Ncert Solutions Chemistry Class 12th

11.46  In this reaction, when propene reacts with the given reagent then the double bond of propene breaks down with charges on them. So, H⁺ gets placed on the carbon which already has two hydrogen atom and OH^- gets substituted on center carbon because it has the more positive charge which attracts OH^-. Thus we get propene-2-or as a

1. In this reaction, when Methyl ( 2-oxocyclohexyl) ethanoate reacts with the given reagent then the double bond between the oxygen atom and cyclohexyl gets breaks down, such that O has a negative charge and that particular carbon will have a positive charge on it. So, to neutralize it, H⁺ gets substituted to that carbon and with O^- to form the structure of alcohol. Thus we get Methyl (2-hydroxycyclohexyl) ethanoate as a
2. In this reaction, when 2-Methylbutanal reacts with the given reagent NaBH₄, then the double bond between carbon and oxygen gets to break down in the above aldehyde compound, where carbon has positive charge and oxygen will have a negative charge on So, H⁺ gets placed on the carbon to complete its octet and H⁺ gets substituted on O^- to form OH, i.e, an alcohol. Thus we get 2-Methylbutan-1-ol as a product.

11.43

The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para- substituted compound.

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

The mechanism is given below:

2- The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

Nitration of anisole gives p-nitroanisole as the major product.

The mechanism is given below:

3- The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

Bromination of anisole in ethanoic acid medium gives p-bromoanisole as the major product.

4-The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being
comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted compound.

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

The Friedel-Craft's acetylation of anisole gives 4- methoxyacetophenone as the major product

The mechanism is given below:

11.42 

The reaction of HI with methoxymethane yields two different sets of products depending upon the initial amount of HI taken.

(i) When equal moles of HI and methoxymethane are taken, a mixture of methyl alcohol and methyl iodide is

The mechanism is given below:

In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.

In the second step of the reaction, the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN² pathway.

(ii) If an excess of HI is used the methyl alcohol formed in Step II is also converted into methyl iodide by the following mechanism : -

In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.

In the second step of the reaction the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN2 pathway.

In the third step of the reaction Methyl alcohol formed above reacts with hydrogen iodide to extract a proton to give the protonated methyl alcohol which finally reacts in the fourth step with the iodide ion via SN² pathway to give methyl iodide and water as the product.