Hydrocarbons

Calcium carbide reacts with water to give ethyne:

$\mathrm{CaC_2 + 2H_2O \rightarrow \underset{(A)}{HC\equiv CH} + Ca(OH)_2}$

Ethyne has acidic hydrogen atoms, so with ammoniacal silver nitrate (Tollens' reagent) it gives a white precipitate of silver acetylide:

$\mathrm{HC\equiv CH + 2[Ag(NH_3)_2]^+ \rightarrow \underset{(B)}{AgC\equiv CAg}\downarrow + 2NH_4^+ + 2NH_3}$

(A) = ethyne (acetylene); (B) = silver acetylide (white precipitate).

Principle: Ethyne (acetylene) is prepared in the laboratory by the action of water on calcium carbide.

$\mathrm{CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2\uparrow}$

Apparatus: A flask fitted with a dropping funnel and a delivery tube is used. Lumps of calcium carbide, mixed with a little sand, are placed in the flask. The sand moderates the vigorous reaction. A trough of water with an inverted gas jar is used to collect the gas.

Procedure:

1. Water (or saturated brine, which makes the reaction gentler) is added drop by drop from the dropping funnel onto the calcium carbide.
2. Ethyne is released at once and is collected by downward displacement of water.

Purification: Commercial carbide contains impurities such as calcium phosphide and sulphide, so the gas has $\mathrm{PH_3}$ and $\mathrm{H_2S}$ with it. These are removed by passing the gas through acidified $\mathrm{CuSO_4}$ solution before collection.

Properties noted: Ethyne is a colourless gas, only slightly soluble in water, and burns with a sooty, luminous flame.

(i) From methylmagnesium bromide, by hydrolysis with water:

$\mathrm{CH_3MgBr + H_2O \rightarrow CH_4 + Mg(OH)Br}$

(ii) From sodium acetate, by heating with soda lime (decarboxylation):

$\mathrm{CH_3COONa + NaOH \xrightarrow[CaO,\ \Delta]{} CH_4 + Na_2CO_3}$

(i) Baeyer's test: Ethene decolourises cold, dilute, alkaline $\mathrm{KMnO_4}$ (pink/purple) and forms a diol, showing unsaturation:

$\mathrm{CH_2{=}CH_2 + H_2O + [O] \xrightarrow{KMnO_4} CH_2OH{-}CH_2OH}$

(ii) Wurtz reaction: An alkyl halide reacts with sodium in dry ether to give a higher alkane:

$\mathrm{2CH_3Cl + 2Na \xrightarrow{dry\ ether} CH_3{-}CH_3 + 2NaCl}$

(i) Markovnikov's rule: when an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part goes to the carbon having fewer hydrogen atoms. $\mathrm{CH_3-CH=CH_2 + HBr \rightarrow CH_3-CHBr-CH_3}$ (2-bromopropane is the major product.)

(ii) Wurtz reaction: an alkyl halide reacts with sodium in dry ether to give a higher alkane. $\mathrm{2CH_3Br + 2Na \xrightarrow{dry\ ether} CH_3-CH_3 + 2NaBr}$

Ozonolysis splits the $\mathrm{C=C}$ bond and puts a $\mathrm{C=O}$ on each carbon. Joining the carbonyl carbons of $\mathrm{CH_3CHO}$ and $\mathrm{HCHO}$ with a double bond gives the alkene:

$\mathrm{CH_3CH=CH_2 \xrightarrow{(i)\,O_3,\ (ii)\,Zn/H_2O} CH_3CHO + HCHO}$

(A) is $\mathrm{CH_3-CH=CH_2}$; IUPAC name: propene.

(i) From sodium ethanoate (decarboxylation with soda lime, $\mathrm{NaOH + CaO}$, on heating):

$\mathrm{CH_3COONa + NaOH \xrightarrow{CaO,\ \Delta} CH_4 + Na_2CO_3}$

(ii) From methyl magnesium iodide (hydrolysis with water):

$\mathrm{CH_3MgI + H_2O \rightarrow CH_4 + Mg(OH)I}$

Laboratory preparation of ethene. Ethene is prepared by dehydration of ethanol with excess concentrated $\mathrm{H_2SO_4}$ at about $170^\circ\mathrm{C}$. A mixture of ethanol and conc. $\mathrm{H_2SO_4}$ (1:2 by volume) with some sand or pumice (to prevent frothing) is heated in a flask:

$\mathrm{CH_3CH_2OH \xrightarrow{conc.\,H_2SO_4,\ 170^\circ C} CH_2=CH_2 + H_2O}$

Ethyl hydrogen sulphate is the intermediate: $\mathrm{C_2H_5OH + H_2SO_4 \rightarrow C_2H_5HSO_4 + H_2O}$, followed by $\mathrm{C_2H_5HSO_4 \xrightarrow{170^\circ C} C_2H_4 + H_2SO_4}$. The gas is passed through NaOH solution to remove the $\mathrm{CO_2}$ and $\mathrm{SO_2}$ formed by side reactions, and collected over water. Temperature is kept at $170^\circ\mathrm{C}$; at $140^\circ\mathrm{C}$ diethyl ether would form.

Test to distinguish ethene from ethane. Pass each gas into bromine water (or shake with alkaline $\mathrm{KMnO_4}$). Ethene, being unsaturated, decolourises the orange-brown bromine water (and the purple $\mathrm{KMnO_4}$ turns colourless with a brown $\mathrm{MnO_2}$ precipitate); ethane, being saturated, does not.

$\mathrm{CH_2=CH_2 + Br_2 \rightarrow CH_2Br-CH_2Br}$

Compound D ($\text{C}_6\text{H}_{14}$) is formed by the Wurtz reaction of an alkyl bromide C, so C contains 3 carbon atoms. C is formed by addition of HBr to the unsaturated hydrocarbon B, so B is propene. Following Markovnikov's rule, the major product is 2-bromopropane.

B to C (addition of HBr):

$\text{CH}_3\text{-CH=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{-CHBr-CH}_3$

C to D (Wurtz reaction, Na in dry ether):

$2\text{CH}_3\text{-CHBr-CH}_3 + 2\text{Na} \xrightarrow{\text{dry ether}} (\text{CH}_3)_2\text{CH-CH}(\text{CH}_3)_2 + 2\text{NaBr}$

So B = propene, C = 2-bromopropane, D = 2,3-dimethylbutane.

From the scheme (propene reacts with HBr to give C, and C reacts with sodium in dry ether to give D, $\text{C}_6\text{H}_{14}$):

• C: $\text{CH}_3\text{-CHBr-CH}_3$, IUPAC name 2-bromopropane
• D: $(\text{CH}_3)_2\text{CH-CH}(\text{CH}_3)_2$, IUPAC name 2,3-dimethylbutane

Compound C is 2-bromopropane, $\text{CH}_3\text{-CHBr-CH}_3$. A positional isomer has the same carbon skeleton and molecular formula ($\text{C}_3\text{H}_7\text{Br}$) but the Br atom at a different position. It is 1-bromopropane:

$\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br}$

Y and Z are functional isomers of formula $\text{C}_3\text{H}_6\text{O}$, so one is the aldehyde propanal and the other the ketone propanone. The alkene that gives these two fragments on cleavage is 2-methylpent-2-ene, $(\text{CH}_3)_2\text{C=CHCH}_2\text{CH}_3$ (X).

Step 1: ozonide formation (X reacts with ozone):

$(\text{CH}_3)_2\text{C=CHCH}_2\text{CH}_3 + \text{O}_3 \rightarrow \text{ozonide}$

Step 2: hydrolysis of the ozonide with Zn and water:

$\text{ozonide} + \text{H}_2\text{O} \xrightarrow{\text{Zn}} (\text{CH}_3)_2\text{C=O} + \text{CH}_3\text{CH}_2\text{CHO} + \text{H}_2\text{O}_2$

Zn removes the $\text{H}_2\text{O}_2$ formed, which prevents oxidation of the aldehyde.

• Y = propanal, $\text{CH}_3\text{CH}_2\text{CHO}$ (aldehyde)
• Z = propanone, $\text{CH}_3\text{COCH}_3$ (ketone)

Structural formulae of Y and Z

The question refers to the pair of functional isomers with molecular formula $\mathrm{C_{3}H_{6}O}$:

Y — Propanal (aldehyde):

        H   H   O
        |   |   ||
   H —  C — C — C — H
        |   |   
        H   H   

$\mathrm{CH_{3}-CH_{2}-CHO}$

Z — Propanone (ketone):

        H   O   H
        |   ||  |
   H —  C — C — C — H
        |       |
        H       H

$\mathrm{CH_{3}-CO-CH_{3}}$

Why they are called functional isomers

Functional isomers are compounds that have the same molecular formula but different functional groups, and hence different chemical properties.

• Both Y and Z have the molecular formula $\mathrm{C_{3}H_{6}O}$.
• Y contains the aldehyde group ($\mathrm{-CHO}$), while Z contains the ketone group ($\mathrm{-CO-}$).
• Since the difference lies in the functional group itself (not merely in the position of the group or the chain), they are called functional isomers.

X is converted into a saturated compound (X undergoes hydrogenation/addition of hydrogen).

Hydrogen gas in the presence of nickel catalyst adds across the double or triple bond of X (an unsaturated compound), converting it into the corresponding saturated alkane. For example, ethene gives ethane:

$\mathrm{CH_{2}=CH_{2} + H_{2} \rightarrow [Ni] CH_{3}-CH_{3}}$

To locate the position of the double bond in an alkene.

Ozonolysis cleaves the C=C bond completely, giving carbonyl compounds (aldehydes/ketones). Since the products retain the positions of the original carbons, identifying them tells us where the double bond was. For example, an alkene that gives only one aldehyde on ozonolysis must be symmetrical, while two different products indicate the double bond lies between those two carbon groups.

Test: Add bromine water (or dilute alkaline KMnO₄) to compound X.

If X is unsaturated, the reddish-brown colour of bromine water is decolourised (or the pink colour of KMnO₄ is discharged), because the π-bond adds bromine across the double/triple bond. A saturated compound gives no such change.