First slide

Alkynes

Question

The ratio of sp, sp2 and sp3 orbitals in the product ‘B’ formed from the reaction below
$C H_{3} - C H = C H - C \equiv C H \to_{- N H_{3}}^{N a N H_{2}} A \to_{- N a B r}^{C H_{3} - B r} B$

Difficult

A

2:3:4
B

1:3:4
C

2:3:10
D

4:3:9

Solution

[Step 1]: terminal alkyne on reaction with sodamide forms sodium acetylide,
[Step 2]: Sodium acetylide on reaction with methyl halide forms compound B.
The steps are respectively
Compound ‘B’ is having 2 sp carbons, 2 sp2 carbons and 2 sp3 carbons
Total sp orbitals are 22=4, total sp2 orbitals are 23=6 and total sp3 orbitals are 2*4=8
Therefore the ratio is 2:3:4

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