Talk:Todd:Chem3x11 ToddL9

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'''Student question 2013''': "When doing an epoxide opening on a trans-decalin derivative, can the trans-diequatorial product form at all, or is it impossible (since ring-flipping a trans-decalin is impossible)?"
'''Student question 2013''': "When doing an epoxide opening on a trans-decalin derivative, can the trans-diequatorial product form at all, or is it impossible (since ring-flipping a trans-decalin is impossible)?"
'''Answer''': The decalin system is locked, as for when you have one ring and a tert-butyl. So yes, no ring flipping of the products. So you will get very significant selectivity for trans-diaxial, which stays trans-diaxial.
'''Answer''': The decalin system is locked, as for when you have one ring and a tert-butyl. So yes, no ring flipping of the products. So you will get very significant selectivity for trans-diaxial, which stays trans-diaxial.
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'''Student question 2013''': I'm a bit confused about one of the diagrams in Lecture 9 for chem3113 and unfortunately i have only just realised it. In the diagram for scheme 5 shouldn't the water add trans to the bromine? I'm just not sure if the absolute stereochemistry has been left out of the final product or if the H2O has added cis as it seems to appear. '''Answer''': Not sure what you mean here. Note the position of the Ph. The water has added to one carbon and the Br is on the other. The water has added to the carbon with the slight positive charge.
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'''Student question 2013''': "If an epoxide opening (or bromination, oxymercuration etc.) reaction were heated, would we see the thermodynamic product (i.e. the trans diequatorial product) favoured? Should we always assume that the diaxial product is favoured under normal conditions?" '''Answer''': Generally speaking (though we're being imprecise and colloquial) when we say we heat something we are implying we access the thermodynamic product (in reality it depends how much we heat something). So if the thermodynamic product can form, it eventually will. Scheme 2 of lecture 9 shows that. Remember to take into account any other features of the molecule when judging whether a conformation is likely - other substituents or any fused rings (e.g. in decalins).

Current revision

Student question 2013: "When doing an epoxide opening on a trans-decalin derivative, can the trans-diequatorial product form at all, or is it impossible (since ring-flipping a trans-decalin is impossible)?" Answer: The decalin system is locked, as for when you have one ring and a tert-butyl. So yes, no ring flipping of the products. So you will get very significant selectivity for trans-diaxial, which stays trans-diaxial.

Student question 2013: I'm a bit confused about one of the diagrams in Lecture 9 for chem3113 and unfortunately i have only just realised it. In the diagram for scheme 5 shouldn't the water add trans to the bromine? I'm just not sure if the absolute stereochemistry has been left out of the final product or if the H2O has added cis as it seems to appear. Answer: Not sure what you mean here. Note the position of the Ph. The water has added to one carbon and the Br is on the other. The water has added to the carbon with the slight positive charge.

Student question 2013: "If an epoxide opening (or bromination, oxymercuration etc.) reaction were heated, would we see the thermodynamic product (i.e. the trans diequatorial product) favoured? Should we always assume that the diaxial product is favoured under normal conditions?" Answer: Generally speaking (though we're being imprecise and colloquial) when we say we heat something we are implying we access the thermodynamic product (in reality it depends how much we heat something). So if the thermodynamic product can form, it eventually will. Scheme 2 of lecture 9 shows that. Remember to take into account any other features of the molecule when judging whether a conformation is likely - other substituents or any fused rings (e.g. in decalins).

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