. A sample of 0.42 gm os (A) with excess of MeMgBr gives 224 ml of CH_4 (g) at S.T.P Give the structure of (A) and write the equations involved.

. A sample of 0.42 gm os (A) with excess of MeMgBr gives 224 ml of CH

1.	. A sample of 0.42 gm os (A) with excess of MeMgBr gives 224 ml of CH_4 (g) at S.T.P Give the structure of (A) and write the equations involved.
Answer» Solution :Reaction of `(A)` with ammoniacal `AgNO_3` (Tollens' reagent, `[AG(NH_3)_2]OH)` shows that `(A)` is either an aldehyde or terminal triple bond. But very slow reaction (or `(A)` does not react appreciably) with Lucas reagent SUGGESTS `(A)` to be a `1^@` alcohol. Since compound `(A)` has only one oxygen atom, it can be either an aldehyde or a `1^@` alcohol. `DU i N (A) = ((2n_C +2)-n_H)/(2) = ((5 xx 2 + 2 - 8))/(2) = 2^@`. Two `DU` in `(A)` can be accounted for one terminal triple that `(A)`with `G.R` to give `CH_4 (g)` suggests acitve `H` atom in `(A)`. ACTIVE `H` atom in `(A)` is due to the terminal triple bond and `1^@` alcohol. Moreover, the formation of n-pentane from `(A)` suggests that `(A)` is a straight-chain compound containing terminal triple bond and `1^@` alcohol group at the other end of the chain. So, the structure is `(A)` is as follows : One mole of compound `(A)` would produce `2 mol` of `CH_4 (g)` (molecular mass of `A(C_5 H_8 O) = 84)`. `:.` 84 gm of (A) gives `2 xx 22.4` litres of `CH_4` (g) at S.T.P 0.42 gm of (A) gives `overset (2 xx 22.4 xx 0.42) underset (84)rarr = 0.224 litres`. =`224 ml`. The volume of `CH_4 (g)` correcponds to the volume given in the problem. Hence, three are two active `H` atoms in `(A)` that react with `G.R`. Reactions : .