Concept Explanation: Dehydrohalogenation Reaction

This question involves a dehydrohalogenation reaction where a haloalkane reacts with an alcoholic base (like KOH in ethanol) to form an alkene. The given compound is ${ClCH}_{2} CHC l_{2}$ (1,1,2-trichloroethane). Let's analyze the reaction step by step.

Step 1: Identify the Reactant

The reactant is ${ClCH}_{2} CHC l_{2}$ (1,1,2-trichloroethane). It has the structure: Cl-CH₂-CHCl₂.

Step 2: Understand the Reaction Conditions

Alcoholic base (e.g., KOH in ethanol) promotes elimination (dehydrohalogenation) rather than substitution. It removes a hydrogen halide (HCl) molecule to form a double bond.

Step 3: Determine Possible Elimination Products

In dehydrohalogenation, the base abstracts a β-hydrogen (hydrogen on carbon adjacent to the carbon with halogen) along with removal of halogen. The reactant has two types of β-carbons:

The carbon with two Cl atoms (CCl₂) has one β-hydrogen (on CH₂).
The carbon with one Cl (CH₂Cl) has no β-hydrogen (since the adjacent carbon is CCl₂, which has no H).

So, only one elimination is possible: removal of HCl from the CH₂ and one Cl from the CHCl₂ group.

Step 4: Write the Elimination Reaction

The reaction is: ${ClCH}_{2} CHC l_{2} \overset{alc . base}{\to} CH = CCl 2 + HCl$

The product is $CH = CCl 2$ (1,1-dichloroethene).

Step 5: Compare with Options

Look at the given options:

CH≡CH (acetylene) – not an alkene from single elimination.
ClCH₂CHO (chloroacetaldehyde) – substitution product, not elimination.
CH₂=CCl₂ (1,1-dichloroethene) – matches our product.
CHCl=CHCl (1,2-dichloroethene) – isomeric but not formed due to less stability.

Final Answer

The compound A is $CH = CCl 2$ (1,1-dichloroethene). So, the correct option is: CH₂ = CCl₂.

Related Topics

Haloalkanes and Haloarenes
Elimination Reactions
Saytzeff's Rule (for preferred alkene formation)
Substitution vs Elimination Competition

Important Formulae and Theory

Dehydrohalogenation Reaction:

General form: $R - CH - CH - R' \overset{alc . KOH}{\to} R - CH = CH - R' + HX$

Saytzeff's Rule: In elimination, the preferred product is the more highly substituted alkene (with more alkyl groups on the double bond carbon).

Note: Alcoholic base favors elimination, while aqueous base favors substitution.