The maximum number of isomeric monochloro der | vedclass

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(A) The structure of $2,2,5,5-$tetramethylhexane is $(CH_3)_3C-CH_2-CH_2-C(CH_3)_3$.To find the number of isomeric monochloro derivatives,we identify

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$3$

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(A) The structure of $2,2,5,5-$tetramethylhexane is $(CH_3)_3C-CH_2-CH_2-C(CH_3)_3$.To find the number of isomeric monochloro derivatives,we identify the non-equivalent hydrogen atoms in the molecule.$1$. The $12$ methyl hydrogens at the ends are equivalent (type $a$). Replacing one gives $1-$chloro-$2,2,5,5-$tetramethylhexane.$2$. The $4$ methylene hydrogens in the middle are equivalent (type $b$). Replacing one gives $3-$chloro-$2,2,5,5-$tetramethylhexane.Since the molecule is symmetric,there are only $2$ distinct sets of equivalent hydrogen atoms.However,the $3-$chloro derivative has a chiral center at the $C-3$ position,leading to a pair of enantiomers $(d$ and $l$ forms).Therefore,the total number of isomeric monochloro derivatives is $1$ (from type $a$) $+ 2$ (enantiomers from type $b$) $= 3$.

Column $I$	Column $II$
$A$. $(CH_3)_3CH + O_2 \xrightarrow{KMnO_4} (CH_3)_3COH$	$i$. $CH_3COOH + H_2O$
$B$. $2CH_4 + O_2 \xrightarrow[100 \ atm]{Cu/523 \ K} 2CH_3OH$
$C$. $CH_4 + O_2 \xrightarrow[\Delta]{Mo_2O_3} HCHO + H_2O$	$iii$. $2CH_3OH$
$D$. $CH_3-CH_3 + \frac{3}{2}O_2 \xrightarrow{(CH_3COO)_2Mn} CH_3COOH + H_2O$	$iv$. $HCHO + H_2O$

The maximum number of isomeric monochloro derivatives which can be obtained from $2,2,5,5-$tetramethylhexane by chlorination is $..............$.

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