Explanation
As refer to the above image, here carbocation is the intermediate. Since it is acid catalysed addition.
Considering the ozonised sites and reduced sites, the most suitable material for the reaction to begin is shown in the above image.
$$CH_3-CH=CH_2 \xrightarrow {I^+- \ Cl^-} CH_3-CHCl-CH_2-I$$
The reaction is electrophilic addition and the product obtained is called Markonikov product.
$$C{H_3} - C{H_3}(g) + C{l_2} \to {C_2}{H_5}Cl(g) + HCl(g)$$
Hence, the correct option will be option A
Hint: The acidic strength of a compound is directly proportional to the stability of conjugate base.
Step 1: Compare stability of conjugate base
$$CHCl_3$$: It break down in solution as-
$$CHC{l_3} \to {H^ + } + CC{l_3}^ - $$
Here, the conjugate base $$CC{l_3}^ - $$ is stabilized by the back bonding of carbanion with vacant d-orbitals of $$Cl.$$
$$CHBr_3$$: It break down in solution as-
$$CHB{r_3} \to {H^ + } + CB{r_3}^ - $$
Here, the conjugate base $$CBr{_3}^ - $$ is stabilized by the back bonding of carbanion with vacant d-orbitals of $$Br.$$
$$CHI_3$$: It break down in solution as-
$$CH{I_3} \to {H^ + } + C{I_3}^ - $$
Here, the conjugate base $$CI{_3}^ - $$ is stabilized by the back bonding of carbanion with vacant d-orbitals of $$I.$$
$$CH(CN)_3$$: It breaks down in solution as-
$$CH{(CN)_3} \to {H^ + } + C{(CN)_3}^ - $$
Here, the conjugate base is stabilized by the $$-M$$ effect of $$CN^-$$.
The mesomeric effect dominates on back bonding. So $$CH(CN)_3$$ has the most stable conjugate base.
Also $$Cl>Br>I$$ ( Extent of back bonding. )
So $$CH{(CN)_3} > CHC{l_3} > CHB{r_3} > CH{I_3}$$ ( Order of stability of conjugate base. )
Step 2: Acidic order
$$CH{(CN)_3} > CHC{l_3} > CHB{r_3} > CH{I_3}$$
Final Step: Correct option (D) $$CH(CN)_3$$.
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