C_8H_{18} (A) on mono-chlorination forms only | vedclass

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(D) For an alkane to form only one type of mono-chlorinated product,all hydrogen atoms in the molecule must be equivalent. In $2,2,3,3$-tetramethylbut

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$2,2,3,3-$tetramethylbutane

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(D) For an alkane to form only one type of mono-chlorinated product,all hydrogen atoms in the molecule must be equivalent. In $2,2,3,3$-tetramethylbutane,all $18$ hydrogen atoms are equivalent because the molecule is highly symmetrical. Therefore,mono-chlorination of $2,2,3,3$-tetramethylbutane yields only one isomer of $C_8H_{17}Cl$.

$C_8H_{18}$ $(A)$ on mono-chlorination forms only one type of $C_8H_{17}Cl$ $(B)$. Hence $(A)$ can be:

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