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Chapter 5: Problem 180

Statement 1 All aldehydes and ketones react with HCN to give a racemic mixture of the corresponding cyanohydrin. and Statement 2 Cyanide ion is an ambident nucleophile.

Short Answer

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Answer: Aldehydes and ketones react with HCN to give a racemic mixture of cyanohydrin products because the cyanide ion is an ambident nucleophile that can attack the carbonyl carbon from either side of the planar C=O double bond, forming both (R) and (S) enantiomers. This results in the formation of a racemic mixture containing equal amounts of both enantiomers.

Step by step solution

01

Statement 1: Reaction of aldehydes and ketones with HCN

Aldehydes and ketones are carbonyl compounds containing a C=O double bond. Upon reaction with hydrogen cyanide (HCN), these compounds form cyanohydrins, which have the general structure R2C(OH)(CN). The reaction occurs via nucleophilic addition, with the HCN acting as a nucleophile attacking the carbonyl carbon.
02

Statement 2: Ambident Nucleophile

An ambident nucleophile is a nucleophile that can attack from two different sites with similar reactivity. In the case of the cyanide ion (CN-), it can attack either from the carbon or nitrogen atom. However, the carbon atom is more nucleophilic due to its higher negative charge compared to nitrogen.
03

Nucleophilic Attack

In the reaction between an aldehyde or ketone and HCN, the cyanide ion (CN-) acts as a nucleophile, attacking the carbonyl carbon. This step results in the opening of the C=O double bond and the formation of an O- anion. Simultaneously, the lone pair on the nitrogen atom of HCN protonates one hydrogen from a CN bond forming a new C-C bond. Now, we have an intermediate with the structure of an alkoxyanion, R2C(O-)(CN).
04

Protonation

In the next step, the O- anion present in the intermediate can capture a proton from the solvent, which is usually an acidic environment. This will result in the formation of a hydroxyl group (OH) at the oxygen atom, thus giving a final product called cyanohydrin, R2C(OH)(CN).
05

Racemic Mixture Formation

During the nucleophilic addition step, the CN- can attack the carbonyl carbon from either side of the planar C=O double bond. This results in the formation of both (R) and (S) enantiomers of the product, giving a racemic mixture. A racemic mixture is a mixture containing equal amounts of both enantiomers and is optically inactive. In conclusion, aldehydes and ketones react with HCN to give a racemic mixture of cyanohydrin products because the cyanide ion is an ambident nucleophile that can attack the carbonyl carbon from either side of the planar C=O double bond, forming both (R) and (S) enantiomers.

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Most popular questions from this chapter

\(2 \mathrm{CH}_{3}-\mathrm{CHO} \longrightarrow \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}+\mathrm{H}_{2} \mathrm{O}\) The above reaction is catalysed by (a) bases only (b) acids only (c) both base and acid (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\)

Bromobenzene can be converted to benzoic acid by (a) treating with \(\mathrm{Mg}\) and then carboxylating. (b) treating with \(\mathrm{NaOH}\) and then carboxylating. (c) treating with KCN and then hydrolysing. (d) oxidation with \(\mathrm{KMnO}_{4}\).

Malonic acid on heating at \(430 \mathrm{~K}\) yields (a) formic acid (b) succinic acid (c) acetic acid (d) lactic acid

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Directions: Each question contains Statement- 1 and Statement- 2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement- 1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 Formic acid produces a white precipitate with \(\mathrm{HgCl}_{2}\). and Statement 2 Mercuric salt of formic acid is insoluble and therefore gets precipitated.
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