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Which of the following reagents is not useful for reduction of aldehydes and ketones? (A) Sodium hydride (B) Sodium borohydride (C) Lithium aluminium hydride (D) Hydrogen gas \(/ \mathrm{Pt}\)

Short Answer

Expert verified
The reagent that is not useful for the reduction of aldehydes and ketones is Sodium hydride (Choice A) as it is a strong base primarily used in deprotonation reactions and not as a reducing agent.

Step by step solution

01

Brief overview of aldehydes and ketones reduction

In general, the reduction of aldehydes and ketones involves the conversion of a carbonyl group (C=O) into an alcohol (O-H). Different reducing agents can be used for this reaction, with each reagent having certain selectivity and reactivity.
02

Evaluate each reagent

(A) Sodium hydride (NaH): Sodium hydride is a strong base and used as a reagent in deprotonation reactions. However, it is generally not used as a reducing agent. (B) Sodium borohydride (NaBH4): Sodium borohydride is a mild reducing agent commonly used for converting aldehydes and ketones into alcohols. It is selective towards aldehydes and ketones and doesn't reduce esters, amides, or carboxylic acids. (C) Lithium aluminium hydride (LiAlH4): Lithium aluminium hydride is a very strong reducing agent which can convert aldehydes and ketones into alcohols. Moreover, it can also reduce esters, amides, and carboxylic acids to alcohols, making it less selective than sodium borohydride. (D) Hydrogen gas / Pt (H2/Pt): Hydrogen gas in the presence of a platinum catalyst is another reducing agent for aldehydes and ketones. It is commonly used as a mild reductant for carbonyl compounds and as a method to hydrogenate carbon-carbon double bonds.
03

Identify the non-useful reagent

Based on the information above, we can conclude that Sodium hydride (Choice A) is not useful for the reduction of aldehydes and ketones, as it is a strong base and not a reducing agent.

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Key Concepts

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

Sodium Hydride
Sodium hydride (NaH) is a chemical compound that is often used in organic synthesis, but perhaps counterintuitively, it is not a suitable reagent for the reduction of aldehydes and ketones. Its primary role is as a strong base, making it excellent for deprotonation reactions, where it can remove a hydrogen atom from molecules to form hydride ions.

For instance, in the reaction of NaH with an alcohol, NaH would abstract the acidic hydrogen to form hydrogen gas and a corresponding alkoxide ion. However, since reduction involves the addition of hydrogen, NaH cannot serve this purpose due to its inability to donate hydrogen atoms. Therefore, despite its potency in other reactions, NaH is not a reducing agent and is not used in the context of transforming carbonyl groups into alcohols.

  • Common usage: deprotonation reactions to form hydride ions.
  • Non-usage: NaH does not engage in the addition of hydrogen atoms to other substances, thus not fulfilling the role of a reducing agent for carbonyl compounds.
Sodium Borohydride
Sodium borohydride (NaBH4) is recognized as a mild and selective reducing agent, especially in the realm of carbohydrate and nucleotide chemistry. It is particularly useful for its ability to reduce aldehydes and ketones to primary and secondary alcohols, respectively.

Chemically, the borohydride ion in NaBH4 easily donates a hydride ion to the carbonyl carbon of aldehydes and ketones, effectively transforming the carbon-oxygen double bond (C=O) into a single bond and thereby creating an alcohol group (O-H). Its selectivity is a major advantage as it typically avoids the reduction of other functional groups like esters, amides, or carboxylic acids, which can be present in the same molecule.

  • Key role: Reduction of carbonyl compounds to alcohols.
  • Selectivity: Preferentially reduces aldehydes and ketones over esters, amides, or carboxylic acids.
Lithium Aluminium Hydride
Lithium aluminium hydride (LiAlH4) is a powerhouse reducing agent in organic chemistry, known for its ability to convert a wide range of functional groups to alcohols. It is considerably more reactive than sodium borohydride and is not as selective.

LiAlH4 is capable of donating hydride ions to aldehydes and ketones, much like NaBH4, but it doesn't stop there. It can also reduce esters, amides, and carboxylic acids aggressively, sometimes making it less desirable when selective reduction is needed. This powerful reactivity comes from the strong nucleophilic character of the hydride ions produced by LiAlH4, allowing it to break multiple types of chemical bonds.

  • Strength: Reduces a wide array of functional groups beyond aldehydes and ketones.
  • Reactivity: More aggressive reducing agent than NaBH4, caution is needed for selective reductions.
Hydrogen Gas Catalytic Reduction
The catalytic hydrogenation of aldehydes and ketones using hydrogen gas (H2) and a noble metal catalyst such as platinum (Pt) is a classical method in organic synthesis. This technique is primarily used to fully or selectively reduce carbon-oxygen double bonds in carbonyl compounds to produce alcohols.

This method of reductions is characterized by its use of molecular hydrogen which, in the presence of a platinum catalyst, dissociates into atomic hydrogen capable of being added across the carbonyl bond. This process is mild compared to the use of metal hydrides and is often employed when a delicate touch is necessary, such as in the hydrogenation of carbon-carbon double bonds without affecting other potentially reactive functional groups.

  • Application: Mild and controlled reduction of carbonyl compounds.
  • Advantages: Selective hydrogenation that can preserve sensitive functional groups.

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Most popular questions from this chapter

Ninhydrin when undergoes hydration, which carbonyl carbon is first attacked by \(\mathrm{H}_{2} \mathrm{O}\) ? (A) 1 (B) 2 (C) 3 (D) None

Which of the following alcohols reacts fastest with concentrated hydrochloric acid in the presence of anhydrous zinc chloride? (A) Butan-2-ol (B) Butan-1-ol (C) 2-Methylpropan-1-ol (D) 2-Methylpropan-2-ol

(P) \(\frac{\text { (i) } \mathrm{Hg}(\mathrm{OAc})_{2} / \mathrm{H}_{2} \mathrm{O}}{\text { (ii) } \mathrm{NaBH}_{4} / \mathrm{OH}}(\mathrm{Q}) \stackrel{\mathrm{O}_{3}}{\mathrm{Zn}}(\mathrm{R})\) Which of the following is not possible as (P)? (A) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (B) \(\mathrm{HC} \equiv \mathrm{CH}\) (C) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\) (D) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)

During oxidation of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{\oplus}, \Delta\) species which does not oxidises is: (A) \(\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{OH}\) (B) \(\mathrm{Cr}(\mathrm{IV})\) (C) \(\operatorname{Cr}(\mathrm{V})\) (D) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})_{2}\)

In the given transformation, which of the following is the most appropriate reagent? CC(=O)C=CC1CCC(O)CC1 CCC=CC1CCC(O)CC1 (A) \(\mathrm{NaBH}_{4}\) (B) (i) \(\mathrm{NH}_{2} \mathrm{NH}_{2}\) (ii) \(\stackrel{\ominus}{\mathrm{O}} \mathrm{H}, \Delta\) (C) \(\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}\) (D) \(\mathrm{Na}\), Liq. \(\mathrm{NH}_{3}\)

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