The dihalogen derivative 'X' of a hydrocarbon with three carbon atoms reacts with alcoholic KOH and produces another hydrocarbon which forms a red precipitate with ammoniacal Cu2Cl2 . 'X' gives an aldehyde on reaction with aqueous KOH . The compound 'X' is

Question

The dihalogen derivative 'X' of a hydrocarbon with three carbon atoms reacts with alcoholic KOH and produces another hydrocarbon which forms a red precipitate with ammoniacal Cu2Cl2 . 'X' gives an aldehyde on reaction with aqueous KOH . The compound 'X' is (A) 1, 3-dichloropropane (B) 1, 2-dichloropropane (C) 2, 2-dichloropropane (D) 1, 1-dichloropropane

Tardigrade Admin · Accepted Answer

X is a three carbon compound with two halogen atom, so its molecular formula is C3H6Cl2 Only terminal alkynes give red ppt with ammoniacal CuCl2, so the hydrocarbon produced by the reaction of X with ale KOH, must be a terminal alkyne (ie, CH3C≡ CH ) C3 H6Cl2 xrightarrow[]Alc KOHCH3C≡ CH xrightarrow[]Amm Cu2Cl2 underset operatornameRed ppt mathopCH3C≡ CCu downarrow Compound (X) gives an aldehyde when reacts with aqueous KOH. This suggests that both the halogens are present on same terminal carbon atom. Thus, the formula of compound (X) is and the reactions are as follows: CH3C≡ CH xrightarrow[Cu2Cl2]Ammoniacal underset operatornameRed ppt mathopCH3C≡ CCu downarrow

$1, 3$ -dichloropropane

$1, 2$ -dichloropropane

$2, 2$ -dichloropropane

$1, 1$ -dichloropropane

$1, 3$ -dichloropropene

Q. The dihalogen derivative ′X′ of a hydrocarbon with three carbon atoms reacts with alcoholic KOH and produces another hydrocarbon which forms a red precipitate with ammoniacal Cu2​Cl2​ . ′X′ gives an aldehyde on reaction with aqueous KOH . The compound ′X′ is

1,3-dichloropropane

1,2-dichloropropane

2,2-dichloropropane

1,1-dichloropropane

1,3-dichloropropene

$1, 3$ -dichloropropane

$1, 2$ -dichloropropane

$2, 2$ -dichloropropane

$1, 1$ -dichloropropane

$1, 3$ -dichloropropene