A hydrocarbon- - X - C 3 H 4 decolourises Br 2 - CCl 4 solution forming -Y- -X- gave a red precipitate of a compound -Z- with ammoniacal cuprous chloride-X- Y- and Z respectively are-A- CH 2- C - CH 2- CH 2 Br - CBr 2- CH 2 Br - CH 3- C - C - Cu -B- CH 3- C - CH - CH 3- CBr 2- CHBr 2- CH 3- C - C - Cu -C- CH 3- C - CH - CH 3- CBr - CHBr - Cu - CH 2- C - CHD- CH 2- C - CH 2- CH 2 Br - CH - CHBr - CH 3- C - C - Cu -

The correct option is B CH_3 C ≡ CH, CH_3 CBr_2 CHBr_2, CH_3 C ≡ C^ Cu^+The molecular formula of compound 'X' is C_3H_4. It decolourises bromine sol ...

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Standard XII

Chemistry

Introduction to Alkanes

A hydrocarbon...

Question

A hydrocarbon, 'X' $(C_{3} H_{4})$ decolourises $B r_{2} / C C l_{4}$ solution forming 'Y'. 'X' gave a red precipitate of a compound 'Z' with ammoniacal cuprous chloride.
X, Y, and Z respectively are:

C H_{2} = C = C H_{2}, C H_{2} B r - C B r_{2} - C H_{2} B r, C H_{3} - C \equiv C^{-} C u^{+}

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C H_{3} - C \equiv C H, C H_{3} - C B r_{2} - C H B r_{2}, C H_{3} - C \equiv C^{-} C u^{+}

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C H_{3} - C \equiv C H, C H_{3} - C B r = C H B r, C u^{+} C H_{2}^{-} - C \equiv C H

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C H_{2} = C = C H_{2}, C H_{2} B r - C H = C H B r; C H_{3} - C \equiv C^{-} C u^{+}

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Solution

The correct option is B $C H_{3} - C \equiv C H, C H_{3} - C B r_{2} - C H B r_{2}, C H_{3} - C \equiv C^{-} C u^{+}$
The molecular formula of compound 'X' is $C_{3} H_{4}$ . It decolourises bromine solution and it gives red precipitate when reacts with ammoniacal cuprous chloride. This confirms the presence of terminal alkyne which has acidic proton.
The only possible structure with molecular formula $C_{3} H_{4}$ for terminal alkyne is $C H_{3} - C \equiv C H$ (X).

When bromine is treated with alkyne in $C C l_{4}$ , they gets added to alkyne to give 1, 1, 2, 2-tetrabromo propane (Y).

When 1-propyne (X) react with ammoniacal cuprous chloride it gives copper acetylide.

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Q. IUPAC name of given compound:

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

C H_{3} - C H = C H_{2} + H B r \to

-------- the product is

$C H_{3} - C | C H_{2} - C H_{3} = C H - C H_{2} - C H_{2} - C H_{3}$

Q. Which amongst the following is a positional isomer of

{CH}_{3} - {CH}_{2} - CH = {CH}_{3}

A straight chain hydrocarbon has the molecular formula C8H10. The hybridisation for the carbon atoms from one end of the chain to the other are respectively $s p^{3}, s p^{2}, s p^{2}, s p^{3}, s p^{2}, s p^{2}, s p$ and $s p$ . The structural formula of the hydrocarbon would be

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Introduction to Alkanes

Standard XII Chemistry

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A hydrocarbon, 'X' (C3H4) decolourises Br2/CCl4 solution forming 'Y'. 'X' gave a red precipitate of a compound 'Z' with ammoniacal cuprous chloride. X, Y, and Z respectively are:

A hydrocarbon, 'X' $(C_{3} H_{4})$ decolourises $B r_{2} / C C l_{4}$ solution forming 'Y'. 'X' gave a red precipitate of a compound 'Z' with ammoniacal cuprous chloride.
X, Y, and Z respectively are: