Question

An alcohol X (C$_4$H$_{10}$O) reacts with Conc.HCl/ZnCl$_2$ to give corresponding chloride. X on dehydration forms Y, which reacts with Baeyer's reagent to give Z. What is Z?

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Tertiary alcohols dehydrate to alkenes; Baeyer’s reagent forms diols from alkenes. Identify alcohol type by chloride formation speed. Draw products to confirm functional groups.

Answer 1

The Correct Option is C

1. X (C$_4$H$_{10}$O) is an alcohol forming a chloride with HCl/ZnCl$_2$, suggesting tertiary or secondary alcohol (primary is slow).
2. Dehydration of X forms alkene Y (C$_4$H$_8$). Baeyer’s reagent (cold dilute KMnO$_4$) converts Y to diol Z.
3. Likely X is (CH$_3$)$_3$COH (tert-butanol); dehydration gives (CH$_3$)$_2$C=CH$_2$ (Y).
4. Baeyer’s reagent adds two OH groups: Z is (CH$_3$)$_2$C(OH)CH$_2$OH (2-methylpropane-1,2-diol), matching (3) as a diol.
5. Thus, the answer is (3) H OH OH (diol structure).