Haloalkanes and Haloarenes

1. (i) $
(CH_3)_2CHCH(Cl)CH_3$ is 2-chloro-3-methylbutane. The carbon bearing Cl is attached to two carbons, so it is a secondary alkyl halide.

2. (ii) $
CH_3CH_2CH(CH_3)CH(C_2H_5)Cl$ is 2-chloro-3-methylpentane. The halogen is on a carbon attached to two other carbons, so it is a secondary alkyl halide.

3. (iii) $
CH_3CH_2C(CH_3)_2CH_2I$ is 1-iodo-3,3-dimethylbutane. The iodine is on a terminal carbon, so it is a primary alkyl halide.

4. (iv) $
(CH_3)_3CCH_2CH(Br)C_6H_5$ is 2-bromo-1,1-dimethyl-3-phenylbutane. The bromine is on a benzylic carbon attached to two carbons, so it is a secondary benzylic halide.

5. (v) $
CH_3CH(CH_3)CH(Br)CH_3$ is 2-bromo-3-methylbutane. The halogen-bearing carbon is attached to two carbons, so it is a secondary alkyl halide.

6. (vi) $
CH_3C(C_2H_5)_2CH_2Br$is **1-bromo-2-ethylbutane**. Since Br is on a$
CH_2$ group, it is a primary alkyl halide.

7. (vii) $
CH_3C(Cl)(C_2H_5)CH_2CH_3$ is 3-chloro-3-methylpentane. The carbon bearing Cl is attached to three carbons, so it is a tertiary alkyl halide.

8. (viii) $
CH_3CH=C(Cl)CH_2CH(CH_3)_2$ is 3-chloro-5-methylhex-2-ene. The halogen is directly attached to a double-bond carbon, so it is a vinyl halide.

9. (ix) $
CH_3CH=CHC(Br)(CH_3)_2$ is 4-bromo-4-methylpent-2-ene. The bromine is on an allylic carbon, so it is an allylic halide.

10. (x) $p$-Cl$
C_6H_4CH_2CH(CH_3)_2$ is 1-chloro-4-isobutylbenzene by common style, but IUPAC is 1-chloro-4-(2-methylpropyl)benzene. Since Cl is directly on the aromatic ring, it is an aryl halide.

11. (xi) $m$-ClCH$_2C_6H_4CH_2C(CH_3)_3$ is 1-(chloromethyl)-3-(2,2-dimethylpropyl)benzene. The Cl is on a benzylic $CH_2$, so it is a benzyl halide.

12. (xii) $o$-Br-$C_6H_4CH(CH_3)CH_2CH_3$ is 1-bromo-2-(butan-2-yl)benzene in common terms, but importantly Br is on the ring, so it is an aryl halide.

The chapter’s answer list gives these structures in the form of names. Writing them concisely:

- (i) 2-Chloro-3-methylpentane
- (ii) 1-Chloro-4-ethylcyclohexane
- (iii) 4-tert-Butyl-3-iodoheptane
- (iv) 1,4-Dibromobut-2-ene
- (v) 1-Bromo-4-sec-butyl-2-methylbenzene

These are the required IUPAC structures/names corresponding to the textbook exercise answer set.

From the chapter’s answer key for Intext Question 6.3, the structures are:

- (i) $
ClCH_2CH_2CH_2Cl$
- (ii) $
ClCH_2CHClCH_3$
- (iii) $
Cl_2CHCH_2CH_3$
- (iv) $
CH_3CCl_2CH_3$

These are the required organic halogen compound structures.

The textbook answer for the order of dipole moment is that $
CH_2Cl_2$ has the highest dipole moment among the given compounds.

A hydrocarbon of formula $
C_5H_{10}$ that does not react with chlorine in the dark but gives only one monochloro product in bright sunlight must be a cycloalkane with all hydrogens equivalent. The chapter’s expected identification is cyclopentane.

The compound $
C_4H_9Br$ has four structural isomers:

1. 1-Bromobutane
2. 2-Bromobutane
3. 1-Bromo-2-methylpropane
4. 2-Bromo-2-methylpropane

These are the isomers required by the exercise.

The chapter asks for the preparation of 1-iodobutane from three starting materials.

### (i) From 1-butanol
Use hydrogen iodide or KI in 95% orthophosphoric acid:

$$CH_3CH_2CH_2CH_2OH + HI \rightarrow CH_3CH_2CH_2CH_2I + H_2O$$

### (ii) From 1-chlorobutane
Use NaI in dry acetone by Finkelstein reaction:

$$CH_3CH_2CH_2CH_2Cl + NaI \rightarrow CH_3CH_2CH_2CH_2I + NaCl$$

### (iii) From but-1-ene
First add HI to but-1-ene:

$$CH_2=CHCH_2CH_3 + HI \rightarrow CH_3CH(I)CH_2CH_3$$

To obtain 1-iodobutane, a better school-level route is to first make 1-butanol by anti-Markovnikov hydration is not in this chapter, so using the chapter content, the direct textbook-preferred preparation of alkyl iodides is from alcohols or halides. Thus for but-1-ene, the standard expected pathway is via conversion to 1-butanol by an accepted intermediate, then reaction with HI. If only chapter methods are used, the direct hydrohalogenation of but-1-ene gives the Markovnikov iodide, not 1-iodobutane.

Ambident nucleophiles are nucleophiles that have two different nucleophilic centres and can attack through either of them.

The chapter gives cyanide ion and nitrite ion as examples:

- CN⁻ can attack through carbon to form alkyl cyanides:
$$R-X + KCN \rightarrow R-CN + KX$$
- AgCN gives attack through nitrogen to form isocyanides:
$$R-X + AgCN \rightarrow R-NC + AgX$$

Similarly, NO₂⁻ can link through oxygen to give alkyl nitrites or through nitrogen to give nitroalkanes.

So, an ambident nucleophile is one that can react from two alternative atoms.

For an **$S_N2$ reaction, the faster compound is the one with the better leaving group and less steric hindrance.

- (i) Between $
CH_3Br$** and **$
CH_3I$**, **$
CH_3I$ reacts faster because iodide is a better leaving group than bromide.
- (ii) Between $(CH_3)_3CCl$ and $
CH_3Cl$**, **$
CH_3Cl$ reacts faster because it is least hindered; tert-butyl chloride is highly hindered.

So the answers are $
CH_3I$** and **$
CH_3Cl$**.

The textbook asks for the alkenes formed on dehydrohalogenation with sodium ethoxide in ethanol.

### (i) 1-Bromo-1-methylcyclohexane
Elimination gives mainly 1-methylcyclohexene as the major alkene.

### (ii) 2-Chloro-2-methylbutane
Possible alkenes:
- 2-Methylbut-1-ene
- 2-Methylbut-2-ene

The major alkene is 2-methylbut-2-ene because it is the more substituted alkene.

### (iii) 2,2,3-Trimethyl-3-bromopentane
Elimination can occur from a β-carbon giving the alkene with the double bond between C3 and the adjacent carbon. The major product is the more substituted alkene. In such cases, write all possible alkenes formed by removing a β-hydrogen adjacent to the carbon bearing Br, and choose the most substituted one as major.

The general rule from the chapter is Zaitsev’s rule: the preferred product is the alkene with the greater number of alkyl groups attached to the doubly bonded carbon atoms.

The question asks for the conversion routes; the chapter expects reagents and steps. A concise set of standard school-level conversions is:

- Use appropriate haloalkanes, halogen exchange, elimination, addition, or Grignard/ Wurtz reactions depending on the target.
- For example:
- Alcohol to haloalkane: with HX, PCl5, PBr3, or SOCl2.
- Haloalkane to alcohol: with aqueous KOH.
- Haloalkane to alkene: with alcoholic KOH.
- Haloalkane to higher alkane: Wurtz reaction with Na/dry ether.
- Halide exchange: Finkelstein for iodides, Swarts for fluorides.
- Aryl amine to aryl halide: diazotisation followed by Sandmeyer reaction.

If a specific conversion is requested, the exact reagents are chosen from these chapter methods.