Organic Chemistry 2 Quiz: Hydride Reductions Nabh4 Lialh4 And Selectivity
18 questions · exam conditions
0:00
Hydride Reductions Nabh4 Lialh4 And SelectivityQuestion 1 of 18

The reduction of crotonaldehyde (CH₃CH=CHCHO) with NaBH₄ primarily yields crotyl alcohol (CH₃CH=CHCH₂OH), representing a 1,2-addition. However, under certain conditions, a minor product resulting from 1,4-addition (conjugate addition) is observed. What would be the structure of this minor product after workup?

Butanal
Butan-1-ol
Butan-2-one
But-2-en-1-ol
← Back to quizzes

Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Hydride Reductions Nabh4 Lialh4 And Selectivity

Practice Hydride Reductions Nabh4 Lialh4 And Selectivity in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Hydride Reductions Nabh4 Lialh4 And Selectivity, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The reduction of crotonaldehyde (CH₃CH=CHCHO) with NaBH₄ primarily yields crotyl alcohol (CH₃CH=CHCH₂OH), representing a 1,2-addition. However, under certain conditions, a minor product resulting from 1,4-addition (conjugate addition) is observed. What would be the structure of this minor product after workup?

  1. Butanal (correct answer)
  2. Butan-1-ol
  3. Butan-2-one
  4. But-2-en-1-ol
Explanation: In a 1,4-addition of a hydride to an α,β-unsaturated aldehyde, the hydride attacks the β-carbon (C4 of the conjugate system). This pushes electrons to form an enolate intermediate. During the aqueous workup, this enolate is protonated at the α-carbon. The net result is the reduction of the C=C double bond, leaving the carbonyl group intact initially. The intermediate enol would then tautomerize to the more stable keto form, which in this case is the saturated aldehyde, butanal. Butan-1-ol (B) would require reduction of both the alkene and the aldehyde. Butan-2-one (C) has the wrong carbon skeleton. But-2-en-1-ol (D) is the major 1,2-addition product.

Question 2

During the reduction of ethyl acetate (CH₃COOCH₂CH₃) with LiAlH₄, the reaction proceeds via an addition-elimination-addition mechanism. Which of the following species is a transient intermediate that is formed and then consumed during this process?

  1. Acetaldehyde (Ethanal) (correct answer)
  2. Acetic acid
  3. A hemiacetal
  4. Acetone (Propanone)
Explanation: The reduction of an ester with LiAlH₄ involves two hydride additions. First, a hydride attacks the ester carbonyl, forming a tetrahedral intermediate. This intermediate collapses, eliminating the ethoxide (-OCH₂CH₃) leaving group to form acetaldehyde (ethanal). This intermediate aldehyde is more reactive than the starting ester and is immediately reduced by a second equivalent of hydride to form an ethoxide ion, which is then protonated during workup to yield ethanol. Acetic acid is an oxidation product, not an intermediate. A hemiacetal would require an alcohol to add to a carbonyl. Acetone has the wrong carbon skeleton.

Question 3

When N,N-dimethylbenzamide is treated with an excess of LiAlH₄ followed by an aqueous workup, what is the major organic product?

  1. Benzyl alcohol and dimethylamine
  2. Benzaldehyde and dimethylamine
  3. Benzyldimethylamine (correct answer)
  4. N,N-dimethylcyclohexylmethanamine
Explanation: Unlike esters or acid chlorides, which are reduced to primary alcohols, amides are reduced by LiAlH₄ to amines. The carbonyl group (C=O) is completely removed and replaced with a methylene group (CH₂). In this case, the carbonyl of N,N-dimethylbenzamide is reduced to a CH₂ group, connecting the benzyl ring to the nitrogen atom. This results in the formation of benzyldimethylamine. The products in (A) and (B) would result from cleavage of the C-N bond, which is not the typical pathway for amide reduction by LiAlH₄. The product in (D) involves reduction of the aromatic ring, which LiAlH₄ does not do under these conditions.

Question 4

To monitor the progress of a reaction reducing benzophenone (a ketone) to diphenylmethanol using NaBH₄, a student runs TLC plates. How would the Rf values of the starting material and the product be expected to compare on a silica gel plate?

  1. The product will have a higher Rf value because it is less polar than the starting material.
  2. The product will have a lower Rf value because it is more polar than the starting material. (correct answer)
  3. The product and starting material will have nearly identical Rf values, making TLC an ineffective monitoring technique.
  4. The product will have a higher Rf value because it has a lower molecular weight.
Explanation: Silica gel is a polar stationary phase. Polar compounds interact more strongly with the silica and therefore travel a shorter distance up the plate, resulting in a lower Rf value. The starting material, benzophenone, is a ketone and is moderately polar. The product, diphenylmethanol, is an alcohol. The hydroxyl group of the alcohol can engage in hydrogen bonding with the silica gel, making it significantly more polar than the ketone. Therefore, the product will be more strongly adsorbed to the silica and will have a lower Rf value than the starting material.

Question 5

A researcher has a sample of ethyl levulinate (ethyl 4-oxopentanoate) and wants to synthesize 1,4-pentanediol. Which of the following single reaction sequences will accomplish this transformation?

  1. Excess NaBH₄ in ethanol, followed by H₃O⁺ workup.
  2. Excess LiAlH₄ in THF, followed by H₃O⁺ workup. (correct answer)
  3. 1 equivalent of NaBH₄ in methanol, followed by H₃O⁺ workup.
  4. H₂ (excess), PtO₂, high pressure and heat.
Explanation: The starting material contains both a ketone and an ester functional group. The target product, 1,4-pentanediol, requires the reduction of both of these groups. LiAlH₄ is a powerful reducing agent capable of reducing both ketones and esters to alcohols. Therefore, treating the starting material with excess LiAlH₄ will produce the desired diol. NaBH₄ (A, C) is not strong enough to reduce the ester. Catalytic hydrogenation (D) could potentially work but often requires harsh conditions and may have selectivity issues; LiAlH₄ is the standard, most reliable reagent for this specific transformation in an introductory course context.

Question 6

The reduction of camphor, a bicyclic ketone, with LiAlH₄ gives a mixture of two diastereomeric alcohols, isoborneol and borneol. Isoborneol is the major product. This outcome is the result of what factor?

  1. Thermodynamic control, where the more stable alcohol product is formed preferentially.
  2. Kinetic control, where the reaction proceeds through a lower energy transition state.
  3. Electronic effects from the methyl groups, which direct the hydride to one face.
  4. Steric hindrance, where the hydride attacks the carbonyl from the less hindered face. (correct answer)
Explanation: When you encounter reduction reactions involving cyclic ketones, especially bicyclic systems like camphor, you need to consider the three-dimensional structure and how it affects reagent approach to the carbonyl carbon. In camphor's rigid bicyclic framework, the carbonyl group sits in an environment where one face is significantly more accessible than the other. The bulky hydride reagent LiAlH₄ will preferentially attack from whichever face offers the least steric obstruction. The face opposite to the bridging methyl groups provides this easier access, leading to isoborneol as the major product. This represents steric control of the reaction outcome, making answer D correct. Let's examine why the other options don't apply here. A is incorrect because this isn't thermodynamic control - we're not seeing equilibration to the most stable product, but rather kinetic selectivity based on approach. B mentions kinetic control, which sounds relevant, but it's too vague and doesn't identify the specific factor (sterics) that determines which transition state is lower in energy. C incorrectly attributes the selectivity to electronic effects from methyl groups, when the real issue is their physical bulk blocking one face of the carbonyl. Remember this pattern: in rigid cyclic systems, reduction reactions are typically controlled by steric accessibility rather than electronic effects or thermodynamic stability. Always visualize the three-dimensional structure and identify which face of the carbonyl is less crowded when predicting major products in these transformations.

Question 7

The standard workup for a LiAlH₄ reduction involves the sequential addition of water, followed by aqueous base (e.g., NaOH), and then more water (the Fieser method). What is the primary purpose of adding the aqueous base in this procedure?

  1. To catalyze the protonation of the product alcohol.
  2. To hydrolyze the product alcohol back to the carbonyl.
  3. To convert the aluminum byproducts into a granular, easily filterable precipitate. (correct answer)
  4. To ensure any unreacted LiAlH₄ is completely and safely neutralized.
Explanation: While the initial addition of water quenches the reaction and protonates the alkoxide, it forms gelatinous aluminum hydroxide (Al(OH)₃), which can be very difficult to filter and can trap the product, leading to low yields. The Fieser workup, involving the addition of aqueous base like NaOH, converts these aluminum byproducts into dense, granular inorganic salts (like NaAlO₂) that are much easier to remove from the organic solution by filtration. This greatly simplifies the purification of the desired alcohol product.

Question 8

A student plans to reduce a dicarboxylic acid, adipic acid, to 1,6-hexanediol. They set up the reaction using NaBH₄ in THF with a catalytic amount of BF₃·OEt₂ and gentle heating. What is the likely outcome of this experiment?

  1. A high yield of 1,6-hexanediol will be obtained.
  2. The adipic acid will be dehydrated to form a cyclic anhydride.
  3. Only one of the carboxylic acid groups will be reduced to form 6-hydroxyhexanoic acid.
  4. The reaction will fail, and the adipic acid will be recovered unchanged. (correct answer)
Explanation: When you encounter reduction reactions of carboxylic acids, you need to carefully consider the reducing agent's strength and selectivity. Carboxylic acids are among the most oxidized carbon functional groups and require very strong reducing agents for conversion to alcohols. Sodium borohydride (NaBH4\text{NaBH}_4) is a mild, selective reducing agent that effectively reduces aldehydes and ketones but cannot reduce carboxylic acids under normal conditions. Even with the Lewis acid catalyst BF3⋅OEt2\text{BF}_3 \cdot \text{OEt}_2 and gentle heating, NaBH4\text{NaBH}_4 lacks sufficient reducing power to break the strong C=O and C-O bonds in carboxylic acids. The reaction conditions described are insufficient to activate the carboxylic acid groups enough for reduction by this mild hydride donor. Answer choice A is incorrect because high yields require a much stronger reducing agent like lithium aluminum hydride (LiAlH4\text{LiAlH}_4). Answer B is wrong because anhydride formation from adipic acid would require much harsher dehydrating conditions and typically occurs with dicarboxylic acids that can form five- or six-membered rings more easily. Answer C represents a common misconception—if NaBH4\text{NaBH}_4 could reduce one carboxylic acid group, it would reduce both under the same conditions, but it cannot reduce either. The correct answer is D because the adipic acid will remain unchanged. Study tip: Remember the reducing agent hierarchy: NaBH4\text{NaBH}_4 for aldehydes/ketones, LiAlH4\text{LiAlH}_4 for carboxylic acids/esters. Always match the reducing agent strength to the substrate's resistance to reduction.

Question 9

A compound with the molecular formula C₅H₈O₃ is treated with excess LiAlH₄ in ether, followed by an H₃O⁺ workup, to yield a single organic product: 1,4-pentanediol. What was the structure of the starting material?

  1. 4-oxopentanoic acid (correct answer)
  2. Ethyl 3-oxobutanoate
  3. Glutaric anhydride (a cyclic anhydride)
  4. Methyl 4-oxopentanoate
Explanation: This is a retrosynthesis problem. The product is 1,4-pentanediol. LiAlH₄ reduces both ketones and carboxylic acids to alcohols. Let's analyze the options. A) 4-oxopentanoic acid has 5 carbons. The ketone at C-4 would be reduced to an alcohol, and the carboxylic acid at C-1 would be reduced to a primary alcohol, yielding 1,4-pentanediol. B) Ethyl 3-oxobutanoate has 6 carbons in total and would yield 1,3-butanediol and ethanol. C) Glutaric anhydride has 5 carbons and would be reduced to 1,5-pentanediol. D) Methyl 4-oxopentanoate has 6 carbons in total and would yield 1,4-pentanediol and methanol. Since the problem states a single organic product, the starting material must have had the correct 5-carbon backbone for the diol, which is 4-oxopentanoic acid.

Question 10

The reduction of a carbonyl compound with lithium aluminum hydride (LiAlH₄) is typically performed in an anhydrous ether solvent (like THF) and is followed by a separate aqueous workup step (e.g., adding H₃O⁺). What is the primary chemical reason for this mandatory two-step procedure?

  1. The ether solvent is required to dissolve the aluminum salts produced, and the workup step regenerates the LiAlH₄ for subsequent reductions.
  2. LiAlH₄ reacts violently with protic sources like water to produce H₂ gas, and the workup is needed to protonate the intermediate alkoxide. (correct answer)
  3. The initial reaction forms a stable, neutral intermediate that is insoluble in ether, requiring an acidic workup to induce precipitation and purification.
  4. The workup step is necessary to quench the reaction at the aldehyde stage when reducing an ester, preventing over-reduction to the alcohol.
Explanation: LiAlH₄ is an extremely reactive source of hydride (H⁻) that reacts violently and exothermically with any protic solvent (like water or alcohols) to liberate hydrogen gas, posing a safety hazard. Therefore, the reaction must be run in an anhydrous, aprotic solvent. The initial nucleophilic attack by the hydride on the carbonyl produces a metal alkoxide intermediate. The subsequent aqueous workup serves two purposes: safely quenching any excess LiAlH₄ and, crucially, providing a proton source to convert the alkoxide intermediate into the final alcohol product.

Question 11

A mixture containing one mole of benzaldehyde and one mole of methyl benzoate is treated with one mole of NaBH₄. After the reaction is complete and worked up, what will be the predominant species in the product mixture?

  1. Equal amounts of benzyl alcohol and methyl benzoate. (correct answer)
  2. Equal amounts of benzyl alcohol and benzoic acid.
  3. Mostly unreacted starting materials.
  4. Benzyl alcohol, phenol, and methanol.
Explanation: This is a competition experiment that tests chemoselectivity. Aldehydes are much more reactive towards NaBH₄ than esters. NaBH₄ will selectively reduce the benzaldehyde to benzyl alcohol. Since NaBH₄ does not react with esters under these conditions, the methyl benzoate will remain unchanged. Therefore, the major components of the mixture after the reaction will be the product from the aldehyde reduction (benzyl alcohol) and the unreacted ester (methyl benzoate).

Question 12

A researcher wishes to convert 4-oxobutanoic acid to 4-hydroxybutanoic acid, selectively reducing the ketone while leaving the carboxylic acid untouched. Which set of reagents is most appropriate for this transformation?

    1. LiAlH₄ in THF; 2. H₃O⁺
  2. NaBH₄ in CH₃OH (correct answer)
  3. H₂, Pd/C
    1. Ethylene glycol, H⁺; 2. NaBH₄, CH₃OH; 3. H₃O⁺
Explanation: The goal is to selectively reduce a ketone in the presence of a carboxylic acid. NaBH₄ is the ideal reagent as it readily reduces ketones but is unreactive towards carboxylic acids. LiAlH₄ (A) is too strong and would reduce both the ketone and the carboxylic acid to a diol. H₂/Pd-C (C) can reduce ketones, but conditions can also reduce carboxylic acids (though it's difficult) and is less chemoselective for this pair. Using a protecting group strategy (D) is overly complex and unnecessary, as the required selectivity can be achieved in a single step with NaBH₄.

Question 13

A student attempts a reaction by adding a solution of propanoyl chloride in methanol directly to a flask containing NaBH₄ powder dissolved in methanol at 0 °C. What is the expected major product?

  1. Propan-1-ol
  2. Propanal
  3. Methyl propanoate (correct answer)
  4. Propanoic acid
Explanation: This question tests the relative rates of two competing reactions: reduction and solvolysis. Propanoyl chloride is a highly reactive acyl chloride. While NaBH₄ can reduce it, the methanol solvent is a nucleophile that will rapidly attack the electrophilic acyl chloride to form methyl propanoate via nucleophilic acyl substitution. This esterification reaction is typically much faster than the reduction by NaBH₄. Once formed, the methyl propanoate is unreactive towards NaBH₄. Therefore, solvolysis will dominate over reduction, making methyl propanoate the major product.

Question 14

Which of the following carbonyl compounds would be reduced most rapidly by NaBH₄ in methanol?

  1. Benzophenone (diphenyl ketone)
  2. Acetone (dimethyl ketone)
  3. Formaldehyde (methanal) (correct answer)
  4. Cyclohexanone
Explanation: The rate of nucleophilic addition to a carbonyl group is governed by both electronic and steric factors. Electronically, aldehydes are more reactive than ketones because they have only one alkyl group donating electron density to the carbonyl carbon, making it more electrophilic. Sterically, aldehydes are less hindered than ketones. Comparing the options, formaldehyde is the least sterically hindered and most electrophilic aldehyde, as it has no alkyl groups attached to the carbonyl carbon. Therefore, it will react fastest with the nucleophilic hydride from NaBH₄.

Question 15

A researcher has a molecule containing both an epoxide and a ketone. They wish to reduce the ketone to an alcohol while leaving the epoxide ring intact. Which reagent would be the best choice?

    1. LiAlH₄ in THF; 2. H₃O⁺
    1. DIBAL-H in toluene; 2. H₂O
  3. H₂, Raney Ni
  4. NaBH₄ in CH₃OH at 0°C (correct answer)
Explanation: When you encounter a molecule with multiple functional groups that need different treatments, you're dealing with chemoselectivity - the ability to target one functional group while leaving others untouched. This requires understanding how different reducing agents behave. NaBH4NaBH_4 in methanol at 0°C (option D) is your best choice because it's a mild, selective reducing agent. Sodium borohydride specifically targets carbonyl groups (aldehydes and ketones) while being too weak to open stable three-membered rings like epoxides. The low temperature and protic solvent further ensure controlled, selective reduction of just the ketone to a secondary alcohol. Option A (LiAlH4LiAlH_4) is too powerful and indiscriminate - lithium aluminum hydride will reduce the ketone but will also open the epoxide ring through nucleophilic attack, destroying your desired functional group. Option B (DIBAL-H) is similarly too reactive; while it's more selective than LiAlH4LiAlH_4, it can still attack epoxides, especially under the reaction conditions needed for ketone reduction. Option C (H2H_2 with Raney Ni) uses hydrogenation conditions that will definitely open the epoxide ring through catalytic reduction, plus the harsh conditions aren't suitable for selective reduction. Remember this hierarchy: NaBH4NaBH_4 < DIBAL-H < LiAlH4LiAlH_4 in terms of reactivity. For chemoselectivity problems, always choose the mildest reagent that can still accomplish your primary goal. NaBH4NaBH_4 is often the answer when you need to reduce carbonyls while preserving other sensitive functional groups.

Question 16

Consider the reduction of 4-tert-butylcyclohexanone with NaBH₄ in ethanol. Due to steric hindrance, the reaction is stereoselective. What is the major stereoisomer produced?

  1. cis-4-tert-butylcyclohexanol, where the hydroxyl group is axial.
  2. trans-4-tert-butylcyclohexanol, where the hydroxyl group is equatorial. (correct answer)
  3. A racemic mixture of the cis and trans isomers in a nearly 1:1 ratio.
  4. The reaction will not proceed due to the severe steric hindrance from the tert-butyl group.
Explanation: The large tert-butyl group acts as a conformational lock, forcing it to occupy the equatorial position to minimize 1,3-diaxial interactions. This leaves the axial face of the carbonyl more sterically accessible to the approaching nucleophile (BH₄⁻). Therefore, the hydride attacks preferentially from the axial direction. This 'axial attack' results in the formation of the alkoxide in the equatorial position, which upon protonation gives the equatorial alcohol. The product with both the tert-butyl and hydroxyl groups in the more stable equatorial positions is the trans isomer.

Question 17

Which of the following statements provides the best mechanistic reason why NaBH₄ reduces aldehydes and ketones but not esters under standard conditions?

  1. Esters are significantly more sterically hindered around the carbonyl carbon than aldehydes or ketones, preventing the approach of the borohydride.
  2. The resonance stabilization of the ester group makes its carbonyl carbon much less electrophilic than that of an aldehyde or ketone.
  3. The tetrahedral intermediate formed from an ester can reform the carbonyl by ejecting a good alkoxide leaving group, whereas the aldehyde intermediate cannot. (correct answer)
  4. The sodium ion in NaBH₄ is not a strong enough Lewis acid to coordinate to and activate the ester carbonyl oxygen.
Explanation: While sterics (A), electronics (B), and Lewis acidity (D) all play a role in the overall reactivity difference, the most critical mechanistic distinction is the fate of the tetrahedral intermediate. When a hydride attacks an aldehyde or ketone, the resulting alkoxide intermediate has only hydride or alkyl/aryl groups attached to the carbon, which are terrible leaving groups. The reaction is effectively irreversible. When a hydride attacks an ester, the tetrahedral intermediate has an alkoxide (-OR) group, which is a competent leaving group. The intermediate can collapse, ejecting the -OR group and regenerating a carbonyl (an aldehyde), which is then rapidly reduced. However, this first addition step is reversible and often has a high activation energy, preventing the reaction from proceeding with a mild reducing agent like NaBH₄.

Question 18

Which statement best explains why LiAlH₄ is a significantly more potent reducing agent than NaBH₄?

  1. The Al-H bond is more polar and weaker than the B-H bond, making the hydride more available for nucleophilic attack. (correct answer)
  2. Sodium is more electropositive than lithium, making the borohydride anion a more stable and less reactive nucleophile.
  3. LiAlH₄ reactions are typically run at higher temperatures, which increases the reaction rate compared to NaBH₄ reactions.
  4. Boron can form stable complexes with protic solvents, whereas aluminum cannot, which deactivates the NaBH₄ reagent.
Explanation: The difference in reactivity stems from the nature of the metal-hydrogen bond. Aluminum is less electronegative than boron. This makes the Al-H bond more polar and weaker than the B-H bond. Consequently, the hydrogen atom in the Al-H bond carries a greater partial negative charge and is more 'hydridic' (H⁻-like). This makes the hydride easier to deliver as a nucleophile, rendering LiAlH₄ a much stronger reducing agent than NaBH₄.