CH_3-CH(CH_3)-C equiv CH xrightarrow{excess | vedclass

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(C) The addition of $HBr$ to alkynes follows Markovnikov's rule.In the presence of excess $HBr$,two molecules of $HBr$ add to the triple bond to form

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$CH_3-CH(CH_3)-C(Br)_2-CH_3$

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(C) The addition of $HBr$ to alkynes follows Markovnikov's rule.In the presence of excess $HBr$,two molecules of $HBr$ add to the triple bond to form a geminal dihalide.Step $1$: $CH_3-CH(CH_3)-C \equiv CH + HBr \rightarrow CH_3-CH(CH_3)-C(Br)=CH_2$Step $2$: $CH_3-CH(CH_3)-C(Br)=CH_2 + HBr \rightarrow CH_3-CH(CH_3)-C(Br)_2-CH_3$ ($2,2$-dibromo-$3$-methylbutane).

$CH_3-CH(CH_3)-C \equiv CH \xrightarrow{excess \ HBr}$
The product of the above reaction is

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$CH_3-C \equiv C-CH_3, CH_3-CH_2-C \equiv CH, C_2H_5-C \equiv C-C_2H_5, HC \equiv CH, CH_3-CH=CH_2, CH_2=CH-C \equiv CH$

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$CH_3-CH(CH_3)-C \equiv CH \xrightarrow{excess \ HBr}$ The product of the above reaction is