JEE MAIN - Chemistry (2023 - 13th April Evening Shift - No. 11)
Match List I with List II
1 - Bromopropane is reacted with reagents in List I to give product in List II
| LIST I - Reagent | LIST II - Product | ||
|---|---|---|---|
| A. | $$\mathrm{KOH}$$ (alc) | I. | Nitrile |
| B. | $$\mathrm{KCN}$$ (alc) | II. | Ester |
| C. | $$\mathrm{AgNO_2}$$ | III. | Alkene |
| D. | $$\mathrm{H_3CCOOAg}$$ | IV. | Nitroalkane |
Choose the correct answer from the options given below:
A-I, B-II, C-III, D-IV
A-IV, B-III, C-II, D-I
A-III, B-I, C-IV, D-II
A-I, B-III, C-IV, D-II
Explanation
For the reactions of bromopropane with the reagents in List I, the products formed in List II are as follows:
A. $$\mathrm{KOH}$$ (alc) - Alcoholic KOH is a strong base and will cause elimination of HBr, forming an alkene as the product. So, A-III.
B. $$\mathrm{KCN}$$ (alc) - Potassium cyanide (KCN) will replace the bromine atom in the alkyl halide with a cyanide group, forming a nitrile. So, B-I.
C. $$\mathrm{AgNO_2}$$ - Silver nitrite (AgNO2) will replace the bromine atom with a nitro group, forming a nitroalkane. So, C-IV.
D. $$\mathrm{H_3CCOOAg}$$ - Silver acetate (H3CCOOAg) will replace the bromine atom with an acetate group, forming an ester. So, D-II.
The correct answer is Option C: A-III, B-I, C-IV, D-II.
A. $$\mathrm{KOH}$$ (alc) - Alcoholic KOH is a strong base and will cause elimination of HBr, forming an alkene as the product. So, A-III.
B. $$\mathrm{KCN}$$ (alc) - Potassium cyanide (KCN) will replace the bromine atom in the alkyl halide with a cyanide group, forming a nitrile. So, B-I.
C. $$\mathrm{AgNO_2}$$ - Silver nitrite (AgNO2) will replace the bromine atom with a nitro group, forming a nitroalkane. So, C-IV.
D. $$\mathrm{H_3CCOOAg}$$ - Silver acetate (H3CCOOAg) will replace the bromine atom with an acetate group, forming an ester. So, D-II.
The correct answer is Option C: A-III, B-I, C-IV, D-II.
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