Hydrocarbons Class 11 Notes for IIT JEE and NEET: Alkanes, Alkenes, Alkynes, Benzene, Mechanisms and Name Reactions

Hydrocarbons is Chapter 9 of the current NCERT Class 11 Chemistry book (Chapter 13 in older books). It follows Organic Chemistry: Some Basic Principles and Techniques, whose ideas (electronic effects, carbocation stability and isomerism) are used in every reaction of this chapter. The chapter is part of the organic chemistry syllabus of JEE Main, JEE Advanced and NEET, and its reaction types come back in the Class 12 chapters on haloalkanes, alcohols and carbonyl compounds. Read eSaral’s Organic Chemistry: Some Basic Principles and Techniques notes first if the electronic effects are not clear.

Hydrocarbons Class 11 Notes for IIT JEE and NEET

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Also read, Class 11 Chemistry Chapter 2 Notes

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

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Hydrocarbons Class 11 Notes: Classification and General Formulas

Hydrocarbons are compounds of carbon and hydrogen only.

Nomenclature and Isomerism in Hydrocarbons Class 11 Notes

IUPAC naming: choose the longest chain that contains the multiple bond, number it from the end that gives the lowest locant to the double or triple bond (then to the substituents), and write the substituents in alphabetical order. Use -ane, -ene and -yne as the suffixes.

Number of chain isomers of alkanes: $\mathrm{C_4H_{10}}$ has 2, $\mathrm{C_5H_{12}}$ has 3, $\mathrm{C_6H_{14}}$ has 5 and $\mathrm{C_7H_{16}}$ has 9.

Alkanes in Hydrocarbons Class 11 Notes

Preparation of Alkanes in Hydrocarbons Class 11 Notes

Conformations of ethane

Rotation about the C–C single bond gives different arrangements (conformations). The staggered form (dihedral angle $60^\circ$) is the most stable, and the eclipsed form (dihedral angle $0^\circ$) is the least stable. The difference is about $12.5\ \mathrm{kJ\,mol^{-1}}$, because of torsional strain. Newman and sawhorse projections are used to draw them.

Free Radical Halogenation in Hydrocarbons Class 11 Notes

$$\mathrm{CH_4 + Cl_2 \xrightarrow{h\nu\ \text{or heat}} CH_3Cl + HCl}$$

• Reactivity of the halogen: $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$ (iodination is not feasible).
• Selectivity of the hydrogen: $3^\circ\,\mathrm{H} > 2^\circ\,\mathrm{H} > 1^\circ\,\mathrm{H}$, because the $3^\circ$ radical is the most stable. For chlorination, the relative rates per hydrogen are about $1^\circ : 2^\circ : 3^\circ = 1 : 3.8 : 5$, and bromination is much more selective (about $1 : 82 : 1600$).

Other reactions

• Combustion: $\mathrm{C_nH_{2n+2} + \tfrac{3n+1}{2}O_2 \rightarrow nCO_2 + (n+1)H_2O}$ (heat is released).
• Pyrolysis (cracking): a higher alkane breaks into smaller alkanes and alkenes at a high temperature.
• Physical properties: $\mathrm{C_1}$ to $\mathrm{C_4}$ are gases, $\mathrm{C_5}$ to $\mathrm{C_{17}}$ are liquids, and the higher ones are solids. The boiling point rises with the chain length, and branching lowers it.

Alkenes in Hydrocarbons Class 11 Notes

Preparation of alkenes

• Dehydrohalogenation of alkyl halides with alcoholic KOH. When there is a choice, the more substituted alkene is the major product (Saytzeff’s rule).
• Dehydration of alcohols with conc. $\mathrm{H_2SO_4}$ (heat), or with $\mathrm{Al_2O_3}$.
• Partial reduction of alkynes (see below).

Geometrical isomerism

Rotation about the C=C bond is restricted, so alkenes such as but-2-ene exist as cis and trans isomers. The trans isomer is usually more stable and has a lower heat of hydrogenation. The more substituted an alkene is, the more stable it is (hyperconjugation).

Electrophilic Addition Reactions of Alkenes in Hydrocarbons Class 11 Notes

Markovnikov’s rule: in the addition of HX to an unsymmetrical alkene, the hydrogen goes to the carbon that already has more hydrogens. The mechanism explains why: the $\mathrm{H^+}$ adds first, and it forms the more stable carbocation, and then $\mathrm{X^-}$ attacks it.

$$\mathrm{CH_3CH{=}CH_2 + H^+ \rightarrow CH_3\overset{+}{C}HCH_3 \xrightarrow{Br^-} CH_3CHBrCH_3}$$

A carbocation can rearrange (a hydride or methyl shift) to a more stable one before it is attacked, which changes the product.

Peroxide effect (Kharasch effect): in the presence of peroxides, HBr adds by a free radical chain mechanism, and the bromine radical adds to the less substituted carbon, which gives the more stable radical. It works only for HBr, and not for HCl or HI.

Ozonolysis and oxidation

$$\text{Alkene} \xrightarrow{O_3} \text{ozonide} \xrightarrow{Zn/H_2O} \text{aldehydes and/or ketones}$$

Ozonolysis locates the double bond: each $\mathrm{C{=}}$ carbon becomes a $\mathrm{C{=}O}$ carbon.

Polymerisation: $\mathrm{nCH_2{=}CH_2 \rightarrow -(CH_2{-}CH_2)_n-}$ (polythene).

Alkynes in Hydrocarbons Class 11 Notes

Preparation of alkynes

• From calcium carbide: $\mathrm{CaC_2 + 2H_2O \rightarrow HC{\equiv}CH + Ca(OH)_2}$.
• Dehydrohalogenation of vicinal dihalides with alcoholic KOH (and then $\mathrm{NaNH_2}$).

Acidity of Terminal Alkynes in Hydrocarbons Class 11 Notes

A hydrogen on an $sp$ carbon is acidic, because the $sp$ carbon is the most electronegative. The order of acidity is $\mathrm{HC{\equiv}CH > H_2C{=}CH_2 > H_3C{-}CH_3}$ ($\mathrm{p}K_a$ about 25, 44 and 50). Terminal alkynes react with a strong base and with some metal salts:

$$\mathrm{RC{\equiv}CH + NaNH_2 \rightarrow RC{\equiv}C^-Na^+ + NH_3}$$ $$\mathrm{RC{\equiv}CH + [Ag(NH_3)_2]^+ \rightarrow RC{\equiv}CAg\downarrow\ (\text{white})}$$ $$\mathrm{RC{\equiv}CH + [Cu(NH_3)_2]^+ \rightarrow RC{\equiv}CCu\downarrow\ (\text{red})}$$

Alkynes with the triple bond inside the chain (non-terminal) do not give these precipitates. The sodium alkynide can react with an alkyl halide to lengthen the carbon chain.

Addition reactions of alkynes

Note: “Birch reduction” is the name of a different reaction, the reduction of benzene to 1,4-cyclohexadiene with Na in liquid $\mathrm{NH_3}$ and an alcohol. The reduction of an alkyne to a trans-alkene is called a dissolving-metal reduction.

Aromatic Hydrocarbons in Hydrocarbons Class 11 Notes

Structure of benzene and aromaticity

Benzene is a planar hexagon in which all the C–C bonds have the same length (about 139 pm). It is a resonance hybrid of the two Kekulé structures, and its resonance energy is about $150\ \mathrm{kJ\,mol^{-1}}$. A compound is aromatic if it is cyclic, planar, has a continuous cloud of $\pi$ electrons, and has $(4n + 2)$ $\pi$ electrons ($n = 0, 1, 2, \ldots$).

Electrophilic Aromatic Substitution (EAS) in Hydrocarbons Class 11 Notes

Mechanism: an electrophile $\mathrm{E^+}$ attacks the ring to give a resonance-stabilised carbocation (the arenium ion, or sigma complex). Then a proton is lost, and the aromatic ring is restored.

$$\mathrm{HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-}$$

• Friedel–Crafts alkylation can give a rearranged product (carbocation shift) and more than one alkyl group. Acylation does not rearrange, and it stops after one acyl group.
• Both Friedel–Crafts reactions fail on strongly deactivated rings (for example nitrobenzene), and on rings with $\mathrm{-NH_2}$ (the amine binds to $\mathrm{AlCl_3}$).

Directing Effects of Substituents in Hydrocarbons Class 11 Notes

Order of reactivity in nitration: toluene > benzene > chlorobenzene > nitrobenzene.

Other reactions of benzene and toluene

• Addition (hard): hydrogenation with Ni at 453–473 K gives cyclohexane, and chlorination in sunlight gives hexachlorocyclohexane (BHC). Birch reduction (Na, liquid $\mathrm{NH_3}$, alcohol) gives 1,4-cyclohexadiene.
• Toluene: $\mathrm{Cl_2/FeCl_3}$ gives ring substitution (ortho and para), and $\mathrm{Cl_2}$ with light gives side-chain substitution (benzyl chloride). Hot $\mathrm{KMnO_4}$ oxidises the side chain to benzoic acid.
• Combustion: benzene burns with a sooty flame because of its high carbon content. Polynuclear aromatic hydrocarbons can be carcinogenic.

Important Trends in Hydrocarbons Class 11 Notes

Summary of Reactions in Hydrocarbons Class 11 Notes

IIT JEE and NEET Tips for Hydrocarbons Class 11 Notes

• Count the products of halogenation: group the hydrogens that are equivalent, and write one product for each group. Add the stereoisomers if a product has a chiral centre.
• Check for a carbocation shift whenever an alkene (or an alcohol) reacts through a carbocation.
• Identify the reagent first: Lindlar (cis), Na and liquid $\mathrm{NH_3}$ (trans), peroxide (anti-Markovnikov), cold $\mathrm{KMnO_4}$ (diol), hot $\mathrm{KMnO_4}$ (cleavage).
• For ozonolysis problems, join the carbonyl carbons with a double bond to get back the alkene.
• In EAS problems, find the directing group first. Remember that halogens are deactivating and ortho, para directing.
• Test questions: $\mathrm{Br_2/CCl_4}$ and Baeyer’s reagent show unsaturation, and $\mathrm{AgNO_3/NH_3}$ shows a terminal alkyne.

Solved Examples from Hydrocarbons Class 11 Notes

Example 1 (number of products): How many monochloro products (structural, and with stereoisomers) does 2-methylbutane give? Estimate the percentage of each with the rates $1^\circ : 2^\circ : 3^\circ = 1 : 3.8 : 5$.

There are four types of hydrogen: (a) 6 H on the two $\mathrm{CH_3}$ groups at C-2 (gives 1-chloro-2-methylbutane, chiral), (b) 1 H on C-2 ($3^\circ$, gives 2-chloro-2-methylbutane), (c) 2 H on C-3 ($2^\circ$, gives 2-chloro-3-methylbutane, chiral), and (d) 3 H on C-4 (gives 1-chloro-3-methylbutane). So there are 4 structural isomers, and 6 with stereoisomers (two chiral products, with two enantiomers each). The relative yields are $6\times1 = 6$, $1\times5 = 5$, $2\times3.8 = 7.6$ and $3\times1 = 3$, which are about 28%, 23%, 35% and 14%.

Example 2 (Markovnikov and the peroxide effect): Give the product of propene with HBr, with and without a peroxide.

Without a peroxide, the product is 2-bromopropane ($\mathrm{CH_3CHBrCH_3}$). With a peroxide, it is 1-bromopropane ($\mathrm{CH_3CH_2CH_2Br}$).

Example 3 (carbocation rearrangement): Predict the major product of 3-methylbut-1-ene with HBr.

$\mathrm{H^+}$ adds to $\mathrm{CH_2{=}}$ to give the $2^\circ$ carbocation $\mathrm{(CH_3)_2CH\overset{+}{C}HCH_3}$. A hydride shift gives the $3^\circ$ carbocation $\mathrm{(CH_3)_2\overset{+}{C}CH_2CH_3}$. So the major product is 2-bromo-2-methylbutane.

Example 4 (ozonolysis): An alkene gives $\mathrm{CH_3CHO}$ and $\mathrm{CH_3COCH_3}$ on ozonolysis with $\mathrm{Zn/H_2O}$. Find the alkene.

Join the two carbonyl carbons with a double bond: $\mathrm{CH_3CH{=}C(CH_3)_2}$, which is 2-methylbut-2-ene.

Example 5 (oxidation): Give the products of the hot, acidic $\mathrm{KMnO_4}$ oxidation of 2-methylbut-2-ene.

The $\mathrm{=C(CH_3)_2}$ end gives propanone ($\mathrm{CH_3COCH_3}$), and the $\mathrm{=CHCH_3}$ end gives ethanoic acid ($\mathrm{CH_3COOH}$).

Example 6 (stereochemistry): What do you get from but-2-yne with (a) $\mathrm{H_2}$/Lindlar and (b) Na in liquid $\mathrm{NH_3}$?

(a) cis-but-2-ene (syn addition). (b) trans-but-2-ene (anti addition).

Example 7 (hydration of an alkyne): Give the product of the hydration of propyne with $\mathrm{H_2SO_4/HgSO_4}$.

The enol $\mathrm{CH_3C(OH){=}CH_2}$ is formed first, and it changes into propanone, $\mathrm{CH_3COCH_3}$.

Example 8 (a chemical test): How can you distinguish but-1-yne from but-2-yne?

Add $\mathrm{AgNO_3/NH_3}$. But-1-yne (terminal) gives a white precipitate of the silver alkynide. But-2-yne gives no precipitate.

Example 9 (directing effects): Give the major products of (a) the nitration of toluene, and (b) the bromination of nitrobenzene with $\mathrm{Br_2/FeBr_3}$.

(a) $\mathrm{-CH_3}$ is an ortho, para director, so the products are o-nitrotoluene and p-nitrotoluene. (b) $\mathrm{-NO_2}$ is a meta director, so the product is m-bromonitrobenzene.

Example 10 (aromaticity): Classify the cyclopentadienyl anion, the cyclopropenyl cation and cyclobutadiene.

The cyclopentadienyl anion has 6 $\pi$ electrons ($4n+2$, $n=1$), so it is aromatic. The cyclopropenyl cation has 2 $\pi$ electrons ($n=0$), so it is aromatic. Cyclobutadiene has 4 $\pi$ electrons ($4n$), so it is antiaromatic.

Example 11 (isomers): How many isomers does $\mathrm{C_5H_{12}}$ have, and how many alkenes (with stereoisomers) does $\mathrm{C_4H_8}$ have?

$\mathrm{C_5H_{12}}$: pentane, 2-methylbutane and 2,2-dimethylpropane, so 3. $\mathrm{C_4H_8}$ alkenes: but-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene, so 4.

Example 12 (acidity and chain extension): Arrange ethane, ethene and ethyne in the order of acidity, and show how propyne can be converted into but-2-yne.

Acidity: ethyne > ethene > ethane ($sp > sp^2 > sp^3$). $\mathrm{CH_3C{\equiv}CH + NaNH_2 \rightarrow CH_3C{\equiv}C^-Na^+}$, and then $\mathrm{CH_3C{\equiv}C^-Na^+ + CH_3Br \rightarrow CH_3C{\equiv}CCH_3 + NaBr}$.

Common Mistakes in Hydrocarbons Class 11 Notes Problems

How to Study Hydrocarbons Class 11 Notes

1. Read the NCERT chapter first, and download the textbook from eSaral’s NCERT books for Class 11 page. Revise the electronic effects in the GOC notes.
2. Learn the mechanisms (free radical halogenation, electrophilic addition and EAS) before the reactions, and write each reaction with its reagent and product.
3. Practise one problem each on counting products, Markovnikov and peroxide addition, ozonolysis, alkyne reductions, directing effects and aromaticity.
4. Revise with eSaral’s Organic Chemistry Revision Series (videos and mind maps), and see the next topic in eSaral’s Halogen Derivatives notes.
5. Test yourself with eSaral’s JEE Main chapter-wise PYQs and NEET chapter-wise PYQs.