${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Cl}\underset{\mathrm{KOH}}{\stackrel{\text{ alc }}{⟶}}\mathrm{B}\stackrel{\mathrm{HBr}}{⟶}\mathrm{C}\underset{\mathrm{ether}}{\overset{\mathrm{Na}}{\to }}\mathrm{D}$ In the above sequence the product D is

To understand the sequence of reactions leading to product D, let’s go through each step of the process carefully.

Step 1: Starting Material

The initial compound in the sequence is CH₃CH₂CH₂Cl, which is 1-chloropropane. This is a halogenated alkane, where a chlorine atom is attached to the terminal carbon of a propyl chain.

Step 2: Reaction with Alcoholic KOH

When 1-chloropropane reacts with alcoholic KOH, a dehydrohalogenation reaction occurs. The base (KOH) abstracts a beta-hydrogen from the carbon adjacent to the one bonded to chlorine, which leads to the elimination of hydrogen chloride (HCl) and the formation of a double bond. This reaction produces an alkene, specifically propene (CH₃CH=CH₂).

Mechanism:

The reaction proceeds through an E2 elimination mechanism, where the base removes a hydrogen atom from a carbon atom adjacent to the carbon attached to the chlorine. The chloride ion (Cl⁻) leaves, and a double bond forms between the two carbons, producing propene.

Step 3: Reaction with HBr

In the next step, propene undergoes an addition reaction with HBr (hydrobromic acid). This addition follows Markovnikov's Rule, which states that the proton (H⁺) will add to the carbon of the double bond that has the greater number of hydrogen atoms (the more substituted carbon), and the bromine (Br⁻) will add to the carbon with fewer hydrogen atoms (the less substituted carbon).

Product:

The result of this reaction is 2-bromopropane (CH₃CHBrCH₃). The bromine atom adds to the second carbon of the propene, while a hydrogen atom adds to the first carbon.

Step 4: Reaction with Sodium in Dry Ether

Next, 2-bromopropane undergoes a Wurtz reaction in the presence of sodium metal in dry ether. The Wurtz reaction involves the coupling of two alkyl radicals that are formed by the removal of the bromine atom, resulting in the formation of a new carbon-carbon bond.

Mechanism:

In this reaction, sodium metal donates electrons to the bromine atom, causing it to leave as a bromide ion (Br⁻) and form an alkyl radical. Two such radicals (from two molecules of 2-bromopropane) then combine to form a new C–C bond, producing the alkane product.

Product:

The product formed is n-hexane (C₆H₁₄), which is a straight-chain alkane.

Final Product

Thus, the final product D of this sequence is n-hexane.