Question
Which statements about substitution reactions are correct?
I. The reaction between sodium hydroxide and 1-chloropentane predominantly follows an \({{\text{S}}_{\text{N}}}{\text{2}}\) mechanism.
II. The reaction between sodium hydroxide and 2-chloro-2-methylbutane predominantly follows an \({{\text{S}}_{\text{N}}}{\text{2}}\) mechanism.
III. The reaction of sodium hydroxide with 1-chloropentane occurs at a slower rate than with 1-bromopentane.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
▶️Answer/Explanation
B
The increasing order of the nucleophilic substitution by mechanism is – alkyl halide < alkyl halide < primary alkyl halide.
The halogenoalkane, 1-chloropentane is a primary halogenoalkane. Hence it will undergo SN2 mechanism.
In 2-chloro-2-methylbutane, the chlorine is attached to tertiary carbon hence this halide is a tertiary halide. It will predominantly undergo SN1 reaction
The reaction of 1 – chloropentane occurs at slower rate than 1 – bromopentane; the leaving ability of Bromine is greater than chlorine.
Question
Which reaction occurs via a free-radical mechanism?
A. \({{\text{C}}_2}{{\text{H}}_6} + {\text{B}}{{\text{r}}_2} \to {{\text{C}}_2}{{\text{H}}_5}{\text{Br}} + {\text{HBr}}\)
B. \({{\text{C}}_2}{{\text{H}}_4} + {\text{B}}{{\text{r}}_2} \to {{\text{C}}_2}{{\text{H}}_4}{\text{B}}{{\text{r}}_2}\)
C. \({{\text{C}}_4}{{\text{H}}_9}{\text{I}} + {\text{O}}{{\text{H}}^ – } \to {{\text{C}}_4}{{\text{H}}_9}{\text{OH}} + {{\text{I}}^ – }\)
D. \({{\text{(C}}{{\text{H}}_3})_3}{\text{CI}} + {{\text{H}}_2}{\text{O}} \to {{\text{(C}}{{\text{H}}_3}{\text{)}}_3}{\text{COH}} + {\text{HI}}\)
▶️Answer/Explanation
A
The free radical substitution reaction between Ethane and bromine happens in the presence of ultraviolet light – typically sunlight. This is a good example of a photochemical reaction – a reaction brought about by light. In organic chemistry, free-radical halogenation is a type of halogenation. This chemical reaction is typical of alkanes and alkyl-substituted aromatics under application of UV light.
Question
Halogenoalkanes can undergo \({{\text{S}}_{\text{N}}}{\text{1}}\) and \({{\text{S}}_{\text{N}}}{\text{2}}\) reactions with aqueous sodium hydroxide. Which halogenoalkane will react fastest with a \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) solution of aqueous sodium hydroxide?
A. 2-chloro-2-methylpropane
B. 2-iodo-2-methylpropane
C. 1-chlorobutane
D. 1-iodobutane
▶️Answer/Explanation
B
In primary halide, the halogen is attached to primary carbon. In secondary halide the halogen is attached to secondary carbon and in tertiary halide the halogen is attached to tertiary carbon atoms.
In A and B, the compounds are tertiary halides. Tertiary alkyl halides undergo SN1 reaction.
In D and B, the compounds are primary halides. Primary alkyl halides undergo SN2 reaction.
In the aqueous conditions the SN1 mechanism is much faster than the SN2, so tertiary haloalkanes react much faster than primary.
Also, the leaving ability of iodine is greater than chlorine, hence B undergoes fastest reaction.
Question
Propanenitrile can be prepared by reacting bromoethane with potassium cyanide. Which statement is not correct about the reaction between bromoethane and potassium cyanide?
A. The reaction is bi-molecular.
B. The reaction follows the \({{\text{S}}_{\text{N}}}{\text{2}}\) mechanism.
C. Homolytic fission occurs between the carbon-bromine bond in bromoethane.
D. The cyanide ion, \({\text{:C}}{{\text{N}}^ – }\), acts as a nucleophile.
▶️Answer/Explanation
C
Due to the difference in the electronegativities of and heterolytic fission will result into the formation of a positively and a negatively charges ion. being more electronegative shifts the bond pair of electrons towards itself and on breaking of bond it takes away the bond pair of electrons and acquires a negative charge. Thus, forming an anion. Whereas, being less electronegative than loses the pair of electron and acquires a positive charge, thus, forming a carbocation. Thus, heterolytic fission occurs between the carbon-bromine bond in bromoethane.
Bromoethane is a primary alkyl halide, hence reaction will be SN2.
Question
Which halogenoalkane reacts fastest with sodium hydroxide?
A. 1-iodobutane
B. 1-chlorobutane
C. 2-chloro-2-methylpropane
D. 2-iodo-2-methylpropane
▶️Answer/Explanation
D
In primary halide, the halogen is attached to primary carbon. In secondary halide the halogen is attached to secondary carbon and in tertiary halide the halogen is attached to tertiary carbon atoms.
In C and D, the compounds are tertiary halides. Tertiary alkyl halides undergo SN1 reaction.
In A and B, the compounds are primary halides. Primary alkyl halides undergo SN2 reaction.
SN1 mechanism is much faster than the SN2, so tertiary haloalkanes react much faster than primary.
Also, the leaving ability of iodine is greater than chlorine, hence D undergoes fastest reaction.