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

Which among the following will give a precipitate with ammonical silver nitrate? (a) 2 -butene (b) 2 -butyne (c) Chlorobenzene (d) 3-methyl-1-butyne

Short Answer

Expert verified
3-methyl-1-butyne will give a precipitate with ammonical silver nitrate.

Step by step solution

01

Identify the Test

The ammonical silver nitrate test is used to identify terminal alkynes, which react with silver nitrate to form a precipitate of silver acetylide.
02

Analyze Each Compound

We must determine if each given compound is a terminal alkyne. Only terminal alkynes will react with ammonical silver nitrate to form a precipitate.
03

Check 2-butene

2-butene is an alkene, not an alkyne. Therefore, it does not form a precipitate with ammonical silver nitrate.
04

Check 2-butyne

2-butyne is a non-terminal alkyne. It has a triple bond, but because it is not terminal, it does not form a precipitate with ammonical silver nitrate.
05

Check Chlorobenzene

Chlorobenzene is an aryl halide, which does not react with ammonical silver nitrate to form a precipitate.
06

Check 3-methyl-1-butyne

3-methyl-1-butyne is a terminal alkyne. It will react with ammonical silver nitrate to form a precipitate of silver acetylide.
07

Determine the Solution

Since 3-methyl-1-butyne is the only terminal alkyne among the options, it is the correct answer that forms a precipitate with ammonical silver nitrate.

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.

Terminal Alkynes
Terminal alkynes are a fascinating group of organic compounds characterized by the presence of a triple bond between carbon atoms. The defining feature of terminal alkynes is that this triple bond is located at the end of a carbon chain. This structure allows them to participate in unique chemical reactions not shared by other types of alkynes, such as internal alkynes.

The chemical structure of terminal alkynes can be represented as R-C≡C-H, where the "R" represents a carbon-containing group (it can be a hydrogen atom or a longer carbon chain). Due to the presence of the hydrogen atom directly bonded to the carbon-carbon triple bond, terminal alkynes are more reactive compared to internal alkynes, which lack this hydrogen atom. This increased reactivity plays a crucial role in a variety of organic reactions, including the ammonical silver nitrate test, as terminal alkynes can easily participate in nucleophilic addition reactions.

Notably, this characteristic makes terminal alkynes perfect candidates for identification using specific tests like the ammonical silver nitrate test, which reveals the presence of this functional group by forming a precipitate.
Silver Acetylide Formation
The formation of silver acetylide is a key chemical reaction that occurs when terminal alkynes react with silver ions. In the presence of ammonical silver nitrate, terminal alkynes undergo a chemical transformation to produce silver acetylide, which is an insoluble compound that precipitates out of solution.

Here's a simplified look at what happens during the reaction:
  • A terminal alkyne with the structure R-C≡C-H reacts with ammoniacal silver nitrate.
  • The hydrogen atom directly bonded to the terminal carbon of the alkyne is replaced by the silver ion (Ag+), resulting in the formation of a silver acetylide (R-C≡C-Ag).
  • This silver acetylide precipitates as a gray or white solid.
This reaction is highly specific to terminal alkynes due to the acidic hydrogen available for replacement at the terminal triple bond. Other compounds (such as internal alkynes, alkenes, or aryl halides) lack this specific hydrogen and do not form a precipitate, making this reaction a key test for identifying terminal alkynes in a mixture of diverse organic compounds.
Organic Chemistry Reactions
Organic chemistry is full of fascinating reactions that showcase the versatility and complexity of carbon-based compounds. Among these, the reactivity of terminal alkynes stands out due to their unique ability to participate in specific addition reactions.

Several types of organic chemistry reactions are essential in understanding how terminal alkynes behave:
  • Nucleophilic Addition: This is a common reaction mechanism for terminal alkynes, where a nucleophile attacks the electrophilic carbon at the end of the alkyne. Such reactions are key for forming compounds like silver acetylide during the ammonical silver nitrate test.
  • Hydrogenation: Employed to convert alkynes to alkanes by adding hydrogen across the triple bond.
  • Halogenation: Involves the addition of halogens like chlorine or bromine to the triple bond, which results in the formation of dihalide compounds.
Understanding these reactions not only helps in identifying specific functional groups but also paves the way for diverse applications in organic synthesis, pharmaceuticals, and material sciences. Recognizing how terminal alkynes interact within these reactions underscores their importance and utility in 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

In the reaction of phenol with \(\mathrm{CHCl}_{3}\) and aqueous \(\mathrm{NaOH}\) at \(70^{\circ} \mathrm{C}(343 \mathrm{~K})\), the electrophile attacking the ring is: (a) \(\mathrm{CHCl}_{3}\) (b) \(\mathrm{CHCl}_{2}\) (c) \(\mathrm{CCl}_{2}\) (d) \(\mathrm{COCl}_{2}\)

\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CMgCl}\) on reaction with \(\mathrm{D}_{2} \mathrm{O}\) produces: (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CD}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COD}\) (c) \(\left(\mathrm{CD}_{3}\right)_{3} \mathrm{CD}\) (d) \(\left(\mathrm{CD}_{3}\right)_{3} \mathrm{COD}\)

Match the following: List 1 List 2 (Reagent) (Electrophiles) \(\oplus\) 1\. \(\mathrm{Cl}_{2}+\mathrm{AlCl}_{3}\) (1) \(\mathrm{NO}_{2}\) 2\. \(\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (2) \(\mathrm{Cl} \ldots \mathrm{Cl} \ldots \mathrm{AlCl}_{3}\) or \(\mathrm{Cl}^{+}\) 3\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\) (3) \(\mathrm{SO}_{3} \mathrm{H}\) (or \(\left.\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{SO}_{3}\right)\) 4\. \(\mathrm{Br}_{2}+\mathrm{Fe}\) (4) \(\mathrm{SO}_{3}\) (5) Br...Br...FeBr \(_{3}\) or \(\mathrm{Br}^{+}\) The correct matching is: \(\begin{array}{lll}1 & 2 & 3\end{array}\) 4 (a) (4) (3) (2) (b) (2) (1) (c) (2) (1) (d) (2) (3) (1) (4)

Ozonolysis of 2,3 -dimethyl-1-butene followed by reduction with zinc and water gives: (a) Methanoic acid and 3-methyl-2-butanone (b) Methanal and 3-methyl-2-butanone (c) Methanal and 2-methyl-3-butanone (d) Methanoic acid and 2 -methyl-3-butanone

Carbon black, which is used in making printer's ink is obtained by the decomposition of: (a) Acetylene (b) Benzene (c) Carbon tetrahedral (d) Methane

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