(I) `CH_(3)-CH_(2)-CH_(2)-CH_(2)- overset(overset(O)(||))(C )-H`
(II) `CH_(3)-CH_(2)-CH_(2)- overset(overset(O)(||))(C )-CH_(3)`
(III) `CH_(3)-CH_(2)- underset(underset(O)(||))(C )-CH_(2)-CH_(3)`
(IV) `CH_(3)- underset(underset(CH_(3))(|))(C )H-CH_(2)- underset(underset(O)(||))(C )-H`
Which of the following pairs are not functional group isomers?

Options

A. II and III
B. II and IV
C. I and IV
D. I and II

Answer

A, C

Explanation

(B) (II) `CH_(3)-CH_(2)-CH_(2)- overset(overset(O)(||))(C )-CH_(3)` is ketone.
(IV) `CH_(3)-underset(underset(CH_(3))(|))(C )H-CH_(2)-underset(underset(O)(||))(C )-H` is aldehyde
but, (II) and (IV) are not positionisomer but they are group isomers.
(D) (I) `CH_(3)-CH_(2)-CH_(2)-CH_(2)-overset(overset(O)(||))(C )-H` is aldehyde.
(II) `CH_(3)-CH_(2)-CH_(2)-overset(overset(O)(||))(C )-CH_(3)` is ketone.
Both are group isomer. (III) and (IV) are not position isomer
(A) (II) `CH_(3)-CH_(2)-CH_(2)- overset(overset(O)(||))(C )-CH_(3)` is ketone
(III) `CH_(3)-CH_(2)-underset(underset(O)(||))(C )-CH_(2)-CH_(3)` is ketone
Both are same 5 carbon containing ketone so they are position isomer, but not group isomer.
(C ) (I) and (IV) both are aldehyde but not position isomers. The branch is present in (I), so (I) and (IV) not taken as position isomer.
Inshort, (I) Aldehyde (II) Ketone (III) Ketone (IV) Aldehyde
(A) II, III- They are not group isomers
(B) II, IV- They are group isomers
(C ) I, IV- They are not group isomers
(D) I, II- They are not group isomers