Which reagents shown below would accomplish the following transformation?
$Isopropyl \ bromide \xrightarrow{1. A \ 2. B} Isopropyl \ alcohol$

  • A
    $A = H_3O^{+}; B = BH_3-THF; H_2O_2/NaOH$
  • B
    $A = NaOH; B = BH_3-THF; H_2O_2/NaOH$
  • C
    $A = HBr \ \text{in ether}; B = Hg(OAc)_2/H_2O; NaBH_4$
  • D
    $A = NaNH_2; B = Hg(OAc)_2/H_2O; NaBH_4$

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The product $(B)$ of the reaction is:

Observe the following reactions:
$1$. $C_6H_5-CH=CH_2 + HBr \rightarrow X$
$2$. $C_6H_5-C(CH_3)=CH_2 + HBr \rightarrow Y$
$3$. $C_6H_5-CH=CH_2 + HBr \xrightarrow{(C_6H_5COO)_2} Z$
The correct order of reactivity of $X, Y, Z$ towards $S_N1$ reaction is:

Arrange the following organic halides in the correct order of reactivity towards $S_{N}2$ displacement:
$(P) \ (CH_3)_2C(Br)CH_2CH_3$
$(Q) \ BrCH_2(CH_2)_3CH_3$
$(R) \ CH_3CH(Br)(CH_2)_2CH_3$

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