For Engineer By Engineer

  • Saturday, 5 July 2025

    Reaction Order Identification - Rate Constant Estimation - Batch Reactor Design

    Hi All....!!

    Today i want to post something better for our chemical engineers / process engineers which is a consolidation of reaction engineering and design by grabbing the basics. I'll take a case study on implementing of mechanistic studies for a considered reaction mechanism and using the data to design a batch reactor. This post is applicable for even the chemists who work in chemical manufacturing industries except the oil men.

    Before that, i want to give you guys a warm welcome at Pharmacalculations. We’ll exploring a powerful transformation used in the manufacture of pharmaceutical intermediates — the Ritter Reaction. This acid-catalyzed process enables the conversion of alcohols or alkenes with nitriles into tertiary amides, which are vital building blocks in API and agrochemical synthesis.

    But we won’t stop at just the chemistry. This post will walk you through:

    • The reaction mechanism

    • How to identify the rate-determining step

    • How to derive the rate law

    • How to determine the reaction order

    • How to design a batch reactor using Levenspiel’s performance equations

    Let’s base this on a real manufacturing case: the production of Tert-butylacetamide.

    Before jumping into the post, i want to dedicate this to a section of Chemical Engineers who feel that chemistry is not our part to work on. Any Reaction that is going to be designed in a chemical manufacturing industry requires two hands for the upliftment i.e., a proper chemistry and also the engineering for its successful establishment. Its never bad to grab few Chemistry insights while working in Chemical manufacturing industry. I would like to thank Mr. P Jagadish Kumar for motivating me to understand chemistry before taking up the process detailing part.


    Anyhow, lets jump into our show.

    Today, i've planned to take a complex Ritter reaction.


    Industrial Case: Ritter Reaction for tert-Butylacetamide Synthesis

    Reaction:

    (CH₃)₃COH+CH₃CN+H₂SO₄(CH₃)₃C–NHCOCH₃\text{(CH₃)₃COH} + \text{CH₃CN} + \text{H₂SO₄} \rightarrow \text{(CH₃)₃C–NHCOCH₃}

    • Alcohol: tert-Butyl alcohol (TBA)

    • Nitrile: Acetonitrile

    • Catalyst: Sulfuric acid

    • Product: tert-Butylacetamide

    • Reactor: Glass-lined batch reactor

    • Temperature: 60–70°C

    • Target Conversion: ≥90% in ≤1 hour


    Before getting into the main case, lets check few basic question & answers to understand case better.

    What is the Ritter reaction used for?
    For making amides from carbocations and nitriles — widely used in pharma.

    Why is tert-butyl alcohol selected?
    It forms a stable tertiary carbocation, ensuring fast and selective reaction.

    Why sulfuric acid?
    It acts both as a catalyst and proton source to initiate carbocation formation.

    Why excess acetonitrile?
    To drive the reaction forward and simplify the rate law (pseudo-first-order in CH₃CN).

    What are typical side reactions?
    Nitrile hydrolysis and unconverted alcohol due to incomplete conversion.

    What kind of reactor is preferred?
    A glass-lined batch reactor to handle the acidic conditions and control the exotherm.

    Is the reaction exothermic?
    Yes — particularly during carbocation formation. Controlled dosing and external cooling are recommended.

    What is the rate-determining step?
    The attack of nitrile on the carbocation intermediate.

    What is the reaction order?
    Typically 2nd order (1st in TBA, 1st in H⁺) when CH₃CN is in excess.

    How do we derive the rate law?
    Using the reaction mechanism and applying the steady-state approximation to the carbocation.

    Reaction Mechanism of the Ritter Reaction

    Let:

    • A=A = tert-butyl alcohol

    • B=B = acetonitrile

    • C+=C^+ = carbocation

    • I=I = iminium intermediate

    • P=P = amide product

    Stepwise Mechanism:

    Step 1: Protonation of Alcohol (Fast, Reversible)

    A+H+k1k1C++H2OA + H^+ \xrightleftharpoons[k_{-1}]{k_1} C^+ + H_2O

    Step 2: Attack of Nitrile (Slow, Rate-Determining)

    C++Bk2IC^+ + B \xrightarrow{k_2} I

    Step 3: Hydrolysis of Iminium (Fast)

    I+H2Ok3PI + H_2O \xrightarrow{k_3} P


    How to Identify the Rate-Determining Step (RDS)

    Knowing the RDS is critical to deriving the right rate expression.

    1. Look for the slow step

    Mechanisms often label the slow step — here, Step 2 (nitrile attacking carbocation) is the slowest.

    2. Analyze intermediate buildup

    The carbocation forms quickly (Step 1), but reacts slowly with CH₃CN. This indicates [C⁺] accumulates, confirming Step 2 is RDS.

    3. Validate with experimental kinetics

    Derived rate law:

    r=k1k2[A][H+][B]k1+k2[B]r = \frac{k_1 k_2 [A][H^+][B]}{k_{-1} + k_2 [B]}

    If [B] is in large excess, this simplifies to:

    r=k1[A][H+]r = k_1 [A][H^+]

    This matches observed 2nd order kinetics, further confirming Step 2 is the RDS.


    Rate Law Derivation Using Steady-State Approximation

    From the mechanism:

    Step 1

    r1=k1[A][H+],r1=k1[C+]r_1 = k_1 [A][H^+], \quad r_{-1} = k_{-1}[C^+]


    Step 2 (RDS)

    r2=k2[C+][B]r_2 = k_2 [C^+][B]

    Assume steady state on [C⁺]:

    d[C+]dt=0=k1[A][H+]k1[C+]k2[C+][B]\frac{d[C^+]}{dt} = 0 = k_1 [A][H^+] - k_{-1}[C^+] - k_2 [C^+][B]

    Solving:

    [C+]=k1[A][H+]k1+k2[B][C^+] = \frac{k_1 [A][H^+]}{k_{-1} + k_2 [B]}

    Rate of product formation (RDS):

    r=k2[C+][B]=k1k2[A][H+][B]k1+k2[B]r = k_2 [C^+][B] = \frac{k_1 k_2 [A][H^+][B]}{k_{-1} + k_2 [B]}

    Reaction Order Identification

    Case 1: [B] is limited

    r=k1k2k1[A][H+][B](3rd order overall)r = \frac{k_1 k_2}{k_{-1}} [A][H^+][B] \quad \text{(3rd order overall)}

    Case 2: [B] in large excess

    rk1[A][H+](2nd order overall)r \approx k_1 [A][H^+] \quad \text{(2nd order overall)}

    In industry, CH₃CN is used in excess — so the working rate law becomes:

    rA=k[A][H+]-r_A = k [A][H^+]


    Batch Reactor Design Using Performance Equation

    From Levenspiel (Ch. 1 & 3) — for 2nd order in A and H⁺:

    t=1kCH+(1CA1CA0)t = \frac{1}{k C_{H^+}} \left( \frac{1}{C_A} - \frac{1}{C_{A0}} \right)

    Assumed Lab Data:



    Estimate k from lab trial:

    X=0.90CA=0.02 mol/LX = 0.90 \Rightarrow C_A = 0.02 \text{ mol/L}
    t=1k0.15(10.0210.20)=60k=5.0 L/(mol.min)t = \frac{1}{k \cdot 0.15} \left( \frac{1}{0.02} - \frac{1}{0.20} \right) = 60 \Rightarrow k = 5.0\ \text{L/(mol·min)}

    Batch Time at Different Conversions (using k=5.0)



    Higher conversions = exponentially more time due to second-order kinetics.


    Thats it....!!

    Hope you understood everything clearly.

    Any queries, feel free to reach us at pharmacalc823@gmail.com

    Comments are most appreciated ....!!


    🗳️ What Should I Post Next?



    Poll Maker






    A
    bout The Author


    Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
    Follow Me on Twitter AjaySpectator & Computer Innovations

    No comments:

    Post a Comment

    This Blog is protected by DMCA.com

    ABOUT ADMIN


    Hi! I am Ajay Kumar Kalva, owner of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.

    Like Us On Facebook