NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

NEET 2017

MCQ (Single Correct Answer)
The correct statement regarding electrophile is
A
electrophile is a negatively charged species and form a bond by accepting a pair of electrons from another electrophile
B
electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
C
electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
D
electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile.

Explanation

Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile.
2

NEET 2017

MCQ (Single Correct Answer)
The IUPAC name of the compound
A
5-formylhex-2-en-3-one
B
5-methyl-4-oxohex-2-en-5-al
C
3-keto-2-methylhex-5-enal
D
3-keto-2-methylhex-4-enal

Explanation

3

NEET 2016 Phase 1

MCQ (Single Correct Answer)
For the following reactions :

(A)  CH3CH2CH2Br + KOH $$ \to $$
                 CH3CH $$=$$ CH2 + KBr + H2O

Which of the following statements is correct?
A
(A) is elimination, (B) and (C) are substitution reactions.
B
(A) is substitution, (B) and (C) are addition reactions.
C
(A) and (B) are elimination reactions and (C) is addition reaction.
D
(A) is elimination, (B) is substitution and (C) is addition reaction.

Explanation

(A) Saturated compound is converted into unsaturated compound by removal of group of atoms hence, it is an elimination reaction.

(B) —Br group is replaced by —OH group hence, it is a substitution reaction.

(C) Addition of Br2 converts an unsaturated compound into a saturated compound hence, it is an addition reaction.
4

NEET 2016 Phase 2

MCQ (Single Correct Answer)
Which among the given molecules can exhibit tautomerism?
A
III only
B
Both I and III
C
Both I and II
D
Both II and III

Explanation

$$\alpha $$-hydrogen at bridge carbon never participates in tautomerism. Thus, only (III) exhibits tautomerism.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12