Question

Count the total number of electrons present in lone pair and \( \pi \) bond.

• A. 12
• B. 14
• C. 16
• D. 18

Hint:

To count electrons in lone pairs and \( \pi \)-bonds, remember that each lone pair on atoms like oxygen and halogens contribute 2 electrons, and each \( \pi \)-bond contributes 2 electrons as well.

Answer 1

The Correct Option is B

Step 1: Analyzing the structure of Aldohexose.
The structure of Aldohexose contains a six-membered carbon chain with hydroxyl groups (-OH) at the \( C_2 \), \( C_3 \), and \( C_4 \) positions, and an aldehyde group at \( C_1 \). The reaction shown involves a few steps of reduction and substitution. Now, we focus on counting the electrons from lone pairs and \( \pi \)-bonds in the molecule:
- The aldehyde group (–CHO) has one \( \pi \)-bond between carbon and oxygen, contributing 2 electrons. - The hydroxyl group (-OH) has lone pairs on oxygen, contributing 2 electrons per hydroxyl group. With 4 hydroxyl groups, we get 8 electrons from lone pairs on oxygen. - The \( \pi \)-bond of the C=C in the product will contribute 2 electrons. - The chloride (Cl) from the substitution at the \( C_6 \) position (phenyl carbon) has a lone pair, contributing 2 electrons.
Step 2: Summing the electrons.
The total electrons from lone pairs and \( \pi \)-bonds are: 2 (π bond from C=O) + 8 (\textlone pairs on 4 OH groups) + 2 (π \textbond from C=C) + 2 (\textlone pair from Cl) + 4 \text(lone pairs on the carbonyl group (C=O)) = 18 \textelectrons. Thus, the total number of electrons present in lone pairs and \( \pi \)-bonds is \( \boxed{18} \).
Final Answer: (D) 18