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)?

C H_{3} (C H_{2})_{3} O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} C H (O H) C H_{2} C H_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(C H_{3})_{3} C O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} C H (O H) C H_{2} C H_{2} C H_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $C H_{3} C H (O H) C H_{2} C H_{3}$
$R O H + N a \to R O N a + 1 / 2 H_{2} ↑$
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
$\frac{11200}{1120} = \frac{7.4}{M M} \Rightarrow M M = 74$
$C_{n} H 2 n + 1 O H = 74 \to C_{n} H_{2 n + 1} = 7417 = 57$
$\Rightarrow C_{n} H_{2 n} = 57 - 1 = 56 i . e . 12 n + 2 n = 14 n = 56$
$\Rightarrow n = 56 / 14 = 4$
Thus molecular formula of (A) is $C_{4} H_{9} O H$ . As (A) gives Lucas test within 5 min., thus $2^{0}$ alcohol corresponding to molecular formula $C_{4} H_{9} O H$ is $C H_{3} C H (O H) C H_{2} C H_{3}$ (butan-2-ol).

Suggest Corrections

Similar questions

C H_{4}

C H_{3} M g I

H_{2} S O_{4}

C_{4} H_{8}

C_{2} H_{4} O

C H_{3}

C H_{4}

C H_{3}

C H_{4}

Join BYJU'S Learning Program