All Free Chemistry MCQs with Answers

191. The rate of an SN1 reaction depends on the concentration of

• A. both the alkyl halide and the nucleophile
• B. the alkyl halide only
• C. the nucleophile only
• D. the solvent only

Explanation: The slow, rate determining step is the ionisation of the alkyl halide into a carbocation and a halide ion, and the nucleophile is not involved until the fast second step, so it does not appear in the rate equation. This is why the reaction is called unimolecular. Detecting first order kinetics is the standard experimental evidence for the SN1 pathway.

192. Reaction of an alkyl halide with alcoholic potassium hydroxide gives mainly

• A. an alcohol
• B. an alkane
• C. a carboxylic acid
• D. an alkene, by elimination of hydrogen halide

Explanation: In ethanol the hydroxide ion acts as a base rather than as a nucleophile, removing a hydrogen from the carbon next to the one bearing the halogen, so a double bond forms and the hydrogen halide is eliminated. Heat favours this route as well. The contrast with aqueous conditions, which give the alcohol, is the essential comparison in this topic.

193. According to Saytzeff's rule, the major product of an elimination reaction is

• A. the less substituted, less stable alkene
• B. always the cis isomer
• C. the more highly substituted and therefore more stable alkene
• D. an alkane

Explanation: When more than one alkene can form, the preferred product is the one with the greater number of alkyl groups attached to the doubly bonded carbons, because such alkenes are thermodynamically more stable. So 2-bromobutane gives mainly but-2-ene rather than but-1-ene. Using a bulky base can override this preference and favour the less substituted product instead.

194. Elimination reactions of alkyl halides are favoured over substitution by

• A. a high concentration of a strong base, a non aqueous solvent and a higher temperature
• B. dilute aqueous conditions at room temperature
• C. the use of a weak nucleophile in water
• D. cooling the mixture in ice

Explanation: Elimination has the higher activation energy and produces more particles, so heat and a strong base in ethanol push the reaction that way, while water and mild conditions favour substitution. The two pathways always compete, and a given reaction usually produces some of both products. Tertiary halides eliminate most readily because the carbon is too crowded for substitution.

195. Reaction of bromoethane with alcoholic ammonia under pressure gives

• A. ethanol
• B. ethylamine
• C. ethanenitrile
• D. ethane

Explanation: Ammonia has a lone pair on nitrogen and acts as a nucleophile, displacing bromide to give ethylamine, though the product is itself nucleophilic so further substitution to secondary and tertiary amines readily follows. Using a large excess of ammonia limits this. Ethanenitrile would be formed with potassium cyanide instead.

196. The reaction of an alkyl halide with potassium cyanide is synthetically useful because it

• A. shortens the carbon chain by one atom
• B. produces an alkene
• C. lengthens the carbon chain by one carbon atom
• D. removes the halogen without adding anything

Explanation: The cyanide ion substitutes for the halogen, so the nitrile formed has one more carbon than the starting halide, and that nitrile can then be hydrolysed to a carboxylic acid or reduced to an amine. Building the chain one carbon at a time is a standard strategy in organic synthesis. Alcoholic conditions are used to favour substitution over elimination.

197. A Grignard reagent is formed when an alkyl halide reacts with

• A. sodium metal in dry ether
• B. magnesium in dry ether
• C. zinc in aqueous solution
• D. copper in ethanol

Explanation: Magnesium inserts itself into the carbon to halogen bond to give an alkylmagnesium halide, and the ether must be perfectly dry because even a trace of water destroys the reagent, converting it into an alkane. The carbon attached to magnesium is strongly nucleophilic, which makes Grignard reagents extremely versatile for building carbon to carbon bonds. Sodium with an alkyl halide gives the Wurtz reaction instead.

198. Alkyl halides have higher boiling points than the corresponding alkanes because

• A. they contain hydrogen bonds
• B. they are ionic
• C. their molecules are lighter
• D. they are polar and have greater molar mass, so intermolecular forces are stronger

Explanation: The polar carbon to halogen bond adds dipole to dipole attraction on top of the dispersion forces, and the heavy halogen atom increases those dispersion forces as well, so more energy is needed to separate the molecules. Boiling point rises from the fluoride to the iodide for the same alkyl group. Hydrogen bonding is not possible because hydrogen is not bonded to fluorine, oxygen or nitrogen here.

199. Which alkyl halide would react fastest by the SN1 mechanism?

• A. CH3Br
• B. CH3CH2Br
• C. (CH3)2CHBr
• D. (CH3)3CBr

Explanation: SN1 rates follow the stability of the carbocation formed, and that increases from primary through secondary to tertiary because each alkyl group releases electron density and spreads the charge. The tertiary bromide therefore ionises most readily. The order is exactly reversed for SN2, where the methyl halide reacts fastest because it is the least hindered.

200. In an SN2 reaction the configuration at the reacting carbon atom is

• A. inverted, because the nucleophile attacks from the side opposite the leaving group
• B. retained completely
• C. randomised, giving a racemic mixture
• D. unchanged because no bond is broken

Explanation: Back side attack turns the three remaining groups inside out like an umbrella in the wind, so an optically active substrate gives a product of opposite configuration, an effect called Walden inversion. An SN1 reaction gives a racemic mixture instead, because the planar carbocation can be attacked equally from either face. Stereochemistry is therefore the clearest experimental way to distinguish the two mechanisms.