Question 1

WHICH SUBSTRATE REACTS FASTEST WITH SODIUM CYANIDE IN DMSO?

Accepted Answer

1 BROMOBUTANE. Cyanide is a strong nucleophile and DMSO is polar aprotic, so this is SN2. SN2 rate collapses with steric bulk at the electrophilic carbon, so primary beats secondary beats tertiary. Bromobenzene is an aryl halide and cannot undergo SN2 at all.

Question 2

(R) 3 BROMO 3 METHYLHEXANE IS HEATED IN AQUEOUS ETHANOL. WHAT IS THE STEREOCHEMICAL OUTCOME AT CARBON THREE?

Accepted Answer

A RACEMIC MIXTURE OF BOTH ENANTIOMERS. A tertiary halide in a weakly nucleophilic protic solvent ionises first, giving a planar carbocation. Solvent attacks either face with near equal probability, so the substitution product is racemic.

Question 3

AN SN2 REACTION PROCEEDS WITH COMPLETE ___ OF CONFIGURATION AT THE ELECTROPHILIC CARBON.

Accepted Answer

INVERSION. The nucleophile attacks the face opposite the leaving group, so the other three bonds flip through like an umbrella in the wind. One clean stereochemical outcome, every time.

Question 4

2 BROMO 2 METHYLBUTANE IS TREATED WITH POTASSIUM TERT BUTOXIDE. WHICH ALKENE DOMINATES?

Accepted Answer

2 METHYLBUT 1 ENE. Tert butoxide is bulky, so it cannot reach the crowded internal beta hydrogens and pulls one from the accessible methyl group instead. That gives the less substituted Hofmann alkene rather than the Zaitsev one.

Question 5

A TERMINAL ALKYNE HAS A PKA OF ABOUT 25. WHICH BASE WILL DEPROTONATE IT ESSENTIALLY COMPLETELY?

Accepted Answer

SODIUM AMIDE, PKA OF AMMONIA 38. An acid base reaction runs downhill toward the weaker acid. Only amide has a conjugate acid weaker than the alkyne, so only sodium amide drives the equilibrium far to the right.

Question 6

E2 WITH A SMALL BASE SUCH AS ETHOXIDE USUALLY GIVES THE MORE SUBSTITUTED ALKENE, CALLED THE ___ PRODUCT.

Accepted Answer

ZAITSEV. Alkene stability rises with substitution, and a small base can reach the crowded internal beta hydrogen. Swap in a bulky base and the selectivity flips to the less substituted alkene.

Question 7

WHICH REAGENT SET CONVERTS 1 METHYLCYCLOHEXENE INTO 2 METHYLCYCLOHEXAN 1 OL?

Accepted Answer

BH3 IN THF, THEN H2O2 WITH NAOH. The hydroxyl ends up on the less substituted alkene carbon, which is anti Markovnikov, and only hydroboration oxidation does that. Acid catalysed hydration and oxymercuration both give the Markovnikov product 1 methylcyclohexanol, and HBr with peroxides gives a bromide.

Question 8

HYDROBORATION OXIDATION ADDS H AND OH TO THE SAME FACE OF THE ALKENE, SO IT IS A ___ ADDITION.

Accepted Answer

SYN. Boron and hydrogen are delivered together through a four centre transition state, so both new bonds form on one face. Bromination, by contrast, goes through a cyclic ion and adds anti.

Question 9

3 METHYLBUT 1 ENE IS TREATED WITH HBR. WHY IS 2 BROMO 2 METHYLBUTANE THE MAJOR PRODUCT?

Accepted Answer

A SECONDARY CATION REARRANGES BY A HYDRIDE SHIFT TO A TERTIARY CATION. Markovnikov protonation gives a secondary cation at carbon two. A hydride slides over from carbon three to give the more stable tertiary cation, and bromide then traps that.

Question 10

WHY DOES SWITCHING THE SOLVENT FROM METHANOL TO DMSO SPEED UP AN SN2 REACTION OF AN ANIONIC NUCLEOPHILE?

Accepted Answer

DMSO CANNOT DONATE HYDROGEN BONDS, SO THE NUCLEOPHILE STAYS UNCAGED. Methanol wraps an anionic nucleophile in a hydrogen bonded shell that must be stripped away before attack. DMSO solvates the counter cation but has no acidic hydrogen for the anion, leaving it far more reactive.

Question 11

2 CHLORO 2 METHYLPROPANE IS WARMED IN ETHANOL WITH NO ADDED BASE. WHICH RATE LAW IS OBSERVED?

Accepted Answer

RATE EQUALS K TIMES THE HALIDE CONCENTRATION ONLY. The slow step is ionisation of the carbon chlorine bond, which involves the substrate alone. Ethanol is present in vast excess and only enters after the rate determining step, so it never appears in the rate law.

Question 12

CIS 1 BROMO 4 TERT BUTYLCYCLOHEXANE ELIMINATES WITH ETHOXIDE FAR FASTER THAN THE TRANS ISOMER. WHY?

Accepted Answer

THE TERT BUTYL GROUP LOCKS THE BROMINE AXIAL, SO A BETA HYDROGEN IS ANTI PERIPLANAR. The bulky tert butyl group is pinned equatorial, which fixes the ring. In the cis isomer that forces bromine axial, giving the trans diaxial arrangement E2 needs, while in the trans isomer bromine is equatorial and no beta hydrogen can line up.

Question 13

TRANS 2 BUTENE IS TREATED WITH BR2 IN DICHLOROMETHANE. WHAT IS THE PRODUCT?

Accepted Answer

MESO 2 3 DIBROMOBUTANE. Bromide opens the bromonium ion from the opposite face, so the two bromines add anti. Applied to the trans alkene that produces the achiral meso diastereomer, whereas the cis alkene would give the racemic pair.

Question 14

PHENOL HAS A PKA NEAR 10 WHILE CYCLOHEXANOL IS NEAR 16. WHAT EXPLAINS THE GAP?

Accepted Answer

THE PHENOXIDE CHARGE IS DELOCALISED INTO THE AROMATIC RING. Acidity is decided by how well the conjugate base carries the negative charge. Phenoxide spreads it over three ring carbons by resonance, while the cyclohexanoxide charge is stuck on one oxygen.

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