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Can you tell? · Q10

Q.Why are Propan-1-ol and 2-Methylpropan-1-ol not prepared by this method?

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
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✓ Free question

Step 1. Recall the method's selectivity. Section 15.2.4 describes cold concentrated H2SO4 adding across an alkene to give an alkyl hydrogen sulfate, which hydrolyses on dilution/heating to the alcohol; this addition follows Markovnikov's rule, so the -OSO3H (and hence eventually -OH) group attaches to the carbon bearing FEWER hydrogens, i.e. the more substituted carbon.

Step 2. Apply this to propene. Propene, CH3-CH=CH2, has one CH (fewer H) and one CH2 (more H) on its double bond; Markovnikov addition puts -OH on the CH carbon, giving propan-2-ol (a secondary alcohol), never propan-1-ol (a primary alcohol, which would need -OH on the terminal CH2).

Step 3. Apply this to 2-methylprop-1-ene. Similarly, 2-methylprop-1-ene, (CH3)2C=CH2, has a fully-substituted carbon and a terminal CH2; Markovnikov addition puts -OH on the more-substituted carbon, giving 2-methylpropan-2-ol (tertiary), never 2-methylpropan-1-ol (primary, needing -OH on the terminal CH2).

Step 4. State the alternative. Making these primary alcohols instead requires the ANTI-Markovnikov hydroboration-oxidation route (diborane/THF, then alkaline H2O2), which specifically places -OH on the less-substituted (terminal) carbon.

✓Final answer

Because the H2SO4/water method follows Markovnikov's rule, so -OH always ends up on the more-substituted carbon; propan-1-ol and 2-methylpropan-1-ol are primary alcohols that would need -OH on the LESS-substituted (terminal) carbon, which this route cannot deliver -- the anti-Markovnikov hydroboration-oxidation route (section 15.2.4) is needed instead.

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