Allyl chloride on dehydrochlorination gives :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Propadiene
Allyl chloride is 3-chloroprop-1-ene, \( \ce{CH2=CH-CH2Cl} \). Dehydrochlorination (β-elimination with a strong base) removes Cl from C-3 and the vinylic hydrogen from the neighbouring C-2, forming a second double bond at C-2:
\( \displaystyle \ce{CH2=CH-CH2Cl + KOH} \xrightarrow{\text{alc.}} \ce{CH2=C=CH2 + KCl + H2O} \)
The product, with two cumulated double bonds, is propadiene.
Why the other options are incorrect
Option (2) Propylene — Incorrect. \( \ce{CH3CH=CH2} \) would be formed by replacing Cl with H (reduction), not by losing HCl.
Option (3) Alkyl alcohol — Incorrect. An alcohol forms by substitution with aqueous alkali (allyl alcohol), not by elimination of HCl.
Option (4) Acetyl chloride — Incorrect. \( \ce{CH3COCl} \) is an acid chloride with two carbon atoms and oxygen; it cannot come from losing HCl.
Note: The question is from beyond the NCERT textbook.