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Predict the product C obtained in the following reaction of butyne-1.
$CH_3CH_2-C\equiv CH + HCl \rightarrow B \xrightarrow{HI} C$
$CH_3CH_2-C\equiv CH + HCl \rightarrow B \xrightarrow{HI} C$
A
$CH_3-CH_2-CH_2-CH(I)Cl$
B
$CH_3-CH_2-CH(I)-CH_2Cl$
C
$CH_3CH_2-C(I)(Cl)-CH_3$
D
$CH_3-CH(Cl)-CH_2CH_2I$
Detailed Solution
The reaction occurs according to Markovnikov's rule: when an unsymmetrical alkene or alkyne undergoes hydrohalogenation, the negative part of the reagent goes to that carbon atom which carries the lesser number of hydrogen atoms.
Step 1: addition of HCl to but-1-yne. H adds to the terminal carbon and Cl to the inner carbon.
$CH_3-CH_2-C\equiv CH + HCl \rightarrow CH_3-CH_2-C(Cl)=CH_2$ (B, 2-chlorobut-1-ene)
Step 2: addition of HI to B. Again H adds to the $CH_2$ carbon (which has more hydrogens) and I adds to the carbon carrying Cl; the intermediate carbocation is stabilised by the lone pair of chlorine.
$CH_3-CH_2-C(Cl)=CH_2 + HI \rightarrow CH_3-CH_2-C(I)(Cl)-CH_3$ (C)
The product C is 2-chloro-2-iodobutane, $CH_3CH_2-C(I)(Cl)-CH_3$.
Step 1: addition of HCl to but-1-yne. H adds to the terminal carbon and Cl to the inner carbon.
$CH_3-CH_2-C\equiv CH + HCl \rightarrow CH_3-CH_2-C(Cl)=CH_2$ (B, 2-chlorobut-1-ene)
Step 2: addition of HI to B. Again H adds to the $CH_2$ carbon (which has more hydrogens) and I adds to the carbon carrying Cl; the intermediate carbocation is stabilised by the lone pair of chlorine.
$CH_3-CH_2-C(Cl)=CH_2 + HI \rightarrow CH_3-CH_2-C(I)(Cl)-CH_3$ (C)
The product C is 2-chloro-2-iodobutane, $CH_3CH_2-C(I)(Cl)-CH_3$.