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.


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