Question

The major products from the following reaction sequence are product A and product B.

The total sum of \(\pi\) electrons in product A and product B are ______ (nearest integer)

Answer 1

Correct Answer: 8

The reaction proceeds as follows:

Cyclohexane $\xrightarrow{\text{Br}_2}$ C$_6$H$_{10}$Br$_2$
1. Product A: Bromination followed by elimination with alcoholic KOH gives a conjugated diene:
\[\text{C}_6\text{H}_{10}\text{Br}_2 \xrightarrow{\text{alc. KOH (3 eq.)}} \text{C}_6\text{H}_6 \, (\text{benzene}).\]
- Benzene contains 6 $\pi$ electrons.
2. Product B: Bromination followed by reaction with sodium hydroxide gives an enolate ion:
\[\text{C}_6\text{H}_{10}\text{Br}_2 \xrightarrow{\text{Na}^+/\text{O}^-} \text{O-CH}_2\text{CH=CH}_2.\]
- This product contains 2 $\pi$ electrons.
Total $\pi$ electrons: $6 + 2 = 8$

Answer 2

The problem asks for the total sum of π electrons in the major products A and B, which are formed from cyclohexene through two different reaction sequences.

Concept Used:

This problem involves several key concepts in organic chemistry:

1. Electrophilic Addition: Alkenes react with Br₂ to undergo electrophilic addition, forming a vicinal dihalide. Cyclohexene forms 1,2-dibromocyclohexane.
2. Elimination Reaction (E2): In the presence of a strong base like alcoholic KOH, alkyl halides undergo dehydrohalogenation. Vicinal dihalides can undergo double elimination to form conjugated dienes or alkynes.
3. Nucleophilic Substitution (S_N2): A strong nucleophile can displace a leaving group from a carbon atom. An alkoxide, such as sodium propargyl oxide, can act as a nucleophile to form an ether (Williamson Ether Synthesis). For secondary halides, S_N2 and E2 are competing reactions. The nature of the major product depends on the specific reagents and conditions.
4. Counting π Electrons: A carbon-carbon double bond (C=C) consists of one σ bond and one π bond, containing 2 π electrons. A carbon-carbon triple bond (C≡C) consists of one σ bond and two π bonds, containing 4 π electrons.

Step-by-Step Solution:

Step 1: Determine the structure of Product B. The reaction starts with cyclohexene. The first step common to both pathways is the addition of Br₂ to cyclohexene, which forms 1,2-dibromocyclohexane.

\[ \text{Cyclohexene} \xrightarrow{\text{Br}_2} \text{1,2-Dibromocyclohexane} \]

This intermediate is then treated with three equivalents of alcoholic KOH. Alcoholic KOH is a strong, non-bulky base that promotes E2 elimination. Since an excess of the base is used (3 eq.), a double dehydrohalogenation occurs.

First elimination: \[ \text{1,2-Dibromocyclohexane} \xrightarrow{\text{alc. KOH}} \text{3-Bromocyclohexene} + \text{H}_2\text{O} + \text{KBr} \] Second elimination: \[ \text{3-Bromocyclohexene} \xrightarrow{\text{alc. KOH}} \text{Cyclohexa-1,3-diene} + \text{H}_2\text{O} + \text{KBr} \]

Thus, Product B is cyclohexa-1,3-diene.

Step 2: Count the number of π electrons in Product B. Product B, cyclohexa-1,3-diene, contains two C=C double bonds. Each double bond has 2 π electrons.

\[ \text{Number of } \pi \text{ electrons in B} = 2 \times (\text{number of double bonds}) = 2 \times 2 = 4 \]

Step 3: Determine the structure of Product A. The intermediate 1,2-dibromocyclohexane is treated with one equivalent of sodium propargyl oxide (HC≡C-CH₂O^-Na⁺). This reagent can act as both a strong base (leading to E2 elimination) and a strong nucleophile (leading to S_N2 substitution). For a secondary halide like 1,2-dibromocyclohexane, S_N2 is a significant pathway. The S_N2 reaction results in the displacement of one of the bromide ions by the propargyl oxide nucleophile, forming an ether.

\[ \text{1,2-Dibromocyclohexane} + \text{HC}\equiv\text{C-CH}_2\text{O}^-\text{Na}^+ \xrightarrow{\text{S}_\text{N}2} \text{2-Bromo-1-(prop-2-yn-1-yloxy)cyclohexane} + \text{NaBr} \]

Thus, Product A is 2-bromo-1-(prop-2-yn-1-yloxy)cyclohexane.

Step 4: Count the number of π electrons in Product A. Product A contains a propargyl group (HC≡C-CH₂-), which has one C≡C triple bond. A triple bond consists of two π bonds.

\[ \text{Number of } \pi \text{ electrons in A} = 2 \times (\text{number of } \pi \text{ bonds in C}\equiv\text{C}) = 2 \times 2 = 4 \]

Step 5: Calculate the total sum of π electrons in Product A and Product B. The total number of π electrons is the sum of the π electrons in product A and product B.

\[ \text{Total } \pi \text{ electrons} = (\pi \text{ electrons in A}) + (\pi \text{ electrons in B}) \] \[ \text{Total } \pi \text{ electrons} = 4 + 4 = 8 \]

The total sum of π electrons in product A and product B is 8.