
An unknown compound \[(A)\]has a molecular formula \[{C_4}{H_6}.\]When \[(A)\]is treated with excess of \[B{r_2}\] a new substance\[(B)\] with formula \[{C_4}{H_6}B{r_4}\] is formed. \[(A)\] forms a white ppt. with ammoniacal silver nitrate solution.\[(A)\] may be:
A. But-1-yne
B. But-2-yne
C. But-1-ene
D. But-2-ene
Answer
516.6k+ views
Hint: \[{C_4}{H_6}\] is alkyne. Alkynes are unsaturated hydrocarbons having triple bonds between at least two adjacent carbon atoms.
It’s general formula is \[{C_n}{H_{2n - 2}}\]
Complete step by step answer:
Let us form an equation according to the given condition
\[{C_4}{H_6}\]is butyne. It can either be But-1-yne or But-2-yne.
\[C{H_3} - C{H_2} - C \equiv CH\] But-1-yne
\[C{H_3} - C \equiv C - C{H_3}\] But-2-yne
If the triple bond is present with the carbon atom \[1\] and \[2\], then the hydrogen atom is present at C-atom \[1\] so it is acidic in nature.
\[C{H_3} - C{H_2} - C \equiv CH\]is acidic in nature.
Since Ammoniacal silver nitrate is reducing agent, it reduces butyne, and butyne oxidize ammonical silver nitrate solution.
We know that oxidation and Reduction occur together.
But-1-yne reduces by tollen’s reagent and form white precipitate
\[C{H_3} - C{H_2} - C \equiv {C^ - }{H^ + }\]
So, the correct answer is “Option A”.
Additional Information:
Tollen’s Reagent is an ammoniacal solution of \[AgN{O_3}\]. It is prepared by adding \[N{H_4}OH\] solution to\[AgN{O_3}\] solution till the precipitate of \[A{g_2}O\] is formed which just gets dissolved. It has the formula \[{[Ag{(N{O_3})_2}]^ + }O{H^{^ - }}\]
$2AgN{O_3} + 2N{H_4}OH \to A{g_2}O + 2N{H_4}N{O_3} + {H_2}O $
$A{g_2}O + 4N{H_4}OH \to 2{[Ag{(N{H_3})_2}]^ + }O{H^ - } + {H_2}O $
We can define oxidation and reduction as follows:
Addition of oxygen or removal of hydrogen is called Oxidation.
Addition of hydrogen or removal of oxygen is called Reduction.
Since But-1-yne is acidic in nature. So it can easily remove H-atoms and oxidates, while Tollen’s reagent reduces. But-2-yne does not have acidic hydrogen, so it will not react with tollen’s reagent.
Note:
Oxidation and reduction occurs simultaneously. This reaction is a Redox reaction in which But-1-yne reduces and Tollen’s reagent oxidises.
It’s general formula is \[{C_n}{H_{2n - 2}}\]
Complete step by step answer:
Let us form an equation according to the given condition

\[{C_4}{H_6}\]is butyne. It can either be But-1-yne or But-2-yne.
\[C{H_3} - C{H_2} - C \equiv CH\] But-1-yne
\[C{H_3} - C \equiv C - C{H_3}\] But-2-yne
If the triple bond is present with the carbon atom \[1\] and \[2\], then the hydrogen atom is present at C-atom \[1\] so it is acidic in nature.
\[C{H_3} - C{H_2} - C \equiv CH\]is acidic in nature.
Since Ammoniacal silver nitrate is reducing agent, it reduces butyne, and butyne oxidize ammonical silver nitrate solution.
We know that oxidation and Reduction occur together.
But-1-yne reduces by tollen’s reagent and form white precipitate
\[C{H_3} - C{H_2} - C \equiv {C^ - }{H^ + }\]
So, the correct answer is “Option A”.
Additional Information:
Tollen’s Reagent is an ammoniacal solution of \[AgN{O_3}\]. It is prepared by adding \[N{H_4}OH\] solution to\[AgN{O_3}\] solution till the precipitate of \[A{g_2}O\] is formed which just gets dissolved. It has the formula \[{[Ag{(N{O_3})_2}]^ + }O{H^{^ - }}\]
$2AgN{O_3} + 2N{H_4}OH \to A{g_2}O + 2N{H_4}N{O_3} + {H_2}O $
$A{g_2}O + 4N{H_4}OH \to 2{[Ag{(N{H_3})_2}]^ + }O{H^ - } + {H_2}O $
We can define oxidation and reduction as follows:
Addition of oxygen or removal of hydrogen is called Oxidation.
Addition of hydrogen or removal of oxygen is called Reduction.
Since But-1-yne is acidic in nature. So it can easily remove H-atoms and oxidates, while Tollen’s reagent reduces. But-2-yne does not have acidic hydrogen, so it will not react with tollen’s reagent.
Note:
Oxidation and reduction occurs simultaneously. This reaction is a Redox reaction in which But-1-yne reduces and Tollen’s reagent oxidises.
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