Q.Identify the products A and B formed in the following reaction:

The key idea is that addition of HCl to an unsymmetrical alkene follows Markovnikov’s rule: the hydrogen adds to the carbon with more hydrogens, and chlorine adds to the more substituted carbon.

Step 1: The alkene is $\mathrm{CH_3-CH_2-CH{=}CH-CH_3}$ (pent-2-ene). The double bond is between C3 and C4 (numbering from left).

Step 2: C3 has one H, C4 has one H — both are equally substituted. However, Markovnikov’s rule still applies: the H⁺ adds to the carbon that yields the more stable carbocation.

Step 3: Protonation at C4 gives a secondary carbocation at C3; protonation at C3 gives a secondary carbocation at C4 — both are equally stable. So both possible products form in roughly equal amounts. …

This is electrophilic addition of HCl to an alkene, but the usual sharp Markovnikov major/minor split only appears when the two alkene carbons are substituted to different DEGREES (e.g. one carbon bearing two alkyl groups, the other bearing none or one). Here that is not the case.

Numbering the chain

Number so the double bond gets the lowest locant (this makes it pent-2-ene, matching how the compound would actually be named): $C1(CH_3)-C2(CH)=C3(CH)-C4(CH_2)-C5(CH_3)$, double bond between C2 and C3.

• C2 is bonded to: C1 (one alkyl group, a methyl), one H, and the double bond.
• C3 is bonded to: C4 (one alkyl group, the start of an ethyl chain), one H, and the double bond.

Both alkene carbons carry exactly one alkyl substituent each. Neither is a terminal $=CH_2$, so there is no way to generate a primary carbocation from this alkene at all.

Both carbocations are secondary

• Protonating C2 places the positive charge on C3, which is then bonded to C2 and C4 — a secondary carbocation.
• Protonating C3 places the positive charge on C2, which is then bonded to C1 and C3 — also a secondary carbocation.

Since both possible carbocations are secondary and structurally very similar (one flanked by a methyl and the chain, the other by an ethyl-chain carbon and the chain), neither is meaningfully more stable than the other. Chloride ion then attacks whichever cation formed, giving: …

Concept: Electrophilic Addition to Unsymmetrical Alkenes (Markovnikov’s Rule)

When an unsymmetrical alkene reacts with HX (like HCl), the hydrogen adds to the carbon that already has more hydrogen atoms, and the halogen adds to the carbon with fewer hydrogen atoms. This is Markovnikov’s rule.

Method: Markovnikov Addition

Step 1 – Identify the double bond and the two alkene carbons

The alkene is:

Number the carbons from left to right:

• C1: $\mathrm{CH_3-}$
• C2: $\mathrm{-CH_2-}$
• C3: $\mathrm{-CH=}$ (one H)
• C4: $\mathrm{=CH-}$ (one H)
• C5: $\mathrm{-CH_3}$

The double bond is between C3 and C4.

Step 2 – Count hydrogens on each alkene carbon

• C3 has 1 hydrogen
• C4 has 1 hydrogen

Both have the same number of hydrogens — so Markovnikov’s rule alone does not give a single product. Both possible additions are equally likely.

Step 3 – Add H⁺ and Cl⁻ in both possible ways

Product B (H⁺ adds to C3, Cl⁻ adds to C4):

$\mathrm{CH_3-CH_2-CH_2-CHCl-CH_3}$ …

Here are the common mistakes students make when solving this reaction, along with how to avoid each.

Mistake 1: Forcing Markovnikov's Rule to Give a Single "Major" Product

The Mistake: Students assume Markovnikov's rule must always identify one clearly major and one clearly minor product, so they label one of 2-chloropentane or 3-chloropentane as "major" without real justification.

Why it's wrong: Markovnikov's rule (in its full form) says H⁺ adds to the carbon that leads to the MORE STABLE carbocation. Here, both possible additions to $\mathrm{CH_3-CH_2-CH=CH-CH_3}$ (pent-2-ene) lead to a carbocation with exactly the same DEGREE of substitution:

• Protonating C3 gives a cation at C4, flanked by a methyl group (C5) and a propyl-length chain (via C3-C2-C1) — a secondary carbocation.
• Protonating C4 gives a cation at C3, flanked by an ethyl group (via C2-C1) and another ethyl-length fragment (via C4-C5) — also a secondary carbocation.

Neither is meaningfully more stabilized than the other (both are simple secondary carbocations with ordinary alkyl substituents, no branching right at the cationic centre to favour one side).

How to Avoid: When protonating an unsymmetrical alkene, always draw BOTH possible carbocations and compare their DEGREE (primary/secondary/tertiary) and any special stabilization (allylic, benzylic, branching at the cationic carbon). If both carbocations come out the same degree with no other stabilizing difference, expect a genuine mixture in comparable amounts — don't force a "major" answer that isn't chemically justified.

Mistake 2: Miscounting Hydrogens and Concluding the Alkene Must Be Symmetrical

The Mistake: Since both alkene carbons (C3, C4) have exactly 1 H each, students sometimes conclude the molecule is symmetrical and expect only one product.

Why it's wrong: Equal H-count on the two alkene carbons does NOT mean the molecule is symmetrical. C3 is attached to an ethyl group (C2-C1) while C4 is attached to a methyl group (C5) — these are different substituents, so the alkene is genuinely unsymmetrical. It just happens that the resulting carbocations, despite this asymmetry, end up comparably stable.

How to Avoid: Check the FULL substituent (not just the H count) on each alkene carbon before concluding anything about symmetry.

Mistake 3: Writing Only One Product (Ignoring the Mixture)

The Mistake: The question asks for A and B, but students write only one product, assuming one is definitively correct.

How to Avoid: When two carbocation pathways are comparably stable, both products form and both must be reported:

• A: $\mathrm{CH_3-CH_2-CHCl-CH_2-CH_3}$ (3-chloropentane)
• B: $\mathrm{CH_3-CH_2-CH_2-CHCl-CH_3}$ (2-chloropentane)

Mistake 4: Miscounting the Chain When Naming the Products …