Alkyl Halides - Practice Questions & MCQ

Halogen atoms are more electronegative than carbon, therefore, carbon-halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge. As we go down the group in the periodic table, the size of halogen atom increases. Fluorine atom is the smallest and iodine atom is the largest. Consequently, the carbon-halogen bond length also increases from  to .

Physical properties of Haloalkanes and Haloarenes:

(1) These are less soluble in  but more soluble in Organic solvents

(2) Their density follows the order: 

    Iodide > Bromide > Fluoride > Chloride

(3) Their boiling point follows the order: 

    Iodide > Bromide > Fluoride > Chloride

(4) Boiling point of Isomeric haloalkanes decreases with the increase in branching

(5) Boiling point of isomeric dihalobenzene is nearly the same and follows the order

    para > ortho > meta

The silver(I) salts of carboxylic acids react with halogens to give unstable intermediates which readily decarboxylate thermally to yield alkyl halides. The reaction is believed to involve homolysis of the C-C bond and a radical chain mechanism. In this reaction, ester is formed as a by-product. The reaction occurs as follows:

$\mathrm{RCOOAg}+\mathrm{X}_2 \xrightarrow{\mathrm{CCl}_4} \mathrm{R}-\mathrm{X}+\mathrm{CO}_2+\mathrm{AgX}$

For example: 

$\mathrm{CH}_3 \mathrm{COOAg}+\mathrm{Br}_2 \xrightarrow{\mathrm{CCl}_4} \mathrm{CH}_3 \mathrm{COOBr}+\mathrm{AgBr}$

Mechanism

Chain Initiation

$\mathrm{CH}_3 \mathrm{COOBr} \xrightarrow{\mathrm{RDS}} \mathrm{CH}_3 \mathrm{COO}^{\bullet}+\mathrm{Br}^{\bullet}$

Chain Propagation

$\mathrm{CH}_3 \mathrm{COO}^{\bullet} \longrightarrow{ }^{\bullet} \mathrm{CH}_3+\mathrm{CO}_2$

$^{\bullet} \mathrm{CH}_3+\mathrm{CH}_3 \mathrm{COOBr} \longrightarrow \mathrm{CH}_3 \mathrm{Br}+\mathrm{CH}_3 \mathrm{COO}^{\bullet}$

Chain Termination

$^{\bullet} \mathrm{CH}_3+\mathrm{Br}{ }^{\bullet} \longrightarrow \mathrm{CH}_3 \mathrm{Br}$ (major)

$\mathrm{CH}_3 \mathrm{COO}^{\bullet}+{ }^{\bullet} \mathrm{CH}_3 \longrightarrow \mathrm{CH}_3 \mathrm{COOCH}_3$ (minor)

$\mathrm{CH}_3 \mathrm{COO}^{\bullet}+\mathrm{Br}^{\bullet} \longrightarrow \mathrm{CH}_3 \mathrm{COOBr}$(minor) 

For example:

The rate of reaction with changing the alkyl group (R) in the above reaction varies as 

$1^{\circ}>2^{\circ}>3^{\circ}$

It is to be noted that with $\mathrm{I}_2$, a silver salt of carboxylic acid gives ester as the main product instead of an alkyl iodide.

$2 \mathrm{RCOOAg}+\mathrm{I}_2 \xrightarrow{\mathrm{CCl}_4} \mathrm{RCOOR}+\mathrm{CO}_2+2 \mathrm{AgI}$

• Reaction with NaCN: NaCN or KCN is ionic in nature i.e, Na⁺CN^-. Thus, the nucleophile in this case is CN^-. The reaction occurs as follows:
              $\mathrm{R}-\mathrm{CH}_2-\mathrm{X} \xrightarrow{\mathrm{NaCN}} \mathrm{R}-\mathrm{CH}_2-\mathrm{CN}+\mathrm{NaX}$
  For example:

                        $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{NaCN} \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}+\mathrm{NaBr}$

• Reaction with AgCN: AgCN is covalent in nature and as a result, the C atom is covalently bonded to Ag. Thus, the attack of the nucleophile in this case will take place from the Lone pairs over N atom

            $\mathrm{R}-\mathrm{CH}_2-\mathrm{X} \xrightarrow{\mathrm{AgCN}} \mathrm{R}-\mathrm{CH}_2-\mathrm{NC}+\mathrm{AgX}$

For example

        $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{AgCN} \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NC}+\mathrm{AgBr}$

• Reaction with NaNO₂: NaNO₂ or KNO₂ is ionic in nature and it breaks into Na⁺ and NO^-₂. Nitrite ion NO^-₂ with structure O=N-O- is an ambient nucleophile with electron pairs on both N & O. NO^-₂ ion having an excess of electrons on O thus, allows it to act as a nucleophile in preference to N.

The reaction occurs as follows:

$\mathrm{CH}_3 \mathrm{Br}+\mathrm{NaNO}_2 \rightarrow \mathrm{CH}_3-\mathrm{O}-\mathrm{NO}+\mathrm{NaBr}$

• Reaction with AgNO₂: AgNO₂ is covalent in nature and it does breaks into Ag⁺ and NO^-₂ ions. The O atom remains bonded to the Ag and hence the attack of the nucleophile takes place by the lone pairs over Nitrogen.

The reaction occurs as follows

$\mathrm{CH}_3 \mathrm{Br}+\mathrm{AgNO}_2 \rightarrow \mathrm{CH}_3-\mathrm{NO}_2+\mathrm{NaBr}$

Finkelstein Reaction
Finkelstein's reaction is a method of preparation of alkyl iodides from alkyl chlorides or alkyl bromides. In this reaction, alkyl chlorides or bromides are treated with NaI in the presence of acetone to form alkyl iodides. The reaction occurs as follows:

$\mathrm{R}-\mathrm{X}+\mathrm{NaI} \rightarrow \mathrm{R}-\mathrm{I}+\mathrm{NaX}$

We use NaI because it is soluble in acetone as it is covalent in nature. All other sodium halides are ionic in nature and thus not soluble. 

For example:

    $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow{\mathrm{NaI} / \text { Acetone }} \mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{I}$

Swarts Reaction

Halide exchange is also used for the preparation of alkyl fluorides by Swarts Reaction. Alkyl chloride/bromide is heated in presence of AgF, Hg₂F₂, CoF₂ or SbF₃ to give alkyl fluoride.

For example:

    $\mathrm{CH}_3-\mathrm{Br}+\mathrm{AgF} \longrightarrow \mathrm{CH}_3-\mathrm{F}+\mathrm{AgBr}$

The reaction of alcohols ROH with PCl₅ and PCl₃ yields an alkyl halide RCl. The reactions of alcohols with PCl₅, PCl₃ and SOCl₂ occurs as follows:

$\begin{aligned} & \mathrm{PCl}_5 \rightarrow \mathrm{POCl}_3+\mathrm{HCl}+\mathrm{RCl} \\ & \mathrm{PCl}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_3+\mathrm{HCl}+\mathrm{RCl} \\ & \mathrm{SOCl}_2 \rightarrow \mathrm{SO}_2+\mathrm{HCl}+\mathrm{RCl}\end{aligned}$

POCl₃ and H₃PO₃ are generated in liquid phase and hence they are very hard to separate while SO₂ and HCl are gases and thus they are easy to remove. Hence, for chlorination, we always use SOCl₂ as the best option among the given reagents.

Mechanism

The reactions occurs as follows:

$\begin{aligned} & \mathrm{RCOOH}+\mathrm{PCl}_3 / \mathrm{PCl}_5 / \mathrm{SOCl}_2 \rightarrow \mathrm{RCOCl} \\ & \mathrm{R}-\mathrm{OH}+\mathrm{PCl}_3 / \mathrm{PCl}_5 / \mathrm{SOCl}_2 \rightarrow \mathrm{R}-\mathrm{Cl}\end{aligned}$