Chemistry MDCAT MCQs

Organic, inorganic & physical chemistry MCQs

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Showing 305 MCQs in Chemistry
Chemistry
A. Tertiary > Secondary > Primary
B. Primary > Secondary > Tertiary
C. Secondary > Primary > Tertiary
D. All equal
Answer: E2 reactivity increases with alkyl substitution because more substituted alkenes (Zaitsev product) are more stable, and tertiary substrates have more beta-hydrogens.
Chemistry
A. IR spectroscopy only
B. 1H NMR spectroscopy
C. UV-Vis spectroscopy
D. Flame test
Answer: 1H NMR gives different splitting patterns: 1-propanol shows a triplet for the CH3, while 2-propanol shows a distinctive septet for the CH and a doublet for two equivalent CH3 groups.
Chemistry
A. mu = sqrt(n(n+2)) BM
B. mu = n(n+2) BM
C. mu = sqrt(n+2) BM
D. mu = 2sqrt(n) BM
Answer: The spin-only formula gives magnetic moment mu = sqrt(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons.
Chemistry
A. A ligand that forms one bond to the metal
B. A ligand that forms two bonds to the metal
C. A ligand that forms six bonds to the metal
D. A ligand that carries a -2 charge
Answer: A bidentate ligand has two donor atoms that simultaneously coordinate to the metal center, forming a chelate ring (e.g., ethylenediamine, oxalate).
Chemistry
A. lambda = h/mv
B. lambda = h x mv
C. lambda = mv/h
D. lambda = h/2mv
Answer: The De Broglie equation states lambda = h/p = h/(mv), where h is Planck\'s constant, m is mass, and v is velocity, predicting wave-particle duality.
Chemistry
A. Enthalpy when 1 mole burns in excess oxygen at standard conditions
B. Enthalpy when all moles burn completely
C. Enthalpy of formation of combustion products
D. Enthalpy per gram of fuel burned
Answer: Standard enthalpy of combustion is the enthalpy change when 1 mole of a substance undergoes complete combustion in excess O2 under standard conditions (298 K, 1 atm).
Chemistry
A. All electrons are paired
B. Zero pairing energy because all t2g and eg orbitals are singly occupied
C. Maximum splitting
D. Delta_o equals the pairing energy
Answer: In high-spin d5, each of the five d-orbitals gets one electron (t2g3 eg2) with no pairing, so no pairing energy is required.
Chemistry
A. 3-Bromo-1-butene
B. 2-Bromo-3-butene
C. 1-Bromo-2-butene
D. 3-Bromo-2-butene
Answer: Numbering gives the double bond the lowest locant: C1=C2-C3(Br)-C4. Bromine is on C3, so the name is 3-bromobut-1-ene.
Chemistry
A. n -> pi*
B. pi -> pi*
C. sigma -> sigma*
D. n -> sigma*
Answer: Sigma bonds are the strongest, so sigma -> sigma* transitions require the highest energy (shortest wavelength), typically in the far UV region.
Chemistry
A. t1/2 = 0.693/k
B. t1/2 = [A]0/2k
C. t1/2 = 1/(k[A]0)
D. t1/2 = 2.303/k
Answer: For first-order reactions, t1/2 = ln(2)/k = 0.693/k, which is independent of the initial concentration.
Chemistry
A. Nucleophilic attack on the carbocation
B. Loss of the leaving group to form the carbocation
C. Deprotonation
D. Formation of the transition state
Answer: The slowest step in SN1 is ionization: the departure of the leaving group to form a carbocation intermediate. This step has the highest activation energy.
Chemistry
A. Rate constant decreases exponentially
B. Rate constant increases exponentially
C. Rate constant remains the same
D. Rate constant increases linearly
Answer: As T increases, Ea/RT decreases, making e^(-Ea/RT) larger, so k increases exponentially with temperature.
Chemistry
A. Polyethylene
B. Polypropylene
C. Polytetrafluoroethylene (PTFE/Teflon)
D. PVC
Answer: Tetrafluoroethylene (CF2=CF2) undergoes addition polymerization to form PTFE (Teflon), where the double bond opens to form long chains.
Chemistry
A. Kp = Kc
B. Kp = Kc(RT)^2
C. Kp = Kc(RT)^(-2)
D. Kp = Kc(RT)^4
Answer: Kp = Kc(RT)^(Delta n). Delta n = 2 - (1+3) = -2, so Kp = Kc(RT)^(-2) = Kc/(RT)^2.
Chemistry
A. Cyclohexanol
B. Adipic acid (hexanedioic acid)
C. Cyclohexanone
D. No reaction
Answer: Hot, concentrated KMnO4 is a strong oxidizing agent that cleaves the C=C double bond in cyclohexene, oxidizing it completely to adipic acid (HOOC-CH2-CH2-CH2-CH2-COOH).
Chemistry
A. Planar (trigonal)
B. Trigonal bipyramidal (pentacoordinate carbon)
C. Tetrahedral
D. Linear
Answer: The SN2 transition state has a pentacoordinate carbon with the nucleophile and leaving group on opposite sides, forming a trigonal bipyramidal arrangement.
Chemistry
A. m = (M x I x t) / (n x F)
B. m = (n x F) / (M x I x t)
C. m = (M x F) / (n x I x t)
D. m = (I x t) / (M x n x F)
Answer: Faraday\'s first law: mass deposited (m) = (M x I x t)/(n x F), where M is molar mass, I is current, t is time, n is electrons transferred, and F = 96485 C/mol.
Chemistry
A. Unimolecular
B. Bimolecular
C. Termolecular
D. Quadrimolecular
Answer: Molecularity is the number of molecules that come together in an elementary step. Three simultaneous collisions give termolecular (trimolecular) molecularity.
Chemistry
A. Less than 7
B. Equal to 7
C. Greater than 7
D. Depends on the indicator
Answer: The salt formed (e.g., NaCl) is neutral, so the pH at equivalence is exactly 7 for strong acid-strong base titrations.
Chemistry
A. Bromination of benzene
B. Addition of HBr to propene
C. Chlorination of methane
D. Hydrolysis of an ester
Answer: HBr adds across the C=C double bond in propene. The electrophilic H+ attacks the pi bond first, followed by nucleophilic Br- attack.