) from m-toluic acid requires the formation of an amide bond. What is the most common and effective intermediate derived from m-toluic acid to achieve this transformation?
Answer: The corresponding acyl chloride, formed by reaction with thionyl chloride (SOCl₂).
Amide formation directly from a carboxylic acid and an amine requires very high temperatures and is often inefficient. A key retrosynthetic step is to recognize that the carboxylic acid must be 'activated'. The most common way to do this is to convert the carboxylic acid into a more reactive acyl derivative. The acyl chloride is the most common and reactive choice. Reacting m-toluic acid with thionyl chloride (SOCl₂) or oxalyl chloride ((COCl)₂) creates m-toluoyl chloride. This highly electrophilic species readily reacts with diethylamine, even at low temperatures, to form the amide bond of DEET. Reacting an ester with an amine is possible but much slower, and the carboxylate salt is unreactive toward nucleophilic acyl substitution.
- The Robinson annulation is a powerful ring-forming method. A retrosynthetic disconnection of a typical product, like a substituted cyclohexenone, reveals two key bond-forming reactions. What are these two reactions in the order they occur in the forward synthesis?
Answer: A Michael addition followed by an intramolecular aldol condensation.
The Robinson annulation builds a six-membered ring onto an existing ketone. Retrosynthetically, this involves two disconnections. The first disconnection is of the double bond and the adjacent C-C bond, which corresponds to an intramolecular aldol condensation (which produces the α,β-unsaturated system). This reveals a 1,5-dicarbonyl intermediate. The second disconnection breaks the bond between the β- and γ-carbons (relative to one carbonyl), which corresponds to a Michael addition. In the forward synthesis, the sequence is reversed: first, a Michael addition between an enolate and an α,β-unsaturated ketone to form a 1,5-dicarbonyl intermediate, followed by an intramolecular aldol condensation and dehydration to form the final cyclohexenone product.
- A retrosynthetic disconnection of an ester, such as isopropyl benzoate, breaks the acyl C-O bond. This leads to an acyl synthon and an alcohol synthon. Which set of reagents best represents these synthons in a forward reaction?
Answer: Benzoyl chloride and isopropanol.
The most common and efficient methods for ester synthesis in a lab setting involve nucleophilic acyl substitution. A retrosynthetic disconnection of the ester's acyl C-O bond breaks the molecule into the part that came from the carboxylic acid and the part that came from the alcohol. For isopropyl benzoate, this gives a benzoyl group and an isopropoxy group. In the forward synthesis, the benzoyl group is typically supplied as an activated carboxylic acid derivative, most commonly benzoyl chloride. The isopropoxy group is supplied as the alcohol, isopropanol. The alcohol acts as a nucleophile, attacking the highly electrophilic acyl chloride to form the ester. This is a more effective method than Fischer esterification (using benzoic acid itself), especially with a secondary alcohol like isopropanol. Choice B represents an SN2 reaction which is not the standard way to make esters.
