We need to find an alkene that meets three criteria: (A) Molecular formula is C
8H
16.
(B) Gives acetone (CH
3COCH
3) upon ozonolysis (followed by reductive workup, usually implied).
(C) Is optically active (possesses a chiral center and is not a meso compound). Let's analyze each option:
- (A) 2,3-Dimethylhex-2-ene:
Formula: C8H16 (Correct).
Ozonolysis: Cleaves the double bond. C(CH3)2= gives acetone. =C(CH3)CH2CH2CH3 gives 2-pentanone. So, it yields acetone.
Optical Activity: No chiral center. Carbon atoms are either sp2 hybridized or have two identical groups attached (e.g., CH3 groups on C2, H atoms on CH2s). It is optically inactive. - (B) 3,4-Dimethylhex-2-ene:
Formula: C8H16 (Correct).
Ozonolysis: C(CH3)2= gives acetone. =CHCH(CH3)C2H5 gives 3-methylpentan-2-al. So, it yields acetone.
Optical Activity: The carbon atom at position 4 (C4) is bonded to four different groups: H, CH3, C2H5 (ethyl), and the C=CH(C(CH3)2) group.
Thus, C4 is a chiral center. The molecule is optically active. - (C) 3,4-Dimethylhex-3-ene:
Formula: C8H16 (Correct).
Ozonolysis: Cleaves the double bond. C(C2H5)(H)= gives propanal. =C(CH3)CH(CH3)2 gives 3-methylbutan-2-one. Does not yield acetone. - (D) 3-Ethyl-2-methylpent-2-ene:
Formula: C8H16 (Correct).
Ozonolysis: C(CH3)2= gives acetone. =C(C2H5)2 gives 3-pentanone. So, it yields acetone.
Optical Activity: No chiral center. The molecule is optically inactive.
Only option
(B) satisfies all three conditions: has the formula C
8H
16, yields acetone upon ozonolysis, and is optically active due to the chiral center at C4.
