Question

Give reasons for the following:
(1) Phenol is more acidic than methanol.
(2) The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle (${109^\circ}28'$).
(3) $(C{H_3})C - O - C{H_3}$ on reaction with HI gives ${(C{H_3})_3}C - I$ and $C{H_3} - OH$ as the main products and not ${(C{H_3})_3} - OH$ and $C{H_3} - I$.

Answer 1

Hint: For (1): First identify the quality required by any compound to be acidic and also pay attention to the delocalisation of electrons in the entire structure.
For (2): Draw the structure of the C-O-H bond and focus on the distribution of bonds and lone pairs.
For (3): First identify the nucleophile and then try to put together the reaction by seeing the most stable intermediate that can be formed.

Complete step by step answer:
(1)Phenol is more acidic than methanol.
To check which one of them is more acidic we need to find that after losing an ${H^ + }$ ion which one of them will be more stable because the compound which gains more stability after losing an ${H^ + }$ ion has more tendency to lose it and hence is more acidic.
We first start with phenol. Phenol loses an ${H^ + }$ ion to form a phenoxide ion which is highly stable due to delocalisation of electrons, which means it is resonance stabilised.

Now let’s talk about methanol. After methanol loses an ${H^ + }$ ion it forms a methoxide ion which is not stable and thus readily takes up ${H^ + }$ ion again to form alcohol. The methoxide ion is unstable because there is no space for the electrons to be delocalised and hence no resonance.

                                      

Now after this above discussion we have seen that phenol is more acidic than methanol.

(2) The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle (${109^ \circ }28'$):
In C-O-H the oxygen atom contains 2 bond pairs and 2 lone pairs which are oriented in a tetrahedral shape around oxygen. These 2 lone pairs repel each other pushing the C-O bond closer to the O-H bond thus reducing the bond angle between them to ${108^ \circ }9'$ from the expected tetrahedral angle of ${109^ \circ }28'$ .
Bond pair-bond pair repulsion also works but it is quite less as compared to lone pair-lone pair repulsion.

                        

So, we conclude that the lone pair-lone pair repulsion causes the C-O-H bond to be slightly lesser than the tetrahedral angle.

(3) $(C{H_3})C - O - C{H_3}$ on reaction with HI gives ${(C{H_3})_3}C - I$ and $C{H_3} - OH$ as the main products and not ${(C{H_3})_3} - OH$ and $C{H_3} - I$.
When $C{H_3}C - O - C{H_3}$ reacts with HI, ${I^ - }$ acts as a nucleophile which attacks on ${(C{H_3})_3}^ + $ (tertiary) carbocation to form ${(C{H_3})_3} - I$ and $C{H_3}OH$ .

It does not form ${(C{H_3})_3} - OH$ and $C{H_3} - I$ as product because for such a product carbocation formed should be $C{H_3}^ + $ (primary). But in actuality the carbocation formed is ${(C{H_3})_3}^ + $ .

But we all know that tertiary carbocation is more stable than primary carbocation due to 3 times more inductive effect than primary carbocation. So, when tertiary carbocation of $(C{H_3}){C^ + }$ is formed the products are ${(C{H_3})_3}C - I$ and $C{H_3} - OH$ and not ${(C{H_3})_3} - OH$ and $C{H_3} - I$.

Note: For (1): Remember that a more stable compound will not lose hydrogen ion easily, so it won't be acidic. But a compound that gains stability after losing a hydrogen ion is acidic. Both of these are different things, don’t get confused.
For (2): Keep in mind that lone pair-lone pair repulsion is higher than bond pair-bond pair and so the angle decreases.
For (3): In such questions always remember what type of intermediate ion will be more preferred to form the final products. Tertiary carbocation is more stable than primary carbocation and not the other way round.