All reactions of Oxalic acid

Reactions of Oxalic Acid (H₂C₂O₄) with Specific Reagents

Oxalic acid is an important organic acid studied in Class 11 and 12 Chemistry. It is a dibasic carboxylic acid and shows acidic as well as reducing properties. In this article, we will study all the important reactions of oxalic acid with specific reagents, which are frequently asked in CBSE board exams, practical exams, and competitive exams.

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1. Reaction with Alkalis (Neutralization Reaction)

Oxalic acid reacts with alkalis like sodium hydroxide to form acid salt and normal salt. Since oxalic acid is dibasic, the reaction occurs in two steps.

(a) With Sodium Hydroxide (NaOH)

Step 1 (Formation of Acid Salt):

H₂C₂O₄ + NaOH → NaHC₂O₄ + H₂O

Step 2 (Formation of Normal Salt):

NaHC₂O₄ + NaOH → Na₂C₂O₄ + H₂O

This reaction proves the dibasic nature of oxalic acid.

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2. Reaction with Carbonates and Bicarbonates

(a) With Sodium Carbonate (Na₂CO₃)

H₂C₂O₄ + Na₂CO₃ → Na₂C₂O₄ + CO₂ ↑ + H₂O

Carbon dioxide gas is evolved with brisk effervescence.

(b) With Sodium Bicarbonate (NaHCO₃)

H₂C₂O₄ + 2NaHCO₃ → Na₂C₂O₄ + 2CO₂ ↑ + 2H₂O

This reaction confirms the acidic nature of oxalic acid.

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3. Reaction with Calcium Chloride (Confirmatory Test)

Oxalic acid reacts with calcium chloride solution to form a white precipitate.

H₂C₂O₄ + CaCl₂ → CaC₂O₄ ↓ + 2HCl

The white precipitate of calcium oxalate is insoluble in water. This reaction is used as a confirmatory test for oxalic acid.

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4. Reaction with Potassium Permanganate (KMnO₄)

In acidic medium, oxalic acid acts as a strong reducing agent and decolourises potassium permanganate solution.

2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 10CO₂ ↑ + 8H₂O

The purple colour of KMnO₄ disappears. The reaction is slow at room temperature and becomes fast on heating.

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5. Reaction with Concentrated Sulphuric Acid

When oxalic acid is heated with concentrated sulphuric acid, it decomposes to give carbon monoxide and carbon dioxide.

H₂C₂O₄ → CO + CO₂ + H₂O

Carbon monoxide burns with a blue flame. This reaction shows the reducing nature of oxalic acid.

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6. Reaction with Alcohols (Esterification)

Oxalic acid reacts with alcohols in the presence of concentrated sulphuric acid to form esters.

With Ethanol

H₂C₂O₄ + 2C₂H₅OH → (COOC₂H₅)₂ + 2H₂O

The product formed is diethyl oxalate, which has a pleasant fruity smell.

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7. Reaction with Metals

Oxalic acid reacts with active metals like zinc to liberate hydrogen gas.

H₂C₂O₄ + Zn → ZnC₂O₄ + H₂ ↑

This reaction confirms the acidic nature of oxalic acid.

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8. Reaction with Ammonia

Oxalic acid reacts with ammonia to form ammonium oxalate.

H₂C₂O₄ + 2NH₃ → (NH₄)₂C₂O₄

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9. Thermal Decomposition

On strong heating, oxalic acid decomposes into carbon monoxide, carbon dioxide and water.

H₂C₂O₄ → CO + CO₂ + H₂O

This reaction again shows the reducing nature of oxalic acid.

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10. Reaction with Ferric Ions (Fe³⁺)

Oxalic acid reduces ferric ions to ferrous ions.

2Fe³⁺ + H₂C₂O₄ → 2Fe²⁺ + 2CO₂ + 2H⁺

This reaction is important in redox chemistry.

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Conclusion

Oxalic acid shows a wide range of reactions due to its acidic and reducing properties. The reactions with calcium chloride and potassium permanganate are especially important for practical and board examinations. A proper understanding of these reactions helps students score well in both theory and practical chemistry.

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Amino acid and it's different types

Amino Acids: Definition, Structure, Types and Importance

Amino acids are one of the most important topics in Class 12 Chemistry and basic biochemistry. Questions related to amino acids are frequently asked in CBSE board examinations, NEET, JEE and even in practical viva. Amino acids are the fundamental building blocks of proteins and play a vital role in the structure and functioning of living organisms.

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What are Amino Acids?

Amino acids are organic compounds that contain both an amino group (–NH₂) and a carboxyl group (–COOH) in the same molecule. Because they contain both acidic and basic functional groups, amino acids show unique chemical behavior.

General definition:
Amino acids are compounds which contain one amino group and one carboxylic acid group attached to the same carbon atom.

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General Structure of Amino Acids

The general structure of an amino acid is:

    H
    |
H₂N – C – COOH
    |
    R

Here, R represents the side chain. The nature of the R-group determines the properties and type of the amino acid.

General formula:
H₂N – CH(R) – COOH

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Zwitterion Nature of Amino Acids

In aqueous solution, amino acids exist as zwitterions. A zwitterion is a molecule that carries both positive and negative charges but is electrically neutral overall.

Zwitterion form:

H₃N⁺ – CH(R) – COO⁻

Because of this zwitterionic nature:

• Amino acids have high melting points
• They are crystalline solids
• They behave as both acids and bases (amphoteric nature)

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Classification of Amino Acids

Amino acids are classified in different ways based on nutrition, structure, side chain properties, and metabolic behavior.

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1. Classification Based on Nutritional Requirement

(a) Essential Amino Acids

Essential amino acids are those which cannot be synthesized by the human body and must be obtained through diet.

Examples of essential amino acids:

• Valine
• Leucine
• Isoleucine
• Lysine
• Methionine
• Phenylalanine
• Threonine
• Tryptophan

These amino acids are very important for growth and tissue repair.

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(b) Non-Essential Amino Acids

Non-essential amino acids are those which can be synthesized by the human body.

Examples:

• Glycine
• Alanine
• Serine
• Aspartic acid
• Glutamic acid

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2. Classification Based on Side Chain Polarity

(a) Non-Polar Amino Acids

These amino acids have non-polar side chains and are hydrophobic in nature.

• Glycine
• Alanine
• Valine
• Leucine
• Isoleucine

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(b) Polar but Uncharged Amino Acids

These amino acids have polar side chains but no net charge.

• Serine
• Threonine
• Asparagine
• Glutamine
• Tyrosine

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(c) Charged Amino Acids

Positively charged (basic) amino acids:

• Lysine
• Arginine
• Histidine

Negatively charged (acidic) amino acids:

• Aspartic acid
• Glutamic acid

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3. Classification Based on Nature of Side Chain

(a) Aliphatic Amino Acids

• Glycine
• Alanine
• Valine
• Leucine

(b) Aromatic Amino Acids

• Phenylalanine
• Tyrosine
• Tryptophan

(c) Sulphur-Containing Amino Acids

• Cysteine
• Methionine

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4. Classification Based on Metabolic Fate

(a) Glucogenic Amino Acids

These amino acids are converted into glucose during metabolism.

• Alanine
• Glycine
• Aspartic acid

(b) Ketogenic Amino Acids

These amino acids are converted into ketone bodies.

• Leucine
• Lysine

(c) Both Glucogenic and Ketogenic

• Isoleucine
• Phenylalanine
• Tyrosine
• Tryptophan

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Important Properties of Amino Acids

• Amino acids are colorless, crystalline solids
• They are soluble in water
• They have high melting points
• They show amphoteric behavior

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Biological Importance of Amino Acids

Amino acids are extremely important for living organisms.

• They are building blocks of proteins
• They help in enzyme formation
• They are required for growth and tissue repair
• They play a role in hormone synthesis
• They are essential for neurotransmitter formation

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Conclusion

Amino acids are the fundamental units of life. Their structure, classification, and properties are very important for understanding proteins and biological processes. A strong understanding of amino acids helps students score high marks in CBSE boards and competitive examinations.

Aldehyde preparation and it's reactions

Preparation and Reactions of Aldehydes – Class 12 Chemistry

Aldehydes form one of the most important chapters in Class 12 Organic Chemistry. Questions from aldehydes are frequently asked in CBSE board examinations, JEE, NEET, and practical viva. This article explains the preparation and reactions of aldehydes in a very simple, readable, and exam-oriented manner.

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What are Aldehydes?

Aldehydes are organic compounds containing the functional group –CHO. In aldehydes, the carbonyl carbon is bonded to one hydrogen atom and one alkyl or aryl group.

General structural formula:

R – C (=O) – H
    ||
    O

General molecular formula:
R – CHO

Examples:
Formaldehyde → H – CHO
Acetaldehyde → CH₃ – CHO
Benzaldehyde → C₆H₅ – CHO

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Preparation of Aldehydes

Several laboratory and industrial methods are used to prepare aldehydes. Only the most important methods required for CBSE and competitive exams are discussed below.

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1. Preparation from Primary Alcohols (Controlled Oxidation)

Primary alcohols on controlled oxidation form aldehydes. Strong oxidation must be avoided, otherwise carboxylic acids are formed.

Reaction diagram:

R – CH₂ – OH   + [O]
            ↓
R – CHO   + H₂O

Oxidizing agents used:
PCC (preferred reagent)
Cu at 573 K
Acidified K₂Cr₂O₇ (controlled)

Example:
CH₃ – CH₂ – OH → CH₃ – CHO

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2. From Acid Chlorides (Rosenmund Reduction)

Acid chlorides on reduction with hydrogen in the presence of a poisoned catalyst give aldehydes. This reaction is called Rosenmund reduction.

Reaction diagram:

R – COCl  + H₂
          ↓ (Pd / BaSO₄)
R – CHO  + HCl

The poisoned catalyst prevents further reduction of aldehyde to alcohol.

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3. From Nitriles (Stephen Reaction)

Nitriles are converted into aldehydes using the Stephen reaction.

Reaction diagram:

R – C ≡ N
    ↓ (SnCl₂ / HCl)
R – CH = NH
    ↓ (H₂O)
R – CHO

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4. From Alkynes (Hydroboration–Oxidation)

Terminal alkynes give aldehydes on hydroboration followed by oxidation.

Reaction diagram:

R – C ≡ CH
    ↓ (BH₃ / THF)
    ↓ (H₂O₂ / OH⁻)
R – CH₂ – CHO

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5. From Alkenes (Ozonolysis)

Alkenes on ozonolysis followed by reduction form aldehydes.

Reaction diagram:

R – CH = CH – R
    ↓ (O₃)
    ↓ (Zn / H₂O)
R – CHO + R – CHO

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Chemical Reactions of Aldehydes

Due to the presence of the polar carbonyl group, aldehydes are highly reactive. They undergo oxidation, reduction, nucleophilic addition, and condensation reactions.

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1. Oxidation Reactions

Aldehydes are easily oxidized to carboxylic acids.

(a) Tollens’ Test (Silver Mirror Test)

Reaction diagram:

R – CHO + [Ag(NH₃)₂]⁺
    ↓
R – COO⁻ + Ag ↓ (silver mirror)

This test is used to distinguish aldehydes from ketones.

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(b) Fehling’s Test

Aliphatic aldehydes give a brick-red precipitate of Cu₂O. Aromatic aldehydes do not respond to this test.

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2. Reduction Reactions

Aldehydes on reduction give primary alcohols.

Reaction diagram:

R – CHO
    ↓ (NaBH₄ / LiAlH₄)
R – CH₂ – OH

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3. Nucleophilic Addition Reactions

(a) Reaction with Hydrogen Cyanide

Reaction diagram:

R – CHO + HCN
    ↓
R – CH(OH) – CN

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(b) Reaction with Sodium Bisulphite

Reaction diagram:

R – CHO + NaHSO₃
    ↓
R – CH(OH) – SO₃Na

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4. Aldol Condensation

Aldehydes containing α-hydrogen undergo aldol condensation.

Reaction diagram:

2 CH₃ – CHO
    ↓ (dil. NaOH)
CH₃ – CH(OH) – CH₂ – CHO

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5. Cannizzaro Reaction

Aldehydes without α-hydrogen undergo Cannizzaro reaction.

Reaction diagram:

2 H – CHO + NaOH
    ↓
H – COONa + CH₃OH

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Important CBSE Exam Points

• Aldehydes are more reactive than ketones
• Aldehhydes give Tollens’ test
• Aldehydes are easily oxidized
• Aldol reaction requires α-hydrogen
• Cannizzaro reaction occurs when α-hydrogen is absent

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Conclusion:
Aldehydes are extremely important compounds in organic chemistry. Mastering their preparation methods and reactions helps students score high marks in CBSE boards and competitive examinations.

Test of primary , secondary and tertiary Amines

Identification of Primary, Secondary and Tertiary Amines

Amines are organic compounds derived from ammonia (NH₃) in which one or more hydrogen atoms are replaced by alkyl or aryl groups. Based on the number of carbon groups attached to nitrogen, amines are classified as primary, secondary, and tertiary amines. This topic is very important for CBSE Class 12 Chemistry as well as competitive examinations.

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1. Classification of Amines

Primary Amine (1°)

In primary amines, nitrogen is attached to one alkyl or aryl group and contains two hydrogen atoms.

General Formula: R–NH₂

Example: CH₃NH₂ (Methyl amine)

Secondary Amine (2°)

In secondary amines, nitrogen is attached to two carbon groups and contains one hydrogen atom.

General Formula: R–NH–R'

Example: (CH₃)₂NH (Dimethyl amine)

Tertiary Amine (3°)

In tertiary amines, nitrogen is attached to three carbon groups and has no hydrogen atom.

General Formula: R–N(R')–R''

Example: (CH₃)₃N (Trimethyl amine)

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2. Laboratory Methods to Distinguish Amines

A. Hinsberg Test

This is the most important test for distinguishing primary, secondary, and tertiary amines.

Reagent Used: Benzenesulphonyl chloride (C₆H₅SO₂Cl) and NaOH

Reaction Diagram:

Primary Amine:

R–NH₂ + C₆H₅SO₂Cl → R–NH–SO₂–C₆H₅ (Soluble in alkali)

Secondary Amine:

R₂NH + C₆H₅SO₂Cl → R₂N–SO₂–C₆H₅ (Insoluble)

Tertiary Amine:

No reaction with benzenesulphonyl chloride

Observation Table:

Amine Type	Observation
Primary	Forms soluble sulphonamide
Secondary	Forms insoluble sulphonamide
Tertiary	No reaction

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B. Carbylamine Test (Isocyanide Test)

This test is specific only for primary amines.

Reagents: Chloroform (CHCl₃) and alcoholic KOH

Reaction Diagram:

R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O

The formation of a foul-smelling isocyanide confirms the presence of a primary amine.

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C. Nitrous Acid Test

Nitrous acid is prepared in situ by reacting sodium nitrite with dilute hydrochloric acid.

Primary Aliphatic Amine:

R–NH₂ + HNO₂ → R–OH + N₂↑ + H₂O

Secondary Amine:

Forms yellow oily nitrosoamine

Tertiary Amine:

No gas evolved, forms salt

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3. CBSE Practical Style Experiment

Aim:

To distinguish between primary, secondary, and tertiary amines using Hinsberg test.

Apparatus and Chemicals:

Test tubes, dropper, NaOH solution, benzenesulphonyl chloride, given amine sample

Procedure:

1. Take a small amount of given amine in a test tube.
2. Add aqueous NaOH solution.
3. Add few drops of benzenesulphonyl chloride.
4. Shake the mixture well.
5. Observe solubility or formation of precipitate.

Observation:

If the product dissolves in alkali, the amine is primary. If it is insoluble, the amine is secondary. If no reaction occurs, the amine is tertiary.

Result:

The given organic compound is identified based on the observation.

Precautions:

• Use freshly prepared reagents.
• Avoid inhaling fumes.
• Handle chemicals carefully.

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4. Multiple Choice Questions (MCQs)

Q1. Which test is used to distinguish all three types of amines?

A. Carbylamine test
B. Nitrous acid test
C. Hinsberg test
D. Lassaigne test

Q2. Carbylamine test is given by:

A. Primary amine only
B. Secondary amine only
C. Tertiary amine only
D. All amines

Q3. Which amine does not react with benzenesulphonyl chloride?

A. Methyl amine
B. Dimethyl amine
C. Trimethyl amine
D. Ethyl amine

Q4. Yellow oily compound is formed when secondary amine reacts with:

A. NaOH
B. Nitrous acid
C. Chloroform
D. HCl

Answers:

Q1. C
Q2. A
Q3. C
Q4. B

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Conclusion

Primary, secondary, and tertiary amines can be easily distinguished using chemical tests such as Hinsberg test, Carbylamine test, and Nitrous acid test. These tests form an essential part of CBSE chemistry practicals and conceptual understanding of organic chemistry.

Different systems in a thermodynamics

Thermodynamics: Types of Systems and Entropy

Three different thermodynamic systems are under study. Their analysis is given below:

(a) Type of system in Case 1

Case 1: Water is boiling in an open container with continuous supply of heat.

Type of system: Open system

Reason: An open system allows exchange of both matter and energy with surroundings. Heat enters the system continuously and water vapour escapes from the container.

(b) Case in which entropy increases continuously

Answer: Case 1

During boiling, liquid water changes into vapour. Continuous heat supply and phase change increase molecular randomness, hence entropy increases continuously.

(c) Expression for internal energy change in Case 1

For an open system, internal energy change is given by:

dU = δQ − δW + Σh_indm_in − Σh_outdm_out

Where δQ is heat supplied, δW is work done, h is enthalpy and dm represents mass flow.

(d) System in which entropy decreases with time

Answer: Case 3

In freezing, liquid changes into solid. Molecular motion decreases and order increases, therefore entropy decreases with time.

Summary Table

Case	Type of System	Entropy Change
Case 1 (Boiling, open)	Open system	Increases continuously
Case 2 (Boiled water, closed)	Closed system	Nearly constant
Case 3 (Freezing, closed)	Closed system	Decreases

Suzuki and Negishi reaction

SN

Suzuki & Negishi Coupling Reactions — Clear Guide for Students

Simple explanations, examples, mechanism highlights, and a comparison table — ready for Blogger.

Summary: Suzuki and Negishi reactions are powerful palladium-catalyzed cross-coupling methods that form new carbon–carbon bonds. The Suzuki reaction uses boronic acids/esters, while the Negishi reaction employs organozinc reagents. Both are widely used in pharmaceuticals, material science, and advanced organic synthesis because of their reliability and functional-group tolerance.

Suzuki Reaction (Suzuki–Miyaura Coupling)

What it does: Couples an aryl or vinyl halide with an aryl or vinyl boronic acid/ester to form a new C–C bond.

Typical reagents & conditions: Pd(0) or Pd(II) catalyst (e.g., Pd(PPh3)4), base (K2CO3, NaOH), solvent (ethanol, toluene, DMF, water mixtures), room temperature to 80 °C.

Ar–X + Ar'–B(OH)2  —(Pd catalyst, base)—>  Ar–Ar'  (X = Br, Cl, I)

Mechanistic highlights: oxidative addition of Ar–X to Pd(0), transmetallation with boron species (assisted by base), and reductive elimination giving the biaryl product while regenerating Pd(0).

Why Suzuki is popular:

• Boronic acids/esters are stable and easy to handle.
• Reaction tolerates many functional groups (alcohols, ethers, esters).
• Works well for sp2–sp2 couplings (biaryls, styrenes).

Negishi Reaction

What it does: Couples an organozinc reagent with an aryl, vinyl, or alkyl halide under Pd or Ni catalysis to form a C–C bond.

Typical reagents & conditions: R–ZnX (prepared from R–Li or R–MgBr and ZnCl2), Pd or Ni catalyst, mild temperatures (often 0–50 °C), solvents like THF or toluene.

R–ZnX + R'–X  —(Pd or Ni catalyst)—>  R–R'  

Mechanistic highlights: oxidative addition of R'–X to Pd(0), transmetallation from the organozinc to Pd, then reductive elimination to form R–R'. Organozinc reagents are more nucleophilic than boronic acids and often react faster.

Strengths of Negishi:

• Organozinc reagents are reactive and enable sp3–sp2 and sp3–sp3 couplings that can be challenging by other methods.
• High chemoselectivity in many cases.

Practical comparison (quick table)

Suzuki (at a glance)
Organometallic partner	Boronic acids/esters
Moisture sensitivity	Low — tolerant to water
Functional group tolerance	Very good
Typical use	Biaryl formation, pharmaceuticals

Negishi (at a glance)
Organometallic partner	Organozinc (R–ZnX)
Moisture sensitivity	Higher — requires dry conditions
Functional group tolerance	Good, but preformed R–Zn may require precautions
Typical use	sp3–sp2 and sp3–sp3 couplings, complex molecule building

Examples

Suzuki example: Synthesis of biphenyl from bromobenzene and phenylboronic acid.

Ph–Br + Ph–B(OH)2 —(Pd(0), K2CO3)—> Ph–Ph (biphenyl)

Negishi example: Coupling of ethylzinc bromide with 1-bromobenzene to form ethylbenzene.

Et–ZnBr + Ph–Br —(Pd or Ni)—> Ph–Et (ethylbenzene)

Teaching tips & exam points

• Draw and label the three key steps: oxidative addition, transmetallation, reductive elimination.
• Ask students to list why Suzuki is preferred in industry (stable reagents, green solvent options, scalability).
• Pose a problem: plan a synthesis of 4-phenylbenzaldehyde using Suzuki coupling — what protecting groups (if any) are needed?

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Corey–House Reaction

Corey–House Reaction (Corey–Posner–Whitesides–House)

Introduction: The Corey–House reaction is a reliable method for making carbon–carbon single bonds by coupling alkyl fragments using organocuprate reagents (Gilman reagents). The reaction is useful for forming higher alkanes.

General Reaction

2 R–Li + CuI  →  R2CuLi  (Gilman reagent)

R2CuLi + R'–X → R–R' + R–Cu + LiX

(X = Br, Cl, I)

Mechanism (Short)

The reaction follows an SN2-type nucleophilic substitution on an alkyl halide. One alkyl group of the Gilman reagent couples with the electrophilic carbon, forming a new C–C bond.

• Formation of R2CuLi from organolithium and CuI
• Nucleophilic substitution on R'–X
• New C–C bond formed

Scope

• Best with primary alkyl halides
• Secondary moderate
• Tertiary not suitable (elimination)
• Aryl/vinyl halides usually unreactive

Example:

C2H5Li + CuI → (C2H5)2CuLi

(C2H5)2CuLi + C3H7Br → C5H12

Advantages

• Forms C–C bonds under mild conditions
• Good for making larger alkanes

Limitations

• Organolithiums are moisture sensitive
• Poor functional group tolerance
• Modern Pd-couplings have broader scope

Teaching Question

• Compare Corey–House vs Suzuki/Negishi for different substrates.