Organic Chemistry - FAQ

Explain why alkanes have low boiling points.

Alkanes have simple molecular structures and are non-polar, hence they are held together by instantaneous dipole-induced dipole (id-id) attractions. Little energy is required to overcome the weak id-id attractions between alkane molecules.

Explain why melting and boiling points of alkanes increase with Mr.

As Mr increases, electron cloud size increases, resulting in greater polarization of the electron cloud. More energy is required to overcome the stronger and more extensive instantaneous dipole-induced dipole attractions between molecules.

Explain the difference in boiling points of branched and unbranched alkane isomers.

Branched alkane isomers have lower boiling points than unbranched isomers. Branched isomers have a more spherical shape, reducing the surface area of contact between molecules. Less energy is required to overcome the weaker and less extensive instantaneous dipole-induced dipole attractions between branched than unbranched isomers.

Explain why alkanes are generally unreactive.

A large amount of energy is required to overcome the high bond energy of the C-C and C-H bonds. The C-C and C-H bonds are non-polar due to the similarity in electronegativity between C and H, hence they are unreactive towards polar reagents. Alkanes may only undergo combustion and free radial substitution.

Explain why alkenes are more reactive than alkanes.

The C=C double bond of alkenes have high electron densities. Alkenes act as Lewis bases and are susceptible to reaction with electrophiles.

Explain why CHBr=CHBr exhibits cis-trans isomerism.

There is restricted rotation about the C=C double bond. Each C atom of the C=C bond is bonded to two different substituents, hence it exhibits cis-trans isomerism.

CH2CHCH2CH3 reacts with HBr.
Give the major product and explain why it is the major product.

CH3CHBrCH2CH3.
The reaction involves the formation of a carbocation. The secondary carbocation is bonded to two electron-donating alkyl groups which disperses the positive charge more than the primary carbocation, which is bonded to only one alkyl group. The secondary carbocation is more stable and forms faster, resulting in CH3CHBrCH2CH3.

CH2CHCH2CH3 reacts with HBr.
Explain why an optically inactive racemic mixture is formed.

The intermediate formed in step 1 of the mechanism has a planar arrangement about the carbocation. The carbocation can be attacked by Br- from either the top or bottom of the plane with equal probability, leading to the formation of 2 mirror images that are non-superimposable. As equal quantities of each enantiomer is formed, an optically inactive racemic mixture is formed.

Explain the similarities and differences in reactions of alkenes and benzenes.

In both alkenes and benzene, pi electrons are loosely held and are available to react with electrophiles, hence both undergo electrophilic reactions.
Resonance stabilization in benzene causes the pi electron cloud to be less reactive than C=C and requires a stronger electrophile to react. Benzene undergoes substitution rather than addition due to the resonance stabilization from the delocalization of the pi electron cloud. Addition would destroy the stable aromatic structure of the ring.

Explain the difference between the boiling points of alcohols and alkanes of comparable Mr.

Boiling points of alcohols are higher than alkanes of comparable Mr. More energy is required to overcome the stronger hydrogen bonds between alcohol molecules than the weak instantaneous dipole-induced dipole attractions between alkanes.

Compare and explain the differences in acidity of carboxylic acids, phenols and alcohols.

The negative charge of the carboxylate ion is delocalized over the 2 oxygen atoms, dispersing the negative charge and stabilizing the carboxylate ion. Hence, carboxylic acids have a greater tendency to dissociate to give H+ and are stronger acids than phenols and alcohols.
The lone pair of electrons on the O atom of the phenoxide ion is delocalized into the benzene ring, dispersing the negative charge and stabilizing the phenoxide ion. Hence, phenols have a greater tendency to dissociate to give H+ and are stronger acids than alcohols.
The negative charge is localized on the O atom of the alkoxide ion. The electron-donating alkyl groups increase the intensity of the negative charge, destabilizing the alkoxide ion. Hence, alcohols have a lower tendency to dissociate to give H+ and are weaker acids than carboxylic acids and phenols.

Explain why phenol is first neutralized with NaOH before forming ester.

Phenol is neutralized with NaOH to form phenoxide ions. The oxygen atom of phenoxide is negatively charged, hence phenoxide ions are better electron donors and stronger nucleophiles than phenols. Thus, phenoxide ions will give a better yield of ester than phenol.

Explain why phenol requires milder conditions for electrophilic substitution than benzene.

In phenol, the lone pair of electrons on the oxygen atom is delocalized into the benzene ring, increasing the electron density of the ring. This increases phenol’s reactivity towards electrophilic substitution reactions.

Explain the boiling points of carbonyl compounds in comparison to alkanes and alcohols of comparable Mr.

Carbonyl compounds have higher boiling points than alkanes and lower boiling points than alcohols of comparable Mr. Due to the large difference in electronegativity, the C=O bond in carbonyl compounds is polar. More energy is required to overcome the permanent dipole-permanent dipole (pd-pd) attractions between carbonyl compounds than the instantaneous dipole-induced dipole attractions between alkanes. Less energy is required to overcome the pd-pd attractions between carbonyl compounds than the hydrogen bonds between alcohols.

Explain why carbonyl compounds undergo nucleophilic addition reactions.

The highly electronegative oxygen atom draws electron density towards itself. This polarizes the C=O bond, making the carbonyl C atom electron deficient, hence it is susceptible to nucleophilic attack. The carbonyl C is sp2 hybridized and unsaturated, hence it undergoes addition reactions.

Explain why ketones are less reactive than aldehydes.

The carbonyl C of ketones are bonded to 2 electron-donating R groups, hence the carbonyl C is less electron deficient and less susceptible to nucleophilic attack than that of aldehydes, which are bonded to only 1 R group.
The carbonyl C of ketones are bonded to 2 bulky R groups that creates steric hindrance around the carbonyl C, hence it is more difficult for nucleophiles to approach than in aldehydes, which are bonded to only 1 R group.

Carbonyl compounds react with HCN with KCN(aq), at 10-20 degrees Celsius.
Explain why KCN is required.

HCN alone is not a suitable reagent as it is a weak acid and only generates CN- in very low concentrations. KCN is a soluble salt and fully dissociates to produce sufficient CN- for the nucleophilic attack on the electron-deficient carbonyl C.

Carbonyl compounds react with HCN with KCN(aq), at 10-20 degrees Celsius.
Explain why the reaction is controlled at 10-20 degrees Celsius.

The temperature cannot be high to prevent poisonous HCN gas from escaping the reaction. The temperature cannot be too low to ensure the rate of reaction is reasonably high.

Explain the boiling points of carboxylic acids in comparison to alcohols of comparable Mr.

Carboxylic acids have higher boiling points than alcohols of comparable Mr. Carboxylic acids form dimers held together by strong hydrogen bonds. The electron-withdrawing C=O group causes the O-H bond to be more polarized, hence the hydrogen bonds between carboxylic acid molecules are stronger than those between alcohol molecules. The dimers increase the electron cloud size of the carboxylic acid molecules, resulting in greater distortion of the electron cloud, hence there are stronger instantaneous dipole-induced dipole (id-id) attractions between dimers. More energy is required to overcome the stronger hydrogen bonds and id-id attractions between carboxylic acids than alcohols.

Compare and explain the difference in acidity of CH3COOH, CH2FCOOH and CH2ClCOOH.

The electron-withdrawing F and Cl atoms disperse the negative charges on the carboxylate ions, stabilizing them. CH2FCOO- and CH2ClCOO- are more stable than CH3COO-, hence CH2FCOOH and CH2ClCOOH are stronger acids than CH3COOH.
F is more electronegative than Cl, hence it has a greater electron-withdrawing effect and disperses the negative charge of CH2FCOO- to a greater extent. CH2FCOO- is more stable than CH2ClCOO-, hence CH2FCOOH is a stronger acid than CH2ClCOOH.

Compare and explain the difference in acidity of CH3CHClCOOH and CH2ClCH2COOH.

The electron-withdrawing Cl atoms disperse the negative charges on the carboxylate ions, stabilizing them. The electron-withdrawing effect of Cl decreases with increasing distance from the -COOH group. Hence, CH3CHClCOOH is a stronger acid than CH2ClCH2COOH.

Compare and explain the difference in acidity of CH3COOH and HOOCCH2COOH.

In HOOCCH2COOH, the intramolecular hydrogen bonding between the dissociated COO- group and remaining unionised COOH group of the monoanion HOOCCH2COO- stabilizes the monoanion. Hence, HOOCCH2COOH is a stronger acid than CH3COOH.

Explain why phenol cannot react with carboxylic acid directly to form ester when alcohols can.

Phenol is a weaker nucleophile and is less reactive than alcohols, hence the reaction with carboxylic acid will be slow and give a poor yield of ester.

Suggest how to form phenylbenzoate from phenol and benzoic acid.

Add NaOH to phenol to form phenoxide ions. Add PCl5 to benzoic acid to form benzoyl chloride. React phenoxide ions and benzoyl chloride at room temperature.

Compare and explain the difference in reactivity of acyl chlorides, and esters and amides.

The electron-withdrawing Cl atom results in the carbonyl C being the most electron-deficient, hence acyl chlorides are the most susceptible to nucleophilic attack.
The delocalization of the lone pair of electrons on oxygen or nitrogen into the C=O bond causes the carbonyl C to be less electron-deficient, hence esters and amides are less susceptible to nucleophilic attack.

Explain why amides are neutral.

The lone pair of electrons on the N atom is delocalized over the O-C-N bond, hence it is not available for donation to a proton.

Explain the boiling points of halogenoalkanes in comparison to alkanes of comparable Mr.

The boiling points of halogenoalkanes are higher than alkanes of comparable Mr. The C-X bond is polar, hence halogenoalkanes are held together by permanent dipole-permanent dipole (pd-pd) attractions. More energy is required to overcome the stronger pd-pd attractions between halogenoalkane molecules than weaker instantaneous dipole-induced dipole attractions between non-polar alkane molecules.

Compare and explain the difference in reactivity towards nucleophilic substitution of R-F, R-Br, R-Cl, and R-I.

Nucleophilic substitution involves the breaking of the C-X bond. Down the Group, Mr increases and electron size increases, resulting in less effective orbital overlap between the carbon and halogen atom, hence bond strength of the C-X bond decreases. Less energy is required to overcome the weaker C-X bond, thus R-F is the least reactive and R-I is the most reactive.

Explain why the optical isomerism of the product of SN2 is reversed.

The nucleophile attacks the C of the C-X bond from the side opposite the halogen atom. If the C is chiral, the product would have an inversion of configuration compared to the reactant.

Explain why primary halogenoalkanes undergo SN2 while tertiary halogenoalkanes undergo SN1.

Primary halogenoalkanes have the least number of bulky R groups bonded to the C atom of the C-X bond, resulting in less steric hindrance. Hence, it is easier for the nucleophile to attack the C atom from the side opposite the halogen atom.
Tertiary halogenoalkanes have the most electron-donating R groups bonded to the C atom of the C-X bond, which disperses the positive charge of the carbocation to the greatest extent. Hence, the tertiary carbocation is the most stable and forms the fastest.

A tertiary chiral halogenoalkane undergoes nucleophilic substitution.
Explain why an optically inactive racemic mixture is formed.

The intermediate formed in step 1 of the mechanism has a planar arrangement about the carbocation. The carbocation can be attacked by the nucleophile from either the top or bottom of the plane with equal probability, leading to the formation of 2 mirror images that are non-superimposable. As equal quantities of each enantiomer is formed, an optically inactive racemic mixture is formed.

Explain why (chloromethyl)benzene undergoes SN1 mechanism.

The carbocation is resonance stabilized by the delocalization of pi electrons from the benzene ring over the carbocation, hence it forms more quickly for SN1 mechanism.

Compare and explain the differences in reactivity of acyl chlorides, alkyl chlorides and aryl chlorides / chlorobenzenes.

In acyl chlorides, the carbonyl C atom is bonded to 2 electron-withdrawing O and Cl atoms, hence the carbonyl C is very electron-deficient and the most susceptible to nucleophilic attack. There is a trigonal planar arrangement around the carbonyl C (120 degrees), resulting in the least steric hindrance. Thus, acyl chlorides are more reactive than alkyl chlorides and chlorobenzenes.
In alkyl chlorides, the carbonyl C atom is bonded to only 1 electron-withdrawing Cl atom, hence the carbonyl C is less electron-deficient and less susceptible to nucleophilic attack. There is a tetrahedral arrangement around the carbonyl C (109.5 degrees), result in in more steric hindrance. Thus, alkyl chlorides are less reactive than acyl chlorides.
In chlorobenzene, the lone pair of electrons on the Cl atom is delocalized into the benzene ring. This gives the C-Cl bond partial double bond character and strengthens the bond, hence it is difficult to break for nucleophilic substitution to occur. The steric hindrance caused by the bulky benzene ring makes it difficult for the nucleophile to approach the C of the C-Cl bond, hence nucleophilic substitution does not occur under normal conditions.

Explain the boiling points of amines in comparison to alkanes and alcohols of comparable Mr.

Amines have higher boiling points than alkanes and lower boiling points than alcohols of comparable Mr. More energy is required to overcome the stronger hydrogen bonds between amines than the weaker instantaneous dipole-induced dipole attractions between alkanes. The N-H bond is less polar than the O-H bond, hence less energy is required to overcome the weaker hydrogen bonds between amines than the stronger hydrogen bonds between alcohols.

Compare and explain the differences in basicity of NH3, CH3CH2NH2 and C6H5NH2.

In CH3CH2NH2, the electron-donating alkyl group increases the electron density on the N atom, making the lone pair more available for dative bonding with a proton. Hence, CH3CH2NH2 is a stronger base than NH3.
In C6H5NH2, the lone pair of electrons on the N atom is delocalized into the benzene ring, making it less available for dative bonding with a proton. Hence, C6H5NH2 is a weaker base than NH3.

Compare and explain the difference in basicity of phenylamine, 4-methylphenylamine, 4-nitrophenylamine.

In 4-methylphenylamine, the -CH3 group is electron-donating and reduces the extent to which the lone pair of electrons on the N atom is delocalized into the benzene ring, making it more available for dative bonding to a proton. Hence, 4-methylphenylamine is a stronger base than phenylamine.
In 4-nitrophenylamine, the -NO2 group is electron-withdrawing and increases the extent to which the lone pair of electrons on the N atom is delocalized into the benzene ring, making it less available for dative bonding to a proton. Hence, 4-nitrophenylamine is a weaker base than phenylamine.

Compare and explain the differences in basicity of ammonia, primary, secondary, and tertiary amines in gaseous phase.

The number of alkyl groups increases from ammonia to primary, secondary, and tertiary amines. The greater the number of electron-donating alkyl groups, the more available the lone pair of electrons on the N atom is for donation to an electron-deficient species, hence the stronger the Lewis base. Thus, ammonia is the least basic while tertiary amines are the most basic.

Explain why CFCs are a major environmental concern.

The weak C-Cl bond in CFCs undergo homolytic fission by ultraviolet radiation, producing Cl radicals. These free radicals will initiate a chain reaction of destruction of ozone molecules, hence damaging the ozone layer.