The amount (in mol) of bromoform (CHBr3) produced when 1.0 mole of acetone reacts completely with 1.0 mole of bromine in the presence of aqueous NaOH is

Question

The amount (in mol) of bromoform (CHBr3) produced when 1.0 mole of acetone reacts completely with 1.0 mole of bromine in the presence of aqueous NaOH is (A) (1/3) (B) (2/3) (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

underset textAcetoneCH3 - underset overset||OC - CH3 + 3Br2 + 4NaOH → CH3 overset underset||OC overset-ON overset+a + 3NaBr + 3H2O + CHBr3 3 moles of bromine produces 1 mole of CHBr3. ∴ 1 mole of bromine produces (1/3)mole of CHBr3.

Q. The amount (in mol) of bromoform $(C H B r_{3})$ produced when $1.0$ mole of acetone reacts completely with $1.0$ mole of bromine in the presence of aqueous $N a O H$ is

$\frac{1}{3}$

$\frac{2}{3}$

1

2

$Acetone C H_{3} - O ∣∣ C - C H_{3} + 3 B r_{2} + 4 N a O H$
$\to C H_{3} C ∣∣ O ON - a + + 3 N a B r + 3 H_{2} O + C H B r_{3}$
$3$ moles of bromine produces $1$ mole of $C H B r_{3}$ .
$∴ 1$ mole of bromine produces $\frac{1}{3}$ mole of $C H B r_{3}$ .

Q. The amount (in mol) of bromoform (CHBr3​) produced when 1.0 mole of acetone reacts completely with 1.0 mole of bromine in the presence of aqueous NaOH is

31​

32​

1

2

AcetoneCH3​−O∣∣C​−CH3​​+3Br2​+4NaOH →CH3​C∣∣O​ON−a++3NaBr+3H2​O+CHBr3​ 3 moles of bromine produces 1 mole of CHBr3​. ∴1 mole of bromine produces 31​mole of CHBr3​.

Q. The amount (in mol) of bromoform $(C H B r_{3})$ produced when $1.0$ mole of acetone reacts completely with $1.0$ mole of bromine in the presence of aqueous $N a O H$ is

$\frac{1}{3}$

$\frac{2}{3}$

$Acetone C H_{3} - O ∣∣ C - C H_{3} + 3 B r_{2} + 4 N a O H$
$\to C H_{3} C ∣∣ O ON - a + + 3 N a B r + 3 H_{2} O + C H B r_{3}$
$3$ moles of bromine produces $1$ mole of $C H B r_{3}$ .
$∴ 1$ mole of bromine produces $\frac{1}{3}$ mole of $C H B r_{3}$ .