The correct state of hybridisation of C2, C3, C5 and C6 of the hydrocarbon
$$ \overset{1}{ ext{HC}} \equiv \

Question

The correct state of hybridisation of C2, C3, C5 and C6 of the hydrocarbon 
$$ \overset{1}{	ext{HC}} \equiv \overset{2}{	ext{C}} - \overset{3}{	ext{CH}}_2 - \overset{4}{	ext{CH}} = \overset{5}{	ext{CH}} - \overset{6}{	ext{CH}}_3 $$
is in the sequence:

Accepted Answer

1. **Determine the Structure and Numbering:** According to IUPAC nomenclature rules, the triple bond gets priority for the lowest number if it allows for a lower locant set or if positions are symmetrical with the double bond (though double bond has priority in naming 'ene' before 'yne', numbering follows the lowest locant rule). Here, numbering from the left gives locants 1 (yne) and 4 (ene). Numbering from the right gives 2 (ene) and 5 (yne). The set (1, 4) is lower than (2, 5). 
   - Structure: $ \overset{1}{	ext{C}}	ext{H} \equiv \overset{2}{	ext{C}} - \overset{3}{	ext{C}}	ext{H}_2 - \overset{4}{	ext{C}}	ext{H} = \overset{5}{	ext{C}}	ext{H} - \overset{6}{	ext{C}}	ext{H}_3 $

2. **Analyze Hybridisation:**
   - **C2:** It is part of a triple bond ($ 	ext{C} \equiv 	ext{C} $). It has 2 sigma bonds and 2 pi bonds. Hybridisation is **sp**.
   - **C3:** It forms 4 single sigma bonds (2 with C, 2 with H). As referenced in Class 12 Chemistry, Unit 6 , a carbon bonded to 4 atoms via single bonds is **sp³** hybridized.
   - **C5:** It is part of a double bond ($ 	ext{C} = 	ext{C} $). It has 3 sigma bonds and 1 pi bond. As referenced in Class 12 Chemistry, Unit 6 , vinylic carbons are **sp²** hybridized.
   - **C6:** It forms 4 single sigma bonds (1 with C, 3 with H). Hybridisation is **sp³**.

3. **Sequence:** C2 (sp), C3 (sp³), C5 (sp²), C6 (sp³).

Step-by-Step Solution

Practice All CHEMISTRY Questions