halogen derivatives

Question 1

why is boiling point of R-X higher than their corresponding alkanes?

Answer 1

1. both have simple molecular structures.
2. alkanes are non-polar with only idid while halogenoalkanes ar epolar with stronger pd-pid.
3. moreover, no. of electrons/size of electorn cloud is greater, hence extent of distortion of electron cloud is greater.
4. strength of id-id interaction between molecules is greater.
5. more energy required to overcome id-id interactions.

Question 2

why do the boiling points of R-X with similar no. of carbon atoms increase in the following order: R-F < R-Cl < R-Br < R-I?

Answer 2

id-id electron cloud explanation.

Question 3

why are halogenoalkanes insoluble in water.

Answer 3

1. cannot form intermolecular hydrogen bonds with water molecules, thus no strong interactions between halogenoalkanes and water molecules
2. insufficient energy is released to overcome the strong hydrogen bonds between water molecules for mixing to occur.

Question 4

why are halogenoalkanes soluble in non-polar solvents?

Answer 4

1. energy released from instantaneous id-id interactions formed between non-polar alkyl groups of halogenoalkanes and non-polar molecules is sufficient to overcome similar id-id interactions between halogenoalkane molecules and between non-polar solvent molecules.

Question 5

are halogenoalaknes denser than water?

Answer 5

yes

Question 6

describe nucleophilic substitution in words.

Answer 6

1. C-X bond in a halogenoalakne molecule is polarised by the electronegative halogen atom which draws the shared pair of electrons away from C.
   2.The carbon atom in the C-X bond is associated with a partial positive charge, thus it is susceptible to attack by nucleophiles.
2. The halogen atom is displaced as a halide anioin, which is known as a good leaving group.

Question 7

what are nucleophiles?

Answer 7

they are electron-pair donors attracted to an electron deficient atom/region of low electron density.

Question 8

describe an Sn2 mechanism.

Answer 8

1. it is a one-step mechanism.
2. The C-X bond is polar.
3. In a one step process, the nucleophile attacks the (+)carbon atom from the opposite side of the halogen atom.
4. A pentavalent transition state is formed.
5. During this transition state, both the nucleophile and X- are partially bonded to the same C atom.
6. Both C-X bond-breaking and C-N-(nucleophile) bond forming process are taking place simultaneously.

Question 9

what happens if carbon atom of halogenoalkane is chiral in a Sn2 mechanism?

Answer 9

1. Inversion of stereochemical configuration occurs as nucleophile attacks the (+)carbon from the opposite side of the halogen atom.
2. if (+)carbon atom of halogenalkane is achiral, then there is no stereochemical configuration/inversion to do.

Question 10

what happens in a Sn1 mechanism?

Answer 10

1. it is a 2-step reaction.
2. In the first step(slow step — rate determining step), the C-X bond breaks heterolytically to form a carbocation intermediate and a halide anion, the electron-donating alkyl groups help to stabalise the carbocation intermediate by dispersing the positive charge on the carbon. carbon is sp2 hybridized.
3. in the 2nd step, the nucleophile then attacks the very reactive carbocation intermediate to form the alcohol.
4. the second step is fast due to the attraction between opposite charges of carvocation and the nucleophile.

Question 11

what happens to an enantiomerically-pure chiral reactant in a Sn1 reaction>

Answer 11

1. since carbocation is trigonal planar with respect to the electron deficient carbon, the nucleophile attacks the carbocation from the top and the bottom of the plane with equal probability.
2. if the carbon with + charge becomes a chiral carbon after the reaction, a reacemic mixture, which is optically inactive is formed as both mirror-image enantiomers are formed in equal quantities.

Question 12

when is Sn1 mechanism preferred?

Answer 12

1. preferred for tertiary halogenoalkanes
2. they give a stable tertiary carbocation intermediate as electron-donating alkyl groups disperse the psotive charge on the carbocation intermediate.

methyl carbocation and primary carbocation are less stabilised.

Question 13

when is Sn2 Mechanism preferred?

Answer 13

1. when it is a methyl/primary halogenoalkane with no or only 1 akyl group bonded to the (+) carbon atom, which allows easy approach of the nucleophile.

**a tertiary halogenoalkanes has 3 alkyl groups which can sterically hinder the approach of nucleophile to the (+) carbon atom.

Question 14

what are the exception to the preferred mechanisms?

Answer 14

1. (CH3)3CCH2Cl
   -Sn1
   - steric hindrance due to bulky (CH3)3C- group makes it difficult for nucleophile to attack from opposite-side of leaving group.
2. C6H5CH2Cl
   - Sn1
   - resonance stablisation of benzyl carbocation by delocalisation from benzene ring.

Question 15

describe the formation of alcohols.

Answer 15

reagents and conditions: NaOH(aq) and heat under reflux

type of reaction: nucleophilic substitution

Nucleophile: OH-

general eq: Rx + NaOH(aq) = ROH + NaX

Question 16

describe the formation of nitriles.

Answer 16

1. reagents and conditions: KCN in ethanol and heat under reflux.
2. type of reaction: nucleophilic substitution.
3. Nucleophile: CN-.
4. general eq: RX + KCN = RCN + KX.

Question 17

describe the formation of carboxylic acid from nitrile.

Answer 17

reagents and conditions: heating with dilute acid under reflux.

type of reaction: acid hydrolysis

general eq: RCN + H+ + 2H2O = RCOOH + NH4+

Question 18

desctibe the formation of carboxylate salt.

Answer 18

reagents and conditons: heating with dilute alkali(under reflux).

type of reaction: alkaline hydrolusis.

general equation: RCN + OH- +H2O = RCOO- + NH3

Question 19

describe one of the methods used to form primary amine from nitrile.

Answer 19

reagents and conditions: H2(g), Ni catalyst.

type of reaction: reduction.

general eq: RCN + 2H2 = RCH2NH2.

Question 20

describe the other method used to form primary amine from nitrile.

Answer 20

reagents and conditon: LiAlH4 in dry ether.

type of reaction: reduction.

general eq: RCN + 4[H] = RCH2NH2.

Question 21

describe the formation of primary amines from a halogenoalkane.

Answer 21

1. reagents and conditons: excess NH3 in ethanol heated under pressure.
2. type of reaction: nucleophilic substitution.
3. general eq: RX + 2NH3 = RNH2 + NH4X.
4. Normally through Sn2 mechanism.

Question 22

what happens in the formation of primaty amines from halogenoalkanes if ammonia is not in excess.

Answer 22

1. further nucleophilic substitution can occur to form a mixture of 1, 2, 3* amines* and quaternary ammonium salt(the final product).

RNH2 - R2NH - R3N - R4N+X-.

As alkyl groups are electron-donating, the electorn density on the alkylamine’s nitrogen is greater than the nitrogen of ammonia, Correspondingly, 1, 2, 3* amines* are more nucleophilic than ammonia. Hence, the alkylamine will compete with ammonia for RX.

Question 23

describe the elimination of halogen atom from halogenoalkanes.

Answer 23

1. reagents and conditons: NaOH in ethanol, heat under reflux.
2. elimination reaction.
3. general eq: CH3CH2X + OH- = CH2CH2 + X- + H20.

Question 24

why are fluroalkanes and flurohalogenoalkanes often used as inert materials?

Answer 24

1. generally stable and unreactive due to strong carbon-fluorine bond
2. good solvents with low boiling temperatueres.
3. Non-flammable, non-toxic and odourless.

Question 25

what are the physical properties of halogenoarenes.

Answer 25

1. dense 2. colourless 3. insoluble in water, unable to form hydrogen bonds. 4. soluble in non-polar solvents due to id-id interactions.

Question 26

# there are 2 reasons. why do halogenoarenes usually not undergo nucleophilic substitution, unless when subjected to very drastic and vigorous conditions?

Answer 26

reason 1: - the p orbital of the halogen atom overlaps side-on with the p orbitals of the six carbon atoms of the benzene ring. - the lone pair of electrons in the p-orbital of the halogen atom can delocalise into the benzene ring to form a delocalised pi electron cloud. - this results in a partial double bond character in the C-X bond, requiring more energy to break the stronger C-X bond. reason 2: - sterically, the rear side of the C-X bond in halogenoarenes is blocked by the bulky benzene rinh. - the pi electorn cloud of the benzene ring will repel the lone pair of electrons of an incoming nucleophile, rendering attack of the nucleophile difficult.

Question 27

how to test for halogen derivatives?

Answer 27

add an equal volume of NaOH(aq) to RX and heat in a water bath. Cool the mixture and add excess HNO3(aq). add AgNO3(aq).

Question 28

what are the observations for different types of halogen derivatives?

Answer 28

2-iodobutane - yellow ppt 2-bromobutane - pale cream ppt 2-chlorobutane - white ppt 2-flurobutane - no ppt chlorobenzene - no ppt.

Question 29

what is the rationale behind the reagents and conditions required?

Answer 29

1. Halogenoalkanes react with the nucleophile, OH-, in nucleophilic substitution to produce halide anions. - RX + NaOH = ROH + NaX. 2. The addition of aqueous HNO3 is important as it **prevents unreacted hydroxide ions from reacting with Ag+ ions to give ppt Ag2O**. 3. Halide anion produced would form a ppt with silver nitrate - Ag+(aq) + X-(aq) = AgX(s). **chlorobenzene gives no AgX ppt as it does not undergo nucleophilic substitution**.