Question

i. `27.8gm` mixture of alkyne and alkane (both containing same number of carbon atoms) is dissolved in `1000gm` of benzene. The solution freezes at `2.45^@C` (lower than that of benzene). Another `27.8gm` mixture requires `0.6` mol of `H_2` for complete hydrogenation. Calculate the chemical formula of alkyne and alkane (`K_f` for `C_6H_6=4.9`).
ii. Alkyne on hydrogenation with `H_2+Pt` gives the same alkane.
Alkyne does not react with ammoniacal `AgNO_3` solution. Give the structures of both alkyne and alkane.

Answer 1

i. Formula of alkyne `=C_nH_(2n-2)`, Alkane `=C_nH_(2n+2)`
(Same number of C atoms).
Let weight of alkyne = `x gm` and weight of alkane
`=27.8-x gm`
Molecular weight `(MW)` of alkyne
`=(12n+2n-2)=(14n-2)`
Molecular weight (MW) of alkane
`=(12n+2n+2)=(14n+2)`
`R-C-=C-Roverset(2H_2//Ni)rarrR-CH_2-CH_2-R`
2 mol of `H_2=1` mol of alkyne
`0.6` mol of `H_2=0.3` mol of alkyne
`:. 0.6` mol of `H_2impliesxgm` of alkyne
2 mol of `H_2implies(x)/(0.6)xx2=(x)/(0.3)gm` of alkyne
`=MW` of alkyne
`:. (x)/(0.3)=(14n-2)` `:. ximplies0.3(14n-2)` ....(i)
`DeltaT_f=K_f(m_1+m_2)` (Let `m_1implies` molality of alkyne, `m_2implies` molality of alkane)
`2.45=4.9(m_1+m_2)`
`:. m_1+m_2=(2.45)/(4.9)=0.5`
`:. m_1+m_2=0.5`
`m_1=("Moles of alkyne" xx 1000)/(W_1(wt. "of solvent"))=(0.3xx1000)/(1000)`
`m_1=0.3, m_1+m_2=0.5` `:. m_2implies0.2`
`m_2=(Wt. "of alkane" xx1000)/(M "of alkane" xx1000)=0.2=(27.8-8)/((14n+2))`
Substitute the value of x from (i).
`0.2=(27.8-0.3(14n-2))/((14n+2))`, solving we get `n=4`
Formula of alkyne `=C_4H_6`
Formula of alkane `=C_4H_(10)`
ii. Alkyne is internal as it does not react with ammoniacal `AgNO_3` solution `[Ag(NH_3)_2]^(o+)overset(Ө)OH`.
So the structures of alkyne and alkanes are: