1. Write the substrate

3-chloro-1-butene has the structure

$\mathrm{CH_2=CH-CH(Cl)-CH_3}$

We add $\mathrm{HCl}$ through carbocation formation, so first protonation of the double bond occurs, giving a carbocation, and then $\mathrm{Cl^-}$ attacks.

2. Possible protonation modes of the double bond

For the alkene $\mathrm{CH_2=CH-CH(Cl)-CH_3}$, proton can add to either carbon of the double bond.

Case A: Proton adds to terminal carbon $C_1$

Then carbocation forms at $C_2$:

$\mathrm{CH_3-CH^+-CH(Cl)-CH_3}$

This is a secondary carbocation.

Now $\mathrm{Cl^-}$ attacks $C_2$, giving:

$\mathrm{CH_3-CH(Cl)-CH(Cl)-CH_3}$

This is 2,3-dichlorobutane.

Its stereochemical possibilities are:

• $(2R,3R)$ and $(2S,3S)$ : one enantiomeric pair
• meso form $(2R,3S)$ = $(2S,3R)$ : one distinct isomer

So from this product, number of isomeric products = 3.

Case B: Proton adds to $C_2$

Then carbocation forms at terminal carbon $C_1$:

$\mathrm{CH_2^+-CH_2-CH(Cl)-CH_3}$

This is a primary carbocation, so it rearranges.

A 1,2-hydride shift from $C_2$ to $C_1$ gives back the more stable carbocation at $C_2$:

$\mathrm{CH_3-CH^+-CH(Cl)-CH_3}$

So again attack by $\mathrm{Cl^-}$ gives the same constitutional product:

$\mathrm{CH_3-CH(Cl)-CH(Cl)-CH_3}$

No new constitutional isomer arises here.

3. Check for neighboring halogen participation / allylic rearrangement

Since the question explicitly says through carbocation formation, we consider carbocation intermediates accessible from protonation of the alkene.

The key cation is

$\mathrm{CH_3-CH^+-CH(Cl)-CH_3}$

Attack of $\mathrm{Cl^-}$ on this planar carbocation gives the stereoisomers of 2,3-dichlorobutane already counted.

No additional distinct constitutional product is formed under this simple carbocation-addition treatment.

4. Total number of isomeric products

Thus the products are the stereoisomers of 2,3-dichlorobutane:

• one meso isomer
• one pair of enantiomers

Total distinct isomeric products:

$3$

5. Comparison with stored answer

Stored correct answer = $4$

My derived answer = $3$

So I do not agree with the stored answer.

A count of $4$ would require one additional distinct product, but under standard ionic addition of $\mathrm{HCl}$ via carbocation formation for $\mathrm{CH_2=CH-CH(Cl)-CH_3}$, the product framework is 2,3-dichlorobutane, which has only 3 distinct stereoisomers.