True or false: Nucleophilicity is the thermodynamic tendency of a species to bind H⁺.
01 Mechanistic Foundations and Reactive Intermediates Online Quiz Questions
Use this free practice quiz with 20 questions to review 01 Mechanistic Foundations and Reactive Intermediates, test your knowledge, and prepare for your next test or exam.
True or false: A transition state is an isolable species formed in one reaction step and consumed in a later step.
- A
True
- B
False
Which statement correctly describes a single-headed (fishhook) curved arrow?
- A
Movement of an electron pair in an ordinary polar step
- B
Movement of one electron in a radical step
- C
Movement of an atom from one reactant to another
A species is produced in one step of a mechanism and consumed in a later step. What is it called?
- A
Reaction intermediate
- B
Transition state
- C
Reactant
Which sequence orders simple alkyl carbocations from most stable to least stable?
- A
Methyl, primary, secondary, tertiary
- B
Tertiary, primary, secondary, methyl
- C
Tertiary, secondary, primary, methyl
- D
Primary, tertiary, methyl, secondary
Under neutral or basic conditions, which of these is commonly a good leaving group?
- A
Hydroxide
- B
Iodide
- C
Amide anion
Select all factors that can stabilize a conjugate base and thereby generally increase the acidity of its conjugate acid.
- A
Delocalization of charge by resonance
- B
An electron-donating group near the negative charge
- C
An electron-withdrawing group near the negative charge
- D
Greater s-character for a carbon-bound negative charge
When checking a proposed polar mechanism, select all statements that should be true of its electron bookkeeping.
- A
Second-row atoms such as carbon, nitrogen, and oxygen obey normal valence limits
- B
Formal charges reflect the electron movement
- C
Every step increases the total charge of the system
- D
Overall charge is conserved
For a negative charge located on carbon, which of the hybridizations sp, sp², or sp³ generally provides the greatest stabilization? Enter the hybridization label.
What is the name of the resonance-stabilized species formed when an α-hydrogen is removed from a carbonyl compound?
In a proton-transfer step, a curved arrow from a base’s lone pair points to the .
Under acidic conditions, protonation can improve an alcohol’s leaving-group ability because the –OH group can leave as .
A bulky species is a strong base but reacts slowly with a crowded electrophilic carbon. Explain how this can happen by distinguishing basicity from nucleophilicity, and identify at least one factor besides basicity that affects nucleophilicity.
When hydroxide deprotonates an acid, where does the full-headed arrow showing hydroxide’s role begin and end?
- A
Its tail begins at the acidic hydrogen and its head points to the hydroxide oxygen
- B
Its tail begins at a hydroxide oxygen lone pair and its head points to the acidic hydrogen
- C
Its tail begins at the acid’s O–H bond and its head points to the hydroxide oxygen
True or false: When a nucleophile attacks a carbonyl carbon, movement of the C=O π electrons onto oxygen commonly accompanies the attack.
- A
True
- B
False
What type of curved arrow represents the movement of one electron in a radical reaction?
What property helps explain why larger atoms can stabilize negative charge effectively down a group?
Under acidic conditions, how can an alcohol’s poor –OH leaving group be made more suitable for departure?
- A
Deprotonate the alcohol so it leaves as hydroxide.
- B
Protonate the alcohol so it can leave as water.
- C
Remove a proton from the adjacent carbon so the alcohol leaves as an alkoxide.
- D
Convert the alcohol’s oxygen into a negatively charged leaving group.
Which is more acidic: acetic acid or ethanol? Choose the explanation that best accounts for the difference.
- A
Acetic acid, because its conjugate base delocalizes negative charge over two oxygen atoms.
- B
Ethanol, because its conjugate base has its negative charge localized on one oxygen atom.
- C
They have equally resonance-stabilized conjugate bases.
- D
Ethanol, because its conjugate base distributes negative charge across two oxygen atoms.
For a negative charge on carbon, which order gives decreasing stability based on hybridization?
- A
sp3>sp2>sp
- B
sp2>sp3>sp
- C
sp>sp2>sp3
- D
sp3>sp>sp2