Question

Identify the correct decreasing order of precipitation power of flocculating ion added, from following.

• A. $\text{Al}^{3+} \gt \text{Na}^+ \gt \text{Ba}^{2+}$
• B. $\text{Ba}^{2+} \gt \text{Al}^{3+} \gt \text{Na}^+$
• C. $\text{Al}^{3+} \gt \text{Ba}^{2+} \gt \text{Na}^+$
• D. $\text{Na}^+ \gt \text{Ba}^{2+} \gt \text{Al}^{3+}$

Hint:

The precipitation power increases dramatically with charge, not just linearly! For instance, a trivalent ion like $\text{Al}^{3+}$ can be up to 500 to 1000 times more effective at coagulating a negative sol than a monovalent ion like $\text{Na}^+$. Always sort by the highest absolute charge first.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The task is to determine the final product 'B' when two molecules of acetaldehyde react sequentially with dilute NaOH and then heat.

Step 2: Key Formula or Approach:
An aldehyde with α-hydrogens undergoes Aldol Condensation, forming a β-hydroxyaldehyde intermediate, which then dehydrates upon heating to an α,β-unsaturated aldehyde.

Step 3: Detailed Explanation:
Two acetaldehydes react via dilute NaOH to give 3-hydroxybutanal (Product A). Heating 'A' eliminates water, producing a conjugated system. The resulting four-carbon chain with a double bond at C-2 and an aldehyde group is named but-2-enal.

Step 4: Final Answer:
Product 'B' is but-2-enal, matching option (C).