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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.

Explanation

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`.
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