Match the starting materials given in Column I with the products formed by these (Column II) in the reaction with HI.

Question

Philoid · Accepted Answer

(i) CH₃—O—CH₃ - (d) CH₃-OH + CH₃-I

When methoxymethane(CH₃OCH₃) reacts with HI, methanol(CH₃OH) and methyl iodide(CH₃I) are produced.

If we want to see how it happened then here it goes. CH₃OCH₃ , the reactant breaks into CH₃O^- and CH₃⁺ (due to more electronegative Oxygen atom, negative charge is acquired) . Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen). Now, CH₃O^- attacks H⁺ to form CH₃OH and CH₃⁺ attacks I^- to form CH₃I as unlike charges attract. This is the explanation to how this reaction proceeds.

(ii)

- (e)

When 2-methoxypropane() reacts with HI, 2-hydroxypropane() and methyl iodide(CH₃I) is produced.

Now lets see what happens in the reaction. 2-methoxypropane() breaks into a stable 2° carbanion() and CH₃⁺ . This is because due to the –I inductive effect of the –CH₃ group the negative charge on the electronegative Oxygen atom increases. Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen). Now, the carbanion attracts the H⁺ to form 2-hydroxypropane() and CH₃⁺ attacks I^- to form CH₃I as unlike charges attract. This is the explanation to how this reaction proceeds.

(iii)

- (b)

When tert-butyl methyl ether() reacts with HI, tert-butyl iodide ()is produced along with methanol (CH₃OH).

Now, lets see what happens in this reaction.

tert-butyl methyl ether() breaks into a very stable tertiary(3°) carbocation, which is the tert-butyl carbocation() and methoxy ion(CH₃O^-). Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen).

As we know –CH₃ groups show –I inductive effect, due to the presence of 3 such methyl groups the delta positive charge on the carbocation starts diminishing due to which the carbocation becomes highly stable. Thus it would now prefer the attack of the I^- ion and it would result in the formation of tert-butyl iodide (). Along with that the methoxy ion(CH₃O^-) reacts with H⁺ to form methanol (CH₃OH) as unlike charges attract. This is the explanation to how this reaction proceeds.

Note:

tert stands for tertiary.

(iv)

- (a)

When anisole () reacts with HI, phenol () and methyl iodide (CH₃I) are produced.

Lets see how the reaction goes. Anisole breaks down into phenoxide ion (C₆H₅O^-). As we know, phenoxide ion is resonance stabilized , the formation of the phenoxide ion is highly preferred. So we get to know that any substance always want to acquire a state where it gains maximum stability.

Along with that CH₃⁺ carbocation is also formed. Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen).

Now, the phenoxide ion() reacts with H⁺ to form a well known compound called phenol (). Also CH₃⁺ carbocation andI^- interact amongst themselves to form methyl iodide(CH₃I) as unlike charges attract. This is the explanation to how this reaction proceeds.