Match the reactions in List-$I$ with the major products in List-$II$:
| List-$I$ | List-$II$ (Major Product) |
| :--- | :--- |
| $(A)$ $CH_3-CHBr-CH_2Br \xrightarrow{KOH/C_2H_5OH}$ | $(I)$ $1^{\circ}$-alkyl bromide |
| $(B)$ $CH_3-CH_2-CH=CH_2 \xrightarrow{HBr, (C_6H_5CO)_2O_2, \Delta}$ | $(II)$ $2^{\circ}$-alkyl bromide |
| $(C)$ $CH_3CH_2CH_3 \xrightarrow{Br_2, h\nu}$ | $(III)$ Allyl bromide |
| $(D)$ $CH_3-CH=CH_2 \xrightarrow{NBS, \Delta}$ | $(IV)$ Alkenyl bromide |

  • A
    $A-IV, B-I, C-II, D-III$
  • B
    $A-I, B-II, C-III, D-IV$
  • C
    $A-IV, B-II, C-I, D-III$
  • D
    $A-III, B-I, C-II, D-IV$

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Which of the following compounds is a tertiary alkyl halide?

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Predict the major product of the reaction:

Assertion : Alkylbenzene is not prepared by Friedel-Crafts alkylation of benzene.
Reason : Alkyl halides are less reactive than acyl halides.

Match the reactions given in Column-$I$ with their names in Column-$II$:
Column-$I$ (Reactants) Column-$II$ (Name)
$A$. $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$ $i$. Electrophilic addition
$B$. $H_3C-Br + AgF \to H_3C-F + AgBr$ $ii$. Finkelstein reaction
$C$. $H_3C-CH_3 + Cl_2 \xrightarrow{hv} H_3C-CH_2Cl + HCl$ $iii$. Swarts reaction
$D$. $CH_2=CH_2 + Br_2 \to BrCH_2-CH_2Br$ $iv$. Free radical substitution

Match the List-$I$ with List-$II$:
List-$I$ Name reaction List-$II$ Product obtainable
$A.$ Swarts reaction $I.$ Ethyl benzene
$B.$ Sandmeyer's reaction $II.$ Ethyl iodide
$C.$ Wurtz Fittig reaction $III.$ Cyanobenzene
$D.$ Finkelstein reaction $IV.$ Ethyl fluoride

Choose the correct answer from the options given below:

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