Hydrocarbons Notes, Formulas, Traps & PYQs | ExamiFly

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Hydrocarbons Notes, Formulas, Traps & PYQs | ExamiFly | Examifly

Key Formulas & High-Yield Equations

Wurtz Reaction

$2R-\text{X} + 2\text{Na} \xrightarrow{\text{dry ether}} R-R + 2\text{NaX}$

Strictly valid for synthesizing symmetrical alkanes with an even number of carbon atoms (

R-R

). Unsuitable for unsymmetrical alkanes due to inseparable mixtures of

R-R

R'-R'

, and

R-R'

Markovnikov's Addition Rule

$R-\text{CH}=\text{CH}_2 + \text{H-X} \rightarrow R-\text{CH}(\text{X})-\text{CH}_3$

Applies to unsymmetrical alkenes. The electrophile (

\text{H}^+

) attaches to the carbon with more hydrogen atoms to form the more stable carbocation intermediate. Carbocation rearrangements (

1,2

-hydride/methyl shifts) must always be evaluated.

Anti-Markovnikov Addition (Kharasch Effect)

$R-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{R_2\text{O}_2} R-\text{CH}_2-\text{CH}_2\text{Br}$

Valid strictly for

\text{HBr}

in the presence of organic peroxides (

R_2\text{O}_2

) via a free radical chain mechanism. Fails for

\text{HCl}

(strong

\text{H-Cl}

bond) and

\text{HI}

(iodine radicals rapidly dimerize to

\text{I}_2

Reductive Ozonolysis

$R_1\text{CH}=\text{CH}R_2 \xrightarrow[2.\, \text{Zn}/\text{H}_2\text{O}]{1.\, \text{O}_3} R_1\text{CHO} + R_2\text{CHO}$

Completely cleaves the

\text{C}=\text{C}

double bond to yield carbonyl compounds without further oxidation to carboxylic acids, provided reducing agents like

\text{Zn}

(\text{CH}_3)_2\text{S}

are present.

Kolbe's Electrolytic Synthesis

$2R\text{COONa} + 2\text{H}_2\text{O} \xrightarrow{\text{Electrolysis}} R-R + 2\text{CO}_2 + \text{H}_2 + 2\text{NaOH}$

Anodic oxidation produces alkane (

R-R

) and

\text{CO}_2

; cathodic reduction releases

\text{H}_2

gas and forms

\text{NaOH}

. Methane (

\text{CH}_4

) cannot be prepared by this method.

Electrophilic Aromatic Substitution (EAS)

$\text{Ar-H} + \text{E}^+ \rightarrow \text{Ar-E} + \text{H}^+$

Governed by ring activation/deactivation. Strong electron-donating groups (

-\text{OH}

-\text{NH}_2

-\text{CH}_3

) direct incoming electrophiles to ortho/para positions, while electron-withdrawing groups (

-\text{NO}_2

-\text{CHO}

) are meta-directing.

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2023

An alkene 'A' on reaction with $\text{O}_3$ followed by $\text{Zn}/\text{H}_2\text{O}$ gives propanone and ethanal in an equimolar ratio. What is the IUPAC name of alkene 'A'?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Identify the carbonyl products formed during reductive ozonolysis: propanone, $(\text{CH}_3)_2\text{C}=\text{O}$ , and ethanal, $\text{CH}_3\text{CHO}$ .

Step 2

Reconstruct the original alkene by removing the oxygen atoms from both carbonyl fragments and connecting the carbonyl carbon atoms with a double bond: $(\text{CH}_3)_2\text{C}=\text{CH-CH}_3$ .

Step 3

Number the carbon backbone according to IUPAC nomenclature rules starting from the end closest to the double bond, yielding $\text{2-methylbut-2-ene}$ .

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Question #2JEE Main 2024

The major product formed when $3$ -methylbutan- $2$ -ol is treated with concentrated $\text{H}_2\text{SO}_4$ at $443\text{ K}$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Protonation of the hydroxyl group in $3$ -methylbutan- $2$ -ol followed by loss of $\text{H}_2\text{O}$ forms a secondary carbocation: $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}^+-\text{CH}_3$ ( $2^\circ$ carbocation).

Step 2

A spontaneous $1,2$ -hydride shift occurs from the adjacent tertiary carbon atom to convert the $2^\circ$ carbocation into a significantly more stable tertiary carbocation: $\text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3$ ( $3^\circ$ carbocation).

Step 3

Elimination of a proton ( $\text{H}^+$ ) from the adjacent carbon following Saytzeff's rule yields the most substituted, highly stable alkene: $\text{2-methylbut-2-ene}$ ( $(\text{CH}_3)_2\text{C}=\text{CH-CH}_3$ ).

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Frequently Asked Questions

Why is the staggered conformation of ethane more stable than the eclipsed conformation by

12.5\text{ kJ/mol}

The higher stability of the staggered conformation arises from minimal torsional strain and stabilizing hyperconjugative interactions. In the staggered arrangement, the electron density of a filled $\text{C-H}$ $\sigma$ -bonding orbital overlaps constructively with the adjacent empty $\text{C-H}$ antibonding $\sigma^*$ orbital. In contrast, the eclipsed conformation suffers from maximum electrostatic repulsion between adjacent parallel $\text{C-H}$ electron clouds.

Why do terminal alkynes exhibit acidic character while alkanes and alkenes do not?

Terminal alkynes possess a hydrogen atom attached to an $sp$ -hybridized carbon. An $sp$ -hybridized carbon features $50\%$ $s$ -character, making it significantly more electronegative than $sp^2$ ( $33.3\%$ $s$ ) and $sp^3$ ( $25\%$ $s$ ) carbons. This high electronegativity stabilizes the resulting conjugate base, the alkynide anion ( $\text{R-C}\equiv\text{C}^-$ ), enabling quantitative proton abstraction by strong bases such as $\text{NaNH}_2$ or liquid $\text{NH}_3$ .

How do I rapidly determine whether a substituent on a benzene ring is ortho/para or meta directing?

Inspect the atom directly bound to the aromatic ring (the key atom). If this atom possesses lone pairs (e.g., $-\text{OH}$ , $-\text{NH}_2$ , $-\text{Cl}$ ) or alkyl groups capable of hyperconjugation ( $-\text{CH}_3$ ), it donates electron density into the $\pi$ -system, activating ortho and para positions. If the key atom is bonded to a more electronegative atom via a multiple bond (e.g., $-\text{NO}_2$ , $-\text{CHO}$ , $-\text{COOH}$ ), it withdraws $\pi$ -electron density, deactivating the ring and leaving the meta position relatively less electron-deficient.

Mastering Hydrocarbons: Formulas, Traps & Exam Strategy | ExamiFly

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