# Conversion

Two isomeric alkyl bromides (A) & (B), $\ce{C5H11Br}$ yield the following results in the laboratory. (A), on treatment with alcoholic $\ce{ KOH}$ gives (C) & (D). $\ce{C5H10}$. (C), on Ozonolysis gives HCHO and 2-Methyl propanal. (B) on treatment with alcoholic KOH gives (D) & (E), $\ce{C5H10}$. All the compounds (C), (D), (E) on catalytic hydrogenation give (F), $\ce{C5H12}$. Deduce the structures from (A) to (F).

Note by Rajdeep Bharati
3 years, 11 months ago

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Okay, so the key step here is the ozonolysis of C. Can you figure out the structure of C from the given products? Once we have C, D is simple to find. Both C and D have a double bond, but in different positions. As C, D are formed from E2 elimination in A, the double bonds must be in positions adjacent to each other (in C and D). The same logic enables us to figure out E, and hence, B and F. If you require a full solution, tell me.

- 3 years, 11 months ago

All right, could you provide a full solution, just to be sure.

- 3 years, 11 months ago

Sorry for the late reply. Here it is- From left to right, A to E I used this to generate the image. Really great tool!

- 3 years, 11 months ago

Yes! Thank you.

- 3 years, 11 months ago

Are you sure the question is right? There is no mention of B beyond the first sentence. Also, ozonolysis on C gives no oxygen in the products D and E? Which compounds do the molecular formulas represent?

- 3 years, 11 months ago

Thank you for pointing out, sir. I have edited the question.

- 3 years, 11 months ago