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Q. An alkane $C_{6}H_{14}\left(X\right)$ can be obtained by reduction of five different alkyl halides $C_{6}H_{13}Cl$ (only structural isomers) with $Zn—CH_{3}COOH$. The IUPAC name of $X$ satisfying above condition is

Hydrocarbons

Solution:

$2-$methyl pentane has five types of hydrogen, therefore, it can be obtained by reduction of five different halides as:
image
All of the above chloroalkanes on reduction with
$ Zn/CH_{3}COOH$ would give the same $2-$methyl pentane