Answer:
Explanation:
The equation is given as:
CH3CHOHC2H4CHO + CH3OH --> CYCLIC ACETAL + H2O
This above equation is carried out in the presence of a strong acid. There are five mechanisms employed and they are:
Step 1:
Initial formation of the hemiacetal which takes several steps
Step 2:
Addition of a proton. The hemicetal is protonated on the hydroxyl group (-OH group)
Step 3:
As seen a bond is broken to give the H2O molecule and a resonance stabilized cation.
The carbonyl group on the cation is enriched with the oxygen-18 got from the H2O molecule as seen in the mechanism.
Step 4:
An attraction occurs between electrophile and nucleophile i.e the stabilised cation and the lone paids of the methanol.
Step 5:
Finally, a proton (+) is removed from the molecule by a lone pair of electron on the methanol.
Attached are the Steps 1 - 5 mechanism below
Answer:
Explanation:
Hello there!
In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:
It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:
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