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MHT-CET Chemistry · Basic Principles of Organic Chemistry

Electronic Effects, Hybridisation, Intermediates and General Reactions

Carbon is sp³ with four single bonds, sp² at a double bond and sp at a triple bond; substituents push or pull electrons by induction and resonance, which is what ranks carbocations tertiary > secondary > primary > methyl and sorts groups into donors and withdrawers.

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Why this matters

24 PYQs, 2 HARD. Of the eleven set from 2021 to 2025, four count sp³ or sp² carbons, atoms or moles from a formula (one is an empirical-formula derivation); three ask which group donates or withdraws electrons or which carbocation is least stable; three are named reagents — the Grignard reagent, its reaction with dry ice, the Clemmensen reduction. Three cards.

Concept 1 of 3: Hybridisation of Carbon and Counting From a Formula

Count what a carbon is bonded to: four single bonds → sp³; one double bond → sp²; a triple bond or two doubles → sp. A saturated formula has only sp³ carbons. For molecular formulas, work from percentages to the empirical formula and scale by the molar mass.

Definition

  • sp³: alkanes, any carbon with four single bonds — isopentane has five sp³ carbons and zero sp². HO(CH2)3CH(CH3)CH(CH3)2\text{HO(CH}_2)_3\text{CH(CH}_3)\text{CH(CH}_3)_2: eight sp³ carbons.
  • sp²: alkene and carbonyl carbons, aromatic ring carbons. sp: alkyne carbons, allene's central carbon.
  • Moles of an element in n mol of compound = n × (atoms per molecule): but-2-ene C4H8\text{C}_4\text{H}_8 gives 8n mol H. Burning 612=0.5\tfrac{6}{12} = 0.5 mol C gives 0.5 mol CO₂ = 11.2 dm³ at STP.
  • Empirical formula: divide each mass % by the atomic mass, then by the smallest. C 42.8/12 = 3.57, H 7.2, N 50/14 = 3.57 → 1 : 2 : 1 → CH2N\text{CH}_2\text{N} (28). Molecular formula = (empirical) × (M / 28).

Hybridisation from bonding

4 σ⇒sp3;3 σ+1 π⇒sp2;2 σ+2 π⇒sp4\ \sigma \Rightarrow sp^3;\qquad 3\ \sigma + 1\ \pi \Rightarrow sp^2;\qquad 2\ \sigma + 2\ \pi \Rightarrow sp

Worked example

Count the sp³, sp² and sp carbons in CH3-CH=CH-C≡C-CH3\text{CH}_3\text{-CH=CH-C≡C-CH}_3.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q69Easy

Example 1 · Basic Principles of Organic Chemistry · Electronic Effects, Hybridization, Intermediates and General Reactions

What is the number of moles of sp² hybrid carbon atoms present in nn moles of isopentane?

Hearing 'isopentane' as an alkene

The iso- prefix marks a branch, not a double bond. Isopentane is 2-methylbutane, fully saturated: every carbon sp³.
Drill 4 more on hybridisation of carbon and counting from a formula

Concept 2 of 3: Inductive and Resonance Effects; Carbocation Stability

Alkyl groups push electrons (+I) and their C–H bonds hyperconjugate into an empty p orbital, so the more alkyls on a positive carbon, the steadier it is: 3° > 2° > 1° > CH₃⁺. Groups with a lone pair on the atom attached to a π-system donate by resonance (+R: −NH₂, −NHR, −OH, −OR); groups with a multiple bond to an electronegative atom withdraw (−R, −I: −NO₂, −CN, −COOH, −COOR, −CHO).

Definition

  • +I (electron-releasing): alkyl groups, increasing with size. −I (withdrawing): −NO₂, −CN, −COOH, −F, −Cl, −Br, −OH, −OR (through the σ bond).
  • +R (donating by resonance): −NH₂, −NHR, −NR₂, −OH, −OR, −Cl (lone pair conjugated). −R (withdrawing): −NO₂, −CN, −CHO, −COR, −COOH, −COOR.
  • Attached to a π-bond, −COOH is a clean electron WITHDRAWER (−I and −R); −OH and −OR are net donors; −Cl is −I but +R, a weak net withdrawer.
  • Carbocation stability: (R)3C+>(R)2CH+>RCH2+>CH3+(\text{R})_3\text{C}^+ > (\text{R})_2\text{CH}^+ > \text{RCH}_2^+ > \text{CH}_3^+ — the methyl cation is LEAST stable. Reverse order for carbanions; free radicals follow the carbocation order.

Carbocation order

3∘>2∘>1∘>CH3+(+I and hyperconjugation)3^\circ > 2^\circ > 1^\circ > \text{CH}_3^+ \quad (\text{+I and hyperconjugation})

Worked example

Rank the stability of (CH3)3C+(\text{CH}_3)_3\text{C}^+, CH3CH2+\text{CH}_3\text{CH}_2^+ and CH2=CH-CH2+\text{CH}_2\text{=CH-CH}_2^+, and say which of −OCH₃ and −NO₂ shows +R.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q83Easy

Example 2 · Basic Principles of Organic Chemistry · Electronic Effects, Hybridization, Intermediates and General Reactions

Which from following carbocations is least stable?

Calling −OH a withdrawer because oxygen is electronegative

Through the σ bond it is (−I), but attached to a π-system the lone pair donates by resonance and +R wins: −OH and −OR are net DONORS. −COOH, with no lone pair to give and a C=O to pull, withdraws by both routes.
Drill 2 more on inductive and resonance effects; carbocation stability

Concept 3 of 3: Grignard Reagent, Dry Ice and the Clemmensen Reduction

Three named items recur. The Grignard reagent R–Mg–X is the chapter's carbon nucleophile; with dry ice (solid CO₂) in dry ether and then dilute acid it adds one carbon and gives a carboxylic acid. Clemmensen (Zn–Hg, conc. HCl) and Wolff–Kishner (hydrazine, KOH) both turn C=O into CH₂ — by different reagents.

Definition

  • Grignard reagent: R-Mg-X\text{R-Mg-X} (R alkyl or aryl, X a halogen), made from R–X and Mg in dry ether; destroyed by water.
  • R-MgBr+CO2→R-COOMgBr→dil. HClR-COOH\text{R-MgBr} + \text{CO}_2 \to \text{R-COOMgBr} \xrightarrow{\text{dil. HCl}} \text{R-COOH}: ethylmagnesium bromide gives propanoic acid (one carbon more than R).
  • Clemmensen: R-CHO→Zn-Hg / conc. HClR-CH3\text{R-CHO} \xrightarrow{\text{Zn-Hg / conc. HCl}} \text{R-CH}_3; R-CO-R→R-CH2-R\text{R-CO-R} \to \text{R-CH}_2\text{-R}. Wolff–Kishner: the same change with NH2NH2/KOH\text{NH}_2\text{NH}_2 / \text{KOH}. LiAlH4\text{LiAlH}_4 gives the ALCOHOL, not the alkane.
  • Stephen reduction: R-CN→SnCl2/HClR-CHO\text{R-CN} \xrightarrow{\text{SnCl}_2 / \text{HCl}} \text{R-CHO}.

Grignard with dry ice

R-MgX→ (i) CO2, dry ether; (ii) H3O+ R-COOH\text{R-MgX} \xrightarrow{\ \text{(i) CO}_2\text{, dry ether; (ii) H}_3\text{O}^+\ } \text{R-COOH}

Worked example

Which acid forms from isopropylmagnesium bromide and dry ice, and what does Clemmensen reduction do to propanone?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q94Moderate

Example 3 · Basic Principles of Organic Chemistry · Electronic Effects, Hybridization, Intermediates and General Reactions

Which of the following reactions represents Clemmensen reduction?

Forgetting the extra carbon from CO₂

Ethyl-MgBr does not give ethanoic acid. The CO₂ carbon joins the chain: ethyl becomes propanoic acid. Count R plus one.
Drill 2 more on grignard reagent, dry ice and the clemmensen reduction

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Hybridisation of Carbon and Counting From a Formula

    Hybridisation from bonding

    4 σ⇒sp3;3 σ+1 π⇒sp2;2 σ+2 π⇒sp4\ \sigma \Rightarrow sp^3;\qquad 3\ \sigma + 1\ \pi \Rightarrow sp^2;\qquad 2\ \sigma + 2\ \pi \Rightarrow sp
  • Inductive and Resonance Effects; Carbocation Stability

    Carbocation order

    3∘>2∘>1∘>CH3+(+I and hyperconjugation)3^\circ > 2^\circ > 1^\circ > \text{CH}_3^+ \quad (\text{+I and hyperconjugation})
  • Grignard Reagent, Dry Ice and the Clemmensen Reduction

    Grignard with dry ice

    R-MgX→ (i) CO2, dry ether; (ii) H3O+ R-COOH\text{R-MgX} \xrightarrow{\ \text{(i) CO}_2\text{, dry ether; (ii) H}_3\text{O}^+\ } \text{R-COOH}

Watch out for (3)

Test yourself on Basic Principles of Organic Chemistry

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