Test: Reactions of Alcohols - JEE MCQ

Step 2 involves the formation of carbonium ion by the loss of weakly basic molecule. It is slowest step. Step 4 involves the conversion of an unstable (or intermediate) into a quite stable product, hence it is fastest step.

or

The reaction involves the formation of carbocation as intermediate. Hence more the stability of the carbocation, more will be the rate of reaction. Let us draw the structure of the corresponding carbocation and observe the relative stability of the four benzyl carbocations. The relative stability of the four carbocations is

Phenol is most acidic because its conjugate base is stabilised due to resonance, while the rest three compounds are alcohols, hence, their corrosponding conjugate bases do not exhibit resonance.

tert - Butyl alcohol is preferentially protonated because its corresponding carbocation is highly stable than that of

The tert-butyl cation is preferentially attacked by rather than bulky

Such questions can be solved by considering the relative basic character of their conjugated bases which for and are More the possibility for the dispersal of the negative charge, weaker will be the base. Thus the relative basic character of the four bases is

Due to resonance Thus the acidic character of the four corresponding acids will be

Reagents used in the various steps indicate that the compound has an alcoholic group. This set of reactions is possible only when is .

In options and cannot be converted back into by the given series of reactions.

Electron releasing groups etc) intensify the negative charge of phenoxide ion, i.e., destabilizes it hence decreases ionization of parent phenol, therefore decreases acidity. While electron withdrawing groups etc. increases acidity. is an electron-donating group so its acidity is the least. Salicylic acid has two electron-donating groups so its acidity is the highest.
Therefore, the correct order of increasing order of acidity is:

Compound is a Sec. alcohol since it gives blue colour in Victor-Meyer test and Sec alcohol are obtained by the action of on an aldehyde other than formaldehyde.
So A is Acetaldehyde and is Isopropyl Alcohol

Nitro group is strongly electron withdrawing group. At ortho position intra-molecular hydrogen bond reduces the acidic strength.

Ethyl Alcohol gives iodoform test
(yellow ppt)
No ppt.

and .

All OH, SH and terminal alkyne will react with Grignard reagent to produce .

Dehydration of cyclopentyl carbinol with conc. forms a cyclic compound. Write the structure of the product.

alcohol oxidizes but does not.

When the vapour of a primary alcohol are passed over heated copper at dehydrogenation takes place and an aldehyde is formed. In case of tertiary alcohols dehydration take place.