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# 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
7 months, 1 week 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. · 7 months, 1 week ago

All right, could you provide a full solution, just to be sure. · 7 months, 1 week 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! · 7 months ago

Yes! Thank you. · 7 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? · 7 months, 1 week ago