The major product $(A)$ formed in the reaction given below is:

  • A
    $CH_3-CH_2-CH(C_6H_5)-CH_2Br$
  • B
    $CH_3-CH_2-C(C_6H_5)=CH_2$
  • C
    $CH_3-CH_2-CH(C_6H_5)-CH_2OH$
  • D
    $CH_3-CH_2-CH(C_6H_5)-CH_2OCH_3$

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Which among the following compounds has the highest boiling point?

Which of the following compounds is $NOT$ optically active?

For the following compounds,the correct statement$(s)$ with respect to nucleophilic substitution reactions is(are):
$I$: Benzyl bromide $(C_6H_5CH_2Br)$
$II$: Cyclohexylmethyl bromide $(C_6H_{11}CH_2Br)$
$III$: tert-Butyl bromide $((CH_3)_3CBr)$
$IV$: $1-$Phenylethyl bromide $(C_6H_5CH(CH_3)Br)$
$A$. $I$ and $III$ follow $S_{N}1$ mechanism
$B$. $I$ and $II$ follow $S_{N}2$ mechanism
$C$. Compound $IV$ undergoes inversion of configuration
$D$. The order of reactivity for $I$,$III$ and $IV$ is: $IV > I > III$

Which information below regarding this reaction is applicable?
$CH_3-O-CH_2-Cl \xrightarrow{aq. OH^-, \Delta} CH_3-O-CH_2-OH$

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