Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?

First we recall the general statement of Markovnikov’s rule: on electrophilic addition of $$\mathrm{HX}$$ to an unsymmetrical alkene, the proton $$\left(\mathrm{H^+}\right)$$ attaches to that double-bonded carbon which already bears the larger number of hydrogen atoms, because this mode of attack generates the more stable carbocation. Put algebraically, if the alkene is $$\mathrm{C_a=C_b}$$, then in the Markovnikov pathway $$\mathrm{H^+}$$ goes to the carbon rich in hydrogens and the intermediate carbocation appears on the other carbon.

The stability of the intermediate carbocation is decided mainly by two electronic effects: the inductive effect $$\left(-I\ \text{or}\ +I\right)$$ and the mesomeric or resonance effect $$\left(-M\ \text{or}\ +M\right)$$ of the substituent already attached to that carbon. A group which donates electron density by $$+I$$ or $$+M$$ stabilises the positive charge, whereas a group which withdraws electron density by a strong $$-I$$ effect destabilises it. Hence, if a substituent has a very powerful $$-I$$ effect and lacks any meaningful $$+M$$ donation, the carbocation situated adjacent to it becomes highly unstable, and the reaction may proceed by the opposite orientation, i.e. the anti-Markovnikov mode.

Now we analyse each given alkene one by one.

Option A : $$\mathrm{CH_3O{-}CH=CH_2}$$

The methoxy group $$\left(\mathrm{-OCH_3}\right)$$ exerts a $$+M$$ resonance donation because of the lone pairs on oxygen. In the Markovnikov pathway the proton adds to the terminal $$\mathrm{CH_2}$$ carbon and the carbocation appears on the carbon bearing $$\mathrm{-OCH_3}$$:

$$\mathrm{CH_3O{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ CH_3O{-}\overset{+}{\mathrm{CH}}{-}CH_3$$

The positive charge is strongly stabilised by resonance with the oxygen lone pair, so this path is preferred. Hence the product is Markovnikov, not anti-Markovnikov.

Option B : $$\mathrm{Cl{-}CH=CH_2}$$

Chlorine possesses a weak $$+M$$ effect as well as a $$-I$$ effect. When the carbocation is placed on the chlorine-bearing carbon (Markovnikov orientation) the lone pairs of chlorine can offer some resonance stabilisation:

$$\mathrm{Cl{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ Cl{-}\overset{+}{\mathrm{CH}}{-}CH_3}$$

The $$+M$$ relief outweighs the inductive withdrawal; consequently the Markovnikov product remains major. No anti-Markovnikov preference is observed.

Option C : $$\mathrm{H_2N{-}CH=CH_2}$$

The amino group $$\left(\mathrm{-NH_2}\right)$$ is a very strong $$+M$$ donor. Exactly as in Option A, the Markovnikov pathway places the positive charge adjacent to the donor nitrogen, which is highly stabilising by resonance:

$$\mathrm{H_2N{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ H_2N{-}\overset{+}{\mathrm{CH}}{-}CH_3}$$

Therefore this alkene also obeys Markovnikov’s rule.

Option D : $$\mathrm{CF_3{-}CH=CH_2}$$

The trifluoromethyl group $$\left(\mathrm{-CF_3}\right)$$ is one of the strongest electron-withdrawing groups known; it shows an intense $$-I$$ effect and has no $$+M$$ resonance donation, because the fluorines are not conjugated with the carbocation centre through a lone-pair bridge as chlorine can be. Examine the two possible orientations:

Markovnikov attack (proton to the terminal $$\mathrm{CH_2}$$ carbon):

$$\mathrm{CF_3{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ CF_3{-}\overset{+}{\mathrm{CH}}{-}CH_3}$$

The carbocation $$\mathrm{CF_3{-}\overset{+}{CH}{-}CH_3}$$ is directly attached to the very powerful $$-I$$ group $$\mathrm{CF_3}$$, so the positive charge is pulled even further electron-deficient and becomes extremely unstable.

Anti-Markovnikov attack (proton to the $$\mathrm{CF_3{-}CH}$$ carbon):

$$\mathrm{CF_3{-}CH=CH_2\ \overset{H^+}{\underset{\text{anti-M}}{\rightarrow}}\ CF_3{-}CH_2{-}\overset{+}{\mathrm{CH}}_2$$

Now the positive charge resides on the terminal carbon, which is not adjacent to the $$\mathrm{-CF_3}$$ group, so it does not suffer the strong inductive withdrawal. Even though both carbocations are primary, the one formed in the anti-Markovnikov pathway is appreciably more stable because it is farther from the electron-withdrawing group. Therefore the reaction proceeds predominantly by this orientation, giving the anti-Markovnikov product $$\mathrm{CF_3{-}CH_2{-}CH_2Cl}$$.

Thus, among the four alkenes, only the trifluoromethyl-substituted alkene in Option D furnishes the anti-Markovnikov addition product with $$\mathrm{HCl}$$.

Hence, the correct answer is Option D.