On boiling 2,2 -dichloropropane with aqueous alkali, the product obtained is (a) 2,2-propanediol (b) acetone (c) propanal (d) 2 -chloro-2-propanol

Short Answer

Expert verified
Answer: Acetone

Step by step solution

01

Draw the structure of 2,2-dichloropropane

Start by drawing the structure of 2,2-dichloropropane, where there are two chlorine atoms attached to the second carbon of the propane chain. The structure would be: H-C-C(Cl)-C(H)(Cl)-H
02

Identify the E2 reaction mechanism

The reaction involves aqueous alkali, which indicates an elimination reaction. The E2 (bimolecular elimination) mechanism involves the use of a strong alkali as the base, such as OH-, to remove the two halide groups from the molecule.
03

Perform the E2 reaction

In the E2 mechanism, the base (OH-) will remove a proton from a neighboring carbon atom while the carbon-chlorine bond is broken, and the Cl- leaves the molecule. In our case, the OH- removes a proton from the first carbon atom, and the chloride ion leaves from the second carbon atom. The same process will happen for the other chlorine atom: H-C(-)-C(Cl)=C(H)(Cl)-H -> H-C=C(Cl)=C(-)-H As a result, we have removed both chloride ions and formed a double bond between the first and second carbons, giving us the final product: H-C=C-C(-)-H
04

Identify the final product

The final product, H-C=C-C(-)-H, is identical to the structure of acetone. Thus, the correct answer is: (b) Acetone

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Most popular questions from this chapter

The product \(\mathrm{P}\) in the following reaction is \(\mathrm{ClCH}_{2}-\mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{P}\)

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