# What will the final product of this synthetic sequence be?

What is the major product formed in the following reaction?

NBS should do allylic bromination but where will it attack in the larger ring or smaller one ?

## 1 Answer

Potassium carbonate is not going to give you hydrolysis of a tertiary bromide. It will however close the halohydrin to form an epoxide (1). This intermediate undergoes Lewis acid catalyzed rearrangement to form the spiro compound that is choice (D).

Note that 1 or the carbocation could undergo certain elimination reactions to form (B) and (C). These are probably minor products.

• Yes, this was what I initially thought too. The proposed SN1 reaction on the bromide does not work: the intermediate carbocation (if it is even formed to begin with) will be snapped up by the oxygen anyway. – orthocresol Dec 20 '16 at 15:38
• looks very much reasonable.Thanks for pointing out @Zhe. I didn't consider epoxide formation because I thought steric hinderance might create some problem. – Amritansh Singhal Dec 20 '16 at 16:28
• @Zhe I still am not clear why allylic bromination doesn't take place. My thought process was as follows: 1. Co-ordination of bromine of NBS and $\pi-$bond. 2. Formation of bromonium carbocation, with succinimide as a nucleofuge. 3. Attack by water on the bromonium resulting in the anti-stereochemistry. 4. Deprotonation by the extra pair on the succinimide. Is this correct? – Eashaan Godbole Dec 22 '16 at 12:57
• @EashaanGodbole That doesn't give an allylic bromide. You're generating one of the two side products if I'm understanding your mechanism. In this reaction, the double bond is just more reactive, and bromonium formation will outcompete radical hydrogen abstraction. – Zhe Dec 22 '16 at 13:35
• Is NBS alone different from NBS in water? Here, I'd expect allylic bromination (proceeding through radical mechanism) to occur, instead of the observed HOBr addition. Please explain. – arya_stark Feb 28 '18 at 11:16