4-hydroxypentanal reacts with one equivalent of methanol to form a cyclic acetal.
1 answer:
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
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Answer:
Explanation:
The missing image is attached below.
The objective of this question is to draw the major product formed from the diagram attached below.
From the diagram attached, we will see the reaction of a tertiary alkyl halide together with a weak nucleophile (ch3ch2oh) undergoing a nucleophilic substitution (SN₁) mechanism to yield a racemic mixture(i.e., compound that is not optically active but contains an equal amount of dextrorotatory and levorotatory stereoisomers) as a product.
