Aromatic Substitution Reactions

Step 1: Reactivity with KOH.
KOH will break down a carbonyl compound into simpler products. The molar ratio of C, O, and H indicates the intermediate structure is likely an alkene with a hydroxyl group.
Step 2: Acidification leads to product formation.
After treatment with acid, the likely structure will be an aldehyde or ketone.
Final Answer:

Concept: Electrophilic aromatic substitution occurs preferentially at positions activated by electron donating groups. The substituent present is an amide linkage: The group donates electron density to the aromatic ring by resonance, activating ortho and para positions.
Step 1:Analyze substituent effects} The carbonyl group withdraws electron density from the first ring, making it less reactive. However the group donates electron density into the second ring.
Step 2:Determine activated positions} Thus electrophilic substitution occurs mainly at the ortho and para positions of the activated ring. These correspond to:

Step 1: Nature of group.
The group is an electron-withdrawing group, which deactivates the benzene ring and makes it less reactive towards electrophilic substitution reactions. Step 2: Directive influence of .
The group directs incoming electrophiles to the meta position, as it deactivates the ortho and para positions due to its electron-withdrawing nature. Step 3: Conclusion.
Thus, the correct answer is that is a deactivating group and a meta director. Final Answer:

Step 1: Analyze the reaction.
The reaction involves the addition of Br to a cycloalkene. The C : Br ratio is 3 : 1, indicating that three carbon atoms are involved for every one bromine atom in the product. Step 2: Calculate the percentage of Br in the product.
For each mole of X, one mole of Br is consumed, and each molecule of the product contains 1 bromine atom for 3 carbon atoms. Therefore, the percentage of bromine in the product is: Final Answer:

Step 1: Understand the reactions.
- A. Br /water reacts with glucose to form gluconic acid (iv). - B. Acetic anhydride (excess) reacts with glucose to form glucose pentaacetate (iii). - C. Conc. HNO reacts with glucose to form saccharic acid (ii). - D. NH OH reacts with glucose to form glucose oxime (i). Step 2: Conclusion.
The correct matching is A – iv, B – iii, C – ii, D – i. Final Answer: