The f.p constant of benzene is 4.9 and its m.p is `5.51^(@)C`. A solution of 0.816g of compound (A) when dissolved in 7.5g of benzene freezes at `1.59^(@)C`. The compound (A) has C, 70.58% and H, 5.88%. Determine the molecular weight and molecular formula of (A). Compound (A) on heating with soda lime gives another compound (B) which on oxidation and subsequent acidification gives an acid (C ) of equivalent weight 122. (C ) on heating with soda lime gives benzene. Identify (A), (B) and ( C) and explain the reaction involved.

Explanation

We have `Delta T_(f)= K_(f).m`
`(5.51-1.59)= 4.90 xx ((0.816)/(M) xx (1000)/(7.5))`
M= 136 [M is mol. Wt. of compound (A)]
Moles of `C: H: O= (70.58)/(12): (5.88)/(1): (23.54)/(16)`
`=5.88 : 5.88: 1.471`
`=4:4:1`
`therfore` empirical formula of (A) `=C_(4)H_(4)O` (68)
As the molecular weight of (A) is 136, molecular formula of (A) is `C_(8)H_(8)O_(2)`
Now, from the given reaction sequence: