Consider the following four reactions: $(I) C_6H_5-CHCl-CH_3 \xrightarrow{95\% \text{ acetone}, 5\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(II) C_6H_5-CHCl-CH_3 \xrightarrow{90\% \text{ acetone}, 10\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(III) C_6H_5-CHCl-CH_3 \xrightarrow{80\% \text{ acetone}, 20\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(IV) C_6H_5-CHCl-CH_3 \xrightarrow{100\% \text{ water}} C_6H_5-CH(OH)-CH_3$ Arrange these reactions in decreasing order of the proportion of inverted product and select the correct answer from the codes given below:

  • A
    $I > II > III > IV$
  • B
    $II > I > III > IV$
  • C
    $III > II > I > IV$
  • D
    $IV > III > II > I$

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