<span>Answer:
H-C-N H-N-C C-H-N
Notice that C-H-N is the same as N-H-C just written backwards. ( i.e. they have the same connectivtiy.) You can exclude the last one with H in the middle since H has two bonds and 4 electrons around it. At this point you couldn't differentiate between the first two, so I would give you the connectivity in such a problem, which in this case is H-C-N.</span>
So I’m not 100% sure what you’re asking but I’m going to give it a go. The elimination reaction is a term used in organic chemistry that describes a type of reactions. The name kinda tells you what’s going to happen. Something is going to be removed/eliminated from initial reactant/substrate and as a result, an alkene (double bond containing compound) will form.
In elimination reactions a hydrogen atom is first removed (as a H+) from the beta carbon. As a result, the left behind electrons create a pi bond between the beta carbon and the neighboring alpha carbon. This promotes the electronegative atom, on the alpha carbon, to leaves the substrate taking both electrons from the shared sigma bond with the alpha carbon.
The answer is A, do you want me to explain it? It’s pretty simple, you just need to follow all the signs in brackets and match them in those in the answer
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A
Answer:
Explanation:
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