Question

Consider the following reactions

'A' is :

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

B gives iodoform and Tollens tests (CH3CHO); C is a ketone without a CH3CO group whose reduction product is a secondary alcohol (Lucas turbidity in about 5 minutes), so C = pentan-3-one.

Answer 1

The Correct Option is B

'A' (C₇H₁₄) is an alkene and ozonolysis gives two carbonyl compounds B + C.
B: positive iodoform test and silver mirror (Tollens) means B is an aldehyde with a CH₃CO- group: CH₃CHO (acetaldehyde).
C: no iodoform, so no CH₃CO- or CH₃CH(OH)- group. LiAlH₄ reduces C to D, which gives turbidity with Lucas reagent (anhydrous ZnCl₂/conc. HCl) within 5 minutes, indicating a secondary alcohol. So C is a ketone, C₅H₁₀O: pentan-3-one, CH₃CH₂COCH₂CH₃ (pentan-2-one would give iodoform). D = pentan-3-ol.
Joining CH₃CH= (from B) and =C(CH₂CH₃)₂ (from C): A = CH₃CH=C(CH₂CH₃)₂, 3-ethylpent-2-ene (C₇H₁₄), which is option (2).
Answer: (2) 3-ethylpent-2-ene, CH₃CH=C(C₂H₅)₂