Key idea: orientation of addition of HCl to an alkene

Normally, addition of $HX$ to an unsymmetrical alkene follows Markovnikov's rule:

• $H^+$ adds first to form the more stable carbocation.
• Then $Cl^-$ attacks that carbocation.

So, to get an anti-Markovnikov product as the major product, the reaction must proceed through a pathway where the less substituted carbocation-like center is favored, or where electronic effects reverse the usual polarization.

General substrate form

All options are of the type: $X-CH=CH_2$

On adding HCl, there are two possibilities:

• Markovnikov addition: $X-CH=CH_2 \xrightarrow[HCl]{} X-CH(Cl)-CH_3$ via protonation at terminal carbon giving carbocation at the carbon bearing $X$.

• Anti-Markovnikov addition: $X-CH=CH_2 \xrightarrow[HCl]{} X-CH_2-CH_2Cl$ via protonation at the $X$-substituted carbon giving carbocation at terminal carbon.

We must see which substituent $X$ makes the second pathway more favorable.

Effect of substituent $X$

The stability of the carbocation formed adjacent to $X$ is crucial.

• If $X$ is electron-donating ($+M$ or $+I$), it stabilizes a carbocation on the adjacent carbon, so Markovnikov addition is favored.
• If $X$ is strongly electron-withdrawing ($-I$), it destabilizes a carbocation on the adjacent carbon, so formation of that carbocation is disfavored. Then protonation occurs the other way, giving the anti-Markovnikov product.

Evaluate each option

Option A: $Cl-CH=CH_2$

Chlorine has strong $-I$ effect, but it can also donate by resonance ($+M$) when adjacent to a double bond. The net effect is not as strongly withdrawing as $CF_3$. So anti-Markovnikov tendency is not strongest here.

Option B: $F_3C-CH=CH_2$

The $CF_3$ group is a very strong electron-withdrawing group due to a powerful $-I$ effect. It strongly destabilizes a carbocation on the adjacent carbon: $F_3C-\overset{+}{C}H-CH_3$ Therefore, the usual Markovnikov pathway becomes unfavorable.

Instead, protonation occurs so that the positive charge develops on the terminal carbon: $F_3C-CH_2-CH_2^+$ followed by attack of $Cl^-$, giving: $F_3C-CH_2-CH_2Cl$ This is the anti-Markovnikov product.

Hence, B is favored.

Option C: $CH_3O-CH=CH_2$

The $OCH_3$ group has strong $+M$ electron-donating effect. It stabilizes a carbocation adjacent to oxygen: $CH_3O-\overset{+}{C}H-CH_3$ Therefore Markovnikov addition is strongly favored.

Option D: $H_2N-CH=CH_2$

The $NH_2$ group is also strongly electron-donating by resonance. It stabilizes the adjacent carbocation, so Markovnikov addition is favored.

Conclusion

The alkene that gives majorly an anti-Markovnikov product with HCl is: $\boxed{F_3C-CH=CH_2}$

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.