JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 22)
In the following reaction :
Aldehyde + Alcohol $$\buildrel {HCl} \over \longrightarrow $$ Acetal
The best combination is
Aldehyde + Alcohol $$\buildrel {HCl} \over \longrightarrow $$ Acetal
| Aldehyde | Alcohol | ||
|---|---|---|---|
| HCHO | tBuOH | ||
| CH3CHO | MeOH |
The best combination is
CH3CHO and MeOH
HCHO and MeOH
CH3CHO and tBuOH
HCHO and tBuOH
Explanation
In Formaldehyde (HCHO) both side of the C atom H presents but in Acetaldehyde (CH3CHO) one side of C atom -CH3 group and other side H atom presents. So Formaldehyde is less crowded then it is more reactive.
CH3OH is 1o alcohol and tBuOH is 3o alcohol. As we know on 3o alcohol steric hindrance is very high compare to 1o alcohol and due to steric hindrance reactivity of 3o alcohol is less compare to 1o alcohol.
CH3OH is 1o alcohol and tBuOH is 3o alcohol. As we know on 3o alcohol steric hindrance is very high compare to 1o alcohol and due to steric hindrance reactivity of 3o alcohol is less compare to 1o alcohol.
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