
A compound A having molecular formula ${C_4}{H_9}Br$ on reaction with alcoholic $KOH$ gives a compound B. Bromination of B gives compound C. Compound C on treatment with soda lime gives a gaseous compound D. The gas D when passed through an ammoniacal silver nitrate solution forms white precipitate. Identify A, B, C and D and write the reactions involved.
Answer
221.1k+ views
Hint: The compound involves a series of reactions including qualitative analysis of organic compounds. As per the given question, compound D on reaction with ammoniacal silver nitrate gives white ppt which is a qualitative test for terminal alkyne i.e., compound D can be identified as terminal alkyne.
Complete Step by Step Solution:
Qualitative Analysis of terminal alkyne:
Terminal alkynes i.e., the alkynes which consist of a triple bond at first carbon react with ammoniacal silver nitrate to form white precipitate. This test is confirmed by the compound D which is given in the question. This means, the compound A is an aliphatic saturated carbon chain consisting of bromine at first carbon.
Thus, compound A will be : $C{H_3} - C{H_2} - C{H_2} - C{H_2} - Br$
As given in the question, compound A on reaction with alcoholic $KOH$ gives compound B as follows:
$C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-Br\xrightarrow[-HBr]{alc.KOH}\underset{B}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}}}\,$
On further bromination of compound B gives compound C as per the following reaction:
Therefore, the compound C is 1,2-dibromobutane which in reaction with soda lime undergoes an elimination reaction to form alkynes. Mostly ${E_2}$ elimination takes place in such cases and the reaction is carried out as follows:
But-1-yne is a terminal alkyne which on reaction with ammoniacal silver nitrate gives white ppt as per following reaction:
$C{{H}_{3}}-C{{H}_{2}}-C\equiv CH\xrightarrow[liq.N{{H}_{3}}]{AgN{{O}_{3}}}\underset{\text{white ppt}\text{.}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C\equiv {{C}^{-}}A{{g}^{+}}}}\,$
Hence, the structures A, B, C and D are 1-bromobutane, but-1-ene, 1,2-dibromobutane and but-1-yne respectively.
Note: It is important to note that the ammoniacal silver nitrate is also known as Tollen’s reagent. Other than terminal alkynes, it is also used for the qualitative analysis of the aldehydes which on reaction gives silver coloured precipitate and hence also known as the silver mirror test.
Complete Step by Step Solution:
Qualitative Analysis of terminal alkyne:
Terminal alkynes i.e., the alkynes which consist of a triple bond at first carbon react with ammoniacal silver nitrate to form white precipitate. This test is confirmed by the compound D which is given in the question. This means, the compound A is an aliphatic saturated carbon chain consisting of bromine at first carbon.
Thus, compound A will be : $C{H_3} - C{H_2} - C{H_2} - C{H_2} - Br$
As given in the question, compound A on reaction with alcoholic $KOH$ gives compound B as follows:
$C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-Br\xrightarrow[-HBr]{alc.KOH}\underset{B}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}}}\,$
On further bromination of compound B gives compound C as per the following reaction:
Therefore, the compound C is 1,2-dibromobutane which in reaction with soda lime undergoes an elimination reaction to form alkynes. Mostly ${E_2}$ elimination takes place in such cases and the reaction is carried out as follows:
But-1-yne is a terminal alkyne which on reaction with ammoniacal silver nitrate gives white ppt as per following reaction:
$C{{H}_{3}}-C{{H}_{2}}-C\equiv CH\xrightarrow[liq.N{{H}_{3}}]{AgN{{O}_{3}}}\underset{\text{white ppt}\text{.}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C\equiv {{C}^{-}}A{{g}^{+}}}}\,$
Hence, the structures A, B, C and D are 1-bromobutane, but-1-ene, 1,2-dibromobutane and but-1-yne respectively.
Note: It is important to note that the ammoniacal silver nitrate is also known as Tollen’s reagent. Other than terminal alkynes, it is also used for the qualitative analysis of the aldehydes which on reaction gives silver coloured precipitate and hence also known as the silver mirror test.
Recently Updated Pages
The hybridization and shape of NH2 ion are a sp2 and class 11 chemistry JEE_Main

What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main

Aromatization of nhexane gives A Benzene B Toluene class 11 chemistry JEE_Main

Show how you will synthesise i 1Phenylethanol from class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

Which of the following Compounds does not exhibit tautomerism class 11 chemistry JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Other Pages
NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

NCERT Solutions ForClass 11 Chemistry Chapter Chapter 5 Thermodynamics

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

