JEE MAIN - Chemistry (2024 - 4th April Evening Shift - No. 13)

$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br}+\mathrm{NaOH} \xrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \text { Product 'A' }$$

JEE Main 2024 (Online) 4th April Evening Shift Chemistry - Haloalkanes and Haloarenes Question 17 English

Consider the above reactions, identify product B and product C.

$$\mathrm{B}=\mathrm{C}=1$$-Propanol

$$\mathrm{B}=1$$-Propanol $$\mathrm{C}=2$$-Propanol

$$\mathrm{B}=\mathrm{C}=2$$-Propanol

$$\mathrm{B}=2$$-Propanol $$\mathrm{C}=1$$-Propanol

Product '$$\mathrm{A}$$' is $$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$$

JEE Main 2024 (Online) 4th April Evening Shift Chemistry - Haloalkanes and Haloarenes Question 17 English Explanation