Question

An alkene (A) $C_{16}{H}_{16}$ on ozonolysis gives only one product (B) $C_8{H}_{8O}$. Compound (B) on reaction with $NaOH/I_2$ yields sodium benzoate. Compound (B) reacts with $KOH/N{H}_{2}N{H}_2$ yielding a hydrocarbon (C) $C_8{H}_{10}$. Write the structures of compounds (B) and (C). Based on this information two isomeric structures can be proposed for alkene (A). Deduce their structures. The isomer of (A) which on catalytic hydrogenation ($H_2/Pd-C$) gives a racemic mixture is:

• A. cis-2,3-diphenylbut-2-ene
• B. trans-2,3-diphenylbut-2-ene
• C. Both the above options are appropriate candidates for "A"; both isomers on catalytic hydrogenation gives a racemic mixture
• D. Neither trans-2,3-diphenylbut-2-ene nor cis-2,3-diphenylbut-2-ene

Answer 1

The correct option is B trans-2,3-diphenylbut-2-ene

Let us just depict the information as reactions for a better perspective:

The product has one additional oxygen atom and exactly half the number carbon and hydrogen atoms as (A).

From what is given, (B) undergoes iodoform reaction to yield sodium benzoate (which has 7 carbon atoms). So B is a methyl ketone or a compound that gets oxidized to a methyl ketone.
Also, (B) gives (C) - a hydrocarbon - $C_8{H}_{10}$ with Wolf-Kishner reduction. This all but confirms that (B) is a methyl ketone. In the iodoform reaction, the product has a benzene ring, which should have been a part of the structure of (B). Piecing all these information together, we get (B) and (C) as:


From here on, things have to be worked backwards. Given there is only one ozonolysis product, it implies that the carbon skeleton on either side of alkene double bond of (A) should be similar if not exactly same.


Since catalytic hydrogenation using $H_2/Pd–C$ is syn-addition, E or cis isomer will give meso products. Hence, the structure that yields a mixture of enantiomers is the trans isomer.