Question

A dihalogen derivative x of a hydrocarbon has three carbon atoms in its molecule. It reacts with alcoholic koh giving another hydrocarbon that gives red precipitate with ammonical cu2cl2. When treated with aqueous koh, x gives an aldehyde. Compound x is select one: a. 2,2-dichloropropane b. 1,3-dichloropropane c. 1,2-dichloropropane d. 1,1-dichloropropane

Answer 1

In the given organiccompound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are sp,sp, sp3, sp3, sp2,and sp2 hybridized respectively. Thus, the pair ofhybridized orbitals involved in the formation of C2-C3bond is sp – sp3.

Answer 2

Answer:

The correct answer is option d

Explanation:

• $X$ is three-carbon compound with two halogen atom, so its molecular formula is $C_{3} H_{6} C l_{2}$ Only terminal alkynes give red ppt with ammoniacal $C u C l_{2}$, so the hydrocarbon produced by the reaction of $X$ with ale $\mathrm{KOH}$, must be a terminal alkyne
• Compound (X) gives an aldehyde when reacts with aqueous $\mathrm{KOH}$. 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:
• Red precipitate with ammoniacal $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ is a characteristic reaction of alk-1yne. That means the hydrocarbon is 1 - propyne.
• Reaction with aqueous $\mathrm{KOH}$ gives aldehyde that means the two halides are attached to the same carbon that is carbon number one.
• The above reason indicates that the compound $(\mathrm{x})$ is 1,1 - Dichloropropane