Choose the correct reason:
o-hydroxybenzaldehyde is a liquid at room temperature while $p$-hydroxy-benzaldehyde is a high melting solid, because
o-hydroxybenzaldehyde shows only weak Vanderwal's forces of attraction while in $p$ hydroxy benzaldehyde there is strong Vanderwal's forces of attraction in solid state
o-hydroxybenzaldehyde shows association of the molecules while in $p$-hydroxy benzaldehyde there is no association among the molecules
o-hydroxybenzaldehyde shows intramolecular hydrogen bonding while in $p$-hydroxy benzaldehyde there is no intramolecular hydrogen bonding
p-hydroxybenzaldehyde, shows intramolecular hydrogen bonding while in o-hydroxy benzaldehyde there is stronger intermolecular hydrogen bonding
The number of bond pairs and lone pairs of electrons in the molecule $I F_5$ is:
5,1
6,0
4,2
4,1
For an ideal gas undergoing an isothermal change, there is $\_\_\_\_$
no change in Internal energy of the system and heat released by the system is equal to the work done by the system
an increase in Internal energy of the system and heat absorbed by the system is greater than the work done on the system
a decrease in Internal energy of the system and heat released by the system is equal to the work done by the system
no change in Internal energy of the system and heat absorbed by the system is equal to the work done by the system
Which one of the following is the correct statement?
Acetone reacts with $\mathrm{NH}_2-\mathrm{NH}_2 / \mathrm{KOH}$ in Glycol on heating to form the saturated hydrocarbon, Butane
Acetone undergoes reaction in presence of $\mathrm{Ba}(\mathrm{OH})_2$ on heating to form 4-Methylpent-3-en-2-one
Acetophenone on reaction with $\mathrm{I}_2 / \mathrm{Na}_2 \mathrm{CO}_3(\mathrm{aq})$ does not undergo any reaction due to the presence of deactivating $-\mathrm{COCH}_3$ group
Acetophenone cannot be prepared on reacting Benzoyl chloride with Dimethyl cadmium in ether solvent
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