Can someone please check these for me? I may need some help on number 4.

asked by @sarahe43 • over 1 year ago • Organic Chemistry • 5 pts

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2 answers

Hi Sarah! Let’s start from the reaction in the upward direction. This reaction is an Oxymerc-reduction reaction which yields us an alcohol (in the more substituted C) which is anti to the newly added H. You can put the -OH to any of the Cs in the former double bonds at the top part. For the double bond at the bottom part, we will put the -OH to the left C of the C=C bond. Also, don’t forget to re-draw the double bond in the ring! You can watch the video about it here:

For the reaction in the right direction, this an acid catalyzed addition reaction where we would expect a carbocation shift in either of the two double bonds at the top part. This also follows the Markovnikov product where the -OCH3 goes to the more substituted C just like the oxymerc reaction. The major product here would be a tertiary allylic methoxy group (where the 1-butenyl substituent attaches to the ring). You can watch the video here:

The reaction in the downward direction follows the Halohydrin formation where the Br and -OCH2CH2CH3 is in the anti-configuration. The PENTOXY, not —OH group, will go to the most substituted C (which is where you put the —OH group). Check this video: For the last reaction in the leftward direction, this is a halogenation reaction where we would expect that two Cls will attach to the Cs in an anti-configuration. Here’s the video for it:

Remember that you should react all of the reagents in all present double bonds—except for the left-direction probem because you only have 1 equivalent of dihalide—because I noticed that you only consider one double bond for the reaction. The leftward reaction’s product should likely be anti, vicinal dihalides where the double bond was. I hope those videos will help you

answered by @reyn • over 1 year ago

@ reyn would it look something like this enter image description here

answered by @sarahe43 • over 1 year ago