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).

#Chemistry

Easy Math Editor

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Comments

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 - 5 years, 3 months ago

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

Rajdeep Bharati - 5 years, 3 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!

Ameya Daigavane - 5 years, 3 months ago

@Ameya Daigavane – Yes! Thank you.

Rajdeep Bharati - 5 years, 3 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?

Ameya Daigavane - 5 years, 3 months ago

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

Rajdeep Bharati - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$