Haloalkanes and HaloarenesNEET Chemistry · Class 12 · NCERT Chapter 6

Solution

The reaction of ethane (${\text{CH}}_3{\text{CH}}_3$) with bromine in the presence of ${\text{FeBr}}_3$ results in the formation of bromoethane (${\text{CH}}_3{\text{CH}}_2\text{Br}$). Since the reaction is a substitution, the correct product is ${\text{CH}}_3{\text{CH}}_2\text{Br}$, and the correct option is (c).

Solution

Racemisation occurs in compounds with a chiral carbon, where the hydroxide ion can attack from either side, leading to a mixture of enantiomers. Among the given options, only (iii) and (iv) have chiral carbons, so option (c) is correct.

Solution

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Solution

I and IV can be used due to presence of anhydrous ZnCl2 (III) gives alkyl halide due to formation of more stable carbocation.

Solution

Benzene reacts with chlorine in presence of anhydrous FeCl 3 to give chlorobenzene

Solution

The reaction sequence involves the formation of a bromobenzene intermediate from benzene diazonium chloride with ${\text{Cu}}_2{\text{Br}}_2/\text{HBr}$, followed by conversion to phenylmagnesium bromide with $\text{Mg}/\text{dry ether}$, and finally hydrolysis to give benzene. NCERT XII chapter Haloalkanes and Haloarenes covers this sequence, so option (c) is correct.

Solution

Reactivity towards SN1 depends upon stability of carbocation. Order of stability is Hence is most reactive - 25 - NEET (UG)-2024 (Code-Q1)

Solution

Tertiary alcohols react instantaneously with Lucas reagent and gives immediate turbidity. In case of tertiary alcohols, they form halides easily with Lucas reagent (conc. HCl and ZnCl2)

Solution

These reactions are preparation of haloalkanes from alcohols. 3 3 3 (A) 3ROH PCl 3RCl H PO+ ⎯⎯ → + 53 (B) ROH PCl RCl HCl POCl+ ⎯⎯ → + + A and B are H3PO3 and POCl3 respectively.

Solution

• Step-I is SN reaction with nucleophile. • Step-II will give amide. • Step-III is Hoffmann bromamide degradation reaction.

Solution

Sol. Rate of $S_{N}2$ reaction of ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{I}$ is faster than ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{Cl}$.

Because iodine is a good leaving group due to large size of iodine. Which stabilises the ${\text{I}}^{-}$ ion.