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3 years ago

Draw the products of the acid-base reaction between stearic acid and OH-

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

Products are stearate anion and water.

Explanation:

Stearic acid is a 18-carbon chain molecule containing -COOH group. IUPAC name of stearic acid is octadecanoic acid.

Molecular formula of stearic acid is $C_{17}H_{35}CO_{2}H$ .

When $OH^{-}$ is added into stearic acid, $OH^{-}$ removes a proton ( $H^{+}$ ) from acidic -COOH group and forms stearate anion and water as products.

The balanced acid-base reaction is given as:

$C_{17}H_{35}CO_{2}H+OH^{-}\rightleftharpoons C_{17}H_{35}CO_{2}^{-}+H_{2}O$

Structure of products are given below.

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MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

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Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

Where K' = K1 * K2

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just mechanism A and B are consistent with observed rate law

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan

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