Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The correct statement is/are (A) Pgives white ppt. with ammonical \(\mathrm{AgNO}_{3}\) (B) Qgives orange colour with 2,4-DNP (C) Qgivescrystalline compound with \(\mathrm{NaHSO}_{3}\) (D) Pgives red colour of \(\mathrm{Cu}_{2} \mathrm{O}\) with ammonical \(\mathrm{Cu}_{2} \mathrm{Cl}\),

Short Answer

Expert verified
The correct statements are (A), (B), and (C). Statement (A) suggests P contains a halide ion as it forms a white precipitate with ammoniacal silver nitrate, statement (B) indicates that Q is a carbonyl compound since it produces an orange color with 2,4-DNP, and statement (C) implies that Q is an aldehyde as it forms a crystalline compound with NaHSO3.

Step by step solution

01

Statement A: P gives white ppt. with ammonical AgNO3

Ammoniacal silver nitrate is commonly used to test for the presence of halide ions (such as chloride, bromide, and iodide) in a compound. When a halide is present, a white precipitate of silver halide generally forms. Since this statement suggests that P forms a white precipitate with ammoniacal silver nitrate, it is likely that P contains a halide ion. Therefore, statement A is a possible correct statement.
02

Statement B: Q gives orange colour with 2,4-DNP

2,4-Dinitrophenylhydrazine (2,4-DNP) is a reagent used to test for the presence of carbonyl compounds, such as aldehydes and ketones. When a carbonyl compound reacts with 2,4-DNP, it forms an orange-red precipitate. Given that Q produces an orange color with 2,4-DNP, it appears that Q is a carbonyl compound. This makes statement B a possible correct statement.
03

Statement C: Q gives crystalline compound with NaHSO3

Sodium bisulfite (NaHSO3) is another reagent used to test for the presence of aldehydes. When an aldehyde reacts with NaHSO3, a crystalline bisulfite addition product forms. Since Q forms a crystalline compound with NaHSO3, this suggests that Q is an aldehyde. This evidence supports statement B and makes statement C a possible correct statement as well.
04

Statement D: P gives red colour of Cu2O with ammonical Cu2Cl

Ammoniacal copper(II) chloride is a Fehling's reagent, commonly used to test for reducing sugars. If P is a reducing sugar, it would react with ammoniacal copper(II) chloride to produce a red precipitate of copper(I) oxide (Cu2O). However, statement A indicates that P contains a halide ion, which is inconsistent with P being a reducing sugar. This inconsistency makes statement D incorrect. #Conclusion# Based on our analysis, the correct statements are: (A) P gives white ppt. with ammonical AgNO3 (B) Q gives orange colour with 2,4-DNP (C) Q gives crystalline compound with NaHSO3

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Halide Ion Testing
Testing for halide ions is a fundamental concept in organic chemistry. Halide ions include chloride, bromide, and iodide ions. To test for their presence in a compound, ammoniacal silver nitrate (AgNO₃) is used.

When halide ions are present, they react with silver nitrate, resulting in the formation of a visible precipitate. Chloride ions form a white precipitate of silver chloride (AgCl). This test is crucial because it helps identify halide-containing compounds simply and effectively.
  • Chloride ions: white precipitate
  • Bromide ions: cream precipitate
  • Iodide ions: yellow precipitate
Recognizing these color changes is essential in laboratory scenarios, as they provide vital clues about the compound under investigation.
Carbonyl Compound Identification
Carbonyl compounds, such as aldehydes and ketones, contain a carbon-oxygen double bond. These compounds are pivotal in organic chemistry due to their reactivity. Identifying them often involves specific tests.

One of the traditional methods includes the reaction with 2,4-Dinitrophenylhydrazine (2,4-DNP). This test distinguishes carbonyl compounds by forming colored precipitates: an orange-red color indicates a positive test.

Recognizing carbonyl compounds is important for many chemical processes, given their diverse roles in synthesis and reactivity in forming other chemical products.
Aldehyde Reactions
Aldehydes are a subgroup of carbonyl compounds characterized by the presence of at least one hydrogen atom linked to the carbonyl carbon. They are significant in organic reactions because of their ability to undergo oxidation and reduction reactions.

Key tests for identifying aldehydes include their reaction with sodium bisulfite (NaHSO₃), forming a crystalline bisulfite addition product. This reaction is an important method for distinguishing aldehydes from other carbonyl-containing molecules, as not all carbonyl compounds form such addition products.

By understanding these reactions, chemists can identify aldehydes and tailor their chemical synthesis accordingly.
Fehling's Reagent
Fehling's reagent is a chemical solution used to differentiate between aldehyde functional groups and ketonic groups. It is specifically efficient in identifying reducing sugars and aldoses, which are types of aldehydes.

The reagent reacts with these compounds, often producing a red precipitate of copper(I) oxide (Cu₂O) when a reducing sugar is present. This reaction forms the basis for distinguishing reducing sugars from non-reducing sugars, aiding in the analysis of organic materials.
2,4-Dinitrophenylhydrazine Test
The 2,4-Dinitrophenylhydrazine (DNPH) test is a widespread qualitative test for carbonyl groups in aldehydes and ketones. It hinges on the ability of DNPH to form a hydrazone with the carbonyl group, producing an intensely colored precipitate.

The core reaction involves nucleophilic addition of the hydrazine nitrogen to the carbonyl carbon, generating the corresponding hydrazone. An orange-red precipitate confirms a positive result, distinguishing aldehydes and ketones from other compounds.

This test is crucial because it offers a straightforward, visual means of identifying carbonyl compounds, which constitute a significant portion of organic chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be (A) Primary structure of proteins (B) Secondary structure of proteins (C) Tertiary structure of proteins (D) Quaternary structure of proteins

During configuration determination of D-Glucose, there was a doubt between the following Isomers. A scientist thought that following reaction sequence help to find out correct D-glucose. If \(\mathrm{R}\) is also a Aldohexose different from sample, then correct configuration of D-Glucose concluded by scientist is (A) I (B) II (C) III (D) Can't differentiate between I and III

Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components of photochemical smog, identify it. (A) \(\mathrm{NO}_{2}\) (B) \(\mathrm{O}_{3}\) (C) \(\mathrm{SO}_{2}\) (D) Unsaturated hydrocarbon

Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a (A) Monosaccharide (B) Disaccharide (C) Reducing sugar (D) Non-reducing sugar

The Hinsberg test of a compound, \(\mathrm{C}_{5} \mathrm{H}_{14} \mathrm{~N}_{2}\) produces a solid that is insoluble in \(10 \%\) aq. \(\mathrm{NaOH}\). This solid derivative dissolves in \(10 \%\) aqueous sulphuric acid. Which of the following would best describe these facts? (A) \(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}\) (B) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NHCH}_{3}\) (C) \(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (D) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NCH}_{2} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free