From amongst the following alcohols the one that would react fastest with conc .HCl and anhydrous ZnCl2, is

  • Option 1)

    1-Butanol

  • Option 2)

    2-Butanol

  • Option 3)

    2-Methylpropan-2-ol

  • Option 4)

    2-Methylpropanol

 

Answers (3)

As we learnt in 

Test of alcohol Lucas test -

ZnCl_{2}+conc.HCl\:\:is\:called\:Lucas\:reagent

- wherein

1^{\circ}alcohol\rightarrow Cloudiness\:appears\:on\:heating

2^{\circ}alcohol\rightarrow Cloudiness\:appears\:after\:5\:minutes

3^{\circ}alcohol\rightarrow Cloudiness\:appears\:immediately

 

 Tertiary alcohol reacts fastest with Lucas reagent.


Option 1)

1-Butanol

This option is incorrect

Option 2)

2-Butanol

This option is incorrect

Option 3)

2-Methylpropan-2-ol

This option is correct

Option 4)

2-Methylpropanol

This option is incorrect

The reagent, conc.HCl and anhydrous ZnCl2 is Lucas reagent, which is used to distinguish between 1°, 2° and 3° alcohols.

3° alcohol + Lucas reagent → Immediate turbidity.
2° alcohol + Lucas reagent → Turbidity after 5 mins.

1° alcohol + Lucas reagent → No reaction.
Thus, the required alcohol is 2-methylpropan-2- ol, i.e.

option 2

 

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions