Class 11 Chemistry – Chapter 21: Organic Synthesis (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Chapter 21 – Organic Synthesis strictly according to the Federal Board (FBISE) syllabus. The chapter focuses on developing logical thinking and stepwise synthesis skills in organic chemistry.
Major topics include introduction to organic synthesis, functional group interconversions, synthesis pathways, reagents and conditions, reaction planning, and stepwise synthesis of organic compounds. Flow charts and exam-oriented methods are emphasized for better understanding.
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Study Notes
Core Reaction Pathways
Organic synthesis involves converting one functional group to another using specific reagents and conditions. Key transformations include:
- Alkanes to Halogenoalkanes: Free radical substitution using $X_{2}$ (halogens) and UV light.
- Alkenes to Alcohols: Hydration using steam and phosphoric acid ($H_{3}PO_{4}$) catalyst under pressure.
- Alcohols to Aldehydes/Carboxylic Acids: Oxidation using acidified potassium dichromate(VI) ($K_{2}Cr_{2}O_{7}/H^{+}$). Distil for aldehydes; reflux for carboxylic acids.
- Alcohols to Halogenoalkanes: Reaction with $KX$ and concentrated $H_{2}SO_{4}$.
- Esters: Formed via esterification of alcohols and carboxylic acids (heat with acid catalyst); reversed by acid hydrolysis.
Increasing Carbon Chain Length
To add a carbon atom to a molecule, the Nitrile group ($-CN$) is commonly used:
- Step: Heat an alkyl halide with ethanolic $KCN$.
- Reaction: $RBr + KCN \rightarrow RCN + KBr$
- Further Reactivity: The nitrile ($RCN$) can be hydrolyzed to a carboxylic acid or reduced to an amine using $LiAlH_{4}$.
Aromatic Synthesis
Friedel-Crafts Reaction: Adds an alkyl or acyl side-chain to a benzene ring using an $AlCl_{3}$ catalyst.
- Example: $C_{6}H_{6} + CH_{3}CH_{2}Cl \xrightarrow{AlCl_{3}} C_{6}H_{5}CH_{2}CH_{3} + HCl$
Concept Assessment & Solutions
Exercise 21.1, Question 1a
Route: $C_{2}H_{5}COOC_{3}H_{7} \xrightarrow{NaOH} A \xrightarrow{HCl} B \xrightarrow{PCl_{5}} C \xrightarrow{NH_{3}} D$
| Product | Chemical Name | Formula |
|---|---|---|
| A | Sodium Propanoate | $C_{2}H_{5}COONa$ |
| B | Propanoic Acid | $C_{2}H_{5}COOH$ |
| C | Propanoyl Chloride | $C_{2}H_{5}COCl$ |
| D | Propanamide | $C_{2}H_{5}CONH_{2}$ |
Exercise 21.1, Question 1b
i. Devise a route from Benzene to Benzenediazonium Chloride:
- Nitration: Benzene + conc. $HNO_{3}$ / conc. $H_{2}SO_{4}$ $\rightarrow$ Nitrobenzene.
- Reduction: Nitrobenzene + $Sn$ / conc. $HCl$ (then $NaOH$) $\rightarrow$ Phenylamine (Aniline).
- Diazotization: Phenylamine + $NaNO_{2}$ / $HCl$ (below $10$°C) $\rightarrow$ Benzenediazonium Chloride.
ii. Conversion to an orange dye:
React the benzenediazonium chloride with an alkaline solution of a phenol (coupling reaction) at low temperature.
Example: Methane to Ethanoic Acid
This multi-step route demonstrates adding a carbon atom via a nitrile intermediate.
- Step 1: $CH_{4} + Cl_{2} \xrightarrow{UV} CH_{3}Cl + HCl$ (Free Radical Substitution)
- Step 2: $CH_{3}Cl + KCN \rightarrow CH_{3}CN + KCl$ (Nucleophilic Substitution)
- Step 3: $CH_{3}CN + 2H_{2}O \xrightarrow{HCl} CH_{3}COOH + NH_{3}$ (Hydrolysis)
Notes: Identification of Functional Groups
1. Unsaturated Hydrocarbons
- Bromine Water Test: Used for alkenes ($C=C$) and alkynes ($C \equiv C$).
- Observation: The reddish-brown color of bromine fades away (decolorization).
2. Haloalkanes (Silver Nitrate Test)
- Reagent: Aqueous silver nitrate in ethanol.
- Observations:
- Chloroalkane: White precipitate of $AgCl$.
- Bromoalkane: Cream precipitate of $AgBr$.
- Iodoalkane: Yellow precipitate of $AgI$.
3. Carbonyl Compounds (Aldehydes & Ketones)
- 2,4-DNPH Test: Identifies the presence of a carbonyl group.
- Observation: Yellow/orange precipitate of dinitro hydrazone.
- Iodoform Test: Specific for methyl ketones ($CH_{3}COR$) and specific alcohols.
- Positive for: Ethanol, secondary alcohols with $OH$ on carbon 2, ethanal, and methyl ketones.
- Observation: Yellow precipitate of triiodomethane ($CHI_{3}$).
- Distinguishing Aldehydes from Ketones:
- Fehling’s Test: Aldehydes produce a red precipitate of $Cu_{2}O$. Ketones do not react.
- Tollen’s Test (Silver Mirror): Aldehydes produce a silver mirror of $Ag$. Ketones do not react.
4. Alcohols (Potassium Dichromate Test)
- Primary Alcohol: Orange solution turns green; oxidizes to aldehyde ($RCHO$) then carboxylic acid ($RCOOH$).
- Secondary Alcohol: Orange solution turns green; oxidizes to ketone ($RCOR$).
- Tertiary Alcohol: No color change (resists oxidation).
5. Carboxylic Acids
- Sodium Carbonate Test: $Na_{2}CO_{3}$ is added to the acid.
- Observation: Effervescence due to the release of $CO_{2}$ gas.
Relevant Questions & Answers
Q1: How can you distinguish between Propanal and Propanone?
A: Use Tollen’s reagent. Propanal (an aldehyde) will form a silver mirror on the walls of the test tube, while Propanone (a ketone) will show no reaction. Alternatively, Fehling’s solution can be used, where Propanal forms a red precipitate of $Cu_{2}O$.
Q2: What is the ionic equation for the reaction of ethanal with Fehling's solution?
A: $$CH_{3}CHO + 2Cu^{2+} + 5OH^{-} \rightarrow Cu_{2}O + CH_{3}COO^{-} + 3H_{2}O$$
Q3: Why do tertiary alcohols not change the color of acidified potassium dichromate?
A: Tertiary alcohols resist oxidation because the carbon atom holding the $-OH$ group does not have a hydrogen atom attached to it, which is necessary for the oxidation process to occur under these conditions.
Q4: A compound gives a yellow precipitate with 2,4-DNPH but does not react with Fehling's solution. What is the functional group?
A: The yellow precipitate with 2,4-DNPH indicates a carbonyl group (aldehyde or ketone). Since it does not react with Fehling's solution, it must be a ketone.
Q5: What observation confirms the presence of a carboxylic acid when reacting with sodium carbonate?
A: The observation of brisk effervescence (bubbles), which is caused by the evolution of Carbon Dioxide ($CO_{2}$) gas. The reaction is: $$2CH_{3}COOH + Na_{2}CO_{3} \rightarrow 2CH_{3}COONa + H_{2}O + CO_{2}$$
21.1.4 Types of Reaction Given by Functional Groups
1. Reactions of Alkanes
Alkanes primarily undergo Free Radical Substitution. In this process, a hydrogen atom of the alkane is replaced by a free radical (such as $Cl^{\bullet}$).
- General Reaction: $Alkane + Cl_2 \rightarrow Chloroalkane + HCl$
- Condition: Diffused sunlight.
2. Reactions of Unsaturated Hydrocarbons (Alkenes/Alkynes)
Unsaturated hydrocarbons undergo two main types of reactions:
i. Electrophilic Addition Reactions
A $\pi$ bond in the carbon-carbon double bond breaks, and new single bonds form on each of the two carbon atoms. A single product is always formed.
- Reaction with $HBr$, $H_2O$, or $HOX$ (where $X$ is a halogen).
- Reaction with $H_2$ or $Br_2$.
ii. Oxidation Reaction
- Hydroxylation: Alkenes react with cold dilute alkaline $KMnO_4$ to produce diols (e.g., propene $\rightarrow$ 1,2-propanediol).
- Ozonolysis:
- Alkenes: The $C=C$ bond breaks to form aldehydes and ketones.
- Alkynes: Ozonolysis yields carboxylic acids.
3. Reactions of Alkyl Halides ($R-X$)
Alkyl halides undergo two primary types of reactions:
- Nucleophilic Substitution Reaction: The halogen group is displaced by nucleophiles like $OH^-$, $NH_3$, or $CN^-$.
General Equation: $R-X + Nu^- \rightarrow R-Nu + X^-$ - Elimination Reactions: Hydrogen and halogen are removed to form an alkene.
Reagent: Ethanolic $NaOH$ / $KOH$ with heat ($\Delta$).
4. Reactions of Carbonyl Compounds
- Nucleophilic Addition: Forms products such as hydroxyl nitriles.
- Oxidation: Aldehydes oxidize to carboxylic acids; ketones resist oxidation.
- Reduction: Aldehydes reduce to primary alcohols; ketones reduce to secondary alcohols.
5. Reactions of Carboxylic Acids
- Acid-Base Reactions: React with bases, metals, and metal carbonates.
- Esterification: Reaction with alcohol to form an ester.
Review Questions & Answers
Q1: What is the necessary condition for the chlorination of alkanes?
A: The reaction requires diffused sunlight to initiate the free radical substitution mechanism.
Q2: How can you distinguish between the ozonolysis of an alkene and an alkyne based on the products?
A: Ozonolysis of an alkene produces carbonyl compounds (aldehydes and/or ketones), whereas ozonolysis of an alkyne yields carboxylic acids.
Q3: What determines if an alkyl halide undergoes substitution or elimination when reacting with a base?
A: While both are possible, the use of ethanolic $NaOH/KOH$ and heat specifically promotes elimination to form an alkene.
Q4: What are the reduction products of aldehydes and ketones respectively?
A: Aldehydes are reduced to primary alcohols, while ketones are reduced to secondary alcohols.
Q5: Describe the visual change when propene reacts with cold dilute alkaline $KMnO_4$.
A: This is an oxidation reaction that forms 1,2-propanediol. (Note: In a lab setting, this is often used as a test for unsaturation as the purple $KMnO_4$ decolors and a brown precipitate forms).
1. Oxidising Agents
Oxidising agents are used to increase the oxygen content or decrease the hydrogen content of a molecule.
- Reagents: Acidified potassium dichromate ($K_{2}Cr_{2}O_{7}/H^{+}$) or acidified potassium permanganate ($KMnO_{4}/H^{+}$).
- Primary Alcohols: Oxidise to aldehydes and then further to carboxylic acids.
- Secondary Alcohols: Oxidise to ketones.
2. Reducing Agents
Reducing agents add hydrogen or remove oxygen from a functional group.
| Reagent | Function |
|---|---|
| $NaBH_{4}$ | Reduces a carbonyl group (aldehydes/ketones) to an alcohol. |
| $LiAlH_{4}$ | A stronger reducing agent; reduces carboxylic acids to alcohols. |
| $H_{2}$ and Nickel catalyst | Reduces $C=C$ double bonds (hydrogenation). |
| $Sn/HCl$ | Reduces nitro groups ($-NO_{2}$) to amines ($-NH_{2}$). |
3. Dehydrating Agents
These agents are used to remove water ($H_{2}O$) from a molecule.
- Common reagents: $Al_{2}O_{3}$ and Concentrated $H_{2}SO_{4}$.
Concept Assessment & Practice
Exercise 21.3 - Question 1 Analysis
Scenario: Compound $X$ ($C_{4}H_{8}$) reacts with $H_{2}/Ni$ to form methylpropane.
(a) Suggest a synthesis of the starting material i.e. $C_{4}H_{8}$.
Answer: Since the hydrogenated product is methylpropane, $X$ must be methylpropene (isobutylene). It can be synthesized via the dehydration of 2-methylpropan-1-ol or 2-methylpropan-2-ol using concentrated $H_{2}SO_{4}$ at $170^{\circ}C$.
(b) What is the formula of compound Z?
Answer:
$X (C_{4}H_{8}) + Br_{2} \rightarrow Y (C_{4}H_{8}Br_{2})$.
$Y$ reacts with $aq. NaOH/heat$ (substitution of halides for hydroxyl groups).
Therefore, $Z$ is $C_{4}H_{10}O_{2}$ (specifically 2-methylpropane-1,2-diol).
Question 2: Prepare ethanal and propanone starting from 2-methyl but-2-ene.
Method: This requires Ozonolysis (or vigorous oxidation with $KMnO_{4}$). The double bond in $CH_{3}-C(CH_{3})=CH-CH_{3}$ cleaves. The $CH-CH_{3}$ fragment becomes ethanal ($CH_{3}CHO$) and the $C(CH_{3})_{2}$ fragment becomes propanone ($CH_{3}COCH_{3}$).
Question 3: Devise the best synthetic route to prepare 2-hydroxy propandioic acid starting from ethanal.
- Step 1: React ethanal ($CH_{3}CHO$) with $HCN$ to form a cyanohydrin: $CH_{3}CH(OH)CN$.
- Step 2: Acid hydrolysis of the nitrile group to form 2-hydroxypropanoic acid (lactic acid).
- Note: To reach a propandioic (dicarboxylic) structure, further substitution or oxidation steps would be required on the terminal methyl group.