
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
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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.
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