See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

We analyze each statement individually: Statement (a): (CH3)2C=CH-CH2-Cl is an allylic chloride where the double bond is conjugated with the leaving group. The resulting carbocation upon ionization is allylic and additionally stabilized by the two methyl groups (tertiary allylic carbocation). Such substrates undergo solvolysis dramatically faster than simple alkyl chlorides due to the combined resonance stabilization and hyperconjugation. A rate enhancement of over 6000 times relative to a simple alkyl chloride at 25°C is consistent with established experimental data for allylic/tertiary systems. Statement (a) is correct. Statement (b): CH3-CH=CH-CH2-OH is 2-buten-1-ol (a primary allylic alcohol). When it reacts with HBr, protonation of the OH gives a primary allylic carbocation (or the reaction proceeds via an allylic intermediate) with resonance structures placing positive charge at C1 (primary) and C3 (secondary). Attack of Br- at C1 gives 1-bromo-2-butene (the direct product) and attack at C3 gives 3-bromo-1-butene (the allylic rearrangement product). Both products arise from the delocalized allylic cation, so a mixture of 1-bromo-2-butene and 3-bromo-1-butene is indeed obtained. Statement (b) is correct. Statement (c): 3-buten-2-ol (CH2=CH-CH(OH)-CH3) in aqueous sulphuric acid undergoes protonation of OH to form a secondary allylic carbocation. The resonance form places the positive charge on C4 (primary allylic position), which upon attack of water gives 2-buten-1-ol (CH3-CH=CH-CH2-OH). Over a long time (one week), equilibrium is established between 3-buten-2-ol and 2-buten-1-ol via this allylic cation intermediate. Thus the solution contains both isomers. Statement (c) is correct. Since all three statements (a), (b), and (c) are individually correct, option (d) 'All of these' is the correct answer. Therefore, the correct answer is D.