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Chapter 5: Problem 50

Tollens' reagent is reduced by (a) methanoic acid (b) ethanal (c) benzaldehyde (d) all the above

Short Answer

Expert verified
(a) methanoic acid (b) ethanal (c) benzaldehyde Answer: Both (b) ethanal and (c) benzaldehyde can reduce Tollens' reagent because they contain aldehyde functional groups (-CHO).

Step by step solution

01

Identify the functional groups

First, let's identify the functional groups in each of the given compounds: methanoic acid, ethanal, and benzaldehyde. This will help us determine whether these compounds can reduce Tollens' reagent or not. - Methanoic acid has a carboxylic acid functional group (-COOH). - Ethanal has an aldehyde functional group (-CHO). - Benzaldehyde also has an aldehyde functional group (-CHO).
02

Check for aldehyde groups

Now that we know the functional groups of each compound, let's remember that Tollens' reagent can be reduced by aldehydes. This means that we are looking for compounds with the aldehyde functional group (-CHO).
03

Determine which compounds can reduce Tollens' reagent

Looking at the functional groups of each compound: - Methanoic acid does not have an aldehyde group; therefore, it cannot reduce Tollens' reagent. - Ethanal has an aldehyde group, so it can reduce Tollens' reagent. - Benzaldehyde also has an aldehyde group, so it too can reduce Tollens' reagent.
04

Choose the correct answer

Of the given options: (a) methanoic acid (b) ethanal (c) benzaldehyde (d) all the above The correct answer is (b) ethanal and (c) benzaldehyde, as they have aldehyde groups and can reduce Tollens' reagent. However, since there is no option for selecting both (b) and (c), the exercise seems to have a slight issue. In this case, it would be best to explain the situation to the teacher and discuss the correct answer.

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Most popular questions from this chapter

Match the elements of Column I to elements of Column II. There can be single or multiple matches. Column I (a) Cannizzaro reaction (b) Perkin reaction (c) Kolbe reaction (d) \(\mathrm{CH}_{3}-\mathrm{CHO}+\mathrm{NaOH}\) Column II (p) Crotonaldehyde (q) Carbanion intermediate (r) Electrophilic substitution (s) Disproportionation (t) Formation of new carbon-carbon bond.

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.

The product (P) formed in the following reaction is \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH} \frac{\text { (i) NaNH }_{2}}{\text { (ii) HCHo }}(\mathrm{P})\) (a) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{COONa}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CHO}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\)

The IUPAC name of is CCC(C=O)CC(C)C (a) 2 -ethyl-4-methylpentanal (b) 4 -methyl 2 -ethylpentanal (c) 2 -methylhexanal (d) 2 -ethylhexanal

Acetaldehyde when heated with \(\mathrm{NaOH}\) gives (a) but-2-enal (b) but- 3 -enal (c) propenal (d) butanal
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