Answer :


(a) When phenol is reacted with bromine water, such is the effect of –OH group on benzene that instead of producing mono bromo derivative on the ring, the reaction readily proceeds to form tri bromo derivative that is 2,4,6-tribromophenol which is observed as a white precipitate.



Whereas, no such perceptible changes are found when primary alcohols like propanol reacts with bromine water.


(b) Ethanol being an alcohol having the acidic hydrogen (directly attached to electronegative O atom) reacts with sodium metal to give sodium ethoxide along with the liberation of a colourless, odourless gas which rekindles a glowing splinter with a characteristic ‘pop’ sound. The gas liberated is H2.



No such observation is found when dimethyl ether is reacted with sodium metal.


(c) 1° , 2° , 3° alcohols are distinguished by Lucas test.


Here, the alcohols are mixed in an equimolar mixture of ZnCl2 and HCl is the reagent.


If the alcohol is primary, no visible reaction or observation is found.


If the alcohol is secondary, an oily layer is formed in 3-5 minutes.


If the alcohol is tertiary, the oily layer is formed instantaneously.


From the , we know that, propanol is a 1° alcohol and 2-methyl-2-propanol is a 3° alcohol.


So, propanol would give no perceptible changes at room temperature to Lucas test, whereas 2-methyl-2-propanol forms an oily layer instantly.



OR


(i) CH3-CH2-O-CH3 + HI


Now lets see what happens in the reaction. CH3-CH2-O-CH3breaks into a stable CH3-CH2-O- and CH3+ . This is because due to the –I inductive effect of the –2CH3 group the negative charge on the electronegative Oxygen atom increases. So,CH3-CH2-O- formation is favoured.


Similarly HI breaks into H+ and I-(as Iodine is more electronegative than Hydrogen). Now, CH3-CH2-O- attracts the H+ to form and CH3+ attacks I- to form CH3I as unlikecharges attract. This is the explanation to how this reaction proceeds.


(ii) If we treat phenol with carbon disulphide(CS2) at 273 K, the reaction is somewhat suppressed and instead of a tribromo derivative (formed when it reacts with bromine water), a monobromo derivative is produced because CS2 is a less polar solvent as compared to bromine water. p-bromophenol is produced as a major product and o-bromophenol is also produced but as a minor product.



(iii) Lets understand the mechanism of the reaction first.


When a +R (Positive resonance effect) showing functional group joins benzene ring, these groups have their lone pairs with them and they participate in resonance or in delocalization. They show electron releasing tendency. The electrons are released towards benzene ring through delocalization. Eg. –OCH3,–NH2, -OR, R, OH, where R is an alkyl group. These are called ortho, para directing groups.


We know most of the reactions in benzene proceed via electrophilic substitution reaction.


So, how can we generate this electrophile?


For the generation of electrophile, we need Lewis acid(AlCl3).Since the Lewis acid need electrons, it gets them when the CH3CO-Cl bond breaks and AlCl3 accepts the Cl- and we get AlCl4- along with the generated electrophile CH3CO+.


Now, this electrophile attacks the benzene ring and as already an ortho –para directing group, -OCH3 is present, the electrophile attaches itself to these positions and through resonance by removal of a H+ from the ring, we get p-methoxy-acetophenone (major product) and o-methoxy-acetophenone (minor product).


This reaction is termed as Fridel Craft Acylation as CH3CO+ was the acyl group.


Now, we are left with a H+ and AlCl4- . They combine among themselves thereby forming AlCl3 and HCl.


Since AlCl3 is formed again, it is said to be a catalyst.



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