1.

An alcohol (A) gives Lucas test within 5 min. 7.4 g of alcohol when treated with sodium metal liberates 1120 mL of H_2 at STP. What will be alcohol (A)?

Answer»

`CH_(CH_2)_3OH`
`CH_3CH(OH)CH_2CH_3`
`(CH_3)_3OH`
`CH_3CH(OH) CH_2CH_2CH_3`

Solution :`ROH+Na rarr RONa+1//2H_(2)uarr`
We have to get molecular mass of alcohol corresponding to half mole of `H_(2)` only.
No .of moles of `H_(2)` =No.of moles of alcohol
`1120/11200=7.4/(MM) rArr MM=74`
`C_(n)H_(2N+1) OH=rarr C_(n)H_(2n+1)=74-17=57`
`rArr C_(n)H_(2n)=57-1=56``i.e.12n+2n=14n=56`
`rArr n=56//14=14` Thus molecular formula of (A) is `C_(4)H_(9)OH`. As (A) gives Lucas test within 5 min, thus `2^@`alcohol corresponding to molecular formula `C_(4)H_(9)OH` is `CH_(3)CH(OH)CH_(2)CH_(3)` (butan -2-ol).


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