NEET Practice Questions (MCQs) with Answers & Solutions

(c) The given reaction follows S_N2 mechanism and S_N2 reactions are favored in polar aprotic medium like DMSO, DMF.. etc.

   CH₃CH₂CH₂Br + NaCN $\stackrel{\mathrm{DMF}}{\to }$ CH₃CH₂CH₂CN + NaBr

So, the correct option is (c).

Key Idea S_N1 reaction involves the formation of carbocation intermediate. More the stability of carbocation, more is the reactivity of alkyl/aryl halides towards S_N1 reaction.

The intermediate carbocations formed by given halides are as:

(a) C₆H₅CH(C₆H₅)Br $\to ({\mathrm{C}}_6{\mathrm{H}}_5{)}_2\stackrel{+}{\mathrm{C}}\mathrm{H}$ + Br^-

(b) C₆H₅CH(CH₃)Br $\to {\mathrm{C}}_6{\mathrm{H}}_5\stackrel{+}{\mathrm{C}}\mathrm{H}\left({\mathrm{CH}}_3\right)$ + Br^-

(c) C₆H₅C(CH₃)(C₆H₅)Br $\to \left({\mathrm{C}}_6{\mathrm{H}}_5{)}_2\stackrel{+}{\mathrm{C}}\right({\mathrm{CH}}_3)$ + Br^-

(d) C₆H₅CH₂Br $\to {\mathrm{C}}_6{\mathrm{H}}_5\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2$ + Br^-

The order of stability of these carbocations is as

$\left({\mathrm{C}}_6{\mathrm{H}}_5{)}_2\stackrel{+}{\mathrm{C}}\right({\mathrm{CH}}_3)>({\mathrm{C}}_6{\mathrm{H}}_5{)}_2\stackrel{+}{\mathrm{C}}\mathrm{H}>{\mathrm{C}}_6{\mathrm{H}}_5\stackrel{+}{\mathrm{C}}\mathrm{H}\left({\mathrm{CH}}_3\right)>{\mathrm{C}}_6{\mathrm{H}}_5\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2$

Thus, C₆H₅(CH₃)(C₆H₅)Br is most reactive towards S_N1 reaction

Anisole is produced by the reaction of phenol (C6H5OH) with sodium hydroxide (NaOH) and methyl iodide (CH3I). This is an example of the Williamson ether synthesis, where an alkoxide ion (from phenol and NaOH) reacts with an alkyl halide to form an ether.

Sodium phenoxide reacts with ethyl iodide in a nucleophilic substitution reaction to form phenetole (C₆H₅OC₂H₅). The phenoxide ion acts as a nucleophile, attacking the ethyl group and displacing the iodide ion.

(a) CH₃CH₂CH=CH₂$\underset{\mathrm{Anti}-\mathrm{Markownikoff}\text{'}\mathrm{s} \mathrm{rule}}{\overset{\mathrm{HBr}/{\mathrm{H}}_2{\mathrm{O}}_2}{\to }}$

      CH₃CH₂CH₂CH₂Br$\to$Bromo-butane (1$°$  product)

      CH₃CH₂CH₂CH₂Br$\underset{\mathrm{SN}2 \mathrm{reaction}}{\overset{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{ONa}}{\to }}$CH₃CH₂CH₂CH₂OC₂H₅  (williamson's synthesis)

Ethyl formate on reaction with excess of CH₃MgI, followed by hydrolysis gives a secondary alcohol i.e. isopropyl alcohol.

(a) The equimolar mixture of concentrated hydrochloric acid and anhydrous ZnCl₂ is called Lucas reagent. Lucas reagent is used to distinguish between 1$°$, 2$°$ and 3$°$ alcohols.

(d) In Lucas test when Lucas reagent is treated with 1$°$ ,2$°$  and 3$°$  alcohols, then turbidity appears, if turbidity is appeared immediately, then alcohol is tertiary. 2-methyl propan-2-ol is a tertiary alcohol. Hence, it reacts fastest with Lucas reagent.

The phenoxide ion (C₆H₅O^-) is stabilized by delocalization of the negative charge over the aromatic ring, making it more stable than the ethoxide ion (CH₃CH₂O^-). This increased stability results in a higher ionization constant for phenol.