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STD 11 - 9. Hydricarbon — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 9. Hydricarbon4 Marks
MCQ
An unknown compound $A$ has a molecular formula ${C_4}{H_6}.$ When $A$ is treated with an excess of $B{r_2}$ a new substance $B$ with formula ${C_4}{H_6}B{r_4}$ is formed. $A$ forms a white precipitate with ammoniacal silver nitrate solution. $A$ may be
  • ✓
    Butyne$-1$
  • B
    Butyne$-2$
  • C
    Butene$-1$
  • D
    Butene$-2$
✓

Answer

Correct option: A.
Butyne$-1$
a
(a) $\mathop {C{H_3} - C{H_2} - C}\limits_{{\rm{1}} - {\rm{Butyne}}} \equiv CH + 2B{r_2} \to C{H_3} - C{H_2} - \mathop C\limits_{\mathop |\limits_{\,\,Br} }^{\mathop |\limits^{\,\,Br} } - \mathop C\limits_{\mathop |\limits_{\,\,Br} }^{\mathop |\limits^{\,\,Br} } - H$

Since the molecule takes $2$ moles of $B{r_2}$. Therefore it is alkyne. Also it gives white ppt with Tollen’s reagent therefore acidic $H$ is present. Hence it is 1-Butyne.

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