Question No. 6 Which of the following represents a method of preparing propene in one step?
CH3 - CH2 - CH2 - Cl + aq. KOH
O
CI – CH2 – CH = CH. + alc. KOH
CH3 - CH2 - CH2 – Cl + alc. KOH
Br – CH2 - CH2 - CH2 – Br + aq. KOH

Option C: CH3 - CH2 - CH2 – Cl + alc. KOH

Explanation:

When we heat an Alkyl halide (R - X) in a solution of potassium hydroxide dissolved in alcohol (alcoholic potash), it eliminates one molecule of halogen acid to form alkenes. This reaction process is called dehydrohalogenation and it is an example of a beta elimination reaction.

Now in the question for us to prepare propene, we will do that by boiling 1-chloropropane in alcoholic potassium hydroxide(KOH) and from earlier statement ,dehydrohalogenation will take place to form propene.

The reaction is;

CH3 - CH2 - CH2 – Cl + alc. KOH ⟶ CH3-CH=CH2 + KCl + H20

Thus, the correct answer is;

CH3-CH2-CH2–Cl + alc. KOH