The blue arrow is indicating that a lone pair on oxygen is being transformed int
ID: 733053 • Letter: T
Question
The blue arrow is indicating that a lone pair on oxygen is being transformed into a bond between C-1 and the O. The red arrow is indicating that the sigma bonding electrons between C-1 and C-2 are moving to become sigma bonding electrons between C-2 and N. How did I know C-2 was going to end up being singly bonded to N? Remember that we cannot "over octet" carbons (the maximum # of bonds carbon forms is 4 for our consideration). C-1 would be "over octeted" when the lone pair electrons from the O made a pi bond, unless another bond left C-1. The arrows are indicating that the C-2 bonded to C-1 is leaving and bonding to N. Like C, N is not "over-octeted" either. Since a new bond is being formed between N and C-2, another bond to the N must be broken. The green arrow is indicating that the sigma bonding electrons between N and Br are being transformed into a lone pair on the Br. See if you can draw the indicated species. Note: 1) You cannot show lone pairs in the JME editor. 2) You can show formal charges - see JME input instructions. 3) Do not include the numbers 1 and 2 to designate C-1 and C-2.
Explanation / Answer
The Br was originally neutral, but picks up an extra electron from the N and becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double bonds on both sides of C1, the O, C1, N backbone is linear.
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