Reactions of Alcohols - Free MCQ Practice Test with solutions, JEE Chemistry

Step 2 involves the formation of carbonium ion by the loss of weakly basic H₂O 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 C₂H₅OH

The tert-butyl cation is preferentially attacked by CH₃CH₂OH rather than bulky (CH₃)₃COH

Such questions can be solved by considering the relative basic character of their conjugated bases which for H₂O, C₂H₂, H₂CO₃, and C₆H₅OH are:

-OH, HC ≡ C⁻, HCO₃⁻, C₆H₅O⁻

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:

HCO₃⁻ < C₆H₅O⁻ < -OH < HC ≡ C⁻

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

H₂CO₃ > C₆H₅OH > H₂O > HC ≡ CH

Reagents used in the various steps indicate that the compound Z has an alcoholic group. This set of reactions is possible only when Z is CH₃CH(OH)CH₃.

In options a and c,Y cannot be converted back into d 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. CH₃ 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: a < c < b

Compound B is a Sec. alcohol since it gives blue colour in Victor-Meyer test and Sec alcohol are obtained by the action of CH₃MgI on an aldehyde other than formaldehyde.
So A is Acetaldehyde and B is Isopropyl Alcohol

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

CH₃OH + CH₃Mg-X → CH₄ + CH₃O-Mg-X

Methane is formed. 

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

C₂H₅OH + R-Mg-X → RH + C₂H₅OMgX (Alkane)

C₃H₇OH + R-Mg-X → RH + C₃H₇OMgX

P/Cl₂, SOCl₂, PCl₃, and PCl₅.

All OH, SH and terminal alkyne will react with Grignard reagent to produce CH₄.

Dehydration of cyclopentyl carbinol with conc. H₂SO₄ forms a cyclic compound. Write the structure of the product.

2° alcohol oxidizes but 3° does not.

When the vapour of a primary alcohol are passed over heated copper at 573 K dehydrogenation takes place and an aldehyde is formed. In case of tertiary alcohols dehydration take place.