ExamsMy practice
Ques. 1JEE Main 2024, 30 January (Morning Shift)
Compound A formed in the following reaction reacts with B to give the product C. Find out A and B. CH3-C CH + Na A B CH3- (C)C C-CH2- CH3 |CH2 + NaBr
- A = CH3-C -C Na+, B = CH3-CH2-CH2-Br
- A = CH3-CH=CH2, B = CH3-CH2-CH2-Br
- A = CH3-C -C Na+, B = CH3-CH2-CH3
- A = CH3-CH2-CH3, B = CH3-C CH
Answer
(1) A = CH3-C -C Na+, B = CH3-CH2-CH2-Br
Solution
Step 1 — terminal ≡C–H is acidic (sp carbon):
CH3-C CH + Na (A) sodium propynideCH3-C -C Na+ + 12H2
Step 2 — propynide ion attacks a 1° alkyl halide (SN2):
CH3-C -C Na+ + (B)CH3CH2CH2Br
CH3-C C-CH2CH2CH3 + NaBr
∵ C needs a 3-carbon chain added and NaBr is formed ⇒ B = 1-bromopropane
Answer: (1)