Conversion

Two isomeric alkyl bromides (A) & (B), CX5HX11Br\ce{C5H11Br} yield the following results in the laboratory. (A), on treatment with alcoholic KOH\ce{ KOH} gives (C) & (D). CX5HX10\ce{C5H10} . (C), on Ozonolysis gives HCHO and 2-Methyl propanal. (B) on treatment with alcoholic KOH gives (D) & (E), CX5HX10\ce{C5H10} . All the compounds (C), (D), (E) on catalytic hydrogenation give (F), CX5HX12\ce{C5H12} . Deduce the structures from (A) to (F).

Note by Rajdeep Bharati
3 years, 8 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.

Ameya Daigavane - 3 years, 8 months ago

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

Rajdeep Bharati - 3 years, 8 months ago

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

Ameya Daigavane - 3 years, 7 months ago

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@Ameya Daigavane Yes! Thank you.

Rajdeep Bharati - 3 years, 7 months ago

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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?

Ameya Daigavane - 3 years, 8 months ago

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

Rajdeep Bharati - 3 years, 8 months ago

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