Answer:
G : pent-2-ene
H : H2o
I : pent-1-ene
Explanation:
The OH would combine with H from H-O4SH
Then the O attached to the carbon would get detached taking the electrons with it, leaving carbon an positively charged.
The OH now combined with H-O4SH would take the H away from the acid forming H2O and (O4SH)-
The negative ion of (O4SH)- would then take the H from positively charged C forming the sulphuric acid back. It does not take the electrons with it*
Which is why the carbon can then form a double bond with the excess electron (2). the major product would be pent-2-ene and minor would be pent-1-ene.
This is because pent-2-ene is more stable than pent-1-ene. (Due to secondary carbon and primary carbon factor)
This is the whole backstage story though.
Answer:
yeah.. u cant only if heated the iced tea, breaking the sugar cube up has no effect on the dissolving process.
Answer:
Butanoic acid and 2-propanol reacts to form isopropyl butyrate.
Explanation:
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Answer:
Chlorination of a 1° alcohol using Thionyl Chloride - an SN2 process
Alkene formation via POCl3 reaction with a 2º alcohol - an El process
Explanation:
For a primary alcohol, the chlorination occurs by SN2 mechanism. Remember that the order of SN2 mechanism is methyl > primary > secondary > tertiary. This means that a primary alkyl halide will undergo nucleophillic substitution by SN2 mechanism.
For a secondary alkyl halide, we normally expect that the mechanism will be E2. When we use POCl3 and pyridine, the alkyl halide passes through a carbocation intermediate which is characteristic of an E1 mechanism.