Part of OC-02 — Hydrocarbons: Alkanes, Alkenes & Alkynes

Hydrocarbons: NEET PYQ Patterns and Exam Strategy

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Top 5 NEET PYQ Patterns from OC-02

PYQ Pattern 1 (Highest Frequency): Anti-Markovnikov HX Specificity

  • Typical question: "Anti-Markovnikov addition of HX to alkene with peroxide is possible with: (a) HCl (b) HBr (c) HI (d) Both HBr and HCl"
  • Correct answer: (b) HBr only
  • Frequency: Appears almost every year in some form
  • Key phrase: "Peroxide = anti; only HBr — HCl too lazy, HI too crazy"

PYQ Pattern 2: Lindlar vs Na/NH3NH_{3} Geometry

  • Typical question: "But-2-yne on treatment with Lindlar's catalyst gives: (a) cis-but-2-ene (b) trans-but-2-ene (c) butane (d) but-1-ene"
  • Correct answer: (a) cis-but-2-ene
  • Trap: Students swap the answers for the two reagents

PYQ Pattern 3: Ozonolysis Product Prediction

  • Typical question: Given an alkene (or SMILES), predict carbonyl products after O3O_{3}/Zn/H2OH_{2}O
  • Method: Cleave C=C; add =O to each former C=C carbon; –CH= → CHO; –C(R)= → CO
  • Example: 2-methylpropene → acetone + formaldehyde

PYQ Pattern 4: Butane Conformational Stability Order

  • Typical question: Arrange anti, gauche, eclipsed, fully eclipsed in decreasing stability
  • Correct: Anti > Gauche > Eclipsed > Fully eclipsed
  • Trap: Placing gauche as least stable

PYQ Pattern 5: Alkyne Acidity and s-Character

  • Typical question: Assertion-Reason about why terminal alkynes are more acidic
  • Correct answer: 50% s-character → more electronegative C → more acidic H
  • Second question type: "Which reacts with NaNH2NaNH_{2}: alkyne/alkene/alkane?" → Terminal alkyne only

Exam Strategy for OC-02

  1. Always check for the word "peroxide" before predicting HBr addition product.
  2. Write out the Newman projection mentally for butane conformation questions.
  3. For ozonolysis: work backwards — identify each C of C=C and what groups are on it.
  4. For acidity: higher s-character → more acidic → can be deprotonated by NaNH2NaNH_{2}.
  5. For alkyne reduction: Lindlar's = L = cis; Na/NH3NH_{3} = N = trans.

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