Daily Olympiad: Chemistry - Organic Chemistry [20260512]

Challenge yourself with today's NEET practice! This test covers 'Organic Chemistry' for Chemistry (NEET - Dropper). Level: Hard | Duration: 45 mins.

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1. Which of the following compounds will undergo Friedel-Crafts alkylation most readily with CH3Cl and AlCl3?

A. p-Nitroanisole B. p-Bromoanisole C. p-Chloroanisole D. p-Fluoroanisole

Solution

Correct: D

p-Fluoroanisole (C6H5OCH3 with -F) has the least deactivating group due to +R effect of -OCH3 and -I effect of -F being weaker than other halogens. Electron-rich aromatic rings undergo Friedel-Crafts more easily.

2. The correct IUPAC name of the compound CH2=C(Cl)-CH3 is:

A. 3-Chlorobut-1-ene B. 2-Chloropropane C. 1-Chloroprop-2-ene D. None of the above

Solution

Correct: C

The correct name is 1-chloroprop-2-ene, assigning the lowest number to the double bond (numbering from left gives position 1 for Cl and 2 for double bond).

3. In the Cannizzaro reaction, the correct order of reactivity for the following species is:

A. Formaldehyde > Benzaldehyde > Acetophenone B. Acetophenone > Benzaldehyde > Formaldehyde C. Benzaldehyde > Formaldehyde > Acetophenone D. All have equal reactivity

Solution

Correct: A

Formaldehyde is most reactive due to lack of alpha-hydrogen, allowing self-oxidation-reduction. Benzaldehyde is less reactive than formaldehyde but more than ketones.

4. Which of the following will show the fastest rate of SN1 reaction with AgNO3?

A. (CH3)3CCl B. CH2=CHCH2Cl C. C6H5CH2Cl D. C6H5CHClC6H5

Solution

Correct: A

(CH3)3CCl forms a stable tertiary carbocation (tert-butyl carbocation) which is the rate-determining step in SN1.

5. Which of the following will give a positive iodoform test?

A. CH3COCH3 B. C6H5CHO C. CH3CH2CH2OH D. CH3CH2CH2NH2

Solution

Correct: A

CH3COCH3 (acetone) gives a positive iodoform test due to the presence of CH3-CO- group. Alcohol needs to be secondary (like CH3CH(OH)CH3) to give iodoform.

6. The correct order of basicity in liquid NH3 is:

A. NH2^- > (CH3)2NH > CH3NH2 B. (CH3)2NH > NH2^- > CH3NH2 C. NH2^- > CH3NH2 > (CH3)2NH D. All have equal basicity

Solution

Correct: A

Amide ion (NH2^-) is strongest base in non-aqueous solvents like liquid NH3. Aliphatic amines (especially tertiary) have higher basicity than ammonia.