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Chapter 4: Problem 66

The product 'P' in the following reaction is CCCC=O \(\stackrel{\mathrm{NaHSO}_{3}}{\longrightarrow}(\mathrm{P})\) (A) Non-crystalline CCCC(O)OC(=O)[O-] , Crystalline (C) CCCC(O)OC(=O)[O-] , Non-crystalline (D) CCCC(O)S(=O)(=O)[O-]

Short Answer

Expert verified
The correct product 'P' in the reaction is option (D) with the SMILES notation CCCC(O)S(=O)(=O)[O-].

Step by step solution

01

Understand the reaction and properties of sodium bisulfite

Sodium bisulfite (NaHSO3) is a weak acid that can act as a reducing agent in some reactions. In this case, we have an aldehyde group in the reactant, represented by the SMILES notation "CCCC=O". Sodium bisulfite can react with aldehydes to form bisulfite adducts.
02

Write balanced equation for the reaction

When the aldehyde represented by "CCCC=O" reacts with sodium bisulfite (NaHSO3), it forms a bisulfite adduct. The balanced equation for this reaction can be written as: \[ \mathrm{CCCC=O + NaHSO_3 \longrightarrow CCCC(O)S(=O)(=O)[O-] + Na^+} \]
03

Compare the options and determine the correct product

Now, we will compare the products given in the options with the bisulfite adduct formed in the reaction: (A) Non-crystalline: CCCC(O)OC(=O)[O-] - this product is not a bisulfite adduct (B) Crystalline: CCCC(O)OC(=O)[O-] - this product is not a bisulfite adduct (C) Non-crystalline: CCCC(O)OC(=O)[O-] - this product is not a bisulfite adduct (D) CCCC(O)S(=O)(=O)[O-] - this product is a bisulfite adduct Based on the comparison, the correct product 'P' in the reaction is given by option (D), which has the SMILES notation CCCC(O)S(=O)(=O)[O-].

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Most popular questions from this chapter

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. \(\mathrm{NH}_{3}\)

For the complete reduction of \(5.8 \mathrm{~g}\) of acetone to isopropyl alcohol, the quantity of \(\mathrm{LiAlH}_{4}\) required (assuming chemical yield to be \(100 \%\) ) is approximately [Mass: \(\mathrm{Li}=6.9, \mathrm{Al}=27\) ] (A) \(3.8 \mathrm{~g}\) (B) \(0.95 \mathrm{~g}\) (C) \(1.9 \mathrm{~g}\) (D) \(15.2 \mathrm{~g}\)

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