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NEET CHEMISTRYEasy

The correct state of hybridisation of C2, C3, C5 and C6 of the hydrocarbon HC1≡C2−CH32−CH4=CH5−CH63\overset{1}{\text{HC}} \equiv \overset{2}{\text{C}} - \overset{3}{\text{CH}}_2 - \overset{4}{\text{CH}} = \overset{5}{\text{CH}} - \overset{6}{\text{CH}}_3 is in the sequence:

A

sp, sp³, sp² and sp³

B

sp³, sp², sp² and sp

C

sp, sp², sp² and sp³

D

sp, sp², sp³ and sp²

Step-by-Step Solution

  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: C1H≡C2−C3H2−C4H=C5H−C6H3\overset{1}{\text{C}}\text{H} \equiv \overset{2}{\text{C}} - \overset{3}{\text{C}}\text{H}_2 - \overset{4}{\text{C}}\text{H} = \overset{5}{\text{C}}\text{H} - \overset{6}{\text{C}}\text{H}_3
  1. Analyze Hybridisation:
  • C2: It is part of a triple bond (C≡C\text{C} \equiv \text{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 (C=C\text{C} = \text{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.
  1. Sequence: C2 (sp), C3 (sp³), C5 (sp²), C6 (sp³).
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