An alcohol $X (C_{7} H_{16} O)$ shows immediate turbidity with $H C l / Z n C l_{2}$ and can be resolved into enantiomers. Also, X has tertiary carbon. How many $C H_{3}$ - groups are present in X?

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Solution

Due to the formation of 3 degree carbocation turbidity form immediately. Also as it is given that it can be resolved into enantiomers, the structure for $X$ is as given above and hence, there are 4 methyl groups in X.

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