201. Ordinary acid-catalysed hydration is unsuitable when the desired conversion is propene to propan-1-ol because:
ⓐ. propene cannot form a carbocation
ⓑ. mainly forms propan-2-ol by Markovnikov addition
ⓒ. dilute acid converts propene directly into propane
ⓓ. water removes one carbon atom from propene
Correct Answer: mainly forms propan-2-ol by Markovnikov addition
Explanation: Protonation of propene preferentially generates a secondary carbocation rather than a primary one. Water attacks this secondary carbocation and forms propan-2-ol after deprotonation. Propan-1-ol would require hydroxyl placement at the less substituted terminal carbon. That orientation is obtained more effectively through hydroboration-oxidation. The limitation concerns regioselectivity rather than failure of the alkene to react.
202. Hydroboration-oxidation of an alkene uses the reagent sequence:
ⓐ. \(\mathrm{H_2O_2/OH^-}\), followed by \(\mathrm{BH_3}\)
ⓑ. dilute \(\mathrm{H_2SO_4}\), followed by \(\mathrm{NaBH_4}\)
ⓒ. \(\mathrm{BH_3}\), followed by \(\mathrm{H_2O_2/OH^-}\)
ⓓ. \(\mathrm{Br_2}\), followed by alcoholic \(\mathrm{KOH}\)
Correct Answer: \(\mathrm{BH_3}\), followed by \(\mathrm{H_2O_2/OH^-}\)
Explanation: In the first stage, borane adds across the carbon-carbon double bond. Boron becomes bonded preferentially to the less substituted alkene carbon. The second stage uses alkaline hydrogen peroxide to replace the carbon-boron bond with a carbon-oxygen bond. The sequence produces an alcohol with overall anti-Markovnikov orientation. Reversing the two stages would not perform the required transformation.
203. In the hydroboration-oxidation of an unsymmetrical alkene, the hydroxyl group is generally found on:
ⓐ. the carbon that initially bears the greater positive charge in a free carbocation
ⓑ. the more substituted carbon in every case
ⓒ. both alkene carbons to form a diol
ⓓ. the less substituted carbon of the original double bond
Correct Answer: the less substituted carbon of the original double bond
Explanation: Boron adds preferentially to the less substituted alkene carbon during hydroboration. The subsequent oxidation step replaces boron with hydroxyl at the same carbon position. The net result is anti-Markovnikov placement of \(\mathrm{-OH}\). Only one oxygen-containing group is introduced across the double bond in this reaction. The outcome does not depend on formation of a freely rearranging carbocation.
204. The statement that hydroboration is a syn addition means that:
ⓐ. hydrogen and boron initially add to opposite faces of the double bond
ⓑ. hydrogen and boron initially add to the same face of the double bond
ⓒ. two hydroxyl groups are added to adjacent carbons
ⓓ. the carbon skeleton must undergo a methyl shift
Correct Answer: hydrogen and boron initially add to the same face of the double bond
Explanation: Syn addition describes the relative spatial direction from which two groups add to an alkene. During hydroboration, hydrogen and boron are delivered to the same face of the double bond through a coordinated addition process. Oxidation then replaces boron with hydroxyl while retaining the carbon position. The overall addition of hydrogen and hydroxyl is therefore recognised as syn. This stereochemical term is separate from the anti-Markovnikov description of which carbon receives \(\mathrm{-OH}\).
205. Assertion: Hydroboration-oxidation normally proceeds without carbocation rearrangement.
Reason: The hydroboration stage does not generate a free carbocation intermediate.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Rearrangements such as hydride and methyl shifts are characteristic of reactions involving suitable carbocations. Hydroboration occurs through a coordinated addition rather than through a freely existing carbocation. There is therefore no carbocation centre that can rearrange to seek greater stability. The original carbon skeleton is normally retained. The Reason directly accounts for the absence of the rearrangement stated in the Assertion.
206. Identify the row in which the alkene, reaction route, and major product are correctly matched.
| Row | Alkene | Reaction route | Major product |
|---|
| P | Propene | Dilute aqueous acid | Propan-2-ol |
| Q | Propene | \(\mathrm{BH_3}\), then \(\mathrm{H_2O_2/OH^-}\) | Propan-2-ol |
| R | But-1-ene | Dilute aqueous acid | Butan-1-ol only |
| S | Ethene | \(\mathrm{BH_3}\), then oxidation | Propan-1-ol |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Acid-catalysed hydration of propene follows Markovnikov orientation and gives propan-2-ol, so row P is consistent. Hydroboration-oxidation of propene gives propan-1-ol rather than propan-2-ol. Acid hydration of but-1-ene mainly gives butan-2-ol. Ethene contains two carbon atoms and gives ethanol, so it cannot produce propan-1-ol without an additional carbon source.
207. Hydroboration-oxidation of \(\mathrm{CH_2=CHC(CH_3)_3}\) gives the unrearranged alcohol:
ⓐ. 2,3-dimethylbutan-2-ol
ⓑ. 3,3-dimethylbutan-1-ol
ⓒ. 2,2-dimethylbutan-2-ol
ⓓ. 3,3-dimethylbutan-2-ol
Correct Answer: 3,3-dimethylbutan-1-ol
Explanation: Boron attaches to the terminal, less substituted carbon of the double bond. Oxidation subsequently replaces the carbon-boron bond with a carbon-oxygen bond. No free carbocation forms, so the neighbouring highly substituted carbon does not trigger a methyl shift. The original carbon skeleton is preserved throughout the reaction. The product is therefore \(\mathrm{HOCH_2CH_2C(CH_3)_3}\), named 3,3-dimethylbutan-1-ol.
208. A terminal alkene of the form \(\mathrm{RCH=CH_2}\) generally gives which type of alcohol after hydroboration-oxidation, provided no other functional group interferes?
ⓐ. a tertiary alcohol
ⓑ. a phenol
ⓒ. a vicinal diol
ⓓ. a primary alcohol
Correct Answer: a primary alcohol
Explanation: In \(\mathrm{RCH=CH_2}\), the terminal \(\mathrm{CH_2}\) carbon is the less substituted alkene carbon. Hydroboration places boron mainly at this terminal position. Oxidation converts that carbon-boron bond into a carbon-oxygen bond. The product has the form \(\mathrm{RCH_2CH_2OH}\). Since the hydroxyl-bearing carbon is attached to only one other carbon group, the resulting alcohol is primary.
209. Which comparison between acid-catalysed hydration and hydroboration-oxidation is correct?
ⓐ. acid hydration may rearrange; hydroboration gives anti-Markovnikov syn addition
ⓑ. Acid hydration is always syn, while hydroboration has no stereochemical preference
ⓒ. Both reactions necessarily give Markovnikov alcohols through free carbocations
ⓓ. Hydroboration gives a diol, whereas acid hydration gives an ether
Correct Answer: acid hydration may rearrange; hydroboration gives anti-Markovnikov syn addition
Explanation: Hydration in an acidic medium commonly proceeds through a carbocation and generally gives Markovnikov orientation. A suitable carbocation can undergo hydride or alkyl shifts before water attacks. Hydroboration occurs through coordinated addition, places boron on the less substituted carbon, and avoids a free carbocation. Oxidation replaces boron with hydroxyl, producing overall anti-Markovnikov syn hydration. The two methods can therefore give different constitutional and stereochemical outcomes from the same unsymmetrical alkene.
210. Bromoethane is heated with aqueous potassium hydroxide. The principal organic product is:
ⓐ. ethene
ⓑ. ethanol
ⓒ. ethoxyethane
ⓓ. ethanal
Correct Answer: ethanol
Explanation: Aqueous potassium hydroxide supplies hydroxide ions, which act mainly as nucleophiles under these conditions. The hydroxide ion replaces bromide at the carbon bearing the halogen. The reaction is represented by \(\mathrm{CH_3CH_2Br+KOH_{(aq)}\rightarrow CH_3CH_2OH+KBr}\). The carbon skeleton remains unchanged during the substitution. Ethene would be favoured more strongly when alcoholic potassium hydroxide and heating promote elimination.
211. Which condition most strongly favours conversion of 2-bromopropane into propan-2-ol rather than propene?
ⓐ. Concentrated alcoholic \(\mathrm{KOH}\) with strong heating
ⓑ. Sodium metal in dry ether
ⓒ. Concentrated \(\mathrm{H_2SO_4}\) at high temperature
ⓓ. dilute aqueous \(\mathrm{KOH}\) favouring substitution
Correct Answer: dilute aqueous \(\mathrm{KOH}\) favouring substitution
Explanation: Water solvates the ions and provides a medium in which hydroxide can act as a nucleophile. Nucleophilic substitution replaces bromide by \(\mathrm{-OH}\), producing propan-2-ol. Alcoholic potassium hydroxide and higher temperature increase the tendency toward beta-elimination and propene formation. Concentrated sulfuric acid is not the standard reagent for hydrolysing a haloalkane to an alcohol. The solvent and temperature therefore help determine whether substitution or elimination predominates.
212. In a pure \(\mathrm{S_N2}\) hydrolysis of an optically active alkyl halide, attack by hydroxide ion at the chiral carbon usually produces:
ⓐ. complete retention of configuration
ⓑ. a carbocation followed by total racemisation
ⓒ. inversion of configuration
ⓓ. no change at the carbon-halogen centre
Correct Answer: inversion of configuration
Explanation: An \(\mathrm{S_N2}\) reaction occurs through backside attack by the nucleophile. Hydroxide approaches from the side opposite the leaving halide ion. Bond formation and carbon-halogen bond breaking occur in a single concerted step. The arrangement around the reacting chiral carbon is therefore turned inside out. Racemisation would instead suggest formation of a planar carbocation through an \(\mathrm{S_N1}\) pathway.
213. Hydrolysis of a tertiary alkyl halide in a polar aqueous medium commonly proceeds through:
ⓐ. a primary carbanion
ⓑ. a tertiary carbocation
ⓒ. a carbon-carbon triple bond
ⓓ. an aryl radical
Correct Answer: a tertiary carbocation
Explanation: The carbon-halogen bond of a suitable tertiary substrate can ionise in a polar aqueous medium. This produces a tertiary carbocation and a halide ion. Water then attacks the carbocation, and deprotonation gives the alcohol. The pathway is classified as \(\mathrm{S_N1}\) because the ionisation step controls the rate. The same mechanism is much less favourable for an ordinary primary haloalkane because a primary carbocation is unstable.
214. Chlorobenzene does not readily form phenol when treated with aqueous potassium hydroxide under ordinary mild conditions mainly because:
ⓐ. chlorine electronegativity prevents all aromatic substitution
ⓑ. phenol would lack a stable carbon–oxygen bond under these conditions
ⓒ. hydroxide cannot approach any aromatic carbon as a nucleophile
ⓓ. the aryl carbon–chlorine bond has partial double-bond character
Correct Answer: the aryl carbon–chlorine bond has partial double-bond character
Explanation: In chlorobenzene, the chlorine lone pair interacts with the aromatic ring and gives the carbon-chlorine bond partial double-bond character. The bond is consequently shorter and stronger than a typical alkyl carbon-chlorine bond. The carbon bearing chlorine is also \(\mathrm{sp^2}\)-hybridised, which prevents ordinary backside \(\mathrm{S_N2}\) attack. Phenol preparation from chlorobenzene therefore requires much harsher conditions. The behaviour of haloarenes should not be inferred directly from that of haloalkanes.
215. Complete hydrolysis of 1-bromobutane produces butan-1-ol in a (1:1) mole ratio:
[
\mathrm{CH_3CH_2CH_2CH_2Br+KOH\rightarrow CH_3CH_2CH_2CH_2OH+KBr}
]
What mass of butan-1-ol can be obtained theoretically from (27.4,\mathrm{g}) of 1-bromobutane? Use (M(\text{1-bromobutane})=137,\mathrm{g,mol^{-1}}) and (M(\text{butan-1-ol})=74,\mathrm{g,mol^{-1}}).
ⓐ. (7.4,\mathrm{g})
ⓑ. (20.3,\mathrm{g})
ⓒ. (14.8,\mathrm{g})
ⓓ. (27.4,\mathrm{g})
Correct Answer: (14.8,\mathrm{g})
Explanation: ( \textbf{Amount of 1-bromobutane:} )
[
n=\frac{m}{M}
]
[
n=\frac{27.4,\mathrm{g}}{137,\mathrm{g,mol^{-1}}}
]
[
n=0.200,\mathrm{mol}
]
( \textbf{Reaction ratio:} )
One mole of 1-bromobutane gives one mole of butan-1-ol.
[
n(\text{butan-1-ol})=0.200,\mathrm{mol}
]
( \textbf{Product mass:} )
[
m=nM
]
[
m=0.200,\mathrm{mol}\times74,\mathrm{g,mol^{-1}}
]
[
m=14.8,\mathrm{g}
]
( \textbf{Final answer:} ) The theoretical mass of butan-1-ol is (14.8,\mathrm{g}). The mass of the isolated organic product differs from the haloalkane mass because bromine is replaced by a hydroxyl group.
216. Identify the row in which the reaction conditions and principal process are correctly matched.
| Row | Substrate and reagent | Principal process | Expected organic product |
|---|
| P | Bromoethane with aqueous \(\mathrm{KOH}\) | Elimination | Ethene only |
| Q | Bromoethane with alcoholic \(\mathrm{KOH}\) and heat | Substitution | Ethanol only |
| R | Chloroethane with moist \(\mathrm{Ag_2O}\) | Reduction | Ethane |
| S | Bromoethane with aqueous \(\mathrm{KOH}\) | Nucleophilic substitution | Ethanol |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Aqueous potassium hydroxide promotes replacement of bromide by hydroxyl, so row S correctly predicts ethanol. Alcoholic potassium hydroxide with heating more strongly favours elimination to ethene. Moist silver oxide can provide hydroxide-containing conditions for alcohol formation rather than reducing a haloalkane to an alkane. The same substrate can therefore give different products when the solvent and reaction conditions are changed. Reagent names alone are insufficient unless the medium is also specified.
217. Reduction of an aldehyde of the general form \(\mathrm{R-CHO}\) gives:
ⓐ. a primary alcohol, \(\mathrm{R-CH_2OH}\)
ⓑ. a secondary alcohol, \(\mathrm{R_2CHOH}\)
ⓒ. a tertiary alcohol, \(\mathrm{R_3COH}\)
ⓓ. an ether, \(\mathrm{R-O-R}\)
Correct Answer: a primary alcohol, \(\mathrm{R-CH_2OH}\)
Explanation: An aldehyde carbonyl carbon is bonded to one carbon group and one hydrogen atom. Reduction adds hydrogen across the carbonyl group and converts \(\mathrm{C=O}\) into \(\mathrm{C-OH}\). The product carbon remains attached to only one other carbon group. It therefore becomes the hydroxyl-bearing carbon of a primary alcohol. The general transformation is \(\mathrm{R-CHO+2[H]\rightarrow R-CH_2OH}\).
218. Reduction of propanone produces:
ⓐ. propan-1-ol
ⓑ. propan-2-ol
ⓒ. propane-1,2-diol
ⓓ. 2-methylpropan-2-ol
Correct Answer: propan-2-ol
Explanation: Propanone is a ketone with the structure \(\mathrm{CH_3COCH_3}\). Reduction converts its carbonyl group into a hydroxyl group without changing the carbon skeleton. The product is \(\mathrm{CH_3CH(OH)CH_3}\). The hydroxyl-bearing carbon is attached to two methyl groups, so the product is a secondary alcohol. Propan-1-ol would require the hydroxyl group to appear at a terminal carbon and cannot arise from simple carbonyl reduction of propanone.
219. Which reagent can reduce many aldehydes and ketones to the corresponding alcohols under suitable conditions?
ⓐ. Aqueous \(\mathrm{NaCl}\)
ⓑ. Bromine water
ⓒ. \(\mathrm{NaBH_4}\)
ⓓ. Concentrated \(\mathrm{HNO_3}\)
Correct Answer: \(\mathrm{NaBH_4}\)
Explanation: Sodium borohydride is a commonly used reducing agent for aldehydes and ketones. It supplies hydride character to the electrophilic carbonyl carbon. Subsequent protonation produces the alcohol. Bromine water and nitric acid are associated with oxidation or electrophilic reactions rather than carbonyl reduction. Sodium chloride does not provide the reducing power required to convert \(\mathrm{C=O}\) into \(\mathrm{C-OH}\).
220. Which alcohol cannot be prepared by simple reduction of an aldehyde or ketone alone?
ⓐ. 2-methylpropan-2-ol
ⓑ. ethanol
ⓒ. propan-2-ol
ⓓ. 2-methylpropan-1-ol
Correct Answer: 2-methylpropan-2-ol
Explanation: Aldehydes reduce to primary alcohols, while ketones reduce to secondary alcohols. Ethanol can be obtained from ethanal, and propan-2-ol can be obtained from propanone. Reduction of 2-methylpropanal gives 2-methylpropan-1-ol. A tertiary alcohol has three carbon groups attached to the hydroxyl-bearing carbon. Simple reduction of an aldehyde or ketone does not introduce the additional carbon-carbon bond needed to create that tertiary centre.
