Chemistry · Carboxylic Acids and Derivatives

JEE Main 2024 — 5 April, Shift 2 — Question 86

The product (C) in the following sequence of reactions has $\_\_\_\_\_\_$ $\pi$ bonds.

Answer: 4

Numerical answer — enter this value.

The reaction sequence is:

\begin{array}{cccccc} \mathrm{C_6H_5CH_2CH_2CH_3} & \xrightarrow[\Delta]{\text{alk. }KMnO_4} & \mathrm{C_6H_5COOK} & \xrightarrow{H^+} & \mathrm{C_6H_5COOH} & \xrightarrow{Br_2/FeBr_3} \mathrm{m\!-\!BrC_6H_4COOH} \\ \text{Propyl\;benzene} & & \text{Potassium\; benzoate (A)} & & \text{Benzoic \;acid (B)} & \;\text{m-Bromobenzoic \;acid (C)} \end{array}

The product (C) is m-bromobenzoic acid.

Number of $\pi$ bonds in product (C) :

Benzene ring: $3$ $\pi$ bonds

Carboxylic acid group (COOH): $1$ $\pi$ bond (C=O)

So, total number of $\pi$ bonds $= 3 + 1 = 4$

Answer key and solution verified before publishing.

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