Q.HCl is added to a hydrocarbon 'A' (C4H8) to give a compound 'B' which on hydrolysis with aqueous alkali forms tertiary alcohol 'C' (C4H10O). Identify 'A', 'B' and 'C'.
Step 1. Identify C from its formula and description. C is described as a tertiary alcohol with formula C4H10O. The only tertiary alcohol with this formula is tert-butanol (2-methylpropan-2-ol), (CH3)3C-OH.
Step 2. Work backwards to B. C is formed from B by hydrolysis with aqueous alkali (an SN substitution, Table 10.3 entry 1), so B must be the corresponding halide with the SAME carbon skeleton and the halogen in place of the -OH: B = (CH3)3C-Cl, tert-butyl chloride (2-chloro-2-methylpropane).
Step 3. Work backwards to A. B is formed by adding HCl to a hydrocarbon A with formula C4H8 (an alkene). For HCl's addition to place the chlorine on the tertiary carbon of (CH3)3C-Cl (ordinary Markovnikov addition, halogen to the more substituted carbon), A must be 2-methylpropene (isobutylene), (CH3)2C=CH2, C4H8: (CH3)2C=CH2 + HCl -> (CH3)3C-Cl.
Step 4. Verify the whole sequence. (CH3)2C=CH2 (A, C4H8) + HCl -> (CH3)3CCl (B) --aq. alkali--> (CH3)3COH (C, C4H10O) -- every formula and every step matches the problem statement.
A = 2-methylpropene (isobutylene); B = tert-butyl chloride (2-chloro-2-methylpropane); C = tert-butyl alcohol (2-methylpropan-2-ol).
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