An alcohol (A) gives Lucas test within 5 min. 7.4 g of alcohol when treated with sodium metal liberates 1120 mL of H2 at STP. What will be alcohol (A)?

Question

An alcohol (A) gives Lucas test within 5 min. 7.4 g of alcohol when treated with sodium metal liberates 1120 mL of H2 at STP. What will be alcohol (A)? (A) CH3(CH2)3OH (B) CH3CH(OH)CH2CH3 (C) (CH3)3COH (D) CH3CH(OH)CH2CH2CH3

Tardigrade Admin · Accepted Answer

R O H+N a arrow K O N a+1 / 2 H2 uparrow We have to get molecular mass of alcohol corresponding to half mole of H2 only. No. of moles of H 2= No. of moles of alcohol. (1120/112000)=(7.4/M M) ⇒ M M=74 Cn H2 n+1 O H=74 arrow Cn H2 n+1=74-17=57 ⇒ Cn H2 n=57-1=56 i.e. 12 n+2 n=14 m=56 ⇒ n=56 / 14=4 Thus molecular formula of (A) is C4 H9 O H . 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).

Q. An alcohol $(A)$ gives Lucas test within $5 min$ . $7.4 g$ of alcohol when treated with sodium metal liberates $1120 m L$ of $H_{2}$ at $STP$ . What will be alcohol $(A)$ ?

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

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

$(C H_{3})_{3} CO H$

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

Q. An alcohol (A) gives Lucas test within 5min. 7.4g of alcohol when treated with sodium metal liberates 1120mL of H2​ at STP. What will be alcohol (A)?

CH3​(CH2​)3​OH

CH3​CH(OH)CH2​CH3​

(CH3​)3​COH

CH3​CH(OH)CH2​CH2​CH3​

Q. An alcohol $(A)$ gives Lucas test within $5 min$ . $7.4 g$ of alcohol when treated with sodium metal liberates $1120 m L$ of $H_{2}$ at $STP$ . What will be alcohol $(A)$ ?

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

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

$(C H_{3})_{3} CO H$

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