Question

1-methylcyclopentene is treated with HCl in stoichiometric amounts. Assuming the reaction goes to completion, how many different structural and stereoisomers (geometrical only) of the products will be formed?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3 isomers

Answer 2

The alkene 1‑methylcyclopentene (with the double bond between C1 and C2, where C1 already bears a –CH₃ group) may in principle react via two routes when treated with HCl:

1. Markovnikov path:

   • H⁺ adds to C2 (which has two –H’s) giving a carbocation at C1.
   • Chloride then attacks C1 giving 1‑chloro‑1‑methylcyclopentane.
   • At C1 the two ring fragments (going clockwise and anticlockwise) are equivalent in an unsubstituted ring so this product is achiral (a unique constitutional isomer).

2. Anti–Markovnikov path (minor):

   • H⁺ adds to C1 (less favored) generating a carbocation at C2.
   • Cl⁻ then attacks C2 to afford 2‑chloro‑1‑methylcyclopentane.
   • Here, the –CH₃ at C1 and –Cl at C2 lie on adjacent ring carbons. They can be arranged either on the same side (cis) or on opposite sides (trans) relative to the mean plane of the ring. Thus this constitutional isomer has two geometrical (stereochemical) forms.

Thus, if the reaction is taken to completion so that even the less‐favoured path gives product, a total of 1 (from Markovnikov) + 2 (cis and trans from anti–Markovnikov) = 3 isomers are formed.

Minimal Explanation:

• Protonation of 1‑methylcyclopentene can occur in two ways.
• Markovnikov addition gives 1‑chloro‑1‑methylcyclopentane (one product).
• Anti–Markovnikov addition gives 2‑chloro‑1‑methylcyclopentane, which exists as cis and trans isomers (two products).
• Total = 3 isomers.