A (C 6 H 12 ) 
B + C
(C 6 H 13 CI)
B 
D (isomer of A)
D 
E (it gives negative test with Fehling solution but responds to iodoform test).
A 
F + G (both give positive Tollen’s test but do not give iodoform test).
F + G 
HCOONa + a primary alcohol
Identify to A to G.
Text Solution
Verified by ExpertsA
A = 
B = 
C = 
D = 
E = 
F & G are 
Sol. In F and G, both are aldehyde because they give positive Tollen’s test and do not give iodoform test. These aldehydes give Cross Cannizzaro’s reacton, so they do not have α -hydrogen atoms. In Cross Cannizzaro’s reaction, HCOONa is formed along with p-alcohol, So in these, an aldehyde is HCHO and another is (CH 3 ) 3 CCHO, F and G are obtained by ozonolysis of A, therefore, compound ‘A’ is CH 2 =CH–C(CH 3 ) 3 . Compound ‘A’ on reaction with HCI gives compound B and C which have C 6 H 13 CI molecular formula. Thus
+ 
Comound ‘A’ Compound ‘C’ Compound ‘B’
Compound ‘B’ gives ‘D’ on dehydrohalogenations with alc. KOH
Compound ‘E’ has methyl ketonic grooup (–COCH 3 ), so it gives positive iodoform test and does not give the test with Fehling solution due to absence of —CHO group.
Compound ‘A’ on ozonolysis to give compounds F and G as follows :
+ 
Compound ‘A’ Compound ‘G’ Compound ‘F’
Compound G and F give crossed Cannizzaro’s reaction with conc. NaOH solution.
+ CH 2 O + conc. NaOH 
Compound ‘G’ Compound ‘F’ Primary alcohol
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