Hydrocarbon $(A) (C = 87.2 %)$ on hydrogenation forms $(B) (C = 84.1 %)$ . Ozonolysis of $(A)$ forms acetic acid, acetone and pyruvic acid $⎛ ⎜ ⎝ C H_{3} - O ∥ C - C O O H ⎞ ⎟ ⎠$ . What are $(A)$ and $(B)$ ?

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Solution

Percentage composition shows that $(A)$ and $(B)$ has empirical formula $(A) C_{4} H_{7}$ and $(B) C_{4} H_{9}$ .
$C o m p o u n d (A) O z o n o l y s i s - ----- \to C H_{3} - O ∥ C - O H + C H_{3} - O ∥ C - C H_{3} + C H_{3} - O ∥ C - O ∥ C - O H$
Thus, compound $(A)$ should have eight carbons.
$∴$ Molecular formula of $(A) w i l l b e$ C_{8}H_{14} $a n d t h a t o f$ (B) $w i l l b e$ C_{8}H_{18} $. D e g r e e o f u n s a t u r a t i o n i n c o m p o u n d$ (A) = \dfrac {2n_{1} - n_{2} + 2}{2} = \dfrac {2\times 8 - 14 + 2}{2} = 2 $T h u s, c o m p o u n d$ (A) $s h o u l d h a v e t w o d o u b l e b o n d s . O n t h e b a s i s o f o z o n o l y s i s p r o d u c t s s t r u c t u r e o f$ (A)$ can be guessed as,

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