Alkyl Halides - Result Question 42

####42. List-I contains reactions and List-II contains major products.

Q. >ONa

Match each reaction in List-I with one or more products in List-II and choose the correct option.

(2018 Adv.)

(a) P1,5;Q2;R3;S4

(b) P1,4;Q2;R4;S3

(c) P1,4;Q1,2;R3,4;S4

(d) P4,5;Q4;R4;S3,4

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Answer:

Correct Answer: 42. (a,d)

Solution:

For this reaction 1 and 4 are probable products.

Product 1 i.e., OHOH is formed due to substitution while product 4 i.e., 1 is formed due to elimination. A tertiary carbocation i.e, + formed during the reaction. Remember for 3 carbocation ions elimination product predominates.

For Q, i.e. χOMe+HBr

Correctly matched product for this reaction is 2 i.e., >Br. The reaction proceeds as

OMe+HBrMeOH+Br

For R i.e., χBr+NaOMe

Correctly matched product is 4 i.e., 1 . It is a normal elimination reaction and proceeds as

χBr+++ (Alkali in   alcohol) NaOMe

Column I Column II Explanation
P. NaOEt(2) O Et (strong
nucleophile) causes
dehydrohalogenation
of 3 alkyl halide
Q. Ia EtBr(3) 3 butoxide undergoes
SN reaction with 1
alkyl halide

3 alkyl halide preferes elimination.

For S i.e., χONa+MeBr

The correct match is 3 i.e., OMe. The reaction proceeds as



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