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Chapter 5: Problem 21

When \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CHO}\) reacts with HCHO in \(\mathrm{NaOH}\) followed by acidification, the products are (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{HCOONa}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COONa}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{HCOOH}\)

Short Answer

Expert verified
Answer: The products of this reaction are (CH3)3C-CH2OH and HCOOH.

Step by step solution

01

Identify the given reactants

The given reactants are \((CH_3)_3C-CHO\) and HCHO. \((CH_3)_3C-CHO\) is an aldehyde, and HCHO is also an aldehyde known as formaldehyde. The base NaOH is used in the reaction followed by acidification, which means we will first react the given compounds under basic (alkaline) conditions and then under acidic conditions.
02

Reaction under basic conditions

When aldehydes react with HCHO in the presence of NaOH, it undergoes the Cannizzaro reaction. In this reaction, one molecule of aldehyde is reduced (gain electron) to an alcohol, and another molecule is oxidized (lose electron) to a carboxylic acid or its sodium salt. The reducing agent used in this reaction is the formaldehyde (HCHO). Since \((CH_3)_3C-CHO\) is the only aldehyde available in the reaction mixture, one molecule of it will be reduced to alcohol to get \((CH_3)_3C-CH_2OH\). On the other hand, the formaldehyde will get oxidized to formic acid (\(HCOOH\)) or its sodium salt, depending on the step of the reaction.
03

Acidification

After the formation of alcohol and formate ion, acidification takes place. Acidification consists of adding an acid to the reaction medium, protonating the formate ion to form formic acid (\(HCOOH\)). Now, we have \((CH_3)_3C-CH_2OH\) and \(HCOOH\) as reaction products.
04

Match the products with the given options

The products obtained from our calculations are \((CH_3)_3C-CH_2OH\) and \(HCOOH\). Look at the given options and find the one that matches these products: (a) \((CH_3)_3C-COOH\) and \(CH_3OH\) (b) \((CH_3)_3C-CH_2OH\) and \(HCOONa\) (c) \((CH_3)_3C-COONa\) and \(CH_3OH\) (d) \((CH_3)_3C-CH_2OH\) and \(HCOOH\) The correct option is (d) as it matches the products obtained from our analysis.

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Most popular questions from this chapter

When acetic acid is treated with aqueous sodium bicarbonate, \(\mathrm{CO}_{2}\) gas is evolved. The 'Carbon' of \(\mathrm{CO}_{2}\) comes from (a) carboxyl group (b) methyl group (c) bicarbonate (d) both carboxyl group and bicarbonate

Identify the correct statements among the following. (a) Formaldehyde can react with both \(-\mathrm{NH}_{2}\) groups in semicarbazide. (b) In mixed Cannizzaro reaction involving HCHO, it is always the intermediate formed from HCHO which transfers its hydride to the other aldehyde. (c) Benzaldehyde reduces both Tollens' reagent and Fehling's solution. (d) In Fehling's solution, Cu(II) ions are kept in soluble form of a complex in alkaline medium.

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

Which of the following reactions will give acetophenone as a product? (a) Acid catalysed hydration of styrene followed by oxidation with hot alkaline \(\mathrm{KMnO}_{4}\). (b) Phenyl acetylene \(\stackrel{\mathrm{d} \mathrm{H}, \mathrm{SO}_{4}^{\prime} \mathrm{H}_{\mathrm{E}}^{2+}}{\longrightarrow}\) Aldehydes, Ketones and Carboxylic Acids (c) Reaction of benzoyl chloride with dimethyl cadmium. (d) Dry distillation of a mixture of calcium acetate and calcium benzoate.

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